J.-M. Berthelot Mechanics of Rigid BodiesBerthelot+Mechanics+of+Rigid+Bodies.p… · reference...

Jean-Marie Berthelot

Mechanics of Rigid Bodies

Les Clousures At the Bottom of Écrins Vallouise, France 4102 m

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Mechanics of Rigid Bodies

Jean-Marie Berthelot is an Honorary Professor of Maine Univerty. He took part to

the installation of the Institute for Advanced Materials and Mechanics (ISMANS),

Le Mans, France. His current research is on the mechanical behaviour of

composite materials and structures. He has published extensively in the area of

composite materials and is the author of numerous international papers and

textbooks, in particular a textbook entitled Composite Materials, Mechanical

Behavior and Structural Analysis published by Springer, New York, in 1999.

see www.compomechaclimb.com..


Mechanichs of Rigid Bodies

Les Clousures At the Bottom of Écrins Vallouise, France 4102 m

Preface

The objective of this book is to develop the fundamental statements of the

Mechanics of Rigid Bodies. The text is designed for undergraduate courses of

Mechanical Engineering. The basic mathematical concepts are covered in the first

part, thereby making the book self-contained. The different parts of the book are

carefully developed to provide continuity of the concepts and theories. Finally the

text has been established so as to construct chapter after chapter a unified proce-

dure for analysing any mechanical system constituted of rigid bodies.

The first part, Mathematical Basics, introduces the usual concepts needed in

the study of mechanical systems: vector space R3, geometric space, vector deriva-

tives, curves. A chapter is devoted to torsors whose concept is the key of the book.

The general notion of “measure centre” is introduced in this chapter.

The second part, Kinematics, begins with the analysis of the motion of a point

(kinematics of point). Particular motions are next considered, with a chapter

related to motions with central acceleration. Next, the kinematics of a rigid body

is studied: parameter of situation, kinematic torsor, analysis of particular motions.

The change of reference system, which introduces the notion of “entrainment” has

been excluded deliberately from this part. The notion of “entrainment” is not

really assimilated by the studients at this level of the text. In fact this notion is

implicitly introduced by using the concept of kinematic torsor. The change of

reference system will be considered as a whole within the frame of Kinetics (Part

4). The last chapter analyses the kinematics of rigid bodies in contact.

The third part, Mechanical Actions, introduces first the general concepts of the

mechanical actions exerted on a rigid body or on a system of rigid bodies.

Represented by torsors, the mechanical actions have general properties which are

derived from the concepts considered previously for torsors. Thus, mechanical

actions are classified as forces, couples and arbitrary actions. Gravitation and

gravity are analysed. A chapter is devoted to the mechanical actions involved by

the connections between rigid bodies, whose concept is the basis of the techno-

logical design of mechanical systems. The introduction of the power developed by

a mechanical action simplifies greatly the restrictions imposed in the case of

perfect connections (connections without friction). In the last chapter, the investi-

gation of some problem of Statics will grow the reader familiar with the analysis

of mechanical actions exerted on a body or a system of bodies.

The fourth part, Kinetics of Rigid Bodies, introduces the tools needed to

analyse the problems of Dynamics: operator of inertia, kinetic torsor, dynamic

torsor and kinetic energy. Next, the problem of the change of reference system is

considered.

At this step, the reader has acquired the whole elements needed to analyse the

problems of Dynamics of a rigid body or a system of rigid bodies. This analysis is

developed in the fifth part Dynamics of Rigid Bodies. First, the general process

for analysing a problem of Dynamics is established. Next, particular problems are

considered. The process of analysis is always the same: kinematic analysis, kinetic

analysis, investigation of the mechanical actions, deriving the equations of Dyna-

mics as a consequence of the fundamental principle of dynamics, assumptions

vi Preface

on the physical nature of connections between bodies, solving the equations of

motion and the equations of connections. The designer will have to take an

interest in the parameters of the motion as well as in the mechanical actions

exerted at the level of connections to design the mechanical systems. The appli-

cation of the fundamental principle of dynamics allows us to derive the whole

equations of dynamics (equations of motion and equations of mechanical actions

at the level of connections). However, designer which takes an interest only in the

equations of motion needs a systematic tool for deriving these equations: the

Lagrange’s equations which are considered in the last chapter of part V.

In general, the equations of motions of a body or of a system of rigid bodies are

complex, and most of these equations can not be solved using an analytical

process. Now, mechanical engineers dispose of numerical tools (numerical pro-

cesses and microcomputers) needed to solve the motion equations, whatever the

complexity of these equations may be. The sixth part, Numerical procedures for

the Resolution of Motion Equations, is an introduction to the numerical processes

used to solve equations of motion. Examples are considered.

The correction of the exercises is reported at the end of the textbook. The

writing has been developed extensively and structured in such a way to improve

the capacity of the comprehension of the reader.

At the end of the textbook, the designer will have all the elements which allow

him to implement a complete and structured analysis of mechanical systems.

June 2015, Vallouise, Jean-Marie BERTHELOT

Note. The development of this textbook is based on a generalized use of the

concept of “torseur” (in French). We think that this concept is not really used in

the English textbooks. We will call this concept as “torsor”. In the textbook, the

English formulation was thus transposed from the French formulation.

The author would be highly grateful with whoever would bring any element likely

to be able to make progress the development, and thus the comprehension, of the

textbook.

Contents

Preface v

PART I Mathematical Basic Elements 1

Chapter 1 Vector Space 3 3

1.1 Definition of the Vector Space 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.1 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.2 Vector Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.3 Multiplication by a Scalar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Linear Dependence and Independence. Basis of 3 . . . . . . . . . . . . . . . . . 5

1.2.1 Linear Combination. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2.2 Linear Dependence and Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2.3 Basis of the Vector Space 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2.4 Components of a Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.3 Scalar Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.3.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.3.2 Magnitude or Norm of a Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.3.3 Analytical Expression of the Scalar Product in an Arbitrary Basis . . . . . . 9

1.3.4 Orthogonal Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.3.5 Orthonormal Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.3.6 Expression of the Scalar Product in an Orthonormal Basis . . . . . . . . . . . . 10

1.4 Vector Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.4.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.4.2 Analytical Expression of the Vector Product in an Arbitrary Basis . . . . . . 11

1.4.3 Direct Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.4.4 Expression of the Vector Product in a Direct Basis. . . . . . . . . . . . . . . . . . 12

1.4.5 Mixed Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.4.6 Property of the Double Vector Product . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.5 Bases of the Vector Space 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.5.1 Canonical Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.5.2 Basis Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

Chapter 2 The Geometric Space 18

2.1 The Geometric Space Considered as Affine to the Vector Space 3 . . . . . 18

Contents ix

2.1.1 The Geometric Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.1.2 Consequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.1.3 Distance between Two Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.1.4 Angle between Two Bipoints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.1.5 Reference Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.2 Subspaces of the Geometric Space: Line, Plane . . . . . . . . . . . . . . . . . . . . . 222.2.1 Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.2.2 Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.2.3 Lines and Planes with Same Directions . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.2.4 Orthogonal Lines and Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.3 Characterization of the Position a Point of the Geometric Space . . . . . . . 262.3.1 Coordinate Axes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.3.2 Direct Orthonormal Reference System . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.3.3 Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.4 Plane and Line Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.4.1 Cartesian Equation of a Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.4.2 Cartesian Equation of a Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.5 Change of Reference System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.5.1 General Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.5.2 Refernce Systems with a Same Axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.5.3 Arbitrary Reference Systems with the Same Origin . . . . . . . . . . . . . . . . . 34

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

Chapter 3 Vector Function. Derivatives of a Vector Function 40

3.1 Vector Function of One Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 3.1.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.1.2 Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.1.3 Properties of the Vector Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.1.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3.2 Vector Function of Two Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443.2.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3.2.2 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3.2.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.3 Vector Function of n Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453.3.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.3.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

Chapter 4 Elementary Concepts on Curves 50

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 4.2 Curvilinear Abscissa. Arc Length of a Curve . . . . . . . . . . . . . . . . . . . . . . . 51 4.3 Tangent. Normal. Radius of Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 4.4 Frenet Trihedron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

Comments 54

x Contents

Chapter 5 Torsors 55

5.1 Definition and Properties of the Torsors . . . . . . . . . . . . . . . . . . . . . . . . . . 555.1.1 Definitions and notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

5.1.2 Properties of the Moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

5.1.3 Vector Space of Torsors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

5.1.4 Scalar Invariant of a Torsor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5.1.5 Product of Two Torsors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

5.1.6 Moment of a Torsor about an Axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

5.1.7 Central Axis of a Torsor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

5.2 Particular Torsors. Resolution of an Arbitrary Torsor . . . . . . . . . . . . . . . 605.2.1 Slider . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

5.2.2 Couple-Torsor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

5.2.3 Arbitrary Torsor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

5.2.4 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

5.3 Torsors associated to a Field of Sliders Defined on a Domain of the Geometric Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 645.3.1 Torsor Associated to a Finite Set of Points . . . . . . . . . . . . . . . . . . . . . . . . 64

5.3.2 Torsor Associated to a Infinite Set of Points . . . . . . . . . . . . . . . . . . . . . . 65

5.3.3 Important Particular Case. Measure Centre . . . . . . . . . . . . . . . . . . . . . . . 67

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

PART II Kinematics 73

Chapter 6 Kinematics of Point 75

6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 6.2 Trajectory and Kinematic Vectors of a Point . . . . . . . . . . . . . . . . . . . . . . . 756.2.1 Trajectory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

6.2.2 Kinematic Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

6.2.3 Tangential and Normal Components of Kinematic Vectors . . . . . . . . . . . 78

6.2.4 Different Types of Motions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

6.3 Expressions of the Components of Kinematic Vectors as Functions of Cartesian and Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

6.3.1 Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

6.3.2 Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82


Chapter 7 Study of Particular Motions 84

7.1 Motions with Rectilinear Trajectory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 7.1.1 General Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

7.1.2 Uniform Rectilinear Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

7.1.3 Uniformly Varied Rectilinear Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

7.1.4 Simple Harmonic Rectilinear Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

7.2 Motions with a Circular Trajectory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 7.2.1 General Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

Contents xi

7.2.2 Uniform Circular Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

7.2.3 Uniformly Varied Circular Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

7.3 Motions with a Contant Acceleration Vector . . . . . . . . . . . . . . . . . . . . . . . 90 7.3.1 General Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

7.3.2 Study of the case where the Trajectory is Rectilinear . . . . . . . . . . . . . . . . 91

7.3.3 Study of the case where the Trajectory is Parabolic . . . . . . . . . . . . . . . . . 92

7.4 Helicoidal Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 947.5 Cycloidal Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

Chapter 8 Motions with Central Acceleration 100

8.1 General Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1008.1.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

8.1.2 A Motion with a Central Acceleration is a Plane Trajectory

Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

8.1.3 Areal Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

8.1.4 Area Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

8.1.5 Expression of the Kinematic Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

8.1.6 Polar Equation of the Trajectory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

8.1.7 Motions for which ( ) 2( , )Ta M t OMω= −

. . . . . . . . . . . . . . . . . . . . . . . . . . 103

8.2 Motions with Central Acceleration for which ( )

3( , )T OM

a M t KOM

= −

. . . . 104

8.2.1 Equations of the Trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

8.2.2 Study of the Trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

8.2.3 Velocity Magnitude at a Point of the Trajectory . . . . . . . . . . . . . . . . . . . . 107

8.2.4 Elliptic Motion. Kepler’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

Chapter 9 Kinematics of Rigid Body 111

9.1 General Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1119.1.1 Notion of Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

9.1.2 Locating a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

9.2 Relations between the Trajectories and the Kinematic Vectors of Two Points Attached to a Solid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

9.2.1 Relation between the Trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

9.2.2 Relation between the Velocity Vectors. . . . . . . . . . . . . . . . . . . . . . . . . . . 114

9.2.3 Expression of the Instantaneous Vector of Rotation . . . . . . . . . . . . . . . . . 115

9.2.4 Kinematic Torsor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

9.2.5 Relation between the Acceleration Vectors . . . . . . . . . . . . . . . . . . . . . . . 117

9.3 Generalization of the Composition of Motions . . . . . . . . . . . . . . . . . . . . . . 1189.3.1 Composition of Kinematic Torsors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

9.3.2 Inverse Motions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

9.4 Examples of Solid Motions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1219.4.1 Motion of Rotation about a Fixed Axis . . . . . . . . . . . . . . . . . . . . . . . . . . 121

9.4.2 Translation Motion of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

xii Contents

9.4.3 Motion of a Body Subjected to a Cylindrical Joint . . . . . . . . . . . . . . . . . . 125

9.4.4 Motion of Rotation about a Fixed Point . . . . . . . . . . . . . . . . . . . . . . . . . . 127

9.4.5 Plane Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129


Chapter 10 Kinematics of Rigid Bodies in Contact 137

10.1 Kinematics of Two Solids in Contact . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 10.1.1 Solids in Contact at a Point. Sliding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

10.1.2 Spinning and Rolling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

10.1.3 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

10.1.4 Solids in Contact in Several points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

10.2 Transmission of a Motion of Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14010.2.1 Général Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

10.2.2 Transmission by Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

10.2.3 Gear Transmission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

10.2.4 Belt Transmission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148


PART III The Mechanical Actions 153

Chapter 11 General Elements on the Mechanical Actions 155

11.1 Concepts Relative to the Mechanical Actions . . . . . . . . . . . . . . . . . . . . . . . 15511.1.1 Notion of Mechanical Action . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

11.1.2 Representation of a Mechanical Action . . . . . . . . . . . . . . . . . . . . . . . . . . 155

11.1.3 Classification of the Mechanical Actions . . . . . . . . . . . . . . . . . . . . . . . . . 156

11.1.4 Mechanical Actions Exerting between Material Sets . . . . . . . . . . . . . . . . 158

11.1.5 External Mechanical Actions Exerting on a Material Set . . . . . . . . . . . . . 158

11.2 Different Types of Mechanical Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15911.2.1 Physical Natures of the Mechanical Actions . . . . . . . . . . . . . . . . . . . . . . . 159

11.2.2 Environnement and Effective Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

11.3 Power and Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16011.3.1 Definition of the Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

11.3.2 Change of Reference System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

11.3.3 Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

11.3.4 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

11.3.5 Power and Work of a Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

11.3.6 Set of Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

Chapter 12 Gravitation. Gravity. Mass Centre 169

12.1 Phenomenon of Gravitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16912.1.1 Law of Gravitation 169

Contents xiii

12.1.2 Gravitational Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170

12.1.3 Action of gravitation induced by a Solid Sphere . . . . . . . . . . . . . . . . . . . 170

12.1.4 Action of gravitation induced by the Earth . . . . . . . . . . . . . . . . . . . . . . . . 172

12.2 Action of Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17312.2.1 Gravity Field Induced by the Earth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

12.2.2 Action of Gravity Exerted on a Material System . . . . . . . . . . . . . . . . . . . 174

12.2.3 Power Developed by the Action of Gravity . . . . . . . . . . . . . . . . . . . . . . . 175

12.3 Determination of Mass Centres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17712.3.1 Mass Centre of a Material System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

12.3.2 Mass Centre of the Union of Two Sets . . . . . . . . . . . . . . . . . . . . . . . . . . 178

12.3.3 Mass Centre of a Homogeneous Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179

12.3.4 Homogeneous Bodies with Geometrical Symmetries . . . . . . . . . . . . . . . . 180

12.4 Examples of Determination of Mass Centres . . . . . . . . . . . . . . . . . . . . . . . 18112.4.1 Homogeneous Solid Hemisphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181

12.4.2 Homogeneous Solid with Complex Geometry . . . . . . . . . . . . . . . . . . . . . 182

12.4.3 Non-Homogeneous Solid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

Chapter 13 Actions of Contact between Solids. Connections 186

13.1 Laws of Contact between Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 13.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

13.1.2 Contact in a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

13.1.3 Couples of Rolling and Spinning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191

13.2 Connections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19213.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

13.2.2 Classification of Connections. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

13.2.3 Actions of Connection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197

13.2.4 Connection without Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198

13.2.5 Connection with Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202

Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203

Chapter 14 Statics of Rigid Bodies 204

14.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 14.2 Law of Statics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20414.2.1 Case of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204

14.2.2 Case of a Set of Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205

14.2.3 Mutual Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206

14.3 Statics of Wires or Flexible Cables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20714.3.1 Mechanical Action Exerted by a Wire or a Flexible Cable . . . . . . . . . . . . 207

14.3.2 Equation of Statics of a Wire . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

14.3.3 Wire or Flexible Cable Submitted to the Gravity . . . . . . . . . . . . . . . . . . . 209

14.3.4 Contact of a Wire with a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210

14.4 Examples of Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21214.4.1 Case of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212

14.4.2 Case of a System of Two Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . 217

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 Comments 223

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PART IV Kinetics of Rigid Bodies 225

Chapter 15 The Operator of Inertia 227

15.1 Introduction to the Operator of Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . 22715.1.1 Operator Associated to a Vector Product . . . . . . . . . . . . . . . . . . . . . . . . . 227

15.1.2 Extending the Preceding Concept . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228

15.1.3 The Operator of Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229

15.2 Change of Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23015.2.1 Change of Origin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230

15.2.2 Relations of Huyghens . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232

15.2.3 Diagonalisation of the Matrix of Inertia . . . . . . . . . . . . . . . . . . . . . . . . . 232

15.2.4 Change of Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233

15.3 Moments of Inertia with respect to a point, an axis, a plane . . . . . . . . . . . 23415.3.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234

15.3.2 Relations between the Moments of Inertia . . . . . . . . . . . . . . . . . . . . . . . . 235

15.3.3 Case of a Plane Solid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235

15.3.4 Moment of Inertia with respect to an Arbitrary Axis . . . . . . . . . . . . . . . . 236

15.4 Determination of Matrices of Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23715.4.1 Solids with Material Symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237

15.4.2 Solids having a Symmetry of Revolution . . . . . . . . . . . . . . . . . . . . . . . . 239

15.4.3 Solids with Spherical Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241

15.4.4 Associativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242

15.5 Matrices of Inertia of Homogeneous Bodies . . . . . . . . . . . . . . . . . . . . . . . . 24415.5.1 One-Dimensional Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244

15.5.2 Two-Dimensional Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245

15.5.3 Three-Dimensional Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249


Chapter 16 Kinetic and Dynamic Torsors. Kinetic Energy 255

16.1 Kinetic Torsor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25516.1.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255

16.1.2 Kinetic Torsor Associated to the Motion of a Body . . . . . . . . . . . . . . . . . 256

16.1.3 Kinetic Torsor for a Set of Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257

16.2 Dynamic Torsor. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25816.2.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258

16.2.2 Dynamic Torsor Associated to the Motion of a Body . . . . . . . . . . . . . . . 258

16.2.3 Dynamic Torsor for a Set of Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259

16.2.4 Relation with the Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260

16.3 Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26016.3.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260

16.3.2 Kinetic Energy of a Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261

16.3.3 Kinetic Energy of a Set of Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262

16.3.4 Derivative of the Kinetic Energy of a Solid with respect to Time . . . . . . 262


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Chapter 17 Change of Reference System 265

17.1 Kinematics of Change of Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26517.1.1 Relation between the Kinematic Torsors . . . . . . . . . . . . . . . . . . . . . . . . . 265

17.1.2 Relation between the Velocity Vectors. Velocity of Entrainment . . . . . . 266

17.1.3 Composition of Acceleration Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . 268

17.2 Dynamic Torsors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26917.2.1 Inertia Torsor of Entrainment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270

17.2.2 Inertia Torsor of Coriolis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271

17.2.3 Relation between the Dynamic Torsors Defined relatively

to Two Different References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272

Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273

PART V Dynamics of Rigid Bodies 275

Chapter 18 The Fundamental Principle of Dynamics and its Consequences 277

18.1 Fundamental Principle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 18.1.1 Statement of the Fundamental Principle of Dynamics . . . . . . . . . . . . . . . 277

18.1.2 Class of Galilean Reference Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 277

18.1.3 Vector Equations Deduced from the Fundamental Principle . . . . . . . . . . 278

18.1.4 Scalar Equations Deduced from the Fundamental Principle . . . . . . . . . . . 279

18.2 Mutual Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28018.2.1 Theorem of Mutual Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280

18.2.2 Transmission of Mechanical Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . 281

18.3 Theorem of Power-Energy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 18.3.1 Case of One Solid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281

18.3.2 Case of a Set of Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282

18.3.3 Mechanical Actions with Potential Energy . . . . . . . . . . . . . . . . . . . . . . . 283

18.4 Application of the Fundamental Principle to the Study of the Motion of a Free Body in a Galilean Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . 284

18.4.1 General Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284

18.4.2 Particular Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286

18.5 Application to the Solar System. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288 18.5.1 Galilean Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288

18.5.2 Motion of Planets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290

18.5.3 The Earth in the Solar System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290

Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291

Chapter 19 The Fundamental Equation of Dynamics in Different References 293

19.1 General Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293 19.1.1 Fundamental Equation of Dynamics in a Non Galilean Reference . . . . . 293

19.1.2 The Reference Systems used in Mechanics . . . . . . . . . . . . . . . . . . . . . . . 294

19.2 Fundamental Relation of Dynamics in the Geocentric Reference . . . . . . . 29519.2.1 General Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295

Contents xvi

19.2.2 Case of a Solid Located at the Vicinity of the Earth . . . . . . . . . . . . . . . . 297

19.3 Fundamental Relation in a Reference Attached to the Earth . . . . . . . . . . 29819.3.1 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298

19.3.2 Action of Earthly Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299

19.3.3 Conclusions on the Equations of Dynamics in a Reference

Attached to the Earth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300

19.4 Equations of Dynamics of a Body with respect to a Reference whose the Motion is Known Relatively to the Earth . . . . . . . . . . . . . . . . 301

Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303

Chapter 20 General Process for Analysing a Problem of Dynamics of Rigid Bodies 304

20.1 Dynamics of Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304 20.1.1 General Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304

20.1.2 General Process of Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305

20.2 Dynamics of a Set of Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306 20.3 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308

Chapter 21 Dynamics of Systems with One Degree of Freedom Analysis of Vibrations 309

21.1 General Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309 21.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309

21.1.2 Parameters of Situation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310

21.1.3 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310

21.1.4 Kinetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310

21.1.5 Mechanical Actions Exerted on the Solid . . . . . . . . . . . . . . . . . . . . . . . . 311

21.1.6 Application of the Fundamental Principle . . . . . . . . . . . . . . . . . . . . . . . . 311

21.2 Vibrations without Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313 21.2.1 Equation of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313

21.2.2 Free Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313

21.2.3 Forced Vibrations. Steady State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314

21.3 Vibrations with Viscous Damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31821.3.1 Equation of Motion with Viscous Damping . . . . . . . . . . . . . . . . . . . . . . 318

21.3.2 Free Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318

21.3.3 Vibrations in the case of a Harmonic Disturbing Force . . . . . . . . . . . . . . 324

21.3.4 Forced Vibrations in the case of a Periodic Disturbing Force. . . . . . . . . . 331

21.3.5 Vibrations in the case of an Arbitrary Disturbing Force . . . . . . . . . . . . . . 332

21.3.6 Forced Vibrations in the case of a Motion Imposed to the Support . . . . . 333

21.4 Vibrations with Dry Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33621.4.1 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336

21.4.2 Free Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337

21.5 Equivalent Viscous Damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33921.5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339

21.5.2 Energy Dissipated in the case of Viscous Damping . . . . . . . . . . . . . . . . 340

21.5.3 Stuctural Damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340

Contents xvii

21.5.4 Dry Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342

21.5.5 Fluid Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343

21.5.6 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345


Chapter 22 Motion of Rotation of a Solid about a Fixed Axis 347

22.1 General Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 22.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347


22.1.3 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349

22.1.4 Kinetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 350

22.1.5 Mechanical Actions Exerted on the Sold . . . . . . . . . . . . . . . . . . . . . . . . . 351

22.1.6 Application of the Fundamental Principle of Dynamics . . . . . . . . . . . . . . 352

22.2 Examples of Motions of Rotation about an Axis . . . . . . . . . . . . . . . . . . . . 35422.2.1 Solid in Rotation Submitted only to the Gravity . . . . . . . . . . . . . . . . . . . 354

22.2.2 Pendulum of Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356

22.3 Problem of the Balancing of Rotors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35722.3.1 General Equations of an Unbalanced Solid in Rotation . . . . . . . . . . . . . . 357

22.3.2 Mechanical Actions Exerted on the Shaft of Rotor . . . . . . . . . . . . . . . . . 360

22.3.3 Principle of the Balancing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364

Chapter 23 Plane Motion of a Rigid Body 365

23.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36523.2 Parallelepiped Moving on an Inclined Plane . . . . . . . . . . . . . . . . . . . . . . . 36523.2.1 Parameters of Situation and Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . 365

23.2.2 Kinetics of the Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366

23.2.3 Mechanical Actions Exerted on the Parallelepiped . . . . . . . . . . . . . . . . . 367

23.2.4 Equations Deduced from the Fundamental Principle . . . . . . . . . . . . . . . . 368

23.2.5 Motion without Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369

23.2.6 Motion with Dry Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370

23.2.7 Motion with Viscous Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371

23.3 Analysis of Sliding and Rocking of a Parallelepiped on an Inclined Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372

23.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372

23.3.2 Parameters of Situation and Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . 373

23.3.3 General Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374

23.3.4 Analysis of the Different Motions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375

23.3.5 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379

23.4 Motion of a Cylinder on an Inclined Plane . . . . . . . . . . . . . . . . . . . . . . . . . 38023.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380


Contents xviii

23.4.3 Mechanical Actions Exerted on the Cylinder. . . . . . . . . . . . . . . . . . . . . . 382

23.4.4 General Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383


23.5 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388

Chapter 24 Other Examples of Motions of Rigid Bodies 389

24.1 Solid in Translation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38924.1.1 General Expressions of a Solid in Translation . . . . . . . . . . . . . . . . . . . . . 389

24.1.2 Free Solid in Translation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391

24.2 Motion of a Solid Placed on a Wagon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39224.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392


24.2.3 Kinetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394

24.2.4 Analysis of the Mechanical Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394

24.2.5 Equations of Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395


24.3 Coupled Motions of Two Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40224.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402


24.3.3 Kinetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 404

24.3.4 Analysis of the Mechanical Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406

24.3.5 Equations Deduced from the Fundamental Principle of Dynamics . . . . . . 408

24.3.6 Analysis of the Equations Deduced from the Fundamental Principle . . . . 409


Chapter 25 The Lagrange Equations 413

25.1 General Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41325.1.1 Free Body and Connected Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413

25.1.2 Partial Kinematics Torsors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413

25.1.3 Power Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415

25.1.4 Perfect Connections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415

25.2 Lagrange Equations Relative to a Rigid Body . . . . . . . . . . . . . . . . . . . . . . 416 25.2.1 Introduction to the Lagrange Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 416

25.2.2 Lagrange Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417

25.2.3 Case where the Mechanical Actions Admit a Potential Energy . . . . . . . . 418

25.3 Lagrange Equations for a Set of Rigid Bodies . . . . . . . . . . . . . . . . . . . . . 41925.3.1 Lagrange Equations for Each Solid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419

25.3.2 Lagrange Equations for the Set (D) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 420

25.3.3 Case where the Parameters of Situation are Linked . . . . . . . . . . . . . . . . . 421

25.4 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42225.4.1 Motion of a parallelepiped Moving on an Inclined Plane . . . . . . . . . . . . . 422

25.4.2 Coupled Motions of Two Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423

25.4.3 Double Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425

A.25 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431

Contents xix


PART VI Numerical Methods for Solving Differential Equations. Application to Equations of Motion 435

Chapter 26 Numerical Methods for Solving First Order Differential Equations 437

26.1 General Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43726.1.1 Problem with Given Initial Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . 437

26.1.2 General Method of Resolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438

26.1.3 Euler Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438

26.2 Single-Step Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44026.2.1 General Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 440

26.2.2 Methods of Runge-Kutta Type . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442

26.2.3 Romberg Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 446

26.3 Multiple-Step Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44926.3.1 Introduction to the Multiple-Step Methods . . . . . . . . . . . . . . . . . . . . . . . 449

26.3.2 Methods based on the Newton interpolation . . . . . . . . . . . . . . . . . . . . . . 450

26.3.3 Generalization of the Multiple-Step Methods . . . . . . . . . . . . . . . . . . . . . 452

26.3.4 Examples of Multiple-Step Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453

26.3.5 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454


Chapter 27 Numerical Procedures for Solving the Equations of Motions 457

27.1 Equation of Motion with One Degree of Freedom . . . . . . . . . . . . . . . . . . . 457

27.1.1 Form of the Equation of Motion with One Degree of Freedom . . . . . . . . 457

27.1.2 Principle of the Numerical Resolution . . . . . . . . . . . . . . . . . . . . . . . . . . . 457

27.1.3 Application to the case of the Motion of a Simple Pendulum . . . . . . . . . 458

27.2 Equations of Motions with Several Degrees of Freedom . . . . . . . . . . . . . . 461

27.2.1 Form of the Equations of Motions with Several Degrees of Freedom . . . 461

27.2.2 Principle of the Numerical Resolution . . . . . . . . . . . . . . . . . . . . . . . . . . . 462

27.2.3 Trajectories and Kinematic Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462

27.3 Motions of Planets and Satellites . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463

27.3.1 Motion of a Planet about the Sun . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463

27.3.2 Motion of a Satellite around the Earth . . . . . . . . . . . . . . . . . . . . . . . . . . 467

27.3.3 Launching and Motion of a Moon Probe . . . . . . . . . . . . . . . . . . . . . . . . 468

27.4 Motion of a Solid on an Inclined Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46927.5 Coupled Motion of Two Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47127.5.1 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471

27.5.2 Analytical Solving in the case of Low Amplitudes and

in the Absence of Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474

27.5.3 Numerical Computation of the Equations of Motion . . . . . . . . . . . . . . . . 476

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 479 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 480

PART VII Solutions of the Exercises 481

Chapter 1 Vector Space 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483

Chapter 2 The Geometric Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486

Chapter 4 Elementary Concepts on Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . 492

Chapter 5 Torsors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 494

Chapter 6 Kinematics of Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 500

Chapter 7 Study of Particular Motions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505

Chapter 9 Kinematics of Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 509

Chapter 10 Kinematics of Rigid Bodies in Contact . . . . . . . . . . . . . . . . . . . . . . 516

Chapter 11 General Elements on the Mechanical Actions . . . . . . . . . . . . . . . . . 523

Chapter 12 Gravitation. Gravity. Mass Centre . . . . . . . . . . . . . . . . . . . . . . . . . 531

Chapter 14 Statics of Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 538

Chapter 15 The Operator of Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 548

Chapter 16 Kinetic and Dynamic Torsors. Kinetic Energy . . . . . . . . . . . . . . . . 559

Chapter 21 Dynamics of Systems with One Degree of Freedom Analysis of Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 567

Chapter 22 Motion of Rotation of a Solid about a Fixed Axis . . . . . . . . . . . . . . 571

Chapter 24 Other Examples of Motions of Rigid Bodies . . . . . . . . . . . . . . . . . . 577

Chapter 25 The Lagrange Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 596

Part I

Mathematical Basic Elements

This part introduces the principal mathematical elements needed for

implementing the various concepts used in Mechanics of Rigid

Bodies. The vector space 3 constitutes the basis of these concepts.

This vector space then allows us to formulate the surrounding phy-

sical space, the geometric space, and to derive formulation of its pro-

perties. The foundation of the development of the present book is

based on the formalism of the torsors. Hence, a particular attention

has to be drawn to this notion.

CHAPTER 1

Vector Space R3

1.1 DEFINITION OF THE VECTOR SPACE R3

1.1.1 Vectors

The vector space 3 may be defined as being the space of triples (C1, C2, C3)

where C1, C2, C3 are three ordered real numbers. The triples thus defined are

called vectors and denoted by V

. Hence:

( )1 2 3, , V C C C=

. (1.1)

The real numbers C1, C2, C3 are the components of the vector V

.

In order to deal with the vectors it is necessary to define laws of composition as

the vector addition and the scalar multiplication.

1.1.2 Vector addition

The first law of composition is the vector addition which associates to the

vectors V

and V ′ a vector sum denoted as V V ′+

:

3 , V V ′∀ ∈

3V V ′+ ∈

.

If ( )1 2 3, , V C C C=

and ( )1 2 3, , V C C C′ ′ ′ ′=

are the two vectors of 3. The

vector sum is derived by the relation:

( )1 1 2 2 3 3, , V V C C C C C C′ ′ ′ ′+ = + + +

. (1.2)

vector addition

4 Chapter 1 Vector Space 3

The neutral element, denoted as 0

and called the zero vector or the null vector,

is defined as:

( )0 0, 0, 0=

. (1.3)

The properties of the vector addition are the following ones:

1. The vector addition is commutative:

1 2 2 1V V V V+ = +

. (1.4)

2. The vector addition is associative:

( ) ( )1 2 3 1 2 3V V V V V V+ + = + +

. (1.5)

3. The neutral element is such as:

0 .V V+ =

(1.6)

4. For each vector V

, corresponds an opposite vector, denoted by V−

, with the

property that:

( ) 0V V+ − =

. (1.7)

1.1.3 Multiplication by a Scalar

The second law of composition is the multiplication by a scalar or multipli-

cation by a real number. If α is a real number and V

a vector, the multiplication

by a scalar associates to V

a vector W

noted Vα

:

3, Vα∀ ∈ ∀ ∈

3W Vα= ∈

.

The vector W

is said to be collinear to the vector V

. If the vector V

is defined

by its components ( )1 2 3, , V C C C=

, the vector W

is defined by:

( )1 2 3, , W C C Cα α α=

. (1.8)

The multiplication by a scalar satisfies the following properties:

1. Distributivity for the addition of scalars:

( )1 2 1 2V V Vα α α α+ = +

. (1.9)

2. Distributivity for the vector addition:

( )1 2 1 2V V V Vα α α+ = +

. (1.10)

3. Associativity for the multiplication by a scalar:

( ) ( )1 2 1 2V Vα α α α=

. (1.11)

multiplication by a scalar

1.2 Linear Dependence and Independence. Basis for 3 5

1.2 LINEAR DEPENDENCE AND INDEPENDENCE

BASIS FOR THE VECTOR SPACE R3

1.2.1 Linear Combination

Consider 1 2, , . . . , , . . . , ,i pV V V V

p vectors of the space 3 and p real numbers:

1 2, , . . . , , . . . , i pα α α α . The vectors 1 21 2, , . . . , , . . . , ,i pi pV V V Vα α α α

are

vectors of the vector space 3 , as well as their sum which defines the vector V

:

1 21 2

1

. . .

p

p ip i

i

V V V V Vα α α α=

= + + + =

. (1.12)

The vector V

thus defined is called the linear combination of the vectors 1 2,V V

,

. . . , .pV

1.2.2 Linear Dependence and Independence

1.2.2.1 Definition

In the vector space 3 , p vectors 1 2, , . . . , ,pV V V

are linearly independent if

and only if the equality

1 21 2

1

. . . 0

p

i pi p

i

V V V Vα α α α=

= + + + =

(1.13)

involves obligatorily:

1 20, 0, . . . , 0pα α α= = = . (1.14)

All the coefficients αi are zero.

If it is not the case, the vectors are said to be linearly dependent.

1.2.2.2 Properties

a. About the independence

1. A non zero vector V

is by itself linearly independent.

2. For a collection of independent vectors, no vector is the null vector. Indeed,

if we had, for example, 0kV =

, Relation (1.13) would be satisfied with

0kα ≠ .


3. In a set of independent vectors, every subspace taken from these vectors is

independent.

b. About the dependence

4. If p vectors are dependent, at least one of these vectors is a linear combi-

nation of the others.

Indeed, consider p vectors 1 2, , . . . , pV V V

. If these vectors are linearly inde-

pendent, the relation:

1

0

p

ii

i

Vα=

=

(1.15)

involves that at least one of the real numbers αi is non zero: α1 for example. The

preceding relation is written:

( )1 21 2 . . . ppV V Vα α α= − + +

, (1.16)

and it is possible to divide by α1 (different from zero) and to express 1V

in the

form:

11

2

1p

ii

i

V Vαα

=

= −

. (1.17)

We say then that 1V

depends linearly of the vectors 2 3, , . . . , .pV V V

5. If 1 2, , . . . , pV V V

are linearly dependent, the vectors 1 2, , . . . , ,pV V V

1 , . . . , ,p p rV V+ + are also dependent whatever are the vectors 1 ,pV +

. . . , .p rV +

6. Theorem

In the subspace generated by p linearly independent vectors, every vector can

be expressed in a unique way as a linear combination of these p vectors.

Let 1 2, , . . . , ,pV V V

be p linearly independent vectors. Every vector V

is

written in a unique way as:

1

p

ii

i

V Vα=

=

. (1.18)

From this theorem is deduced the following important result:

A vector equality between p independent vectors of the form:

1 1

p p

i ii i

i i

V Vα α= =

′=

(1.19)

1.2 Linear Dependence and Independence. Basis for 3 7

is equivalent to p scalar equalities between the real numbers:

1 1 2 2, , . . . , p pα α α α α α′ ′ ′= = = . (1.20)

This property is no more true if the vectors are dependent.

1.2.3 Basis of the Vector Space R3

Searching for sets of independent vectors in the vector space 3 can be imple-

mented in the following way.

We have noted previously that a non zero vector V

is by itself linearly inde-

pendent. Thus, we choose a non zero vector 1V

of 3 . Then, we search for a

vector 2V

such as 1V

and 2V

are linearly independent; and then a vector 3V

such

as 1V

, 2V

, 3V

are linearly independent; etc. So, we observe that it is possible to

obtain a set of 3 linearly independent vectors (there exists an infinity of such

sets), and if we add a fourth vector 4V

, the four vectors 1V

, 2V

, 3V

and 4V

are

linearly independent whatever the vector 4V

is. Thus, the vector space is a space

of dimension 3.

Every set of 3 linearly independent vectors is then called a basis of the vector

space 3 .

It results from the properties reported previously:

1. Every vector of 3 is expressed (in a unique form) as a linear combination

of the 3 vectors of the basis.

2. The whole set of the linear combinations of the 3 vectors of the basis

generates the vector space 3 .

The vector space 3 is thus determined entirely when a basis is given.

1.2.4 Components of a Vector

Let 1 2 3, , e e e

be three vectors of 3 which are linearly independent. Their set

( )1 2 3( ) , , b e e e=

constitutes a basis of the space 3 . According to the previous

properties, every vector V

of 3 is written in a unique way as follows:

1 1 2 2 3 3V C e C e C e= + +

. (1.21)

The real numbers (C1, C2, C3) are then called the components of the vector with

respect to the basis (b). Ci is the component along ie

.


1.3 SCALAR PRODUCT

1.3.1 Definition

We call scalar product of two vectors V

and W

a law of external composition

which associates to these two vectors a real number (said a scalar) denoted by

V W⋅

:

3 , V W∀ ∈

V W ∈⋅

,

having the following properties:

( )2 21 1 ,V V W V W V W+ = +⋅ ⋅ ⋅

(1.22)

( ) ( ) ,V W V Wα α=⋅ ⋅

(1.23)

,V W W V=⋅ ⋅

(1.24)

0 si 0 .V V V> ≠⋅

(1.25)

The first two properties express the linearity of the scalar product with respect

of the vector .V

In particular 0 0V⋅ =

.

The third property expresses that the scalar product is symmetric with respect

to V

and W

. It results that the scalar product is also linear with respect to .W

These properties may be summarized by saying that the scalar product of two

vectors ,V

W

is a symmetric linear form associated to the vectors V

and .W

1.3.2 Magnitude or Norm of a Vector

We call magnitude or norm of the vector V ,

that we shall denote by V

, the

positive square root of the scalar product of the vector by itself.

Thus:

2

,V V V V= =⋅

(1.26)

by denoting:

2.V V V=⋅

(1.27)

In particular, we have:

V Vα α=

, (1.28)

1 2 1 2 1 2V V V V V V− ≤ + ≤ +

. (1.29)

This last inequality is called triangle inequality.

scalar product

1.3 Scalar Product 9

1.3.3 Analytical Expression of the Scalar Product in an Arbitrary Basis

Consider two vectors V

and .V ′ Their expressions in the basis ( )1 2 3, , e e e

of

the space 3 are:

1 1 2 2 3 3V C e C e C e= + +

, (1.30)

1 1 2 2 3 3V C e C e C e′ ′ ′ ′= + +

. (1.31)

The scalar product of these two vectors is written as:

( ) ( ) 1 1 2 2 3 3 1 1 2 2 3 3V V C e C e C e C e C e C e′ ′ ′ ′= + + + +⋅ ⋅

. (1.32)

By considering the properties (1.22) to (1.24), the preceding expression may be

written:

( )( )

( )( ) ( )( )

2 2 21 1 1 2 2 2 3 3 3 1 2 2 1 1 2

2 3 3 2 2 3 3 1 1 3 3 1 .

V V C C e C C e C C e C C C C e e

C C C C e e C C C C e e

′ ′ ′ ′ ′ ′= + + + +

′ ′ ′ ′+ + + +

⋅ ⋅

⋅ ⋅

(1.33)

This relation expresses the scalar product of the two vectors V

and V ′ in an arbi-

trary basis. This expression simplifies by considering particular bases that we

introduce hereafter.

1.3.4 Orthogonal Vectors

We say that two vectors are orthogonal if and only if their scalar product is

zero.

Thus:

and orthogonal 0.V W V W⇔ =⋅

(1.34)

Theorem: If n non zero vectors (n = 2 or 3) are pairwise orthogonal, they are

linearly independent. If n = 3, the vectors constitute an orthogonal basis of 3 .

1.3.5 Orthonormal Basis

A basis is orthonormal, if the vectors which constitute this basis are pairwise

orthogonal (orthogonal basis) and if their norms are equal to 1 (basis normed to

1).

If the basis ( )1 2 3, , e e e

is orthonormal, we have then:


1 2 2 3 3 10, 0, 0,e e e e e e= = =⋅ ⋅ ⋅

(1.35)

2 2 21 2 31, 1, 1.e e e= = =

(1.36)

1.3.6 Expression of the Scalar Product in an Orthonormal Basis

In the case of an orthonormal basis, Expression (1.33) of the scalar product

simplifies and reduces to:

1 1 2 2 3 3V V C C C C C C′ ′ ′ ′= + +⋅

. (1.37)

The scalar product with respect to an orthonormal basis is then equal to the

sum of the product of the corresponding components of the vectors.

The norm of a vector is written:

2 2 21 2 3V C C C= + +

. (1.38)

1.4 VECTOR PRODUCT

1.4.1 Definition

We call vector product of two vectors V

and W

a law of internal composition

in 3 , which associates to these two vectors a vector denoted by V W×

and

which is an antisymmetric bilinear law:

3 , V W∀ ∈

3 .V W× ∈

From this definition, it results that:

1. The vector product is distributive on the left and on the right for the vector

sum:

( )1 2 1 2V V W V W V W+ × = × + ×

, (1.39)

( )1 2 1 2V W W V W V W× + = × + ×

. (1.40)

2. The vector product is associative for the multiplication by a real number:

( ) ( ) ,V W V Wα α× = ×

(1.41)

( ) ( ).V W V Wα α× = ×

(1.42)

3. The vector product is antisymmetric:

( )V W W V× = − ×

. (1.43)

vector product

1.4 Vector Product 11

The last property, applied to the vector product of a vector by itself, involves

that:

( )V V V V× = − ×

.

Thus it results from this the property:

0V V× =

. (1.44)

From this property, we deduce the following theorem: Two non zero vectors

are collinear if and only if their vector product is the null vector.

In fact:

( ) ( ) collinear to 0W V W V W V V V V Vα α α⇔ = ⇔ × = × = × =

.

1.4.2 Analytical Expression of the Vector Product in an Arbitrary Basis

Consider again Expressions (1.30) and (1.31) of the two vectors V

and

V ′ expressed in the basis ( )1 2 3, , e e e

. The vector product of the two vectors is

written:

( ) ( ) 1 1 2 2 3 3 1 1 2 2 3 3V V C e C e C e C e C e C e′ ′ ′ ′= + + × + +×

. (1.45)

By considering the properties of distributivity and associativity of the product

vector, we obtain:

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

1 1 1 1 1 2 1 2 1 3 1 3

2 1 2 1 2 2 2 2 2 3 2 3

3 1 3 1 3 2 3 2 3 3 3 3 .

V V C C e e C C e e C C e e

C C e e C C e e C C e e

C C e e C C e e C C e e

′ ′ ′ ′= ∧ + ∧ + ∧

′ ′ ′+ ∧ + ∧ + ∧

′ ′ ′+ ∧ + ∧ + ∧

∧

By using the property of antisymmetry, this expression is reduced to the form:

( ) ( ) ( ) ( )

( )( )1 2 2 1 1 2 1 3 3 1 1 3

2 3 3 2 2 3 .

V V C C C C e e C C C C e e

C C C C e e

′ ′ ′ ′ ′= − ∧ + − ∧

′ ′+ − ∧

∧

(1.46)

This relation expresses the vector product of two vectors in an arbitrary basis.

Hereafter, we introduce particular bases which allow to simplify this expression.

1.4.3 Direct Basis

We call direct basis, a basis such as:

1 2 3 2 3 1 3 1 2, , .e e e e e e e e e× = × = × =

(1.47)

The basis is said to be oriented in the direct sense.


Thus, a direct basis is such as the vector product of two vectors give the third

one in the order 1, 2, 3, 1, 2, etc.

1.4.4 Expression of the Vector Product in a Direct Basis

In the case of a direct basis, Expression (1.46) of the vector product is reduced

to:

( ) ( ) ( )2 3 3 2 1 3 1 1 3 2 1 2 2 1 3V V C C C C e C C C C e C C C C e′ ′ ′ ′ ′ ′ ′= − + − + −×

. (1.48)

The preceding expression can be easily derived by expressing the vector

product in the form of a determinant (from a formalism viewpoint this writing is

however incorrect):

1 2 3

1 2 3

1 2 3

e e e

V V C C C

C C C

′ =

′ ′ ′

×

.

By expanding this determinant according to the first row, we obtain Expression

(1.48) effectively.

Furthermore, from Expression (1.48) it is easily derived that: The vector

product of V

and V ′ is a vector orthogonal to vector V

and vector .V ′

1.4.5 Mixed Product

We call mixed product of the three vectors 1 2 3, , ,V V V

considered in this

order, the real number defined by:

( )1 2 3V V V×⋅

. (1.49)

It is easy to show that, in a direct orthonormal basis, the mixed product is an

invariant in circular permutation of the three vectors.

( ) ( ) ( )1 2 3 2 3 1 3 1 2V V V V V V V V V× = × = ×⋅ ⋅ ⋅

. (1.50)

1.4.6 Property of the Double Vector Product

The double vector product of three vectors can be expressed by the relation:

( ) ( ) ( )1 2 3 1 3 2 1 2 3V V V V V V V V V× × = −⋅ ⋅

. (1.51)

1.5 Bases of the Vector Space 3 13

This equality can be easily verified by expressing the components of

( )1 2 3V V V× ×

, then the ones of ( ) ( )1 3 2 1 2 3V V V V V V−⋅ ⋅

, and then by veri-

fying that these components are equal.

1.5 BASES OF THE VECTOR SPACE R3

1.5.1 Canonical Basis

The basis of the space 3 which is the most used is the canonical basis defined

as the set of the three vectors:

( ) ( ) ( )1, 0, 0 , 0, 1, 0 , 0, 0, 1 ,i j k= = =

(1.52)

considered in this order.

We verify easily that the set ( ), , i j k

constitutes a direct orthonormal basis:

— orthonormal basis:

0, 0, 0,i j j k k i= = =⋅ ⋅ ⋅

(1.53)

2 2 21, 1, 1,i j k= = =

(1.54)

— direct basis:

, , .i j k j k i k i j× = × = × =

(1.55)

The demonstration assumes that the basis is expressed (1.52) in a basis which is

itself a direct orthonormal basis.

Afterwards, we shall denote by X, Y, Z the components of a vector V

with

respect to the canonical basis:

V X i Y j Z k= + +

. (1.56)

1.5.2 Basis Change

In this subsection, we derive, first considering an example, the relations of

basis change in the space 3 and in the case of direct orthonormal bases. Then,

the relations obtained will be generalized.

1.5.2.1 Example of a Basis Change

We consider the direct orthonormal basis ( ) ( )1 1 1 1, , b i j k=

and we derive from

this basis the set of the three vectors ( )2 2 2, , i j k

defined in the following way:


( )

( )

( )

2 1 1 1

2 1 1 1

2 2 2 1 1

1 2 ,6

1 ,3

1 .2

i i j k

j i j k

k i j j k

= − +

= − − +

= ∧ = − −

(1.57)

We verify easily that the set (b2) of these three vectors constitutes a direct ortho-

normal basis.

Relations (1.57) may be written in a practical form, derived from the matrix

notation, as follows:

2 1

2 1

2 1

2 1 1

6 6 6

1 1 1

3 3 3

1 102 2

i i

j j

k k

− = − −

− −

, (1.58)

or in condensed form:

2 1

2 1

2 1

i i

j j

k k

=

A

, (1.59)

by introducing the matrix of the basis change:

2 1 1

6 6 6

1 1 1

3 3 3

1 102 2

− = − −

− −

A . (1.60)

We find easily the following properties of the matrix of the basis change:

— the determinant of A is equal to 1 ;

— If we express ( )1 1 1, , i j k

as a function of ( )2 2 2, , i j k

according to Rela-

tions (1.57), we obtain:

column matrix

of the basis (2) matrix of

basis change

column matrix

of the basis (1)

1.5 Bases of the Vector Space 3 15

1 2

1 2

1 2

2 1 06 3

1 1 1

6 3 2

1 1 1

6 3 2

i i

j j

k k

− = − − −

−

. (1.61)

The matrix inverse of A is equal to the matrix transposed of A:

1 t− =A A . (1.62)

Consider now the relations which exist between the components of a vector V

expressed in the two bases under consideration:

— in the basis (b1), we have:

(1) (1) (1)1 1 2 1 3 1V C i C j C k= + +

, (1.63)

— in the basis (b2), we have:

(2) (2) (2)1 2 2 2 3 2V C i C j C k= + +

, (1.64)

By substituting Relation (1.61) into Expression (1.63), we obtain:

(1) (1)1 2 2 2 2 2 2

(1)3 2 2 2

2 1 1 1 1

6 3 6 3 2

1 1 1 ,6 3 2

V C i j C i j k

C i j k

= − + − − −

+ + −

hence:

(1) (1) (1)1 2 3 2

(1) (1) (1) (1) (1)1 2 3 2 2 3 2

2 1 1

6 6 6

1 1 1 1 1 .3 3 3 2 2

V C C C i

C C C j C C k

= − +

+ − − + + − −

By comparing this result with Expression (1.64), we derive:

(2) (1) (1) (1)1 1 2 3

(2) (1) (1) (1)2 1 2 3

(2) (1) (1)3 2 3

2 1 1 ,6 6 6

1 1 1 ,3 3 3

1 1 .2 2

C C C C

C C C C

C C C

= − +

= − − +

= − −

(1.65)

By introducing the column matrices of the components in the basis (b2) and in the

basis (b1), Expression (1.65) is then written as:


(2) (1)1 1

(2) (1)2 2

(2) (1)3 3

C C

C C

C C

=

A . (1.66)

In the same way, the inverse relation is written:

(1) (2)1 1

(1) t (2)2 2

(1) (2)3 3

C C

C C

C C

=

A . (1.67)

1.5.2.2 Generalizing

The results established in the preceding subsection for a particular case can be

generalized and expressed in the following way.

Any transformation from a direct orthonormal basis to another direct ortho-

normal basis is characterized by a square matrix, such as the determinant is equal

to 1 and the inverse matrix is the transposed matrix. Reciprocally, every matrix

which has these properties represents a change of direct orthonormal bases.

If ( )1 1 1, , i j k

and ( )2 2 2, , i j k

are two direct orthonormal bases, the basis

change is expressed in the practical form:

2 1 1 2t

2 1 1 2

2 1 1 2

, .

i i i i

j j j j

k k k k

= =

A A

(1.68)

Between the components of a vector in the two bases, we have analogous expres-

sions:

(2) (1) (1) (2)1 1 1 1

t2 2 2 2

3 3 3 3

, .

C C C C

C C C C

C C C C

= =

A A (1.69)

EXERCISES

1.1 Derive the unit vectors collinear to a given vector. Apply to the case of the

vector of components (2, –5, 3) in the canonical basis.

1.2 Determine the parameter α, in such a way as the vectors ( )1 5, 4, 3V =

and

( )2 , 2, 1V α= −

are orthogonal. The components of the vectors are given in an

orthonormal basis.

Comments 17

1.3 Derive the unit vectors orthogonal to two given vectors.

Apply to the case of the vectors of components (2, –5, 3) and (–2, 1, –3) with

respect to the canonical basis.

1.4 Expand the scalar product ( ) ( )1 2 1 2V V V V+ −⋅

; then the vector product

( ) ( )1 2 1 2V V V V+ × −

.

1.5 A vector V

has for components (4, –9, 3) in the basis ( ) ( )1 1 11 , , i j k=

. We

consider the basis ( ) ( )2 2 22 , , i j k=

deduced from (1) by the relations:

2 1 2 1 2 12 , 2 , i i j j k k= = = −

.

Express the components of V

in the basis (2).

1.6 The vectors 1V

and 2V

are two given vectors. Derive the vectors V

such as:

1 2 1V V V V× = ×

.

Apply to the case where: 1 4V i j= −

and 2 5 6 2V i j k= + −

.

COMMENTS

The vector space 3 is the space whose the vectors are characterized by

their three components which are ordered real numbers. The vector space 3 is a mathematical space with abstract feature which can not be

represented in a concrete way. However, different operations have been

defined on this space that the reader must perfectly handle: vector addition,

scalar product, vector product. The scalar product leads to the notion of

orthogonality between two vectors and the vector product to the notion of

collinear vectors. The vector space 3 is generated from a basis

×constituted of three linearly independent vectors. The basis which is the

most used is the canonical basis This basis is direct and orthonormal. Any

other direct orthonormal basis can be derived from the canonical basis

introducing a square matrix, the determinant of which is equal to 1 and the

inverse matrix is the transposed matrix.

CHAPTER 2

The Geometric Space

2.1 THE GEOMETRIC SPACE CONSIDERED AS

SPACE AFFINE TO THE VECTOR SPACE R3

2.1.1 The Geometric Space

The geometric space allows us to characterize the surrounding physical space.

This space is constituted of points, called geometric points. The affinity allows us

“to formulate” the physical space (Figure 2.1), by reducing the operations in the

geometric space to operations in the vector space 3, already considered in the

preceding chapter.

Thus, the geometric space is the affine space associated to the vector space 3.

It is then denoted by ( )3 and it is related to the space 3 in the following way.

1. An application f is defined which associates to every ordered couple (A, B)

of two geometric points of ( )3 a vector V

of 3 and only one:

( )

( )

3

3

A

B

∀ ∈

∀ ∈

(A, B) 3V ∈ .

We have then:

( ),V f A B=

. (2.1)

That means that the vector V

is the result of the application f to the couple of

points (A, B). The ordered couple (A, B) is called bipoint of origin A and end B.

Lastly, there is a contraction of the notation, since it is usual to write:

V AB=

instead of ( ),V f A B=

. (2.2)

However, it is necessary to keep in mind that the notation V AB=

means that V

f

2.1 The Geometric Space Considered as Space Affine to the Vector Space 3 19

FIGURE 2.1. Formulation of the geometric space.

is the image in the space 3 of the bipoint (A, B) of the geometric space.

The bipoint (A, B) is represented conventionally according to the scheme of

Figure 2.2 distinguishing the origin A and the end B of the bipoint.

2. The application f is such as, for any points A, B, C of the geometric space,

we have the relation:

( ) ( ) ( ) , , ,f A B f B C f A C+ =

, (2.3)

or in contracted notation:

AB BC AC+ =

, (2.4)

This relation is known as Chasles relation.

2.1.2 Consequences

1. If the points A and B are the same, Expression (2.4) leads to:

0AB =

.

2. If the points A and B are different, 0AB ≠

.

3. If the points A and C are the same, Expression (2.4) leads to:

0AB BA+ =

hence BA AB= −

. (2.5)

FIGURE 2.2. Bipoint of origin A and end B.

Vector Space 3

1 2 3 . . . .V V V

formulation

Geometric Space

geometric point vector

A

B

AB V=

20 Chapter 2 The Geometric Space

4. It results that Chasles relation is written in the equivalent forms:

BC AC AB= −

, (2.6)

0AB BC CA+ + =

. (2.7)

5. Mid-point of a bipoint. The point I is the mid-point of bipoint (A, B) or seg-

ment AB if and only if:

AI IB=

. (2.8)

It results that, if O is a point of the geometric space, we have:

( )12

OI OA OB= +

. (2.9)

6. Equipollent bipoints. Two bipoints are equipollent if and only if they have

the same image in the space 3.

(A, B) equipollent to (C, D) AB CD⇔ =

. (2.10)

The quadrilateral ABCD is a parallelogram.

2.1.3 Distance between Two Points

We call distance between two points A and B or length of the segment AB, the

norm of the vector AB

.

The distance between the points A and B is denoted by d(A, B) and we have:

( )2

,d A B AB AB AB= = =

. (2.11)

The properties of the distance result from the ones of the scalar product and

norms of two vectors of 3:

— ( ), 0d A B = ⇔ A and B are the same points,

— ( ) ( ) , ,d A B d B A= ,

— ( ) ( ) ( ) , , ,d A B d A C d C B≤ + , the equality is satisfied only if the point C

is a point of the segment AB.

2.1.4 Angle between Two Bipoints

The notion of angle associated with the notion of distance allows us to localize

every point of the geometric space ( )3 .

The angle γ (Figure 2.3) between the two bipoints (A, B) and (A, C) with the

same origins and considered in this order, also called angle between the vectors

AB

and AC

, is denoted by:

( ),AB ACγ =

. (2.12)

2.1 The Geometric Space Considered as Space Affine to the Vector Space 3 21

FIGURE 2.3. Angle between two bipoints.

This oriented angle is defined by its cosinus and its sinus which occur in the

expressions of the scalar product and vector product of the vectors AB

and AC

in the following way:

— scalar product:

cos cosAB AC AB AC AB ACγ γ= =⋅

, (2.13)

— vector product:

sin sinAB AC u AB AC u AB ACγ γ× = =

, (2.14)

where u

is the unit vector associated (Figure 2.4) to the unit bipoint (A, U) (or to

an equipollent bipoint) orthogonal to the plane (ABC) and such as an observer,

placed the feet in A and the head at the end U, must move from its right to its left

to direct its glance from the end B of the first bipoint to the end C of the second

one.

The expression of the vector product orientates the geometric space.

2.1.5 Reference Systems

The problem to be solved is that of the determination of the position of every

point M of the geometric space.

We choose a particular point O of the geometric space as reference point

(Figure 2.5). To every point M of the geometric space corresponds then in a bi-

unique way a vector OM

of 3 image of the bipoint (O, M). Thus, the vector

OM

allows us to characterize in a unique way the position of the point M. This

FIGURE 2.4. Orientation.

A

B

C

A B

C

U


FIGURE 2.5. Characterization of the position of a point.

vector is called position vector of the point M. Next, this vector is characterized

by its components in a basis (b).

The data of the point O and the basis (b) thus makes it possible to characterize

the position of any point M of the geometric by reporting the ordered set of the

three components the vector OM

in the basis (b).

The unit constituted by a reference point O of the geometric space and by a

basis (b) of the vector space 3 is called reference system of the geometric space.

We shall denote by (O/b).

The point O is called origin of the reference system. The components of the

position vector OM

in the basis (b) are called the coordinates of the point M in

the reference system (O/b).

2.2 SUBSPACES OF THE GEOMETRIC SPACE

LINE, PLANE

2.2.1 Line

A line (D), denoted by ( )1,A V

is the set (D) of points M of the geometric

space, such as the vectors AM

are collinear to the vector 1V

(Figure 2.6).

( ) 1 , M D AM Vα α∈ ⇔ = ∀ ∈

. (2.15)

The straight line (D) passes through the point A. The vector 1V

is called

direction vector of the line (D). We say that (D) is the line that passes through the

point A and with direction 1V

. A line (D) is only defined by a point of the line

and a direction vector.

We call axis a line on which we have chosen a one-dimensional reference: a

point O for the origin and a direction vector .V

M (arbitrary

point)

O (point of

reference)

geometric space

2.2 Subspaces of the Geometric Space. Line, Plane 23

FIGURE 2.6. Straight line.

We shall denote such an axis by ( ),Ox O V=

. The conventional representation

of an axis (Figure 2.7) will report the origin O and the bipoint that has the vector

V

for image and the point O for origin. The real number α defining the position

of the point M on the axis:

OM Vα=

(2.16)

is called the abscissa of the point M on the axis Ox

.

The length of the segment OM is equal to α . The bipoint (O, M) is directed

in the positive sense if 0α > , in the negative sense if 0α < .

2.2.2 Plane

A plane (P), denoted by ( ) 1 2, ,A V V

is the set (P) of the points M of the geo-

metric space, such as the vectors AM

are linear combinations of the vectors

1V

and 2 .V

( ) 1 21 2 1 2 , ,M P AM V Vα α α α∈ ⇔ = + ∀ ∈

. (2.17)

We say that (P) is the plane that passes through the point A and with direction

( )1 2,V V

.

It results from the various notions introduced previously that:

1. α1 and α2 are the components of the vector AM

in the two-dimensional

basis ( )1 2,V V

. They are also the coordinates of the point M of the plane (P) in

the reference system ( )1 2/ ,O V V

;

2. 11Vα

and 22Vα

are respectively the projections of the vector AM

in the

directions defined by 1V

et 2V

;

FIGURE 2.7. Axis.

A

(D)

M

1AM Vα=

O (D)

M

x

V


FIGURE 2.8. Decomposition of a bipoint.

3. if we introduce the point N such as:

1 21 2, AN V NM Vα α= =

, (2.18)

Relation (2.16) is written:

AM AN NM= +

. (2.19)

Whence the construction of the point N in Figure 2.8.

The bipoint (A, N) is the projection of the bipoint (A, M) on the axis ( )1,A V

;

the bipoint (N, M) is the projection on the axis ( )2,N V

.

Generally in the construction (Figure 2.9), we introduce the projection (A, P) of

the bipoint (A, M) on the axis ( )2,A V

, bipoint of origin A and equipollent to (N,

M).

In the case where the vectors 1V

and 2V

are orthogonal, the projections consi-

dered are orthogonal projections.

2.2.3 Lines and Planes with Same Directions

2.2.3.1 Lines with Same Direction

Two lines ( )1,A V

and ( )2,B V

are of the same direction (or are parallel), if

and only if the vectors 1V

and 2V

are collinear.

FIGURE 2.9. Projections on the axes.

A

1V

M

(P) N

2V

A

1V

M

(P) N

2V

P

2.2 Subspaces of the Geometric Space. Line, Plane 25

The two lines ( )1,A V

and ( )2,B V

have the same direction if and only if:

1 2V Vλ=

or 1 2 0V V× =

. (2.20)

If the points A and B are different, the lines have no common point. If the

points A and B are the same, the two lines are identical.

2.2.3.2 Planes with Same Direction

Two planes ( ) 1 2, ,A V V

and ( ) 1 2, ,B V V′ ′ are of the same direction (or are

parallel), if and only if the vector subspaces having ( )1 2,V V

and ( )1 2,V V′ ′ for

bases are identical.

Thus, the two planes have the same direction if and only if the vectors 1V ′ and

2 ,V ′ for example, are linearly dependent on the vectors 1V

and 2 :V

1 1 2 ,1 2

2 1 21 2 .

V V V

V V V

λ λ

µ µ

′ = +

′ = +

(2.21)

If the points A and B are different, the planes have no common point. If the

points A and B are the same, the two planes are identical.

2.2.3.3 Line Parallel to a Plane

The line ( ),A V

and the plane ( ) 1 2, ,B V V

are parallel if and only if V

is

linearly dependent on 1V

and 2 ,V

hence if and only if:

1 2V V Vλ µ= +

. (2.22)

2.2.4 Orthogonal Lines and Planes

2.2.4.1 Orthogonal Lines

Two lines are orthogonal if and only if the direction vectors are orthogonal.

The line ( )1,A V

is orthogonal to the line ( )2,B V

1 2 0V V⇔ =⋅

. (2.23)

2.2.4.2 Orthogonal Lines and Planes

The line ( ),A V

is orthogonal to the plane ( ) 1 2, ,B V V

if and only if the vector

V

is orthogonal to the vector 1V

and to the vector 2V

.

Thus:

1 20, 0.V V V V= =⋅ ⋅

(2.24)


2.2.4.3 Perpendicular Planes

Two planes are perpendicular if and only if a line of one of the planes is

orthogonal to the other plane.

2.3 CHARACTERIZATION OF THE POSITION

OF A POINT OF THE GEOMETRIC SPACE

2.3.1 Coordinate Axes

We have considered (Subsection 2.1.5) that any point M of the geometric space

could be characterized with respect to a reference system (O/b). The basis (b) is

constituted of three vectors 1 2 3, ,V V V

of the space 3, linearly independent. The

position of the point M is then characterized by the position vector OM

of the

space 3 associated to the bipoint (O, M). This vector is written:

1 2 31 2 3OM V V Vα α α= + +

. (2.25)

The parameters α1, α2, α3 are the components of the position vector OM

in the

basis ( ) 1 2 3, ,V V V

or the coordinates of the point M in the reference system

( )1 2 3/ , ,O V V V

.

The considerations developed in the previous subsections lead to the following

constructions (Figure 2.10). The reference system ( )1 2 3/ , ,O V V V

is represented

the three axes ( )1/O V

, ( )2/O V

and ( )3/O V

. On each axis, we report the

points N, P, Q with respective abscissae α1, α2, α3; hence such as:

1 2 31 2 3, , .ON V OP V OQ Vα α α= = =

(2.26)

We construct then the end R of the bipoint (N, R) equipollent to the bipoint (O,

P). From this it results:

1 21 2OR ON NR ON OP V Vα α= + = + = +

. (2.27)

Thus, the point M is the end of the bipoint (R, M) equipollent to the bipoint (O,

Q). We have well:

1 2 31 2 3OM OR RM OR OQ V V Vα α α= + = + = + +

. (2.28)

The bipoint (O, R) is the projection of the bipoint (O, M) in the plane

( )1 2/ ,O V V

. The bipoints (O, N), (O, P) and (O, Q) are the projections respect-

tively on the axes ( )1/O V

, ( )2/O V

and ( )3/O V

.

In the case where the vectors 1V

, 2V

and 3V

are orthogonal, the projections are

orthogonal projections.

2.3 Characterization of a Point of the Geometric Space 27

FIGURE 2.10. Projections of a point.

2.3.2 Direct Orthonormal Reference System

The reference system ( )1 2 3/ , ,O V V V

is a direct orthonormal reference system

if the vectors 1 2 3, ,V V V

constitute a direct orthonormal basis.

We have then:

1. 2 2 2

1 2 31, 1, 1.V V V= = =

The vectors are unit vectors.

2. 1 2 2 3 3 10, 0, 0V V V V V V⋅ = ⋅ = ⋅ =

. The axes ( )1/O V

, ( )2/O V

and

( )3/O V

are pairwise orthogonal. We say that the set of the three axes is a tri-

rectangular system.

3. 1 2 3 2 3 1 3 1 2, , V V V V V V V V V× = × = × =

. The system is oriented in

the direct sense: an observer, placed the feet at the point O and the head at the end

of the axis 3O

, must move from its right to its left to direct its glance from the

end of the 1-axis to the end of the 2-axis (Figure 2.11). The orientation of the

system is unchanged in a circular permutation of the indexes. It is also said that

the reference system is a right-handed system.

2.3.3 Cartesian Coordinates

The reference systems used are usually direct orthonormal reference systems

of which the basis is the canonical basis of the space 3, hence:

1 2 3, , V i V j V k= = =

. (2.29)

In the following, the axes will be denoted by:

( ) ( ) ( ) , , , , , ,Ox O i Oy O j Oz O k= = =

(2.30)

O

1V

M

N

2V

P

3V

R

Q


FIGURE 2.11. Direct or right-handed system.

and the reference system is denoted by:

( ) ( ) / , ,Oxyz O i j k=

. (2.31)

The points N, P, Q (Figure 2.12) considered in the subsection 2.3.1 have

respective abscissae on the axes which are denoted by x, y, z and are called

Cartesian coordinates of the point M.

The vector image of the bipoint (O, M) is written:

OM x i y j z k= + +

. (2.32)

The Cartesian coordinates of the point M are the components, in the canonical

basis of 3, of the vector .OM

FIGURE 2.12. Cartesian system.

3V

O 1V

2V

1

2

3

left

right

2

1

O

i

j

k

x

y

z

Q

P

R N

M

2.4 Plane and Line Equations 29

2.4 PLANE AND LINE EQUATIONS

2.4.1 Cartesian Equation of a Plane

We are searching for the Cartesian equation of the plane ( ) 1 2, ,A V V

:

— passing through the point A of Cartesian coordinates xA, yA, zA with respect

to the Cartesian system ( ) / , ,O i j k

;

— of direction defined by the vectors 1V

and 2V

of respective components

(X1, Y1, Z1) and (X2, Y2, Z2) in the canonical basis ( ) , ,i j k

.

Thus, we have:

1 1 1 1

2 2 2 2

,

,

.

A A AOA x i y j z k

V X i Y j Z k

V X i Y j Z k

= + +

= + +

= + +

(2.33)

The plane ( ) 1 2, ,A V V

is the set of the points M such as:

1 21 2 1 2, ,AM V Vα α α α= + ∀ ∈

. (2.34)

The Cartesian equation of the plane is the relation which allows us to express

the Cartesian coordinates (x, y, z) of the point M:

OM x i y j z k= + +

. (2.35)

By expressing the vector AM

, we have:

( ) ( ) ( )A A AAM OM OA x x i y y j z z k= − = − + − + −

. (2.36)

By substituting then this expression into (2.34), and then by equating the respec-

tive components for i

, j

and k

, we obtain:

1 1 2 2

1 1 2 2

1 1 2 2

,

,

.

A

A

A

x x X X

y y Y Y

z z Z Z

α α

α α

α α

− = +

− = +

− = +

(2.37)

These equations are the parametric equations of the plane.

The Cartesian equation is obtained by eliminating the parameters α1 and α2.

Thus:

( )( ) ( ) ( ) ( )( )1 2 1 2 1 2 1 2 1 2 1 2 0A A AZ Y Y Z x x X Z Z X y y Y X X Y z z− − + − − + − − = .

(2.38)

The Cartesian equation of a plane is then of the form:

1 2 1 2 1 2 1 2

1 2 1 2

0,

with

, ,

, .A A A

ax by cz d

a Z Y Y Z b X Z Z X

c Y X X Y d ax by cz

+ + + =

= − = −

= − = − − −

(2.39)


Plane passing through three non-aligned points

To find the equation of the plane passing through the points A, B, C of res-

pective coordinates (xA, yA, zA ), (xB, yB, zB), (xC, yC, zC), we search for the plane

passing through the point A and of direction defined, for example, by the vectors

AC

and AB

:

( ) ( ) ( )

( ) ( ) ( )

1

2

,

.

C A C A C A

B A B A B A

V AC x x i y y j z z k

V AB x x i y y j z z k

→ = − + − + −

→ = − + − + −

(2.40)

By substituting the components of these vectors into Equation (2.38), we obtain

the equation of the plane.

Particular planes

— Plane ( ) ( ) , ,Oxy O i j=

: the vectors 1V

and 2V

are the vectors i

and .j

The

vector equation of the plane is written:

1 2 1 2, ,x i y j z k i jα α α α+ + = + ∀ ∈

, (2.41)

and the parametric equations are:

1 2 1 2, , 0, ,x y zα α α α= = = ∀ ∈ . (2.42)

— We find analogous equations for the planes (Oyz) and (Oxz).

2.4.2 Cartesian Equations of a Line

We search for the equation of the line ( )1,A V

passing through the point A and

of direction 1.V

With notations already used, the vector equation (2.15) leads to

the three parametric equations:

1

1

1

,

,

.

A

A

A

x x X

y y Y

z z Z

α

α

α

− =

− =

− =

(2.43)

If 1 1 1, and X Y Z are non zero, these equations lead to one of the following pairs of

equations:

( ) ( ) ( )

( ) ( ) ( )

1 1 1

1 1 1

1 1 1

1 1 1

, , ,

, , ,

A A A A A A

A A A A A A

Y Z Xy y x x z z y y x x z z

X Y Z

Z X Xz z x x x x y y y y z z

X Y Z

− = − − = − − = −

− = − − = − − = −

(2.44)

equations which may be written in the form:

1 1 1

AA Ay yx x z z

X Y Z

−− −= = . (2.45)

A line is then defined by two Cartesian equations.

2.5 Change of Reference System 31

O11i

1j1k

y1

z1 M

x1

z2

y2

O2

x2

2j

2i

2k

Particular cases

— If X1 = 0, the equations of the line are:

( )1

1

0,

.

A

A A

x x

Yy y z z

Z

− =

− = − (2.46)

This is the equation of a line contained in the plane Ax x= .

— We obtain similar equations in the case of lines contained in the plane

1 ( 0)Ay y Y= = or in the plane 1 ( 0)Az z Z= = .

2.5 CHANGE OF REFERENCE SYSTEM

In this section we consider only the case of direct orthonormal reference

systems.

2.5.1 General Case

We consider two reference systems (Figure 2.13):

( ) ( ) ( )

( ) ( ) ( )

1 1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2 2

/ , , ,

/ , , .

T O x y z O i j k

T O x y z O i j k

= =

= =

The problem to be solved is:

Starting from the coordinates with respect to the reference system (T2) of a

point M of the geometric space, we have to find the coordinates of M expressed

with respect to the reference system (T1).

FIGURE 2.13 Change of reference system .


The coordinates of M with respect to (T1): ( ) ( ) ( )(1) (1) (1), , ,x M y M z M are the

components in the basis ( ) 1 1 1, ,i j k

of the position vector 1O M

, hence:

( ) ( ) ( )(1) (1) (1)1 1 1 1O M x M i y M j z M k= + +

. (2.47)

The coordinates of M with respect to (T2): ( ) ( ) ( )(2) (2) (2), , ,x M y M z M are

the components in the basis ( ) 2 2 2, ,i j k

of the position vector 2O M

, hence:

( ) ( ) ( )(2) (2) (2)2 2 2 2O M x M i y M j z M k= + +

. (2.48)

Between the vectors 1O M

and 2O M

, we have the relation:

1 1 2 2O M O O O M= +

. (2.49)

By introducing the coordinates with respect to the system (T1) of the point O2

origin of the system (T2): ( ) ( ) ( )(1) (1) (1)2 2 2, , ,x O y O z O the vector 1 2O O

is written

( ) ( ) ( )(1) (1) (1)1 2 2 1 2 1 2 1O O x O i y O j z O k= + +

. (2.50)

By substituting the expressions of the vectors 1O M

, 2O M

and 1 2O O

into

Relation (2.49), we observe that, for expanding this relation, it is necessary to

apply the basis change (1.67) to the components of the vector 2O M

. All the

components are then expressed in the basis ( ) 1 1 1, ,i j k

. Thus, Relation (2.49) leads

to the relation for the coordinate transformation:

( )

( )

( )

( )

( )

( )

( )

( )

( )

(1) (1) (2)2

(1) (1) t (2)2

(1) (1) (2)2

x M x O x M

y M y O y M

z M z O z M

= +

A , (2.51)

coordinates of coordinates of transposed coordinates of

the point M the point O2 matrix of the the point M

expressed in (T1) expressed in (T1) basis change expressed in (T2)

where A is the matrix of the basis change defined by Expression (1.62).

If the reference systems (T1) and (T2) have the same origin, the point O1 and

O2 coincides and Relation (20.51) is the same as Expression (1.63). Thus, it is

only necessary to derive the matrix of the basis change in the case of systems

having the same origin.

2.5.2 Reference Systems with a Same Axis

Consider the system ( ) ( ) 1 1 1 1/ , , .T O i j k=

We transform (Figure 2.14) this


FIGURE 2.14. Reference systems with a same axis.

reference system (T1) under a rotation through an angle γ in the direct sense about

the direction 1k

. We obtain the system ( ) ( ) 2 2 2 2/ , ,T O i j k=

. We shall denote:

( ) 1 1 1/ , ,O i j k

( ) 2 2 2/ , ,O i j k

.

Between the basis vectors, we have linear relations of the type:

2 11 1 12 1 13 1

2 21 1 22 1 23 1

2 1

,

,

.

i a i a j a k

j a i a j a k

k k

= + +

= + +

=

(2.52)

We search for the expressions of the coefficients aij, considering that the bases

( ) 1 1 1, ,i j k

and ( ) 2 2 1, ,i j k

are orthonormal and direct. Thus:

1 2

1 2

1 2 1

1 2 1

cos ,

cos ,

sin ,

sin .

i i

j

i i k

j k

j

j

γ

γ

γ

γ

=

=

× =

× =

⋅

⋅

(2.53)

The development of 1 2i i⋅

, taking account of (2.52), leads to:

( ) 1 2 1 11 1 12 1 13 1 11i i i a i a j a k a= + + =⋅ ⋅

.

Thus, by comparing to (2.53):

11 cosa γ= . (2.54)

We obtain similarly:

1 2 22 cosj aj γ= =⋅

. (2.55)

O

1i

1k

y1

z1

x1

y2

x2

2j

( )1,k γ

1j

2i


12

1 2 12 1 13 1 113

sin ,sin , thus

0.

ai i a k a j k

a

γγ

=× = − =

=

(2.56)

21

1 2 21 1 23 1 123

sin ,sin , thus

0.

aj j a k a j k

a

γγ

= −× = − + =

=

(2.57)

Relations (2.52) are thus written:

2 1 1

2 1 1

2 1

cos sin ,

sin cos ,

.

i i j

j i j

k k

γ γ

γ γ

= +

= − +

=

(2.58)

The matrix of the basis change is:

cos sin 0

sin cos 0

0 0 1

γ γ

γ γ

= −

A . (2.59)

The relation inverse of the basis change is written by transposing Expression

(2.58); We obtain:

1 2 2

1 2 2

1 2

cos sin ,

sin cos ,

.

i i j

j i j

k k

γ γ

γ γ

= −

= +

=

(2.60)

2.5.3 Arbitrary Reference Systems with the Same Origin

Hereafter, we show that it is always possible to transform a reference system

( )1 1 1Ox y z into a system ( )2 2 2Ox y z with the same origin but arbitrary with

respect to the first one, by applying three successive rotations (Figure 2.15).

1. The first rotation, through an angle ψ about the direction 1k

, transforms the

initial system ( ) 1 1 1/ , ,O i j k

into the reference system ( ) 3 3 1/ , ,O i j k

:

( ) 1 1 1/ , ,O i j k

( ) 3 3 1/ , ,O i j k

.

The basis change is written:

3 1 1

3 1 1

1

cos sin ,

sin cos ,

,

i i j

j i j

k

ψ ψ

ψ ψ

= +

= − +

(2.61)

or

3 1

3 1

1 1

i i

j j

k k

ψ

=

A

, (2.62)

( )1,k ψ



cos sin 0

sin cos 0

0 0 1

ψ

ψ ψ

ψ ψ

= −

A . (2.63)

2. The second rotation, through an angle θ about the direction 3i

, leads next to

the reference system ( ) 3 4 2/ , ,O i j k

:

( ) 3 3 1/ , ,O i j k

( ) 3 4 2/ , ,O i j k

.


3

4 3 1

2 3 1

,

cos sin ,

sin cos ,

i

j j k

k j k

θ θ

θ θ

= + = − +

(2.64)

or

3 3

4 3

2 1

i i

j j

k k

θ

=

A

, (2.65)

introducing the matrix of the basis change:

FIGURE 2.15 Eulerian angles.

( )3,i θ

z1

x1

O1i

1k

y1

y3

3j

3i

x3

1j

2k

2i

4j

x2

2j

y4

y2

z2


1 0 0

0 cos sin

0 sin cos

θ θ θ

θ θ

= −

A . (2.66)

The system ( )3 4 2Ox y z is not arbitrary with respect to the first one, since the axis

3Ox

is contained in the plane ( )1 1Ox y of the first system. So, it is necessary to

consider a third rotation to obtain an arbitrary system.

3. The third rotation, through an angle ϕ around the direction 2 ,k

leads to the

second system ( )2 2 2 ,Ox y z which is arbitrary with respect to the first one:

( ) 3 4 2/ , ,O i j k

( ) 2 2 2/ , ,O i j k

.


2 3 4

2 3 4

2

cos sin ,

sin cos ,

,

i i j

j i j

k

ϕ ϕ

ϕ ϕ

= +

= − +

(2.67)

or

2 3

2 4

2 2

i i

j j

k k

ϕ

=

A

, (2.68)


cos sin 0

sin cos 0

0 0 1

ϕ

ϕ ϕ

ϕ ϕ

= −

A . (2.69)

The three rotation angles then introduced are called the Eulerian angles: ψ is the

precession angle, θ the nutation angle, ϕ is the proper rotation angle or the spin

angle. They completely characterize the situation of the second system.

The relation of the basis change which expresses ( ) 2 2 2, ,i j k

as a function of

( ) 1 1 1, ,i j k

introduces the matrix A of basis change:

2 1

2 1

2 1

i i

j j

k k

=

A

. (2.70)

This relation can be obtained by combining Relations (2.61), (2.64) and (2.67).

So, this relation is deduced from the matrix (2.62), (2.65) and (2.68). Indeed, we

may write from these relations:

2 3 3 1

2 4 3 1

2 2 1 1

( ) ( ( ))

i i i i

j j j j

k k k k

ϕ ϕ θ ϕ θ ψ

= = =

A A A A A A

,

( )2 ,k ϕ

Exercises 37

or by taking into account the associativity of the matrix product:

( )2 1

2 1

2 1

i i

j j

k k

ϕ θ ψ

=

A A A

. (2.71)

The comparison between relations (2.70) and (2.71) leads to:

ϕ θ ψ=A A A A . (2.72)

The matrix of the basis change is equal to the product of the three matrices in the

order: 3rd rotation, 2nd rotation, 1st rotation. Calculation leads to:

cos cos sin cos sin sin cos cos cos sin sin sin

cos sin sin cos cos sin sin cos cos cos sin cos

sin sin cos sin cos

ψ ϕ ψ θ ϕ ψ ϕ ψ θ ϕ θ ϕ

ψ ϕ ψ θ ϕ ψ ϕ ψ θ ϕ θ ϕ

ψ θ ψ θ θ

− +

= − − − +

−

A .

(2.73)

EXERCISES

2.1 Derive the coordinates of the orthogonal projection H of a point M on the line

(D) (Figure 2.16). The coordinates x, y, z of the point M are given and the line (D)

passes through the origin point O and has V

for direction vector.

Apply the result to the case where the direction vector V

has (1, –2, 3) for

components in the canonical basis.

2.2 Derive the equation of the line passing through the point A (–1, 2, 1) and or-

thogonal to the plane passing through the three points A, B (2, 3, –1), C (–3, 4, –2).

2.3 Show that the triangle having for vertices the points of Cartesian coordinates:

( ) ( ) ( ) 2, 0, 2 , 1, 2, 1 , 1, 2, 1 ,A B C− − − −

is an isoscele triangle rectangular at A.

FIGURE 2.16. Orthogonal projection of a point on a line.

M (x, y, z)

H

OV

(D)


FIGURE 2.17. Parallelepiped.

2.4 Derive the area of the triangle ABC, the vertices of which have the Cartesian

coordinates:

A (–1, –2, –1) , B (2, 2, –1) , C (3, 2, 1) .

2.5 Calculate the volume of a parallelepiped (Figure 2.17), constructed on the

bipoints (A, B), (A, C) and (A, D).

Apply the result to the case of the points of Cartesian coordinates:

A (0, 0, 0) , B (3, 2, 1) , C (1, 1, 2) , D (–1, –1, 2) .

2.6 Calculate the distance from the point D to the plane that passes (Figure 2.18)

through three points A, B and C.

Apply the results to the case where the Cartesian coordinates are:

A (0, 0, 0) , B (1, 2, 3) , C (2, 1, 1) , D (–2, –1, –3) .

2.7 Find the necessary and sufficient condition for which the four points A, B, C

and D are contained in the same plane.

2.8 We consider a direct orthonormal system ( ) ( ) 1 1 1 1/ , ,T O i j k=

. This system is

transformed using a rotation through an angle of 30° around the axis ( )1,O i

: we

obtain the system ( ) 1 3 2/ , ,O i j k

. Next, we apply to this new system a rotation

FIGURE 2.18. Distance from a point to a plane.

AB

C

D

H

A B

C

D

Comments 39

through an angle of 45° around the axis ( )2,O k

: we obtain the reference system

( ) ( ) 2 2 2 2/ , ,T O i j k=

.

1. Derive the relations which express the basis vectors ( ) 2 2 2, ,i j k

of the

system ( )2T as functions of the basis vectors ( ) 1 1 1, ,i j k

of the system ( )1T .

2. A point M has for Cartesian coordinates (–1, 2, 4) with respect to the system

( )1 1 1Ox y z . Derive the Cartesian coordinates with respect to the system ( )2 2 2Ox y z .

3. A point N has for Cartesian coordinates (3, –4, 8) with respect to the system

( )2 2 2Ox y z . Derive the coordinates with respect to the system ( )1 1 1Ox y z .

COMMENTS

The notion of geometric space allows us to characterize the surrounding

physical space. This space is then a concrete space constituted of geometric

points, and its formulation is obtained by reducing the operations in the

geometric space to operations in the vector space 3, operations introduced

in the previous chapter.

An arbitrary solid of the geometric space is defined by the set of its

points of which the positions are determined by their coordinates. The

coordinates that are the most used are the Cartesian coordinates defined

with respect to a Cartesian reference system constituted of an origin O and

three tri-rectangular axes Ox

, Oy

and Oz

, with unit direction vectors

, and ,i j k

the vectors of the canonical basis. When the set of the coordi-

nates of the points of the solid are obtained, all information about the solid

is known and the figure can be deleted. Next, all the properties or

transformations of the solid are obtained from operations in the vector

space 3, implemented on the image vectors of the bipoints of the solid.

CHAPTER 3

Vector FunctionDerivatives of a Vector Function

3.1 VECTOR FUNCTION OF ONE VARIABLE

3.1.1 Definition

If, to any value of a real variable q, there corresponds a vector ,V

then V

is

called a vector function of the scalar variable q.

We shall denote such a function by ( )V q

:

q ∈ ( ) 3V q ∈

. (3.1)

If the vector ( )V q

is defined by its components in a given basis, these compo-

nents are real functions of the variable q. The vector function will be then deter-

mined by reporting the three components of the ( )V q

: X(q), Y(q), Z(q), for

example in the basis ( ) ( ) 1 2 3, ,b e e e=

:

1 2 3( ) ( ) ( ) ( )V q X q e Y q e Z q e= + +

. (3.2)

3.1.2 Derivative

The first derivative of the vector function ( )V q

with respect to the variable q

and in the basis ( ) ( ) 1 2 3, ,b e e e=

, that we shall denote by:

( )dd

b

Vq

or

( )dd

b Vq

, (3.3)

is defined by:

3.1 Vector Function of One Variable 41

( )

1 2 3d d d dd d d d

b X Y ZV e e eq q q q

= + +

. (3.4)

In the problems where only one basis is considered, the vector obtained is simply

called the derivative vector of ( )V q

with respect to q and it will be denoted by

ddVq

. Thus, it is implied that the derivative is concerned in the considered basis. In

the case where several bases are considered (case of the Mechanics of Solids), it is

necessary to specify the basis in which the derivation is implemented.

For example, if the vector ( )V q

is defined:

— in the basis ( ) ( ) 1 1 11 , ,i j k=

by:

(1) (1) (1)1 1 1( ) ( ) ( ) ( )V q i X q j Y q k Z q= + +

, (3.5)

— in the basis ( ) ( ) 2 2 22 , ,i j k=

by:

(2) (2) (2)2 2 2( ) ( ) ( ) ( )V q i X q j Y q k Z q= + +

, (3.6)

we shall distinguish:

— the vector (1)d

dV

q

, derivative of V

in the basis (1):

(1) (1) (1) (1)

1 1 1d d d dd d d d

X Y ZV i j kq q q q

= + +

, (3.7)

— and the vector (2)d

dV

q

, derivative of V

in the basis (2):

(2) (2) (2) (2)

2 2 2d d d dd d d d

X Y ZV i j kq q q q

= + +

. (3.8)

Generally, these two vectors are different.

3.1.3 Properties of the Vector Derivative

If the vectors ( ) ( ) ( )1 2 3, et V q V q V q

are vector functions of the same variable

q, we have in a given basis:

1. ( )

1 21 2

d ddd d d

V VV V

q q q+ = +

. (3.9)

This relation can be extended to the case of an arbitrary number of vectors.

2. ( )

1 21 2 2 1

d ddd d d

V VV V V V

q q q= +⋅ ⋅ ⋅

, (3.10)

42 Chapter 3 Vector Function. Derivatives of a Vector Function

with in particular:

2d d2d d

VV Vq q

= ⋅

. (3.11)

3. ( )

1 21 2 2 1

d ddd d d

V VV V V V

q q q× = × + ×

. (3.12)

4. ( ) ( )

31 21 2 3 2 3 1 3 1 2

dd ddd d d d

VV VV V V V V V V V V

q q q q

× = × + × + ×

⋅ ⋅ ⋅ ⋅

.

(3.13)

5. If f(q) is a real function of the variable q:

( )

dd dd d d

f Vf V V fq q q

= +

. (3.14)

In particular, if ( )f q k= independent of q:

( )

d dd d

VkV kq q

=

. (3.15)

6. If q is a function of the variable p: q(p)

dd dd d d

qV Vp q p

=

. (3.16)

7. The differential of the vector function V

is defined as:

dd dd d dd d d

qV VV q pq q p

= =

. (3.17)

As the derivative, the differential depends on the basis under consideration.

3.1.4 Examples

3.1.4.1 First Example

We consider, in the basis ( ) ( ) 1 , ,i j k=

, the vector:

( ) cos sinu i jα α α= +

. (3.18)

The derivative with respect to α in the basis (1) is:

(1)d ( ) sin cos cos ( ) sin ( )d 2 2

u i j i jπ πα α α α αα

= − + = + + +

.

Whence the important relation:

(1)d ( ) ( )d 2

u u πα αα

= +

. (3.19)

3.1 Vector Function of One Variable 43

Similarly, we obtain:

(1)d ( ) ( ) ( )d 2

u u uπα α π αα

+ = + = −

. (3.20)

Derivatives with respect to α, are equivalent to addition of / 2π to the angle α.

Moreover, we find easily that the vectors:

( ), ( ), ,2

u u kπα α +

(3.21)

constitute a direct orthonormal basis, that we shall denote by (2) in the following

example.

3.1.4.2 Second Example

We consider the function [ ]( ) sinV a u kα α= +

. We search for the derivatives

of V

with respect to α in the bases (1) and (2).

1. In the basis (2) ( ), ( ), 2

u u kπα α = +

.

The derivative is deduced by considering the previous expression of V

:

(2)d cosd

V a k αα

=

.

2. In the basis ( ) ( ) 1 , ,i j k=

.

— Ist method

We express the vector V

in the basis (1):

( )cos sin sinV a i j kα α α= + +

,

next, we deduce the derivative from this expression:

( )(1)d sin cos cosd

V a i j kα α αα

= − + +

.

Thus: (1)d ( ) cosd 2

V a u kπα αα

= + +

.

— 2nd method

We keep the expression of V

in the form written in the basis (2) and we

deduce from this expression the derivative in the basis (1):

[ ]

(1) (1) (1)d d d( ) sin ( ) cosd d d

V a u k a u kα α α αα α α

= + = +

.

Thus, taking account of the results obtained in the first example we obtain: (1)d ( ) cosd 2

V a u kπα αα

= + +

.


3.1.4.3 Third Example

Obtain the derivative of 2cos sin 2V a i b j c kα α= + +

with respect to α in

the basis ( ) , ,i j k

, where a, b and c are real number independent of α.

The derivative is easily obtained as:

d 2 cos sin 2 cos 2d

V a i b jα α αα

= − +

.

3.2 VECTOR FUNCTION OF TWO VARIABLES

3.2.1 Definition

If, to any pair of two real independent variables q1 and q2, there corresponds a

vector ,V

then V

is called a vector function of the scalar variables q1 and q2.

We denote such a function by 1 2( , )V q q

. The components of this vector are

functions of q1 and q2, and in the basis ( ) ( ) 1 2 3, , ,b e e e=

we have:

1 2 1 2 1 1 2 2 1 2 3( , ) ( , ) ( , ) ( , )V q q X q q e Y q q e Z q q e= + +

. (3.22)

3.2.2 Partial Derivatives

The partial derivatives of the function 1 2( , )V q q

in the basis (b) are defined as

follows:

— derivative with respect to q1 :

( )

1 2 31 1 1 1

b V X Y Ze e eq q q q

∂ ∂ ∂ ∂= + +∂ ∂ ∂ ∂

, (3.23)

— derivative with respect to q2 :

( )

1 2 32 2 2 2

b V X Y Ze e eq q q q

∂ ∂ ∂ ∂= + +∂ ∂ ∂ ∂

. (3.24)

When only one basis is considered, the partial derivatives are simply denoted

by 1

Vq

∂∂

and 2

Vq

∂∂

.

The differential of the vector function 1 2( , )V q q

is defined by:

( ) ( )( )

1 21 2

d d db b

b V VV q qq q

∂ ∂= +∂ ∂

. (3.25)

3.2 Vector Function of n Variables 45

If q1 and q2 are functions of the same variable p, the derivative of V

with respect

to p is given by:

( ) ( ) ( )1 2

1 2

d ddd d d

b b bq qV V Vp q p q p

∂ ∂= +∂ ∂

. (3.26)

3.2.3 Examples

Obtain the partial derivatives and the differential in the basis ( ) , ,i j k

of the

function:

( ) 2 2

1 2 1 1 2 1 2( , ) 2V q q a q i q q j q q k = + + +

.

We obtain easily:

( ) ( )

( ) ( )

1 2 1 21 2

1 2 1 1 2 2

2 2 , 2 ,

d 2 2 d 2 d .

V Va q i q j k a q j q kq q

V a q i q j k q q j q k q

∂ ∂= + + = +∂ ∂

= + + + +

3.3 VECTOR FUNCTION OF n VARIABLES

3.3.1 Definitions

The previous considerations can be extended to the case of an arbitrary number

of variables.

A vector function of the variables 1 2, , . . . , nq q q , associates to any set of the

values of these n variables a vector denoted by 1 2( , , . . . , )nV q q q

.

The components of the vector 1 2( , , . . . , )nV q q q

are real functions of the n

variables, and in the basis ( ) ( ) 1 2 3, , ,b e e e=

we have:

1 2 1 2 1 1 2 2 1 2 3( , , . . . , ) ( , , . . . , ) ( , , . . . , ) ( , , . . . , ) .n n n nV q q q X q q q e Y q q q e Z q q q e= + +

(3.27)

The partial derivative of the function V

with respect to the variable qi

( 1, 2, . . . , )i n= in the basis (b) is defined by:

( )

1 2 3

b

i i i i

V X Y Ze e eq q q q

∂ ∂ ∂ ∂= + +∂ ∂ ∂ ∂

, (3.28)

and the differential of V

in the basis (b) is written:

( )( )

1

d d

n bb

ii

i

VV qq

=

∂=∂

. (3.29)


If 1 2, , . . . , nq q q are functions of the same variable p, the derivative of V

with

respect to p in the basis (b) is:

( ) ( )

1

ddd d

nb bi

ii

qV Vp q p

=

∂=∂

. (3.30)

3.3.2 Examples

3.3.2.1 Example 1. Cylindrical Coordinates

Let M be a point of the geometric space localized by its Cartesian coordinates

(x, y, z). The position vector of the point M is:

OM x i y j z k= + +

. (3.31)

The position of the point M can be also characterized (Figure 3.1) by the para-

meters:

( ), , , -abscissa of ,r OH i OH z z Mα= = =

(3.32)

where H is the orthogonal projection of the point M in the plane (Oxy). The

position parameters (r, α, z) are called the cylindical coordinates of the point M.

The position vector (3.31) is then written:

cos sinOM r i r j z kα α= + +

,

or

( )OM r u z kα= +

. (3.33)

This is the expression of the position vector, when the point M is localized by its

cylindrical coordinates.

FIGURE 3.1. Cylindrical coordinates.

k

O

i

j

x

y

z

z

H x

M

( )u α

( )2

u πα +

y

r

3.2 Vector Function of n Variables 47

The vector ( )u α

(3.18) is the unit direction vector of the line OH. Similarly,

the vector ( )2

u πα +

is the unit direction vector of the line orthogonal to OH

(Figure 3.1). The system ( )/ ( ), ( ), 2

O u u kπα α +

is a direct orthonormal system.

Consider the partial derivatives of the position vector OM

with respect to r, α,

z, in the basis ( ) , ,i j k

. We have:

( ), ( ), ,2

OM OM OMu r u kr k

πα αα

∂ ∂ ∂= = + =∂ ∂ ∂

and the differential of the position vector is written:

d ( ) d ( ) d d2

OM u r r u k zπα α α= + + +

. (3.34)

If for example r, α and z are functions of the time t, the derivative with respect

to t in the basis ( ) , ,i j k

is the velocity vector of the point M with respect to the

reference system (T) = (Oxyz) and is written according to (3.30):

( )( ) d d d d( , ) ( ) ( )

d d d 2 d

TT OM r zM t u r u k

t t t tα πα α= = + + +

. (3.35)

The components of the velocity vector in the basis ( ), ( ), 2

u u kπα α +

are then:

( )

d d d, , d d d

r zrt t t

α . (3.36)

3.3.2.2 Example 2. Basis Change

Let ( ) 1 1 1, ,i j k

and ( ) 2 2 2, ,i j k

be two bases, of which the transformation from

one basis to the other is obtained by using the Eulerian angles (ψ, θ, ϕ). We wish

to express the partial derivatives, in the basis (1) and with respect to ψ, θ and ϕ, of

the vectors 2 2 2, ,i j k

. In this way, we consider again the three rotations introduced

in Subsection 2.5.3.

— 1st rotation

( ) 1 1 1/ , ,O i j k

( ) 3 3 1/ , ,O i j k

.

We have:

3 1 1

3 1 1

( ) cos sin ,

( ) sin cos .2

i u i j

j u i j

ψ ψ ψ

πψ ψ ψ

= = +

= + = − +

Whence

(1) (1)3 3

3 3, .i j

j iψ ψ

∂ ∂= = −

∂ ∂

(3.37)

( )1,k ψ


— 2nd rotation

( ) 3 3 1/ , ,O i j k

( ) 3 4 2/ , ,O i j k

.

We have:

4 3 1

2 3 1

( ) cos sin ,

( ) sin cos .2

j u j k

k u j k

θ θ θ

πθ θ θ

= = +

= + = − +

From this we deduce: (1) (1)

4 43 2

(1) (1)2 2

3 4

cos , ,

sin , .

j ji k

k ki j

θψ θ

θψ θ

∂ ∂= − =

∂ ∂

∂ ∂= = −

∂ ∂

(3.38)

— 3rd rotation

( ) 3 4 2/ , ,O i j k

( ) 2 2 2/ , ,O i j k

.

We have:

2 3 4

2 3 4

( ) cos sin ,

( ) sin cos .2

i u i j

j u i j

ϕ ϕ ϕ

πϕ ϕ ϕ

= = +

= + = − +

Whence the searched results:

(1) (1) (1)2 2 2

3 3 2 2

(1) (1) (1)2 2 2

3 3 2 2

(1) (1) (1)2 2 2

3 4

cos cos sin , sin , ,

sin cos cos , cos , ,

sin , , 0.

i i ij i k j

j j jj i k i

k k ki j

ϕ θ ϕ ϕψ θ ϕ

ϕ θ ϕ ϕψ θ ϕ

θψ θ ϕ

∂ ∂ ∂= − = =

∂ ∂ ∂

∂ ∂ ∂= − − = = −

∂ ∂ ∂

∂ ∂ ∂= = − =

∂ ∂ ∂

(3.39)

It follows from this that the differential in the basis (1) of 2i

is written:

( )

(1)2 3 3 2 2d cos cos sin d sin d di j i k jϕ θ ϕ ψ ϕ θ ϕ= − + +

, (3.40)

or

( )

(1)2 1 3 2 2d d d di k i k iψ θ ϕ= + + ×

. (3.41)

If the angles ψ, θ, ϕ (and consequently the vectors 2 2 2, ,i j k

) depend on the

variable p, we may write:

(1)2

2d

dp

ii

pω= ×

, (3.42)

by introducing:

1 3 2d ddd d d

p k i kp p p

ψ ϕθω = + +

. (3.43)

( )3,i θ

( )2 ,k ϕ

Comments 49

Similarly, we find:

(1) (1)2 2

2 2d d

, .d d

p pj k

j kp p

ω ω= × = ×

(3.44)

Important application

We seek for the expression of the derivative with respect to the variable p and

in the basis (1) of a vector V

whose components in the basis (2) are independent

of the parameter p , for example, the position vector of a point fixed in the

reference system ( ) 2 2 2/ , ,O i j k

. We have:

(2) (2) (2)2 2 2V X i Y j Z k= + +

, (3.45)

where the components X(2), Y(2), and Z(2) are independent of parameter p. The

derivative in the basis (1) of the vector V

is written:

(1)(1) (1)(1)(2) (2) (2)22 2dd dd

d d d d

ji kV X Y Zp p p p

= + +

, (3.46)

hence from (3.42) and 3.44):

( )

(1)(2) (2) (2)

2 2 2d

dp

V X i Y j Z kp

ω= × + +

. (3.47)

Whence the result: (1)dd

pV V

pω= ×

. (3.48)

This result will be used in Kinematics of Rigid Body (Chapter 9).

COMMENTS

The concepts of derivatives will be used in Kinematics (Part II). We

will have to express the velocity vectors and acceleration vectors of the

points of a rigid body. These vectors will be deduced from the derivatives

of the position vectors with respect to the time and in different reference

systems, that will lead to consider the derivatives in different bases. The

notions developed in Subsections 3.1, 3.2 and 3.3 must thus be perfectly

assimilated. The examples considered in this chapter are sufficient to illus-

trate the use of the vector derivatives.

The result (3.48) of Subsection 3.3.2.2 is an important result that will be

used in Kinematics of Rigid Body (Chapter 9). This result is interesting

owing to the fact that the derivative operation is replaced by a vector

product that is easier to implement, in particular in the case of a numerical

application.

CHAPTER 4

Elementary Concepts on Curves

4.1 INTRODUCTION

A curve (C) (Figure 4.1) may be defined in a reference system (T), as the set of

points M of the system whose the position vectors are determined by a vector

function of a parameter q: ( ),OM V q=

O being a reference point of the system

(T).

If the position vector is defined in the basis ( ) ( ) , , ,b i j k=

we have:

( ) ( ) ( )OM X q i Y q j Z q k= + +

. (4.1)

The components X(q), Y(q), Z(q) of the position vector are also the coordinates of

the point M in the reference system ( ) ( ) / , , .T O i j k=

Furthermore the curve (C) has a tangent at the point M of direction vector ( )dd

b Vq

or more generally

( ) ( ) dd d ,d d d

b b qV Vp q p

=

if q is a function of the variable p.

FIGURE 4.1. Curve.

O

M(q) (C)

4.2 Curvilinear Abscissa. Arc Length of a Curve 51

4.2 CURVILINEAR ABSCISSA

ARC LENGTH OF A CURVE

Among all the variables q which allow us to characterize the position of the

point M on the curve (C), a particular variable has been chosen, which will be

denoted by s, such as the vector ( )d

d

b OMs

is a unit vector:

2( ) ( )d d1 or 1d d

b bOM OMs s

= =

. (4.2)

Let M ′ be a point infinitely close to the point M (Figure 4.2) obtained by

increasing the variable s by the value d .s We have:

( )( ) dd d

d

bb OMMM OM OM OM s

s′ ′= − = =

. (4.3)

The arc length of the curve between the two points and M M ′ is the same as the

length MM ′ . Thus:

( ) ( )d dd dd d

b bOM OMMM MM s ss s

′ ′≈ = = ±

. (4.4)

Whence the result:

ds MM ′= ± . (4.5)

If M0 and M are two arbitrary points of the curve (C), the previous relation is

written as:

0 0( ) ( )s M s M M M− = ± . (4.6)

The variable s thus introduced measures the arc length of the curve. Its sign

depends on the orientation of the curve. We shall write for example:

0( ) ( )s M s M− = . (4.7)

The curve is thus oriented in the sense of the increase of s. The variable s is called

the curvilinear abscissa of the point M.

FIGURE 4.2. Arc length.

M0M

O

( )M s(C)

( d )M s s′ +

M0

52 Chapter 4 Elementary Concepts on Curves

If the point M0 is taken as the origin of the curvilinear abscissae, it follows

from this that: 0( ) 0,s M = and Relation (4.6) is written:

0( )s M M M= ± . (4.8)

4.3 TANGENT. NORMAL. RADIUS OF CURVATURE

From the definition of the curvilinear abscissa, it follows that the vector:

( )dd

b

tOMes

=

(4.9)

is a unit vector. The vector te

is thus the unit direction vector of the tangent to the

curve (C) at the point M, orientated in the sense of the increasing s. The orientated

tangent is the axis ( ), tM e

.

Since 2 1te =

, we obtain by considering the derivative of this expression with

respect to s and in the basis (b):

( )d0

d

bt

te

es

=⋅

. (4.10)

The vector ( )d

d

bte

s

is thus orthogonal to the vector te

, and we state:

( )d

d

bt ne e

s=

, (4.11)

where by definition:

— ne

is the unit vector of the principal normal direction of the curve (C) at

the point M ;

— is a positive scalar called the radius of curvature of the curve (C) at the

point M.

4.4 FRENET TRIHEDRON

The two vectors te

and ne

constitute the first two vectors of a direct ortho-

normal basis. The third vector, called the unit vector of the binormal direction of

the curve (C) at the point M, is defined by the relation:

b t ne e e= ×

. (4.12)

The basis thus obtained is called the basis of Frenet. It is a function of the

curvilinear abscissa s, hence of the point M. The moving reference system (Figure

4.3) ( )/ , , t n bM e e e

is called the Frenet trihedron.

4.4 Frenet Trihedron 53

FIGURE 4.3. Frenet trihedron.

The plane ( )/ ,t nM e e

is the osculating plane at M of the curve (C), the plane

( )/ ,n bM e e

is the normal plane at M of the curve (C), and the plane ( )/ ,b tM e e

is the rectifying plane at M of the curve (C).

The derivative of the position vector OM

with respect to the parameter q and

in the basis (b) is written:

d d d dd d d d

tOM OM s se

q s q q= =

. (4.13)

And considering the second derivative, we obtain:

( )

22 2 2

2 2 2

d dd d d d dd d d dd d d

t tt t

e eOM s s s se eq q s qq q q

= + = +

, (4.14)

or taking account of (4.11):

( )

22 2

2 2

d d ddd d

nt

eOM s seqq q

= +

. (4.15)

From this relation, it follows that:

1. The bipoint of origin M and image 2

2

d

d

OM

q

is contained in the osculating

plane of the curve (C) at M (Figure 4.4).

2. The scalar product:

( )22

2

d 1 ddd

nOM se

qq=⋅

, (4.16)

is always positive. The vector 2

2

d

d

OM

q

has thus always a positive component

in the direction defined by ne

and this component defines the convexity of

the curve (C) at the point M.

Lastly, the point D defined by:

nMD e=

(4.17)

is called the centre of curvature of the curve (C) at the point M (Figure 4.4).

te

M

(C)

be

ne

54 Chapter 4 Elementary Concepts on Curves

FIGURE 4.4. Convexity and curvature.

EXERCISE

4.1 In the reference system ( ) / , ,O i j k

, we define the curve (C) as the set of the

points M of Cartesian coordinates:

3 3sin , cos , cos 2 ,x a q y a q z a q= = = − with 0 and 02

a q π> < < .

Derive the unit direction vector of the tangent, the curvilinear abscissa, the

vector of the principal normal, the radius of curvature and the basis of Frenet at an

arbitrary point of the curve (C).

COMMENTS

The present chapter introduces the elementary concepts relative to the space

curves. Exercise 4.1 illustrates how these concepts can be used simply

starting from their definitions.

te

M

(C)

be

ne

Dosculating plane

normalplane

ddOM

q

2

2

d

d

OM

q

CHAPTER 5

Torsors

5.1 DEFINITION AND PROPERTIES OF THE TORSORS

5.1.1 Definitions and Notations

A torsor, which we shall denote by may be defined as the set of two fields

of vectors, defined on the geometric space or on a subspace (D) of the geometric

space and having the following properties.

1. The first field of vectors associates to every point P of the space (D) a vector

R

of 3, independent of the point under consideration:

( )P D∀ ∈ 3R ∈ . (5.1)

The vector R

is called the resultant of the torsor . We shall denote this

resultant by R

or .R

2. The second field of vectors associates to every point P of the space (D) a

vector P

which depends on the point P:

( )P D∀ ∈ 3P ∈

. (5.2)

The vector P

is called the moment vector at the point P or moment at the

point P of the torsor . We will denote the moment by P

or P

.

Between the moment vectors at two points P and Q of the space (D), there

exists the relation:

Q P R PQ= + × . (5.3)

This fundamental relation can be considered as the relation of definition of the

field of moment vectors, and by extension as the relation of definition of the

torsors:

The torsor is the set of the two fields of vectors: resultant and moment

56 Chapter 5 Torsors

defined on the space (D), satisfying Relation (5.3) at every point P of this space.

The two vectors R

and P

are called the elements of reduction at the point P

of the torsor or the vector components at the point P of the torsor .

Usually, we shall denote these elements by , P PR . The importance of the

elements of reduction at a point results from the following theorem:

If R

and P

are two given vectors, and if P is a given point, there exists one

torsor and only one having R

and P

as elements of reduction at P. From this

theorem, it results that a torsor is defined in a unique way if the elements of

reduction are given at a point.

The six real numbers X, Y, Z and LP, MP, NP, respective components of R

and

P

in a given basis, are called the components at P of the torsor . Usually,

we shall denote these components by , , , , , P P P PX Y Z L M N .

5.1.2 Properties of the Moments

The two moment vectors P

at the point P and Q

at the point Q have the

same projection on the line PQ: we say that the field of the moment vectors is

equiprojective.

The projection of the moment vector P

on PQ (or more generally along the

direction PQ

) is given according to the definition by the scalar product PPQ ⋅

(except for a multiplicative factor). Considering Expression (5.3), we may write:

( )Q PPQ PQ PQ R PQ= + ×⋅ ⋅ ⋅ . (5.4)

The two vectors PQ

and R PQ×

being orthogonal, the previous relation is redu-

ced to:

Q PPQ PQ=⋅ ⋅ . (5.5)

This relation expresses that the vectors P

and Q

have the same projection

on the line PQ.

5.1.3 Vector Space of Torsors

The set of the torsors defined on a space (D) constitutes a vector space.

5.1.3.1 Equality of Two Torsors

Two torsors are equal (we say also equivalent), if and only if there exists a

5.1 Definition and Properties of the Torsors 57

point at which the elements of reduction of the two torsors are equal.

The equality between two torsors:

1 2= (5.6)

is thus equivalent to the set of the two vector equalities:

1 2

1 2

,

and for example .P P

R R =

=

(5.7)

5.1.3.2 Sum of Two Torsors

The torsor sum of the two torsors 1 and 2 , which we shall denote by:

1 2= + (5.8)

has for elements of reduction at a point P:

1 2

1 2

,

.P P P

R R R = +

= +

(5.9)

5.1.3.3 Multiplication by a Scalar

The torsor:

1λ= , (5.10)

where λ is a real number, has for elements of reduction at a point P:

1

1

,

.P P

R Rλ

λ

=

=

(5.11)

5.1.3.4 Null Torsor

The null torsor or zero torsor, denoted by ,0 is the neutral element for the

addition of the two torsors. Its elements of reduction at any point are:

( )

0,

0, .

0

0P

R

P D

=

= ∀ ∈

(5.12)

5.1.4 Scalar Invariant of a Torsor

The scalar invariant of a torsor is by definition the scalar product of the

elements of reduction at an arbitrary point of the torsor under consideration.

The scalar invariant is independent of the point chosen, that justifies the inte-

rest of the definition. Indeed, if we consider the torsor , the scalar invariant


is given for example by the expression:

QI R= ⋅

, (5.13)

or considering the point P (Relation (5.3)):

( )PI R R R PQ= + ×⋅ ⋅

.

Hence:

PI R= ⋅

. (5.14)

The scalar invariant is quite independent of the point.

5.1.5 Product of Two Torsors

We call product of the two torsors 1 and 2 , the real number defined as

follows:

1 2 1 2 1 2P PR R= +⋅ ⋅ ⋅

. (5.15)

This definition is independent of the point P chosen, as it can be easily shown

by considering Relation (5.3).

5.1.6 Moment of a Torsor about an Axis

We consider the torsor and the axis ( ) ( ), P u∆ =

passing through the

point P and of unit direction vector u

(Figure 5.1). Let Q be an arbitrary point of

the axis (∆). We have:

,

, .

Q P R PQ

PQ uα α

= + ×

= ∀ ∈

It follows from this that:

Q Pu u=⋅ ⋅ . (5.16)

The scalar product is independent of the point Q, when Q moves on the axis (∆).

This is the property of equiprojectivity (Subsection 5.1.2).

FIGURE 5.1 Projection of a moment on an axis.

P

()

Q

u

5.1 Definition and Properties of the Torsors 59

The scalar product Pu ⋅ is called the moment of the torsor about

the axis ( ), P u

. It is independent of the point chosen on the axis.

Note. Not to confuse:

— the moment of a torsor about an axis which is a real number ;

— and the moment of a torsor at a point which is a vector.

5.1.7 Central Axis of a Torsor

We consider a given torsor with a nonzero resultant. The set of the points of the

geometric space at which the moment of the torsor is collinear to its resultant is a

line which has this resultant as direction vector. This line is called the central axis

of the torsor.

Thus:

Central axis of the torsor ( ), PP Rα α= = ∀ ∈

. (5.17)

Let us show this theorem. Let be then a given torsor and we search for the

set (∆) of the points P such as P

is collinear to ,R

or what is equi-

valent such as:

0PR × = . (5.18)

Let O be a reference point of the geometric space. Expression (5.3) is written:

P O R OP= + × . (5.19)

The condition (5.18) of collinearity is then written:

( ) 0OR R R OP× + × × = , (5.20)

or taking account of the property (1.51) of the double vector product:

( ) 20OR R OP R R OP× + − =⋅

. (5.21)

From this expression, we derive:

2 2OR R OP

OP RR R

×= +

⋅

. (5.22)

The first term is a vector independent of the point P:

0 2OR

VR

×=

. (5.23)

The second term depends on the point P, and we introduce the real number λdepending on the point P:

2

R OP

Rλ =

⋅

(5.24)


FIGURE 5.2. Central axis.

The position vector of the point P is thus written:

0OP V Rλ= +

. (5.25)

This result expresses that the set (∆) of the points P is a line of direction vector R

,

resultant of the torsor under consideration.

To determine the central axis of the torsor, it is sufficient (knowing a direction

vector) to find a particular point of the axis. As particular point, we seek for the

point P0 such as the position vector 0OP

is orthogonal to the central axis. We have

then:

0 0R OP =⋅

, (5.26)

and Expression (5.25) is written:

0 0 2OR

OP VR

×= =

. (5.27)

The central axis is the axis ( )0 , P R

.

5.2 PARTICULAR TORSORS

RESOLUTION OF AN ARBITRARY TORSOR

5.2.1 Slider

5.2.1.1 Definition

A torsor of nonzero resultant is a slider, if and only if its scalar invariant is

zero.

The definition of a slider can thus be formulated as:

is a slider ⇔

0, ,

with 0.

PI R P

R

= = ∀

≠

⋅

(5.28)

P0

central axis ( )0 , P R

O R

5.2 Particular Torsors. Resolution of an Arbitrary Torsor 61

The scalar invariant being independent of the point P at which it is determined,

it is equivalent to say:

A torsor is a slider if and only if there exists at least a point at which the

moment of the torsor is zero.

5.2.1.2 Moment at a Point of a Slider

We consider the slider . There exists at least a point at which the moment

of the slider is zero. Let A be this point:

0A = . (5.29)

The moment at an arbitrary point P is written:

P A R AP= + × . (5.30)

Hence:

P R AP= × . (5.31)

This relation expresses the moment vector at an arbitrary point P of a slider of

which the moment is zero at the point A.

5.2.1.3 Axis of a Slider

Let be a slider and A a point at which the moment of the torsor is zero.

We search for the set of the points P at which the moment of the slider is zero.

From (5.31) the set of these points satisfies the relation:

0 with 0R AP R× = ≠ . (5.32)

This relation shows that AP

is collinear to the resultant, hence the point P is a

point of the line passing through the point A and which has the resultant of the

slider as direction vector.

This line is called the axis of the null moments of the slider or in a contracted

state: the axis of the slider. This axis is the central axis of the slider.

5.2.1.4 Conclusions

1. A torsor , of nonzero resultant, is a slider if and only if the scalar inva-

riant is zero.

2. A slider is entirely determined when are given:

— its resultant: ,R

— a point A at which its moment is null: 0A = .

3. A slider has an axis of null moments: the axis ( ), A R .

4. If Q is an arbitrary point of this axis, the moment at the arbitrary P is

expressed as:

P R QP= × . (5.33)


5.2.2 Couple-Torsor

5.2.2.1 Definition

A nonzero torsor is a couple-torsor if and only if the resultant of this torsor is

zero.

Hence:

is a couple-torsor

0,

a point such as 0.P

R

P

=⇔

∃ ≠

(5.34)

5.2.2.2 Property of the Moment-Vector

It follows from Expression (5.3) that a couple-torsor is such as:

P Q= = , (5.35)

where is a vector independent of the points P and Q.

The moment-vector is independent of the point under consideration.

5.2.2.3 Resolution of a Couple-Torsor

Let c be a couple-torsor 0, . This couple-torsor can be resolved into

the sum of two sliders 1 and 2 :

1 2 c = + , (5.36)

where the sliders are defined as follows:

1 2

1 2

1 2

0,

, being an arbitrary point,

0, 0.

P P

R R

P

I I

+ =

+ = = =

(5.37)

The first relation shows that there exists an infinity of couples of sliders equi-

valent to a given couple-torsor. The sliders which constitute one of these have

opposite resultants. Thus, the axes of the sliders are parallel.

One of these equivalent couples can be obtained in the following way.

1. We choose the slider 1 while giving us:

— its resultant 11R R= ;

— its axis (∆1) determined by a point P1: ( ) ( )11 1, P R∆ =

.

At this stage there is thus a “double” infinity of possible choices. When these

choices are done, the resolution is unique.

5.2 Particular Torsors. Resolution of an Arbitrary Torsor 63

2. The slider 2 is then defined as follows:

— its resultant is 12R R= − ;

— its axis (∆2) is determined, if we know one of the points of this axis: P2 for

example. The point P2 is such as:

2 2 21 2 1P P P+ = = . (5.38)

Whence from (5.33):

1 1 2R P P× =

. (5.39)

This relation determines the point P2 in a unique way.

5.2.3 Arbitrary Torsor

5.2.3.1 Définition

A torsor is arbitrary if and only if its scalar invariant is nonzero.

is an arbitrary torsor 0I⇔ ≠ . (5.40)

5.2.3.2 Resolution of an Arbitrary Torsor

An arbitrary torsor can be resolved into a sum of a slider and a couple-torsor;

and this by an infinity in ways.

The resolution of an arbitrary torsor is implemented as follows.

1. We choose a point P where the elements of reduction of the torsor are

known, thus:

and .PR (5.41)

There is an infinity of possible choices for the point P. This choice will depend

on the easiness to express the elements of reduction of the torsor at such or such

point. Once the choice of P made, the resolution of the torsor is unique.

2. The slider 1 is such as:

— its resultant is equal to the resultant of the torsor :

1 ,R R=

(5.42)

— its axis passes through the point P chosen.

3. the couple-torsor 2 is such as:

2 20, .PR = =


We obtain well thus:

1 2= + . (5.44)

At each point P chosen is associated a couple: slider/couple-torsor, and only

one. The sliders of all the couples equivalent to a given arbitrary torsor have the

same resultant. They differ by their axes which have however the same direction,

given by the resultant of the torsor.

5.2.4 Conclusions

Let be a torsor of elements of reduction at P: R and .P

1. If 0PI R= =⋅

1.1 If 0R ≠ , the torsor is a slider.

1.2 If 0R =

— If 0 P P= ∀ , the torsor is the null torsor.

— If 0P ≠ , the torsor is a couple-torsor, which may be resolved into a

sum of two sliders of opposite resultants.

2. If 0PI R= ≠⋅

, the torsor is an arbitrary torsor. The torsor

may be resolved, at a point P, into a slider of resultant R and of axis

( ), P R and a couple-torsor of moment vector .P

5.3 TORSORS ASSOCIATED TO A FIELD

OF SLIDERS DEFINED ON A

DOMAIN OF THE GEOMETRIC SPACE

Afterwards, we shall be brought to consider torsors associated to fields of

torsors defined on particular subspaces of the geometric space: for example, the

set of the points of a rigid body, the set of the points of several bodies, etc. We

shall note (D) this subspace which could be a curve, an area or a volume.

Moreover, this domain could be finite (if there exist a one-to-one mapping of the

points onto the natural numbers), or infinite otherwise.

5.3.1 Torsor Associated to a Finite Set of Points

Let us consider a finite set (D) = (M1, M2, . . . , Mi, . . . , Mn) constituted of n

points. On this domain (D) we define a field of sliders which associates to each

5.3 Torsors Associated to a Field of Sliders Defined on a Domain of the Geometric Space 65

point Mi of the domain (D) a slider i of axis passing through the point Mi:

( )iM D∀ ∈ .i (5.45)

The torsor i is a slider of resultant iR

and axis passing through Mi:

, 1, 2, . . . , .

0,i

ii

M i

R Ri n

==

=

(5.46)

We call torsor associated to the domain (D) and to the field of sliders i

defined on (D), the torsor ( ) D sum of the sliders i . Whence:

( ) 1

n

i

i

D

=

= . (5.47)

It follows from this definition that:

— the resultant of the torsor ( ) D is:

( ) 1 1

n n

ii

i i

R D R R

= =

= =

, (5.48)

— the moment at an arbitrary point P of the geometric space of the torsor

( ) D is given by the expression:

( ) 1 1

n n

i iP P i

i i

D PM R

= =

= = × . (5.49)

5.3.2 Torsors Associated to an Infinite Set of Points

Let (D) be an infinite domain of the geometric space (Figure 5.3). On this

domain (D), we consider a field of sliders which associates to every point M of the

domain (D) a slider, denoted by ( ) d M , of axis passing through M, defined in

the following way:

( )M D∀ ∈ ( ) d .M (5.50)

The torsor ( ) d M is a slider of resultant d ( )R M

and axis passing though M:

( )

( )

d d ( ),

d 0.M

R M R M

M

=

=

(5.51)

The resultant d ( )R M

may be put in the form:

d ( ) ( ) d ( )R M R M e M=

, (5.52)


FIGURE 5.3. Infinite domain.

where d ( )e M is an element of the domain (D) surrounding the point M: element

of curve, area or volume, according as the domain is a curve, an area or a volume.

The vector ( )R M

is called the vector density of the field of sliders. The torsor

associated to the field of sliders (5.50) is derived as an extension to an infinite

domain of Expressions (5.47) to (5.49).

The torsor associated to the domain (D) and to the field of sliders ( ) d M

defined on (D) is the torsor ( ) D that we admit to write:

( ) ( ) ( )

dD

D M= . (5.53)

It follows from this expression and by extension of (5.48) and (5.49) that:

— the resultant of the torsor ( ) D is:

( )

( ) ( )

d ( ) ( ) d ( )D D

R D R M R M e M= = , (5.54)

— the moment at the arbitrary point P of the geometric space of the torsor

( ) D is expressed as:

( )

( ) ( )

d ( ) ( ) d ( )PD D

D PM R M PM R M e M= × = × . (5.55)

The integrals which occur in the preceding expressions will be curve, area or

volume integrals according to the nature of the domain (D): curve, area or

volume.

Relations (5.54) and (5.55) are well suited to a method of literal calculation of

the integrals. However, it is always possible to reduce the case of an infinite

domain to the case of a finite domain. In this way, the domain (D) is divided into

n elements (Figure 5.4). The element (i) is then referred by the point Mi “centre”

of this element. Next, it is considered that the vector density ( )R M

it constant

inside the element (i):

( ) ( ), ( )iR M R M M i= ∀ ∈

. (5.56)

(D)

M

de(M)


FIGURE 5.4. Discretization of an infinite domain.

Next, it is considered that the slider of axis passing through Mi and associated to

the element (i) has for resultant:

( ) ( )

( ) d ( ) di i i i ii i

R R M e R M e= = .

Hence:

( )i i iR R M e= , (5.57)

where ei is the length, the area or the volume of the element (i), according as the

domain (D) is a curve, an area or a volume. We are brought back to a finite

domain, constituted of the points Mi, with which is associated the slider field:

Mi ( )i iR M e

.

Whence from (5.48) and (5.49), the resultant and the moment of the associated

torsor:

( )

( )

1

1

( ) ,

( ) .

n

i i

i

n

iP i i

i

R D R M e

D PM R M e

=

=

=

= ×

(5.58)

5.3.3 Important Particular Case. Measure Centre

In the general case, the vector density ( )R M

introduced in (5.52) is a vector

function of the point M and may be written in the form:

( ) ( ) ( )R M f M u M=

, (5.59)

where ( )u M

is a unit vector and ( )f M a positive real number equal to the norm

of ( )R M

. In the general case, the norm and the direction of the vector density are

both depending on the point M.

In this subsection, we consider the particular case where the vector ( )u M

is a

(D)

Mi


unit vector u

independent of the point M. The vector density is:

( ) ( ) R M f M u=

. (5.60)

The field of sliders defined on the domain (D) associates then at each point M a

slider of resultant:

( )M D∀ ∈ d ( ) ( ) d ( )R M f M u e M=

. (5.61)

The axes of the sliders defined on the domain (D) have thus all the same direction,

whatever the point M may be.

The elements of reduction, at an arbitrary point P of the geometric space, of the

torsor associated with such a field of sliders, are from (5.54) and (5.55):

( ) ( )

( ) d ( )D

R D u f M e M= , (5.62)

( ) ( )

( ) d ( )PD

D PM f M e M u

= ∧

. (5.63)

The moment vector ( ) P D is orthogonal to u

, hence to the resultant

( ) R D . The scalar invariant of the torsor ( ) D :

( ) ( ) ( ) PI D R D D= ⋅

(5.64)

is thus zero. It follows that the torsor ( ) D is either the null torsor, or a couple-

torsor, or a slider.

1. If ( )

( ) d ( ) 0D

f M e M = and ( )

( ) d ( ) 0D

PM f M e M u

× =

, the torsor

( ) D is the null torsor.

2. If ( )

( ) d ( ) 0D

f M e M = and ( )

( ) d ( ) 0D

PM f M e M u

× ≠

, the torsor

( ) D is a couple-torsor.

3. If ( )

( ) d ( ) 0D

f M e M ≠ , the torsor is a slider.

Hereafter in this subsection, we study the case where the torsor is a slider:

( )

( ) d ( ) 0D

f M e M ≠ . (5.65)

The resultant (5.62) of the slider can then be written in the form:

( ) ( ) R D D uµ= , (5.66)

on introducing the quantity:


( )

( ) ( ) d ( )D

D f M e Mµ = . (5.67)

The quantity µ(D) thus defined is called the measure of the domain (D), associa-

ted to the field of sliders considered on (D). The quantity f(M) is the specific

measure (curvilinear, area or volume measure) at the point M.

Different fields of sliders can be associated to a same domain (D). Thus, it

follows that different measures will be associated to a same domain: volume (area

or length), mass, pressure, intensity of the gravitation field, intensity of the

electrostatic field, etc.

The torsor ( ) D considered being a slider, it has an axis of null moments

the points P of which, from (5.63), satisfy the relation:

( ) ( )

( ) d ( ) 0PD

D PM f M e M u

= × =

. (5.68)

Thus, the points of the axis are such as the vector ( )

( ) d ( )D

PM f M e M

is

collinear to u

. In particular, there exists a point of this axis, which we shall

denote by H such as the previous vector is zero. Whence:

( )

( ) d ( ) 0D

HM f M e M =

. (5.69)

The point H plays an important role and is called the measure centre related to the

field of the sliders considered on (D). We will say more briefly (but in an incur-

rect way) that H is the measure centre of the domain (D).

The position of the point H in the geometric space can be defined with respect

to a reference point O, by searching for the position vector OH

. Relation (5.69) is

written:

( )( )

( ) d ( ) 0D

OM HM f M e M− =

, (5.70)

or

( ) ( )

( ) d ( ) ( ) d ( )D D

OH f M e M OM f M e M=

. (5.71)

The position vector of the measure centre H is thus expressed in the form:

( )

1 ( ) d ( )( ) D

OH OM f M e MDµ

=

. (5.72)

Finally, the slider associated to the domain (D) and the field of sliders of

vector density ( ) ( ) R M f M u=

has a resultant given by Expression (5.66) and

an axis ( ), H u

of direction u

passing through the measure centre, defined by

(5.69) or (5.72).


In the case where the specific measure f(M) is independent of the point M:

( ) Constantf M k= = , (5.73)

Expression (5.72) is reduced to:

( )

1 d ( )( ) D

OH OM e Me D

=

, (5.74)

where e(D) is the length, the area or the volume of the domain (D). Relation

(5.74) shows that in this case the measure centre H is the same as the centroid of

the length, of the area or of the volume of the domain (D).

EXERCISES

5.1 Let (D) be the domain constituted of four points M1, M2, M3 and M4:

(D) = (M1, M2, M3 M4).

On this domain, we define a field of sliders, such as the resultants of the sliders

associated to each point are:

M1 (2, –2, 3) 1R

(5, 0, 0),

M2 (–4, 2, –1) 2R

(0, –2, 0),

M3 (5, –2, 3) 3R

(0, 0, 3),

M4 (0, 2, 0) 4R

(3, 4, 1).

The coordinates of the points are the Cartesian coordinates with respect to a refe-

rence system of origin O. The components of the resultants of the sliders are given

in the canonical basis.

Derive the resultant of the torsor associated to this field, its moment at the

point O. Characterize the torsor. Derive the moment of the torsor at an arbitrary

point P. Find the equations of the axis of the torsor.

5.2 We consider the same domain (D) as the one considered in the previous exer-

cise, but with a different field of sliders defined as follows:

M1 (2, –2, 3) 1R

(100, 0, 0),

M2 (–4, 2, –1) 2R

(0, 200, 50),

M3 (5, –2, 3) 3R

(–100, 0, –50),

M4 (0, 2, 0) 4R

(0, –200, 0).

Derive the resultant of the torsor associated to this field, its moment at O.

Characterize the torsor. Resolve the torsor at the origin point.

Exercises 71

5.3 Always to the domain (D), defined in Exercise 5.1, we associate the field of

sliders:

M1 (2, –2, 3) 1R

(5, – 4, 1),

M2 (–4, 2, –1) 2R

(0, –2, 0),

M3 (5, –2, 3) 3R

(0, 0, 3),

M4 (0, 2, 0) 4R

(3, 4, 1).

Derive the resultant of the torsor associated to this field, its moment at the

origin point O. Characterize the torsor. Derive the moment of the torsor at an

arbitrary point P. Find the equations of its central axis. Resolve the torsor at the

origin point.

5.4 Torsor associated to an infinite domain. We consider a domain (D) consti-

tuted of a rectangular surface ABCD (Figure 5.5). As reference system, we choose

the reference system (Axyz), of which the axes Ax

and Az

are respectively along

the sides AB and AD. To every area element dS(M) surrounding an arbitrary point

of the domain (D), we associate the slider of vector density ( ) p M i

. The field of

sliders thus defined on the domain (D) associates to every point M a slider of

resultant:

d ( ) ( ) d ( )R M p M i S M=

,

and axis ( ), M i

.

Derive the torsor associated to this field of sliders.

FIGURE 5.5. Rectangular domain.

(D)

A B

C D

M

dS(M)

x

z

y


COMMENTS

The formalism of torsor constitutes the key of the concepts which will

be introduced in the continuation of this textbook. Thus, the reader will

have to study thoroughly all the elements considered in the present chapter.

The concept of vector makes it possible to work simultaneously on three

real numbers. The concept of torsor makes it possible to operate simulta-

neously on two vectors, the resultant of the torsor and its moment.

Three types of torsors exist: slider, couple-torsor and arbitrary torsor.

The type of torsor is characterized by its scalar invariant. The slider consti-

tutes the fundamental type of torsor, since a couple-torsor can be resolved

into the sum of a couple of two sliders, and an arbitrary torsor can be

resolved into a slider and a couple-torsor.

The reader will pay a particular attention to Subsection 5.3.3 which

considers the very important case for which there is a measure centre.

The exercises suggested illustrate simply the whole of the concepts

which are introduced into the chapter.

Part II

Kinematics

Kinematics is part of Mechanics whose the object is the study of

the motion of a physical system: point or particle, rigid body, set of

rigid bodies, etc., without reference to the cause of the motion. Kine-

matics is used to relate displacement, velocity, acceleration, and time.

Causes of the motions will be considered as part of Dynamics (Part

IV).

Kinematics of point (or particle) consists in characterizing the

motion of a point with respect to a reference system: first, the place

(the trajectory) where the point is moving; then, how it is moving on

this trajectory (quickly, slowly, more and more quickly, less and less

quickly, etc.). The way of moving on the trajectory will be charac-

terized by the kinematic vectors of the point: velocity vector and acce-

leration vector.

The object of the Kinematics of rigid body is to establish the rela-

tions between the motions of all the points of a given rigid body. The

relation between the trajectories of the points is brought back to the

problem of the change of coordinates, problem which was considered

previously in Chapter 2. The relations between the velocity vectors

will be characterized by introducing the notion of kinematic torsor.

Kinematics of two solids in contact needs a particular analysis

which introduces the notion of sliding, rolling and spinning.

CHAPTER 6

Kinematics of Point

6.1 INTRODUCTION

To translate into equations the motions of physical systems, an observer has to

schematize the space which surrounds him. The assumed schematization is that of

the geometric space (Chapter 2). In particular, the observer will be brought to

choose a reference system (apparatus frame, part of the Earth surface, moving

train, etc.), to which the observer will link reference axes and with respect to

which the observer will analyse the motions.

With the schematization of the space, the observer must add the notion of time.

This notion makes it possible to give an account of the simultaneity of two events,

of the order of succession of these two events and the duration of the interval

which separates them. The measurement of a duration is related to the choice of a

measurement unit (the international unit is the second: s). This measurement is

implemented using clocks (watch, Earth motion, etc.) associated with a calendar

With the indications provided by a clock and a calendar, we associate the

numerical value of a variable called time variable, and which we will denote by t.

We call date of an instantaneous event then, the numerical value t at the moment

when the event occurs. If t1 is the date of an event E1, and t2 that of an event E2,

the duration of the interval which separates them is 2 1t t− . By convention, it has

been chosen to take 2 1 0,t t− > when E2 is posterior to E1. It is said that the time

scale is increasing in the succession of the events.

6.2 TRAJECTORY AND

KINEMATIC VECTORS OF A POINT

Let (T) be a reference system that we will simply call reference, and let (D) be

a set (Figure 6.1) of which we want to study the motion with respect to reference

(T). The reference system (T) is such as each one of its points is motionless with

76 Chapter 6 Kinematics of Point

FIGURE 6.1. Motion of a set (D) with respect to a reference (T).

respect to the other points. It is the same for the set (D). We will call M an arbi-

trary point of the set (D) and O a point of the reference system (T), chosen as refe-

rence point

6.2.1 Trajectory

The position of point M with respect to the reference (T) is defined by the

vector OM

. If the set (D) is moving with respect to (T), the vector OM

is a

vector function of the time:

( )OM V t=

. (6.1)

When t varies, the point M follows a curve (Chapter 4) defined by the preceding

relation and called the trajectory of the point M with respect to the reference (T).

The trajectory is the set of the points of the reference (T) with which the point M

comes in coincidence at every moment (Figure 6.2).

FIGURE 6.2. Trajectory in a given reference.

M

(D)

(T)

O

trihedron attached

to reference (T)

reference system

(reference)

O

M

reference (T )

trajectory in the

reference (T )

points of

reference (T )

6.2 Trajectory and Kinematic Vectors of a Point 77

The trajectory depends on the reference system. For example, if we consider a

traveller sitting in a train moving with a rectilinear motion, the trajectory of this

traveller with respect to a reference linked to the Earth is a line segment, whereas

the trajectory with respect to the coach is reduced to a point.

6.2.2 Kinematic Vectors

The trajectory is not sufficient to characterize the motion of a point completely.

In addition to the geometrical nature, it is necessary to specify the motion of the

point on this trajectory. The study of this motion is conducted by studying the

vector function ( )V t

(Relation 6.1), which leads to introduce the first and second

derivatives with respect of the time of the position vector OM

. These two vectors

are called the kinematic vectors of the point M and make it possible to entirely

characterize the motion of the point on its trajectory.

6.2.2.1 Velocity Vector

The first kinematic vector is the velocity vector, vector derivative, with respect

to the time and relatively to the reference (T), of the position vector of the point

M, hence:

( )( )( )

d,d

TT M t OM

t=

. (6.2)

The notation ( )( ),T M t expresses the fact that we consider the velocity vector,

relatively to the reference (T), of the point M, at the date t. This notation will be

simplified when there will be no possible confusion: ( ) ( )( ) ( ) , , ,TM t M M or

. The notation

( )dd

T

t expresses the fact that the derivation with respect to the

time is carried out in a basis related to the reference (T).

The velocity is a continuous function of the time function, except at the times of

collisions between bodies. Such events must then be the subject of particular

studies.

6.2.2.2 Acceleration Vector

The second kinematic vector is the acceleration vector, second derivative

vector, with respect to the time and relatively to the reference (T), of the position

vector of the point M:

( )( )( ) ( )

( )( )

2

2

d d, ,dd

T TTM t OM M t

tt= =

. (6.3)

As in the case of the velocity vector, the notation of the acceleration vector could

be simplified.


The magnitudes of the velocity and acceleration vectors are respectively called

the velocity and the acceleration of the point under consideration.

Note. It results from the definitions of the kinematic vectors that: (1) the

magnitude and the components of the velocity vector have the physical dimension

of length divided by time (in the International System of Units, they are expressed

in m s–1); (2) those of the acceleration vector have the dimension of length divi-

ded by time square (in m s–2 in the International System).

6.2.3 Tangential and Normal Components of the Kinematic Vectors

6.2.3.1 Velocity Vector

From Relation (3.16), we may write:

( )( )( ) ( )

d d d,d d d

T TT OM sM t OM

t s t= =

, (6.4)

by introducing the curvilinear abscissa s (Section 4.2) of the point M along its

trajectory relatively to the reference (T). From Relation (4.9), ( )

dd

T OMs

is the unit

vector te

of the tangent to the trajectory at the point M. The real number:

dd

st

= (6.5)

is called the instantaneous algebraic velocity of the point M at the moment t and

relatively to the reference (T). The velocity vector is then written in the form:

( )( ),TtM t e=

. (6.6)

The velocity vector is collinear to the unit vector of the tangent. It is said by

language misuse that “the velocity is supported by the tangent to the trajectory” at

the point under consideration.

6.2.3.2 Acceleration Vector

Starting from Expression (6.6), the acceleration vector is expressed as:

( )( )( )

( )( )( )

dd d, ,d d d

TTT t

te

M t M t et t t

= = +

, (6.7)

with, from (4.11): ( ) ( )

d d dd d d

T Tt t

ne e s e

t s t= =

. (6.8)

Whence the expression of the acceleration vector:

( )( )2d,

dt nM t e e

t= +

. (6.9)

6.2 Trajectory and Kinematic Vectors of a Point 79

We call tangential acceleration vector, the vector:

dd

t ta et

= , (6.10)

and normal acceleration vector, the vector: 2

n na e=

. (6.11)

Relation (6.9) is then written: ( )( ), t nM t a a= +

, (6.12)

or ( )( ) , t t n nM t a e a e= +

, (6.13)

where at and an are the tangential and normal components of the acceleration

vector: 2d ,

dt na a

t= =

. (6.14)

The normal component is always positive. The tangential component is positive if

the algebraic velocity is increasing, and negative on the contrary.

6.2.3.3 Representation of the Kinematic Vectors

The velocity vector and the acceleration vector are vectors of the space 3. It

happens however that, by convention, one represents, in the geometric space, the

bipoints having for origin the position of the point M at the date t and having for

images the kinematic vectors of the point M (Figure 6.3).

6.2.4 Different Types of Motions

6.2.4.1 Definitions

We shall say that the motion of the point M is:

— accelerated at time t, if the magnitude of the velocity vector ( )( ),T M t is

increasing at this moment;

FIGURE 6.3. Symbolic representation of the kinematic vectors.

te M(t)

(C)

ne

( )( ),T M t

dd

tat

=

( )( ),M t2

na =


— decelerated at time t, if the magnitude of the velocity vector ( )( ),T M t is

decreasing at this moment;

— uniform in the time interval [t1, t2], if the velocity vector ( )( ),T M t has a

constant magnitude in this interval.

The magnitude of the velocity vector ( )( ),T M t varies in the same sense as

2, and the type of motion depends on the sign of

2dd t

. In fact, we have:

( )( )

2d d2 2 , with ,d d

a a M tt t

= = =⋅ ⋅

. (6.15)

So, it results that the type of motion depends on the sign of the scalar product of

the two kinematic vectors.

In the case of a curvilinear trajectory:

— the motion is accelerated if and only if 0a >⋅ ;

— the motion is decelerated if and only if 0a <⋅ ;

— the motion is uniform if and only if 0a =⋅ (the kinematic vectors are ortho-

gonal).

In the case of a rectilinear trajectory, the radius of curvature is infinite. It

results that the kinematic vectors are collinear. Whence the different types of

motions in the case of a rectilinear trajectory:

— the motion is accelerated if and only if and a

are of the same sign;

— the motion is decelerated if and only if and a

are of opposite signs;

— the motion is uniform if and only if does not depend on the time, the vector

a

is then the null vector.

6.2.4.2 Remark

According to the definition of a uniform motion, the magnitude of the velocity

vector keeps a constant value during the motion, thus:

0d( )d

stt

= = . (6.16)

So, the expression of the curvilinear abscissa as a function of time is deduced as:

( )0 0 0( )s t t t s= − + , (6.17)

where s0 is the value of the curvilinear abscissa at time t0:

( )0 0s s t= . (6.18)

In the case of a uniform motion, the curvilinear abscissa is a first degree function

of the time variable.

Furthermore, if the motion is rectilinear, the acceleration is null:

0a =

. (6.19)

On the other hand, if the motion is curvilinear, only the tangential acceleration is

zero. Whence:

6.3 Expressions of the Components of the Kinematic Vectors 81

20

n na a e= =

. (6.20)

These results call the following remark. When one speaks about “velocity”, it is

necessary to specify if it is the velocity, the algebraic velocity, or the magnitude

of the velocity vector, which is concerned. Thus, we have just seen that a motion

with a constant magnitude of the velocity vector is, either rectilinear, or curvi-

linear. On the other hand, a motion with constant velocity vector is necessarily

rectilinear. This remark also applies to the “acceleration”.

6.3 EXPRESSIONS OF THE COMPONENTS OF THE

KINEMATIC VECTORS AS FUNCTIONS OF THE

CARTESIAN AND CYLINDRICAL COORDINATES

6.3.1 Cartesian Coordinates

We consider the case where the point M in the reference (T) is characterized by

the Cartesian coordinates (x, y, z) relatively to the coordinate system (Oxyz) fixed

with respect to the reference (T) (Figure 2.12). The position vector is written:

OM x i y j z k= + +

, (6.21)

where ( ) , ,i j k

is the canonical basis of the vector space 3.

According to the relations of definitions (6.2) and (6.3) of the kinematic

vectors, and the concepts introduced in Chapter 3, the expressions of the kine-

matic vectors are written as:

( )( )( )

dd d d,d d d d

TT yx zM t OM i j k

t t t t= = + +

, (6.22)

( )( )( )

( )( )

22 2

2 2 2

dd d d, ,d d d d

TT yx zM t M t i j k

t t t t= = + +

. (6.23)

It is usual to introduce in the expressions for the derivatives as functions of

time the condensed notations:

— ( )f t or f for the first derivative with respect to time of the function f(t);

— ( )f t or f for the second derivative.

With such notations, Expressions (6.22) and (6.23) are written:

( )( ) ,T M t x i y j z k= + +

, (6.24)

( )( ) ,M t x i y j z k= + +

. (6.25)


6.3.2 Cylindrical Coordinates

The cylindrical coordinates (r, α, z) of a point M with respect to the coordinate

system Oxyz have been introduced in Subsection 3.3.2.1 (Figure 3.1). It has been

shown that:

( )OM r u z kα= +

, (6.26)

( )d ( ) ( )d 2

T

u u πα αα

= +

,

( )d ( ) ( ) ( )d 2

T

u u uπα α π αα

+ = + = −

(6.27)

( )d d d d( ) ( )d d d 2 d

T OM r zu r u kt t t t

α πα α= + + +

, (6.28)

where ( )u α

is the unit vector of the projection of the bipoint (O, M) in the plane

Oxy, and ( )2

u πα +

is the unit vector of the direction of the plane Oxy orthogonal

to the preceding direction (Figure 3.1).

The cylindrical coordinates make it possible to express the kinematic vectors

in the basis ( ), ( ), 2

u u kπα α +

. Thus, from (6.28) we have:

( )( , ) ( ) ( )

2T M t r u r u z kπα α α= + + +

. (6.29)

The components of the velocity vector in the basis ( ), ( ), 2

u u kπα α +

are:

( ), , r r zα . (6.30)

To obtain the expression of the acceleration vector, it is necessary to derive

with respect to time Expression (6.29) of the velocity vector taking account of the

relations:

( ) ( )d d d( ) ( ) ( )d d d 2

T T

u u utα πα α α α

α α= = +

, (6.31)

( ) ( )d d d( ) ( ) ( )d 2 d 2 d

T T

u u ut t

π π αα α α αα

+ = + = −

. (6.32)

We obtain then:

( )( ) ( ) ( ) 2, ( ) 2 ( )

2M t r r u r r u z kπα α α α α= − + + + +

. (6.33)

The components of the acceleration vector in the basis ( ), ( ), 2

u u kπα α +

are:

( )2 , 2 , r r r r zα α α− + . (6.34)

In the case where the trajectory of point M is plane, it is possible to choose the

coordinate system (Oxyz) so that the plane (Oxy) contains the trajectory. The

cylindrical coordinates of point M are then (r, α, 0), and the parameters (r, α) are

called the polar coordinates of point M. The expressions of the kinematic vectors

Exercises 83

are deduced from Relations (6.29) and (6.33), with 0z z= = . We obtain:

( )( , ) ( ) ( )

2T M t r u r u πα α α= + +

, (6.35)

( )( ) ( ) ( )2, ( ) 2 ( )2

M t r r u r r u πα α α α α= − + + +

. (6.36)

EXERCISES

6.1 The motion of a point M is defined by its Cartesian coordinates as functions

of time:

( )2 2 3 33 , 3 , 0,x a t y a t t zω ω ω= = − =

where a and ω are positive constants (a is a length and ω is the inverse of time).

1. Plot the trajectory of the point M for 0t ≥ .

2. Derive as functions of time t: the velocity vector; the instantaneous algebraic

velocity; the acceleration vector and its tangential and normal components; the ra-

dius of curvature of the trajectory.

6.2 Two cities A and B are distant of 160 km. A cyclist leaves A at 8 h, and

moves toward B at the average speed of 30 km/h. At 9 h a car leaves A in

direction of B, with an average speed of 85 km/h. Lastly, a truck starts at 9h30

from B towards A, with an average speed of 60 km/h.

1. Establish the motion equations for the cyclist, the car, the truck.

2. Derive the places and the dates at which:

— the car draws ahead of the cyclist;

— the truck meets the car, the cyclist.

COMMENTS

The motion of a point of a rigid body is defined by the place where it

moves: the trajectory, and by the way in which it moves on this trajectory:

quickly, slowly, more and more quickly, more and more slowly. The way

in which the point moves on its trajectory is characterized by its kinematic

vectors of the point: velocity vector and acceleration vector. Trajectory and

kinematic vectors depend of the reference system in which the motion is

observed. The expressions of the kinematic vectors are obtained simply as

functions of the Cartesian coordinates. The reader will give a particular

attention to the determination of the kinematic vectors as functions of the

cylindrical coordinates.

O

(D)

M

x

M

(D)

y

z

CHAPTER 7

Study of Particular Motions

7.1 MOTIONS WITH RECTILINEAR TRAJECTORY

7.1.1 General Considerations

The trajectory of a point M in the reference (T) is rectilinear, if the point M

moves along a straight line belonging to (T) (Figure 7.1). We may choose a

coordinate system (Oxyz) fixed to the reference (T), such as the axis Ox

coincides

with the line (D). The position vector is then written as:

.OM x i=

(7.1)

The velocity vector of the point M is:

,x i= (7.2)

and its acceleration vector is:

.a x i= (7.3)

FIGURE 7.1 Rectilinear trajectory.

7.1 Motions with Rectilinear Trajectory 85

7.1.2 Uniform Rectilinear Motion

The motion of a point M is rectilinear uniform, if and only if the velocity vector

is constant:

0 0cst i= = =

(7.4)

where 0 is independent of time. The acceleration vector is null.

From (7.2), we have:

0dd

x xt

= = . (7.5)

Hence while integrating:

0 cstx t= + . (7.6)

If at the time t = t0, the point M is at M0, such as 0 0 ,OM x i=

we obtain:

( )0 0 0x t t x= − + . (7.7)

In the particular case where the point M is at the origin O at the time origin 0,t =

Equation (7.7) is reduced to:

0x t= . (7.8)

7.1.3 Uniformly Varied Rectilinear Motion

The motion of a point M is rectilinear uniformly varied, if and only if the acce-

leration vector is constant:

0a a i=

(7.9)

where a0 is independent of time.

We have: 2

02

d

d

x x at

= = . (7.10)

By integrating twice, we obtain:

( )0 0 0a t t= − + , (7.11)

( ) ( )200 0 0 0

2

ax t t t t x= − + − + , (7.12)

the point M being at the time t0 at M0, such as 0 0 ,OM x i=

with a velocity 0 .i

In the particular case where the point M is at the origin O with a zero velocity

at the time origin 0,t = the motion equations (7.11) and (7.12) are reduced as:

0a t= , 20

2

ax t= . (7.13)

Between the variables x and , there exists the general relation obtained by eliminating the time in (7.11) and (7.12):

( ) ( )2 20 0 0

12

a x x− = − . (7.14)

86 Chapter 7 Study of Particular Motions

From the results derived in Subsection 6.2.4.1, we deduce that the motion is:

— uniformly accelerated, if 0 0a > ;

— uniformly decelerated or retarded, if 0 0a < .

7.1.4 Simple Harmonic Rectilinear Motion

The rectilinear motion of a point M is a simple harmonic motion, if the motion

is described by the law:

cos sinx A t B tω ω= + , (7.15)

or

( )cosmx x tω ϕ= − . (7.16)

Between the constants A, B, xm and ϕ, we have the relations:

2 2 1

cos , sin ,

, tan , with cos .

m m

mm

A x B x

B Ax A BA x

ϕ ϕ

ϕ ϕ−

= =

= + = = (7.17)

Without restricting the generality of the study, the constant ω is taken positive.

The algebraic velocity is:

sin cosx A t B tω ω ω ω= − + ,

or (7.18)

( )sinmx x tω ω ϕ= − − .

The acceleration vector has for component:

2 2cos sinx A t B tω ω ω ω= − − ,

or (7.19)

( )2 cosmx x tω ω ϕ= − − .

From these expressions, we draw the relation:

2 2 ou x x a OMω ω= − = −

, (7.20)

and the expressions of the constants:

00

22 10 0 0

0 20

, ,

, tan , with cos ,mm

A x B

xx x

x x

ω

ϕ ϕωω

−

= =

= + = =

(7.21)

where x0 and 0 are the respective values of the variables x and at time t = 0.

The variations of x are reported in Table 7.1. The point M oscillates indefini-

tely between the extreme points xm and –xm, with the period 2T π ω= . The quan-

7.2 Motions with a Circular Trajectory 87

TABLE 7.1. Variation of the abscissa of a point having a simple harmonic motion.

tϕω

− 0 4T

2T 3

4T T

x 0 − mxω− − 0 + mxω + 0

x mx 0 mx− 0 mx

The motion is: accelerated retarded accelerated retarded

tity xm is called the amplitude of the vibratory motion; the point O is the centre of

the oscillation. The point M has an accelerated motion if it moves toward O and a

retarded motion if it moves away.

7.2 MOTIONS WITH A CIRCULAR TRAJECTORY

7.2.1 General Equations

The motion of a point M is circular in the reference (T) if the point M moves on

a circle belonging to (T).

We choose the coordinate system (Oxyz) fixed to the reference (T), so that the

plane (Oxy) coincides with the plane of the circle and that O is the centre of the

circle (Figure 7.2). If a is the radius of the circle, the polar coordinates of the point

M are (a, α).

The position vector is written:

( )OM a u α=

. (7.22)

FIGURE 7.2. Circular motion.

x

M

y

a

( )2

a u πω α= +

( )2

a u πω α +

2 ( )a uω α−

( )u α( )

2u πα +

aO


Differentiating this expression, we obtain successively the velocity and acce-

leration:

( )( , ) ( )

2T M t a u πα α= +

, (7.23)

( ) 2( , ) ( ) ( )2

Ta M t a u a u πα α α α= − + +

. (7.24)

The parameter α is called the angular velocity (measured in rad s–1) of the point

M, at the time t. It is generally denoted by ω:

ddtαω α= = . (7.25)

The parameter α is the angular acceleration (measured in rad s–2) of the point M:

ddtωα ω= = . (7.26)

The expressions of the kinematic vectors can then rewritten while introducing

the angular velocity as:

( )( , ) ( ),

2T

M t a u πω α= + (7.27)

( ) 2( , ) ( ) ( ).2

Ta M t a u a u πω α ω α= − + +

(7.28)

The algebraic velocity of the point M at the time t is:

aω= . (7.29)

The acceleration vector has:

— a tangential component:

ta aω= , (7.30)

— a normal component:

2na aω= − . (7.31)

The acceleration vector na

is always of opposite sign to the position vector OM

:

2na OMω= −

. (7.32)

Lastly, the motion is: accelerated, if 0 ;ωω > retarded, if 0 ;ωω < uniform, if

0.ω =

7.2.2 Uniform Circular Motion

A circular motion is uniform, if its angular frequency is independent of time.

Whence:

0cstω ω= = . (7.33)

The kinematic vectors (7.27) and (7.28) are reduced to:

7.2 Motions with a Circular Trajectory 89

( ) 0( , ) ( )

2T

M t a u πω α= + , (7.34)

( ) 20( , ) ( )Ta M t a uω α= −

. (7.35)

The tangential component of the acceleration vector is zero. Whence it results that

the acceleration vector is collinear to the position vector:

( ) 20( , )Ta M t OMω= −

. (7.36)

In addition, the motion law is written as:

( )0 0 0t tα ω α= − + ,

where α0 is the value of the angle α at the time t0.

Lastly, in a uniform circular motion, the actual parameter is the number of

revolution N per unit of time. The angular velocity is then expressed by the

relation:

0 2 Nω π= . (7.37)

7.2.3 Uniformly Varied Circular Motion

A circular motion is uniformly varied, if its angular acceleration is indepen-

dent of time. Thus:

0d cstdtωα ω= = = . (7.38)

Expressions (7.27) and (7.28) of the kinematic vectors are written:

( )( , ) ( )

2T

M t a u πω α= + , (7.39)

( ) 20( , ) ( ) ( )

2Ta M t a u a u πω α ω α= − + +

. (7.40)

The tangential component of the acceleration vector is constant.

The motion law is written as:

( )

( ) ( )

0 0 0

200 0 0 0

,

,2

t t

t t t t

α ω ω ω

ωα ω α

= = − +

= − + − +

(7.41)

where ω0 and α0 are the respective values of ω and α at the time t0.

Lastly, the motion is uniformly:

— accelerated, if 0 0 ;ωω >

— retarded, if 0 0 .ωω <


7.3 MOTIONS WITH A CONSTANT

ACCELERATION VECTOR


The motion of a point M with a constant acceleration vector is such as:

( )0( , )Ta M t a=

(7.42)

where the vector 0a

is a vector independent of time.

We choose (Figure 7.3), as coordinate reference, the system (Oxyz) so that the

vector 0a

is the direction vector of the axis Oz

( 0a

is collinear to k

). In addition,

to adapt the investigation to the analysis of the motion of projectiles in the vicinity

of the Earth surface, we take the determination:

0 0a a k= −

(7.43)

with 0 0a > .

By integrating the expression:

( )( )

( ) 0

d( , ) ( , )d

TT Ta M t M t a k

t= = −

, (7.44)

we obtain: ( )

0 0( , )T M t a t k= − +

, (7.45)

introducing the velocity vector 0 at the time 0t = :

( )0 ( , 0)T M t= = . (7.46)

We characterize the direction given by the vector 0 , by introducing (Figure

7.3) the angle 2π ϕ+ between the axis ( )0,O

and the axis ( )0,O a

. The axis Oy

is then chosen so that the plane (Oyz) contains the axis ( )0,O and so that this

axis forms an angle ϕ with the axis Oy

. The coordinate system is then entirely

FIGURE 7.3. Motion with a constant acceleration vector.

0

0ax

y

z

O

k

7.3 Motions with a Constant Acceleration Vector 91

defined. This particular choice of the coordinate system is well adapted to the

analysis of the motions of projectiles in the vicinity of the Earth surface. By intro-

ducing the magnitude 0 of the velocity vector 0 at the time 0t = , we can write:

( )0 0 cos sinj kϕ ϕ= +

. (7.47)

Thus, by substituting into Expression (7.45):

( )( )

( )0 0 0d( , ) cos sin

d

TT OMM t j k a t

tϕ ϕ= = + −

. (7.48)

By integrating twice with respect to time, we obtain:

( )200 0 0cos sin

2

aOM OM j t k t tϕ ϕ= + + −

, (7.49)

the point M being at point M0 at the time 0.t =

If we choose a coordinate system, such as the point M is at the origin O at the

time 0,t = the motion equation is reduced to:

( )200 0cos sin

2

aOM j t k t tϕ ϕ= + −

. (7.50)

The Cartesian coordinates of the point M relatively to the system (Oxyz) are then:

200 00, cos , sin .

2

ax y t z t tϕ ϕ= = = − (7.51)

The trajectory is contained into the plane (Oyz). If 2πϕ = ± , the trajectory is recti-

linear and supported by the axis Oz. If 2πϕ ≠ ± , the trajectory is a parabola.

7.3.2 Study of the case where the Trajectory is Rectilinear

7.3.2.1 Case where2πϕ =

From (7.51), the point M has for coordinates:

2000, 0, .

2

ax y z t t= = = − (7.52)

The velocity vector ( )

( , )T M t has for components:

0 00, 0, .x y z a t= = = − (7.53)

We deduce from these results the table 7.2 of the variations. The point M

moves away from O along the half-axis Oz

of positive abscissae, with a retarded


TABLE 7.2. Variations of z-coordinate in the case 2πϕ = .

t 0 0

0a

0

0

2

a

+∞

z 0 + 0 − 0− − −∞

z 0 2

0

02a

0 −∞

The motion is: retarded accelerated accelerated

motion up to the point of abscissa 2

0

02a

. This point being reached, the motion

becomes accelerated. The point M returns towards O, passes through O with a

velocity 0− , then moves away from O along the half-axis Oz

of negative abs-

cissae. An example of a motion of this type is that of a projectile launched verti-

cally towards the sky.

7.3.2.2 Case where 2πϕ = −

The point M has for coordinates:

2000, 0, .

2

ax y z t t= = = − − (7.54)

The velocity vector ( )

( , )T M t has for components:

0 00, 0, .x y z a t= = = − − (7.55)

The point M moves indefinitely away from the point O along the half-axis Oz

of

negative abscissae, with an accelerated motion. An example of such a motion is

that of a projectile launched vertically in a well.

7.3.3 Study of the case where the Trajectory is Parabolic

The coordinates of the point M are written (7.51):

200 00, cos , sin ,

2

ax y t z t tϕ ϕ= = = − (7.56)

and the components of the velocity vector are (7.48): 2

0 0 00, cos , sin .x y z a tϕ ϕ= = = − (7.57)

7.3.3.1 Case where 02πϕ< <

In the case where 02πϕ< < (or

2π ϕ π< < ), we have sin ϕ > 0. So, it results

7.3 Motions with a Constant Acceleration Vector 93

TABLE 7.3. Variations of z-coordinate in the case where 0 .2πϕ< <

t 0 0

0

sina

ϕ 0

0

2 sina

ϕ

+∞

z 0 sinϕ + 0 − 0 sinϕ− − −∞

z 0 2

20

0

sin2a

ϕ

0 −∞

y 0 2

0

0

sin 22a

ϕ

20

0

sin 2a

ϕ

+∞

the motion is: retarded accelerated accelerated

from (7.56) that at the beginning of the motion z is positive, then passes by a

maximum, is zero for 0

0

2 sinta

ϕ=

and becomes negative. The variations are

reported in Table 7.3. The trajectory is drawed in Figure 7.4. The motion of the

point M is retarded on the arc of parabola OH , H being the summit of the para-

bola of coordinates: 2 2

20 0

0 0

0, sin 2 , sin2 2

H H Hx y za a

ϕ ϕ= = =

. (7.58)

The motion is then accelerated, the point M passing through the point P (called

the horizontal range) of the axis Oy

:

20

0

0, sin 2 , 0P P Px y za

ϕ= = =

, (7.59)

FIGURE 7.4. Parabolic trajectory in the case 0 .2πϕ< <

0

0 0a a k= −

y

z

O

k

H

P 20

0

sin 22a

ϕ

20

0

sin 2a

ϕ

220

0

sin2a

ϕ

retarded

accelerated


with a velocity of components:

0 00, cos , sinP P Px y zϕ ϕ= = = − . (7.60)

In a general way, we observe that the magnitude of the velocity is the same one

at two points of the same z-coordinates (the components along k

being opposite).

Moreover, the distance from O to P is maximum and equal to 2

0

0a

, if

4πϕ = .

An example of such a motion is that of a projectile launched in the direction

which forms an angle ϕ with the Earth surface (the plane (Oxy) is then the hori-

zontal plane). The angle ϕ is the fire angle, the z-coordinate zH is the maximum

altitude reached by the projectile, and the distance from O to P is the horizontal

range of firing.

7.3.3.2 Case where 02π ϕ− < <

In the case where 02π ϕ− < < (or 3

2ππ ϕ< < ), we have sin 0ϕ < . It results

from (7.56) that z is always negative. As in the preceding case, the trajectory is an

arc of parabola tangent at the point O to the line ( )0, O , and contained in the

plane ( 0, 0y z≥ ≤ ) (Figure 7.5). The motion of the point M is constantly accele-

rated.

An example of such a motion is given by the motion of a projectile launched

with a negative fire angle, for example from a tower or a plane.

7.4 HELICOIDAL MOTION

A point M moves with a helicoidal motion in a given reference, if its trajectory

is a right circular helix, drawn on a right cylinder (Figure 7.6).

FIGURE7.5. Parabolic trajectory in the case 02π ϕ− < < .

0

0 0a a k= −

y

z

O

accelerated

7.4 Helicoidal Motion 95

In the coordinate system (Oxyz), the parametric representation of the helix can

be written, for the Cartesian coordinates of the point M, in the form:

cos , sin , ,x a y a z bα α α= = = (7.61)

where α is the angle of the cylindrical coordinates (an arbitrary function of time).

The quantity a is the radius of the cylinder on which the helix is drawn. The

parameter b is the reduced pitch of the helix: increasing the angle α by 2π leads

to translate the helix of 2 bkπ

since:

( 2 ) ( ),

( 2 ) ( ),

( 2 ) ( ) 2 .

x x

y y

z z b

α π α

α π α

α π α π

+ =

+ =

+ = +

(7.62)


( )OM a u b kα α= +

. (7.63)

Thence we deduce the kinematic vectors:

( ) ( , ) ( )

2M t a u b kπα α α= + +

, (7.64)

( )

2( , ) ( ) ( )2

Ta M t a u a u b kπα α α α α= − + + +

. (7.65)

The ratio of the components of the velocity vector is:

b ba aαα

=

(7.66)

which is independent of α. So, it results that the tangent at any point M of the helix

forms a constant angle with the axis ( ),M k

, parallel to the axis of the cylinder.

FIGURE 7.6. Helicoidal motion.

y

z

M

D

O a

x

k

( )u α

( )2

u πα +

( )2

a u πα α +

b kα

( )( , )T M t

helix


The helicoidal motion is uniform if:

0α ω= (7.67)

where ω0, the angular velocity of rotation, is independent of time. The kinematic

vectors are written in this case:

( ) 0 0( , ) ( )

2M t a u b kπω α ω= + +

, (7.68)

( )

20( , ) ( )Ta M t a uω α= −

. (7.69)

The algebraic velocity is deduced from (7.68) as:

2 20a b ω= + . (7.70)

For a uniform motion, the acceleration vector has only a normal component (6.9).

Hence it results that the principal normal at the point M of the helix is the normal

to the cylinder at this point :

( )ne u α= −

. (7.71)

Relations (7.68) and (7.69) compared to the expression (6.9) make it possible to

derive the radius of curvature of the helix. Thus:

2baa

= + . (7.72)

The curvature centre D (4.17) is defined by the relation:

( )MD u α= −

. (7.73)

7.5 CYCLOIDAL MOTION

A point M moves with a cycloidal in a given reference, if its trajectory is a

cycloid. The Cartesian coordinates of the point M relatively to a coordinate

system linked to this reference are written as:

( ) ( )sin , 1 cos , 0,x a q q y a q z= − = − = (7.74)

where q is a parameter which is a function of time.

An example of cycloidal motion is given by the motion of a point M of a disk

or wheel of radius a, rolling on the axis Ox

(Figure 7.7). By taking account of the

equality OH HM= , we obtain really Expressions (7.74) of the coordinates of the

point M.

If we increase the angle q by 2π, Expressions (7.74) show that:

( 2 ) ( ) 2 ,

( 2 ) ( ),

0.

x q x q a

y q y q

z

π π

π

+ = +

+ =

=

(7.75)

7.5 Cycloidal Motion 97

FIGURE 7.7. Cycloidal motion.

Increasing q by 2π leads to translate the curve of 2 a iπ

. The study of the trajec-

tory can thus be limited to the interval 0 2q π≤ ≤ .


( ) ( ) sin 1 cosOM i a q q j a q= − + −

. (7.76)

Hence we deduce the expressions of the kinematic vectors:

( ) ( ) ( , ) 1 cos sinT M t i aq q j aq q= − + , (7.77)

( ) ( ) ( )

2 2( , ) 1 cos sin sin cosTa M t i a q q q q j a q q q q = − + + + . (7.78)

The velocity vector can be put in the form:

( )

2( , ) 2 sin 2 sin cos2 2 2

T q q qM t i aq j aq= +

,

or

( ) ( , ) 2 sin ( )

2 2 2T q q

M t aq u π= −

. (7.79)

While comparing with Expression (6.6), we deduce then the unit vector of the

direction of the tangent:

( ) sin cos2 2 2 2

tq q q

e u i jπ= − = +

, (7.80)

and the algebraic velocity:

d 2 sind 2

qs aqt

= = . (7.81)

The expression of the vector ne

of the principal normal and that of the curvature

radius can be obtained by expressing the derivative of te

with respect to s:

d d d d 1 1( )d d d d 2 2

2 sin2

t te e q qt uqs q t s

a

= = −

. (7.82)

y

O

x

M

H

y(q)

x(q)

a q

t = 0


Hence by comparing to (4.11), we obtain:

( ) cos sin2 2 2

nq q q

e u i j= − = −

, (7.83)

4 sin , avec 0 2 .2

qa q π= ≤ ≤ (7.84)

The expression of the acceleration in the Frenet basis is from (6.9):

( ) ( )

2 2( , ) 2 sin cos sin2 2 2

Tt n

q q qa M t a q q e aq e= + +

. (7.85)

This expression can be found from (7.77).

EXERCISES

7.1 Performances relative to the accelerations of a car are the following ones:

a. initial accelerations time (in s)

de 0 à 60 km/h 6.4

de 0 à 80 km/h 10.5

b. acceleration stages

from 30 to 100 km/h in 4th 21.6

in 5th 30.0


in 5th 26.4

from 80 to 100 km/h in 3rd 5.7

in 4th 6.9

in 5th 9.5


in 5th 18.4

7.1.1. Give comments about these performances established on a rectilinear car-

track, and derive the average accelerations for each performance by assuming that

the motions are uniformly accelerated.

7.2.2. The car moves with the following stages:

a. acceleration from 0 to 80 km/h with the characteristics derived in 7.1.1;

b. from 80 to 100 km/h in 3rd;

c. from 100 to 120 km/h in 4th;

d. beyond 120 km/h in 5th (with the acceleration derived in 7.1.1 between 80

and 120 km/h ).

By assuming that the different stages are rectilinear and uniformly accelerated

motions of characteristics given or obtained in 7.1.1, derive:

Comments 99

7.2.2.1 the time and the distance necessary to reach the speed of 100 km/h,

of 120 km/h ;

7.2.2.2 the time and the speed reached after 1 000 m.

7.2 We consider the motion with constant acceleration studied in Section 7.3. We

study here the trajectories of the projectile M for a given initial speed 0 and when

the angle ϕ varies.

7.2.1. Derive the set of points which can be reached by the projectile M when the

angle ϕ varies.

7.2.2. Show that there exists two values of the angle ϕ which make it possible to

reach a given point Q inside of the set of points obtained previously.

COMMENTS

The motions studied in the present chapter are simple elementary motions

and do not call particular comments. The reader will be interested by all the

different types of motions.

CHAPTER 8

Motions with Central Acceleration

8.1 GENERAL PROPERTIES

8.1.1 Definition

The motion of a point M is a motion with central acceleration in the reference

system (T), if and only is there exists a point O in (T), such as the position vector

OM

of the point M is collinear to the acceleration vector of the point M:

( ) ( , ) ( )Ta M t M OMλ=

, (8.1)

where λ(M) is a real number dependent or independent of the point M.

8.1.2 A Motion with Central Acceleration is a Plane Trajectory Motion

It results from the definition (8.1) that a motion is a motion with central accele-

ration, if and only if: ( )

( , ) 0TOM a M t× =

. (8.2)

And, we have the relation:

( )( ) ( )

d ( , ) ( , )d

TT TOM M t OM a M t

t × = ×

. (8.3)

Comparison of (8.2) and (8.3) shows that the motion is a motion with central

acceleration, if and only if: ( )

( , )TOM M t C× =

, (8.4)

where C

is a vector independent of time.

8.1 General Properties 101

If C

is different from the null vector, the preceding expression shows that the

point M moves in a plane that passes through the point O and with direction

orthogonal to the vector C

.

If C

is the null vector, the trajectory is supported by the line that passes

through the point O.

8.1.3 Areal Velocity

The motion of the point M being a plane motion, it is possible to locate the

point M by its polar coordinates (r, α) in this plane (Figure 8.1a). Then let us

consider (Figure 8.1b) two infinitely close positions M(t) and M(t + dt) of the

point M on the trajectory. We have:

( ) ( )OM t r u α=

, (8.5)

and ( ) ( )

( ) ( d ) d ( ) d ( ) d ( )

d ( ) ( )d .2

T TM t M t t OM t r u r u

r u r u

α α

πα α α

+ = = +

= + +

(8.6)

The area swept by the segment OM is equal to the area of the surface dσ of the

triangle OM(t)M(t + dt). Thus:

21 1d ( ) d ( ) d2 2

OM t OM t rσ α= ∧ =

. (8.7)

The quantity σ represents the area swept between a date taken as the time origin

and the date t. Its derivative with respect to time σ is called the areal velocity of

the motion at the date t:

2d 1d 2

rtσσ α= = . (8.8)

The areal velocity represents the area swept by time unit.

FIGURE 8.1. Polar coordinates and swept area.

O

M(t + dt)

M(t)

(b) (a)

x

M

y

r

( )u α( )

2u πα +

O

102 Chapitre 8 Motion with Central Acceleration

8.1.4 Area Law

In the case of a plane motion, the velocity vector of the point M is written from

(6.35):

( ) ( , ) ( ) ( )

2T M t r u r u πα α α= + +

. (8.9)

It results that Relation (8.4) leads to the following expression of the vector C

:

2C r k C kα= =

, (8.10)

while stating:

2C r α= . (8.11)

The vector C

being independent of time, it results from it that C is also inde-

pendent of time. Moreover, by comparing with Expression (8.8), we obtain:

dd 2

Ctσσ = = . (8.12)

The constant C is then called the area constant.

For a plane motion with central acceleration of centre O, the areal velocity

relatively to the point O is constant.

8.1.5 Expressions of the Kinematic Vectors

The kinematic vectors (6.29) and (6.33) can be expressed while introducing the

area constant C expressed by (8.11). We have:

( )

( ) ( )

2

2 2

2 2 2

22

2

d d d d d d 1 ,d d d d d d

d d d d 1 d 1 ,d d d d d

, .

r r r C rr Ct t rr

r r C Cr Ct r rr r

C Cr rr r

α αα α α α

αα α α α

α α

= = = = = −

= = = − = −

= =

(8.13)

Hence:

( ) ( )

d 1( , ) ( ) ( )d 2

T CM t C u ur r

πα αα

= − + + , (8.14)

( ) ( )

2 2

2 2

1 d 1( , ) ( )d

T Ca M t ur rr

αα

= − +

. (8.15)

8.1.6 Polar Equation of the Trajectory

In the case of a motion with central acceleration, Expression (8.1) of the

acceleration vector is written as:

8.1 General Properties 103

( ) ( , ) ( , ) ,Ta M t r OMλ α=

(8.16)

where λ is a real number which depends a priori of r and α. The equations of

Dynamics (Part V) will make it possible to derive the expression for λ.

By introducing, into Relation (8.16), Expressions (8.5) of the position vector

and (8.15) of the acceleration vector, we obtain the differential equation which

relates the variables r and α :

( )2

3

2 2

d 1 1 0d

rr rC

λα

+ + = . (8.17)

This equation allows us to obtain r as a function of α, thus:

( )r f α= , (8.18)

when λ is known. The time law of the motion along the trajectory is then derived

from (8.11) in the form:

[ ]221 1d d ( ) dt r fC C

α α α= = . (8.19)

If α0 is the value of α at the date t0, the expression of t is obtained as:

[ ]0

20

1 ( ) dt t fC

α

αα α− = . (8.20)

8.1.7 Motions with Central Acceleration for which ( ) 2( , )Ta M t OMω= −

The case of rectilinear motions has been studied in Subsection 7.1.4, and we

thus study in this subsection only the case of curvilinear motions. Let (x, y, 0) be

the Cartesian coordinates of the point M in the plane (Oxy). The coordinates (x, y)

check the relations:

2 2 and x x y yω ω= − = − . (8.21)

Whence, the equations of motion:

cos sin ,

cos sin ,

0.

x A t B t

y D t E t

z

ω ω

ω ω

= +

= +

=

(8.22)

We choose a time scale such as at the date 0,t = the point M is in M0 such as:

0 0 ,OM x i=

(8.23)

with a velocity:

0 0 0 .x i y j= + (8.24)


The constants A, B, D and E are deduced as functions of the initial conditions x0,

0x and 0y . Hence it results that the motion equations are written as:

00

0

cos sin ,

sin ,

0, (by assuming 0).

xx x t t

yy t

z

ω ωω

ωω

ω

= +

=

= >

(8.25)

The trajectory is then an ellipse of focus O of equation:

( )2 2

02

0 00

1 1x

x y yy yx

ω − + =

. (8.26)

The trajectory is a circle if 0 0x = and 0 0y xω= ± .

Whatever the trajectory, the motion is periodic, of period:

2T πω

= . (8.27)

The area constant is:

0 0C x y= . (8.28)

8.2 MOTIONS WITH CENTRAL ACCELERATION

FOR WHICH ( )

3( , )T OMa M t K

OM= −

We study in this section the motions with central acceleration for which the

acceleration vector can be expressed in the form:

( )

3( , )T OMa M t K

OM= −

, (8.29)

where K is a real number independent of the point M.

8.2.1 Equation of the Trajectories

The equation of the trajectories is derived from Relation (8.17) which is

written here as:

( )2

2 2

d 1 1 0d

Kr r Cα

+ − = .

(8.30)

The general solution of this equation is:

( )02

1 cosK Ar C

α α= + − , (8.31)


3( , )

T OMa M t KOM

= −

105

where A and α0 are positive or negative constants determined by the initial condi-

tions (conditions at a given date). The preceding equation can be rewritten in the

form:

( )2

02

1 1 cosK ACr KC

α α = + −

. (8.32)

We observe then that the form of this equation is not changed, when we substitute

for the couple of constants ( )0, A α the couple ( )0, A α π− + . Without restricting

the generality of the study, it is then possible to choose the quantity 2AC

K as

positive. We state: 2

, with 0ACe eK

= ≥ . (8.33)

The equation of the trajectory is thus written finally as:

( )[ ]02

1 1 cosK er C

α α= + − . (8.34)

The trajectory of Equation (8.34) is derived from the curve of polar equation:

[ ]2

1 1 cosK er C

α= + , (8.35)

by applying to it a rotation of centre O and angle α0. Equation (8.35) is the polar

equation of a conic (ellipse, parabola, or hyperbola) of eccentricity e and para-

meter: 2Cp

K= . (8.36)

The origin O is one of the foci of the conic and the axis Ox

is the axis of the

conic. Equation (8.34) thus represents a conic of focus O, the axis of which forms

an angle α0 with the axis Ox

. However the condition 0r > imposes some res-

trictions according to the sign of K.

8.2.2 Study of the Trajectories

8.2.2.1 Case where K > 0

The parameter of the conic is then written as:

2CpK

= , (8.37)

and Equation (8.34) of the trajectory becomes:

( )01 cos

pr

e α α=

+ −. (8.38)


FIGURE 8.2. Trajectories in the case where K > 0.

1. If 1e > , the trajectory is the branch of hyperbola which turns its concavity

towards O (Figure 8.2a). The point P of smaller polar radius is called the peri-

centre:

1p

pOP r

e= =

+. (8.39)

2. If 1e = , the trajectory is a parabola (Figure 8.2b). The pericentre is then

defined by:

2p

pOP r= = . (8.40)

3. If 0 1e< < , the trajectory is an ellipse (Figure 8.2c). The pericentre is given

by:

1p

pOP r

e= =

+. (8.41)

The point A the most far from O is called the apocentre :

1A

pOA r

e= =

−. (8.42)

4. If 0e = , the trajectory is a circle of centre O.

conicaxis

x

y

P

p

O 0

asymptote

X

(a) 1e >

y

conicaxis

x

X P

p

O 0

(b) 1e =

x

conic axis

y

X P

p

O

0

A

C

Y

(c) 0 1e< <


3( , )

T OMa M t KOM

= −

107

FIGURE 8.3. Trajectory in the case where K < 0.

8.2.2.2 Case where K < 0

When K is negative, the parameter of the conic is:

2CpK

= − , (8.43)

and Equation (8.34) of the trajectory is put in the form:

( )01 cos

pr

e α α−

=+ −

. (8.44)

The condition that r is positive imposes that the eccentricity e is higher than 1.

The trajectory is the branch of hyperbola (figure 8.3), which turns its convex part

towards O. The pericentre is defined by:

1p

pOP r

e= =

−. (8.45)

8.2.3 Velocity Magnitude at a Point of the Trajectory

We denote by the velocity vector

( )( , )T M t

at a point of the trajectory.

From Expression (8.14), we have:

( )2 2

2 2

2

d 1d

CCr rα

= +

. (8.46)

Whence, while introducing Expression (8.34) of 1r

, we obtain:

( )2

2 2

22 1K K e

r C= + −

, (8.47)

conic axis

x

y

X P O

0

asymptote


or

( )2 2 K Er

= + , (8.48)

setting:

( )2

2

2

1 12

KE eC

= − . (8.49)

We thus find that the quantity 2

2Kr

− remains constant during the motion. Thus:

2

2K Er

− = . (8.50)

The sign of E depends (8.49) only on the eccentricity of the conic:

— if the trajectory is an ellipse, 0E < ;

— if the trajectory is a parabola, 0E = ;

— if the trajectory is a hyperbola, 0E > .

8.2.4 Elliptic Motion. Kepler’s Laws

8.2.4.1 Characteristics of the Elliptic Trajectory

In the case of an elliptic trajectory, the equation of the trajectory is given by

Relation (8.38), with 0 1e≤ ≤ . The distance between the pericentre and the

apocentre is equal to the major axis 2a of the ellipse. Whence, from (8.41) and

(8.42) :

21

pa

e=

−. (8.51)

The distance c between the centre C of the ellipse (Figure 8.2c) and the focus O,

called focal distance is:

21

pec a OP

e= − =

−. (8.52)

Whence, the expression of the eccentricity:

cea

= . (8.53)

In the axis system (CXY) of the ellipse (Figure 8.2c), Equation (8.35) of the ellipse

is written as:

( )1

2 2 2p eX X Y= + + , (8.54)

or while expending:

( )2 2

2 21X ae Y

a b

+ + = , (8.55)


3( , )

T OMa M t KOM

= −

109

with

( )2 2 21b a e= − . (8.56)

The parameter b represents the semi-minor axis of the ellipse.

Lastly, Expression (8.49) shows that the constant E is expressed in the case of

an elliptic trajectory in the form:

2KEa

= − . (8.57)

Hence it results that the magnitude of the velocity (8.48) is written as:

( )2 2 1Kr a

= − . (8.58)

The velocity is thus maximum at the pericentre (point nearest to the focus) and

minimum to the apocentre (the most distant point).

8.2.4.2 Periodic Time

The areal velocity being constant, the motion of the point M along the elliptic

trajectory is periodic of period T, equal to the time required by the point to

describe its trajectory. Thus, from Expression (8.12):

12

ab C Tπ = . (8.59)

By taking account of Relations (8.37), (8.51) and (8.56), the period of revolution

is written:

3/22T aK

π= . (8.60)

8.2.4.3 Kepler’s Laws

The Kepler’s laws regroup some of the results established in this chapter, and

may be stated as follows:

If, relatively to a reference system, a point M has a central acceleration

relatively to a point O fixed in this reference and if the trajectory of M has no

infinite branch, it results that:

1. The trajectory of the point M is an ellipse with the point O located at one of

its foci.

2. The areal velocity of the point relatively to the point O is constant.

3. The square of the periodic time is proportional to the cube of the semi-major

axis of the ellipse.

These laws were formulated in equivalent forms by the astronomer Kepler

(1571-1630), starting from astronomical observations of the motions of the

planets.


COMMENTS

A particularly important application of the motions with central accele-

ration is that of the motions of the planets and the motion of the Earth.

These motions of which the trajectories are ellipses are governed by the

Kepler’s laws introduced in this chapter. The reader will be thus more

particularly interested by the results established in Section 8.2.4. These

results will be used in Chapter 19.

CHAPTER 9

Kinematics of Rigid Body

9.1 GENERAL CONSIDERATIONS

9.1.1 Notion of Rigid Body

The purpose of the Mechanics of Rigid Bodies is to study the motions of

bodies, such that the distance between two arbitrary points of a given body is

independent of time or at least varies very little according to time. Such bodies are

called solid bodies, rigid bodies, or simply solids, and do not deform. Actual

structures or machines, however, are never absolutely rigid and deform under the

loads to which they are subjected. But these deformations are usually small and do

not appreciably affect the conditions of motions. The Mechanics of Rigid Bodies

allows us to describe the global behaviour of solids. The analysis of deformations

then requires us to consider theories which take into account the deformability of

bodies (resistance of materials, mechanics of deformable solids, etc.).

A rigid body will thus be described as a set of points constituting a geometric

space (Chapter 2) and such that the distance between two arbitrary points (P and

Q for example) of the body is independent of time:

( ) ( ), ( ) ( ) ( ) constantd P t Q t P t Q t= =

. (9.1)

9.1.2 Locating a Rigid Body

To know the motion of a rigid body relatively to a reference system, that is to

know the motion (position and kinematic vectors at any time) of each point of the

body. This problem constitutes the object of the Kinematics of Rigid Body.

To solve this problem, it is first necessary to express how it is possible to des-

cribe the situation of the body considered.

To situate a solid (S) relatively to a given reference system (T) (Figure 9.1), it

is necessary to know in the general case:

— the position, relatively to the reference (T), of a particular point of solid (S),

— the orientation of the solid (S) relatively to the reference (T).


FIGURE 9.1. Determination of the situation of a solid (S) relatively to the reference (T).

In this way (Figure 9.1), we associate first to the reference (T) a coordinate

system ( ) ( )/ , , Oxyz O i j k=

.

1. We choose then a particular point of OS of the body. The position of the

point OS to every time is given by the position vector SOO

which will be expres-

sed either as a function of the Cartesian coordinates of the OS relatively to the

system (Oxyz), or as a function of its cylindrical coordinates or of other coordi-

nates. The coordinates of the point OS depending on time are called parameters of

translation or degrees of freedom in translation of the solid. The choice of OS is

not arbitrary. It is necessary to choose the point or the one of the points having the

smallest number ( 3≤ ) of coordinates depending of time.

2. Lastly, we attach to the solid (S) a coordinate system ( )/ , , S S S SO i j k

. The

orientation is then determined by the knowledge of the matrix of basis change

allowing us to express ( ), , S S Si j k

as a function of ( ), , i j k

.The parameters

( 3≤ ), necessary to express this matrix and depending on time, are called the

parameters of rotation or degrees of freedom in rotation. For example, the matrix

of basis change is expressed (Subsection 2.5.3) as a function of the Eulerian

angles. The angles depending on time will be the parameters of rotation. The

orientation of the solid (S) does not depend on the choice of the point OS.

The set of the parameters of translation and rotation constitutes the parameters

of situation or degrees of freedom of the solid (S) relatively to the reference (T). If

the number of these parameters is equal to 6 (3 in translation and 3 in rotation),

we say that the solid is free in the reference (T). If this number is lower than 6,

some of these parameters of situation are “locked” (these parameters cannot vary

any more during time). We say then that, relatively to the reference (T), the solid

is jointed or subjected to joints.

Oi

j

k

y

z

x

(T )

P zS

yS

OS

xS

Sj

Si

Sk

M (S )

9.2 Relations between the Trajectories and the Kinematic Vectors of Two Points 113

9.2 RELATIONS BETWEEN THE TRAJECTORIES

AND THE KINEMATIC VECTORS OF TWO POINTS

ATTACHED TO A SOLID

We shall denote by x, y, z, the Cartesian coordinates of a point relatively to the

axis system ( )Oxyz attached to the reference (T) and xS, yS, zS, the coordinates of

this point relatively to the system ( )S S SOx y z attached to the solid (S). We shall

call P and M (figure 9.1) two arbitrary fixed points of solid (S).

9.2.1 Relation between the Trajectories

The problem to be solved is the following one. We know the trajectory in the

reference (T) of the point P of the solid (S). This trajectory for example is deter-

mined by the knowledge of the Cartesian coordinates, referred to the reference

(T), of the point P as functions of time: x(P, t), y(P, t), z(P, t). The position of the

point P in the solid (S) is known by the data of its Cartesian coordinates referred

to the coordinate system ( )S S SOx y z : ( ) ,Sx P ( ) ,Sy P ( ) .Sz P We search for the

trajectory of the point M, of which the position in the solid (S) is defined by its

Cartesian coordinates relatively to the system ( )S S SOx y z : ( ) ,Sx M ( ),Sy M

( ) .Sz M The points P and M being fixed in the solid (S), their coordinates relati-

vely to ( )S S SOx y z are independent of time.

To know the trajectory of the point M in the reference (T), it is necessary to

express, for example, the Cartesian coordinates: x(M, t), y(M, t), z(M, t) of the

point M, relatively to the system (Oxyz). These coordinates are the components of

the position vector OM

in the basis ( ), , i j k

. This vector is written:

OM OP PM= +

, (9.2)

with

[ ] [ ] [ ] ( ) ( ) ( ) ( ) ( ) ( )S S S S S S S S SPM x M x P i y M y P j z M z P k= − + − + −

. (9.3)

The exploitation of Relation (9.2) requires to express the vector PM

in the basis

( ), , i j k

, by introducing the matrix A(t) of basis change which relates the basis

( ), , S S Si j k

as a function of the basis ( ), , i j k

:

( )

S

S

S

i i

j t j

k k

=

A

. (9.4)

Taking into account (1.69), Relation (9.2) leads then to the relation:


t

( , ) ( , ) ( ) ( )

( , ) ( , ) ( ) ( ) ( )

( , ) ( , ) ( ) ( )

S S

S S

S S

x M t x P t x M x P

y M t y P t t y M y P

z M t z P t z M z P

− = + − −

A . (9.5)

trajectory of M trajectory of P transposed matrix coordinates of

in the reference (T) in the reference (T) of basis change M and P in (S)

This relation is the same as Relation (2.51) of reference change.

9.2.2 Relation between the Velocity Vectors

The velocity vector of the point M relatively to the reference (T) is:

( )( )

d( , )d

TT M t OM

t=

. (9.6)

In the same way, the velocity vector of the point P relatively to the reference (T) is

written as:

( )( )

d( , )d

TT P t OP

t=

. (9.7)

By substituting Expression (9.2) of OM

into Expression (9.6) of the velocity

vector, we obtain:

( )( ) ( )

( )( )

d d d( , ) ( , )d d d

T T TT TM t OP PM P t PM

t t t= + = +

. (9.8)

From expression (9.3), we have:

( )

[ ]( )

[ ]( )

[ ]( )

d d d( ) ( ) ( ) ( )d d d

d( ) ( ) .d

T T T

S S S S S S

T

S S S

PM x M x P i y M y P jt t t

z M z P kt

= − + −

+ −

(9.9)

Expressions (3.42) and (3.44) derived in Chapter 3 show that it is possible to write

the derivatives of the vectors of the basis in the forms:

( )( )

( )( )

( )( )

d d d, , .d d d

T T TT T T

S S S S S S S S Si i j j k kt t t

ω ω ω= × = × = ×

By substituting these expressions into (9.9), we obtain:

( )( )d

d

TT

SPM PMt

ω= ×

. (9.10)

Whence, the relation between the velocity vectors:

( ) ( ) ( ) ( , ) ( , )T T T

SM t P t PMω= + ×

. (9.11)

The vector ( )TSω

is called the instantaneous rotation vector relatively to the

motion of solid (S) with respect to the reference (T).


FIGURE 9.2. Orientation of a solid defined by the Eulerian angles.

9.2.3 Expression of the Instantaneous Vector of Rotation

In the case where the orientation of the solid (S) with respect to (T) is defined

at every instant by the Eulerian angles (Subsection 2.5.3), the expression of the

instantaneous vector of rotation is deduced from Expression (3.43). Thus: ( )

3T

S Sk i kω ψ θ ϕ= + + , (9.12)

where ψ, θ and ϕ are the three Eulerian angles which define in the general case

the orientation of the solid (S) relatively to the reference (T) (Figure 9.2). The

Eulerian angles are the angles of the three successive rotations (Subsection 2.5.3)

which make it possible to move from the coordinate system ( )/ , , SO i j k

to the

system ( )/ , , S S S SO i j k

:

— a rotation of angle ψ about the direction k

:

3

3

cos sin ,

sin cos ,

,

i i j

j i j

k

ψ ψ

ψ ψ

= +

= − +

(9.13)

— a rotation of angle θ about the direction 3i

:

3

4 3

3

,

cos sin ,

sin cos ,S

i

j j k

k j k

θ θ

θ θ

= + = − +

(9.14)

x

Oi

k

y

y3

3j

3i

x3

j

Sk

Si

4j

xS

Sj

y4

yS

zS

z


— a rotation of angle ϕ about the direction Sk

:

3 4

3 4

cos sin ,

sin cos ,

,

S

S

S

i i j

j i j

k

ϕ ϕ

ϕ ϕ

= +

= − +

(9.15)

The components (9.12) of the rotation vector ( )TSω

thus correspond to the three

rotations.

The expression of the rotation vector in the basis ( ), , i j k

is deduced from

(9.12) by expressing the vectors 3i

and Sk

. Thus, from (9.13) and (9.14):

( )3 cos sin ,

sin cos sin cos .S

i i j

k i j k

ψ ψ

ψ ψ θ θ

= +

= − + +

Whence, the expression of the rotation vector:

( ) ( ) ( ) ( )cos sin sin sin cos sin cos .TS i j kω θ ψ ϕ ψ θ θ ψ ϕ ψ θ ψ ϕ θ= + + − + +

(9.16)

In the same way, it is possible to express the rotation vector in the basis

( ), , S S Si j k

. We obtain:

( ) ( ) ( ) ( )cos sin sin cos sin sin cos .TS S S Si j kω θ ϕ ψ ϕ θ ψ ϕ θ θ ϕ ϕ ψ θ= + + − + +

(9.17)

9.2.4 Kinematic Torsor

Let us compare Relation (9.11) of the velocity vectors of the points M and P:

( ) ( ) ( ) ( , ) ( , )T T T

SM t P t PMω= + ×

,

with the relation expressing the moments of a torsor at the points M and P:

M P R PM= + ×

.

We observe that there is identity of the structure of these relations. This identity

shows that it is thus possible to consider:

The velocity vector of a point M as being the moment at the point M of a

torsor, which will denote by ( ) TS , of which the resultant is the instantaneous

rotation vector ( )TSω

relative to the motion of the solid (S) with respect to the

reference (T).

The torsor ( ) TS thus introduced is called the kinematic torsor or velocity

distributor torsor, relative to the motion of the solid (S) with respect to the

reference (T). Its elements of reduction at point M are:


( )( )T TSSR ω=

, instantaneous rotation vector; (9.18)

( )( )( , )

T TM S M t=

, velocity vector of the point M of (S). (9.19)

The relation between the velocity vectors is then obtained in the inverse way

according to the relation of the moments of a torsor:

( ) ( ) ( )T T TM PS S SR PM= + ×

, (9.20)

or taking account of (9.18) and (9.19):

( ) ( ) ( ) ( , ) ( , )T T T

SM t P t PMω= + ×

, (9.21)

which is the initial relation (9.11). The two formalisms, one of mathematical

nature and the other of mechanical nature, will not have to be mixed within a same

relation.

If the elements of reduction of the kinematic torsor are expressed at the

particular point OS chosen to define the parameters of situation of the solid (Sub-

section 9.1.2), its resultant (rotation vector) depends only on the parameters of

rotation and its moment (velocity vector of the point OS) depends only on the

parameters of translation. There is decoupling between the parameters of trans-

lation and rotation. Moreover, the kinematic torsor entirely characterizes the

motion of the solid (S) relatively to the reference (T), with regard of the velocity

vectors, whence its interest.

9.2.5 Relation between the Acceleration Vectors

The relation between the acceleration vectors is obtained by deriving Expres-

sion (9.11). That is:

( )( )

( )

( )( )

( )( ) ( )

( )

d( , ) ( , )d

d d d( , ) .d d d

TT T

T T TT T T

S S

a M t M tt

P t PM PMt t t

ω ω

=

= + × + ×

(9.22)

The first term is the acceleration vector of the point P. The third term is expressed

using Relation (9.10). Furthermore, it is possible to show that:

( )( )

( )( ) ( )

d d , vector which we shall denote by .d d

T ST T T

S S St t

ω ω ω=

(9.23)

The rotation vector thus has the same derivative relatively to (T) and (S). Finally,

Expression (9.22) leads to the relation between the acceleration vectors:

( ) ( ) ( ) ( ) ( )( ) ( , ) ( , )T T T T TS S Sa M t a P t PM PMω ω ω= + × + × ×

. (9.24)


FIGURE 9.3. Composition of motions.

9.3 GENERALIZATION OF THE COMPOSITION

OF MOTIONS

9.3.1 Composition of Kinematic Torsors

9.3.1.1 Problem

We consider the case of two rigid bodies (S1) and (S2) moving relatively to the

reference (T), and moving the one relatively to the other (Figure 9.3).

The motion of solid (S1) with respect to the reference (T) is characterized by its

kinematic torsor ( ) 1

TS of elements of reduction at the point O1 of the solid (S1):

( ) ( )

( ) ( )

( ) ( ) ( )

( )

1 1

1 1

1

1

1 1

, rotation vector relatively to the motion of

the solid with respect to the reference ;

( , ), velocity vector with respect to

of the point of the solid .

T TS S

T TO S

R

S T

O t T

O S

ω =

=

(9.25)

The motion of solid (S2) with respect to the reference (T) is characterized by its

kinematic torsor ( ) 2

TS of elements of reduction at the point O2 of the solid (S2):

( ) ( )

( ) ( )

( ) ( ) ( )

( )

2 2

2 2

2

2

2 2


the solid with respect to the reference ;

( , ), velocity vector relatively to


T TS S

T TO S

R

S T

O t T

O S

ω =

=

(9.26)

Oi

j

k

y

z

x

(T )

z2

y2

O2x2

2j

2i

2k

(S2 )

M

z1

y1

O1

1j

1i

1k

(S1 )

x1

9.3 Generalization of the Composition of Motions 119

Lastly, the motion of the solid (S2) relatively to the solid (S1) is characterized

by its kinematic torsor ( ) 1

2

SS of elements of reduction at the point O2 of the

solid (S2):

( ) ( )

( ) ( )( ) ( ) ( )

( )

1 1

2 2

1 12 2

2 1

2 1

2 2


the solid with respect to the solid ;

( , ), velocity vector relatively to


S SS S

S SO S

R

S S

O t S

O S

ω =

=

(9.27)

9.3.1.2 Relation between the Moment Vectors

The velocity vector relatively to the reference (T) of the point O2 of the solid

(S2) is written:

( )( ) ( )

( ) ( )( )

2 12 1 1 12 2d d d( , ) ( , )d d d

T T TT TO t OO OO O O O t O O

t t t= = + = +

. (9.28)

If we introduce the coordinates x1(O2, t), y1(O2, t) and z1(O2, t) of the point O2

with respect to a coordinate system attached to (S1) , we have:

1 1 2 1 1 2 1 1 2 12 ( , ) ( , ) ( , )O O x O t i y O t j z O t k= + +

. (9.29)

Whence:

( )

( ) ( ) ( )

1 1 2 1 1 2 1 1 2 12

11 11 2 1 2 1 2

d( , ) ( , ) ( , )

d

dd d ( , ) ( , ) ( , ) ,

d d d

T

TT T

O O x O t i y O t j z O t kt

ji kx O t y O t z O t

t t t

= + +

+ + +

or by analogy with (9.9) and (9.10)

( )( ) ( )

1

11 2 12 2

d( , )

d

TS T

SO O O t O O

tω= + ×

. (9.30)

Thus, by substituting into Relation (9.28:

( ) ( ) ( ) ( )

1

12 2 1 1 2( , ) ( , ) ( , )T S T T

SO t O t O t O Oω= + + ×

. (9.31)

However, we have the relation: ( ) ( ) ( ) 21 1

1 1 2( , )T T TOS S

O t O Oω+ × =

. (9.32)

Relation (9.31), by taking account of (9.26), (9.27) and (9.32), is thus written

finally in the form: ( ) ( ) ( ) 1

2 2 22 2 1

ST TO O OS S S

= +

. (9.33)

9.3.1.3 Relation between the Resultants

Let us consider a point M of the solid (S2) (Figure 9.3). We may write:


( ) ( ) ( ) ( )

2 22 2( , ) ( , )T T T T

M S SM t O t O Mω= = + ×

, (9.34)

and

( ) ( ) ( ) ( )

1 11 1

2 22 2( , ) ( , )

S SS SM S S

M t O t O Mω= = + × . (9.35)

Furthermore, by applying Relation (9.31) at the point M of the solid (S2) (instead

of the point O2), we obtain:

( ) ( ) ( ) ( )

1

11 2( , ) ( , ) ( , )

ST T TS

M t M t O t O Mω= + + ×

. (9.36)

Combination of Expressions (9.36) and (9.31) leads to:

( ) ( ) ( ) ( ) ( )

1 1

12 2 2( , ) ( , ) ( , ) ( , )

S ST T TS

M t O t M t O t O Mω− = − + ×

, (9.37)

or by taking account of Relations (9.34) and (9.35):

( ) ( ) ( )1

2 2 12 2 2

ST TS S S

O M O M O Mω ω ω× = × + ×

.

We deduce from that the relation between the rotation vectors:

( ) ( ) ( )1

2 2 1

ST TS S S

ω ω ω= +

. (9.38)

9.3.1.4 Relation between the Kinematic Torsors

From the relations between the resultants (9.38) and moments (9.33) of the

kinematic torsors, we deduce the relation of the composition of motions:

( ) ( ) ( ) 1

2 2 1

ST TS S S

= + . (9.39)

motion of (S2) motion of (S2) motion of (S1)

relatively to (T) relatively to (S1) relatively to (T)

The preceding relation expresses the combination of motions. This relation can

be extended to an arbitrary number of rigid bodies:

( ) ( ) ( ) ( ) 1 1

2 1 . . . n

n n

S ST TS S S S

−= + + + . (9.40)

9.3.2 Inverse Motions

The motion of solid (S1) relatively to solid (S2) is called the inverse motion of

the motion of (S2) relatively to solid (S1).

By identifying the solid (S2) with the reference (T), Relation (9.39) is written:

( ) ( ) ( ) 1 2 2

2 1 20S S S

S S S+ = = . (9.41)

9.4 Examples of Solid Motions 121

Whence, the relation between the inverse motions:

( ) ( ) 1 2

2 1

S SS S

= − . (9.42)

In particular, it results that:

( ) ( )1 2

2 1.

S S

S Sω ω= −

(9.43)

The instantaneous rotation vectors of the two inverse motions are opposite.

9.4 EXAMPLES OF SOLID MOTIONS

9.4.1 Motion of Rotation about a Fixed Axis

9.4.1.1 Definition and Parameters of Situation

The motion of a solid (S) relatively to a reference (T) is a motion of rotation

about an axis, if and only if two points A and B, distinct, of the solid (S) remain

fixed in (T) during the motion (Figure 9.4a).

Because of the invariance of the distances, all the points of the body located on

the line AB remain also fixed. The line ( )AB ∆= is the axis of rotation of the

motion.

The examples of motions of rotation about an axis are multiple: rotor, wheel,

winch, pendulum, etc.

To locate the solid (S) relatively to the reference (T), we attach first the coor-

dinate system (Oxyz) to the reference (T) such as the axis Oz

coincides with the

axis (∆) of rotation (Figure 9.4b), then we search for the parameters of situation.

1. We choose a particular point of the solid (S): a point of the axis of rotation,

for example the point O. This point being fixed, the motion does not have any

parameter of translation.

2. We attach to the solid (S) a coordinate system: ( ) ,S SOx y z of the same

origin and with the same axis Oz

as the system ( ).Oxyz The orientation of this

coordinate system is characterized by the angle of rotation ψ about the axis Oz

:

( ) ( ) ( ) , ,S St i i j jψ = =

. (9.44)

The motion is finally characterized by one parameter of rotation ψ. The basis

change between the two coordinate systems is written as:

cos sin ,

sin cos ,

.

S

S

i i j

j i j

k

ψ ψ

ψ ψ

= +

= − +

(9.45)

We have:

( ), ( ),2

S Si u j uπψ ψ= = +

(9.46)


FIGURE 9.4. Motion of Rotation about a Fixed Axis.

and ( ) ( )

d d, ,

d d

T TS S

S Si j

j it t

ψ ψ= = −

(9.47)

9.4.1.2 Kinematic Torsor

The kinematic torsor ( ) TS relatively to the motion of rotation of the solid (S)

with respect to (T) is defined by its elements of reduction at the point O:

( ) ( )T TS SR kω ψ= =

, (9.48)

( ) ( )( , ) 0T T

O S O t= = . (9.49)

9.4.1.3 Kinematic Vectors of an Arbitrary Point

Let M be an arbitrary point of the solid (S) (Figure 9.4). Its position in (S) is

determined by the data of its Cartesian coordinates xS, yS, zS, relatively to the

system ( )S SOx y z :

S S S S SOM x i y j z k= + +

. (9.50)

The velocity vector of the point M can be obtained:

— either by deriving directly the position vector and using Relations (9.47):

( )( )

d( , )d

TT

S S S SM t OM x j y it

ψ ψ= = −

, (9.51)

y

A

B

(S )

( )

(a)

Oi

j

k

z

x (T )

zS

yS

Sj

Si

M

yS

(S )

( )

(b)

xS

xS


— or by using Relation (9.21):

( ) ( ) ( )

( )

( , ) ( , )

.

T T TS

S S S S S

M t O t OM

k x i y j z k

ω

ψ

= + ×

= × + +

(9.52)

Whence:

( ) ( , )T

S S S SM t x j y iψ ψ= − .

The velocity vector can then be expressed in the basis ( ), , i j k

by considering

the basis change (9.45). Thus:

( ) ( ) ( ) ( , ) sin cos cos sinTS S S SM t x y i x y jψ ψ ψ ψ ψ ψ= − + + −

. (9.53)

The acceleration vector can then be obtained in different ways.

1. Starting from Relation (9.24) which is written:

( ) ( ) ( ) ( ) ( )( ) ( , ) ( , )T T T T TS S Sa M t a O t OM OMω ω ω= + × + × ×

,

with ( )

( ) ( )( ) ( )( ) ( )

2 2

( , ) 0,

,

T

TS S S S S S S S S S

T TS S S S S S S S S S

a O t

OM k x i y j z k x j y i

OM k x j y i x i y j

ω ψ ψ ψ

ω ω ψ ψ ψ ψ ψ

=

× = × + + = −

× × = × − = − −

Whence:

( ) ( ) ( )

2 2( , )TS S S S S Sa M t y x i x y jψ ψ ψ ψ= − + + −

. (9.54)

2. By deriving Expression (9.51) of the velocity vector:

( )( )

( )

2 2d( , ) ( , )d

TT T

S S S S S S S Sa M t M t x j x i y i y jt

ψ ψ ψ ψ= = − − − . (9.55)

We find Relation (9.54) again. The acceleration vector can thus be obtain from

(9.45) in the basis ( ), , i j k

:

( ) ( ) ( )

( ) ( )

2 2

2 2

( , ) cos sin

sin cos .

TS S S S

S S S S

a M t y x x y i

y x x y j

ψ ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ ψ

= − + + −

+ − + + −

(9.56)

3. Lastly, ( )

( , )Ta M t

can be deduced by deriving directly Expression (9.53):

( ) ( ) ( )

( ) ( )

2

2

( , ) sin cos cos sin

cos sin sin cos .

TS S S S

S S S S

a M t x y x y i

x y x y j

ψ ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ ψ

= − + + −

+ − − +

We obtain Relation (9.56) again.


9.4.2 Translation Motion of a Rigid Body


A solid (S) has a translation motion relatively to the reference (T) if the solid

(S) has an invariable orientation during the time relatively to the reference (T).

It is equivalent to say that the basis ( ), , S S Si j k

attached to the solid (S) is

independent of time. It is then possible to choose the coordinate systems attached

to (S) and (T) so that their axes remain parallel (Figure 9.5) during the motion of

the solid (S):

, , .S S Si i j j k k= = =

(9.57)

To locate the solid (S) relatively to the reference (T), it is thus necessary and

sufficient to determine the position of a point OS of the solid (S), thus 3 para-

meters. The translation motion is a motion with 3 degrees of freedom in trans-

lation.

Note. We will denote in a similar way the axes of same direction. Thus, the

coordinate systems respectively attached to (T) and (S) are noted here by (Oxyz)

and (OSxyz). They differ simply by their origins.

The position relatively to the reference (T) of any point M of the solid (S) is

given by the position vector:

S SOM OO O M= +

, (9.58)

where SO M

is an invariable vector during the motion of the solid (S) relatively to

(T). Thus, it results that the trajectory of the point M in the reference (T) is

deduced from the one of OS, by the translation of vector SO M

: the trajectories of

FIGURE 9.5. Translation motion.

Oi

j

k

y

z

x

(T )

j

k

i OS (S )

M

z

y

x


all the points of (S) are superposable curves. If the trajectory of the point OS is a

straight line, it is said that the solid (S) has a motion of rectilinear translation (the

number of parameters is reduced to 1). If its trajectory is a curved line, the motion

of the solid (S) is a curvilinear translation with 2 or 3 parameters de translation,

according as the curve is plane or not.

The examples of translation motion are numerous: slides (shaper, planer, etc.),

tables of machines-tools (milling machine, etc.), pistons, elevators, coupling

crank, etc.


The elements of reduction of the kinematic torsor( ) TS are written at the point

OS as: ( ) ( )

0T TS SR ω= =

, (9.59)

( ) ( )

( , )S

T TO S SO t=

. (9.60)

The expression of the velocity vector of the point OS depends on the translation

motion under consideration.

The translation motion is characterized by a null rotation vector and conse-

quently by a kinematic torsor which is a couple-torsor.


The kinematic torsor being a couple-torsor, its moment is the same at any point

M of the solid (S). So, it results that:

( ) ( ) ( , ) ( , ), ( ).T T

SM t O t M S= ∀ ∈ (9.61)

It is the same for the acceleration vector:

( ) ( ) ( , ) ( , ), ( ).T T

Sa M t a O t M S= ∀ ∈

(9.62)

In a motion of translation all the points of the rigid body have the same kinematic

vectors.

9.4.3 Motion of a Body Subjected to a Cylindrical Joint


During its motion, a solid (S) is subjected to a cylindrical joint relatively to the

reference (T), if and only if a straight line attached to the solid (S) remains in

geometrical coincidence with a line attached to the reference (T).

The line (∆) attached to the reference (T) is called the axis of the cylindrical

joint. To define the parameters of situation, we associate the system (Oxyz) to the


FIGURE 9.6. Solid (S) with cylindrical joint of axis (∆).

reference (T) so that the axis Oz

coincides with the axis of the cylindrical joint

(Figure 9.6). Then the parameters of situation are defined as follows.

1. As particular point of the solid (S), we choose a point OS of the axis of the

cylindrical joint. The coordinates of OS relatively to the system (Oxyz) are:

( )0, 0, SO z . (9.63)

The motion thus has 1 parameter of translation: z.

2. As coordinate system attached to the solid (S), we choose the system

( )S S SO x y z , thus having the z-axis common with the system ( )Oxyz . The orien-

tation of this system is characterized by the angle of rotation ψ about the axis Oz

:

( ) ( )( ) , , S St i i j jψ = =

. (9.64)

The motion thus has 1 parameter of rotation: ψ. The basis change between the two

coordinate system is written:

cos sin ,

sin cos ,

.

S

S

i i j

j i j

k

ψ ψ

ψ ψ

= +

= − +

(9.65)

Finally, the motion of the solid (S) relatively to the reference (T) has 2 para-

meters of situation: z, ψ.

z

OS

i

j

k

y

x

(T )

zS

yS

Sj

Si

M

yS

(S )

( )

x

yO

xSxS



The kinematic torsor ( ) TS relatively to the motion of the solid (S) with

respect to the reference (T) has for elements of reduction at the point O:

( ) ( ),T T

S SR kω ψ= = (9.66)

( ) ( ) ( , ) .S

T TO S SO t z k= =

(9.67)


By comparing the kinematic torsor of a motion of a rigid body subjected to a

cylindrical joint with the one of the motion of rotation about a fixed axis (Rela-

tions (9.48) and (9.49)), we observe that these relations differ by the moment at

the point OS. It results from the expression of the moment at any point M:

( ) ( ) ( ) ( , ) ( , ) ,T T T

S S SM t O t O Mω= + ×

that the kinematic vectors of the motion with a cylindrical joint are deduced from

the relations obtained in the case of the rotation about an axis, by adding the trans-

lation terms:

( ) ( ) ( , ) , ( , )T T

S SO t z k a O t z k= = . (9.68)

Hence the expressions of the kinematic vectors:

( ) ( , )T

S S S SM t y i x j z kψ ψ= − + +

, (9.69)

( ) ( ) ( ) ( , ) sin cos cos sinT

S S S SM t x y i x y j z kψ ψ ψ ψ ψ ψ= − + + − +

, (9.70)

( ) ( ) ( )

2 2( , )TS S S S S Sa M t y x i x y j z kψ ψ ψ ψ= − + + − +

, (9.71)

( ) ( ) ( )

( ) ( )

2 2

2 2

( , ) cos sin

sin cos .

TS S S S

S S S S

a M t y x x y i

y x x y j z k

ψ ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ ψ

= − + + −

+ − + + − +

(9.72)

9.4.4 Motion of Rotation about a Fixed Point


The motion of a solid (S) relatively to the reference (T) is a motion of rotation

about a fixed point, if and only if a point A of the solid (S) remain fixed in the

reference (T) during the motion.

The coordinate systems associated to (S) and (T) are chosen so that their origins

O and OS coincide with the point A (Figure 9.7). To locate the solid (S) relatively

to the reference (T), it is necessary and sufficient to determine the orientation of

(S), defined by the three Eulerian angles: ψ, θ and ϕ as functions of time. A


FIGURE 9.7. Body (S) in rotation about the point A.

motion of rotation about a point is then a motion with 3 degrees of freedom in

rotation. Any point M of the solid (S) has a trajectory supported by a sphere of

centre A.

As examples of motions of rotation about a point, we can quote: coupling of

caravan, gyroscope, cardan joint, etc.


At the point A fixed, the elements of reduction of the kinematic torsor are

written:

( ) ( ) 3

T TS S SR k i kω ψ θ ϕ= = + +

, (9.73)

( ) ( )( , ) 0T T

A S A t= = . (9.74)

The Eulerian angles are defined in Figure 9.2.

It results from (9.73) and (9.74) that the torsor is a slider. It has an axis of null

moments: at all the points of this axis, at a given instant, the velocity vectors are

null. This axis is called the instantaneous rotation axis. The axis passes through

the point A and has the vector ( )TSω

as direction vector.


The kinematic vectors of the point A being null, Expressions (9.21) and (9.22)

lead to: ( ) ( )

( , )T TSM t AMω= ×

, (9.75)

( ) ( ) ( ) ( )( )( , )T T T TS S Sa M t AM AMω ω ω= × + × ∧

. (9.76)

Ay

z

x

(T )

zS

yS

xS

M

xS

yS

(S )

zS


The vector AM

is a vector independent of time in the system ( )S S SAx y z :

S S S S S SAM x i y j x k= + +

, (9.77)

where xS, yS and zS are the Cartesian coordinates of the point M relatively to the

reference ( ).S S SAx y z The product vectors are thus obtained simply by expres-

sing ( )TSω

in the basis ( ), , S S Si j k

. Hence:

( ) 1 2 3

TS S S Si j kω ω ω ω= + +

, (9.78)

with, from (9.17):

1

2

3

cos sin sin ,

cos sin sin ,

cos .

ω θ ϕ ψ ϕ θ

ω ψ ϕ θ θ ϕ

ω ϕ ψ θ

= +

= − = +

(9.79)

Whence:

( ) ( ) ( ) ( ) 2 3 3 1 1 2( , )T

S S S S S S S S SM t z y i x z j y x kω ω ω ω ω ω= − + − + −

. (9.80)

The determination of ( )

( , )Ta M t

needs to express ( )TSω . We obtain:

( ) 1 2 3

TS S S Si j kω ω ω ω= + +

, (9.81)

with

1

2

3

cos sin sin sin cos sin sin cos ,

cos sin sin sin cos cos sin cos ,

cos sin .

ω θ ϕ θϕ ϕ ψ ϕ θ ψϕ ϕ θ ψθ ϕ θ

ω ψ ϕ θ ψϕ ϕ θ ψθ ϕ θ θ ϕ θϕ ϕ

ω ϕ ψ θ ψθ θ

= − + + +

= − + − − = + −

(9.82)

It results from this that the acceleration vector is expressed in the form:

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

2 22 3 1 2 3 1 2 2

2 23 1 2 3 1 1 2 3

2 21 2 1 3 2 2 3 1

( , )

.

TS S S S

S S S S

S S S S

a M t x y z i

y z x j

z x y k

ω ω ω ω ω ω ω ω



= − + + − + +

+ − + + − + +

+ − + + − + +

(9.83)

9.4.5 Plane Motion


The motion of a solid (S) relatively to the reference (T) is a plane motion, if an

only if, a plane (PS) attached to the solid (S) remains in coincidence with a plane

(P) attached to the reference (T).

It is then always possible to choose (figure 9.8) the coordinate systems atta-

ched to (S) and (T), so that the plane ( )S SOx y is the plane (PS) and the plane

( )Oxy is the plane (P). The axes Oz

and SO z

have then the same direction k

.


The situation of the solid (S) is determined by:

— the position of the point OS in the plane (P) defined by its coordinates x and

y:

SOO x i y j= +

, (9.84)

— the orientation of the coordinate system ( )S S SO x y z with respect to the

system ( )SO xyz defined by the angle of rotation ψ about the direction k

:

( ) ( ) , ,S Si i j j ψ= =

. (9.85)

The basis change between the two system is written:

cos sin ,

sin cos ,

.

S

S

i i j

j i j

k

ψ ψ

ψ ψ

= +

= − +

(9.86)

The plane motion is a motion with 3 degrees of freedom: x, y, ψ, (2 degrees of

freedom in translation and 1 degree in rotation).

As examples of plane motions, we quote:

— Solid sliding on a plane (Figure 9.9a): the plane of contact (PS) of the solid

moves on the plane (P).

— Cylinder rolling on a plane (Figure 9.9b), in the case where the axis of the

cylinder (∆) remains parallel to itself: a cross-section (PS) moves on the plane (P).

— Articulated systems with plane symmetry: slider-crank system (Figure

9.9c), cams (Figure 9.9d). These systems make it possible to transform a rotation

motion into an alternative rectilinear motion.

FIGURE 9.8. Plane motion.

z

OS

i

j

k

y

x

zS

yS

Sj

Si

M

yS

(PS )

(S )

xSxS

x

(P )

yO

z

(T )


FIGURE 9.9. Examples of plane motions.

9.4.5.2 Trajectories of an Arbitrary Point

If M is an arbitrary point of the solid (S) and H its orthogonal projection in the

plane ( )S S SO x y (Figure 9.8), the position of M with respect to the reference (T) is

given by:

SOM OH HM OH z k= + = +

, (9.87)

where zS, the z-coordinate of M, is independent of time. Whence the result:

The trajectory of the point M of z-coordinate zS is deduced from the trajectory

of the point H, projection of M in the plane (PS), by the translation of vector Sz k

:

the trajectory is thus plane and located in the plane of coordinate zS parallel to

the plane (PS). It is then sufficient to know the trajectories of the points of the

plane (PS).


The elements of reduction of the kinematic torsor ( ) TS relative to the motion

of (S) with respect to the reference (T) are at the point OS:

( ) ( )T TS SR kω ψ= =

, (9.88)

( ) ( ) ( , )S

T TO S SO t x i y j= = +

. (9.89)

(P)

(PS)

(a) (b)

(c) (d)

rod

piston

(P)

(PS)



Let M be an arbitrary point of the solid (S) of Cartesian coordinates xS, yS, zS,

relatively to the coordinate system ( )S S SO x y z :

S S S S SOM x i y j z k= + +

. (9.90)

The velocity vector is written:

( ) ( ) ( ) ( , ) ( , )T T T


. (9.91)

Thus: ( )

( , )TS S S SM t x i y j x j y iψ ψ= + + −

.

Or by taking account of (9.86):

( ) ( )[ ] ( )[ ] ( , ) sin cos cos sinTS S S SM t x x y i y x y jψ ψ ψ ψ ψ ψ= − + + + −

.

(9.92)

The acceleration vector is deduced from the preceding relation by derivation:

( ) ( ) ( )

( ) ( )

2

2


cos sin sin cos .

TS S S S

S S S S

a M t x x y x y i

y x y x y j

ψ ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ ψ

= − + − −

+ + − − +

(9.93)

9.4.5.5 Instantaneous Centre of Rotation

The scalar invariant of the kinematic torsor of the plane motion is from (9.88)

and (9.89): ( ) ( ) ( )

( , ) 0T T TS S SI O tω= =⋅

. (9.94)

The kinematic torsor is thus a slider which has an instantaneous axis of rotation,

the points of which have null velocity vectors. This axis has ( )TSω

for direction

vector. It is thus orthogonal to the plane (Oxy), at a point I which depends of the

motion. This point I, belonging to the planes (P) and (PS), is called the

instantaneous centre of rotation of the motion of (S) with respect to (T) (or of the

plane (PS) with respect to the plane (P)). Its position is given by Expression (5.27)

which is written here:

( )

2

( , )TS

Sk O t

O Iψ

ψ

×=

. (9.95)

We have:

( )( ) ( )

d d( , )

d d

T TT

S SSO t OO OOt

ψψ

= =

.

It results that the position of the point I is expressed according to the relation: ( )

d

d

T

SSO I k OOψ

= ×

. (9.96)


9.4.5.6 Space Centrode and Body Centrode

The instantaneous centre of rotation is a mobile point relatively to the refe-

rence (T) and relatively to the solid (S). We call then:

— space centrode of the motion of the plane (PS) on (P), the set of points of

the plane (P) in coincidence at every time with the instantaneous centre of rota-

tion: that is the trajectory of the centre of rotation in the reference (T);

— body centrode of the motion of the plane (PS) on (P), the set of points of the

plane (PS) in coincidence at every time with the instantaneous centre of rotation:

that is the trajectory of the centre of rotation in the reference (S).

Let us express the velocity vector of the point I relatively to the reference (T):

( )( )

d( , )

d

TT I t OI

t=

, (9.97)

with ( ) ( )

( ) ( )( )

d d d( , )

d d d

T T TT

S S S SOI OO O I O t O It t t

= + = +

. (9.98)

By introducing the coordinates xS (I, t), yS (I, t) and zS (I, t) of the centre of rotation

relatively to the coordinate system ( ) ,S S SO x y z we have:

( , ) ( , ) ( , )S S S S S SO I x I t i y I t j z I t k= + +

, (9.99)

and by analogy with (9.30):

( )( ) ( )d

( , )d

TS T

S S SO I I t O It

ω= + ×

. (9.100)

Hence finally: ( ) ( ) ( ) ( )

( , ) ( , ) ( , )T S T TS S SI t I t O t O Iω= + + ×

, (9.101)

or ( ) ( ) ( ) ( , ) ( , )T S T

I SI t I t= + . (9.102)

Taking into account the definition of the centre of rotation, the preceding relation

is reduced to: ( ) ( )

( , ) ( , )T SI t I t= . (9.103)

The velocity vector of the instantaneous centre of rotation is the same in the refe-

rences (T) and (S). It results that the space centrode and the body centrode are tan-

gent at point I. Relation (9.103) shows also that the body centrode rolls without

sliding on the space centrode (see Sections 10.1.2 and 10.1.3 of Chapter 10).

Lastly, for two inverse plane motions, the instantaneous centres of rotation

coincide. The space centrode and the body centrode are inverted.

9.4.5.7 Equations of the Space and Body Centrodes

The position of the point OS, relatively to the reference (T), was defined (9.84)

by its coordinates (x, y, 0). Expression (9.96) is then written:


d d

d dS

y xO I i j

ψ ψ= − +

. (9.104)

The position vector of the point I in the reference (T) is expressed by the relation:

d d

d dS S

y xOI OO O I x i y j

ψ ψ = + = − + +

, (9.105)

which thus defines the trajectory of the centre of rotation in the reference (T):

d ( , ) ,

d

d( , ) ,

d

0.

yx I t x

xy I t y

z

ψ

ψ

= − = +

=

(9.106)

To obtain the trajectory in the reference (S), we may apply to Expression (9.104)

the inverse relation of basis change:

cos sin ,

sin cos .

S S

S S

i i j

j i j

ψ ψ

ψ ψ

= −

= +

(9.107)

We obtain:

d dd dsin cos cos sin

d d d dS S S

y yx xO I i jψ ψ ψ ψ

ψ ψ ψ ψ = − + +

. (9.108)

Hence the equation of the trajectory in the reference (S):

dd ( , ) sin cos ,

d d

dd( , ) cos sin ,

d d

( , ) 0.

S

S

S

yxx I t

yxy I t

z I t

ψ ψψ ψ

ψ ψψ ψ

= − = +

=

(9.109)

EXERCISES

9.1 Implement the kinematics of the motion of a parallelepiped (S) relatively to

the plane (T), such as the plane ABCD of the solid (S) remains in contact with the

plane (T) (Figure 9.10).

9.2 Implement the kinematics of the motion of a cylinder (S) on a plane (T), when

a generator of the cylinder remains in contact with the plane (T) (Figure 9.11).

9.3 We consider the mechanical system schematized in Figure 9.12. A rigid body

(S1) is connected to the support (T) so that its motion is a rectilinear translation

motion. Moreover, the solid (S1) is connected to the support through a spring (R).

The solid (S2) is connected to the solid (S1) so that the motion is a motion of

Exercises 135

FIGURE 9.10. Parallelepiped on a plane.

rotation about an axis (∆2) orthogonal to (∆1) and A1A2. The point A2 of the solid

(S2) is distant of a from the point A1 common to (S1) and (S2).

1. Determine the parameters of situation.

2. Inplement the kinematic analysis.

3. Express the kinematic vectors of point A2.

FIGURE 9.11 Cylinder on a plane.

FIGURE 9.12. Mechanical system of two rigid bodies.

(T)

(S)A

B

C

DA'

B'

C'

D'

(T)

(S)

(T)

(R) (S1)

(S2)

A2

A1

(1)

(2)


COMMENTS

The chapter introduces the very important concept of kinematic torsor

relatively to the motion of a rigid body with respect to another one. The

expression of its moment allows us to derive the relation between the

velocity vectors of two points of a rigid body in motion. Its resultant is the

instantaneous rotation vector which is simply expressed as a function of

the angle of rotations applied to the rigid body considered to characterize

its orientation. Then, its expression allows us to express the rotation vector

either in a basis associated to the rigid body itself or in a basis associated to

the reference system.

The concepts of the kinematics of solid are then applied to elementary

examples of motions of a solid, which the reader will study carefully. Also,

the reader will note that the analysis of the kinematics of a solid is always

implemented by the same process: determination of the parameters of

situation, determination of the kinematic torsor of which the elements of

reduction are evaluated at the point where the parameters of translation

were determined, determination of the kinematic vectors (velocity vector

and acceleration vector) of an arbitrary point of the solid.

The notations used:( )

d,

d

T

OMt

( )( , ) ,T M t

( )( , ) ,Ta M t

( )

( )d( , ) ,

d

TT M t

t

( ) TS ,

( ) TM S

, ( )TSω

, etc., can be appear somewhat heavy to operate.

They were chosen in such a way to render a best account of the concepts

that the notations represent, while unifying the notations. Thus, the

reference (T) with respect to which are referred the quantities is always in

the same position in the various notations. The derivative ( )

d

d

T

t means that

the derivation is implemented in a basis associated to (T), basis which is

considered as being independent of time. The kinematic vectors ( )

( , )T M t

and ( )

( , )Ta M t

point out that the velocity quantity or the acceleration a

is considered relatively to the reference (T). The kinematic torsor ( ) TS

and the rotation vector ( )TSω

are related to the motion of the solid (S) with

respect to the reference (T). Lastly, in the notation ( ) ,T

M S

the point

M must belong to the solid (S).

CHAPTER 10

Kinematics of Rigid Bodies in Contact

10.1 KINEMATICS OF TWO SOLIDS

IN CONTACT

10.1.1 Solids in Contact at a Point. Sliding

Let (S1) and (S2) be two solids in contact, at time t, at the point M (Figure

10.1). If the solids remain in point contact during a given time interval, the point

M describes, during this interval, a trajectory C(1)(M) drawn on the surface (Σ1)

bounding the solid (S1) and a trajectory C(2)(M) drawn on the surface (Σ2)

bounding the solid (S2).

The motion of the solid (S2) with respect to the solid (S1) is characterized by

FIGURE 10.1. Solids in contact at a point.

M

(S2)

(S1)

(1)

(2) C(2)(M)

C(1)(M)

138 Chapter 10 Kinematics of Rigid Bodies

the kinematic torsor ( ) 12 of resultant

( )12ω

which is the instantaneous rotation

vector of the motion. The point M being in motion relatively to the solids (S1) and

(S2), we obtain by analogy with Relation (9.100):

( ) ( ) ( )

1 2 12( , ) ( , ) MM t M t= +

, (10.1)

where the vectors ( )1 ( , )M t

and ( )2 ( , )M t

are the velocity vectors of the point

of contact M respectively with respect to the solid (S1) and with respect to the

solid (S2).

The moment at the point M of the kinematic torsor ( ) 12 introduced in the

preceding expression is called the velocity vector of sliding of the solid (S2) on the

solid (S1) at the point of contact M at time t. Hence:

( ) ( ) 1 12 2( , ) Mg M t =

. (10.2)

From (10.1), we have:

( ) ( ) ( )

1 1 22( , ) ( , ) ( , )g M t M t M t= −

. (10.3)

This relation shows that the sliding velocity vector is a direction of the plane of

direction vectors ( )1 ( , )M t

and ( )2 ( , ) ,M t

plane which is in coincidence with

the plane (T) tangent at M to the two surfaces (Σ1) and (Σ2) (Figure 10.2).

If the sliding velocity vector is not null, it is said that the solid (S2) slides on the

solid (S1) at the point M and at time t. On the contrary if this vector is null, it is

said that the solid (S2) does not slide on the solid (S1) at the point M and at time t.

10.1.2 Spinning and Rolling

When the two solids (S1) and (S2) are tangent at the point M it is usual to

resolve the rotation vector( )12ω

into the sum of two vectors of rotation:

FIGURE 10.2. Sliding, rolling and spinning.

(S1)

(S2)

M

( )

12 nω

( )

12 tω

( )12ω

( )12 ( , )g M t

tangent plane

10.1 Kinematics of Two Solids in Contact 139

— a vector ( )12 tω

of direction contained in the plane (T) tangent to (Σ1) and to

(Σ2);

— a vector ( )12 nω

of direction orthogonal to the plane (T) thus to the directions ( )1 ( , )M t

and ( )2 ( , )M t

.

Hence: ( ) ( ) ( )

1 1 12 2 2t nω ω ω= +

. (10.4)

These vectors are called rolling rotation vector ( )12 tω

and spinning rotation vector( )12 nω

. If n

is the unit direction vector of the normal at M to the tangent plane:

( ) ( )

( ) ( )

1 2

1 2

( , ) ( , )

( , ) ( , )

M t M tn

M t M t

∧=

∧

, (10.5)

we have:

( ) ( )( )( ) ( )( )

1 12 2

1 12 2

,

.

t

n

n n

n n

ω ω

ω ω

= × ×

= ⋅

(10.6)

If the rotation vector of rolling (or/and of spinning) is not null, it is said that the

solid (S2) rolls (or/and spins) on the solid (S1), at point M at time t. In contrast if

this vector is null, it is said that the solid (S2) does not roll (or/and does not spin).

10.1.3 Conclusions

If two solids (S2) and (S1) are in contact at the point M at time t, the motion of

(S2) with respect to (S1) is characterized by the elements of reduction at the point

M of the kinematic torsor ( ) 12 :

( ) ( )

( ) ( )

( ) ( )

1 12 2

1 12 2

2 1

, instantaneous rotation vector;

( , ) , velocity vector of at in the motion of on .

M g

R

M t sliding MS S

ω = =

(10.7)

The rotation vector is then resolved into two vectors: ( )12 tω

: rotation vector of rolling, of direction contained in the plane tangent at

the point M to the solids (S1) and (S2);

( )12 nω

: rotation vector of spinning, of direction orthogonal to the plane tangent

at the point M.

The nullity or not of these vectors characterizes then the type of motion of the

solid (S2) with respect to the solid (S1) at the point M, in accordance with Table

10.1.


TABLE 10.1 Sliding, rolling and spinning at a point of contact between two solids (S1) and

(S2).

If at a given time

( )12 tω

rolling

( )12 nω

spinning

( ) 12M

sliding

it is said that

0

0

0

(S2) rolls, spins and slides relatively to (S1).

0

0

= 0

(S2) rolls and spins without sliding.

0

= 0

= 0

(S2) rolls without sliding, nor spinning.

= 0

0

= 0

(S2) spins without sliding, nor rolling.

= 0

= 0

0

(S2) slides without rolling, nor spinning.

= 0

0

0

(S2) spins and slides without rolling.

0

= 0

0

(S2) rolls and slides without spinning.

10.1.4 Solids in Contact in Several Points

The preceding considerations can be applied in every point of contact, in the

case where the two solids are in contact in several points. In particular:

— If two solids (S1) and (S2) are in contact in two points and if the sliding

velocity vector is null in these two points, the rotation vector ( )12ω

is direction

vector of the line passing through these two points.

— If two solids (S1) and (S2) are in contact in more than two points and if the

sliding is null in all these points, they are necessarily aligned.

10.2 TRANSMISSION OF A MOTION

OF ROTATION

10.2.1 General Elements

Two solids (S1) and (S2) have rotation motions of respective axes ( )1 1,O k

and

( )2 2,O k

attached to a reference system (T). The problem of the transmission of

the rotation motions consists in finding mechanisms which allow to transform a

rotation motion of (S1) with respect to the support (T) into a rotation motion of

(S2) with respect to the support and such as if the motion of (S1) is uniform, the

10.2 Transmission of a Motion of Rotation 141

motion of (S2) is uniform also.

The motion of the solid (S1) with respect to the support (T) is characterized by

the kinematic torsor ( ) 1T of resultant:

( )1 1 1T kω ω=

. (10.8)

The motion of the solid (S2) with respect to the support (T) is characterized by

the kinematic torsor ( ) 2T of resultant:

( )2 2 2T kω ω=

. (10.9)

The motion of (S2) with respect to (S1) is thus characterized by the kinematic

torsor ( ) 1

2 such as:

( ) ( ) ( ) 1 2 2 1

T T= − . (10.10)

The instantaneous rotation vector of the motion of (S2) with respect to (S1) is:

( )

12 2 2 1 1k kω ω ω= −

. (10.11)

The axes ( )1 1,O k

and ( )2 2,O k

being the axes of rotation attached to the support

(T), the vector( )12ω

is a vector independent of time relatively to a basis of the refe-

rence (T).

10.2.2 Transmission by Friction

In the case of the transmission of rotation motions by friction, the transmission

is obtained through a direct contact between the two solids (S1) and (S2), without

sliding at the points of contact. It results from Section 10.1.4 that the solids must

be in contact through a straight line, which is the instantaneous rotation axis of the

motion of (S2) with respect to (S1) and hence ( )12ω

is direction vector. It is shown

that this axis () is fixed in the reference (T). The surfaces coming in contact are

the surfaces generated by the rotation of () about the axis ( )1 1,O k

for the solid

(S1) and by the rotation of () about the axis ( )2 2,O k

for the solid (S2). These

surfaces of contact which are the axoidal surfaces are thus revolution surfaces

They are right circular cylinders if the axes ( )1 1,O k

and ( )2 2,O k

are parallel,

right circular cones if the axes intersect, revolution hyperboloid if they are not in

the same plane.

At a point M of contact, Expression (10.10) leads to:

( ) ( ) ( ) ( ) 1 12 2 2 1( , ) T T

M M Mg M t = = −

. (10.12)

The condition of non sliding at the point M is thus written as:

( ) ( ) 2 1 0T TM M− =

. (10.13)


10.2.2.1 Cylindrical Wheels

In the case of cylindrical wheels (Figure 10.3), the axes are parallel and the

surfaces of contact are right circular cylinders. The axes being parallel, we have:

1 2k k k= =

. (10.14)

And the condition of non sliding (10.12) is written:

2 2 1 1 0k O M k O Mω ω× − × =

, (10.15)

or

( )2 2 1 1 0k O M O Mω ω× − =

. (10.16)

Whence the relation between the angular velocities:

2 2 1 1O M O Mω ω=

. (10.17)

This relation leads to:

2 1

1 2

R

R

ωω

= ± , (10.18)

with the sign – if the contact operates on the outside of cylinders, and the sign + if

the contact operates inside one of cylinders.

10.2.2.2 Conical Wheels

In the case of conical wheels (Figure 10.4), the axes intersect at the point O and

the surfaces of contact are revolution cones. As origins of the axes we may choose

the point O: the points O1 and O2 coincide with O.

If M is a point of contact, the condition (10.13) of non sliding is written:

1 1 2 2 0k OM k OMω ω× − × =

. (10.19)

FIGURE 10.3. Transmission through cylindrical wheels.

M

O1

O2

R1

R2

(S1)

(S2)

1ω

2ω


FIGURE 10.4. Transmission through conical wheels.

Hence:

( )1 1 2 2 0k k OMω ω− × =

. (10.20)

This expression shows that 1 1 2 2k kω ω−

is a vector collinear to OM

, thus:

1 1 2 2k k OMω ω λ− =

, (10.21)

or by scalar multiplication with the vector u

orthogonal to the vector OM

:

1 1 2 2 0k u k uω ω− =⋅ ⋅

. (10.22)

By introducing the angles 1α and 2α (Figure 10.4), we obtain finally:

1 1 2 2sin sin 0ω α ω α+ = . (10.23)

In the frequent case where the axes are orthogonal (Figure 10.5), we have

1 2 90α α+ = ° and the preceding relation is written as:

21

1

tanω

αω

= − . (10.24)

If R1 and R2 (Figure 10.5) are the middle radii, we obtain:

2 1

1 2

R

R

ωω

= − . (10.25)

10.2.2.3 Variable Speed Transmission

The diagram of a variable speed transmission is given in Figure 10.6. The

driving shaft (∆1) is connected to a plate (S1), in contact with a roller (S2) which

O

1ω

2ω

(S1)

(S2)

1k 2k

12


FIGURE 10.5. Transmission through conical wheels with orthogonal axes.

can move along a radius of the plate. The angular velocity of the output is that of

the roller axis.

If x is the distance from the point of contact to the centre of the plate and if r is

the radius of the roller, we obtain without difficulty the relation:

2 1x

rω ω= − . (10.26)

The translation of the roller thus makes it possible to vary 2ω for a given 1ωangular velocity.

FIGURE 10.6. Variable speed transmission.

O

1ω

2ω

12

2R2

2R1

1ω

2ω

(S2)

(S1)

r

translation

x

(1)

M


10.2.3 Gear Transmission

10.2.3.1 Notion of Gears

The entrainment by friction is used in the case where the mechanical power to

transmit is small. This type of entrainment of easy and thus inexpensive reali-

zation, of noiseless running, however is limited when the power to transmit is

high. It presents then the following disadvantages: running with sliding (in parti-

cular at the starting), attrition of the wheel linings, bending of the shafts resulting

from the need to exert a contact pressure to decrease the tendency to slippage. To

palliate these disadvantages, cogs are carved in the surfaces of contact, that makes

it possible an entrainment by obstacle; gears are thus realized. Instead of being

smooth the surface of each gear wheel includes hollows and teeth which remain

constantly in contact with the teeth of the other gear wheel (Figure 10.7). The

entrainment is thus induced by pushing from a tooth of the driving wheel to the

tooth of the receiving wheel. The transmission of the motion is then possible in a

continuous way if the teeth have profiles said conjugate. The axoidal surfaces

correspond to the surfaces of contact for the entrainment by friction. For the gears,

the axoidal surfaces are called the primitive surfaces.

If (R1) is the wheel which transmits power and (R2) the receiving wheel, it is

said: (R1) engages into (R2), (R1) is the driving wheel, (R2) is the driven wheel.

If n1 and n2 are the numbers of the teeth of the wheels (R1) and (R2), Expres-

sions (10.18) and (10.25) are written:

— for cylindrical gears:

2 1

1 2

n

n

ωω

= ± , (10.27)

with a sign + or – according to whether the contact is internal or external,

FIGURE 10.7. Gear transmission.


— for conical gears with orthogonal axes:

2 1

1 2

n

n

ωω

= − . (10.28)

10.2.3.2 Gear Train

1. Gear train ratio A gear train is symbolized in Figure 10.8. The axes (∆1), (∆2), ... (∆p) of the

different wheels are fixed relatively to a support (T). Each intermediate axis (∆2),

(∆3), ... (∆p – 1) comprises a driving wheel (Ri) and a driven wheel ( iR′ ). The wheel

(R1) is a driving wheel and the wheel ( pR′ ) is a driven wheel. The gears are either

cylindrical, or conical.

It is found without difficulty that:

1 2 1

1 2 3

. . .

. . .

p p

p

n n n

n n n

ω

ω−

= ±′ ′ ′

, (10.29)

with the sign + or – according to the number of external contacts and according to

the number of conical gears. The quantity 1

pr

ω

ω= is called the ratio of the shear

train.

2. Epicyclic Trains The epicyclic trains are gear trains such as those described previously, but the

axes of which are connected to a frame (B) which is itself animated of a rotation

motion of angular velocity Ω about an axis (∆) attached to the reference (T). The

epicyclic train is schematized in Figure 10.9.

If r is the ratio of the gear train, we have:

1

pr

ω

ω= , (10.30)

where pω and 1ω are the angular velocities of the last driven wheel and the first

driving wheel relatively to the frame (B). Let Ωp and Ω1 be the angular velocities

FIGURE 10.8. Gear train.

1( )R

2( )R

3( )R

2( )R′

3( )R′ 4( )R′

1( )∆

2( )∆

3( )∆

4( )∆


FIGURE 10.9. Epyciclic train.

relatively to the reference (T). The law (10.11) of the composition of the rotation

vectors leads, in the case where the axes (∆1), (∆p) and (∆) are parallel, to the

relations:

1 1, .p pω Ω Ω ω Ω Ω= − = −

Whence the expression of the ratio of the gear train:

1

pr

Ω Ω

Ω Ω

−=

−. (10.31)

This relation is called Willis formula.

3. Application to the case of car differential

Along a straight line the two wheels of a car turn at the same angular velocity

of rotation. When cornering, the distance covered by the external wheel is larger

than that covered by the internal wheel. It results from it that the external wheel

must turn more quickly than the internal wheel. This process is obtained using a

differential, of which the general diagram is given in Figure 10.10. A pinion asso-

ciated to the output shaft of the gear box makes turn a toothed wheel connected to

a cage. On the cage a planet (S) is assembled which gears with two driving wheels

(R1) and (R2). The ratio of the gear train (from the wheel (R1) to the wheel (R2)) is

1r = − , and the Willis formula is written:

22 1

1

1 or 2Ω Ω

Ω Ω ΩΩ Ω

−= − + =

−, (10.32)

where Ω is the angular velocity of the rotation of the cage.

When cornering, the internal wheel (the wheel (R1) for example) can turn with

a velocity lower than that of the external wheel. Along a straight line, the velo-

cities are the same: 2 1 2Ω Ω Ω= = .

1( )∆

2( )∆ 3( )∆

( )∆

(B)


FIGURE 10.10. Car differential.

10.2.3 Belt Transmission

The belt transmission is used to link mechanically two or more rotating shafts

when they are distant. Belts are looped over pulleys connected to the rotating

shafts.

10.2.3.1 Right Belt

In the case of a right belt (Figure 10.11), the axes ( )1 1, O k

and ( )2 2, O k

are

parallel: 1 2k k k= =

. The condition of non sliding of the belt on the pulleys is

written:

2 1

1 2

R

R

ωω

= . (10.33)

In fact there exists always a slippage, due in particular to the deformation of the

belt, that leads to a variation of the preceding ratio of about 1 to 3%.

FIGURE 10.11 Belt transmission.

(R2) (R1)

(S)

cage

output shaft

of the gear box

O1 O2

R1

R2

(S1) (S2)

1ω

2ω


FIGURE 10.12. Transmission with crossed belt.

10.2.3.2 Crossed Belt

In the case of a crossed belt (Figure 10.12) the angular velocities of rotation are

opposite:

2 1

1 2

R

R

ωω

= − . (10.34)

10.2.3.3 Set of Belts and Pulleys

In the case of a set of belts and pulleys (Figure 10.13), each intermediate axis

comprises a driving pulley (Si) of radius Ri and a driven pulley ( iS′ ) of radius iR′ . If the system comprises p axes, it is established without difficulty that:

1 2 1

1 2 3

. . .

. . .

p p

p

R R R

R R R

ω

ω−

= ±′ ′ ′

, (10.35)

with the sign + if the number of crossed belts is even, and the sign – if this

number is odd.

Note. To eliminate the slippage of the belts on the pulleys, the pulleys are repla-

ced by toothed wheels and the belts by chains. For example, Relation (10.33) is

then replaced by:

2 1

1 2

n

n

ωω

= , (10.36)

where n1 and n2 are respectively the number of the teeth of the wheels (1) and (2).

FIGURE 10.13. Set of belts and pulleys.

O1 O2

R1R2

(S1) (S2)

1ω

2ω

(S1)

1ω2ω

3ω

(S2)

2( )S ′

3( )S ′


EXERCISES

10.1 A wheel (S) moves along a straight line (D) while remaining in the same

plane and in contact with the line (D) at the point I (Figure 10.14).

10.1.1. Implement the kinematic analysis in the general case where the wheel (S)

rolls and slides on the line (D).

10.1.2. Express the sliding velocity vector at the point I. Deduce from that the

condition of non sliding.

10.2 A cylinder (S) moves inside a cylinder (T) in such a way that the two cylin-

ders remain in contact along a common generator (Figure 10.15).

10.2.1. Implement the kinematic study in the general case where the cylinder (S)

rolls and slides on the cylinder (T).

10.2.2. Express the sliding velocity vector at the point I. Deduce from that the

condition of non sliding.

10.3 Study the conditions of non sliding and non spinning in the case of the

motion of a cylinder on a plane, studied in Exercise 9.2.

FIGURE 10.14. Motion of a wheel on a straight line.

FIGURE 10.15. Motion of a cylinder inside a cylinder.

I

a

(S)

(D)

a

(S)

I

b

(T)

Comments 151

COMMENTS

The kinematics of rigid bodies in contact is first implemented in the

usual way by using the concepts of the kinematics of solids considered in

the preceding chapter. Having determined the kinematic torsor relatively to

the motion of the two bodies in contact, its moment at a point of contact

allows us to derive the sliding velocity vector at this point of contact and

thus to analyze the conditions of non sliding. The instantaneous rotation

vector (resultant of the kinematic torsor) allows us then to analyze the

conditions of spinning and rolling.

Part III

The Mechanical Actions

The equilibrium or the motion of a solid of a set of solids results from

the mechanical actions which are exerted. These actions will be first

considered without being concerned with the physical phenomena by

which they are induced, and they will then be classified as force,

couple and arbitrary action. The actions of gravitation and the actions

of gravity will be studied in a particular chapter. Next, a chapter will

be devoted to the actions induced by connections, of which the

concept is at the basis of the technological design of the mechanical

systems. Lastly, the study of some problems of static equilibrium of

rigid bodies will allow us to understand the structuring of the

mechanical actions exerted on a solid or a set of solids.

CHAPTER 11

General Elements on the Mechanical Actions

11.1 CONCEPTS RELATIVE TO

THE MECHANICAL ACTIONS

11.1.1 Notion of Mechanical Action

The mechanical processes result from mechanical actions of which we have a

usual notion:

— an object, left to itself, falls: the Earth attracts the object;

— the same object, placed on a table, does not fall any more: the table exerts

on the object an action which prevents from falling;

— a child who kneads modelling clay exerts an action which deforms the

paste;

— the cyclist exerts on the pedals an action which produces the displacement

of the bicycle;

— the brake exerts an action which opposes to this displacement;

— etc.

Thus, in a general way, a mechanical action is a process which maintains an

equilibrium, causes a deformation, produces a motion or opposes to a motion.

11.1.2 Representation of a Mechanical Action

If the notion of mechanical action is usual, it is not in fact directly accessible

by measurement. We have knowledge of it only by its consequences: presence or

absence of equilibrium, measure of displacements, measure of deformations,

derivation of laws of motions, etc. Thus, to translate by equations the various


mechanical phenomena, we are brought to state the following axiom:

Any mechanical action which is exerted on a material set may be represented

by a torsor associated to this set.

We understand by material set either a rigid body, or a set of rigid bodies.

11.1.3 Classification of the Mechanical Actions

To each type of torsor corresponds a type of mechanical action whose the

properties are immediate consequences of the results established in Chapter 5.

11.1.3.1 Force

We say that an exerted mechanical action is a force, if and only if the torsor

representing the mechanical action is a slider.

It results from the properties derived in Section 5.2.1, that a force is character-

rized by:

— the resultant of the slider associated to the force, generally called by lan-

guage contraction: the resultant of the force (the norm of the resultant of the force

called intensity or magnitude of the force is expressed in newtons: N);

— the axis of null moments (determined by only one point when the resultant

is known), called support of the force or line of action.

If the mechanical action exerted on the set (D) is a force, it will be possible

symbolically to represent it by making appear the support (∆) of the force and a

bipoint (A, B) of which the image in 3 is the resultant of the force: R

(Figure

11.1a). If we are in the case studied in 5.3.3 of Chapter 5, the force has a measure

centre H defined by (5.69) or (5.72); and we shall take for the point A the measure

centre (Figure 11.1b).

Lastly, let us note that a force tends to move the set, on which it is exerted,

along the direction defined by the resultant, therefore parallel to the support of the

force.

11.1.3.2 Couple

We say that a mechanical action is a couple (couple-action), if and only if the

torsor which represents this action is a couple-torsor.

It results from the properties established in Section 5.2.2 that a couple is cha-

racterized by its moment-vector, independent of the point considered, and of

which the magnitude is expressed in N m. This moment-vector is sometimes

called couple. Let us note however it is necessary to distinguish the couple-

action, from the couple-torsor and from its moment-vector.

Furthermore, it results from Section 5.2.2 that a couple is equivalent to a

couple of two forces of opposite resultants, hence of parallel supports. There

11.1 Concepts Relating to the Mechanical actions 157

FIGURE 11.1. Symbolic representation of a force.

exists an infinity of couples of forces equivalent to a couple; these couples are

obtained in accordance to the results of Subsection. A couple tends to impart a

rotation to the set on which the couple acts, in the direct way about the direction

defined by the moment-vector of the couple (Figure 11.2).

11.1.3.3 Arbitrary Mechanical Action

We say that a mechanical action is arbitrary, if and only if the torsor which

represents this action is an arbitrary torsor.

According to the results established in Section 5.2.3, an arbitrary mechanical

action may be described as being the superposition of a force and a couple. The

mechanical action is then reduced to a force and a couple. There exists an infinity

of force-couple sets equivalent to a given arbitrary mechanical action (Subsection

5.2.3.2).

FIGURE 11.2. The couple-action tends to impart a rotation to the set (D).

A

()

B

(D)

(): support of the force

AB R=

: resultant of the force

(): support of the force

HB R=

: resultant of the force

H: measure centre

(a)

H

()

B

(D) (b)

A

B

(D)

AB =

: moment of the couple


11.1.4 Mechanical Actions Exerting between Material Sets

Let (D1) and (D2) be two material sets. The mechanical actions exerted by (D1)

on (D2) are represented by the torsor which we shall denote by:

1 2D D→ . (11.1)

In the same way, the mechanical actions exerted by (D2) on (D1) are represented

by the torsor denoted by:

2 1D D→ . (11.2)

1. Superposition of the generators of mechanical actions

If the set (D1) is constituted of the union of two disjoint sets ( )1D′ and ( )1D′′ , we

shall write:

1 2 1 1 2 1 2 1 2( )D D D D D D D D D′ ′′ ′ ′′→ = ∪ → = → + → . (11.3)

2. Superposition of the receivers of mechanical actions

If the set (D2) is constituted of the union of two disjoint sets ( )2D′ and ( )2D′′ ,

we shall write in the same way:

1 2 1 2 2 1 2 1 2( )D D D D D D D D D′ ′′ ′ ′′→ = → ∪ = → + → . (11.4)

Relations (11.3) and (11.4) combine and extend to the cases where the sets

considered are the unions of an arbitrary finite number of disjoint sets.

11.1.5 External Mechanical Actions Exerting on a Material Set

The Universe which shall be denoted by (U) is the material set of all the phy-

sical systems which are more or less distant: chair, table, house, city, country,

Earth, planets, Sun, stars, etc. Being given a set (D), we call exterior of the set

(D), which we shall denote by ( )D , the complement of (D) in the Universe; that is

saying all that in the Universe is not (D):

( ) ( ) ( ), D D U D D∪ = ∩ = ∅ . (11.5)

We call mechanical actions exerting on the set (D), or actions external to (D),

the set of mechanical actions exerted on (D) by the exterior of (D). These actions

are represented by the torsor:

D D→ . (11.6)

The exterior of (D) is constituted of disjoint subsets: ( ) ( ) ( ) 1 2, , . . . , nD D D .

11.2 Different Types of Mechanical Actions 159

Hence:

( ) ( )1 2 . . . nD D D D= ∪ ∪ , (11.7)

and the torsor of the actions external to (D) is written:

1

n

i

i

D D D D

=

→ = → . (11.8)

11.2 DIFFERENT TYPES OF MECHANICAL ACTIONS

11.2.1 Physical Natures of the Mechanical Actions

The mechanical actions exerting on the set (D) can be of different natures and

classified according to:

— Either actions of contact, denoted by , which are exerted on the set (D) at

the level of the boundary between (D) and ( )D : action of the wind on a sail,

action of the water on a boat, action of the air on a plane, action of the ground on

a wheel, etc. The actions of contact occur in the contacts between solids, between

solids and fluids, etc. The actions of contact between solids will be studied in

Chapter 13.

— Or actions at distance due to physical phenomena:

• the gravitation which will be studied in Chapter 12;

• the electromagnetism , in which we classify all the electric, magnetic and

electromagnetic processes. For example some materials rubbed with a piece of

wool exert on pieces of paper an electrostatic action; the Earth exerts on a ma-

gnetic needle a mechanical action; a conductor thread traversed by an electric

current exerts on this same needle an electromagnetic action, etc.

The three types of actions are currently the only known ones. Let us note

already that, in addition to their physical natures, the actions of contact and the

actions at distance differ basically by the fact that the actions at distance are

determined by the physical phenomena induced, and the actions are thus calcu-

lable a priori. In contrast, the actions of contact depend, and that contrary to the

preceding ones, upon the other mechanical actions exerted on the material system

under consideration.

11.2.2 Environment and Effective Actions

Each subset of ( D ) (Section 11.1.5) can exert on the set (D) mechanical actions

of each type, represented by the torsors:

, , .i i iD D D D D D→ → →

(11.9)


If the subset ( )iD exerts simultaneously the three types of actions, the actions

exerted by ( )iD on the set (D) are represented by the torsor:

,i ii iD D D DD D D D= + +→ →→ →

(11.10)

and the torsor (11.8) of the mechanical actions exerted on (D) is written:

1

n

ii i

i

D DD D D D D D

=

= + +→→ → → . (11.11)

In practice, by taking account of the expressions of the physical laws of the

actions at distance, it will be possible to neglect such or such action of such or

such set ( )iD of the exterior of the set (D) considered. For example, for a material

system in the vicinity of the Earth, the actions of gravitation exerted by the Earth

are preponderant compared to the gravitation exerted by the Moon, the planets,

the Sun. In the same way, the action of gravitation exerted on a magnetic needle

by a conductor thread crossed by an electric current can be neglected in front of

its electromagnetic action. Thus, Expression (11.11) of the torsor of the mecha-

nical actions exerted on the set (D) can be simplified, in accordance with the

problems considered, by taking into account only the preponderant mechanical

actions. We shall denote generally by ( ) D the torsor (11.11) simplified of the

preponderant actions exerted on the set (D). We shall write:

( ) ( ) j

j

D D= , (11.12)

where ( ) j D is the torsor which represents the preponderant action j.

11.3 POWER AND WORK

11.3.1 Definition of the Power

We call power developed, at time t, relatively to the reference (T), by the

action j acting on solid (S) and represented by the torsor ( ) j S , the scalar:

( ) ( ) ( ) ( ) T Tj j SP S S= ⋅ . (11.13)

If M is a point of the solid (S), the power is written, from (5.15):

( ) ( ) ( ) ( ) ( ) ( ) T T TM Mj j jS SP S R S S R= +⋅ ⋅

, (11.14)

or, by introducing the elements of reduction (9.18) and (9.19) of the kinematic

torsor at the point M:

( ) ( ) ( ) ( )( ) ( ) ( ),T T T

Mj j j SP S R S M t S ω= +⋅ ⋅

. (11.15)

11.3 Power and Work 161

11.3.2 Change of Reference System

Let (1) and (2) be two reference systems. The power developed relatively to

the reference (1) is: ( ) ( ) ( ) ( ) 1 1

j j SP S S= ⋅ , (11.16)

the power developed relatively to the reference (2) is:

( ) ( ) ( ) ( ) 2 2j j SP S S= ⋅ . (11.17)

From these two relations and from the law (9.39) of compositions of motions, we

derive the relation: ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1

2j j jP S P S S= + ⋅ . (11.18)

11.3.3 Potential Energy

It happens that, having derived the developed power, we observe that the

power can be put in the form of a derivative with respect to time of a function

which depends only on the parameters of situation 1 2( , , . . . , )kq q q and eventually

on time. Hence:

( ) ( ) ( ) p 1 2

d ( , , . . . , , )d

T Tj kP S E q q q t

t= − . (11.19)

The function ( )pTE , thus defined except for an additive constant (independent of

the parameters qi and time t), is called potential energy of the solid (S) relatively

to the reference (T). It is thus said that the mechanical action exerted on (S) and

represented by the torsor ( ) j S admits a potential energy in the reference (T).

It should however be noted that this case is not general.

Example. Action exerted by a spring.

Let us consider a spring (R) (Figure 11.3) of negligible mass and of axis AA1

fixed in the reference (T). The experimental study shows that the mechanical

action exerted by the spring (R) on the solid (S) is a force of support AA1, repre-

sented by a slider R S→ of which the resultant is proportional to the

elongation (positive or negative) of A1A from the initial length l0 of the spring

when no action is exerted on the spring: the point A1 is then at A0. For a position

of the point A1, defined by 1AA l= , it is observed experimentally that:

— if 0l l> , the spring is stretched and it tends to draw back the point A1 at A0

by attracting A1 towards A0;

— if 0l l< , the spring is compressed and it tends to push back A1 towards A0.

By taking as axis Ax

the axis of the spring, the experimental results observed

are described while writing:


FIGURE 11.3. Spring of traction-compression.

( )

0 ,

0, point of the axis ,P

R R S k x l i

R S P Ax

→ = − −

→ = ∀

(11.20)

where k is a constant characteristic of the spring, called coefficient of rigidity or

stiffness of the spring, and x is the abscissa of the point A1 on the axis Ax

.

In the case of rectilinear translation motion of axis Ax

, the kinematic torsor

relative to the motion of the solid (S) with respect to reference (T) is written at the

point A1:

( ) ( )

( ) ( )1 1

0,

( , ) .

T TS S

T TA S

R

A t x i

ω = =

= =

(11.21)

By expressing in the present case Relation (11.13), we obtain:

( ) ( ) ( )

20 0

d dd d 2

T x kP R S k x l x lt t

→ = − − = − − . (11.22)

From this we deduce that the action exerted by the spring admits a potential ener-

gy in the reference (T) of the form:

( ) ( )2p 0 cte

2T kE x l= − + . (11.23)

11.3.4 Work

We call work in the reference (T), between the instants t1 and t2, of a mecha-

nical action represented by the torsor ( ) j S the integral:

(T)

A0

(S)

A1

A

11.3 Power and Work 163

( ) ( ) ( ) 2

1

1 2( , ) dt

T Tj

t

W t t P S t= . (11.24)

If the mechanical action admits a potential energy in the reference (T), the

work is written from (11.19):

( ) ( ) ( )1 2 p 1 p 2( , ) ( ) ( )T T TW t t E t E t= − . (11.25)

Work thus depends only upon the initial state and the final state of the solid.

11.3.5 Power and Work of a Force

11.3.5.1 Power

If the action exerting on the solid (S) is a force, the torsor ( ) j S which

represents this action is a slider, and Expression (11.15) is written at a point P of

the support of the force:

( ) ( ) ( ) ( ) ( ),T Tj jP S R S P t= ⋅

. (11.26)

If the position vector of the point P in the reference (T) does not depend expli-

citly on time, we have, by introducing the parameters qi of situation:

( )( )( )

1

d, d

kTT

ii

i

OPM t OP qt q

=

∂= =∂

, (11.27)

where O is a point fixed in reference (T). Expression (11.26) of the power is then

written:

( ) ( ) ( ) 1

kT

j j ii

i

OPP S R S qq

=

∂=∂

⋅

. (11.28)

Case where the force admits a potential energy

In the case where the force admits in the reference (T) a potential energy which

does not depend explicitly on time, Relation (11.19) is written:

( ) ( ) ( ) p 1 2

1

( , , . . . , )

kT T

j k ii

i

P S E q q q qq

=

∂= −∂ . (11.29)

By comparing Expressions (11.28) and (11.29), we obtain:

( ) ( ) ( )pT T

ji i i

OPR S E Uq q q

∂ ∂ ∂= − =∂ ∂ ∂

⋅

, (11.30)

setting: ( ) ( )

pT TU E C= − + ,


where C is a constant independent of the parameters qi and of time. The function

U(T) of the variables 1 2, , . . . , kq q q is called the force function. We say then that

the force derives from a force function in the reference (T). Expression (11.19) of

the power is then written:

( ) ( ) ( )

( ) 1 2

d ( , , . . . , )d

TT T

j kP S U q q qt

= . (11.31)

11.3.5.2 Work

The works between the instants t1 and t2 is, from (11.24) and (11.26):

( ) ( ) ( ) ( )

2

1

1 2( , ) , dt

T Tj

t

W t t R S P t t= ⋅

, (11.32)

or considering (11.28):

( ) ( ) 2

1

1 2

1

( , ) d

kPT

j iiP i

OPW t t R S qq

=

∂=∂

⋅

, (11.33)

where the curvilinear integration is applied to the trajectory 21P P described

between the times t1 and t2.

In the case where the force derives from a force function, the work is:

( ) ( ) ( )1 2 2 1( , ) ( ) ( )T T TW t t U M U M= − . (11.34)

11.3.6 Set of Rigid Bodies

Let (D) be a set constituted (Figure 11.4) of solids ( ) ( ) ( ) 1 2, , . . . , nS S S disjoint:

( ) ( )1 2 . . . nD S S S= ∪ ∪ . (11.35)

The actions exerted on a solid (Si) of the set (D) are classified according to:

— internal actions exerted on (Si) by the other solids of the set (D):

, with , 1, 2, . . . , , and j iS S i j n j i→ = ≠, (11.36)

FIGURE 11.4. Set of solids.

(S1)

(Sj)

(Sj)

(S2)

(Sn)

(D)

Exercises 165

— external actions exerted on (Si) by the exterior of (D):

iD S→ . (11.37)

The resultant mechanical action exerted on the solid (Si) is represented by the

torsor:

1

n

i i i j i

ji

S S D S S S

=≠

→ = → + → , (11.38)

external actions internal actions

and the power developed by the whole of the actions exerted on the solid (Si) is

written:

( ) ( ) ( ) 1

nT T T

i i i j i

ji

P S S P D S P S S

=≠

→ = → + → . (11.39)

We call power developed in the reference (T) on the set (D), the sum of the

powers developed on all the solids. Thus:

( ) ( ) ( ) ( )

1 1 1

n n nT T T

i j i

i i ji

P D P D S P S S

= = =≠

= → + → . (11.40)

The first term can be put in the form:

( ) ( ) ( )

11 1

n n nT T T

i i ii

i i

P D S P D S P D S=

= =

→ = → + →

∪ . (11.41)

Thus finally:

( ) ( ) ( ) ( ) 1 1

n nT T T

i j i

i ji

P D P D S P S S

= =≠

= → + → . (11.42)

set of the set of the


EXERCISES

11.1 In a preliminary design of the calculation of a frame, we have to consider

(Figure 11.5) the field of forces defined on the set of the points Mi reported in the

figure. The supports of the forces i corresponding to each point Mi are contained

in the plane Oxy. The figure shows the direction of the support of each force,


FIGURE 11.5. Frame.

by representing a bipoint of which the image is the resultant iR

of the force i :

— 1 2 3, , R R R

are collinear to the vector j

;

— 4 5 6 7, , , R R R R

are respectively orthogonal to the vectors OA

, AB

,

CD

, DE

;

— the points M4, M5, M6, M7 are the respective middles of the segments OA,

AB, CD, DE;

— the intensities expressed in N of the resultants are:

R1 R2 R3 R4 R5 R6 R7

2000 1000 1500 1500 1000 1500 2000

As well as possible characterize the mechanical exerted on the frame.

11.2 We consider (Figure 11.6) a vertical rectangular barrage of wide a. The

depth of the dam is h.

11.2.1. Determine, as a function of the atmospheric pressure p0, the mass per unit

volume ρ of the water, the intensity g of the field of gravity, the resultant of the

force exerted on an element of surface d S(M) surrounding a point M of the

barrage (force exerted by the water of the dam). Application: 50 m,a =

30 m.h =

11.2.2. Characterize the mechanical action exerted by the dam on the barrage.

11.2.3. Characterize the mechanical action exerted on a circular sluice of diameter

D and located at a distance d from the surface of the water. The sluice is comple-

tely immersed:2Dd ≥ .

11.3 A sphere of radius a is immersed in a liquid of mass ρ per unit volume, so

that its centre O is to a depth h higher than a.

Study the mechanical action exerted by the liquid on the sphere.

4 m 4 m 4 m 4 m 4 m

M4

M5 M6

M7

M1 M2 M3

4R

5R

6R

7R

y

O

A

B C

D

1R

2R

3R

x

4 m 3 m 3 m

Comments 167

FIGURE 11.6. Action exerted on a barrage.

COMMENTS

Any mechanical action exerted on a solid may be represented by a torsor.

The concepts introduced for the torsors (Chapter 5) can thus be transposed

to the mechanical actions. In the case where the mechanical is represented

by a slider, the mechanical action is a force, characterized by its resultant

which is that of the slider and by its support or its line of action which is

the axis of the null moments of the slider. In the case where the mechanical

is represented by a couple-torsor, the mechanical action is a couple-action,

usually called couple. This action is characterized by a null resultant and a

moment independent of the point considered. The action may be resolved

into a couple of two forces of opposite resultant and of parallel supports.

Lastly, when the mechanical action is represented by an arbitrary torsor, the

mechanical action is arbitrary. The action can be resolved into the sum of a

force and a couple.

The mechanical actions exerted on a material set are of different physical

natures and can be classified as actions at distance and actions of contact.

The actions at distances result from physical phenomena: gravitation (or

gravity) and electromagnetism. These actions are entirely determined by

the physical phenomena induced. The actions of gravitation and gravity

will be studied in Chapter 12. The electromagnetic actions are induced by

electrostatic, electric and magnetic phenomena as well as by their

M

dS(M)

y

x

z

h

a

barrage water


couplings. The actions of contact are exerted on a system at the level of its

boundary. In contrast to the actions at distance, the actions of contact

depend on the other actions exerted on the material system. Their complete

characterization can then be obtained only as part of the resolution of the

problem of mechanics.

An important notion introduced is that of the power developed by a

mechanical action exerted on a moving solid. This power is expressed

simply as the product of the torsor which represents the mechanical by the

kinematic torsor relative to the motion of the solid. The power is an instan-

taneous quantity which expresses a state at every moment. The concept of

work which is deduced from it, by integration of the power with respect to

time, is less interesting.

CHAPTER 12

Gravitation. Gravity Mass Centre

12.1 PHENOMENON OF GRAVITATION

12.1.1 Law of Gravitation

The law of gravitation (or Newton’s law) introduces the concept of mass of a

material system, physical quantity which allows us to measure the quantity of

matter of the material system. The law of gravitation may be formulated in the

following way.

Let us consider two material elements of centres M and M', of respective

masses m and m'. Because of its presence, the material element of centre M' exerts

on the set of centre M a mechanical action (the torsor representing this mecha-

nical action will be denoted by M M′ → ). This action is a force of support

MM' and of resultant:

( )3

MMR KmmM M

MM

′′′ =→

′

, (12.1)

where K is the universal constant of gravitation:

K = 6,67 × 10–11 m3 kg–1 s–2.

The law of gravitation is all the more verified as the material elements have

dimensions (Figure 12.1) which are low compared to the distance from M to .M ′All occurs then as if the masses of the material sets were respectively concen-

trated at the points M and .M ′ The points M and M ′ associated to their respective

masses m and m' are called material points.

It results from the statement of the law of gravitation that:

1. The torsor M M′ → which represents the action of gravitation exerted by

M ′ on M is a slider of elements of reduction:

170 Chapter 12 Gravitation. Gravity. Mass Centre

FIGURE 12.1. Material elements.

( )

3,

0, point of the line passing through and .P

MMR KmmM M

MM

P M MM M

′′′ =→

′

′′ = ∀→

(12.2)

2. The action of gravitation exerted by M on M' is a force opposite to the prece-

ding one, that is to say of the same support as the action exerted by M' on M, but

with an opposite resultant.

12.1.2 Gravitational Field

We call vector of the gravitational field at a point of the space, the resultant of

the action of gravitation exerted at this point on the unit mass.

The gravitation field induced by the material point (M', m') is then defined at

the point M by the vector:

( )3( )M

MMG M Km

MM′

′′=

′

, (12.3)

and the resultant of the action of gravitation is written:

( )MR m G MM M ′′ =→

. (12.4)

The magnitude of the gravitation field is expressed in N kg–1.

12.1.3 Action of Gravitation Induced by a Solid Sphere

Let us consider (Figure 12.2) a material point (M, m) located at a distance r of

the centre O of a solid sphere (S) of mass mS. An element of mass d ( )m M ′ sur-

rounding a point M' of the sphere exerts on M a force of resultant:

( )3

d d ( )MM

R Km m MM MMM

′′′ =→

′

, (12.5)

or

( )

3d ( )d ( )

MMR Km M V MM M

MMρ

′′ ′′ =→

′

, (12.6)

M(m)

M'(m')

12.1 Phenomenon of Gravitation 171

FIGURE 12.2. Action of gravitation exerted by a sphere.

where ( )Mρ ′ is the mass per unit volume of the sphere at the point M ′ and

d ( )V M ′ is the element of volume surrounding the point M'. The action of gravi-

tation exerted by the sphere on the point (M, m) is then represented by the torsor

associated to the field of sliders defined by (12.5) or (12.6). Thus, from (5.54) and

(5.55):

( ) ( )

( )

3( )d ( ),

d 0.

S

MS

MMR Km M V MS MMM

MM R M MS M

ρ′ ′ ′=→′

′= ∧ ′ =→→

(12.7)

The action of gravitation is thus a force of support passing through the point M.

In the case where the sphere is homogeneous by concentric layers, that is saying

for a sphere of which the mass per unit volume depends only on the distance R

from the centre of the sphere:

( ) ( )M Rρ ρ′ = , (12.8)

it is shown (Exercise 12.7) that the resultant of the action is expressed as follows:

3S

MOR KmmS M

OM=→

. (12.9)

The action of gravitation exerted by the sphere on the material point is thus a

force of support OM, identical to the action of gravitation which would be exerted

by a material point of mass equal to that of the sphere and placed at the centre of

the sphere.

The gravitational field induced at the point M by the sphere is then:

3( )S S

MOG M Km

OM=

. (12.10)

The expression of the field can be rewritten by introducing the unit direction

M

dm(M')

(S)

O

r

M ′


vector ( )n M

of the line MO orientated from M towards O:

2( ) ( )S

SKm

G M n Mr

=

, (12.11)

where r is the distance from the point M to the centre of the sphere (Figure 12.2).

12.1.4 Action of Gravitation Induced by the Earth

In all the problem of engineering at the surface of the Earth or in its vicinity,

the gravitational actions of the other systems, in particular of the Moon and the

Sun, are negligible compared to the action of gravitation of the Earth. Further-

more, at first approximation, the simplifying assumption of an Earth spherical and

homogeneous by concentric layers is sufficient. It then follows from the results

derived in the preceding subsection that the action of gravitation exerted by the

Earth at the point M is a force of support passing though the point M and the

centre of the Earth, with the gravitational field:

TeTe

2( ) ( )

KmG M n M

r=

, (12.12)

where mTe is the mass of the earth, r is the distance from the point M to the centre

of the Earth and ( )n M

the unit direction vector of the line passing through the

point M and the centre of the Earth, orientated towards the centre of the Earth

(Figure 12.3).

If the point M is located on the surface of the Earth, the gravitational field is

written:

Te( ) ( )G M G n M=

, (12.13)

with

Te2

KmG

R= , (12.14)

FIGURE 12.3. Action of gravitation exerted by the Earth.

M

Earth

OTe

r

R

( )n M

12.2 Action of Gravity 173

where R is the radius of the Earth. If the point M is located at the altitude h, the

gravitational field induced by the Earth is written:

Te( ) ( ) ( )G M G h n M=

, (12.15)

with:

( )2

( )

1

GG hhR

=

+

. (12.16)

We observe that for an altitude much lower than the radius of the Earth, the

intensity G(h) of the gravitational field does not depend on h and coincides with

its value at the surface of the Earth. Moreover, at the scale of the technological

works the unit vector ( )n M

is independent of the point M:

( )n M n=

, (12.17)

where n

is the unit vector of the direction of the place of work to the centre of the

Earth. It results that the gravitational field is uniform at any point of the work:

Te point of the work ( ) .M G M G n∀ ≈

(12.18)

That is the model which we shall consider thereafter for any problem of me-

chanics relative to systems located at the vicinity of the Earth.

12.2 ACTION OF GRAVITY

12.2.1 Gravity Field Induced by the Earth

On the action of gravitation, exerted by the Earth on a material system placed

at its surface or at its vicinity, another mechanical action is superimposed due to

the rotation motion (Figure 12.4) of the Earth around its axis South-North. The

FIGURE 12.4. Action of gravity exerted by the Earth.

M

Earth

OTe

North

South


resultant action is called the action of gravity induced by the Earth. We shall see

(Chapter 19, Section 19.3.2) that the action of gravity differs very little from the

action of gravitation which is preponderant.

In particular, at the surface of the Earth or at its vicinity, the field of gravity,

which we shall denote by g

, is uniform. In practice, its magnitude g at the surface

of the Earth varies slightly owing to the flatness of the Earth at the poles:

Poles Paris Equator

g (N kg–1) 9.8322 9.8066 9.7804

The field of gravity is then written:

g g u=

, (12.19)

where u

is the unit direction vector of the downward vertical (direction given by

a plumb-line) at the place of the analysis, differing very little from the vector n

introduced in (12.17).

Note. A body left at itself at the vicinity of the surface of the Earth is submitted

to an acceleration of value g (Subsection 18.4.2.1 of Chapter 18). It results from

this that the intensity g is usually expresses in m s–2.

12.2.2 Action of Gravity Exerted on a Material System

Let (D) be a material system. A material element (Figure 12.5) surrounding the

point M is characterized by its mass:

d ( ) ( )d ( )m M M e Mρ= , (12.20)

where ( )Mρ is the specific mass (mass per unit volume, surface or line) at the

point M and d ( )e M is the element of volume, of surface or of curve surrounding

the point M, according as the set (D) is a volume, a surface or a curve. If the set

(D) is at the vicinity of the Earth, the action of gravity exerted by the Earth on the

FIGURE 12.5. Material system.

dm(M)

M

(D)

12.2 Action of Gravity 175

element d ( )e M is a force of support ( ),M u

and of resultant:

d ( ) ( )d ( )R M g M m M=

, (12.21)

where ( )g M

is the field of gravity at the point M. For a material set located at the

vicinity of the Earth, the field of gravity is uniform and given by Expression

(12.19). The resultant is thus written as:

d ( ) d ( ) ( )d ( )R M u g m M u g M e Mρ= =

. (12.22)

The action of gravity exerted on the set (D) is thus represented by the torsor

associated to the field of sliders (or the field of forces) defined on the set (D) by

Relation (12.22). We are in the case of the field of sliders having parallel axes

independent of the point M (Section 5.3.3). From the results established in this

section, we deduce the following consequences.

The torsor ( ) e D which represents the action of gravity exerted by the Earth

on the set (D) located on its surface or at its vicinity:

1. has a measure centre G, called the mass centre of the set (D), defined by

one of the equivalent relations deduced from (5.69) and (5.72):

( )

1 d ( )D

OG OM m Mm

=

, (12.23)

or

( )d ( ) 0

D

GM m M =

, (12.24)

where m is the mass of the set (D) ;

2. is a slider of axis ( ),G u

and resultant:

( ) eR D mg u= , (12.25)

where u

is the unit direction vector of the downward vertical at the place of the

study.

The action of gravity exerted by the Earth on the set (D) is thus a force of

which the line of action passes through the mass centre of the set (D) and of which

the direction is given by the downward vertical. The magnitude of the force:

P mg= (12.26)

is called the weight of the set (D).

12.2.3 Power Developed by the Action of Gravity

Let (D) be a material set of mass m and of mass centre G, located on the

surface of the Earth or at its vicinity. The power developed by the action of

gravitation exerted on the set (D) is, from (11.26):


FIGURE 12.6. Power developed by the action of gravity between two positions of the

mass centre of a material set.

( ) ( ) ( )e ( , )T TP D mg u G t= ⋅

. (12.27)

In a reference system (Oxyz) attached to the Earth and such as the axis Oz

is

upward vertical:

k u= −

, (12.28)

the position vector of the mass centre is written:

G G GOG x i y j z k= + +

, (12.29)

by introducing the Cartesian coordinates (xG, yG, zG) of the point G. The velocity

vector of the mass centre is:

( ) ( , )T

G G GG t x i y j z k= + +

, (12.30)

and the power (12.27) is written:

( ) ( ) eTGP D mgz= − . (12.31)

The power developed depends only on the z-coordinate of the mass centre. More-

over, the preceding expression shows that the action of gravitation admits a

potential energy: ( ) ( ) p e cteT

GE D mg z= + . (12.32)

The work of the action of gravity between two positions of the set (D), where the

mass centre is respectively in G1 then in G2 (Figure 12.6) is thus expressed as

follows: ( ) ( )2 11 2( , )T

G GW G G mg z z= − − . (12.33)

O

x

y

z

G1

G2

G

(D)

u

12.3 Determination of Mass Centres 177

FIGURE 12.7. Determination of the mass centre.

12.3 DETERMINATION OF MASS CENTRES

12.3.1 Mass Centre of a Material System

The determination of the position of the mass centre of a material system is

implemented by applying the general expression (12.23) to each particular case of

coordinates.

If for example we operate relatively to a Cartesian reference (Oxyz) (Figure

12.7), the Cartesian coordinates of the mass centre G are expressed according to

(12.23) by the relations:

( )

( )

( )

1 ( ) d ( ),

1 ( ) d ( ),

1 ( ) d ( ).

GD

GD

GD

x x M m Mm

y y M m Mm

z z M m Mm

=

=

=

(12.34)

where x(M), y(M), z(M) are the Cartesian coordinates of the point M, m is the

mass of the material system and the mass d ( )m M of an element is expressed by

Relation (12.20).

The preceding relations are fitted to a method of analytical determination. Ho-

wever, it is always possible to employ a numerical method by replacing in Expres-

sions (12.34) the integrals by sums. In this way the set (D) is divided into n

elements (Figure 12.7). The element i is located by the point Mi centre of the ele-

ment and of Cartesian coordinates (xi, yi, zi). We then assign to the point Mi the

mass mi of the surrounding element. Expressions (12.34) are substituted by the

relations:

dm(M)

M

(D)

y

z

x

O


FIGURE 12.8. Dividing a material set.

1

1 1

1

1 ,

1 , avec .

1 ,

n

G i i

i

n n

G i i i

i i

n

G i i

i

x x mm

y y m m mm

z z mm

=

= =

=

=

= = =

(12.35)

The accuracy of the determination increases with the number of elements used.

12.3.2 Mass Centre of the Union of Two Sets

Let us consider a set (D) constituted (Figure 12.9) of the union of two sets (D1)

and (D2). We search for the position of the mass centre G of the set (D), knowing

those of the mass centres G1 and G2 respectively of the sets (D1) and (D2).

The mass centre of (D1) is defined, relatively to a point O of reference, by:

( )1

11

1 d ( )D

OG OM m Mm

=

, (12.36)

where m1 is the mass of the set (D1). In the same way, the mass centre of (D2) is

defined by:

( )2

22

1 d ( )D

OG OM m Mm

=

, (12.37)

where m2 is the mass of the set (D2). The position of the mass centre of the set (D)

is given by:

(D)

Mi

12.3 Determination of Mass Centres 179

FIGURE 12.9. Mass centre of the union of two sets.

( )1 2

1 d ( )D D

OG OM m Mm ∪

=

, (12.38)

where m is the mass of (D):

1 2m m m= + . (12.39)

Relation (12.38) leads to:

( ) ( )

1 2

1 d ( ) d ( )D D

OG OM m M OM m Mm

= +

. (12.40)

Taking account of Expressions (12.36) to (12.40), we derive:

( )1 21 21 2

1OG m OG m OGm m

= ++

. (12.41)

If the point of reference O coincides with G1, the preceding expression is written:

21 1 2

mG G G G

m=

. (12.42)

The mass centre G is a point of segment G1G2. If the point O coincides with G,

Expression (12.41) leads to:

1 21 2 0m GG m GG+ =

. (12.43)

12.3.3 Mass Centre of a Homogeneous Set

A material set is of homogeneous mass, when the specific mass is independent

of the point M:

( )( ) , .M M Dρ ρ= ∀ ∈ (12.44)

The mass of the element d ( )e M is:

d ( ) d ( )m M e Mρ= , (12.45)

and the mass of the material set is:

( )m e Dρ= , (12.46)

O

G1 G

(D1)

G2

(D2)


where ( )e D is the volume, the surface or the length of the set (D). Expression

(12.23) of the mass centre is then reduced to:

( )

1 d ( )( ) D

OG OM e Me D

=

. (12.47)

This expression shows that the mass centre of a homogeneous set coincides with

the centroid of the set (D).

The Cartesian coordinates of G are then written:

( )

( )

( )

1 ( ) d ( ),( )

1 ( ) d ( ),( )

1 ( ) d ( ).( )

GD

GD

GD

x x M e Me D

y y M e Me D

z z M e Me D

=

=

=

(12.48)

For a numerical determination, we have:

1

1 1

1

1 ,( )

1 , with ( ) .( )

1 ,( )

n

G i i

i

n n

G i i i

i i

n

G i i

i

x x ee D

y y e e D ee D

z z ee D

=

= =

=

=

= = =

(12.49)

12.3.4 Homogeneous Bodies with Geometrical Symmetries

In the case where a body has geometrical symmetries, Expressions (12.47) and

(12.48), giving the position of the mass centre, show that if a homogeneous body

has:

— a centre of symmetry, the mass centre coincides with the centre of sym-

metry,

— a plane of symmetry, the mass centre is contained in this plane,

— an axis of symmetry, the mass centre is a point of this axis.

These considerations will make easier the determination of the mass centre of a

homogeneous body having geometrical symmetries.

12.4 Examples of Determinations of Mass Centres 181

12.4 EXAMPLES OF DETERMINATIONS

OF MASS CENTRES

12.4.1 Homogeneous Solid Hemisphere

We search for the mass centre of a homogeneous hemisphere (Figure 12.10), of

mass m and radius a. We choose a Cartesian reference (Oxyz) such as the axis Oz

is the axis of symmetry of the half-sphere. The mass centre G is on the axis of

symmetry and its coordinates are:

( )

10, 0, ( )d ( ),G G GS

x y z z M V MV

= = = (12.50)

where V is the volume of the half-sphere, and d ( )V M is the volume of the ele-

ment surrounding the point M of z-coordinate equal to z(M). To calculate the

integral extended over the solid hemisphere (S), it is possible to choose as element

of volume an element such as z(M) does not vary, for any point M of this element:

an element included between the sections z and dz z+ (Figure 12.10). This ele-

ment of volume is a cylinder of height dz and radius:

2 2r a z= − . (12.51)

Its volume is:

( )

2 2 2d ( ) d dV M r z a z zπ π= = − . (12.52)

The z-coordinate of the mass centre is thus obtained, from (12.50), by:

( )

2 2

30

3 d2

a

Gz z a z za

ππ

= − . (12.53)

We obtain:

38

Gz a= . (12.54)

FIGURE 12.10. Homogeneous hemisphere.

(S)

a

r

z

dz

z

y

x

O


12.4.2 Homogeneous Solid with Complex Geometry

We consider the solid of Figure 12.11 constituted of a parallelepiped and a

cylinder of a same homogeneous material.

Parallelepiped

• its volume is: 1V Llh= ;

• its mass centre G1 is the centre of symmetry of the parallelepiped:

1 1 1, ,

2 2 2G G G

L l hx y z= = = .

Cylinder

• its volume is: 22

14

V d cπ= ;

• its mass centre G2 (centre of the cylinder) has for coordinates:

2 2 2, ,

2G G G

cx L a y b z h= − = = + .

Relation (12.41) is written in the case of homogeneous solids of the same mass

per unit volume:

( )1 21 21 2

1OG V OG V OGV V

= ++

. (12.55)

Whence the Cartesian coordinates of the mass centre:

( )

( )

2 2

2

2 2

2

2 2

2

2 ,4

2 ,4

22 .

4

G

G

G

hlL d c L axLlh d c

hl L d bcyLlh d c

ch lL d c hz

Llh d c

ππ

ππ

π

π

+ −=+

+=+

+ +=

+

(12.56)

FIGURE 12.11. Solid with a complex geometry.

L

(S1)

a

z

y

x

d

c

l

b

(S2)

h

12.4 Examples of Determinations of Mass Centres 183

FIGURE 12.12. Non-homogeneous Solid.

12.4.3 Non-Homogeneous Solid

We consider the solid (Figure 12.12), constituted of a half-sphere of mass ρ1

per unit volume and radius a, and of a cylinder of height h, of mass ρ2 per unit

volume and having the same base as the half-sphere.

Half-sphere

• its mass is: 31 1

23

m aπ ρ= ;

• its mass centre G1 was determined in Section 12.4.1. Its coordinates

are:

1 1 1

30, 0, 8

G G Gx y z a= = = − .

Cylinder

• its mass is: 22 2m a hπ ρ= ;

• its mass centre G2 is the centre of symmetry of coordinates:

2 2 20, 0,

2G G G

hx y z= = = .

The mass centre of the solid is deduced from Relation (12.41). We obtain:

2 22 1

1 2

234 2 3

h aOG k

a h

ρ ρρ ρ

−=

+

. (12.57)

(S1)

z

h

a

O

(S2)


EXERCISES

12.1 Determine the mass centre of an arc of circle (Figure 12.13).

12.2 Determine the mass centre of a circular sector (Figure 12.14).

12.3 Determine the mass centre of a circular segment (Figure 12.15).

12.4 Determine the mass centre of a cone (Figure 12.16).

12.5 Determine the mass centre of a spherical segment (Figure 12.17).

12.6 Determine the mass centre of a cylinder of radius a and height h, in which a

cylinder of half-radius is hollowed out (Figure 12.18).

12.7 We consider a solid sphere of centre O and radius a. Derive the action of

gravitation exerted by the sphere at a point M external to the sphere and located at

a distance r of the centre O (r > a), in the case where the sphere is homogeneous

by concentric layers.

FIGURE 12.13. Arc of circle. FIGURE 12.14. Circular sector.

FIGURE 12.15. Circular segment. FIGURE 12.16. Cone.

2

a

2

a

2

a

h

a

Comments 185

FIGURE 12.17. Spherical segment. FIGURE 12.18. Hollowed cylinder.

COMMENTS

The actions of gravitation are induced by the phenomena of attraction

between masses and are characterized by the Newton’s law. This law

allows us to characterize the action of gravitation exerted by the Earth on a

material body. With the action of gravitation a mechanical action is super-

imposed, induced by the motion of rotation of the Earth around the axis of

the poles South-North. The resultant action is the action of gravity exerted

by the Earth which differs little from the action of gravitation. This action

of gravity exerted on a solid or a set of solids is a force of which the

magnitude is the product of the mass of the solid or set of solids by the

magnitude of the field of the Earth gravity and whose support is the down-

ward vertical axis passing through the mass centre.

Note. The action of gravity is denoted by e in reference to the

corresponding French term “action de pesanteur”.

h

a

h

a

CHAPTER 13

Actions of Contact between Solids Connections

13.1 LAWS OF CONTACT BETWEEN SOLIDS

13.1.1 Introduction

To move on the ground a body (cupboard, case (Figure 13.1), etc.), it is neces-

sary to exert a sufficient mechanical action to overcome the action exerted by the

ground on the body, action which is opposed to any motion of the body on the

ground. The actions of contact between solids are of intermolecular nature and

appear only at this scale. They are thus exerted only at extremely small distances,

hence their name of actions of contact. So the actions of contact are very sensitive

to the state of the surfaces in contact. In addition, the actions of contact depend on

the other mechanical actions which are exerted. For example, it is more difficult

to draw the case filled than the empty case. The phenomena of contact are

complex, and the laws of contact that we shall state are only approximate laws.

They constitute however a satisfactory approach in many problems involving

actions of contact between solids.

FIGURE 13.1. Displacement of a case.

13.1 Laws of Contact between Solids 187

13.1.2 Contact in a Point

13.1.2.1 Laws of Contact in one Point

Let (S) and (T) be two solids, in contact at the point P at a given time (Figure

13.2). In fact the contact occurs between surfaces of very small dimensions and

may be considered as a contact in a point. The two bodies being supposed in-

deformable and impenetrable, they are tangent at the point P. We are in the

kinematic model studied in Chapter 10 (Section 10.1.1).

Owing to the contact of the two bodies at the point P, the body (T) exerts on

the solid (S) an action of contact represented by the torsor T S→

. The laws of

contact in a point are the following ones:

1st law

The action of contact exerted by the solid (T) on the solid (S) is a force of

which the line of action passes through the point of contact P.

The torsor T S→ is thus a slider of which the axis passes through the point

of contact P. In particular:

0P T S =→ . (13.1)

The experimental analysis of the phenomena of contact shows that the resultant

of the action of contact exerted by the solid (T), is not, as in the case of the actions

at distance, known or calculable a priori, but depends on the other mechanical

actions exerted on (S). The action of contact must however verify some conditions

expressed in the laws which we state hereafter.

FIGURE 13.2. Solids in contact in one point.

(S)

(T)

P

188 Chapter 13 Actions of Contact between Solids. Connections

The force of contact exerted by the rigid body (T) on the rigid body (S) is

resolved into two forces:

— a force of resultant tR

, called force of resistance to sliding or force of

friction, of which the line of action is contained in the plan tangent at P to the two

solids;

— a force of resultant nR

called normal force of contact, of which the line of

action is the line normal at the point P to the tangent plane.

The resultant of the action of contact is thus written:

t nR R RT S = +→ . (13.2)

2nd law

If the vector n

is the unit vector of the normal direction at the point P orient-

tated from the solid (T) towards the solid (S), in all the cases where (S) and (T) are

not stuck at the point P, we have:

, with 0,n n nR R n R= ≥

(13.3)

where Rn is the component of the normal force of contact. This law expresses the

fact that the normal force is opposed to the penetration of the solid (S) in the solid

(T). The symbolic representation of the force of contact is reported in Figure 13.3.

3rd law or Coulomb’s law

There exists a positive coefficient f called coefficient of friction of (S) on (T),

dependent on the materials of which (S) and (T) are made, dependent on the state

of the surfaces in contact, but independent of the motions or of the equilibrium of

(S) and (T),such that the following condition is verified at any time:

t nR f R≤

. (13.4)

FIGURE 13.3. Normal and tangential components of the force of contact.

(T)

(S)

P

R

tR

nR

n

tangent plane


This law must be specified in the following way:

— If the solid (S) slides on (T), thus if the sliding velocity is not null:

( ) ( ) ( , ) 0T TPg S SP t = ≠

, (13.5)

• on the one hand, that is the equality which is satisfied:

t nR f R=

, (13.6)

• on the other hand, tR

and ( )

( , )Tg S P t are collinear and of opposite signs:

( ) ( )

( , ) 0, ( , ) 0.T Tt tg S g SR P t R P t× = <⋅ (13.7)

— If the solid (S) does not slide on (T), thus if its sliding velocity is null:

( ) ( ) ( , ) 0T TPg S SP t = =

, (13.8)

that is the inequality which is satisfied:

t nR f R<

. (13.9)

The previous formulation may be also translated by saying that, as long as the

inequality (13.9) is satisfied, the solid (S) cannot slide on the solid (T). Sliding

occurs only when the other actions exerted on the solid (S) are large enough to

satisfy Relation (13.6). The solid (S) slides then on (T), the force of friction being

opposed to the velocity vector of sliding at the point P. Moreover, for a given

value of Rn, Equality (13.6) is all the more satisfied as the coefficient f is low.

This result is expressed by saying “the lower the coefficient of friction is, the

more the sliding is easy”. Orders of magnitude can be given for the coefficient of

friction according to the nature of the solids in contact:

wood on wood: 0.3 to 0.5;

steel on wood: 0.25;

bronze on bronze: 0.2;

steel on steel: 0.15;

brake lining on steel drum: 0.4;

tire on roadway: 0.2 to 0.6.

13.1.2.2 Corrections to Coulomb’s Law

The laws of solid friction are only applicable to the case of dry (not lubricated)

friction between two solids. Coulomb’s law provides usually a satisfying quali-

tative approach to the phenomena of dry friction. If the quantitative results which

are deduced from the law are not always in agreement with the measured values,

that results from the fact that the coefficient of friction is very sensitive to the

surface state of the materials in contact, to traces of moisture or lubricant, etc.,

and that varying from one area to the other one of the solids in contact. Moreover,

the coefficient of friction depends upon the temperature of the parts in contact. In


fact the friction increases the temperature of the parts in contact, hence a decrease

of the coefficient of friction. The importance of this effect is highlighted by the

behaviour of car braking, where the friction is higher at the beginning of braking.

The coefficient f also depends to a certain extent on the normal component Rn.

Lastly, the coefficient of friction depends on the sliding velocity.

A way rather simple to take account of the dependence of the coefficient of

friction as a function of speed consists in considering two different values of the

coefficient of friction: a coefficient of static friction fs and a coefficient of

kinematic friction fk, of value lower than that of the static friction. This distinction

between the two conditions of friction then makes it possible to explain some

usual effects. For example, consider a solid resting on an inclined plane. For a

given inclination where the equilibrium is precarious, a very low impulse is

sufficient to break the equilibrium, the body having then an accelerated slippage

motion. If the plane is horizontal, a higher effort is necessary to make move the

solid than that to move it then.

13.1.2.3 Power Developed

The power developed in the reference (T) by the action exerted on the solid

(S), in contact at the point P with (T) is, from (11.13):

( ) ( ) T T

SP T S T S=→ → ⋅ . (13.10)

Thus, when expressed at the point of contact P:

( )

( )( , )T T

g SP R P tT S T S=→ → ⋅

, (13.11)

or since ( )

( , )Tg S P t is orthogonal to nR

:

( )

( )( , )T T

t g SP R P tT S =→ ⋅ . (13.12)

The power developed by the normal force of contact is null. The power is reduced

to that developed by the force of friction. From the Coulomb’s law this power is

negative or null.

13.1.2.4 Contact Without Friction

If friction is necessary in some cases (walk on the ground, entrainment of a car,

etc.), in other cases it is necessary as much as possible to diminish it in order to

decrease the energy dissipated by friction and to avoid a premature attrition of the

parts in contact.

In the extreme case where the coefficient of friction is zero, we will say that the

contact occurs without friction or that the contact is perfect at the point of contact

considered.


In such a description, we have:

0 and t nR R RT S= =→ . (13.13)

The solid (T) exerts on (S) only a normal action of contact. The least action

exerted on the solid (S) will induce sliding of solid (S). Moreover, Expression

(13.12) shows that the power developed is zero.

In conclusion, we will say that the contact between two solids is perfect or

without friction at the point P, if and only if one of the following equivalent con-

ditions is verified:

— the coefficient of friction is zero,

—the action of contact is normal at P to the two solids,

— the power developed by the action of contact is zero.

This model of perfect contact remains however an ideal model, towards which

one tends to approach by polishing the surfaces in contact and lubricating them.

13.1.3 Couples of Rolling and Spinning

13.1.3.1 Introduction

In the preceding section, we studied the case of a contact at a point for which

the action of contact can be reduced to a force of contact. In practice, the contact

between the two solids occurs on a surface localized at the centre P. The action of

contact exerted must then be resolved, at the point P, into a force of contact, of

which the properties were studied in the preceding section 13.1.2, and a couple of

contact of moment-vector

equal to the moment at the point P of the action of

contact:

P T S= → . (13.14)

As the force of contact (Relation (13.2)), the couple is resolved into two

couples:

— a couple of resistance to rolling of moment-vector t

, of which the

direction is contained in the plane tangent in P to the two solids;

— a couple of resistance to spinning of moment-vector n

of direction

orthogonal to the tangent plane.

The moment-vector is thus written:

t n= +

. (13.15)

The properties of the couples of contact are complex. Laws similar to the

Coulomb’s law are however stated for the qualitative analysis of the phenomena

of rolling and spinning.


13.1.3.2 Laws of Rolling

The model usually considered is the following.

— If the solid (S) does not roll on (T), thus if the rotation vector of rolling

(Section 10.1.2) is null: ( )

0TS tω =

, (13.16)

the moment of the couple of resistance to rolling satisfies the inequality:

t nhR<

. (13.17)

— If the solid (S) rolls on (T), thus if: ( )

0TS tω ≠

, (13.18)

• on the one hand:

t nhR=

, (13.19)

• on the other hand t

and ( )TS tω

are collinear and of opposite signs.

The parameter h is called coefficient of resistance to rolling. It has the dimen-

sion of a length.

13.1.3.3 Laws of Spinning

The laws of spinning can be stated in the same way by substituting in the law

of rolling the rotation vector of spinning ( )TS tω

and the moment n

of the couple

of resistance to rolling for ( )TS tω

and t

, respectively, and by introducing a

coefficient of resistance to spinning. Let us note that the resistance to spinning

results from the resistance to sliding of the surfaces in contact. The coefficient of

resistance to spinning is thus a function of the coefficient of friction and of the

dimensions of the surfaces in contact. This function is however difficult to derive.

13.2 CONNECTIONS

13.2.1 Introduction

The motions of a solid (S) with respect to a reference (T), of which we have

studied the kinematics in Chapter 9, are obtained by realizing a connection

between the solid bodies (S) and (T). This connection is realized by putting in

contact surfaces of solids (S) and (T), the contact occurring along an arc of curve

or a surface. The action of contact exerted by the solid (T) on the solid (S) results

from the actions of contact exerted at every point of the arc of curve or the surface

of contact. This action of contact is usually called action of connection. This

action is represented by a torsor which will denote by ( ) T S .

13.2 Connections 193

13.2.2 Classification of Connections

13.2.2.1 Simple Connections

We will say: Two solids (S) and (T) are connected by a simple connection, if

the solids are in contact along two geometrical surfaces, the one being part of (S),

the other being part of (T).

We will restrict the analysis in this chapter to the case of the elementary sur-

faces: plane, revolution cylinder and sphere. These surfaces are simple to realize,

they are not however the only elementary surfaces which are used. By setting

contact between these surfaces, we obtain six simple connections:

Plane Cylinder Sphere

Plane Plane connection

Linear support Point support

Cylinder Cylindrical connection

Bearing connection

Sphere Spherical

connection

The representations of these connections, with their symbols, are reported in

Figures 13.4 to 13.9.

Plane connection (Figure 13.4)

The surfaces in contact are plane. The solid (S) has, with respect to the solid

(T), 3 degrees of freedom: 2 degrees in translation and 1 in rotation.

Linear support (Figure 13.5)

The solids are in contact along a segment of line. The solid (S) has, with

respect to the solid (T), 4 degrees of freedom: 2 degrees in translation and 2 in

rotation.

FIGURE 13.4. Plane connection.

(S)

(T)

(T)

(S)


FIGURE 13.5. Linear support.

FIGURE 13.6. Point support.

Point support (Figure 13.6)

The solids are in contact in a point. The solid (S) has 5 degrees of freedom: 2

in translation and 3 in rotation.

Cylindrical connection (Figure 13.7)

The solids are in contact along a cylinder. The solid (S) has, with respect to

(T), 2 degrees of freedom: 1 in translation and 1 in rotation.

Bearing connection (Figure 13.8)

The solids are in contact along a circle. The solid (S) has 4 degrees of freedom:

1 in translation and 3 in rotation.

Spherical connection (Figure 13.9)

The solids are in contact along a sphere. The solid (S) has 3 degrees of free-

dom in rotation.

(T)

(S)

(T)

(S)

(T)

(S)

(T)

(S)


FIGURE 13.7. Cylindrical connection.

FIGURE 13.8. Bearing connection.

FIGURE 13.9. Spherical connection.

(T)

(S)

(T)

(S)

(T)

(S) (S)

(T)

(T)

(S)

(S)

(T)


FIGURE 13.10. Symbolic representation of a combined connection.

13.2.2.2 Combined Connections

Two solids (S) and (T) are connected by a combined connection, if the con-

nection is realized through several simple connections.

A combined connection can be represented symbolically by the diagram of

Figure 13.10, where l1, l2, l3, ..., are simple connections.

Examples of combined connections — A hinge connection can be realized using for example a cylindrical con-

nection and a spherical connection (Figure 13.11a), or using two spherical con-

nections (Figure 13.11b). The solid (S) has, with respect to the solid (T), 1 degree

of freedom in rotation.

— A prismatic connection can be realized (Figure 13.12) using two plane

supports. The solid (S) has 1 degree of freedom in translation.

13.2.2.3 Complex Connections

Two solids (S) and (T) are connected by a complex connection, if the con-

nection is realized through one or several bodies.

A complex connection is symbolized on the diagram of Figure 13.13a. The

solids (S) and (T) are connected through the solids (S1) and (S2), connected ones to

FIGURE 13.11. Hinge connection.

(S)

(T)

l1

l2l3

(S)

(T)

(S)

(T)

(T)

(S)

(a) (b)

(c)


FIGURE 13.12. Prismatic connection.

the others by connections l1, l2, l3. Figure 13.13b gives an example of complex

connection: the solids (S) and (T) are connected through a cylindrical connection,

a spherical connection and a hinge connection, the axes of cylindrical and hinge

connections intersecting at the centre of the spherical connection.

13.2.3 Actions of Connection

13.2.3.1 General Elements

The elements of reduction at a point P of the action of connection induced by

the solid (T) on the solid (S) can be expressed in a basis ( ), , i j k

as follows:

( ) ( )

,

.

T l l l

P T l l l

R S X i Y j Z k

S L i M j N k

= + +

= + +

(13.20)

The action of connection, and consequently the components Xl, Yl, Zl, Ll, Ml

and Nl depend upon the other mechanical actions exerted on the solid (S).

FIGURE 13.13. Complex connection.

(T)

(S) (T)

(S)

(a)

(T)

(S)

(b)

l1

l2

l3

(S1)

(S2) (S)

(T)


However, to solve the problems of Mechanics of rigid bodies, it is necessary to

introduce assumptions on some of the components according to the physical

nature of the connections: connection without friction, connection with dry fric-

tion or viscous friction.

13.2.3.2 Power Developed by the Actions of Connection

The power developed in the reference (T) by the action of connection exerted

by the solid (T) on the solid (S) is from (11.13):

( ) ( ) ( ) ( ) T TT T SP S S= ⋅ , (13.21)

where ( ) TS is the kinematic torsor relative to the motion of the solid (S) with

respect to the solid (T).

By introducing the elements of reduction at the point P of the action of con-

nection (13.20), the preceding relation is written:

( ) ( ) ( ) ( ) ( ) ( ) T T TP PT T TS SP S R S S R= +⋅ ⋅

, (13.22)

or ( ) ( ) ( ) ( ) ( )

( , )T TPT T T SP S R S P t S ω= + ⋅⋅

, (13.23)

by introducing the veloctity vector of the point P and the instantaneous rotation

vector.

13.2.4 Connection Without Friction

13.2.4.1 Model of Perfect Connection

So as to reduce the dissipated energy and to decrease the attrition of the sur-

faces in contact, it is necessary to obtain surfaces such as the contact in every

point approach as much as possible a perfect contact. We will say that a con-

nection between two solids is perfect, if the contact between the solids is perfect

in every point. By extension of the results established in Subsection 13.1.2.4, we

deduce then:

A connection is perfect, if and only if the power developed by the action of

connection is null.

We will consider this property as the definition of a perfect connection. The

model of perfect connection is however only an idealized model, towards which

we usually tend to approach in the technological realizations.

13.2.4.2 Hinge Connection

In the case of a hinge connection, the solid (S) is animated, with respect to the

reference (T), by a motion of rotation about the axis of the hinge connection. This

motion was studied in Section 9.4.1. The solid (S) has one degree of freedom in


rotation ψ and the kinematic torsor is defined (Section 9.4.1.2) by its elements of

reduction at an arbitrary point OS of the axis of rotation:

( ) ( )

( ) ( )

,

( , ) 0.S

T TS S

T TO S S

R k

O t

ω ψ = =

= =

(13.24)

The power developed, in the reference (T), by the action of connection is from

(13.23): ( ) ( ) ( ) ( )

S

T TOT T S lP S S Nω ψ= ⋅ =

. (13.25)

The condition of perfect connection is then written:

( ) ( ) 0, TT lP S N ψ ψ= = ∀ . (13.26)

Thus:

0lN = . (13.27)

Hence the result:

If the solid (S) is connected to the solid (T) by a perfect hinge connection, of

axis of unit direction vector k

, the action exerted by (T) on (S) is represented by

a torsor having in a basis ( ), , i j k

:

—an arbitrary resultant of components Xl, Yl, Zl ;

— a moment at an arbitrary point of the axis of the hinge connection which is

orthogonal to the direction of this axis, thus of components Ll , Ml, 0.

We write the result in the form:

( ) , , , , , 0 ,SS

T l l l l l OOS X Y Z L M= (13.28)

where OS is an arbitrary point of the axis of the connection. The components Xl,

..., Ml, depend upon the other mechanical actions exerted on the solid (S).

13.2.4.3 Prismatic Connection

In the case of a prismatic connection, the solid (S) moves with a rectilinear

translation motion. If i

is the direction of the prismatic connection, the solid (S)

has one degree of freedom in x (abscissa of an arbitrary point P of the solid (S)).

The elements of reduction at the point P of the kinematic torsor are:

( ) ( )

( ) ( ) ( )

0,

( , ) , .

T TS S

T TP S

R

P t x i P S

ω = =

= = ∀ ∈

(13.29)

The power developed, in the reference (T), by the action of connection is:

( ) ( ) ( ) ( )( , )T T

T T lP S R S P t X x= =⋅

. (13.30)

The condition of perfect connection is thus written:

0lX = . (13.31)


Hence the result:

If the solid (S) is connected to the solid (T) by a perfect prismatic connection of

direction i

, the action exerted by (T) on (S) is represented by a torsor having in

a basis ( ), , i j k

:

— a resultant orthogonal to i

, hence of components 0, Yl, Zl,;

— an arbitrary moment of components Ll, Ml, Nl whatever the point of the

solid (S).

Thus:

( ) 0, , , , , ,T l l l l l PPS Y Z L M N= (13.32)

where P is an arbitrary point of the solid (S).

13.2.4.4 Cylindrical Connection

In the case where the solid (S) is connected to the solid (T) by a cylindrical

connection of direction k

, the solid (S) has (Section 9.4.3) one degree of freedom

in translation z (abscissa of an arbitrary point OS of the axis of the cylindrical

connection) and one degree of freedom in rotation ψ. The elements of reduction at

point OS of the kinematic torsor (Relations (9.66) and (9.67)) are:

( ) ( )

( ) ( )

,

( , ) .S

T TS S

T TO S S

R k

O t z k

ω ψ = =

= =

(13.33)

The power developed, in the reference (T), by the action of connection is from

(13.23): ( ) ( ) T

T l lP S Z z N ψ= + (13.34)

The condition of perfect connection is thus written:

0, , .l lZ z N zψ ψ+ = ∀ (13.35)

Hence:

0, 0.l lZ N= = (13.36)

Hence the result:

If the solid (S) is connected to the solid (T) by a perfect cylindrical connection

of axis of direction k

, the action exerted by (T) on (S) is represented by a torsor

having in a basis ( ), , i j k

:

— a resultant of components Xl, Yl, 0 ;

— a moment of components Ll, Ml, 0, at an arbitrary point of the axis of the

cylindrical connection.

This result may be written in the form:

( ) , , 0, , , 0 ,SS

T l l l l OOS X Y L M= (13.37)

where OS is an arbitrary point of the axis of the connection. The components Xl,

Yl, Ll, and Ml depend upon the other mechanical actions exerted on the solid (S).


13.2.4.5 Spherical Connection

In the case where the solid (S) is connected to the solid (T) by a spherical

connection of centre A, the solid (S) has three degrees of freedom in rotation. The

motion of (S) is a motion of rotation about a point (Section 9.4.4) and the kine-

matic torsor is expressed at A as follows:

( ) ( )

( ) ( )

3 ,

( , ) 0.

T TS S S

T TA S

R k i k

A t

ω ψ θ ϕ = = + +

= =

(13.38)

The condition of perfect connection is written:

( ) ( ) ( ) ( )0T T

AT T SP S S ω= ⋅ = . (13.39)

This condition must be satisfied whatever the motion of rotation of the solid (S),

therefore whatever the rotation vector ( )TSω

. The condition of perfect connection

is thus written here:

( ) 0A T S =

. (13.40)

Hence the result:

If the solid (S) is connected to the solid (T) by a perfect spherical connection of

centre A, the action of connection exerted by (T) on (S) is a force whose the line

of action passes through the centre A of the spherical connection.

The components of the resultant of the force depend upon the other mechanical

actions exerted on the solid (S).

13.2.4.6 Plane Connection

In the case of a plane connection, the solid (S) is animated by a plane motion

(Section 9.4.5) with respect to the solid (T). The solid (S) has two degrees of

freedom in translation x and y (coordinates of an arbitrary point P of the plane of

contact) and one degree of freedom in rotation ψ about the direction orthogonal to

the plane of contact (Figure 13.14).

The elements of reduction, at the point P of the plane of contact, of the kine-

matic torsor are written:

( ) ( )

( ) ( )

,

( , ) .

T TS S

T TP S

R k

P t x i y j

ω ψ = =

= = +

(13.41)

The power developed is:

( ) ( ) TT l l lP S X x Y y N ψ= + + , (13.42)

and the condition of perfect connection is written:

0, 0, 0.l l lX Y N= = = (13.43)


FIGURE 13.14 Plane motion of a rigid body.

We write this result in the form:

( ) 0, 0, , , , 0T l l l PPS Z L M= (13.44)

where P is an arbitrary point of the plane of contact.

13.2.4.7 Conclusions

The examples studied in the preceding subsections show that, in the case of a

connection without friction, the components of the action of connection, which

are associated to the degrees of freedom of the solid (S), are zero: components of

the resultant for the degrees of freedom in translation and components of the

moment for the degrees of freedom in rotation. This property results from Expres-

sion (13.23) of the power and from the condition of connection without friction

which expresses that the power is zero.

13.2.5 Connection With Friction

In practice, it is necessary to take account of the friction between the surfaces

of contact of the solids in connection. In the case of dry friction, it will be pos-

sible to transpose the laws stated in Section 13.1 and to apply these laws to the

action of connection exerted by the solid (T) on the solid (S). In the case of

viscous friction, it is possible to describe the friction by considering the com-

ponents of the action of connection as proportional to the components of the

velocities and of opposite signs. For example:

, , , ,l x l y l z lX f x Y f y Z f z N fψ ψ= − = − = − = − (13.45)

where the coefficients fi (i = x, y, z, ψ) are the coefficients of viscous friction.

z

O

(T )

y

z

P y

x

yS

xS

(S )

x

Comments 203

COMMENTS

Connections have a particular importance within the framework of the

design of mechanical systems. Thus, the reader will have to pay a close

attention to the concepts developed in the present chapter. As application

of the general concepts, this chapter considers the cases of connections

between solids using elementary connections. The reader must have assi-

milated well the elements developed in this context.

Contrary to the actions at distance, the actions of contact depend upon

the other actions exerted on the solid or the system of solids under consi-

deration. These conditions are easily derived, in the case of connections

without friction, by writing the nullity of the power developed in the

motion of the solid in connection. To take account of the conditions of

friction, the model simplest to treat is that of viscous friction where the

components of the actions of connections are proportional to the compo-

nents of the velocities and of opposite signs. Dry friction is usually rather

difficult to analyze. The behaviour is transposed from the friction law of

Coulomb which is stated in the case of two solids in contact in a point and

from the laws of rolling and spinning friction.

Note. The actions of connections are denoted by ( )T S in reference to

the corresponding French term “actions de liaisons”.

CHAPTER 14

Statics of Rigid Bodies

14.1 INTRODUCTION

The purpose of this chapter is to analyze the mechanical actions exerted on a

material system, through the study of the equilibrium of a rigid body or a system

of rigid bodies.

A system of rigid bodies is in equilibrium with respect to a given reference, if

during time every point of the system keeps a fixed position with respect to the

reference.

The laws of statics are a consequence of the fundamental principle of dynamics

which will state in Chapter 18.

14.2 LAW OF STATICS

14.2.1 Case of a Rigid Body

A solid (S) submitted to mechanical actions is in equilibrium, if and only if the

torsor representing the whole of the mechanical actions exerted on the solid is the

null torsor.

Thus:

( ) 0S = , (14.1)

with

( ) S S S= → .

The mechanical actions exerted on a solid can be divided into:

— known or calculable actions (actions of gravitation or gravity, electro-

magnetic actions) represented by the torsor ( ) S ;

14.2 Laws of Statics 205

— actions of connections, depending on the other actions exerted on the solid

(S), represented by the torsor ( ) S .

The law of statics for the solid (S) is thus written:

( ) ( ) 0S S =+ . (14.2)

This relation leads to two vector equations:

— the equation of the resultant:

( ) ( ) 0R S R S =+ , (14.3)

— the equation of the moment at any point P:

( ) ( ) 0P PS S =+ . (14.4)

The equilibrium of a solid thus provides 6 scalar equations of which the reso-

lution will be made easier by a discerning choice of the point P and bases in

which the resultant and the moment will be expressed.

14.2.2 Case of a Set of Rigid Bodies

A set of rigid bodies is in equilibrium if and only if every rigid body is in

equilibrium.

We consider the set (D) constituted of n solids: 1 2( ), ( ), . . . , ( ),iS S S . . . ,

( ), . . . , ( ).j nS S The actions exerted on the solid (Si) are divided into:

— external actions, actions exerted by the exterior of (D):

( ) ( ) ( ) i i i iD S S S S→ = = + , (14.5)

known (or calculable) actions actions of connections

exerted by the exterior of (D) with the exterior of (D)

— internal actions, exerted by the other solids of (D):

( ) ( ) ( ) 1 1 1

n n n

j i j i j i j i

j j ji i i

S S S S S

= = =≠ ≠ ≠

→ = = + . (14.6)

known actions exerted actions of connections

by the other solids (Sj) with the solids (Sj)

The equilibrium of each solid (Si) is thus written in one of the forms:

1

0

n

i j i

ji

D S S S

=≠

→ + → = , (14.7)

or

206 Chapter 14 Statics of Rigid Bodies

( ) ( ) ( ) ( )

1

,0

for 1, 2, . . . , .

n

i i j i j i

ji

S S S S

i n

=≠

+ + + =

=

(14.8)

The equilibrium of the set (D) thus leads to n equations of torsors, 2n vector equa-

tions and 6n scalar equations.

Some equations, linear combinations of the preceding ones, may be obtained

by considering the equilibrium of a part of the set (D). These equations will be

able, in some cases, to replace some of Equations (14.7) or (14.8) advantageously.

In particular, it is possible to write the global equilibrium of the set (D), thus:

0D D→ = , (14.9)

or from (11.4):

1

0

n

i

i

D S

=

→ = . (14.10)

This equation introduces only the actions external to the set (D).

14.2.3 Mutual Actions

We consider two materials set (D1) and (D2) which are disjoint. The mecha-

nical actions exerted on the set (D1) are represented by the torsor:

1 1 2 11 2 1D D D DD D D→ = + →∪ → . (14.11)

The mechanical actions exerted on the set (D2) are:

2 2 1 21 2 2D D D DD D D→ = + →∪ → . (14.12)

The equilibrium of each set (D1) and (D2) is written:

1 1 0D D→ = , (14.13)

2 2 0D D→ = . (14.14)

The equilibrium of the set ( )1 2D D∪ is written:

1 2 1 2 0D D D D =∪ → ∪ , (14.15)

or

1 2 1 1 2 2 0D D D D D D+ =∪ → ∪ → . (14.16)

The association of the preceding relations leads to the relation:

2 1 1 2D D D D→ = − → . (14.17)

14.3 Statics of Wires or Flexible Cables 207

This relation expresses the theorem of mutual actions:

The mechanical action exerted by a material set on another material set is

opposed to the mechanical action exerted by the second on the first.

Relation (14.17) associated to Expression (11.9) of the mechanical actions

exerted on a given set leads to a global relation including actions of gravitation,

actions of contact and electromagnetic actions exerted on the sets. Thus:

2 12 1 2 1

1 21 2 1 2 .

D DD D D D

D DD D D D

→+ + =→ →

→− + +→ →

(14.18)

Relation (14.17) of the theorem of mutual actions in fact is extended to each

type of mechanical actions considered separately. Thus:

2 1 2 1D D D Dϕ ϕ→ →= − , (14.19)

whatever the physical law ϕ induced on the two sets ( ,ϕ = or ) .

14.3 STATICS OF WIRES OR FLEXIBLE CABLES

14.3.1 Mechanical Action Exerted by a Wire or a Flexible Cable

Wires or flexible cables are linear deformable solids, used generally to connect

bodies between them. Consider A and B (Figure 14.1a) two points of a wire or

cable (extensible or not) and M a point located between A and B. In the general

case of a cable having a bending stiffness, the mechanical action exerted by

FIGURE 14.1. Mechanical action exerted on a wire or flexible cable.

M

A

B

M

A

B

M

( )T M

( )T M′

(a) (b)


the part AM on the part MB is arbitrary and can be resolved into a force and a

couple depending on the point M.

We say that wire (or cable) is flexible, if and only if the couple which is exerted

is null at any point of wire (or cable).

The mechanical action exerted by the part AM on the part MB is thus a force,

called tension at the point M, of which the support passes through the point M and

the resultant ( )T M

depends on the point M. Moreover, it is possible to show and

experiment confirms that:

The line of action of the force exerted by the part AM on the part BM coincides

with the tangent at M to the wire or cable, orientated from B towards M (Figure

14.1b).

The roles of A and B can be exchanged, and so the part BM exerts on the part

AM a force of resultant ( )T M′ collinear to the preceding one but of opposite sign:

( ) ( ), with 0T M T Mα α′ = − >

. (14.20)

14.3.2 Equation of Statics of a Wire

Consider an element ds MM ′= of wire (Figure 14.2). The resultant of the

forces of tension which are exerted on this element is:

d( d ) ( ) d d

d

TT s s T s T s

s+ − = =

. (14.21)

The equilibrium of the element is written:

dd d 0

dl

Ts s g

sρ+ =

, (14.22)

by introducing the mass by unit length ρl of the wire and g

the field of gravity

induced by the Earth. Hence the equilibrium equation of the wire:

1 d0

dl

Tg

sρ+ =

. (14.23)

FIGURE 14.2. Mechanical action exerted on an element of wire.

M

A

B

M'

M'

M

T

te

ne


The tension of the wire at the point M is written:

tT T e=

, (14.24)

hence:

d d

d dt n

T T Te e

s s= +

, (14.25)

by introducing the unit vectors ( ), t ne e

of the tangent and principal normal

directions, and the radius of curvature of the wire at the point M. The equi-

librium equation (14.23) of the wire thus can be written in the form:

( )1 d0

dt n

l

T Tg e e

sρ+ + =

. (14.26)

In the case of a wire of negligible mass, the equilibrium equation (14.22) of the

element is reduced to:

d0

d

T

s=

. (14.27)

This relation shows that:

If the tensions at A and B are not null, the tension exerted by the portion AM on

the portion MB has the same resultant, whatever the point M of AB. The portion

AB of the wire is rectilinear.

14.3.3 Wire or Flexible Cable Submitted to the Gravity

We search for the shape which is getting by a wire or a flexible cable of homo-

geneous linear mass submitted to the action of gravity. We choose (Figure 14.3) a

frame of reference (Oxyz) so that the axis Oy

is upward vertical and the points A

and B of the wire are contained in the plane Oxy. The equation of equilibrium

(14.23) leads, while introducing the angle α between the tangent at M to the curve

and the axis Ox

, to two equations:

FIGURE 14.3. Wire submitted to the action of gravity.

M

A B

T

y

xO

a


( )

( )

dcos 0,

d

1 dsin 0.

dl

Ts

g Ts

α

αρ

= − + =

(14.28)

Hence by integration:

1 2cos , sin .lT C T gs Cα α ρ= = + (14.29)

It results that:

( )

11 2 2

1tan et d d

cosl l

Cs C C s

g gα α

ρ ρ α= − = . (14.30)

The coordinates (x, y) of the point M of the wire are expressed as follows:

2

dd d cos ,

cos

sind d sin d ,

cos

x s a

y s a

αααα

α αα

= = = =

(14.31)

setting:

1

l

Ca

gρ= . (14.32)

By integration, we obtain:

0

0

ln tan( ) ,4 2

.cos

x a x

ay y

π α

α

= + + = +

(14.33)

It is possible to exclude α, by taking account of the following relations:

( )01 tan

2exp tan( )4 2

1 tan2

x x

a

απ α

α

+−= + =

−,

( ) ( )2

0 0 0

2

1 tan22cosh exp exp 2

cos1 tan

2

x x x x x x

a a a

α

α α

+− − −= + − = =

−.

We then deduce that:

00 cosh

x xy y a

a

−− = . (14.34)

This equation is the equation of a catenary, reported in Figure 14.3 in the case

where the constants x0 and y0 are taken equal to zero.

14.3.4 Contact of a Wire with a Rigid Body

Consider a wire in contact with a solid (S) (Figure 14.4). The wire is submitted


at its points A and B to tensions TA and TB, respectively. The contact with the solid

(S) occurs between the points M1 and M2. Any element ds of the wire is

submitted to a force of contact, which can be resolved (13.2) into a force of

friction of resultant tR

and a normal force of resultant nR

. In the case where the

action of gravity can be neglected compared to the other actions exerted on the

element of wire, the equation of equilibrium (14.23) is modified as:

d0

dt n

TR R

s+ + =

, (14.35)

or by introducing the components Rt and Rn of the force of friction and the normal

force:

d0

dt t n n

TR e R e

s+ + =

, (14.36)

where te

and ne

are the unit vectors of the tangent and the normal at M to the wire

(Figure 14.4). Taking account of Relation (14.25), the equation of equilibrium

leads to the two equations:

d0

dt

TR

s+ = , (14.37)

d0 or 0

dn n

TR T R

s

α+ = + =

, (14.38)

where T is the magnitude of the tension of the wire at point M and α the angle

between the direction te

and the direction 1AM

of the wire at point M1.

1. In the case where there is no friction with the solid (S): Rt = 0, and Relation

(14.37) shows that the magnitude of the tension is maintained along the wire.

2. In the case where friction is induced between the solid (S) and the wire, cha-

racterized by a coefficient of friction f, Coulomb’s law involves that equilibrium

FIGURE 14.4. Wire in contact with a rigid body.

A

B

AT

BT

te

ne

M

M1M2

2

(S)


is maintained as long as:

t nR f R< , (14.39)

or from (14.37) and (14.38):

d d

d d

Tf T

s s

α< . (14.40)

The ultimate equilibrium is thus obtained when:

d d

d

Tf

T s

α= . (14.41)

Thus, integrating between the points M1 and M2:

2fB AT T e α= , (14.42)

where α2 is the winding angle at the point M2, evaluated from the point M1.

For a coefficient of friction of 0.25, and for 3 winding turns (α2 = 6π), we find

111B AT T≈ . The tension to be exerted at B to cause sliding of the wire on the

solid is thus much higher than the tension exerted at A. This result is extensively

used in practice, for example for mooring of the boats.

14.4 EXAMPLES OF EQUILIBRIUM

14.4.1 Case of a Rigid Body

We consider the device of Figure 14.5. A crank can turn about a horizontal axis

BE. This axis is connected to the frame (T) through two connections of respective

centres C and E. A pulley of centre D, rigidly locked with the crank, is connected

to a mass M through a flexible wire and a second pulley attached to the frame.

The position of the frame is defined by the value α of the angle between AB and

the horizontal direction (Figure 14.5b). To maintain the equilibrium of the crank,

it is exerted at the extremity A a force of magnitude F and of support having a

direction β with respect to the direction BA

(Figure 14.5b). The pulley of centre

D has a radius R and a mass m. The mass of the crank ABE is negligible compared

to the masses M and m. The nature of the connections at points C and E is to be

defined so that the system is entirely determined. Thus, it is said that the system is

isostatic.

14.4.1.1 Analysis of the Mechanical Actions Exerted on the Crank

We denote: AB = a, BC = b, CD = d1, DE = d2, and γ the angle between the

vertical direction and the wire connected to the pulley.

As coordinate system attached to the crank-pulley set (S), we choose the

system (Bxyz), such as the axis Bz

coincides BE the axis Bx

is horizontal. The

14.4 Examples of Equilibrium 213

FIGURE 14.5. Equilibrium of a crank-pulley system.

Cartesian coordinates of the different points are then:

( ) ( ) ( )

( ) ( ) ( )1 1 2 1

cos , sin , 0 , 0, 0, 0 , 0, 0, ,

0, 0, , 0, 0, , cos , sin , .

A a a B C b

D b d E b d d F R R b d

α α

γ γ+ + + +

1. Action of gravity The action of gravity exerted on the pulley is represented by the torsor

( ) e S whose the elements of reduction at the point D are:

( )

( )

e ,

e 0.D

R S mg j

S

= −

=

2. Force exerted at the point A

The force is represented by the torsor ( ) S of elements of reduction:

( ) ( ) ( )

( )

cos sin ,

0.A

R S F i j

S

α β α β = + + +

=

A

B

y

x

C

D

E

F

z

a

b

d1

d2

M(a)

B

A

F

x

y

horizontal (b)


FIGURE 14.6. Tension exerted by the wire on the pulley.

3. Action exerted by the wire at F

The wire being flexible, it transmits entirely the action of gravity exerted by the

mass M. This action is a force of which the direction of the support is given by

that of the wire (Figure 14.6). The action is represented by the torsor ( ) f S :

( ) ( )

( )

sin cos ,

0.

f

F f

R S Mg u Mg i j

S

γ γ = = − +

=

4. Action exerted by the frame at the level of the connection in C

It is represented by the torsor ( ) C S of elements of reduction at the point C:

( ) ( )

,

.

C C C C

C C C C C

R S X i Y j Z k

S L i M j N k

= + +

= + +

The components XC, YC, ..., NC, of connection are to be determined.

5. Action exerted by the frame at the level of the connection in EIt is represented by the torsor ( ) E S of elements of reduction at the point E:

( )

( )

,

.

E E E E

E E E E E

R S X i Y j Z k

S L i M j N k

= + +

= + +

The components XE, YE, ..., NE, of connection are also to be determined.

14.4.1.2 Equilibrium Equation of the Crank-Pulley Set

The equation of equilibrium is written:

( ) ( ) ( ) ( ) ( ) e 0f C ES S S S S+ + + + = .

y

x

D

FR

pulley

u


1. Equation of the resultant The equation is written:

( ) ( ) ( ) ( ) ( ) e 0f C ER S R S R S R S R S+ + + + = ,

and leads to three scalar equations:

( )

( )

cos sin 0,

sin cos 0,

0.

C E

C E

C E

F Mg X X

mg F Mg Y Y

Z Z

α β γ

α β γ

+ − + + =

− + + + + + = + =

2. Equation of the moment The equation of the moment must be written at a same point for all the

moments. Generally, the equation will be simplified by choosing a point of one of

the connections and intermediate to the points where the different moments were

expressed. In the present case, we choose the point C. Hence the equation:

( ) ( ) ( ) ( ) ( ) e 0C C C C Cf C ES S S S S+ + + + = .

Deriving the moments at the point C gives:

( ) ( ) 1e eC S R S DC mgd i= × = ,

( ) ( ) ( ) ( ) sin cos sinC S R S AC F i b j b k aα β α β β = × = + − + +

,

( ) ( ) ( ) 1 1cos sinC f fS R S FC Mg i d j d Rkγ γ= × = − − +

,

( ) ( ) ( )

( )[ ] ( )[ ]1 2 1 2 .

C EE E E

E E E E E

S S R S EC

L d d Y i M d d X j N k

= + ×

= − + + + + +

Hence the equations of the moment at the point C:

( ) ( )( ) ( )

1 1 1 2

1 1 2

sin cos 0,

cos sin 0,

sin 0.

C E E

C E E

C E

mgd bF Mgd L L d d Y

bF Mgd M M d d X

aF MgR N N

α β γ

α β γ

β

+ + − + + − + =

− + − + + + + = + + + =

The equilibrium of the crank-pulley set thus leads to 6 scalar equations:

( )

( )

( ) ( )( ) ( )

1 1 1 2

1 1 2

cos sin 0,

sin cos 0,

0,

sin cos 0,

cos sin 0,

sin 0,

C E

C E

C E

C E E

C E E

C E

F Mg X X

mg F Mg Y Y

Z Z

mgd bF Mgd L L d d Y

bF Mgd M M d d X

aF MgR N N

α β γ

α β γ

α β γ

α β γ

β

+ − + + =

− + + + + + =

+ =

+ + − + + − + =

− + − + + + + =

+ + + =

for 13 unknown to derive: , , . . . , , , , . . . , ,C C C E E EX Y N X Y N and F the magni-

tude of the force necessary to obtain the equilibrium.


14.4.1.3 Choice of the Connections

The choice of the connections must be made so as to find 7 equations of con-

nections, so that the preceding system of equations can be solved. The mechanical

system is then known as “isostatic”. It is necessary in the present case to find in C

and E, two connections which will have 7 degrees of freedom on the whole.

For example, let us put in E a spherical connection (3 degrees in rotation). If

the connection is perfect, we have (Section 12.2.4.5):

0E E EL M N= = = ,

(the components of the moment associated to the 3 rotations about the point E are

zero). It is then needed in C a connection with 4 degrees of freedom. Consider a

bearing connection of axis BE. If the connection is perfect, we have:

0CZ = (component associated to the translation along the axis),

0C C CL M N= = = (associated to the rotations about the point C).

The preceding system of equations of equilibrium is thus written as:

( )

( )

( ) ( )( ) ( )

1 1 1 2

1 1 2

cos sin 0,

sin cos 0,

0,

sin cos 0,

cos sin 0,

sin 0.

C E

C E

E

E

E

F Mg X X

mg F Mg Y Y

Z

mgd bF Mgd d d Y

bF Mgd d d X

aF MgR

α β γ

α β γ

α β γ

α β γ

β

+ − + + =

− + + + + + =

=

+ + − − + =

− + − + + =

+ =

The system can then be solved.

It should be noted that the choice of the connections is not arbitrary. In addition

to that the connections must have a total of 7 degrees of freedom, the choice must

lead to a system of equations which can be solved.

14.4.1.4 Exploiting the Equations of Equilibrium

The sixth equation of the equilibrium equations gives the magnitude F of the

force exerted at A necessary to obtain the equilibrium:

sin

RF Mg

a β= − .

The magnitude F is independent of the inclination α of the crank and indepen-

dent of the angle γ of the wire. Moreover, 0F > imposes sin 0β < , hence

0π β− < < (Figure 14.7). For a given mass M, F is minimum for sin 1β = − , thus

for 2

πβ = − . We have then:

RF Mg

a= .


FIGURE 14.7. Practical orientation of the force exerted at the point A.

The other equations then allow us to determine the components of connections on

which no assumptions were stated:

( )

( )

( )

( )

11 2

1 11 2

cos1sin ,

sin

sin1cos ,

sin

cossin ,

sin

sincos .

sin

E

E

C E

C E

RbX d Mg

d d a

RbY md d M g

d d a

RX Mg X

a

RY m M g Y

a

α βγ

β

α βγ

β

α βγ

β

α βγ

β

+ = − +

+ = − + +

+ = + −

+ = + + −

14.4.2 Case of a Set of Two Solids

A mural bracket (S) is constituted (Figure 14.8) of a beam AC (solid (S1)) in

connection at the point C with the wall, and of a tie-rod AB (solid (S2)). The tie-

rod is connected at the point B with the wall and at the point A with the beam. The

beam is used as rolling track by a monorail which supports a mass of value m. The

mass of the bracket (S) is neglected compared to the mass m.

We will denote:

, , ( point of contact with the beam).BC h CA l CM x M= = =

We choose the coordinate system (Cxyz) so that the axis Cx

passes through the

points C and A, and the axis Cy

passes through the points C and B.

14.4.2.1 Mechanical Actions Exerted on the Beam (S1)

1. Action exerted by the mass m, represented by the torsor ( ) 1S :

( )

( ) 1

1

,

0.M

R S mg j

S

= −

=

B

A

F

x

y

horizontal


FIGURE 14.8. Wall-bracket.

2. Action exerted by the wall at C, represented by the torsor ( ) 1C S :

( )

( )

1

1

,

,

C C C C

C C C C C

R S X i Y j Z k

S L i M j N k

= + +

= + +

where the components XC, YC, ..., NC, are to be determined.

3. Action exerted by the tie-rod (S2) at A, represented by the torsor ( ) 2 1S :

( )

( )

2 1 21 21 21

2 1 21 21 21

,

,A

R S X i Y j Z k

S L i M j N k

= + +

= + +

where the components X21, Y21, ..., N21, are to be determined.

14.4.2.2 Mechanical Actions Exerted on the Tie-rod (S2)

1. Action exerted by the wall at B, represented by the torsor ( ) 2B S :

( )

( )

2

2

,

,

B B B B

B B B B B

R S X i Y j Z k

S L i M j N k

= + +

= + +

where the components XB, YB, ..., NB, are to be determined.

M

A

B

y

xC

x

l

h

m


2. Action exerted by the beam (S1) at A, represented by the torsor ( ) 2 1S :

The properties of mutual actions allow us to write:

( ) ( ) 2 1 1 2S S= − .

14.4.2.3 Equilibrium of the Beam (S1)

The equilibrium equation of the beam (S1) is written:

( ) ( ) ( ) 1 1 2 1 0CS S S+ + = .

1. Equation of the resultant It leads to three scalar equations:

21

21

21

0,

0,

0.

C

C

C

X X

mg Y Y

Z Z

+ =

− + + = + =

2. Equation of the moment This equation may be written at the point A, expressing the moments as:

( ) ( ) ( )1 1A S R S MA mg l x k= × = −

,

( ) ( ) ( )

( ) ( )

1 1 1

.

A CC C C

C C C C C

S S R S CA

L i M lZ j N lY k

= + ×

= + + + −

Hence the three scalar equations of the moment:

( )

21

21

21

0,

0,

0.

C

C C

C C

L L

M lZ M

N lY N mg l x

+ =

+ + = − + + − =

14.4.2.4 Equilibrium of the Tie-Rod (S2)

The equilibrium equation of the tie-rod (S2) is written:

( ) ( ) 2 2 1 0B S S− = .

1. Equation of the resultant It leads to the three scalar equations:

21

21

21

0,

0,

0.

B

B

B

X X

Y Y

Z Z

− =

− = − =

2. Equation of the moment The equation may be written at the point A, with:


( ) ( ) ( )

( ) ( ) ( )

2 2 2

.

A BB B B

B B B B B B B

S S R S BA

L hZ i M lZ j N hX lY k

= + ×

= + + + + − −

Hence the three scalar equations of the moment:

21

21

21

0,

0,

0.

B B

B B

B B B

L hZ L

M lZ M

N hX lY N

+ − =

+ − = − − − =

14.4.2.5 Equilibrium of the Bracket (S)

The equilibrium equation is written:

( ) ( ) ( ) 1 1 2 0C BS S S+ + = .

That is the equation obtained by superimposing the equilibrium equations of the

beam (S1) and tie-rod (S2). This equation is independent of the action of con-

nection between (S1) and (S2). The moments having been derived at the same

point A, the scalar equations of the equilibrium of the bracket are obtained by

superimposing the scalar equations obtained for the equilibrium of the beam ant

the one of the tie-rod. Thus:

( )

0,

0,

0,

0,

0,

0.

C B

C B

C B

C B B

C C B B

C C B B B

X X

mg Y Y

Z Z

L L hZ

M lZ M lZ

N lY mg l x N hX lY

+ = − + + = + =

+ + = + + + =

− + − + − − =

The equations thus obtained are not new equations compared to the equations

obtained for the equilibrium of the beam and the equilibrium of the tie-rod. They

constitute another form of these equations.

14.4.2.6 Choice of the Connections

We have 12 scalar equations (among the equilibrium equations of the beam,

tie-rod or bracket), to derive 18 unknowns: , ,B BX Y . . . , ;BN , , . . . , ;C C CX Y N

21 21 21, , . . . , .X Y N To make the system isostatic, it is necessary to put at points A,

B and C connections which will on the whole have 6 degrees of freedom and

which make it possible to solve the equations of equilibrium.

At point B, we choose a spherical connection. If the connection is perfect, we

have:

0, 0, 0.B B BL M N= = =

At point A, we introduce a hinge connection of axis Az

. If the connection is perfect:

N21 = 0.


Lastly at point C, we choose a cylindrical connection of axis Cz

. If the con-

nection is perfect:

ZC = 0, NC = 0.

In the case of perfect connections, the scalar equations of the equilibrium of the

wall-bracket are thus written:

— Equilibrium of the beam (S1)

( )

21

21

21

21

21

0,

0,

0,

0,

0,

0.

C

C

C

C

C

X X

mg Y Y

Z

L L

M M

lY mg l x

+ =

− + + =

=

+ =

+ =

− + − =

— Equilibrium of the tie-rod (S2)

21

21

21

21

21

0,

0,

0,

0,

0,

0.

B

B

B

B

B

B B

X X

Y Y

Z Z

hZ L

lZ M

hX lY

− =

− =

− =

− =

− =

− − =

— Equilibrium of the beam tie-rod set

( )

0,

0,

0,

0,

0,

0.

C B

C B

B

C

C

C B B

X X

mg Y Y

Z

L

M

lY mg l x hX lY

+ =

− + + =

=

=

=

− + − − − =

The preceding equations can be solved and we obtain:

21 21 21

21

21

0.

0, 0, 0.

0, 0, 0.

, , .

, , .

B

C C C

B C B B

B C B

Z

L M N

Z L M

xX mg X X X X

h

x l xY mg Y mg Y Y

l l

=

= = =

= = =

= = − =

−= = =


EXERCISES

14.1 Two beams of lengths l1 and l2 are connected between them at point B and

connected to a frame at points A and C (Figure 14.9). The nature of the connection

is to be determined. Two masses m1 and m2 are suspended respectively at the

points M1 and M2 distant of α1l1 and α2l2 from points A and C. The masses of the

beams can be neglected compared to the masses m1 and m2.

14.1.1. Analyze the mechanical actions exerted on each beam.

14.1.2. Derive the equations of equilibrium of the system.

14.1.3. Choose the connections so that the system is isostatic.

14.1.4. The connections being chosen, derive the actions of connections.

14.2 A person (P) climbs up a ladder (S). The ladder is supported on a wall at

point B and rests on the ground at point A (Figure 14.10). To treat of the problem,

it will be supposed that there is a plane symmetry. In particular, the person is such

as it is “located” in the plane of symmetry of the ladder.

The person stands on the ladder, with the feet posed on the rung C and the

hands at point D. The mass centre of the person is at point G.

14.2.1. Analyze the mechanical actions exerted on the person, on the ladder.

14.2.2. Study the equilibrium of the ladder-person set.

FIGURE 14.9. System of two beams.

M1

A

C

1l1

M2

B

2l2

l

hhorizontal

Comments 223

FIGURE 14.10 Equilibrium of a ladder.

COMMENTS

The laws of statics result from the fundamental principle of dynamics

(Chapter 18), and the study of statics should thus be implemented after

having stated the principle of dynamics. However, the analysis of the

equilibrium of sets of rigid bodies allows us to get a good understanding of

the mechanical actions exerted on rigid bodies. The two examples consi-

dered in section 14.4 will be studied with the greatest attention.

B

A

C

D

ground

wall

G

Part IV

Kinetics of Rigid Bodies

The kinematics of rigid bodies considers the motion of bodies without

being concerned with masses to move. However it is easier to move at

a given speed a body of low mass. It is thus necessary to introduce

concepts which associate motion of bodies and mass of bodies. These

concepts are based on the introduction of the notions of kinetic torsor,

dynamic torsor and kinetic energy.

CHAPTER 15

The Operator of Inertia

The concept of operator of inertia, that we study in the present chapter, will

allows us to simply express the different torsors (Chapter 16) necessary to the

study of the dynamics of rigid bodies.

15.1 INTRODUCTION TO THE OPERATOR OF INERTIA

15.1.1 Operator Associated to a Vector Product

Consider two vectors a

and V

, of which the components in the basis (b) =

( ), , i j k

are:

, .x y za a i a j a k V X i Y j Z k= + + = + +

(15.1)

The vector product of the two vectors is written:

( ) ( ) ( )y z z x x ya V a Z a Y i a X a Z j a Y a X k× = − + − + −

. (15.2)

If the vector a

is a given vector, we observe that, whatever the vector V

, the

vector V

is derived from the vector a V×

by a linear operation. Indeed, we have:

( ) ( )

( )

3

31 2 1 2 1 2

and , ,

, , .

V a V a V

V V a V V a V a V

λ λ λ∀ ∈ ∀ ∈ × = ×

∀ ∈ × + = × + ×

(15.3)

It comes to the same thing to say that the vector V

is derived from the vector

a V×

, by making act on V

a linear operator and to write:

.a V V× =

(15.4)

In matrix form, Expression (15.2) of the vector product is written in the

228 Chapter 15 The Operator of Inertia

basis ( ), , i j k

as:

0

0

0

y z z y

z x z x

x y y x

a Z a Y a a X X

a X a Z a a Y Y

a Y a X a a Z Z

− − − = − = − −

A , (15.5)

by introducing the antisymmetric matrix:

0

0

0

z y

z x

y x

a a

a a

a a

− = − −

A . (15.6)

A is the matrix which represents the operator (or the vector product a ×

), in the

basis (b) = ( ), , i j k

.

When there is only one basis concerned, the notation A is not ambiguous. But

if there is several bases, it will be necessary to specify the notation, while writing

for example: A(b), matrix representing the operator in the basis (b).

15.1.2 Extending the Preceding Concept

Now, we wish to determine the double vector product ( )a a V× ×

. From the

preceding result, we may write:

( ) ( ) 2a a V a V V V× × = × = =

. (15.7)

The new operator 2 thus introduced is a linear operator. It is represented by

the matrix A2 in the basis ( ), , i j k

:

( )( )

( )

2 2

2 2 2

2 2

y z x y x z

x y x z y z

x z y z x y

a a a a a a

a a a a a a

a a a a a a

− + = − + − +

A . (15.8)

The matrix A2

is a symmetric matrix.

In the same way, we may write:

( ) ( ) 2a V a a a V V V× × = − × × = − =

, (15.9)

where the operator 2= − is represented by the matrix 2= −B A :

2 2

2 2

2 2

y z x y x z

x y x z y z

x z y z x y

a a a a a a

a a a a a a

a a a a a a

+ − −

= − + −

− − +

B . (15.10)

15.1 Introduction to the Operator of Inertia 229

FIGURE 15.1. Rigid body.

15.1.3 Operator of Inertia

In the evaluation (Chapter 16) of the torsors used in dynamics, we shall have to

express vectors of the form:

( )( )

1 d ( )S

W OM V OM m M= × ×

, (15.11)

( )[ ]( )

2 d ( )S

W OM V V OM m M= × × ×

. (15.12)

The integrals are calculated over the solid (S) (curve, surface or volume). The

point M (Figure 15.1) is a variable point of (S), and d ( )m M is the mass of the

element of (S) surrounding the point M. The point O is a point of reference of the

solid (S). The vector V

is independent of the point M.

From the results derived in the preceding subsection, we may write:

( )1 OW S V= , (15.13)

introducing the operator ( )O S , called operator of inertia at point O of solid (S).

This operator is represented in a basis (b) attached to the solid by a matrix ( )( )bO SI , called matrix of inertia, at point O and in the basis (b), of the solid (S).

The matrix is written according to one of the forms:

( )( )Ox Oxy Oxz

bO Oxy Oy Oyz

Oxz Oyz Oz

A F E I P P

S F B D P I P

E D C P P I

− − − − = − − = − − − − − −

I . (15.14)

If (x, y, z) are the Cartesian coordinates of the point M in the coordinate system ( )/O b = (Oxyz), we have:

y

z

x

O

dm(M)

M

k

(S)

j

i


OM x i y j z k= + +

, (15.15)

and Expression (15.10) allows us to write:

( )( ) ( )

( )( ) ( )

( )( ) ( )

2 2

2 2

2 2

d ( ), d ( ),

d ( ), d ( ),

d ( ), d ( ).

Ox OxyS S

Oy OxzS S

Oz OyzS S

I y z m M P xy m M

I x z m M P xz m M

I x y m M P yz m M

= + =

= + =

= + =

(15.16)

The quantities IOx, IOy and IOz are called the moments of inertia of the solid (S)

with respect to the axes , , ,Ox Oy Oz

respectively. The quantities POxy, POyz and

POxz are the products of inertia of the solid (S) with respect to the planes (Oxy),

(Oyz) and (Oxz), respectively.

If (X, Y, Z) are the components of the vector V

in the basis (b), the compo-

nents (X1, Y1, Z1) of the vector 1W

in the basis (b) are derived from (15.13) by the

matrix relation: ( )

( )

( )1

1

1

b b

bO

X X

Y Y

Z Z

=

I . (15.17)

Thus:

( ) ( ) ( )1W AX FY EZ i FX BY DZ j EX DY CZ k= − − + − + + + − − +

.

The vector 2W

(15.12) is expressed in the same way in the form:

( )2 OW V S V= ×

(15.18)

Note. ( )O S V

represents the vector derived from the vector V

by making act the

operator ( )O S . The formulation ( )

O S V

must thus be read “the operator ( )O S

acting on V

”. This formulation is comparable with the writing ( ),f x where ( )f x

represents the value obtained from x by the function f.

15.2 CHANGE OF COORDINATE SYSTEM

The change of coordinate reference can be carried out either by change of its

origin or by change of its basis.

15.2.1 Change of Origin

We search for the effect of a change of origin (Figure 15.2). Let (xO', yO', zO')

be the Cartesian coordinates of the new origin O' with respect to the coordinate

15.2 Change of Coordinate System 231

system (Oxyz). The operator of inertia at the point O' of the solid (S) is repre-

sented in the basis (b) = ( ), , i j k

by the matrix of inertia at the point O':

( )( )O x O xy O xz

bO xy O y O yzO

O xz O yz O z

I P P

S P I P

P P I

′ ′ ′

′ ′ ′′

′ ′ ′

− − = − − − −

I . (15.19)

The elements of this matrix are obtained by substituting, in the results intro-

duced in Section 15.1.3, for the vector OM

the vector:

( ) ( ) ( )O O OO M OM OO x x i y y j z z k′ ′ ′′ ′= − = − + − + −

. (15.20)

For example, we can write:

( ) ( )( )

( )( ) ( ) ( )

( ) ( )

2 2

2 2 2 2

d ( )

d ( ) d ( ) d ( )

2 d ( ) 2 d ( ).

O x O OS

O OS S S

O OS S

I y y z z m M

y z m M y m M z m M

y y m M z z m M

′ ′ ′

′ ′

′ ′

= − + −

= + + +

− −

Thus, introducing the mass m of the solid and the Cartesian coordinates (xG, yG, zG)

of the mass centre G of the solid, expressed in (12.34), we obtain:

( )2 2 2 2O x Ox O O O G O GI I m y z y y z z′ ′ ′ ′ ′ = + + − − . (15.21)

The expressions of IO'y and IO'z are deduced from the preceding expression by

permutation. In the same way, we find:

( )O xy Oxy O G O G O OP P m x x y y x y′ ′ ′ ′ ′= − + − , (15.22)

and analogous relations for PO'xz and PO'yz .

FIGURE 15.2. Change of the origin of the coordinate reference.

y

z

x

O

(S) y

z

x

O'


15.2.2 Relations of Huyghens

In the case where the point O' coincides with the mass centre G of the solid,

Relations (15.21) and (15.22) are simplified and the matrix of inertia at point G in

the basis (b) may be written in the form:

( )( ) ( )( ) ( ) ( )b b bG O OGS S S= −I I D , (15.23)

with

( ) ( )

( )( )

( )

2 2

2 2

2 2

G G G G G G

bOG G G G G G G

G G G G G G

m y z mx y mx z

S mx y m x z my z

mx z my z m x y

+ − − = − + −

− − +

D . (15.24)

Thus, Expression (15.23) allows us to express the matrix of inertia at O as a

function of the matrix of inertia at G, generally easier to derive:

( )( ) ( )( ) ( ) ( )b b bO G OGS S S= +I I D . (15.25)

This expression leads to the six relations of Huyghens between the moments

and products of inertia:

( )( )( )

2 2

2 2

2 2

, ,

, ,

, .

Ox Gx G G Oxy Gxy G G

Oy Gy G G Oxz Gxz G G

Oz Gz G G Oyz Gyz G G

I I m y z P P mx y

I I m x z P P mx z

I I m x y P P my z

= + + = +

= + + = +

= + + = +

(15.26)

15.2.3 Diagonalisation of the Matrix of Inertia

We deduce from the properties of the symmetric linear operators the following

fundamental results.

The operator of inertia ( )O S has at least an orthonormal basis of eigen vectors

( )1 2 3, , u u u

called the principal basis at point O.

The axes ( ), iO u

are called principal axes of inertia at the point O, and the

reference ( )1 2 3/ , , O u u u

is the principal reference of inertia at the point O.

In the basis ( )1 2 3, , u u u

, the matrix of inertia at the point O is a diagonal

matrix, called principal matrix of inertia at the point O. Its non-zero terms are the

principal moments of inertia at the point O.

In the principal basis (p), the matrix is thus written as:

( )( )1

2

3

0 0

0 0

0 0

pO

I

S I

I

=

I , (15.27)

where I1, I2 et I3 are the principal moments of inertia at the point O. We deduce:

15.2 Change of Coordinate System 233

( ) ( ) ( ) 1 1 1 2 2 2 3 3 3, , .O O OS u I u S u I u S u I u= = =

(15.28)

The principal moments of inertia Ii (i = 1, 2, 3) can thus be obtained by expressing

Relations (15.28) in the form:

( )O i i iS u I u=

. (15.29)

In the non principal basis (b), this relation is written:

i i

i i i

i i

A F E u u

F B D v I v

E D C w w

− − − − = − −

, (15.30)

by introducing the components (ui, vi , wi ) in the basis (b) of the eigen vector iu

.

The preceding expression can be rewritten:

0

0

0

i i

i i

i i

A I F E u

F B I D v

E D C I w

− − − − − − = − − −

. (15.31)

The vectors iu

being different from the null vector, this system has solutions if

the determinant is zero:

det 0

i

i

i

A I F E

F B I D

E D C I

− − − − − − = − − −

. (15.32)

This equation allows us to obtain the principal moments I1, I2 and I3. The principal

directions are then determined by substituting I1, I2 and I3 into Relation (15.30).

15.2.4 Change of Basis

Let ( ) ( )1 1 1 1, , b i j k=

and ( ) ( )2 2 2 2, , b i j k=

be two bases related by the basis

change:

2 1

2 1

2 1

i i

j j

k k

=

A

, (15.33)

where A is the matrix of basis change. The expressions of the matrices of inertia

allow us to establish the relation which expresses the matrix of inertia ( )( )2bO SI at

the point O in the basis (b2) as a function of the matrix of inertia in the basis (b1).

This relation is written in the form: ( )( ) ( )( )2 1 tb bO OS S=I A I A , (15.34)

where At is the matrix transposed of the matrix A.


15.3 MOMENTS OF INERTIA WITH RESPECT TO

A POINT, AN AXIS, A PLANE

15.3.1 Definitions

We call moment of inertia of a solid (S) with respect to a point (with respect to

an axis or with respect to a plane) the integral:

( )

2d ( ),S

l m M (15.35)

where l is the distance (for example Figure 15.3) from the variable point M of the

solid (S) to the point under consideration (to the axis or to the plane).

If (x, y, z) are the coordinates of point M relatively to a system of origin O, the

expressions of the moments of inertia of the solid (S) are from (15.35):

1. Moment of inertia with respect to the point O:

( ) ( )( )

2 2 2 d ( )OS

I S x y z m M= + + . (15.36)

2. Moments of inertia with respect to the axes , , ,Ox Oy Oz

(already expressed

in 15.16):

( )( )

( )( )

( )( )

2 2

2 2

2 2

d ( ),

d ( ),

d ( ).

OxS

OyS

OzS

I y z m M

I x z m M

I x y m M

= +

= +

= +

(15.37)

FIGURE 15.3. Distances from an arbitrary point to a point, an axis, a plane.

y

z

x

O

M

l

y

z

x

O

M

l

y

z

x

O

M

l

15.3 Moments of Inertia with respect to a Point, an Axis, a Plane 235

3. Moments of inertia with respect to the planes (Oxy), (Oyz), (Oxz):

( )

( )

( )

2

2

2

d ( ),

d ( ),

d ( ).

OxyS

OyzS

OxzS

I z m M

I x m M

I y m M

=

=

=

(15.38)

15.3.2 Relations between the Moments of Inertia

By addition of the integrals (15.36) to (15.38), we obtain the following pro-

perties:

1. The sum of the moments of inertia of a solid with respect to three trirect-

angular axes intersecting at a same point is equal to twice the moment of inertia of

the solid with respect to this point:

2Ox Oy Oz OI I I I+ + = . (15.39)

2. The sum of the moments of inertia of a solid with respect to two perpen-

dicular planes is equal to the moment of inertia of the solid with respect to the axis

intersection of these two planes:

,

,

.

Oxy Oxz Ox

Oxy Oyz Oy

Oxz Oyz Oz

I I I

I I I

I I I

+ =

+ =

+ =

(15.40)

15.3.3 Case of a Plane Solid

In the case of a plane solid, contained in the plane (Oxy) (Figure 15.4), point M

of the solid has for coordinates (x, y, 0) and the moments of inertia are reduced to:

( )( )

( )

( )

( )( )

2 2

2

2

2 2

d ( ),

d ( ),

d ( ),

d ( ).

OS

OxS

OyS

OzS

I x y m M

I y m M

I x m M

I x y m M

= +

=

=

= +

(15.41)

Between the moments of inertia, we have the relation:

Oz O Ox OyI I I I= = + . (15.42)


FIGURE 15.4. Plane solid.

15.3.4 Moment of Inertia with Respect to an Arbitrary Axis

Let us express the moment of inertia of solid (S) with respect to the axis (∆) of

unit direction vector u

and passing through the point O (Figure 15.5). Hence from

(15.35):

( )

2d ( )S

I HM m M∆ = , (15.43)

where H is the orthogonal projection of point M on the axis (∆). We have then:

HM u OM= ×

. (15.44)

Hence:

( ) ( ) ( )22 22HM z y x z y xβ γ γ α α β= − + − + − , (15.45)

introducing the components (α, β, γ) of the vector u

and the coordinates (x, y, z)

of point M. The components (α, β, γ) of the unit direction vector of the axis (∆)

FIGURE 15.5. Moment of inertia with respect to an arbitrary axis.

y

z

x

OM

(S)

y

z

x

O

(S)

()

M

15.4 Determination of Matrices of Inertia 237

are also called the direction cosines of the axis. Reporting Relation (15.45) into

Expression (15.43), we obtain:

2 2 2 2 2 2Ox Oy Oz Oxy Oyz OxzI I I I P P P∆ α β γ αβ βγ αγ= + + − − − . (15.46)

This relation can also be expressed, introducing the operator of inertia at the

point O, in the form: ( )

OI u S u∆ = ⋅ , (15.47)

or in the matrix form:

[ ]Ox Oxy Oxz

Oxy Oy Oyz

Oxz Oyz Oz

I P P

I P I P

P P I

∆

α

α β γ β

γ

− − = − − − −

. (15.48)

In the case where the operator of inertia is referred to its principal axes, Rela-

tion (15.46) is reduced to: 2 2 21 1 2 2 3 3I I I I∆ α α α= + + , (15.49)

where (α1, α2, α3) are the direction cosines of the axis (∆) with respect to the prin-

cipal axes at point O.

15.4 DETERMINATION OF MATRICES OF INERTIA

15.4.1 Solids with Material Symmetries

In the case where the solids have material symmetries, these symmetries make

easier the research of the principal axes of inertia. From this it results a simpli-

fication for deriving the matrix of inertia.

15.4.1.1 Plane of Symmetry

Suppose that the solid (S) has a plane of material symmetry, for example the

plane (Oxy) (Figure 15.6a). It results that the products of inertia:

( )( )

d ( ) and d ( )Oxz OyzS S

P xz m M P yz m M= = are zero, since it is possible to associate elements which have the same value of x

(or of y) and opposite values of z (Figure 15.6a). Thus, we have:

0 0 0 0

0 0 0 0

0 0 1 1

Ox Oxy

Oxy Oy Oz

Oz Oz

I P

P I I

I I

− − = =

, (15.50)

or

( )O OzS k I k=

. (15.51)


FIGURE 15.6. Material symmetries.

Thus, it results that the axis Oz

is principal axis of inertia. Hence the result:

Any axis orthogonal to a plane of material symmetry is principal axis at each

of the points of the plane.

15.4.1.2 Axis of Symmetry

Suppose that the solid (S) has an axis of material symmetry, for example the

axis Oz

(Figure 15.6b). It results from this that the products of inertia:

( )( )

d ( ) and d ( )Oxz OyzS S

P xz m M P yz m M= = are zero, since it is possible to associate elements which have the same value of z

and opposite values of x (or of y) (Figure 15.6b). As in the preceding subsection,

the axis Oz

is principal axis of inertia. Hence the result:

Any axis of material symmetry is principal axis of inertia at each of the points

of the axis.

15.4.1.3 Consequences

1. Any trirectangular trihedron, of which two of its planes are planes of

material symmetries of a given solid, is principal trihedron of inertia of the solid.

2. Any trirectangular trihedron, of which two of its axes are axes of material

symmetries of a solid, is principal trihedron of inertia of the solid.

axe of

symmetry

y

z

x

O

M

M'

(b)

(a)

y

z

x

O

M

M'(a)


15.4.2 Solids having a Symmetry of Revolution

15.4.2.1 General Properties

In the case of a solid (cylinder, cone, disc, etc.) having an axis of revolution,

for example the axis Oz

, the planes Oxz and Oyz are planes of material symmetry

and the coordinate system (Oxyz) is a principal trihedron of inertia (whatever the

axes Ox

and Oy

). The matrix of inertia is written:

( )( )

0 0

0 0

0 0

Oxb

O Oy

Oz

I

S I

I

=

I , (15.52)

with Ox OyI I= due to the symmetry of revolution. Moreover, it is generally easier

to derive the moment of inertia IOz with respect to the axis Oz

, and then to intro-

duce the moment of inertia IOxy with respect to the plane (Oxy). Indeed, from

(15.40) we have:

2Oz Oxz Oyz Ox Oy OxyI I I I I I= + = + − , (15.53)

thus:

1

2Ox Oy Oxy OzI I I I= = + . (15.54)

In the case of a plane solid of revolution, this relation is reduced from (15.42)

to the relation:

1

2Ox Oy OzI I I= = . (15.55)

15.4.2.2 Matrix of Inertia of a Disc

We determine the matrix of inertia of a disc of radius a and mass m (Figure

15.7a). The moment of inertia with respect to the axis Oz

is written:

( )( )

( )( )

2 2 2 2d ( ) d ( )Oz sS S

I x y m M x y S Mρ= + = + , (15.56)

where ρs is the mass per unit surface of the disc and d ( )S M the area of an

element of surface. The calculation of the integral is made easier by introducing

the polar coordinates (r, α) of point M (Figure 15.7a). The element of surface is

obtained by increasing r by dr and α by dα (Figure 15.7b). So, the integral

(15.56) is written, in the case of a homogeneous disc (ρs independent of point M):

23

0 0

d da

Oz sr

I r rπ

αρ α

= == . (15.57)

Thus:


FIGURE 15.7. Disc.

4 2

2 2Oz s

a aI mρ π= = , (15.58)

introducing the mass m of the disc. The matrix of inertia is thus written:

( )( )

2

2

2

0 04

0 04

0 02

bO

am

aS m

am

=

I . (15.59)

15.4.2.3 Matrix of Inertia of a Cylinder

We consider a cylinder of radius a, height h and mass m (Figure 15.8a). The

moment of inertia with respect to the axis Oz

is expressed as:

( )( )

( )( )

2 2 2 2d ( ) d ( )OzS S

I x y m M x y V Mρ= + = + , (15.60)

where ρ is the mass per unit volume of the cylinder and d ( )V M the volume of an

element of volume. Calculation of IOz is simplified while introducing the cylin-

drical coordinates (r, α, z) of the point M (Figure 15.8a). The element of volume

is obtained by increasing respectively by dr , dα and dz the cylindrical coor-

dinates (Figure 15.8b). Integral (15.60) is then written in the case of a homo-

geneous cylinder as:

23

0 0 0

d d da h

Ozr z

I r r zπ

αρ α

= = == . (15.61)

Thus: 4 2

2 2Oz

a aI h mρπ= = , (15.62)

(b)

y

z

x

O

M

d S(M)

a

(a)

r

x

y

O

d

d S(M) = rddr

r rr + dr


dr

FIGURE 15.8. Cylinder.

introducing the mass m of the cylinder.

The moment of inertia with respect to the plane Oxy is written:

2 22

0 0 0

d d d3

a h

Oxyr z

hI z r r z m

π

αρ α

= = == = . (15.63)

We then deduce, from (15.54):

2 2

4 3Ox Oy

a hI I m

= = +

. (15.64)

15.4.3 Solid with a Spherical Symmetry

15.4.3.1 General Properties

In the case of a body with spherical symmetry (solid sphere, hollow sphere,

etc.) of centre O, any system (Oxyz) is principal trihedron of inertia, and the

moments of inertia with respect to the axes are equal. It is then more convenient to

calculate the moment of inertia IO with respect to the point O and to express the

moments by taking account of Relation (15.39). Thus:

2

3Ox Oy Oz OI I I I= = = . (15.65)

15.4.3.2 Matrix of Inertia of a Solid Sphere

We have to determine the matrix of inertia of a solid sphere of mass m and

radius a (Figure 15.9a). The calculation of the moment of inertia with respect to

(a)

y

z

x

O

r

Mz

(b)

x

yO

d

d V(M) = rddrdz

z

r

dr


FIGURE 15.9 Solid sphere.

the point O is simplified by using the spherical coordinates. The element of volume

is obtained by increasing the spherical coordinates by dR , dα and d ,β respecti-

vely. Thus:

2d ( ) cos d d dV M R Rβ α β= . (15.66)

The moment of inertia with respect to the point O is then expressed as follows:

22 4

0 02

cos d d da

OR

I R R

ππ

πα βρ β α β

= = =−= . (15.67)

In the case of a homogeneous sphere, we obtain:

5 24 3

5 5OI a maπ ρ= = , (15.68)

introducing the mass m of the sphere. We deduce from this the moments of inertia

with respect to the axes:

22

5Ox Oy OzI I I ma= = = . (15.69)

15.4.4 Associativity

In the case where a solid (S) is constituted of the union of several solids (Si), the matrix of inertia at a point is the sum of the matrices of inertia of each solid (Si) at this same point. This property is a consequence of the definition of the moments and products of inertia (property of integration over a domain) and allows us to split up the calculation in the case of complex solids. We thus have the relation:

( )( ) ( )( )1

nb b

O O i

i

S S

=

=I I . (15.70)

(a) (b)

z

y

M

O

R

x

x

yO

d

d V(M) = R

2cos dddR

z

R

d

R cos R cos d


FIGURE 15.10. Cylinder with a cylindrical cavity.

An example of application is that of the calculation of the matrix of inertia of a

cylinder with a cylindrical cavity (Figure 15.10). The solid cylinder (S1) can be

considered as the union of the cylinder with the cavity (S) and of the cylinder (S2)

which was removed. The property of associativity is written:

( )( ) ( )( ) ( )( )1 2b b b

O O OS S S= +I I I .

Hence the matrix of inertia of the cylinder with the cavity:

( )( ) ( )( ) ( )( )1 2b b b

O O OS S S= −I I I . (15.71)

The matrix of inertia of the cylinder (S1) of mass m1 is from Expressions (15.62)

and (15.64):

( )( )

2 2

1

2 2

1 1

2

1

0 04 3

0 04 3

0 02

bO

a hm

a hS m

am

+

= +

I . (15.72)

The matrix of inertia of the cylinder (S2) of mass m2 which was removed, is ex-

pressed from (15.25) by:

( )( )

2 2 2 2

2 2

2 2 2

2 2 2 2

22

2 22

0 0 0 016 3 4 4

0 0 016 3 4 4

00 04 48

bO

a h a hm m

a h h ahS m m m

ah aam mm

+ + = + + −

−

I

(15.73)

h

a

O

x

y

z


In addition, the masses of the cylinders are related to the mass m of the cylinder

hollowed out by the relations:

1 24

, 5 5

mm m m= = . (15.74)

Hence the matrix of the cylinder with the cavity:

( )( )

( )( )

2 2

2 2

2

113 0 0

20 4

150 3

20 4 20

020 4

bO

ma h

m mS a h ah

m aah m

+ = +

I . (15.75)

15.5 MATRICES OF INERTIA OF HOMOGENEOUS

BODIES

We collect in this section the matrices of inertia of various homogeneous solids.

The matrices of inertia are referred to the basis (b) = ( ), , i j k

associated for each

solid to the coordinate system chosen, generally the principal system of inertia.

15.5.1 One-Dimensional Solids

15.5.1.1 Straight Rod (Figure 15.11)

The length of the segment of line is AB = l. The mass centre is given by:

2

lAG i=

.

The matrix of inertia at the point A is:

( )( )2

2

0 0 0

0 03

0 03

bA

lS m

lm

=

I . (15.76)

FIGURE 15.11. Straight rod.

z

A BG

y

x

15.5 Matrices of Inertia of Homogeneous Bodies 245

FIGURE 15.12. Arc of circle.

15.5.1.2 Arc of Circle (Figure 15.12)

The arc of circle is characterized by its radius a and its angle 2α. The position

of the mass centre and the matrix of inertia are expressed by:

sinOG a i

αα

=

, (15.77)

( )( )

( )( )

2

2

2

sin 21 0 0

2 2

sin 20 1 0

2 2

0 0

bO

am

aS m

ma

αα

αα

−

= +

I . (15.78)

Particular cases

— Semicircle: 2

πα =

( )( )

2

2

2

0 02

2, 0 0

2

0 0

bO

am

a aOG i S m

ma

π

= =

I

. (15.79)

— Circle (case of a hoop): α π=The mass centre is at the point O and the matrix of inertia at O has the same

form as the semicircle.

15.5.2 Two-Dimensional Solids

15.5.2.1 Circular Sector (Figure 15.13)

As arc of circle, the circular sector is characterized by its radius and its angle

2α. The position of the mass centre and the matrix of inertia are expressed by:

a

x G

z

y

−O


FIGURE 15.13. Circular sector.

2 sin

3OG a i

αα

=

, (15.80)

( )( )

( )( )

2

2

2

sin 21 0 0

4 2

sin 20 1 0

4 2

0 02

bO

am

aS m

am

αα

αα

−

= +

I . (15.81)

Particular cases

— Half-disc: 2

πα =

( )( )

2

2

2

0 04

4, 0 0

3 4

0 02

bO

am

a aOG i S m

am

π

= =

I

. (15.82)

— Disc: α π=

The mass centre is at the point O and the matrix of inertia at O has the same

expression as that of a half-disc.

— Annulus limited by two concentric circles of radii a1 and a2.

The matrix of inertia is deduced from the property of associativity:

( )( )

( )

( )

( )

2 21 2

2 21 2

2 21 2

0 04

0 04

0 02

bO

ma a

mS a a

ma a

+ = +

+

I . (15.83)

a

G

−

x

y

z

O


FIGURE 15.14. Circular segment.

15.5.2.2 Circular Segment (Figure 15.14)

The circular segment is defined by its radius a and its angle 2α. The position of

the mass centre is given by:

32 sin

3 sin cosOG a i

αα α α

=−

. (15.84)

The coordinate system (Oxyz) is principal trihedron of inertia at O. The princi-

pal moments of inertia are:

2

2

2

2 1sin 2 sin 4

3 6 ,4 sin cos

1sin 4

2 ,4 sin cos

1 1sin 2 sin 4

3 6 .2 sin cos

Ox

Oy

Oz

aI m

aI m

aI m

α α α

α α α

α α

α α α

α α α

α α α

− +=

−

−=

−

− −=

−

(15.85)

15.5.2.3 Rectangle (Figure 15.15)

The mass centre is at the centre O of the rectangle. The matrix of inertia is:

( )( )

( )

2

2

2 2

0 012

0 012

0 012

bO

mb

mS a

ma b

=

+

I . (15.86)

15.5.2.4 Triangle (Figure 15.16)

The triangle is defined by:

, , .OA a i OB bi OC hj= − = =

(15.87)

a

G

−x

y

z

O


FIGURE 15.15. Rectangle.

The position of the mass centre is given by:

3 3

b a hOG i j

−= +

. (15.88)

The matrices of inertia at the point O and at the mass centre are expressed as:

( )( )

( )

( ) ( )

( )

2

2 2

2 2 2

06 12

012 6

0 06

bO

m mh h b a

m mS h b a a ab b

ma ab b h

− − = − − − +

− + +

I , (15.89)

( )( )

( )

( ) ( )

( )

2

2 2

2 2 2

018 36

036 18

0 018

bG

m mh h b a

m mS h b a a ab b

ma ab b h

− = − + +

+ + +

I . (15.90)

FIGURE 15.16. Triangle.

x

a

y

z

O b

A x

y

z

O B

C

G


Particular cases— Isosceles triangle: a = b

( )( )

( )

2

2

2 2

0 06

, 0 03 6

0 06

bO

mh

h mOG j S a

ma h

= =

+

I

. (15.91)

— Rectangle triangle: a = 0

( )( )

( )

2

2

2 2

06 12

, 03 3 12 6

0 06

bO

m mh hb

b h m mOG i j S hb b

mb h

− = + = −

+

I

. (15.92)

15.5.2.4 Ellipse (Figure 15.17)

The mass centre is at the centre of the ellipse, and the matrix of inertia at its

centre is expressed by:

( )( )

( )

2

2

2 2

0 04

0 04

0 04

bO

mb

mS a

ma b

=

+

I . (15.93)

15.5.3 Three-Dimensional Solids

15.5.3.1 Spherical Segment (Figure 15.18)

The spherical segment is located on the sphere of centre C and is defined by its

FIGURE 15.17. Ellipse.

x

y

z

O a

b


FIGURE 15.18. Spherical segment.

height h and the radius a of the sphere. Its volume is:

( )2 33

V h a hπ

= − , (15.94)

and its mass centre is defined by:

( )23 2

4 3

a hCG a k

a h

−=

−

. (15.95)

The system (Oxyz) is principal trihedron of inertia and the moments of inertia are:

( )

22

2 2

,3 4 20

2 3 3.

3 4 20

Ox Oy

Oz

m h ah hI I a

a h

hI m a ah h

a h

= = − +

−

= − +−

(15.96)

Particular cases

— Half-sphere

The points C and O coincide and the radius of the circle of the base is the

radius of the half-sphere.

23 2, .

8 5Ox Oy OzOG a k I I I ma= = = =

(15.97)

— Sphere

The matrix of inertia was derived in Subsection 15.4.3.2. Its expression is the

same as that of the half-sphere.

15.5.3.2 Circular Cone (Figure 15.19)

The cone is defined by its height h and the radius a of the circle of the base.

The mass centre and the matrix of inertia are given by:

3

4OG hk=

. (15.98)

h

a

z

G

y

x

O

C


FIGURE 15.19. Cone.

( )( )

( )

( )

2 2

2 2

2

34 0 0

203

0 4 020

30 0

20

bO

m a h

S m a h

ma

+

= +

I . (15.99)

15.5.3.3 Circular Cylinder (Figure 15.20)

Thee mass centre is at the centre of the cylinder and the matrix of inertia is:

( )( )

2 2

2 2

2

0 04 3

, 0 02 4 3

0 02

bO

a hm

h a hOG k S m

am

+

= = +

I

(15.100)

FIGURE 15.20. Cylinder.

h

a

y

x

z

O

G

h

y

x

z

O

a

G


15.5.3.4 Rectangle Parallelepiped (Figure 15.21)

The mass centre is at the centre of the parallelepiped:

2 2 2

a b cOG i j k= + +

, (15.101)

and the matrix of inertia at the mass centre is:

( )( )

( )

( )

( )

2 2

2 2

2 2

0 012

0 012

0 012

bG

mb c

mS a c

ma b

+ = +

+

I . (15.102)

The matrix of inertia at the point O, one of the vertices of the parallelepiped, is

deduced from the matrix at the centre by applying the relations of Huyghens. We

obtain:

( )( )

( )

( )

( )

2 2

2 2

2 2

3 4 4

4 3 4

4 4 3

bO

m m mb c ab ac

m m mS ab a c bc

m m mac bc a b

+ − − = − + −

− − +

I . (15.103)

FIGURE 15.21 Rectangle parallelepiped.

c

y

x

z

O

G

a

b

Exercises 253

EXERCISES

15.1 Derive the principal matrix of inertia at the centre of a rectangular plate of

low thickness (Figure 15.22). Next, deduce the moment of inertia with respect to

an axis (∆) contained in the plane of the plate and forming an angle θ with the axis

Ox

.

15.2 Express the matrix of inertia of a quarter of disc. Study the variation of the

moment of inertia with respect to an axis contained in the plane of the disc.

15.3 Derive the matrix of inertia of a homogeneous hollowed cylinder, of inner

radius a1, of outer radius a2 and of height h.

15.4 Derive the matrix of inertia of a solid (Figure 15.24) constituted of a cylin-

der of height h and of a solid half-sphere of radius a.

FIGURE 15.22. Rectangular plate.

FIGURE 15.23. Association of a cylinder

and a solid half-sphere.

y

x

a

b O

()

(S1)

z

h

a

O

(S2)

y

x


15.5 Derive the matrix of inertia of a non homogeneous parallelepiped (Figure

15.25) constituted of four parallelepipeds with edges 2a, b, c, and with respective

masses m1 and m2. From this, deduce the moment of inertia with respect to a

diagonal.

15.6 Express the matrix of inertia of a solid sphere with a spherical hole of half-

radius, passing through the centre of the sphere.

15.7 Determine the matrix of inertia of a homogeneous rectangular plate of length

a and width b, with a circular hole at its centre of radius c ( c < b/2 ).

FIGURE 15.24. Non homgeneous rectangular prism.

COMMENTS

The operator of inertia is used to formulate the expressions of the kinetic

torsor and dynamic torsor which will be considered in the following

chapter. The use of the operator of inertia is particularly important. This

operator is represented in a given basis attached to the body considered by

the symmetric matrix of inertia 3 × 3, of which the diagonal terms are the

moments of inertia of the body with respect to three trirectangular axes and

the other terms are the products of inertia of the body with respect to three

perpendicular planes. The reader must have a perfect knowledge of all the

concepts considered in the present chapter.

2c

y

x

z

2a

2b

Om1

m1 m2

m2

CHAPTER 16

Kinetic and Dynamic Torsors Kinetic Energy

16.1 KINETIC TORSOR

16.1.1 Definition

The study of dynamics introduced a first torsor which allows us to express the

kinetic energy (Section 16.3) and whose concept is also used within the frame of

the theory of impacts. This torsor called kinetic torsor, denoted by ( ) T

D and

associated to the motion of a material set (D) relatively to a reference (T), is

defined over the set (D) in the following way.

We call kinetic torsor relative to the motion of the material set (D) with respect

to the reference (T), the torsor defined on this set and associated to the field of

sliders of which the vector density (relative to the mass) in every point of the

material set (D) is equal to the velocity vector of this point with respect to the

reference (T).

The kinetic torsor ( ) T

D is thus associated to the field of sliders (Section

5.3.2.) of resultants:

( )M D∀ ∈ ( )

d ( ) ( , ) d ( )TR M M t m M=

. (16.1)

From (5.54) and (5.55), the elements of reduction of the kinetic torsor at a

point P of the reference (T) are:

( ) ( )

( )( , ) d ( )T T

DD

R M t m M= , (16.2)

( ) ( )

( )( , ) d ( )T T

P DD

PM M t m M= × . (16.3)

256 Chapter 16 Kinetic and Dynamic Torsors. Kinetic Energy

FIGURE 16.1. Motion of the set (D) relatively to a given reference.

16.1.2 Kinetic Torsor Associated to the Motion of a Body

Consider a solid (S) in motion relatively to the reference (T). This motion is

characterized (Chapter 9) by the motion of a particular point P of the solid (S) (or

attached to the solid) and by its rotation vector ( )TSω

. We established (9.11):

( ) ( ) ( ) ( , ) ( , )T T T

SM t P t PMω= + ×

.

The elements of reduction of the kinetic torsor at the point P of the solid (S) are

derived by substituting this expression into Relations (16.2) and (16.3).

1. Resultant of the kinetic torsor


( ) ( ) ( )

( ) ( , ) d ( )T T T

S SS

R P t PM m Mω = + × .

Hence:

( ) ( )

( )

( )

( )

( , ) d ( ) d ( )T T TS S

S S

R P t m M PM m Mω= + × .

Thus, by introducing the mass centre and the mass of the solid (Relation (12.23)),

we obtain:

( ) ( ) ( )( , )T T T

S SR m P t PGω = + × . (16.4)

Or:

( ) ( )( , )T T

SR m G t= . (16.5)

Hence the result:

The resultant of the kinetic torsor associated to the motion of a solid relatively

to a given reference is equal to the product of the mass of the solid by the velocity

vector, in this reference, of the mass centre of the solid.

y

z

x

O

(T) dm(M)

M (D)

P

16.1 Kinetic Torsor 257

2. Moment vector of the kinetic torsor


( )

( )

( ) ( )( )( )

d ( ) ( , ) d ( )T T TP S S

S S

PM m M P t PM PM m Mω

= × + × ×

.

The mass centre can be introduced in the first integral. The second integral is

expressed by introducing the operator of inertia at point P of the body (S). The

moment vector at the point P is thus written in the form:

( ) ( ) ( ) ( ) ( , )T T T

P S P Sm PG P t S ω= × + . (16.6)

This expression is simplified, when the moment-vector is expressed at the mass

centre (P coincides with G):

( ) ( ) ( )T TG S G SS ω=

. (16.7)

This simplification confirms the importance of the concept of the mass centre.

16.1.3 Kinetic Torsor for a Set of Bodies

We call kinetic torsor associated to the motion, relatively to a given reference,

of a set (D), constituted of solids ( ) ( ) ( ) 1 2, , . . . , ,nS S S the torsor obtained by

doing the sum of the kinetic torsors associated to the motions of every solid

relatively to the reference under consideration.

We thus have the relation:

( ) ( ) 1

i

nT T

D S

i=

= . (16.8)

This relation leads to the expressions of the elements of reduction of the kinetic

torsor:

( ) ( ) ( )

1 1

( , )i

n nT T T

D i iS

i i

R R m G t

= =

= = .

Thus:

( ) ( )( , )T T

DR m G t= , (16.9)

where m and G are respectively the mass and the mass centre of the set (D). We

find again the same expression of the resultant as in the case of only one solid

(Relation (16.5)). The moment vector at a point P of reference is written in the

form:

( ) ( ) ( ) ( ) ( )1 1

ii i i

n nT T T T

P P GD iS S S

i i

R G P

= =

= = + ×

,


or introducing Relations (16.5) and (16.7) for every solid:

( ) ( ) ( ) ( )

1

( , )i i

nT T T

P D i i i G i S

i

m PG G t S ω=

= × + . (16.10)

16.2 DYNAMIC TORSOR

16.2.1 Definition

The fundamental principle of the dynamics (Chapter 18) utilizes the dynamic

torsor, denoted by ( ) T

D and defined over the material set as follows.

We call dynamic torsor relative to the motion of the material set (D) with

respect to the reference (T), the torsor defined over this set and associated to the

field of sliders of which the vector density (relative to the mass) in every point of

the material set (D) is equal to the acceleration vector of this point with respect to

the reference (T).

The dynamic torsor ( ) T

D is thus associated to the field of sliders of resul-

tants:

( )M D∀ ∈ ( )

d ( ) ( , ) d ( )TR M a M t m M=

. (16.11)

From (5.54) and (5.55), the elements of reduction at a point P of the dynamic

torsor are:

( ) ( )

( )( , ) d ( )T T

DD

R a M t m M= , (16.12)

( ) ( )

( )( , ) d ( )T T

P DD

PM a M t m M= × . (16.13)

16.2.2 Dynamic Torsor Associated to the Motion of a Body

Consider a solid (S) in motion relatively to the reference (T). We established

(9.24): ( ) ( ) ( ) ( ) ( )( ) ( , ) ( , )T T T T T

S S Sa M t a P t PM PMω ω ω= + × + × × .

The elements of reduction at a particular point P of the solid (S) are derived

while substituting this expression into Relations (16.12) and (16.13).

1. Resultant of the dynamic torsor


( ) ( ) ( ) ( ) ( )( )( )

( , ) d ( )T T T T TS S S S

S

R a P t PM PM m Mω ω ω = + × + × × ,

16.2 Dynamic Torsor 259

or in introducing the mass centre G and the mass m of the solid (Relation (12.23)):

( ) ( ) ( ) ( ) ( )( )( , )T T T T TS S S SR m a P t PG PGω ω ω = + × + × ×

. (16.14)

Thus: ( ) ( )

( , )T TSR ma G t=

. (16.15)

Hence the fundamental result:

The resultant of the kinetic torsor associated to the motion of a solid relatively

to a given reference is equal to the product of the mass of the solid by the

acceleration vector, in this reference, of the mass centre of the solid.

2. Moment-vector of the dynamic torsor


( ) ( )

( )

( )( )( )

( ) ( )( )( )

d ( ) ( , )

d ( )

d ( ).

T TP S

S

TS

S

T TS S

S

PM m M a P t

PM PM m M

PM PM m M

ω

ω ω

= ×

+ × ×

+ × × ×

The first integral introduces the mass centre, and the two last integrals are

expressed by introducing the operator of inertia at P of the solid (S). Hence:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( , )T T T T T

P S P S S P Sm PG a P t S Sω ω ω= × + + × . (16.16)

This expression is simplified, when the moment vector is expressed at the mass

centre (P coincides with G):

( ) ( ) ( ) ( ) ( ) ( )T T T TG S G S S G SS Sω ω ω= + ×

. (16.17)

16.2.3 Dynamic Torsor for a Set of Solids

We call dynamic torsor associated to the motion, relatively to a given

reference, of a set (D), constituted of solids ( ) ( ) ( ) 1 2, , . . . , ,nS S S the torsor

obtained by doing the sum of the dynamic torsors associated to the motions of

every solid relatively to the reference under consideration.

We thus have:

( ) ( ) 1

i

nT T

D S

i=

= . (16.18)

The resultant is written:

( ) ( ) ( )

1

( , )i

nT T T

D i iS

i

R R m a G t

=

= = .


Thus:

( ) ( )( , )T T

DR m a G t= , (16.19)

where m and G are the mass and the mass centre of the set (D). In the same way,

the moment at the point P is expressed as:

( ) ( ) 1

i

nT T

P PD S

i=

=

, (16.20)

where the moments of the dynamic torsors of every solid (Si) can be expressed at

the mass centres Gi:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( , )

i ii i i i

T T T T TP i i i G G iS S S S

m PG a G t S Sω ω ω= × + + × . (16.21)

16.2.4 Relation with the Kinetic Torsor

Expressions (16.5) and (16.15) show that:

( ) ( )

( ) d

d

TT T

S SR Rt

=

. (16.22)

Moreover, it can be verified that:

( ) ( )

( ) d

d

TT T

P PS St

=

. (16.23)

These two relations can be expressed in the form:

( ) ( )

( ) d

d

TT T

S St

= . (16.24)

Hence the result:

The dynamic torsor associated to the motion of a solid relatively to a given

reference is the torsor derivative, with respect to time and relatively to this

reference, of the kinetic torsor.

16.3 KINETIC ENERGY

16.3.1 Definition

We call kinetic energy of a set (D) relatively to the reference (T), the integral:

( ) ( )

( )

2

c1

( , ) d ( )2

T T

D

E M t m M = . (16.25)

16.3 Kinetic Energy 261

16.3.2 Kinetic Energy of a Solid

In the case where the set is a solid (S), the velocity vector at any point M of the

solid is expressed (Relation (9.11)) as a function of the velocity vector of a parti-

cular point P of the solid as:

( ) ( ) ( ) ( , ) ( , )T T T

SM t P t PMω= + ×

.

Thus:

( )

( ) ( ) ( )( ) ( )( ) ( )( )( ) ( ) ( )( ) ( ) ( )( )

2

2

2

( , )

( , ) 2 ( , )

( , ) 2 ( , ) ,

T

T T T T TS S S

T T T T TS S S

M t

P t P t PM PM PM

P t P t PM PM PM

ω ω ω

ω ω ω

= + × + × ×

= + × + × ×

⋅ ⋅

⋅ ⋅

from the property of the double vector product. Relation (16.25) is thus written:

( ) ( )

( )

( ) ( )

( )

( ) ( )( )( )

2

c1( ) ( , ) d ( ) ( , ) d ( )2

1 d ( ).2

T T T TS

S S

T TS S

S

E S P t m M P t PM m M

PM PM m M

ω

ω ω

= + ×

+ × ×

⋅

⋅

By introducing the mass, the mass centre and the operator of inertia at P of the

solid (S), the preceding expression of the kinetic energy is written as:

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

2

c1 1( ) ( , ) ( , ) .2 2

T T T T T TS S P SE S m P t m P t PG Sω ω ω = + × + ⋅ ⋅

(16.26)

This relation is simplified when the point P coincides with the mass centre

according to:

( ) ( ) ( ) ( ) ( )2

c1 1( ) ( , )2 2

T T T TS G SE S m G t Sω ω = + ⋅

. (16.27)

The first term constitutes the kinetic energy of translation of the solid, the second

term represents the kinetic energy of rotation of the solid.

It is possible to find another practical form of the kinetic energy, by noticing

that Expression (16.27) can be written as:

( ) ( ) ( ) ( ) ( ) c1( )2

T T T T TG GS S S SE S R R = + ⋅ ⋅

. (16.28)

Finally, we thus have:

( ) ( ) ( ) c1( )2

T T TS SE S = ⋅ . (16.29)

Hence the result:

In a given reference, the kinetic energy of a solid is equal to half of the scalar

product of the kinetic torsor and the kinematic torsor, expressed in this reference.


16.3.3 Kinetic Energy of a Set of Solids

The kinetic energy of a set (D), constituted of the solids ( ) ( )1 2, ,S S ( ) ... , ,nS

is, relatively to a given reference, the sum of the kinetic energies of every solid

relatively to this reference:

( ) ( )c c

1

( ) ( )

nT T

i

i

E D E S

=

= . (16.30)

16.3.4 Derivative of the Kinetic Energy of a Solid with respect to Time

Differentiating Expression (16.25) for the definition of the kinetic energy in the

case of a solid, we obtain:

( )

( ) ( )

( )

( ) ( ) ( )

cd

( )d

( , ) ( , ) d ( ) ( , ) d ( ).

T

T T T TM S

S S

E St

a M t M t m M a M t m M

=

=⋅ ⋅

By introducing a particular point P of the solid (S), we thus have:

( )

( ) ( ) ( ) ( )

( ) ( )

( )

( ) ( )

( )

( ) ( ) ( ) ( )

cd

( )d

( , ) d ( )

( , ) d ( ) ( , ) d ( )

.

T

T T TP S S

S

T T T TP S S

S S

T T T TP PS S S S

E St

a M t R PM m M

a M t m M R PM a M t m M

R R

=

= + ×

= + ×

= +

⋅

⋅ ⋅

⋅ ⋅

(16.31)

Hence the result:

( ) ( ) ( ) cd

( )d

T T TS SE S

t= ⋅ . (16.32)

A second expression can be obtained while differentiating Relation (16.29).

We obtain:

( )( )

( ) ( ) ( ) ( )

( )

cd d d

2 ( )d d d

T TT T T T T

S S S SE St t t

= +⋅ ⋅ ,

and taking account of Relations (16.24) and (16.32), we have:

( ) ( ) ( )

( )

cd d

( )d d

TT T T

S SE St t

= ⋅ . (16.33)

Exercises 263

FIGURE 16.2. Motion of rotation of a parallelepiped: a) about an axis passing through its

centre and b) about an eccentric axis.

EXERCISES

16.1 Express the kinetic torsor, the dynamic torsor and the kinetic energy of a

homogeneous rectangular parallelepiped, for a motion of rotation about an axis

passing through its centre (Figure 16.2a).


homogeneous rectangular parallelepiped, for a motion of rotation about an ec-

centric axis (Figure 16.2b).


homogeneous rectangular parallelepiped, for a motion on a plane (Figure 16.3).

FIGURE 16.3. Motion of a parallelepiped on a plane.

d

(a) (b)


COMMENTS

Kinetics combines the effects of motions with the repartition of masses

over a body or a set of bodies.

The first torsor which was considered is the kinetic torsor which asso-

ciates the effects of masses and velocities. This torsor allows us to express

easily the kinetic energy of a solid.

The second torsor which was studied is the dynamic torsor which com-

bines the effects of masses and accelerations. This torsor is used in the

statement of the fundamental principle of the dynamics (Chapter 19).

The notations which are used for the kinetic torsor( ) T

S and the

dynamic torsor ( ) ,T

S associated to the motion of the solid (S) relatively

to the reference (T) are similar to the notation introduced for denoting the

kinematic torsor ( ) .T

S

The concepts introduced in the present chapter are very important. The

reader will pay a great attention to the way in which these two torsors are

built. To apply the concepts introduced by these torsors, the reader will

have to know the expressions which express the resultant (16.5) and the

moment (16.6) and (16.7), of the kinetic torsor, as well as the expressions

which give the resultant (16.15) and the moment (16.16) and (16.17), of the

dynamic torsor.

These concepts are applied (Exercises) to the case of three motions of

the same rectangular parallelepiped, allowing us to highlight simply the

importance of the conditions of motions.

CHAPTER 17

Change of Reference System

We consider in this chapter the case of a solid (S), whose we study the motions (Figure 17.1) with respect to a reference system ( )1 1 1(1) Ox y z= and with respect to a reference system ( )2 2 2(2) .Ox y z= The two systems of reference (1) and (2) are in motion the one with respect to the other. M is an arbitrary point of the solid (S).

17.1 KINEMATICS OF CHANGE OF REFERENCE

17.1.1 Relation between the Kinematic Torsors

The motion of the solid (S) relatively to the reference (1) is characterized by its

kinematic torsor ( ) 1S , of elements of reduction at point M of the solid (S):

FIGURE 17.1. Change of reference system.

y1

z1

x1

O1

(1)

(S)

M

y2

z2

x2

O2

(2)

266 Chapter 17 Change of Reference System

( ) ( )

( ) ( )

( ) ( ) ( )( )

1 1

1 1

, rotation vector relative to the motion ofthe solid with respect to reference 1 ;

( , ) , velocity vector with respect to 1of the point of the solid .

S S

M S

RS

M tM S

ω =

=

(17.1)

The motion of the solid (S) relatively to the reference (2) is characterized in the

same way by its kinematic torsor ( ) 2S of elements of reduction at the point M:

( ) ( )

( ) ( )

( ) ( ) ( )( )

2 2

2 2

, rotation vector relative to the motion ofthe solid with respect to reference 2 ;

( , ) , velocity vector with respect to 2of the point of the solid .

S S

M S

RS

M tM S

ω =

=

(17.2)

The relative motions of references (1) and (2) are characterized, for example,

by the kinematic torsor ( ) 12 associated to the motion of reference (2) relatively

to reference (1). Its elements of reduction at the point O2 of the reference (2) are:

( ) ( )

( ) ( )

( ) ( ) ( )( )

2

1 12 2

1 12 2

2

, rotation vector relative to the motion ofreference 2 with respect to reference 1 ;

( , ) , velocity vector with respect to 1of the point of reference 2 .

O

R

O tO

ω =

=

(17.3)

From the law of composition of motions (9.39), we have:

( ) ( ) ( ) 1 2 12S S= + . (17.4)

motion of solid (S) motion of solid (S) motion of reference (2)

with respect to reference (1) with respect to reference (2) with respect to reference (1)

The composition of the rotation vectors are deduced from the preceding law of

composition. Hence:

( ) ( ) ( )1 2 12S Sω ω ω= +

. (17.5)

17.1.2 Relation between the Velocity Vectors. Velocity of Entrainment

The relation between the moment-vectors deduced from (17.4) must be

expressed at the same point. Consider the point M of the solid (S). The relation is

thus written: ( ) ( ) ( ) 1 2 1

2MM M PS S= +

. (17.6)

In the formulation of this relation, the moment of the torsor relatively to the motion

of reference (2) with respect to reference (1) is expressed at a point which has:

— to coincide with the point M of the solid (S),

— to belong to reference (2).

17.1 Kinematics of Change of Reference 267

FIGURE 17.2 Coinciding point.

We have denoted it by .MP This point belongs to reference (2) and coincides at

time t with the point M of the solid (S). It is called coinciding point. The point

MP belonging to reference (2) and which coincides at time t with the point M is

not however identical to the point M (Figure 17.2).


( ) ( ) ( )

1 2 12( , ) ( , ) MPM t M t= +

. (17.7)

We notice that ( ) 12MP

is identified with the velocity vector ( )1 ( , )M t with

respect to reference (1), when the point P is motionless relatively to reference (2).

This vector is called the velocity of entrainment of the point M relatively to the

motion of reference (2) with respect to reference (1). This velocity is denoted by ( )12 ( , )e M t . Hence:

( ) ( )1 12 2 ( , )MP e M t=

. (17.8)

Relation (17.7) is put thus in the form:

( ) ( ) ( )

1 2 12( , ) ( , ) ( , )eM t M t M t= +

. (17.9)

The velocity vector of entrainment can be expressed as a function of the velo-

city vector with respect to reference (1) of any point P2 attached to reference (2):

( ) ( ) ( ) ( ) 2

1 1 1 12 2 2 2 2( , ) MP P Me M t R P P= = + ×

.

Thus: ( ) ( ) ( )

1 1 12 2 2 2( , ) ( , ) Me M t P t P Pω= + ×

. (17.10)

O1

M

reference (1)

trajectory of point M

in reference (2 )

points of

reference (2)

O2

x1

z1

z2

y1

y2

x2

(S)

PM

PMPM


In the case where the point P2 coincides with the origin O2 of reference (2), the

preceding relation is written:

( ) ( ) ( )

1 1 12 2 2 2( , ) ( , ) ,Me M t O t O Pω= + ×

(17.11)

or ( ) ( ) ( )

1 1 12 2 2 2( , ) ( , )e M t O t O Mω= + ×

. (17.12)

Relation (17.7) between the velocity vectors relatively to the two references

may be finally written in the form:

( ) ( ) ( ) ( )

1 2 1 12 2 2( , ) ( , ) ( , )M t M t O t O Mω= + + ×

. (17.13)

From the definition of the velocity vectors, we can write:

( )( ) ( ) ( )

1 1 11

21 1 2d d d

( , )d d d

M t O M O O O Mt t t

= = +

,

or

( ) ( )( )

11 1

2 2d

( , ) ( , )d

M t O t O Mt

= +

. (17.14)

The comparison of Relations (17.13) and (17.14) shows thus the following

property: ( )

( ) ( )1

2 12 2 2

d( , )

dO M M t O M

tω= + ×

. (17.15)

17.1.3 Composition of Acceleration Vectors

17.1.3.1 Relation

The acceleration vector relatively to reference (1) of the point M is:

( )( )

( )

( )( ) ( ) ( )

11 1

12 1 1

2 2 2

d( , ) ( , )

d

d( , ) ( , ) .

d

a M t M tt

M t O t O Mt

ω

=

= + + ×

(17.16)

— ( )

( )1

2d( , )

dM t

t

is obtained by analogy with Relation (17.15):

( )( ) ( ) ( ) ( )

12 2 1 2

2d

( , ) ( , ) ( , )d

M t a M t M tt

ω= + ×

,

— ( )

( ) ( )

11 1

2 2d

( , ) ( , )d

O t a O tt

= is the acceleration vector relatively to reference

(1) of the point O2,

17.2 Dynamic Torsors 269

( )( )( )

( )( ) ( )

( )

( ) ( ) ( ) ( ) ( )( )

1 1 11 1 1

2 2 2 2 2 2

1 1 2 1 12 2 2 2 2 2

d d d

d d d

( , ) .

O M O M O Mt t t

O M M t O M

ω ω ω

ω ω ω ω

−− ∧ = × + ×

= × + × + × ×

Hence the relation giving the acceleration vector:

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )( )

1 2 1 22

1 1 1 12 2 2 2 2 2

( , ) ( , ) 2 ( , )

( , ) .

a M t a M t M t

a O t O M O M

ω

ω ω ω

= + ×

+ + × + × ×

(17.17)

17.1.3.2 Acceleration of entrainment

The acceleration of entrainment, denoted by ( )1

2( , )ea M t

is the acceleration of

the point MP of reference (2) coinciding at the time considered with the point M:

( ) ( )

1 12 ( , ) ( , )Mea M t a P t=

. (17.18)

From Expression (9.24), the acceleration of entrainment of the point M is expres-

sed as a function of the acceleration vector of the point O2 by the relation:

( ) ( ) ( ) ( ) ( )( )

1 1 1 1 12 2 2 2 2 2 2( , ) ( , )ea M t a O t O M O Mω ω ω= + × + × ×

. (17.19)

17.1.3.3 Acceleration of Coriolis

The acceleration of Coriolis is the term ( ) ( )

1 222 ( , )M tω ×

introduced in Rela-

tion (17.17). Thus: ( ) ( ) ( )

1 1 22 2( , ) 2 ( , )ca M t M tω= ×

. (17.20)

The acceleration of Coriolis is null when the velocity vector of the point M is null

relatively to reference (2).

17.1.3.4 Composition of Accelerations

Relation (17.17) expressing the acceleration vector of the point M may thus be

written in the form:

( ) ( ) ( ) ( )

1 2 1 12 2( , ) ( , ) ( , ) ( , )e ca M t a M t a M t a M t= + +

. (17.21)

This relation constitutes the relation of composition of the acceleration vectors of

a point M in motion relatively to the references (1) and (2).

17.2 DYNAMIC TORSORS

In this section, we study how the dynamic torsor of a solid (S) is transformed

when we change a reference (1) for a reference (2). This change of reference


introduces two new torsors: the inertia torsor of entrainment and the inertia

torsor of Coriolis.

17.2.1 Inertia Torsor of Entrainment

The inertia torsor of entrainment of the solid (S) relatively to the motion of the

reference (2) with respect to the reference (1), denoted by ( )( ) 1

2 ,S is the torsor

defined over this solid and associated to the field of sliders whose the vector

density (relative to the mass) at every point of the solid is equal to the acce-

leration vector of entrainment of this point relatively to the motion of the

reference (2) with respect to the reference (1).

The inertia torsor of entrainment ( )( ) 1

2 ,S is thus associated to the field of

sliders of resultants:

( )M S∀ ∈ ( )

12d ( ) ( , )d ( )eR M a M t m M=

, (17.22)

where the acceleration vector of entrainment is expressed by Relation (17.19).

The elements of reduction of the inertia torsor of entrainment are thus:

1. Resultant

( )( ) ( )

( )

1 12 2 ( , ) d ( )

SeR S a M t m M=

. (17.23)

Hence:

( )( ) ( ) ( ) ( ) ( )( )1 1 1 1 12 2 2 2 2 2 2( , )R S m a O t O G O Gω ω ω = + × + × ×

,

or

( )( ) ( )1 12 2 ( , )eR S ma G t=

. (17.24)

2. Moment-vector

Expression (17.19) of the acceleration vector of entrainment of a point M of

the solid leads to express the moment-vector at a point attached to the reference

(2), the point O2 for example:

( )( ) ( )

( )2

1 12 2 2 ( , ) d ( )O

SeS O M a M t m M= ×

. (17.25)

Thus:

( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2

1 1 1 1 12 2 2 2 2 2( , )O O OS m O G a O t S Sω ω ω= × + + ×

, (17.26)

where the operator of inertia ( )2O S has its representative matrix defined from

the matrix of inertia at the mass centre by applying the relations of Huyghens

(Relations (15.25)).

17.2 Dynamic Torsors 271

17.2.2 Inertia Torsor of Coriolis

The inertia torsor of Coriolis of the solid (S) relatively to the motion of the

reference (2) with respect to the reference (1), denoted by ( )( ) 1

2 S , is the torsor

defined over this solid and associated to the field of sliders whose the vector

density (relative to the mass) at every point of the solid is equal to the acce-

leration vector of Coriolis of this point relatively to the motion of the reference

(2) with respect to the reference (1).

This torsor is thus associated to the field of sliders of resultant:

( )M S∀ ∈ ( )

12d ( ) ( , )d ( )cR M a M t m M=

, (17.27)

where the acceleration vector of Coriolis is expressed by (17.20). In this relation,

the velocity vector ( )2 ( , )M t can be expressed as a function of the velocity

vector of a particular point of the solid (S), for example the mass centre. The ace-

leration vector of Coriolis has then for expression:

( ) ( ) ( ) ( )

1 1 2 22 2( , ) 2 ( , )c Sa M t G t GMω ω = × + ×

. (17.28)

The elements of reduction of the inertia torsor of Coriolis are:

1. Resultant

( )( ) ( )

( )

1 12 2 ( , ) d ( )

ScR S a M t m M=

. (17.29)

Hence from (17.28): ( )( ) ( ) ( )

1 1 22 22 ( , )R S m G tω= ×

, (17.30)

or ( )( ) ( )1 1

2 2 ( , )cR S ma G t= . (17.31)

2. Moment-vector

Expression (17.28) of the acceleration vector of Coriolis of a point M of the

solid leads to express the moment vector at the mass centre of the solid:

( )( ) ( )

( )

1 12 2 ( , ) d ( )G

ScS GM a M t m M= ×

. (17.32)

Thus:

( )( ) ( ) ( )( )( )

1 1 22 22 d ( )G S

S

S GM GM m Mω ω = × × × . (17.33)

Considering the property of the double vector product, we have: ( ) ( )( ) ( )( ) ( )1 2 1 22 2S SGM GM GM GMω ω ω ω × × × = × ⋅

,

and ( )( ) ( ) ( ) ( )( )1 1 12 2 2GM GM GM GM GM GMω ω ω× × = −⋅ ⋅

.


Finally:

( ) ( )( )( ) ( )( ) ( )( ) ( )

1 22

2 1 2 1 22 2 .

S

S S

GM GM

GM GM GM

ω ω

ω ω ω ω

× × ×

= × − × × ×

The moment at the mass centre (17.33) is thus written as:

( )( ) ( )

( ) ( ) ( ) ( ) ( )21 1 2 2 12 2 22 d ( ) 2G S S G

S

S GM m M Sω ω ω ω = × + × .

In this expression, the integral is the moment of inertia of the solid (S) with

respect to the mass centre: IG(S). The moment of the inertia torsor of Coriolis is

thus written:

( )( ) ( ) ( ) ( ) ( ) ( )1 1 2 2 12 2 22 ( ) 2G G S S GS I S Sω ω ω ω= × + ×

. (17.34)

Note. The inertia torsors of entrainment and of Coriolis can be generalized to the

case of a system of solids by doing the sum of torsors defined on every solid.

17.2.3 Relation between the Dynamic Torsors Defined Relatively to Two Different References

Express the dynamic torsors relative to the motions of the solid (S) with

respect to two references (1) and (2). We have from (16.14):

— motion of (S) relatively to (1), dynamic torsor ( ) 1

S with:

( ) ( )1 1( , )SR ma G t= , (17.35)

— motion of (S) relatively to (2), dynamic torsor ( ) 2

S with:

( ) ( )2 2 ( , )SR ma G t= . (17.36)

It results from the law of composition of the accelerations (Relation (17.21)) and

from Expressions (17.24) and (17.31) that:

( ) ( ) ( )( ) ( )( ) 1 2 1 12 2S SR R R S R S= + +

. (17.37)

A similar relation exists with the moment vectors. Hence the expression of

reference change for the dynamic torsors:

( ) ( ) ( )( ) ( )( ) 1 2 1 12 2S S S S= + + . (17.38)

This expression gives the relation which exists between the dynamic torsors rela-

tively to the motions of a same solid (S) with respect to the reference (1) and with

respect to the reference (2).

Comments 273

COMMENTS

This chapter is interested in the motions of a solid (S) with respect to

two reference systems (1) and (2). Within this context, the concept of coin-

ciding point and the concept of velocity of entrainment are important

concepts. The expression of the velocity vector of entrainment is obtained

simply by using the concept of kinematic torsor associated to the motion of

one of the references with respect to the other. The relation between the

dynamic torsors associated to the motions of the solid (S) with respect to

the reference (1) and to the reference (2) leads to introduce the inertia

torsors of entrainment and of Coriolis. The concepts of these torsors can

appear somewhat complex to handle. In fact, these concepts are necessary

to formalize the fundamental law of dynamics expressed in different

reference systems (Chapter 19). In practice, these concepts will be

introduced implicitly, without using the relations established in the present

chapter, when the dynamic torsors will be expressed directly with respect to

the different reference systems.

Thus, in a first approach, the reader will simply endeavour the deve-

lopment of the various concepts which are introduced. These notions will

be then studied thoroughly and progressively with the development of the

knowledge of the reader.

Part V

Dynamics of Rigid Bodies

The motions of bodies having been characterized, the mechanical

actions having been analyzed, it now remains to understand how the

mechanical actions act to produce such or such motion. This problem

is solved starting from the fundamental principle of dynamics. This

principle will be applied to the analysis of various elementary motions.

The application of the fundamental principle gives access to the

nature of the motions as well as to the characteristic parameters of the

actions induced by the connections. The application of the equations

of Lagrange will allow us to acquire a systematic tool to derive the

equations of motions.

CHAPTER 18

The Fundamental Principle of Dynamics and its Consequences

18.1 FUNDAMENTAL PRINCIPLE

18.1.1 Statement of the Fundamental Principle of Dynamics

There exists at least a reference system, called Galilean reference, which will be

denoted by (g), such as, at any moment and for any material set (D), the dynamic

torsor associated to the motion of the set (D) relatively to this reference system is

equal to the torsor of mechanical actions which are exerted on the set (D).

This principle is written in the form:

( ) ( )g

D D= , (18.1)

where ( )D is the torsor which represents the whole of the mechanical actions

exerted on the set (D).

The fundamental principle of dynamics, justified by the agreement between the

theoretical results derived from this principle and the experimental results obser-

ved, states the existence of at least a Galilean reference system. We clarify, in the

following subsection, the relations which exist between the Galilean references.

18.1.2 Class of Galilean Reference Systems

Let (1) and (2) be two Galilean references (Figure 18.1). We derived in the

case of the motions of a solid (S) relatively to two references (Relation (17.38)):

( ) ( ) ( )( ) ( )( ) 1 2 1 12 2S S S S= + + .

So that the fundamental relation (18.1) has the same form relatively to the two

278 Chapter 18 The Fundamental Principle and its Consequences

FIGURE 18.1. Galilean reference systems.

references, the torsor of the mechanical actions being unchanged while passing

from the reference (1) to the reference (2), it is necessary and sufficient that:

( )( ) 12 0S = , (18.2)

( )( ) 12 0S = . (18.3)

Expressions (17.30) and (17.34) of the elements of reduction of the torsor of

Coriolis at the mass centre shows that the second relation (18.3) is satisfied if and

only if ( )12 0ω =

. Hence the first result: two Galilean references are animated the

one with respect to the other by a motion of translation.

Expressions (17.24) and (17.26) of the elements of reduction of the torsor of

entrainment at the point O2 shows then that the first relation (18.2) is satisfied if:

( )12( , ) 0a O t =

. (18.4)

The velocity vector at the point O2 is thus a constant vector: the motion of the

reference (2) is a motion of uniform rectilinear translation with respect to the

reference (1).

This result shows that the fundamental principle of dynamics admits the

existence of a class of Galilean references:

There exists an infinity of Galilean references. Two Galilean references move

the one with respect to the other with a motion of uniform rectilinear translation.

The fundamental principle has an invariant expression (18.1) relatively to any of

these Galilean references.

18.1.3 Vector Equations Deduced from the Fundamental Principle

The fundamental relation of dynamics (18.1) leads to the two vector equations:

y1

z1

x1

O1

(1) x2

y2

z2

O2

(2)

(S)

18.1 Fundamental Principle 279

( ) ( )g

DR R D=

, (18.5)

( ) ( )g

P PD D=

, (18.6)

where P is an arbitrary reference point.

1. Equation of the resultantThe resultant of the dynamic torsor expressed in (16.19) leads to rewrite Equa-

tion (18.5) of the resultant in the form:

( ) ( , ) ( )g

ma G t R D= . (18.7)

This result is usually stated in the form:

The general resultant of the mechanical actions exerting on a material set is

equal to the product of the mass of this set by the acceleration vector of the mass

centre of this set relatively to a Galilean reference system.

2. Equation of the momentIn the case of the motion of a solid (S), the equation of the moment is written,

from (16.16):

( ) ( ) ( ) ( ) ( ) ( ) ( , ) ( )g g gg

PP PS S Sm PG a P t S S Sω ω ω× + + × = . (18.8)

This expression is simplified, when the equation of the moment is expressed at the

mass centre:

( ) ( ) ( ) ( ) ( ) ( )g g g

GG GS S SS S Sω ω ω+ × = . (18.9)

In the case of a set of bodies, the moment of the dynamic torsor is expressed

according to Relation (16.20).

Moreover, the equation of the moment may be expressed while taking account

of Relation (16.23) as a function of the kinetic torsor in the form:

( )

( ) d( )

d

gg

G G SSt

=

. (18.10)

18.1.4 Scalar Equations Deduced from the Fundamental Principle

The vector equations of the resultant and of the moment lead each one to 3

scalar equations, that is a total of 6 scalar equations for a given material set. The

choice of the bases (which can be different) to express the equation of the

resultant and that one of the moment, and the choice of the point at which the

equation of the moment will be expressed, will be carried out so as to simplify

calculation as well as possible. In the case of elementary motions, these choices

are generally obvious. In the case of complex motions, these choices are not

always simple to take. They will be generally implemented considering different

successive tests.


18.2 MUTUAL ACTIONS

18.2.1 Theorem of Mutual Actions

The theorem of mutual actions stated within the framework of statics (Section

14.2.3) can be transposed to the context of the dynamics of bodies. We establish

hereafter this relation considering the notations of Section 14.2.3.

Applying the fundamental law to the two material sets (D1) and (D2), we have:

( ) 1

1 1 2 11 2 1g

DD D D DD D D= → = + →∪ → , (18.11)

( ) 2

2 2 1 21 2 2g

DD D D DD D D= → = + →∪ → . (18.12)

Applying the fundamental law of dynamics to the union ( )1 2 ,D D∪ we have:

( ) 1 2 1 2 1 2

g

D D D D D D∪ = ∪ → ∪ , (18.13)

or taking account of properties (11.4) and (16.17):

( ) ( ) 1 2 1 2 1 1 2 2g g

D D D D D D D D+ = +∪ → ∪ → . (18.14)

Comparing the three relations (18.11), (18.12) and (18.14), we obtain well the

relation which translates the theorem of mutual actions:

2 1 1 2D D D D→ = − → . (18.15)

This relation associated to Expression (11.9) of the mechanical actions exerted

on a given material set leads to a global relation of the actions of gravitation,

actions of contact and electromagnetic action exerted on a given set:

2 12 1 2 1

1 21 2 1 2 .

D DD D D D

D DD D D D

→+ + =→ →

→− + +→ →

(18.16)

Relation (18.15) in fact is extended to each type of mechanical actions consi-

dered separately. Thus:

2 1 1 2D D D Dϕ ϕ→ →= − , (18.17)

whatever the physical law ϕ exerted on the two sets ( ,ϕ = or ) .

Thus, the properties of the mutual actions established in the case of statics

(Section 14.2.3) can be transposed to the case of the dynamics of sets of solids.

18.3 Theorem of Power-Energy 281

FIGURE 18.2. Transmission of mechanical actions.

18.2.2 Transmission of Mechanical Actions

Let (D), (D1) and (D2) be three material sets disjoint (Figure 18.2), and suppose

that the mass of the set (D) can be neglected, its dynamic torsor being able to be

considered as the null torsor:

( ) 0g

D= . (18.18)

Moreover, we consider the case where the only mechanical actions exerted on the

set (D) are the actions exerted by the sets (D1) and (D2). From the fundamental

principle of dynamics, we have:

( ) 1 2( ) 0g

DD D D D D= → + → = = , (18.19)

or considering the theorem of mutual actions:

1 2D D D D→ = → . (18.20)

Hence the result:

The torsor of the mechanical actions exerted by the set (D1) on the set (D) is

equal to the torsor of the actions exerted by (D) on (D2).

This result expresses the properties which have the rigid bodies of negligible

masses to transmit entirely the mechanical actions.

18.3 THEOREM OF POWER-ENERGY

18.3.1 Case of a Solid

Let (S) be a solid on which mechanical actions are exerted, represented by the

torsor ( )S . The power developed by these actions in the motion of the solid

(S) relatively to an arbitrary reference (T) is expressed by Relation (11.13):

( ) ( ) ( ) ( )T TSP S S= ⋅ , (18.21)

(D2)

(D1)

(D)


where ( ) TS is the kinematic torsor relatively to the motion of the solid (S) with

respect to the reference (T).

In the case where the reference (T) is a Galilean reference (g), the fundamental

law of dynamics allows us to write:

( ) ( ) ( ) ( )g gg

S SP S = ⋅ . (18.22)

Taking account of Expression (16.32), we obtain:

( ) ( )c

d( ) ( )

d

g gP S E S

t= . (18.23)

Hence the theorem of power-energy:

In any motion of a solid with respect to a Galilean reference, the power deve-

loped by the mechanical actions exerted on this solid is, at any moment, equal to

the derivative with respect to time of the kinetic energy of the solid relatively to

the Galilean reference.

This theorem can take another form while introducing into Expression (18.23)

the work of the mechanical actions exerted on the solid (S) between the instants t1and t2. From the relation of definition (11.24) of work, Expression (18.23) is

written:

( ) ( ) ( )1 2 c 2 c 1( , ) ( ) ( )

g g gW t t E t E t= − . (18.24)

Hence the new statement, known under the name of theorem of kinetic energy:

In any motion of a solid with respect to a Galilean reference, the work, between

two instants, of the mechanical actions acting on the solid is equal to the

variation, during the motion, of the kinetic energy of the solid between these two

instants.

Note. The theorem of power-energy, deduced from the fundamental principle of

dynamics, does not bring any new information. Relation (18.23) leads to an equa-

tion which is a linear combination of the six scalar equations deduced from the

fundamental law (18.1), combination which is sometimes more interesting to use.

18.3.2 Case of a Set of Bodies

Let (D) be a set of n solids (Subection 11.3.6 and Figure 11.4). The actions

exerted on the solid (Si) are (Relation (11.38)):

1

( )

n

i i i i j i

ji

S S S D S S S

=≠

= → = → + → , (18.25)

and the theorem of power-energy (18.22) is written for each solid:

( ) ( )c

d( ) ( )

d

g gi iP S E S

t= . (18.26)

18.3 Theorem of Power-Energy 283

By summation on the set of the solids (Si), we obtain, taking account of Relation

(16.30):

( ) ( ) c

1

d( ) ( )

d

ng g

i

i

E D P St

=

=

. (18.27)

Thus:

( ) ( ) ( ) c

1 1 1

d ( )d

n n ng g g

i j i

i i ji

E D P D S P S St

= = =≠

= → + → , (18.28)


with:

( ) ( )c c

1

( ) ( )

ng g

i

i

E D E S

=

= . (18.29)

Hence the theorem of power-energy for a set of solids:

The derivative with respect to time of the kinetic energy of a set of solids,

relatively to a Galilean reference, is equal to the sum of the powers developed in

this reference by the mechanical actions exerted on each solid by the other solids

and by the material systems external to the set of the solids.

The integration between times t1 and t2 of Relation (18.28) leads to the follo-

wing formulation of the theorem of kinetic energy for a set of solids:

( ) ( ) ( ) ( ) 1 2 1 2

c 2 c 1 , ,

1 1 1

( , ) ( , )

n n ng gg g

i j it t t t

i i ji

E D t E D t W D S W S S

= = =≠

− = → + → . (18.30)

Note. It is important to observe that the theorem of power-energy or that of

kinetic energy considers the external and internal mechanical actions. In contrast,

we shall see (Section 20.2) that the fundamental principle applied to a set of solids

introduces only the external mechanical actions.

18.3.3 Mechanical Actions with Potential Energy

We consider the case where the mechanical actions exerted on a material set

admit a potential energy in the Galilean reference (g). We have then (Relation

(11.19)):

( ) ( )p

d( ) ( , )

d

g gP D E D t

t= − . (18.31)

By substitution into Expression (18.27) of the theorem of power-energy, we obtain:

( ) ( )

c pd d

( , ) ( , ) 0d d

g gE D t E D t

t t+ = . (18.32)

Thus by integration:


( ) ( ) ( ) c p m( , ) ( , ) ( )

g g gE D t E D t E D+ = , (18.33)

where ( )m ( )g

E D is a function independent of time, called mechanical energy of the

set (D) relatively to the Galilean reference (g). Hence the theorem of conser-

vation of energy:

In the case where the mechanical actions exerted on a material set admit

relatively to a Galilean reference a potential energy, the sum of the kinetic energy

and of this potential energy is in a Galilean reference a function independent of

time, called mechanical energy of the material set in the reference under con-

sideration.

18.4 APPLICATION OF THE FUNDAMENTAL

PRINCIPLE TO THE STUDY OF THE MOTION OF A

FREE BODY IN A GALILEAN REFERENCE

The analysis which will be implemented in this section can be applied in

particular to the motions of projectiles, planets, satellites, etc.

18.4.1 General Problem

Consider a solid (S) of mass m and mass centre G, free (Figure 18.3) relatively

to a Galilean reference (g), at which the coordinate system ( )/ , , g g g gO i j k

is

attached. The motion of solid (S) with respect to the reference (g) is defined by:

— the position of a point of the solid: we choose the mass centre G of the solid

of which the position will be characterized relatively to the coordinate system

attached to (g) by three coordinates (Cartesian, cylindrical, spherical or other

ones) which we will denote in a general way by p1, p2, p3 (the position vector

gO G

of the mass centre is a function of the coordinates p1, p2, p3);

— the orientation of the solid with respect to the reference (g): we choose a

trihedron ( )S S SGx y z attached to the solid (S), such that this trihedron is a prin-

cipal trihedron of inertia at the point G. The orientation of this trihedron is

characterized for example by the Eulerian angles ψ, θ, ϕ.

Let ( )S be the torsor which represents the whole of the mechanical actions

exerted on the solid. To study the motion of (S) with respect to the Galilean refe-

rence (g), we have the fundamental law of dynamics:

( ) ( )g

S S= , (18.34)

which leads to the vector equations of the resultant and of the moment.

18.4 Application of the Fundamental Principle to the Study of the Motion of a Free Body 285

FIGURE 18.3. Free solid in a Galilean reference.

1. Equation of the resultant

The equation is written as:

( ) ( , ) ( )g

ma G t R S= . (18.35)

The scalar equations which will be deduced from this equation will depend upon

the type of coordinates used to describe the position of the mass centre G. For

example, in the case where the position of G is characterized by its Cartesian

coordinates (x, y, z) in the system ( ) ,g g g gO x y z the scalar equations of the

resultant will be written as:

,

,

,

mx X

my Y

mz Z

=

=

=

(18.36)

where X, Y and Z are the components in the basis ( ), , g g gi j k

of the resultant of

the mechanical actions exerted on the solid. In the case where this resultant is

independent of the parameters of rotation, solving the system (18.36) allows us to

derive the motion of the mass centre in the reference (g).

2. Equation of the moment

At the mass centre, the equation of the moment is given by Relation (18.9).

The basis ( )( ) , , S S S Sb i j k=

being a principal basis of inertia at the point G, the

matrix of inertia which represents the operator of inertia at G, relatively to this

basis, is diagonal:

Oggi

gj

gk

yg

zg

xg

Galilean reference (g )

zS

yS

G

xS

Sj

Si

Sk

solide (S )


( )0 0

( ) 0 0

0 0

SbG

A

S B

C

=

I . (18.37)

The instantaneous vector of rotation and its derivative with respect to time relative

to the motion of the solid (S) with respect to the reference (g) are expressed, in the

basis (bS), according to Expressions (9.78) and (9.81). Hence:

( ) ( ) 1 2 3

gG S S SSS A i B j C kω ω ω ω= + +

, (18.38)

( ) ( ) ( ) ( ) ( ) ( ) 2 3 1 3 1 2

g gG S S SS SS C B i A C j B A kω ω ω ω ω ω ω ω× = − + − + −

. (18.39)

Expression (18.9) of the moment at the point G leads thus to the scalar equations:

( )

( )

( )

1 2 3

2 1 3

3 1 2

,

,

,

A C B L

B A C M

C B A N

ω ω ω

ω ω ω

ω ω ω

+ − =

+ − =

+ − =

(18.40)

where L, M and N are the components, in the basis (bS), of the moment at G of the

mechanical actions exerted on the solid. In the general case, these components are

functions of the variables , , , , , , ,i ip p ψ θ ϕ ψ θ ϕ . Analytical solutions will be

obtained only for some simple particular cases. If the components L, M and N are

independent of the parameters , i ip p , Equations (18.40) will be decoupled from

Equations (18.35) of the resultant. Solving Equations (18.40) will then allow us to

determine the motion of rotation of the solid.

18.4.2 Particular Cases

18.4.2.1 Case where the Resultant of the Mechanical Actions

is a Constant Vector

We study here the case where the resultant of the mechanical actions exerted

on the solid is a constant vector in the reference (g):

( )R S R=

. (18.41)

Equation (18.35) of the resultant shows that the acceleration vector of the mass

centre is constant:

( )0( , )g R

a G t am

= =

. (18.42)

The motion was studied in Section 7.3 of Chapter “Kinematics of Point”. The

trajectory of the mass centre is either a straight line or a parabola.

An example of this type of motion is given by a solid submitted to the field

18.4 Application of the Fundamental Principle to the Study of the Motion of a Free Body 287

of gravity at the vicinity of the Earth. We have then:

R mg=

.

Thus from (18.42):

( ) ( , )g

a G t g=

. (18.43)

The acceleration of the mass centre coincides with the field of gravity induced

by the Earth, hence the appellation “acceleration of the Earth gravity” for the

field vector of gravity.

18.4.2.2 Case where the Resultant of the Mechanical Actions

is collinear to

gO G

In such a case, the resultant of the mechanical actions is of the form:

( ) ( ) gR S k G O G=

. (18.44)

It results that the acceleration vector of the mass centre is collinear to the position

vector gO G

of the point G. We showed (Chapter 8) that the motion is a plane

motion with central acceleration.

18.4.2.3 Motions with Central Acceleration for which

3

( )g

S

g

O GR S mK

O G= −

where KS is a Constant

The acceleration vector is of the form:

( )3

( , )gg

S

g

O Ga G t K

O G= −

. (18.46)

This type of motion was studied in Section 8.2. The trajectory of the mass centre

is a conic. Furthermore, when the trajectory is an ellipse (KS is then positive), the

motion is governed by the Kepler’s laws (Section 8.2.4).

18.4.2.4 Case where the Mechanical Actions are Equivalent to a

Force whose the Support Passes through the Mass Centre

The moment of the torsor which represents the mechanical actions is then null

at the mass centre:

( ) 0G S =

, (18.47)

and the equation at G of the moment of the fundamental law is written as:

( ) 0g

G S =

. (18.48)


Thus, from (16.23):

( )

( ) d0

d

gg

G St=

. (18.49)

That leads to:

( ) ( )0

g gG S σ=

, (18.50)

where ( )0gσ

is a vector independent of time in the reference (g). This expression is

also written from (16.7) as:

( ) ( ) ( )0

g gG SS ω σ=

. (18.51)

In the particular case where the solid has a spherical symmetry of centre G, the

principal moments of inertia are equal ( A B C= = ), and we have:

( ) ( ) ( )g gG S SS Aω ω=

, hence

( )( )0g

gS A

σω =

. (18.52)

The instantaneous vector of rotation ( )gSω

is thus a vector independent of time in

the reference (g). In the case where the vector ( )0gσ

is the null vector, the vector of

rotation is also null. Thus, we deduce the following results.

The motion, with respect to a Galilean reference, of a solid having a spherical

symmetry and submitted in this reference to mechanical actions equivalent to a

force whose the support passes through the centre of symmetry is:

— either a motion of translation,

— or a motion of uniform rotation about a mobile axis, which passes through

the centre of symmetry and which keeps a direction fixed in the Galilean reference.

The motion of the centre of symmetry depends on the resultant of the equi-

valent force exerted on the solid.

18.5 APPLICATION TO THE SOLAR SYSTEM

18.5.1 Galilean Reference

The solar system is constituted (Figure 18.4) of the Sun and of multiple planets

including nine principal planets: Mercure, Venus, Mars, Jupiter, Saturn, Uranus,

Neptune, Pluto and the Earth. Several planets are accompanied by smaller

satellites which turn around them: thus the Moon is the natural satellite of the

Earth. These planets have dimensions much lower than that of the Sun. The other

planets have dimensions still much lower and are called asteroids. The Sun

represents 99.87% of the total mass of the Sun system.

The distance between the solar system and the nearest star (Proxima Centauri)

is enormous compared to the dimensions of the Sun system. Indeed the light

18.5 Application to the Solar System 289

FIGURE 18.4. The solar system seen from a point of space.

emitted by the Sun reaches the Earth in 8 minutes and Pluto (planet furthest away

from the Sun) in less than 6 hours, whereas it takes 4 years to reach Proxima

Centauri. From this, it results that the solar system can be considered as isolated in

the Universe and that it is possible to neglect the mechanical actions exerted by

the other solar systems on our solar system.

Thus, it results that the torsor of the mechanical actions exerted on the solar

system can be considered as being null, and the equation of the resultant of the

fundamental principle is written as:

( )Sso Sso( ) 0

gm a G =

(18.53)

where mSso and GSso are respectively the mass and the mass centre of the solar

system.

This equation shows that the mass centre of the solar system is either motion-

less, or animated by a rectilinear and uniform motion with respect to a Galilean

reference. So, it results (Section 18.1.2) that the mass centre of the solar system is

itself a point attached to a Galilean reference (g), point which we can take as

origin of the reference. The axes of the reference are then chosen so has to have

fixed directions of the axes with respect to “fixed” stars: stars very far away from

the solar system and appearing for an observer of the solar system under fairly

constant angular distances. The mass of the Sun representing the near totality of

the mass of the solar system, the mass centre of the solar systems coincides

practically with the mass centre of the Sun, itself coinciding fairly with the centre

of the Sun considered as a sphere.

Thus the existence of a Galilean reference (g) is materialized as having:

— for origin, the centre of Sun;

— directions of axes ( ), , g g gi j k

which are fixed with respect to directions of

fixed stars.

Earth

Sun Mercure

Mars

Jupiter

Venus

Saturn

Uranus

Neptune

Pluto


Any other reference (Section (18.1.2), in motion of uniform rectilinear trans-

lation with respect to this reference, is also a Galilean reference.

18.5.2 Motion of Planets

At first approximation, the planets and the Sun can be assimilated to bodies

having a spherical symmetry. Thus, in accordance with the laws of the gravitation

(Chapter 12), with the results of Subsection 18.4.2.3 of the present chapter and

with the results established in Section 8.2 of Chapter 8, the centre of the planets

describe ellipses having the centre of the Sun as focus. These ellipses are fairly

circular and approximately located in the same plane. The orbits of the planets are

described in the same direction and the laws of the motions of the planet centres

are governed by the Kepler’s laws (Section 8.2.4).

Moreover, in accordance with the results of Subsection 18.4.2.4, the planets are

animated by a motion of rotation about a mobile axis keeping a direction fixed in

the Galilean reference (g). This motion of rotation is called motion of sidereal

proper rotation.

18.5.3 The Earth in the Solar System

We specify in this section, the motion of the Earth of which the general cha-

racteristics were considered in the preceding subsection.

The law of the gravitation and its consequences (Chapter 12) allow us to

conclude that the mechanical actions exerted on the Earth are reduced to the

action of gravitation exerted by the Sun. Thus:

(Te) So Te= →. (18.54)

This description comes to neglect on the one hand the gravitational attractions of

the stars other than the Sun and of the planets, and on the other hand the actions

other than the gravitation.

If the Sun and the Earth are assimilated to solid spheres homogeneous by con-

centric layers, it is shown (Exercise 12.7) that the action of gravitation exerted by

the Sun on the Earth is a force of which the support passes through the centre of

the Earth and of resultant:

( )

TeSo Te 3

Te So

So TeSoO O

R Km mO O

=→

, (18.55)

where K is the constant of gravitation. All occurs as if the masses of the Sun and

of the Earth were concentrated respectively at their centres.

This model and the results of Subsections 18.4.2.3 and 18.4.2.4 allow us to

justify the motion of the Earth assimilated to a body with spherical symmetry

(motion similar to the motions of the other planets, Sub-section 18.5.2).

Comments 291

1. The centre of the Earth describes an elliptic trajectory of which one of the

foci is the centre of the Sun and of which the plane is called plane of the ecliptic.

The areal velocity of the centre of the Earth along its trajectory and relatively to

the Sun centre is constant. The motion of the Earth along its trajectory is gover-

ned by the Kepler’s laws.

2. The Earth is animated by a motion of sidereal proper rotation about an axis

( )Te Te, O k

which forms a constant angle of 23°27' with the normal to the plane of

the ecliptic. The axis ( )Te Te, O k

is the axis South-North of the poles. The angular

velocity of sidereal proper rotation is defined by:

( )TeTe

gkω Ω Ω= =

, (18.56)

with

4 120,729 10 rad s

86164

πΩ − −= = × ,

owing to the fact that the Earth achieves a rotation on itself in 23h56min04s.

The motion of the Earth is schematized in Figure 18.5. The inclination of the

axis of sidereal rotation with respect to the normal direction of the plane of the

ecliptic leads to sunning daily durations varying according to the latitude and the

period of the year.

The 24 hours duration for one day combines the effect of the proper rotation

with the motion of the Earth on its trajectory.

FIGURE 18.5. Motion of the Earth.

COMMENTS

The fundamental principle of the dynamics is the key for the analysis of

a problem of Mecanics of Rigid Bodies. Its formulation while considering

the concept of torsors leads to the equality of the dynamic torsor with the

torsor which represents the whole of the mechanical actions exerted on the

material set considered, and this when the motion is analyzed with respect

Earth

Sun

23°27'

axis South-North


to a Galilean reference. The fundamental principle leads thus to two vector

equalities, that is a total of six scalar equations for a given material set.

The fundamental principle was then applied to the analysis of the motion

of free body, with for object to find a Galilean reference system. This

reference has for origin the centre of the solar system, coinciding practi-

cally with the centre of the Sun, and axis directions which are fixed with

respect to fixed stars. The chapter gives then some elements about the

motions of the planets and of the Earth in the solar system.

The reader will pay all his attention on the development implemented

throughout this chapter.

CHAPTER 19

The Fundamental Equation of Dynamics in Different References

19.1 GENERAL ELEMENTS

19.1.1 Fundamental Equation of Dynamics in a Non Galilean Reference

We consider a reference (R) animated with respect to the Galilean reference (g)

by a motion which is known but arbitrary and we seek how the fundamental

principle of dynamics is written in the non Galilean reference (R).

The fundamental principle of dynamics applied to the motion of a solid (S) is

written as:

( ) ( ) ( )g g

S S= , (19.1)

where ( ) ( )g

S is the torsor of the mechanical actions exerted on the solid (S)

and measured in the Galilean reference (g). Thus, it results that the dynamic torsor

associated to the motion of the solid (S) with respect to the reference (R) is written

from Relations (19.1) and (17.38) as:

( ) ( ) ( )( ) ( )( ) ( )g ggR

S R RS S S= − − , (19.2)

or ( ) ( ) ( )R R

S S= , (19.3)

setting: ( ) ( ) ( )( ) ( )( ) ( ) ( )

g ggRR RS S S S= − − . (19.4)

Equation (19.3) thus obtained is similar to the fundamental equation (19.1)

which is expressed in a Galilean reference. Expression (19.4) shows that the

294 Chapter 19 The Fundamental Equation of Dynamics in Different References

torsors ( )( ) gR S− and ( )( ) g

R S− represent physical entities similar to the

entity which the torsor ( ) ( )g

S represents. They are not however mechanical

actions with the proper sense, because they are not exerted by material systems. It

is however usual to say that these torsors represent actions called inertia action of

entrainment for the torsor ( )( ) gR S− and inertia action of Coriolis for the torsor

( )( ) gR S− . These actions must be considered as fictitious actions, which express

the effect of the motion of a non Galilean reference with respect to a Galilean

reference. From the established expressions, these fictitious actions do not depend

on the Galilean reference considered. Expressions (19.3) and (19.4) can then be

stated in the following way.

In a non Galilean reference, the fundamental principle of the dynamics can be

written (19.3) in a similar way as in a Galilean reference, on condition to intro-

duce the torsor sum of the mechanical actions exerted on the solid (S), expressed

in a Galilean reference, and of two torsors representing fictitious actions: the

inertia action of entrainment and the inertia action of Coriolis, fictitious actions

which take account of the motion of the non Galilean reference with respect to a

Galilean reference.

19.1.2 The Reference Systems used in Mechanics

1. In the preceding chapter, we highlighted the existence of a Galilean reference

(g) having the Sun centre for origin and axis directions which are fixed with

respect to directions of fixed stars. Moreover, the plane of the ecliptic is fixed in

this reference. It is thus possible (Figure 19.1) to choose as Galilean trihedron, the

trihedron of origin OSo the Sun centre and of plane ( )So , , g gO i j

coinciding with

the plane of ecliptic.

2. A second reference used is the reference associated to the motion of trans-

lation of the Earth on its trajectory. This reference which we shall denote by (Te)

is called geocentric reference.

At this reference system, we associate the trihedron (Figure 19.1) of which:

— the origin is the centre OTe of the Earth,

— the axis ( )Te Te, O k

is the axis of sidereal proper rotation of the Earth and

the axis ( )Te Te, O i

is contained in the plane of the ecliptic.

So, the trihedron defined has thus directions of the axes fixed with respect to

the reference (g). The reference (Te) is animated with respect to (g) by a motion

of elliptic translation.

3. A third type of reference, the most used by Engineer, is the class of refe-

rences attached to the Earth. Any trihedron attached to the Earth will be cha-

racterized by an origin and axes fixed with respect to the Earth. We shall denote

by (T) the one of these references. They are animated with respect to the

geocentric reference (Te) by a motion of uniform rotation Ω about a fixed axis

19.2 Fundamental Relation of Dynamics in the Geocentric Reference 295

FIGURE 19.1. Geocentric reference.

of direction Tek

. In Figure 19.2 we have represented one of these trihedrons (T),

at the surface of the Earth in a place of latitude β. The selected axes are: Oz

the

direction from the place of the study to the centre of the Earth, Ox

the East

direction and Oy

the North direction.

4. Lastly, we shall have to use references (R) animated by a known motion with

respect to the Earth, hence with respect to one of the preceding references (T).

19.2 FUNDAMENTAL RELATION OF DYNAMICS

IN THE GEOCENTRIC REFERENCE


The motion of the geocentric reference (Te) being a motion of translation with

respect to the Galilean reference (g), it results that:

( ) ( )Te Te0 and 0g gω ω= = . (19.5)

This implies first that, for any motion of a solid (S) with respect to the reference

(Te), we have:

( )( ) Te 0

gS = . (19.6)

The inertia actions of Coriolis exerted on the solid (S) are null. Relation (19.3) is

then reduced, in the geocentric reference, to:

( ) ( ) ( )( ) TeTe( )

ggS S S= − . (19.7)

Relations (19.5) imply next that the acceleration vector of entrainment of the

mass centre of the solid (S) coincides, from (17.19), with the acceleration vector,

with respect to the Galilean reference (g), of the centre of the Earth:

( ) ( )

TeTe ( , ) ( , )g g

ea G t a O t=

. (19.8)

23°27'

OSo

gi

gj

gk

OTe

gi

gj

gk

Tek

Tei

Tej


FIGURE 19.2 Trihedron attached to the Earth.

It follows that the elements of reduction, at the centre of the Earth, of the inertia

torsor of entrainment are written, from (17.24) and (17.26), as:

( )( ) ( )TeTe ( , )

g gR S ma O t= , (19.9)

( )( ) Te

( )Te TeTe ( , )

g gO S m O G a O t= ×

. (19.10)

The calculation of the moment at the mass centre of the solid (S) leads then to:

( )( ) Te 0g

G S =

. (19.11)

The vector equations deduced from Relation (19.7) are thus written as:

( ) (Te) (g)

Te( , ) ( ) ( , )g

m a G t R S m a O t−= , (19.12)

( ) ( ) Te ( )g

G GS S=

. (19.13)

The equation of the moment at the mass centre has the same form in the geo-

centric reference as in the Galilean reference.

Now let us express in the Galilean reference the torsor ( ) ( )g

S of the me-

chanical actions exerted on the solid (S). Taking account of the different types of

the actions, we have (Chapter 11):

( ) ( )g S S SS S S S= + +→→ → . (19.14)

Denote by (E) all that, in the Universe, is not the Earth and the solid (S). We may

then write:

Te S E SS S = +→ →→ . (19.15)

zTe

North

South

x (East)

xTe

yTe

zTe y

OTe

Ω

z

O

19.2 Fundamental Relation of Dynamics in the Geocentric Reference 297

Moreover, the motion of the centre of the Earth, supposed to be submitted to the

only gravitational actions, is defined in the reference (g) by:

( )Te Te( , ) Te Te

gm a O t R= → , (19.16)

where mTe is the mass of the Earth and where the actions of gravitation are:

Te TeTe Te E S= +→ →→ . (19.17)

The actions of gravitation exerted by the solid (S) on the Earth are negligible

compared to the actions of gravitation exerted by (E). Whence:

TeTe Te E= →→ , (19.18)

and the vector equation (19.12) is written:

(Te)

Te

( , ) Te

.Te

m a G t R R R SS SS S

mR RE S E

m

= + + →→→

+ −→ →

(19.19)

The fundamental equations of dynamics, in the geocentric reference, are finally

given by Relations (19.19) for the resultant and (19.13) for the moment at the

mass centre of the solid. To apply Equation (19.19) of the resultant, it is then

necessary to consider specific assumptions adapted to each type of analysis.

19.2.2 Case of a Solid Located at the Vicinity of the Earth

For a solid located at the vicinity of the Earth, the model generally considered

consists first in neglecting the actions of gravitation other than the action of

gravitation exerted by the Sun, and next in neglecting the distance between the

centre of the Earth and the solid, compared to the distance between the centre of

the Earth and the centre of the Sun. The gravitational field is then constant over

the geometric domain constituted of the Earth and of its vicinity, this field being

the same one as that induced by the Sun at the centre of the Earth. Thus:

( ) ( ) ( )So TeTeG E G E S G O→ = → =

. (19.20)

It results that:

( )TeER m G OE S =→ , (19.21)

( )Te TeTe ER m G OE =→ , (19.22)

and Equation (19.19) is reduced to:

(Te)( , ) Tem a G t R R R SS SS S= + + →→→ . (19.23)

For a solid located at the vicinity of the Earth at the altitude h, the results

established in Chapter 12 (Subsection 12.1.4), show that the action of gravitation


exerted by the Earth on the solid is a force whose the support passes through the

mass centre of the solid and whose the resultant is:

( )TeR mG h nS =→

, (19.24)

where ( )G h is the magnitude of the gravitational field induced by the Earth at the

altitude h, expressed by Relation (12.16), and n

is the unit vector of the direction

from the place where the solid is located to the centre of the Earth (Figure 12.3).

Finally, the fundamental relation of dynamics applied to the motion of the solid

(S) with respect to the geocentric reference (Te) is written in the form:

( ) ( ) Te Te ( )S S= , (19.25)

with

( ) Te ( ) TeS S S SS S+ += → →→ . (19.26)

19.3 FUNDAMENTAL RELATION OF DYNAMICS

IN A REFERENCE ATTACHED TO THE EARTH

19.3.1 Equations of Motion

Let (T) be a reference attached to the Earth and (Oxyz) the associated trihedron

(Figure 19.2). For a solid (S) located at the vicinity of the Earth, the fundamental

relation of dynamics (19.3) is written as:

( ) ( ) ( )T TS S= , (19.27)

with ( ) ( ) ( )( ) ( )( ) Te Te Te( ) ( )T

T TS S S S= − − . (19.28)

The torsors ( )( ) TeT S− and ( )( ) Te

T S− represent the inertia actions of entrain-

ment and of Coriolis which result from the motion of rotation of the Earth about

the axis of the poles. This motion of rotation is uniform, with a rotation vector

expressed in (18.56).

Inertia torsor of entrainment

The elements of reduction of the inertia torsor of entrainment at the mass centre

of the solid (S) are from (17.24), (17.26) and (18.56):

( )( ) ( )Te 2Te Te TeTR S m k k O GΩ= × ×

, (19.29)

( )( ) Te 2Te Te( )G T GS k S kΩ= ×

, (19.30)

where OTe is the centre of the Earth.

19.3 Fundamental Relation of Dynamics in a Reference Attached to the Earth 299

Inertia torsor of Coriolis

The elements of reduction are expressed from (17.30), (17.34) and (18.56) as:

( )( ) ( )TeTe2 ( , )T

TR S m k G tΩ= ×

, (19.31)

( )( ) ( ) ( )

TeTe Te2 ( ) 2 ( )T T

G T G S S GS I S k S kΩ ω Ωω= × + ×

. (19.32)

Equation (19.27) leads thus to the two vector equations:

( ) ( )

( )

2Te Te Te Te

( , ) Te

2 ( , ),

Tm a G t R R RS S SS S

m k k O G m k G tΩ Ω

= + +→ →→

− × × − ×

(19.33)

( )

( ) ( )

2Te Te

Te Te

( )

2 ( ) 2 ( ) .

TG G GS G

T TG S S G

k S kS SS S

I S k S k

Ω

Ω ω Ω ω

= + − ×→→

− × − ×

(19.34)

19.3.2 Action of Earthly Gravity

The action of gravity induced by the Earth was described in Section 12.2. We

then simply reported (Subsection 12.2.1) that this action resulted from the super-

position of the action of gravitation exerted by the Earth and the action generated

by the motion of the Earth about its South-North axis. We give now a more

precise definition of this action.

The action of gravity exerted by the Earth on a solid is the force whose the

support passes through the mass centre of the solid and whose the resultant is the

sum of the resultants of the action of gravitation induced by the Earth and of the

inertia action of entrainment.

The action of gravity is represented by a torsor which we denoted by e( )S .

From the preceding definition, we have:

( )

2Te Te Tee( ) ,Te

e( ) 0.G

R S R m k k O GS

S

Ω= − × ×→

=

(19.35)

The resultant of the action of gravity is thus written in the form:

e( )R S mg= , (19.36)

with

( )

2Te Te Teg G n k k O GΩ= − × ×

, (19.37)

where n

is the unit vector of the direction from the mass centre to the centre of

the Earth (Figure 19.3).

Express the vector g

as a function of the latitude β and of the longitude α of


FIGURE 19.3. Direction at a point at the vicinity of the Earth.

the place where the solid is located (Figure 19.3). We have:

Te

Te

sin ( )cos ,

,

n k u

O G Rn

β α β = − +

=

(19.38)

where R is the distance from the mass centre to the centre of the Earth, practically

equal to the radius of the Earth when a solid is at the vicinity of the Earth. Thus, it

results that:

2 ( ) cosg G n R uΩ α β= +

. (19.39)

The numerical application of this expression shows that, in practice, g

differs

very little from G n

(Subsection 12.2.1).

19.3.3 Conclusions on the Equations of Dynamics in a Reference Attached to the Earth

Equation (19.33) of the resultant is written:

( )

( )

( )

Te

( , ) e( )

2 ( , ).

T TSR ma G t R S R R S SS S

m k G tΩ

= = + + →→

− ×

(19.40)

In a great number of problems, the term ( )

Te 2 ( , )m k G tΩ− × will have a negli-

gible influence.

yTe

North

South

xTe

zTe

OTe

G

Tek n

( )u α

19.4 Equations of Dynamics with respect to a Reference of Arbitrary Known Motion 301

With regard to Equation (19.34) of the moment at the mass centre:

— The term 2Te Te( )Gk S kΩ− ×

is null if the solid has a spherical symmetry.

In the other cases, this term is rather low so that it can be neglected being given

the value of Ω2.

— The two other terms in Ω will be with a less degree negligible in many

applications. We keep them in the expression of the moment for a later decision.

Thus:

( )

( ) ( ) Te Te 2 ( ) 2 ( ) .

TG G GS

T TG S S G

S SS S

I S k S kΩ ω Ω ω

= + →→

− × − ×

(19.41)

In all the analyses for which the terms in Ω can be neglected, it is said that the

Earth can be considered as a Galilean reference. On this assumption of “Galilean

Earth”, the fundamental equation of dynamics thus can be written:

( ) ( ) ( )T TS S= , (19.42)

with

( ) ( ) e( )T S S S SS S= + + →→ (19.43)

The exerted actions are reduced to the action of gravity induced by the Earth, to

the electromagnetic actions and to the actions of contact.

This model allows us to solve all the usual problems of engineering, for which

too high speeds do not occur.

19.4 EQUATIONS OF DYNAMICS OF A BODY WITH

RESPECT TO A REFERENCE WHOSE THE MOTION

IS KNOWN RELATIVELY TO THE EARTH

Let (R) be a reference of which the motion is known with respect to a reference

(T) attached to the Earth. This motion will be characterized by the kinematic

torsor, itself defined by its elements of reduction at a point OR of the reference (R):

( ) ( )

( ) ( )

,

( , ).R

T TR R

T TO R R

R

O t

ω=

=

(19.44)

The dynamic torsor relatively to the reference (R) of the solid (S) of which we

study the motion is:

( ) ( ) ( )( ) ( )( ) R T T TS S R RS S= − − . (19.45)


And the fundamental equation of dynamics is written in the reference (R):

( ) ( ) ( )R RS S= , (19.46)

with

( ) ( ) ( )( ) ( )( ) ( ) ( )R T T TR RS S S S= − − . (19.47)

The torsors ( )( ) TR S− and ( )( ) T

R S− represent the inertia actions of entrain-

ment and Coriolis which result from the motion of the reference (R) with respect

to the Earth.

The vector equations of the motion with respect to the reference (R) are

deduced from Expression (19.46), while taking account of (19.40) or (19.41), and

from the expressions of the elements of reduction of the inertia torsors (Relations

(17.24) to (17.34)).

Thus, the equation of the resultant is written:

( ) ( )( , ) ( )R RSR m a G t R e S R R S SS S= = + + →→

( ) ( ) ( )( )( ) ( , )T T T TR R R R R Rm a O t O G O Gω ω ω − + × + × ∧

(19.48)

( ) ( ) ( )

Te 2 ( , ) ( , ) ,TRm k G t G tΩ ω − × + ×

with

( ) ( ) ( ) ( ) ( ) ( )

( , ) ( , )

( , ) ( , ) .

TG R

T TR R R

G t G t

G t O t O Gω

= +

= + + ×

(19.49)

In the same way, the equation of the moment is written in the form:

( ) RG G GS S SS S= + →→

( ) ( ) Te Te2 ( ) 2 ( )T T

G S S GI S k S kΩ ω Ω ω− × − ×

( ) ( ) ( )( ) ( )T T T

G R R G RS Sω ω ω− − ×

(19.50)

( ) ( ) ( ) ( )2 ( ) 2 ( ) .T R R T

G R S S G RI S Sω ω ω ω− × + ×

The two expressions of the resultant (19.48) and of the moment (19.50) have

complex general forms. In practice, they will not be used directly in these forms,

but they will be introduced simply when the relations of composition of motions

will be applied to the motions under consideration.

Comments 303

COMMENTS

In the preceding chapter, the fundamental principle of dynamics was

stated with respect to a Galilean reference of which the origin is the centre

of the Sun and directions are fixed relatively to fixed stars. In practice,

Engineer studies motions with respect to references attached to the Earth.

After having established the general equations of dynamics relatively to

a non Galilean reference, the present chapter derives the fundamental

relation of dynamics in a reference attached to the Earth. The reader will be

interested in the development of the process and will retain the result: the

fundamental equation of dynamics in a reference attached to the Earth has

the same form as the fundamental equation in a Galilean reference, the

exerted mechanical actions being reduced to the action of gravity, the

electromagnetic actions and the actions of contact. The action of gravity

induced by the Earth which was introduced in Chapter 12 is entirely

characterized in the present chapter.

CHAPTER 20

General Process for Analysing a Problem of Dynamics of Rigid Bodies

In this chapter, we study the problem of the dynamics of a rigid body or a set of

rigid bodies. In practice, the results which will be derived in Galilean reference

can be applied to a reference attached to the Earth provided that the induced velo-

cities are not too high (Subsection 19.3.3). This type of reference will be called as

pseudo-Galilean reference

20.1 DYNAMICS OF RIGID BODY


We consider a solid (S) in motion with respect to a Galilean (or pseudo-

Galilean) reference (g). Among the mechanical actions exerted on the solid, we

distinguish, as in Subsection 14.2.1, between:

— The known actions which can be calculated: actions of gravitation, of gravity,

electromagnetic actions. The whole of the actions of this type exerted on the solid

(S) and measured in the Galilean reference (g) are represented by the torsor

denoted by ( ) S .

— The actions induced by the connections (or actions of contact) which are

represented by the torsor S S→

, which we shall denote by ( ) S .

In order to simplify the notations, we denoted by ( ) S and ( ) S , instead

of ( )( ) gS of ( )( ) ,g S the torsors of the actions expressed in the reference (g).

Contrary to the actions of the first type, the actions induced by the connections

are not known a priori. Their determination is part of the problem to be solved. In

order to completely determine the problem of the dynamics of the solid (S), it will

20.1 Dynamics of Rigid Body 305

be necessary to make assumptions on the physical nature of the connections:

perfect connections, connections with viscous friction or connections with dry

friction. The validity of these assumptions will be deduced a posteriori, by con-

fronting the experimental observations with the theoretical results derived starting

from these assumptions.

The equation of dynamics is thus written in the Galilean reference (g) in the

form: ( ) ( ) ( ) g

S S S= + . (20.1)

If m and G are respectively the mass and the mass centre of the solid, the prece-

ding equation leads to the two vector equations of the dynamics of the solid.

Equation of the resultant( ) ( ) ( ) ( ) ( , )g T

SR ma G t R S R S= = + . (20.2)

Equation of the moment at a point P

( ) ( ) ( ) gP P PS S S= +

. (20.3)

Let us recall (Section 16.3) that we have:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( , )

g g g ggP P PS S S Sm PG a P t S Sω ω ω= × + + ×

, (20.4)

and that this expression is simplified when the point P coincides with the mass

centre: ( ) ( ) ( ) ( ) ( ) ( )g g g g

G G GS S S SS Sω ω ω= + × . (20.5)

By choosing an orthonormal basis ( )1 2 3, , u u u

to express the equation of the

resultant and an orthonormal basis ( )1 2 3, , v v v

to express the equation of the

moment at the point P, the vector equations (20.2) and (20.3) lead to 6 scalar

equations. The choices of the point P and of the bases (identical or not) will be

carried out so as to simplify as well as possible the writing of these equations.

In some cases, the theorem of power-energy could be useful. It will then be

written as:

( ) ( ) ( ) ( ) ( ) cd

d

g g gE P S P S

t= + . (20.6)

It does not bring however any new information with respect to the preceding

scalar equations.

20.1.2 General Process of Analysis

The general process for analysing a problem of dynamics of a solid consists to

establish Equation (20.1) of the dynamics. This analysis will be implemented in

the following way.

1. Choose a Galilean (or pseudo-Galilean) reference. For Engineer, this refe-

rence will be generally a reference attached to the Earth.

306 Chapter 20 General Process for Analysing a Problem of Dynamics of Rigid Bodies

2. Find the parameters of situation relative to the motion of the body with

respect to the reference system, taking into account the connections.

3. Implement the kinematic analysis: determine the kinematic torsor; the

conditions of sliding and non-sliding if necessary; the velocity and acceleration

vectors of the mass centre of the body.

4. Implement the kinetic analysis: determine the kinetic torsor, the dynamic

torsor, the kinetic energy, relatively to the motion of the body with respect to the

reference system.

5. Analyse the mechanical actions exerted on the body: torsors representing

these actions, powers developed by these actions. Examine the assumption of

perfect connections.

6. Apply the fundamental principle of dynamics, which leads to six scalar

equations.

6'. In some cases, applying the theorem of power-energy could provide a scalar

equation apt to replace one of the six preceding equations judiciously.

7. To solve the problem of dynamics, it will be then necessary to introduce

assumptions on the physical nature of the connections: perfect connections, con-

nections with viscous or dry friction. These assumptions will lead to additional

scalar equations.

8. Solving the system constituted of the six scalar equations derived from the

fundamental principle and of equations deduced from the assumptions on the phy-

sical nature of connections will then make it possible to find:

— the equations of motion: parameters of situation as a function of time;

— the components of the actions induced by the connections, on which no

assumption will have been considered previously.

20.2 DYNAMICS OF A SET OF BODIES

We consider a set (D) constituted of n solids: (S1), (S2), ..., (Si), ..., (Sj), ..., (Sn).

The mechanical actions which are exerted have been considered in Subsection

14.2.2, in the context of the analysis of the equilibrium of this set of solids. The

actions were then divided into internal actions and external actions. We keep the

notations used in this subsection.

The equation of dynamics applied to the motion of each solid with respect to

the Galilean reference (g) is written:

( ) ( ) ( ) ( ) ( )

1

1, 2, . . . , ,

i

ng

i i j i j iS

ji

S S S S

i n

=≠

= + + +

=

(20.7)

where ( ) i

g

S is the dynamic torsor relatively to the motion of the solid (Si) with

respect to the reference (g).

20.3 Conclusion 307

Equation (20.7) leads to 6 scalar equations for each solid, thus a total of 6n

scalar equations for the set (D).

Other equations with different forms, but depending on the preceding

equations, can be obtained by applying the fundamental principle to any part of

the set (D). This can be applied to the set (D) itself, which leads to:

( ) ( ) ( ) 1

ng

i iD

i

D D S S

=

= → = + , (20.8)

with

( ) ( ) 1

i

ng g

D S

i=

= . (20.9)

Relation (20.8) leads to the elimination of the actions internal to the set (D). This

property is general: the application of the fundamental principle of dynamics to

any part of the set (D) leads to a relation, where the mechanical actions internal to

this part are excluded.

As in the case of one solid, it is necessary to introduce assumptions on the

physical nature of the internal and external connections. These assumptions

associated to Relations (20.7), or to linear combinations, will then make it

possible to solve the problem of dynamics of the set of rigid bodies, by deriving

the equations of motion and characterizing the actions induced by the connections.

20.3 CONCLUSION

The general unity of the development implemented in the present textbook has

consisted in setting up gradually the different concepts necessary to analyse a

problem of mechanics (dynamics or statics) of a rigid body or a set of rigid

bodies. This development finds its conclusion in this chapter. Its objective was to

lay the foundations of a general and systematic process for analysing a problem of

mechanics. The issue of this development shows that actually there exists only

one problem of mechanics, of which the analysis is implemented for each solid

according the general process of analysis reported in Subsection 20.1.2. Each

problem differs by the mechanical system to analyse and the particular points of

interest of Engineer, associated to each problem. In fact, once the mechanical

system is given, the problem is entirely stated and the process of analysis is

determined.

308 Chapter 20 General Process for Analysing a Problem of Dynamics of Rigid Bodies

COMMENTS

The chapter develops the general process for analysing a problem of

dynamics of a rigid body. This analysis is always implemented in the same

way: 1) choose a reference system, 2) derive the parameters of situation, 3)

implement the kinematic analysis, 4) implement the kinetic analysis, 5)

analyse the mechanical actions, 6) apply the fundamental principle of

dynamics.

Then, solving the equations of the problem of dynamics requires to

introduce assumptions on the physical nature of the connections: con-

nections without friction, connections with viscous friction or connections

with dry friction. Next, solving the equations deduced from the funda-

mental principle and associated to the equations on the physical nature of

the connections allows us to derive the equations of motion and to charac-

terize the actions induced by the connections.

For a set of rigid bodies, the preceding process is applied to each body of

the set.

In practice, all the problems of mechanics will be analysed using this

same process.

CHAPTER 21

Dynamics of Systems with One Degree of FreedomAnalysis of Vibrations

21.1 GENERAL EQUATIONS

21.1.1 Introduction

Mechanical vibrations are induced when an elastic system is disturbed from a

position of stable equilibrium. The majority of the vibrations in the machines are

harmful, on the fact they generate higher stresses and energies of which the dissi-

pation can lead to a deterioration in fatigue of the systems. It is thus necessary to

reduce the vibrations as well as possible.

The simplest configuration of a vibrating system is a system with one degree of

freedom where the configuration is described by a single coordinate. The impor-

tance of the analysis of one degree of freedom system lies in the fact that the

results which are established for this system constitute the basics for the analysis

of the mechanical vibrations of complex structures. As one degree of freedom

system, we consider the spring-mass system of Figure 21.1. The results which will

be established for this system can be transposed to every vibrating system with

one degree of freedom.

FIGURE 21.1. Spring-mass system.

(T)

(R) (S) ()

O

G x

y

310 Chapter 21 Dynamics of Systems with One Degree of Freedom. Analysis of Vibrations

The spring-mass system is constituted of a solid (S) linked to a support (T)

through an elastic spring (R). Moreover, the solid is connected to the support by a

prismatic connection of horizontal axis (∆), also axis of the spring. It results that

horizontal displacement of the mass centre G of the solid is possible only along

the axis (∆).

21.1.2 Parameters of Situation

We choose the coordinate system (Oxyz) attached to the support (T) in such a

way that the axis Ox

coincides with the axis (∆), that the point O coincides with

the equilibrium position of the mass G (in the case where there is no friction) and

that the axis Oy

is upward vertical. The orientation of the solid (S) does not

change during the motion, so the motion has one parameter of situation, the

abscissa x of the point G along the axis Ox

: OG x i=

.

21.1.3 Kinematics

The elements of reduction at the mass centre G of the kinematic torsor ( ) TS

are: ( ) ( )

( ) ( )

0,

( , ) .

T TS S

T TG S

R

G t x i

ω= =

= =

(21.1)

All the points of the solid have the same velocity vector and the same acceleration

vector: ( )

( , )Ta G t x i= . (21.2)

21.1.4 Kinetics

The elements of reduction at the mass centre G of the kinetic torsor are:

( ) ( )

( ) ( )

( , ) ,

( ) 0.

T TS

T TG S G S

R m G t mx i

S ω

= =

= =

(21.3)

In the same way, the elements of reduction of the dynamic torsor are:

( ) ( )

( ) ( ) ( ) ( )

( , ) ,

( ) ( ) 0.

T TS

T T T TG S G S S G S

R ma G t mx i

S Sω ω ω

= =

= + × =

(21.4)

Lastly, the kinetic energy is:

( ) ( ) ( ) 2c

1 1( )

2 2T T T

S SE S mx= =⋅ . (21.5)

21.2 Vibrations without Friction 311

21.1.5 Mechanical Actions Exerted on the Solid

The mechanical actions exerted on the solid are reduced to the action of gra-

vity, the action of the spring and the action of the support induced by the prismatic

connection.

1. Action of gravity

The action is represented by the torsor e( )S of which the elements of reduc-

tion at the mass centre are:

e( ) ,

e( ) 0.G

R S mg j

S

= −

=

(21.6)

The power developed by the action of gravity is:

( ) ( ) e( ) e( ) 0T TSP S S= =⋅ . (21.7)

2. Action of the spring

The action exerted by the spring is a force of which the support is the axis of

the spring. The action is represented by the torsor ( )S of which the elements

of reduction at G are:

( ) ,

( ) 0,G

R S kx i

S

= −

=

(21.8)

where k is the constant of rigidity or the stiffness of the spring.

The power developed by the action of the spring is:

( ) ( ) ( ) ( )T TSP S S k x x= = −⋅ . (21.9)

3. Action of the support induced by the prismatic connection

The action of connection exerted by the support is represented by the torsor

( )S of which the elements of reduction at the point G are:

( ) ,

( ) .

l l l

G l l l

R S X i Y j Z k

S L i M j N k

= + +

= + +

(21.10)

The components Xl, Yl, ..., Nl, of the action of connection are to be determined.

The power developed by the action of connection is:

( ) ( ) ( ) ( )T TlSP S S X x= =⋅ . (21.11)

21.1.6 Application of the Fundamental Principle

In the case where the support is a pseudo-Galilean reference (attached to the

Earth), the fundamental principle applied to the spring-mass system is written:


( ) e( ) ( ) ( )TS S S S= + + . (21.12)

This equation leads to two vector equations, the equation of the resultant and the

equation of the moment at the point G:

( ) e( ) ( ) ( )TSR R S R S R S= + +

, (21.13)

( ) e( ) ( ) ( )TG G G GS S S S= + +

. (21.14)

From these equations, we deduce the 6 scalar equations:

,

0 ,

0 ,

0 ,

0 ,

0 .

l

l

l

l

l

l

mx k x X

mg Y

Z

L

M

N

= − +

= − +

=

=

=

=

(21.15)

The theorem of power-energy:

( ) ( ) ( ) ( ) cd

e( ) ( ) ( )d

T T T TE P S P S P St

= + + (21.16)

leads to the equation:

lmxx kxx X x= − + . (21.17)

We find the first of Equations (21.15) again.

Finally, we obtain 6 equations for 7 unknowns: Xl, Yl, Zl, Ll, Ml, Nl, x. An

additional equation will be derived from the physical nature of the connection.

The problem could then be entirely determined.

In fact, five equations of the system (21.15) are already solved:

lY mg= , (21.18)

0lZ = , (21.19)

( ) 0G S =

. (21.20)

It results that the action of connection is a force of which the support passes

through the mass centre G.

It remains to solve the first equation of (21.15):

lmx k x X= − + . (21.21)

The hypothesis on the physical nature of the connection will allows us to express

the component Xl. Next, solving (21.21) will lead to the expression of the motion

x as a function of time. Equation (21.21) is called the equation of motion.

We observe that, in the case of this spring-mass system, the theorem of power-

energy leads to the equation of motion.


21.2 VIBRATIONS WITHOUT FRICTION

21.2.1 Equation of Motion

In the case where the connection is without friction (perfect connection), the

power developed by the action of connection is zero. Thus from (21.11):

0lX = . (21.22)

The equation of motion is reduced to:

mx k x= − . (21.23)

This equation can be written in the reduced form:

20 0x xω+ = , (21.24)

introducing:

20

k

mω = . (21.25)

The quantity 0ω is the natural angular frequency of the spring-mass system

without friction.

21.2.2 Free Vibrations

The free vibrations are the vibrations which are observed when the solid (S) is

displaced from its equilibrium position and released. These vibrations are solu-

tions of Equation (21.24). This equation is satisfied by:

1 0cosx C tω= and 2 0sinx C tω= ,

where C1 and C2 are arbitrary constants. By addition of these solutions, we obtain

the general solution of equation of motion (21.24). Hence:

1 0 2 0cos sinx C t C tω ω= + . (21.26)

The vibratory motion represented by this equation is a simple harmonic

motion, where the constants C1 and C2 are deduced from the initial conditions.

We consider that, at the initial instant ( 0)t = , the solid has a displacement x0 from

its equilibrium, then the solid is released with a velocity 0x . By substituting 0t =

into Equation (21.26), we obtain:

1 0C x= . (21.27)

Deriving Equation (21.26) with respect to time, then substituting 0t = , we have:

02

0

xC

ω=

. (21.28)

Thus, the expression of the free vibrations of the solid (S) is written:


FIGURE 21.2 Free vibrations as function of time.

00 0 0

0

cos sinx

x x t tω ωω

= +

. (21.29)

This expression can also be written in the form:

( )m 0cosx x tω ϕ= − , (21.30)

with

22 0

m 00

xx x

ω = +

, (21.31)

and

1 0

0 0

tanx

xϕ

ω−=

. (21.32)

The displacement x as function of time t is reported in Figure 21.2. The maximum

displacement xm is called the amplitude of vibration and the angle ϕ is the phase

difference or phase angle. The interval of time T0 for which the motion repeats

itself is the natural period of vibrations and is expressed as:

00

2T

πω

= . (21.33)

The number f0 of cycles per unit of time is the natural frequency of the vibrations:

00

0

1

2f

T

ωπ

= = . (21.34)

21.2.3 Forced Vibrations. Steady State

In numerous practical applications, the solid (S) is subjected to a periodic

disturbing force or a periodic displacement is imposed to the spring support. The

response of the system to these conditions is referred as forced vibrations.

0

-0.00

Time t

Dis

pla

cem

ent

x

x0 xm

T0


We consider the case where the solid (S) of the spring-mass system of Figure

21.1 is subjected to a periodic force f of horizontal component m sinf tω . The

term m sinf tω is called a harmonic forcing function. The equation of motion

(21.23) without friction is then written as:

m sinmx kx f tω= − + . (21.35)

This equation is rewritten in the reduced form as:

20 m sinx x q tω ω+ = , (21.36)

with

mm

fq

m= . (21.37)

A particular solution of Equation (21.36) is:

3 sinx C tω= , (21.38)

where C3 is a constant which must satisfy Equation (21.36). We obtain:

3 2 20

mqC

ω ω=

−. (21.39)

Thus, the particular solution is given by:

2 20

sinmqx tω

ω ω=

−. (21.40)

The general solution of Equation (21.36) is obtained by adding this particular

solution to the general solution (2.7) of the free vibrations. We obtain:

1 0 2 0 2 20

cos sin sinmqx C t C t tω ω ω

ω ω= + +

−. (21.41)

The first two terms of this expression represent the free vibrations which were

considered previously. These free vibrations are also called transient vibrations

since in the practice these vibrations are rapidly damped by the damping forces

(Subsection 21.3.3.1). The third term, depending on the disturbing force, repre-

sents the forced vibrations of the system, obtained in the steady state. These

forced vibrations have the same period 2 /T π ω= as that of the disturbing force.

They can be expressed as:

m2 2 20 0

1sin

1 /

qx tω

ω ω ω=

−. (21.42)

The factor 2m 0/q ω is the displacement that the disturbing force qm would produce

if it was acting as a static force. The term ( )2 201/ 1 /ω ω− accounts for the dyna-

mical effect of the disturbing force. Its absolute value:

2 20

1( )

1 /K ω

ω ω=

−, (21.43)

is usually called the magnification factor. It depends only of the frequency ratio


FIGURE 21.3 Variation of the magnification factor as a function of the frequency.

0/ω ω , ratio of the frequency of the disturbing force to the natural frequency of the

system. The variation of the magnification factor is plotted against the frequency

ratio in Figure 21.3.

When the frequency of the disturbing force is small in comparison with the

frequency of the free vibrations, the magnification factor is approximately equal to

1. The displacements are about the same as in the case of a static disturbing force.

When the frequency of the disturbing force approaches the natural frequency of

the system, the magnification factor and thus the amplitude of the forced vibra-

tions rapidly increase and become infinite when the force frequency exactly coin-

cides with the natural frequency. The system is subjected to the resonance. In

practice, there is a dissipation of energy due to damping and the amplitude of the

vibrations is limited by the damping effects (Section 21.3). However, the system

should not be excited near its natural frequency.

When the frequency of the disturbing force increases beyond the natural

frequency, the magnification factor decreases and approaches zero for high values

of the frequency. The system may be considered as remaining stationary.

Considering the sign of the expression ( )2 201/ 1 /ω ω− , it is observed that for the

case where 0ω ω< this expression is positive. The displacement of the vibrating

mass has the same sign as that of the disturbing force. The vibrations are in phase

with the excitation. In the case where 0ω ω> , the expression is negative and the

displacement of the mass is in the direction opposite to that of the force. The

vibrations are out of phase.

In what precedes, we have considered the case of an excitation of the system by

an imposed force. It is also possible to produce forced vibrations by imposing a

displacement to the end support of the spring (Figure 21.4). In the case of a

harmonic displacement, the support displacement is:

m sins sx x tω= , (21.44)

where xs is the displacement of the support from the equilibrium position. The

displacement of the solid (S) referred to the support (T) is given by:

s rx x x= + , (21.45)

0.0 0.5 1.0 1.5 2.0 2.5 3.00

1

2

3

4

Frequency 0/ω ω

Magn

ific

ation f

acto

r K


FIGURE 21.4. Displacement imposed to the spring end.

introducing the displacement xr referred to the end support of the spring. The

resultant of the force exerted by the spring is transposed of (21.8). Thus:

( ) rR S k x i= − , (21.46)

and the equation of motion (21.23) is modified as:

rmx k x= − . (21.47)

This equation leads to the motion equation of forced vibrations:

m sinsmx k x k x tω+ = − . (21.48)

This motion equation can be written in the reduced form (21.36), setting:

2m m 0 ms s s

kq x x

mω= = . (21.49)

The motion equation is reduced to the case of a disturbing force.

In some other applications, it is more interesting to consider the case where an

acceleration is imposed to the support. The system spring-mass is then used as

accelerometer, device used to measure the acceleration of the support. In the case

of a harmonic acceleration, we have:

m sinsx a tω= . (21.50)

Considering Relations (21.45) and (21.47), the equation of motion is written in the

form:

s r rmx mx k x+ = − ,

or

m sinr rmx kx ma tω+ = − . (21.51)

Hence, the reduced form of motion equation:

20 m sinr rx x q tω ω+ = , (21.52)

setting

m mq a= − . (21.53)

Again, the motion equation is reduced to the form (21.36) of an imposed force.

y

(T)

(R) (S) ()

O

G x

xs x


However, it must be noted that Equation (21.52) is the motion equation expres-

sed in the relative reference associated to the end support of the spring. The forced

vibrations in this reference are transposed from Equation (21.42):

m2 2 20 0

1sin

1 /r

ax tω

ω ω ω= −

−(21.54)

21.3 VIBRATIONS WITH VISCOUS DAMPING

21.3.1 Equation of Motion with Viscous Damping

In previous discussions we did not consider the effects of dissipative forces. In

practice, it is necessary to take into consideration the damping forces which may

arise from several different sources, such as friction between dry sliding surfaces,

friction between lubricated surfaces, air or fluid resistance, internal friction due to

imperfect elasticity of materials, etc. Among all these processes of energy dissi-

pation, the simplest case to deal with mathematically is the case where the

damping force is proportional to the velocity. This damping process is called

viscous damping. The damping processes of complex types are generally re-

placed, for the purpose of the analyses, by an equivalent viscous damping. This

equivalent damping is determined in such a way as to produce the same dissi-

pation of energy by cycle as that produced by the actual damping processes

(Section 21.5).

In the case of a viscous damping of the spring-mass system of Figure 21.1, the

component Xl of connection introduced in (21.10) is opposed to the component x

of the velocity. Hence:

lX cx= − . (21.55)

The coefficient c is the coefficient of viscous damping. The motion equation

(21.21) is then written:

0mx cx k x+ + = . (21.56)

This equation can be rewritten in the reduced form:

202 0x x xδ ω+ + = , (21.57)

setting:

2

c

mδ = . (21.58)

The parameter δ is the damping coefficient. Equation (21.57) is the general reduced

form of the vibrations of a one degree of freedom system with viscous damping.



To solve Equation (21.57) of the free vibrations, we use the usual method for

21.3 Vibrations with Viscous Friction 319

solving the linear differential equations, by searching a solution of the form:

rtx Ce= , (21.59)

where r is a parameter determined by reporting Expression (21.59) into Equation

(21.57). Thus, we obtain the characteristic equation:

2 202 0r rδ ω+ + = . (21.60)

The solutions of this equation are:

1,2r δ ∆′= − ± , (21.61)

where ∆′ is the reduced discriminant of the characteristic equation:

2 20∆ δ ω′ = − . (21.62)

The final form of the solution of Equation (21.57) depends on the sign of ∆′ .

21.3.2.2 Case of Low Damping

In the case of low damping such as:

0δ ω< , (21.63)

the term ∆′ is negative and Equation (21.60) has two conjugated complex roots:

2

1,2 0 20

1r iδδ ωω

= − ± − . (21.64)

These two roots can be put in the form:

1,2 dr iδ ω= − ± , (21.65)

introducing the angular frequency:

2

0 20

1dδω ωω

= − . (21.66)

It is usual to introduce the viscous damping ratio ξ, defined as:

00

or δξ δ ξω

ω= = . (21.67)

It results that:

20 1dω ω ξ= − , (21.68)

and the two roots (21.65) are expressed as:

21,2 0 0 1r iξ ω ω ξ=− ± − . (21.69)

Finally, Equation (21.57) of the free vibrations can be written in the form:


20 02 0x x xξω ω+ + = . (21.70)

The two complex roots (21.65) are thus given by:

1 2, .d dr i r iδ ω δ ω= − + = − − (21.71)

or from (21.69):

( ) ( )2 21 0 2 01 , 1 r i r iω ξ ξ ω ξ ξ= − + − = − − − . (21.72)

Substituting these roots into Expression (21.59), we obtain two solutions of Equa-

tion (21.57) or (21.70). Any linear combination of these solutions is also a

solution. For example:

( )

( )

1 2

1 2

11 1

12 2

cos ,2

sin .2

r t r t td

r t r t td

Cx e e C e t

Cx e e C e t

i

δ

δ

ω

ω

−

−

= + =

= − =

Adding these solutions, we obtain the general solution of Equation (21.57) or

(21.70) in the form:

( )1 2cos sintd dx e C t C tδ ω ω−= + , (21.73)

where C1 and C2 are constants which are determined from the initial conditions.

The factor te δ− in Solution (21.73) decreases with time, and the vibrations

generated by the initial conditions are gradually damped out.

The expression between brackets in Equation (21.73) is of the same form that

the one obtained in the case of vibrations without damping (21.26). It represents a

harmonic function of angular frequency given by Equation (21.66) or (21.68).

This frequency is called the angular frequency of the damped vibrations. The

variation 0/dω ω of this frequency referred to the natural frequency of the free

undamped vibrations is plotted in Figure 21.5 as a function of the damping ratio

0/ξ δ ω= . From this figure, it is observed that the frequency of the damped vibra-

tions is close to the frequency of undamped vibrations, even for notable value of

the damping ratio. For 0.1ξ = , the damped frequency is 00.995dω ω= ; for

0.2ξ = , the frequency is equal to 00.98ω and for 0.3ξ = , the damped frequency

is still 00.95 .ωConstants C1 and C2 in Expression (2.56) are deduced from the initial condi-

tions at time 0t = : the solid is displaced from its equilibrium position by a displa-

cement x0 and the solid is released with a velocity 0x . Thus, reporting these initial

conditions into Equation (21.73) and into the expression of the derivative of the

displacement with respect to time, we obtain:

0 01 0 2,

d

x xC x C

δω+

= =

. (21.74)

Thus, the motion of damped free vibrations of a one degree of freedom system is:

0 00 cos sint

d dd

x xx e x t tδ δ

ω ωω

− + = +

. (21.75)


FIGURE 21.5. Frequency variation as a function of damping.

This expression can be rewritten in the form:

( )m costdx x e tδ ω ϕ−= − , (21.76)

expression in which the maximum value is:

( )22 2 2 0 0

m 1 2 0 2d

x xx C C x

δ

ω

+= + = +

, (21.77)

and the phase angle is given by:

1 1 0 01

2 0

tan tand

x xC

C x

δϕ

ω− − + = =

. (21.78)

Expression (21.76) may be considered as representing a pseudo-harmonic motion,

having an exponentially decreasing amplitude mtx e δ− , a phase angle ϕ and a

pseudo-period:

2d

d

Tπ

ω= . (21.79)

The graph of the motion is plotted in Figure 21.6. The displacement-time curve is

tangent to the envelopes mtx e δ−± at the points m1, 1m′ , m2, 2m′ , etc., at instants

separated by the time interval /2dT . Because the tangents at these points are not

horizontal, the points of tangency do not coincide with the points of extreme

displacements from the equilibrium position. If the damping ratio is low, the

difference in these points may be neglected. For any damping, the time interval

between two consecutive extreme positions is however equal to half the pseudo-

period. Indeed, the velocity of the vibrating solid is derived from (21.76) as:

( ) ( )m mcos sint td d dx x e t x e tδ δδ ω ϕ ω ω ϕ− −= − − − − . (21.80)

The velocity is equal to zero when:

( )tan dd

tδω ϕ

ω− = − , (21.81)

damping ξ 1 0 0

1

dfr

eq

uen

cy

ω


FIGURE 21.6 Pseudo-harmonic motion.

which leads effectively to / /2d dt Tπ ω= = .

The ratio between two successive amplitudes mix and m 1ix + is:

( )m m

m 1 m

id

i d

tTi

t Ti

x x ee

x x e

δδ

δ

−

− ++

= = . (21.82)

The quantity l dTδ δ= is the logarithmic decrement. It is given by:

m

m 1 0

2 2ln i

l di d

xT

x

πδ πδδ δω ω+

= = = ≈ . (21.83)

This equation can be used for an experimental determination of the damping coef-

ficient δ. However, a greater accuracy is obtained by measuring the extreme am-

plitudes separated by n pseudo-cycles. In this case we have:

m

m

dn Ti

i n

xe

xδ

+= , (21.84)

and the logarithmic decrement is obtained as:

m

m

1ln i

li n

x

n xδ

+= . (21.85)

21.3.2.3 Case of High Damping

In the case of high damping such as:

0δ ω> , (21.86)

the term ∆′ is positive and the characteristic equation (21.60) has two roots r1 and

r2 which are real and negative. The general solution of the motion equation

(21.57) is:

0

0

m1

m2

m3 m4

1m′

2m′3m′

/ dϕ ω

m3xm4x

m1x

m2x

dT

mx

0x

0

Time t

Dis

pla

cem

ent

x

tmx e δ−


FIGURE 21.7 Displacement as a function of time in the case of an aperiodic motion.

1 21 2

r t r tx C e C e= + . (21.87)

In this case the viscous damping is such as when the solid is displaced from its

equilibrium position, it does not vibrate but creeps gradually back to that position.

The motion is said aperiodic.

Constants C1 and C2 are deduced from the initial conditions:

( ) ( )0 00 , 0 ,x t x x t x= = = = (21.88)

which lead to:

1 2 0 1 1 2 2 0, .C C x r C r C x+ = + =

We obtain:

0 2 0 1 0 01 2

1 2 1 2

, ,x r x r x x

C Cr r r r

− −= =

− −

(21.89)

and Expression (21.87) is written:

1 20 2 0 1 0 0

1 2 1 2

r t r tx r x r x xx e e

r r r r

− −= +

− −

. (21.90)

The motion depends on the values of 0 0, and x xδ . Figure 21.7 shows examples

of displacement-time curves for a fixed value of the initial displacement x0 and

several values of the initial velocity 0x (positive, zero or negative).

21.3.2.4 Critical Damping

The transition between the pseudo-harmonic motion and the aperiodic motion

corresponds to a viscous damping cδ called critical damping and given by:

0cδ ω= . (21.91)

00

Time t

Dis

pla

cem

ent

x

x0

0 0x >

0 0x =

0 0x <


In this particular case, the characteristic equation (21.60) has a double root:

1 2 0r r ω= = − , (21.92)

and the solution of the motion equation is:

( ) 01 2

tx C C t e ω−= + . (21.93)

Taking the initial conditions into account, we obtain:

1 0 2 0 0 0, ,C x C x xω= = + (21.94)

and the solution of the motion equation (21.57) is written as:

( )[ ] 00 0 0 0

tx x x x t e ωω −= + + . (21.95)

The displacement-time curves are similar to the curves obtained in the case of

aperiodic motion (Figure 21.7), but the solid comes back to the equilibrium posi-

tion more rapidly for the critical damping.

21.3.3 Vibrations in the case of a Harmonic Disturbing Force

21.3.3.1 Time Domain

As in Section 21.2.3, we consider the case where the solid (S) of the spring-

mass system of Figure 21.1 is subjected to a harmonic force of horizontal compo-

nent m cosf tω . Under this condition, the motion equation (21.56) of the forced

vibrations becomes:

m cosmx cx k x f tω+ + = . (21.96)

This equation is written in the reduced form as:

20 m2 cosx x x q tδ ω ω+ + = , (21.97)

with

mm

fq

m= . (21.98)

Equation (21.97) constitutes the general form of the forced vibrations of a system

with one degree of freedom in the case of harmonic disturbing force.

A particular solution of Equation (21.96) is of the form:

cos sinx A t B tω ω= + , (21.99)

where A and B are constants which are determined by substituting Expression

(21.99) of this particular solution into the general equation of motion (21.97). We

obtain:

( ) ( )2 2 2 20 m 02 cos 2 sin 0A B A q t B A B tω δω ω ω ω δω ω ω− + + − + − − + = .


This equation is satisfied for all values of time t if:

2 20 m

2 20

2 ,

2 0.

A B A q

B A B

ω δω ω

ω δω ω

− + + =

− − + =

From which:

( )

( )

2 20

m22 2 2 20

m22 2 2 20

,

4

2.

4

A q

B q

ω ω

ω ω δ ω

δω

ω ω δ ω

−=

− +

=− +

(21.100)

Next, the total solution of Equation (21.97) is obtained by adding the particular

solution (21.99) to the general solution of Equation (21.97) with the second

member equal to zero, thus to the general solution of Equation (21.57) of the free

vibrations.

We consider hereafter the case of low damping for which the damping is lower

than the critical damping. Thus, the solution of Equation (21.97) is given by:

( )1 2cos sin cos sintd dx e C t C t A t B tδ ω ω ω ω−= + + + . (21.101)

The first term represents the damped free vibrations, whereas the last two terms

represent the damped forced vibrations. The free vibrations have the angular fre-

quency dω as determined in Subsection 21.3.2.2, when the forced vibrations have

the angular frequency of the disturbing force. Due to the factor ,te δ− the free

vibrations gradually decrease, then vanish, leaving only the steady forced vibra-

tions. These vibrations are maintained as long as the disturbing force is applied.

We study the forced vibrations hereafter.

In the case of steady-state, the harmonic response (21.99) may be written in the

form:

( )m cosx x tω ϕ= − , (21.102)

with

2 2 1m , tan .

Bx A B

Aϕ −= + =

Hence:

( ) ( ) ( )

2m m 0

m2 2 22 2 2 2 2 2

0 0 0

/

4 1 / 2 /

q qx

ω

ω ω δ ω ω ω ξω ω

= =

− + − +

, (21.103)

and

1 1 02 2 2 20 0

2 /2tan tan

1 /

ξω ωδωϕ

ω ω ω ω

− −= =− −

. (21.104)

When a static load fm is applied to the system, the static displacement xst is


deduced from (21.96) as:

st m mk x f mq= = .

From which:

m mst 2

0

q fx

kω= = . (21.105)

Thus, considering Equations (21.103) and (21.105), the amplitude xm of the displa-

cement may be written in the form:

m st( )x K xω= , (21.106)

in which K(ω) is the magnification factor expressed by:

( ) ( )2 22 2

0 0

1( )

1 / 2 /

K ω

ω ω ξω ω

=

− +

. (21.107)

So, the damped harmonic vibrations can be written as:

( )st ( ) cosx x K tω ω ϕ= − . (21.108)

21.3.3.2 Frequency Domain

The steady state of the harmonic forced vibrations can be studied in the fre-

quency domain by representing the excitation ( )f t and the response ( )x t in com-

plex forms ( ) and ( ) ,i t i tF e X eω ωω ω respectively. The quantities ( )F ω and ( )X ωare the complex amplitudes associated to the excitation and response, respecti-

vely. In the case of the harmonic forced vibrations considered previously, the

complex amplitudes are:

m m( ) , ( ) .iF f X x e ϕω ω −= = (21.109)

Introducing these complex forms into the motion equation (21.97) leads to the

complex equation of motion which may be written in one of the following forms:

( )2 20

12 ( ) ( )i X F

mω ω δω ω ω− + = , (21.110)

or

( )2 20 0

12 ( ) ( )i X F

mω ω ξω ω ω ω− + = . (21.111)

Thus, the response as a function of the excitation in complex form is expressed as:

1( ) ( ) ( )X H F

mω ω ω= , (21.112)

introducing the transfer function of the vibration system expressed by:

2 20 0

1( )

2H

iω

ω ω ξω ω=

− +. (21.113)


This transfer function is sometimes resolved in the form:

( )R R

Hi r i r

ωω ω

= +− −

, (21.114)

where

21 0 0

22 0 0

1 ,

1 ,

d

d

r r i i

r r i i

δ ω ξω ω ξ

δ ω ξω ω ξ

= = − + = − + −

= = − − = − − − (21.115)

and

1 1, .

2 2d d

R Ri iω ω

= = − (21.116)

The conjugate quantities R and R are then called the residues of the transfert

function and the quantities r and r are the poles of the transfer function.

When the frequency approaches zero, the transfer function ( )H ω approaches

201 ω and the function X(ω = 0) is identified with the static response xst introduced

in Equation (21.105). So, Expression (21.112) of the response may be rewritten as:

1( ) ( ) ( )rX H F

kω ω ω= , (21.117)

introducing the reduced transfer function of the vibration system expressed by:

2 20 0

1( )

1 / 2 /rH

iω

ω ω ξω ω=

− +. (21.118)

So, the complex amplitude X(ω) is simply given by:

st( ) ( )rX H xω ω= . (21.119)

Next, the amplitude xm of the harmonic steady-state vibration is deduced from

the previous expression (21.119), by considering the modulus of X(ω ), which

yields:

m st( )rx H xω= , (21.120)

with

( )

22 2 2 2 20 0

1( )

1 / 4 /

rH ω

ω ω ξ ω ω

=

− +

. (21.121)

The modulus of the function Hr(ω ) is identified with the magnification factor in-

troduced in (21.107).

The phase angle ϕ is the opposite of the argument of the transfer function or of

the reduced function Hr(ω ). Thus:

1 02 2

0

2 /arg ( ) tan

1 /rH

ξω ωϕ ω

ω ω

−= − =−

. (21.122)

which is the result expressed in Equation (21.104).


0.0 0.5 1.0 1.5 2.00

1

2

3

4

Frequency 0/ω ω

Mag

nif

icat

ion

fac

tor

K

0ξ =

0.10

0.15

0.20

0.25

0.50

0.70

1.00

FIGURE 21.8. Variation of the reduced amplitude of harmonic vibrations as a function of

the frequency for different values of damping.

21.3.3.3 Effect of the Frequency of the Disturbing Force

The amplitude xm of the harmonic forced vibrations, referred to the static dis-

placement xst, is simply given either by the magnification factor (Relation

(21.106)), or by the modulus of the reduced transfer function (Relation (21.120)):

m

st

( ) ( )rx

H Kx

ω ω= = , (21.123)

where K(ω ) is expressed in (21.107).

Figure 2.8 shows the variation of the magnification factor as a function of the

reduced frequency 0/ω ω for different values of damping. From these curves it is

observed that, when the angular frequency is small compared to the natural fre-

quency, the value of the magnification factor is not greatly different from unity.

Thus, the amplitude of vibrations is approximately the one which would be pro-

duced by a static disturbing force.

When the angular frequency of the excitation is large compared to the natural

frequency, the value of the magnification factor tends toward zero, regardless the

value of damping. So, a high frequency disturbing force induces practically no

forced vibrations of the system.

The curves of Figure 2.8 show that for low values of damping the magni-

fication factor grows rapidly with the frequency, and its value near resonance is

very sensitive to the values of damping. It is also observed that the maximum

value occurs for a value of 0/ω ω less than unity. Setting the derivative of the ma-

gnification factor with respect to 0/rω ω ω= equal to zero, we find that the maxi-


mum occurs for a reduced frequency mrω defined by:

2mm

0

1 2rω

ω ξω

= = − . (21.124)

The maximum amplitude of the amplification factor is then given by:

m m2

1( ) ( )

2 1rK Hω ω

ξ ξ= =

−. (21.125)

For small damping ratios the maximum value of the magnification factor occurs

very near to the undamped natural frequency and the maximum is approximately:

m m1

( ) ( )2

rK Hω ωξ

= ≈ . (21.126)

For example, for 0.20ξ = , the maximum occurs for 00.96 ω and its value is

2.55. Then, when the damping increases, the value of angular frequency mω

decreases and vanishes when 1ξ = .

In the case of low damping, the peak width of the magnification factor can be

evaluated by considering the reduced frequencies rω for which the magnification

factor is reduced by a factor 1/ 2 with respect to the maximum, corresponding to

a reduction of –3 dB. We obtain:

( )2 22 2 2

1 1 1

2 2 11 4r rξ ξω ξ ω

=−− +

. (21.127)

The expansion of this equation leads to:

( )4 2 2 2 42 1 2 1 8 8 0r rω ξ ω ξ ξ− − + − + = . (21.128)

The solutions for this equation are:

( )

( )

2 2 21

2 2 22

2 1 2 2 1 ,

2 1 2 2 1 .

r

r

ω ξ ξ ξ

ω ξ ξ ξ

= − + −

= − − − (21.129)

An approximate solution can be formulated in the case of low values of damping

by expressing that 1rω and 2rω are not greatly different from the frequency mrωof the maximum. Hence:

2 21 2 1 2 1 2 m 1 2( )( ) 2 ( )r r r r r r r r rω ω ω ω ω ω ω ω ω− = + − ≈ − ,

or considering (21.124):

2 2 21 2 2 1 2r r rω ω ξ ∆ω− ≈ − , (21.130)

where r∆ω is the frequency band corresponding to –3 dB reduction centred on


mrω . Considering Equations (21.129), we finally obtain:

2

2

12

1 2r

ξ∆ω ξ

ξ

−=

−. (21.131)

In the case of low values of the damping, the expression of the bandwidth is

simplified as:

2r∆ω ξ≈ . (21.132)

The frequency response of the damped system is also characterized by the

phase angle ϕ expressed by Equations (21.104) and (21.122). Figure 21.9 shows

the variation of the phase angle as a function of the frequency obtained for

different values of the damping. For frequencies much lower than the natural

frequency of the system, the vibrations are in phase with the imposed force

whatever the value of the damping. Then, the phase angle increases, differently

according to the value of the damping, and reaches a phase delay of /2π (thus a

delay of a quarter-cycle) when the frequency of the force of excitation reaches the

value of the natural frequency. Next, the phase angle continues to increase and

tends towards π for high values of the frequency. This value is reached all the

more quickly than the damping is low.

FIGURE 21.9. Variation of the phase angle as a function of the frequency, for different

values of damping.

0.0 0.5 1.0 1.5 2.0 2.5 3.00

20

40

60

80

100

120

140

160

180

0.1

0ξ =

0.2

0.5

1 2 4

Frequency 0/ω ω

Ph

ase

ang

le

(

°)


21.3.4 Forced Vibrations in the case of a Periodic Disturbing Force

In the case where the solid (S) of the spring-mass system of Figure 21.1 is

submitted to a force of horizontal component ( )f t function of time, the motion

equation (21.96) is written:

( )mx cx kx f t+ + = . (21.133)

If the imposed force ( )f t is periodic of period T, the force can be expanded in the

form of Fourier series as:

( )0

1

( ) cos sinn n

n

f t a a n t b n tω ω∞

=

= + + , (21.134)

with 2 /Tω π= . The coefficients a0, an and bn are expressed as:

00

1( ) d

T

a f t tT

= , (21.135)

0

2( )cos d

T

na f t n t tT

ω= , (21.136)

0

2( )sin d

T

nb f t n t tT

ω= . (21.137)

The equation of motion (21.133) is thus written in the reduced form as:

( )20 0 0

1

2 cos sinn n

n

x x x q q n t p n tξω ω ω ω∞

=

+ + = + + , (21.138)

with

00 , , .n n

n na a b

q q pm m m

= = = (21.139)

The general solution of Equation (21.138) consists of the sum of the free vibra-

tions and the forced vibrations. The free vibrations diminish and vanish with

damping. The forced vibrations are obtained by superimposing the steady state

forced vibrations produced by every terms of the second member of Equation

(21.138) These vibrations can be obtained by applying the results obtained in the

previous section (Subsection 21.3.3). In practice, the coefficients of the terms of

the series decrease when n increases. So, the analysis will be limited to a value N

of n for which the terms of upper orders can be neglected. Considering the results

established in Subsection 21.3.3, it can be concluded that forced vibrations with

high amplitudes may occur when the period of one of the terms of series (21.134)

coincides with the period of the natural vibrations of the system, i.e. if the period

T of the disturbing force is equal to, or a multiple of the damped period Td.


21.3.5 Vibrations in the case of an Arbitrary Disturbing Force

We consider, in this subsection, the case where the spring-mass system is

submitted to an arbitrary force ( )f t . The reduced equation of motion is deduced

from Equation (21.133) and is written:

20 02 ( )x x x q tξω ω+ + = , (21.140)

introducing the force q(t) imposed by unit mass:

1( ) ( )q t f t

m= . (21.141)

The force q(t) is arbitrary. Its variation is represented in Figure 21.10.

At any instant t′ , we may consider (figure 21.10) an impulse of height

( )q t q′ = and width d t′ . This impulse imparts to each unit of mass an instant-

taneous acceleration from the instant t′ given by:

d

dx x q

t= =

′ , (21.142)

which leads to an increase in velocity from t′ given by:

d dx q t′= . (21.143)

regardless of what other forces, such as the spring force, may be acting, and

regardless of the displacement and velocity of solid (S) at the instant t′ . Then, the

increment of displacement at instant t posterior to t′ , is deduced from Equation

(21.75) by substituting the velocity increment (21.143) for the initial velocity 0x

(with a zero initial displacement) and substituting the instant t t′− for instant t (in

Equation (21.75), the disturbing force is exerted at instant 0t = , whereas the

force ( )q t′ is applied at t t′= ). We obtain:

( ) dd sin ( )t t

dd

q tx e t tδ ω

ω′− − ′

′= − . (21.144)

FIGURE 21.10. Arbitrary force as a function of time.

t′ dt′ Time t

Red

uce

d f

orc

e

q(t

)

q


Since each impulse ( ) dq t t′ ′ between 0t′ = and t t′ = produces an increment of

displacement given by the preceding expression, the total displacement x(t) which

results from the disturbing force is obtained by integration between 0 and t:

0

( ) ( ) sin ( )dtt

td

d

ex t e q t t t t

δδ ω

ω

−′ ′ ′ ′= − . (21.145)

This form is referred as Duhamel’s integral. It includes both steady state and

transient terms. The integral can be evaluated by an analytical method or a nume-

rical process.

To take account of the effect of possible initial conditions of displacement x0

and velocity 0x , it is necessary to add to the results (21.145) the solution for the

initial conditions considered in Equation (21.75). Thus, the total solution is:

0 00

0 0

1( ) cos sin ( )sin ( )d

tt t

d d dd

x xx t e x t t e q t t t tδ δδ

ω ω ωω ω

′− +′ ′ ′= + + −

.

(21.146)

21.3.6 Forced Vibrations in the case of a Motion Imposed to the Support

21.3.6.1 Equation of Motion

We consider the case where a motion is imposed to the support connected to

one end of the spring (Figure 21.4). The displacement of the solid (S) in the refe-

rence attached to the support is thus expressed by Relation (21.45). The resultant

of the force exerted by the spring is given by Expression (21.46), and the

component of the viscous friction is expressed in (21.55). The motion equation

(21.56) is modified as:

( ) 0smx cx k x x+ + − = , (21.147)

or

smx cx kx kx+ + = . (21.148)

This equation shows that the system is submitted to the imposed force skx .

Equation (21.148) is written in the reduced form:

20 02 sx x x qξω ω+ + = , (21.149)

setting:

20s s s

kq x x

mω= = . (21.150)

Equation (21.149) is identical to the motion equation with an imposed force. We

are brought back to the case studied in Subsections 21.3.3 to 21.3.5.


y

(T)

(R) (S) G

xxs

FIGURE 21.11. Motion imposed to the support-spring-mass set.

In the case of Figure 21.4, the motion is imposed to the one end of the spring,

the solid (S) remaining in contact with the support. Another case can be consi-

dered (Figure 21.11), where the support-spring-mass set has an imposed motion

with respect to the reference (T). In this case, Expression (21.55) of the

component Xl of friction is modified as:

( )l sX c x x= − , (21.151)

and the equation of motion (21.148) becomes:

s smx cx kx kx cx+ + = + . (21.152)

The system is then submitted to two imposed forces: the one skx and the other

scx . The preceding case is thus a particular case of the present general case.

Equation (21.152) is written in the reduced form:

20 0 1 22 s s sx x x q q qξω ω+ + = = + , (21.153)

with

21 0s s s

kq x x

mω= = , (21.154)

2 10

2s s s

cq x q

m

ξω

= = . (21.155)

21.3.6.2 Case of a Harmonic Motion Imposed to the Support

We consider the case where a harmonic motion is imposed to the support:

m coss sx x tω= . (21.156)

We analyse the most general case of the equation of motion (21.153). The

imposed forces are:

21 0 m coss sq x tω ω= , (21.157)

and

2 m0

2 sins sq x tωξ ωω

= − . (21.158)


The total imposed force is:

21 2 m 0

0

cos 2 sins s s sq q q x t tωω ω ξ ωω

= + = −

, (21.159)

which may be put in the form:

( )m coss sq q tω α= − , (21.160)

with

22 2

m m 0 20

1 4s sq xωω ξω

= + , (21.161)

1

0

tan 2ωα ξω

− =

. (21.162)

The equation of motion (21.153) is then written as:

( )20 0 m2 cossx x x q tξω ω ω α+ + = − . (21.163)

It is identical to the motion equation (21.97) obtained in the case of an imposed

force, qm being replaced by msq and the phase angle α being introduced. The

results obtained in Subsection 21.3.3 can then be transposed to the present case.

For the steady state, the response is deduced from Expression (21.102). We obtain:

( )m coss sx x tω α ϕ= − − , (21.164)

with

( ) ( )

2 2 2m 0

m2 22 2

0 0

1 4 /

1 / 2 /

sxx

ξ ω ω

ω ω ξω ω

+=

− +

, (21.165)

1 02 2

0

2 /tan

1 /

ξω ωϕ

ω ω

−=−

. (21.166)

This way of proceeding can also be applied to the case of an arbitrary motion

imposed to the support, while transposing the results derived in Subsection 21.3.5.

21.3.6.3 Case where an Acceleration is Imposed to the Support

We consider the case where a given acceleration sx is imposed at the spring-

mass-support set (Figure 21.11). The equation of motion is given by Equation

(21.152), which is written, introducing the relative displacement xr of the solid

(S), in the form:

r r r smx cx kx mx+ + = − . (21.167)

The second member is equivalent to an imposed force of value smx− . By divi-


ding Equation (21.167) by the mass m, we obtain the reduced equation of the

motion:

20 02r r r srx x x qξω ω+ + = , (21.168)

with

sr sq x= − . (21.169)

Equation (21.169) is identical to the motion equation with an imposed force. We

are brought back to the cases studied in Sections 21.3.3 to 21.3.5.

21.4 VIBRATIONS WITH DRY FRICTION


In the general case, the motion of the solid (S) of Figure (21.1) is characterized

by the first two Equations (21.15):

lmx kx X= − + , (21.170)

lY mg= . (21.171)

In the case of a dry friction between two solids, the components Xl and Yl are

related by the Coulomb’s law (Chapter 13), which introduces the coefficient f of

friction. The Coulomb’s law states that, if the solid (S) is in equilibrium, the

components Xl and Yl are such as:

l lX f Y< ,

Thus, from (21.171):

lX f mg< . (21.172)

If the solid (S) moves, the Coulomb’s law states that components Xl and Yl are

related by the equality:

l lX f Y f mg= = , (21.173)

and that the component Xl is opposed to the velocity of sliding x . The equality

(21.173) can thus be put in the form:

sign( )lX x f mg= − . (21.174)

In the case where the solid is in equilibrium, the equation of motion (21.170)

becomes:

lX kx= . (21.175)

This result associated to the condition (21.172) of friction leads to the relation:

kx f mg< . (21.176)

21.4 Vibrations with Dry Friction 337

Thus, there is equilibrium of the solid if:

2 20 0

1 1fg x fg

ω ω− < < . (21.177)

These inequalities define the limits for the equilibrium of the solid (S).

In the case where the solid (S) moves, Equation (21.170) and the condition

(21.174) of friction leads to the equation of motion:

sign( )mx kx x f mg+ = − , (21.178)

equation which may be rewritten in the reduced form:

20 sign( )x x x f gω+ = − , (21.179)

where 0ω is the natural angular frequency of the spring-mass system without

friction.

Over each interval of time where x keeps a constant sign, the general solution

of the equation of motion is:

1 0 2 0sign( ) cos sinx x f g C t C tω ω= − + + . (21.180)

The values of C1 and C2 depend on the initial conditions for the interval under

consideration. Thus, it results that intervals of time, corresponding to signs of x

which are different, will succeed while satisfying the continuity of the functions

x(t) and ( )x t : the values of the two functions at the end of one interval of time

will provide the initial conditions for the following interval of time.


The motion for the free vibrations of the solid (S) is given by Equation

(21.180). So as to illustrate this motion, we consider the case where, at the initial

instant ( 0t = ), the solid is displaced from its equilibrium position by a displace-

ment x0 and the solid is released with a velocity 0x positive.

Following these initial conditions, an episode of motion occurs with 0x > . The

equation of motion is, from (21.180):

1 0 2 0

0 1 0 0 2 0

cos sin ,

sin cos .

x f g C t C t

x C t C t

ω ω

ω ω ω ω

= − + +

= − + (21.181)

The initial conditions impose:

0 1 0 0 2, ,x fg C x Cω= − + =

and the motion (21.181) is written:

( )

( )

00 0 0

0

0 0 0 0 0

cos sin ,

sin cos .

xx fg x fg t t

x x fg t x t

ω ωω

ω ω ω

= − + + +

= − + +

(21.182)


This episode of motion continues until the instant where x becomes equal to zero,

hence until time t1 such as:

( )1 0

0 10 0

tanx

tx fg

ωω

−=+

. (21.183)

The displacement of the solid reaches thus the value x1:

( ) 01 1 0 0 1 0 1

0

( ) cos sinx

x x t fg x fg t tω ωω

= = − + + +

. (21.184)

At this first episode, an interval of time succeeds with 0x < , of which the

equation of motion starting from the time t1 is:

1 0 1 2 0 1cos ( ) sin ( )x fg C t t C t tω ω= + − + − , (21.185)

with for initial conditions at time t1: 1 1( )x t x= and 1( ) 0x t = . Taking account of

these conditions, the equations of motion are:

( )

( )1 0 1

1 0 0 1

cos ( ),

sin ( ).

x fg x fg t t

x x fg t t

ω

ω ω

= + − −

= − − − (21.186)

This episode of motion continues until the time 2 1 0/2t t T= + 0 0( 2 / ),T π ω= where

the velocity x becomes equal to zero. The displacement of the solid reaches thus

the value x2 given by:

2 1( 2 )x x fg= − − . (21.187)

Two possibilities exist then. Either x2 is included in the limits of equilibrium and

the motion stops. Or x2 is outside these limits, and a new episode of motion occurs

with 0x > .

In the case where the motion continues, the following episode has for equation

of motion, starting from t2:

1 0 2 2 0 2cos ( ) sin ( )x fg C t t C t tω ω= − + − + − , (21.188)

with for initial conditions at time t2: 2 2( )x t x= and 2( ) 0x t = . Hence the equa-

tions of motion:

( )

( )2 0 2

2 0 0 2

cos ( ),

sin ( ).

x fg x fg t t

x x fg t t

ω

ω ω

= − + + −

= − + − (21.189)

This episode continues until time 3 2 0/2t t T= + , where the velocity becomes equal

to zero again. The displacement of the solid reaches then the value x3 given by:

3 2 1( 2 ) 4x x fg x fg= − + = − . (21.190)

At the instant t3, we must consider whether the value of x3 is included or not in

the limits of equilibrium, and so on.

21.5 Equivalent Viscous Damping 339

FIGURE 21.12. Free vibrations of one degree freedom system with dry friction.

The graph of displacement as a function of time is thus constituted, from time

t1, of a succession of descending arcs of half-sinusoids (Equation (21.186)), of

which the points of inflection have as abscissa ,x fg= connected with a hori-

zontal tangent to ascending arcs of half-sinusoids (Equation (21.189)), of which

the points of inflection have as abscissa .x fg= − The extreme elongations

reached at each alternation decrease according to an arithmetic progression. The

number of episodes executed from time t1 and before the stop of the motion is the

greatest integer strictly lower than 10.5 /2x fg+ .

21.5 EQUIVALENT VISCOUS DAMPING

21.5.1 Introduction

As reported in Section 21.3.1, the damping of vibrations can be induced by

different phenomena. The viscous damping has been studied extensively in

Section 21.3. Dry friction has been considered in Section 21.4. The implemented

analysis showed the complexity to take into account this type of friction while

considering the laws of dry friction. This complexity of the analysis is also found

for the other types of friction: internal friction in materials, fluid friction, etc.

These different types of friction can be replaced by an equivalent viscous friction,

in order to lead the analysis back to the analysis implemented Section 21.3. The

equivalent viscous friction is then evaluated so as to produce the same dissipation

of energy per cycle as the actual phenomena of friction.

0-1,5

0,0

1,5

x1

1 4x fg−

1 8x fg−

1( 2 )x fg− −

1( 6 )x fg− −

1( 10 )x fg− −

t1

1 0/2t T+ 1 03 /2t T+ 1 05 /2t T+

1 0t T+ 1 02t T+

x0

0

20

1fg

ω

20

1fg

ω−

Time t

Dis

pla

cem

ent

x


21.5.2 Energy Dissipated in the case of Viscous Damping

The work done per cycle by the disturbing force m( ) cosf t f tω= during the

steady state response is:

0

cos dT

mW f x t tω= . (21.191)

The velocity x may be obtained by differentiating Expression (21.102) of the dis-

placement with respect to time. Hence:

m sin( )x x tω ω ϕ= − − . (21.192)

Combining Relations (21.191) and (21.192) leads to the expression of the work:

m m sinW x fπ ϕ= . (21.193)

Similarly, the energy aU dissipated per cycle by the viscous damping force cx

is given by:

0

dT

aU cxx t= . (21.194)

Hence, taking account of (21.192), then integrating:

2maU cxπ ω= . (21.195)

For a harmonic steady state, the work done by the disturbing force is equal to the

dissipated energy. From which the amplitude of the displacement is deduced as:

mm sin

fx

cϕ

ω= . (21.196)

When the angular frequency is equal to the natural frequency ( 0ω ω= ), the phase

angle ϕ is /2π and the displacement amplitude is:

mm 0

0

( )f

xc

ωω

= . (21.197)

This result coincides with the result (21.126) obtained for low values of damping.

The equivalent viscous damping constant will be obtained by equating Expres-

sion (21.195) of the energy dissipated by viscous damping to the energy dissi-

pated by the actual damping process. We consider different cases in the following

subsections.

21.5.3 Structural Damping

The structural damping is associated to the internal friction in materials which

are not perfectly elastic. For these materials, the loading stress-strain curve for

increasing levels of stress and strain is different from the unloading curve. Figure

21.13 shows the hysteresis loop obtained in the case of one cycle of vibration.


FIGURE 21.13. Stress-strain curve for successive loading and unloading of a material.

The experimental results show that the energy dissipated per cycle is approxi-

mately proportional to the square of the strain amplitude. So, the work sU dissi-

pated by structural damping may be written as:

2ms sU xα= , (21.198)

in which sα is a parameter which characterizes the structural damping of the

material considered. Equating Expressions (21.195) and (21.198) of the dissipated

energies leads to the equivalent viscous damping constant:

eqsc

απω

= . (21.199)

The parameter sα has the dimension of a stiffness k and it is usually replaced by

kη , introducing the dimensionless quantity:

s

k

αη

π= . (21.200)

This quantity is called structural damping factor. Relations (21.199) and (21.200),

associated with Relations (21.58) and (21.67) lead to the expression of the

equivalent viscous friction ratio:

eq eq0 0 0

1 1 1

2 2 2

kc

m m

ωξ η ηω ω ω ω

= = = . (21.201)

By substituting this expression into Equation (21.118), the reduced transfer func-

tion is written:

2 20

1( )

1 /rH

iω

ω ω η=

− +, (21.202)

loading

unloading

Strain

Str

ess


and the magnification factor becomes:

( )

22 2 20

1( ) ( )

1 /

rK Hω ω

ω ω η

= =

− +

. (21.203)

For the natural frequency, the magnification factor is:

0 01

( ) ( )rK Hω ωη

= = , (21.204)

and the amplitude of vibrations is deduced from (21.120):

mm st

1 1 fx x

kη η= = . (21.205)

21.5.4 Dry Friction

In the case of a contact with dry friction between two solids, the process of

friction is generally described by the laws of Coulomb (Relations (21.172) to

(21.174)). These laws introduce the coefficient f of friction. Experiment shows

that, during motion, this coefficient is rather constant, and usually lower than the

coefficient of friction when there is equilibrium.

To determine the equivalent viscous damping, we consider the energy dissi-

pated by the component of friction Xl expressed in (21.173). The energy dissi-

pated fU per cycle is:

f m4 lU f Y x= . (21.206)

In the case of the horizontal spring-mass system (Figure 21.1), we have lY mg= .

In the most general case, lY could be a component of tightening imposed to the

solid (S), orthogonal to the direction of the motion.

By equalling the energy (21.206) to the energy (21.195) dissipated by viscous

friction, we obtain the equivalent friction coefficient:

eqm

4 lf Yc

xπ ω= . (21.207)

In this case, the equivalent friction coefficient depends on the normal component

lY of friction and ω , but also on the amplitude xm of the vibration.

As previously, the equivalent damping ratio eqξ is deduced from Expression

(21.207), associated to Relations (21.58) and (21.67). Thus:

0eq eq

0 m

21

2lf Y

cm x k

ωξ

ω π ω= = . (21.208)

The reduced transfer function (21.118) is written:


2

2m0

1( )

1 4

rl

Hf Y

ix k

ωω

πω

=

− +

. (21.209)

The amplitude given by Expression (21.120) leads to:

stm

2 22

2m0

41 l

xx

f Y

x k

ωπω

=

− +

. (21.210)

From this expression, we derive the amplitude of the motion:

2

mm st 2

20

41

1

lfY

fx x

π

ω

ω

− = ±

−

. (21.211)

The second term of this expression is the magnification factor. This factor has a

real value if:

m 4lf Y

f

π≤ . (21.212)

In practice where low forces of friction are induced, this condition is satisfied. In

this hypothesis, the magnification factor becomes infinite when the frequency

reaches the value of the natural frequency (Relation (21.211)). This result is

explained by the fact that for the value of the natural frequency, the energy dissi-

pated per cycle is lower than the energy brought by the imposed force. In fact,

Relations (21.206) and (21.212) lead to:

f m mU f xπ< . (21.213)

The expression of the work supplied by the imposed force, shows that, for the

natural frequency 0ω , this work is:

0 m m( )W x fω π= . (21.214)

We verify indeed that for the natural frequency:

f 0( )U W ω< . (21.215)

21.5.5 Fluid Friction

As another type of friction, we consider the case of a solid immersed in a fluid

of low viscosity, such as the air for example. In the case where the mass of the

solid is low and its volume is rather high, it is necessary to take account of the

friction induced by the resistance of the fluid. The force of resistance (Figure

21.14) exerted by the fluid on the solid can be evaluated as:


FIGURE 21.14. Fluid friction.

2fl t p

1

2R x C Sρ= , (21.216)

where ρ is the mass per unit volume of the fluid, tC the drag coefficient and pS

the area of the section projected in a plane orthogonal to the direction of the

motion. The force of resistance exerted by the fluid is proportional to the square

of the velocity and is opposite to its velocity. The energy Ufl dissipated per cycle

by this force is:

/4

fl fl fl0 0

d 4 dT T

U R x t R x t= = . (21.217)

Introducing Relations (21.102) and (21.216) into the preceding expression, then

integrating, the dissipated energy is written:

3 2fl fl m

8

3U C x ω= , (21.218)

setting:

fl t p1

2C C Sρ= . (21.219)

While identifying the energy (21.218) with the energy dissipated in the case of

viscous damping (21.195), we obtain the equivalent damping coefficient:

eq fl m8

3c C x ω

π= . (21.220)

As previously, the equivalent viscous damping ratio eqξ is deduced from the

preceding expression, considering Relations (21.58) and (21.67):

fl m 0eq

4

3

C x

k

ω ωξ

π= . (21.221)

k

m

(S)

Sp

x


The reduced transfer function (21.118) is written:

22fl m

20

1( )

1 83

rHC x

ik

ωωω

πω

=

− +

. (21.222)

The amplitude of the vibrations is expressed by (21.120). Thus:

stm

22 22fl m

20

81

3

xx

C x

k

ωωπω

=

− +

. (21.223)

This expression leads to the quadratic equation of which xm is solution:

2 22 24 2 2 2flm m m2

0

81 0

3

Cx k x f

ω ωπ ω

+ − − =

. (21.224)

21.5.6 Conclusion

In conclusion, an equivalent viscous damping may always be considered, what-

ever the dissipative process of energy which is induced. The dissipated energy RU

per cycle is expressed in the form:

0

dT

RU Rx t= , (21.225)

where R is the resultant of the force of resistance to the motion and x is the

velocity of displacement deduced from Expression (21.103).

The equivalent viscous damping coefficient is then evaluated by equalling the

dissipated energy (21.225) with the energy (21.195) induced by the process of

viscous damping. Thus:

eq 2R

m

Uc

xπ ω= . (21.226)

Next, the equivalent viscous damping ratio eqξ is determined by using Relations

(21.58) and (21.67):

eqeq

02

c

mξ

ω= . (21.227)

This viscous damping ratio determines the transfer function ( )rH ω using Relation

(21.118). The modulus of this function relates (21.120) the amplitude of the vibra-

tions to the amplitude of the response obtained in the case of an imposed static

force.

Lastly, let us note that it is also possible to take account simultaneously of

several types of damping.


( )f t

A

−A

T/4

T/2

3T/4 T 2T

t

EXERCISES

21.1 A wheel rolls (Figure 21.15) over an undulated surface with a constant

speed . The undulated surface can be approximated by a sine curve of equation

sin /y d x lπ= , with d = 30 mm and 1 m.l = A mass m equal to 80 kg is

connected to the axle-tree of the wheel through an elastic device of stiffness k

equal to 150 kN/m. Derive the amplitude of the forced vibrations of the mass, as a

function of the speed of the wheel, considering a viscous damping of the vibra-

tions with a coefficient 0.10.ξ =

21.2 A one degree freedom system is submitted to a periodic force ( )f t of which

the variation as a function of time is reported in Figure 21.16. The function is

characterized by its amplitude A and its period T. Study the vibrations in the case

of the steady state.

FIGURE 21.15. Mass-spring-wheel system.

FIGURE 21.16. Excitation in triangle form.

COMMENTS

The study of the vibrations of a system with one degree of freedom is

particularly important, since the derived results are at the foundation of the

modal analysis of the vibrations of a complex structure. Thus, the reader

interested by the vibratory phenomena will have to pay a great attention to

all the concepts introduced in the present chapter.

x

k

m

Ay

CHAPTER 22

Motion of Rotation of a Solid about a Fixed Axis

22.1 GENERAL EQUATIONS

22.1.1 Introduction

The motion of rotation about an axis occurs in many industrial applications:

rotors, wheels, crankshafts, revolving machines, etc. The study of the kinematics

of this motion was implemented in Subsection 9.4.1 of Chapter 9. Within the

framework of the present chapter, we consider (Figure 22.1) the motion of rotation

of a solid (S) about a horizontal axis (∆), obtained through a hinge connection

between the solid and the support (T). The solid (S) has a mass m, a mass centre G

and an arbitrary form. The mass centre is distant of a from the axis of rotation.

FIGURE 22.1. Rotation of a solid (S) about the axis (∆).

(S)

G

a

()

348 Chapter 22 Motion of Rotation about a Fixed Axis

xS

(S)

G

a

()

x

yS

y

z

O

FIGURE 22.2. Choice of the coordinate systems.


We choose the coordinate system (Oxyz) attached to the support (T) such as the

axis Oz

coincides with the axis (∆), the axis Ox

has the downward vertical as

direction and such as the mass centre of the solid is contained in the plane (Oxy)

(Figure 22.2).

The parameters of translation are determined while choosing a particular point

of the solid. We choose a point of the axis of rotation: the point O. This point does

not move during the motion of rotation of the solid. Thus, there does not exist any

parameter of translation.

The parameters of rotation are determined while choosing a coordinate system

attached to the solid (S). We choose the system ( )S SOx y z such as the axis SOx

passes through the mass centre G. (Another possible choice could have been to

choose the axis SOx

coinciding with a principal axis of inertia of the solid (S)).

The orientation of the solid (S) is defined by the angle ψ between the axis Ox

and

the axis SOx

.

Finally, the motion is characterized by one parameter of rotation ψ about the

axis Oz

. Between the unit direction vectors ( ), S Si j

of the axes SOx

and SOy

and the unit direction vectors ( ), i j

of the axes Ox

and Oy

, we have from (9.45)

the relation:

cos sin ,

sin cos .

S

S

i i j

j i j

ψ ψ

ψ ψ

= +

= − +

(22.1)

22.1 General Equations 349

22.1.3 Kinematics


The kinematic torsor ( ) TS associated to the motion of rotation of the solid (S)

with respect to the support (T) is defined by its elements of reduction at the point

O: ( ) ( )

( ) ( )

,

( , ) 0.

T TS S

T TO S

R k

O t

ω ψ= =

= =

(22..2)

22.1.3.2 Kinematic Vectors of the Mass Centre

The position of the mass centre is defined by its position vector:

SOG a i=

. (22.3)

The velocity vector of the mass centre can be derived in two ways, either by

using directly the definition of the velocity vector:

( )( )

d( , )d

TT G t OG

t=

, (22.4)

or by using the relation between the velocity vectors of two points of the solid:

( ) ( ) ( ) ( ) ( , ) ( , )T T T T

S SG t O t OG OGω ω= + × = ×

. (22.5)

In the two cases, we obtain: ( )

( , )TSG t a jψ= . (22.6)

This velocity vector can possibly be expressed in the basis ( ), , i j k

of the refe-

rence (T). We obtain: ( )

( , ) sin cosT G t a i a jψ ψ ψ ψ= − + . (22.7)

The acceleration vector of the mass centre is easily obtained by deriving the

velocity vector:

( )( )

( )

d( , ) ( , )d

TT Ta G t G t

t=

. (22.8)

The application of this relation to Expression (22.6) of the velocity vector leads to:

( )

2( , )TS Sa G t a i a jψ ψ= − +

, (22.9)

which expresses the acceleration vector in the basis ( ), , S Si j k

attached to the solid.

The expression of the acceleration in the basis ( ), , i j k

can then be obtained

either by expanding Expression (22.9) considering Relations (22.1) of basis

change, or by deriving directly Expression (22.7). We obtain:

( ) ( ) ( )

2 2( , ) sin cos cos sinTa G t a i a jψ ψ ψ ψ ψ ψ ψ ψ= − + + −

(22.10)


22.1.4 Kinetics


We have to derive here the elements of reduction of the kinetic and dynamic

torsors. In this way, the question is to know at which point we must determine the

moments of the torsors. In fact, a discerning choice will simplify the resolution of

the equations deduced from the fundamental principle of the dynamics. The

expression of the moment of the dynamic torsor is simpler at the mass centre

(18.9). In a general way, that is this point which will be selected to study the

motion of a solid. However, the fundamental principle of dynamics introduces the

actions of connections, on which it is necessary then to set assumptions on the

physical nature of the induced processes of friction. In this way, the application of

the fundamental principle to the study of the motion of rotation about an axis

shows that the analysis of the problem is made easier while expressing the

moments at a point of the axis of rotation. We choose the point O.

22.1.4.2 Kinetic Torsor

The elements of reduction at the point O of the kinetic torsor ( ) T

S are from

(16.5) and (16.6):

( ) ( ) ( , )T T

S SR m G t ma jψ= =

, (22.11)

( ) ( ) ( ) ( ) ( ) ( ) ( , )T T T T

O S O S O Sm OG O t S Sω ω= × + = . (22.12)

To obtain the moment at the point O, it is necessary to introduce the matrix of in-

ertia( )

( )SbO SI at the point O of the solid (S) expressed in the basis ( ) ( ), , S S Sb i j k=

attached to the solid. Owing to the fact that the axes of the trihedron ( )S SOx y z

are not generally the principal axes of inertia, the matrix of inertia has the general

form:

( )( )Sb

O

A F E

S F B D

E D C

− − = − − − −

I . (22.13)

The expression of the moment is thus:

( )

TO S S SE i D j C kψ ψ ψ= − − +

. (22.14)

22.1.4.3 Dynamic Torsor

The elements of reduction at the point O of the dynamic torsor ( ) T

S are from

(16.15) and (16.16):

( ) ( )( , )T T

SR ma G t= , (22.15)


( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

( , )

.

T T T T TO S O S S O S

T T TO S S O S

m OG a O t S S

S S

ω ω ω

ω ω ω

= × + + ×

= + ×

(22.16)

The resultant of the dynamic torsor is obtained by substituting one of Expres-

sions (22.9) or (22.10) of the acceleration vector into Expression (22.15).

The expression of the moment is written by introducing the matrix of inertia

(22.13). We obtain:

( ) ( ) ( )

2 2TO S S SE D i D E j C kψ ψ ψ ψ ψ= − + − + +

(22.17)

Possibly, the moment can be rewritten in the basis ( ), , i j k

by introducing

Relations (22.1) of basis change into (22.17).

22.1.4.4 Kinetic Energy

The kinetic energy is obtained using the relation:

( ) ( ) ( ) c1( )2

T T TS SE S = ⋅ ,

thus, taking account of (22.2), (22.11) and (22.14):

( ) 2c

1( )2

TE S Cψ= . (22.18)

22.1.5 Mechanical Actions Exerted on the Solid

The mechanical actions exerted on the solid are the action of gravity, the action

of the support induced by the hinge connection and possibly a driving or braking

action.


The action is represented by the torsor ( ) e S of which the elements of

reduction at the mass centre are:

( )

( )

e ,

e 0.G

R S mg i

S

=

=

(22.19)


( ) ( ) ( ) ( ) e eT TSP S S= ⋅ . (22.20)

This expression developed at the point O is written:

( ) ( ) ( ) ( ) ( ) ( ) e e ( , ) eT T T

OSP S R S O t Sω= +⋅ ⋅

. (22.21)

It is necessary to express the moment at O of the action of gravity:

( ) ( ) ( ) ( ) e e e eO GS S R S GO OG R S= + × = ×

. (22.22)


Finally, we have:

( ) e sinO S mgak ψ= − . (22.23)

And Expression (22.21) is written:

( ) ( ) e sinTP S mgaψ ψ= − . (22.24)

2. Action of the support induced by the hinge connection

The action of connection exerted by the support is represented by the torsor

( ) S of which the elements of reduction at the point O are:

( )

( )

,

.

l l l

O l S l S l

R S X i Y j Z k

S L i M j N k

= + +

= + +

(22.25)

The components Xl, Yl, ..., Nl, of the action of connection are to be determined.


( ) ( ) ( ) ( ) T TS lP S S N ψ= =⋅ . (22.26)

3. Driving action or braking action

To put the solid in rotation or to maintain the rotation, it will be necessary to

exert a driving action, which will be reduced to a driving couple. A braking couple

could possibly be applied to stop the rotation. The driving couple or the braking

couple will be represented by a torsor ( )S of which the elements of reduction

at the point O are:

( ) 0,

( ) .O

R S

S N k

=

=

(22.27)

The component N which is imposed is known.

The power developed by this action is:

( ) ( ) ( ) ( )T TSP S S Nψ= =⋅ . (22.28)

22.1.6 Application of the Fundamental Principle of Dynamics

22.1.6.1 General Equations

In the case where the support (T) is attached to the Earth (pseudo-Galilean

reference), the fundamental principle applied to the motion of rotation of the solid

(S) is written:

( ) ( ) ( ) e ( )TS S S S= + + . (22.29)

This equation between torsors leads to the equation of the resultant and to the

equation of the moment at the point O:


( ) ( ) ( ) e ( )TSR R S R S R S= + +

, (22.30)

( ) ( ) ( ) e ( )TO O O OS S S S= + +

. (22.31)

The two vector equations of the resultant and of the moment lead then to the six

following scalar equations:

( )( )

( )

2

2

2

2

cos sin ,

sin cos ,

0 ,

,

,

sin .

l

l

l

l

l

l

ma mg X

ma Y

Z

E D L

D E M

C mga N N

ψ ψ ψ ψ

ψ ψ ψ ψ

ψ ψ

ψ ψ

ψ ψ

− + = +

− + =

=

− + =

− + =

= − + +

(22.32)

The theorem of the power-energy:

( )( ) ( ) ( ) ( ) ( ) ( ) cd e ( )d

T T T TE S P S P S P St

= + + (22.33)


sin lC mga N Nψψ ψ ψ ψ ψ= − + + . (22.34)

We find the sixth one of Equations (22.32) again.

Finally, we obtain 6 equations for 7 unknowns: Xl, Yl, Zl, Ll, Ml, Nl, ψ. An

additional equation will be derived from the physical nature of the connection.

Thus, the problem is entirely determined.

The consideration of the physical nature of the friction of the hinge connection

will allow us to express the component Nl which is introduced in the expression of

the power developed by the action of connection. Thus the last equation of Equa-

tions (22.32) is the equation of motion, of which the resolution allows us to obtain

the expression of the angle ψ as a function of time. Next, the other components of

the action of connection will be determined by substituting the expression of ψinto the other equations. Note that the moment at O of the action of connection

was expressed in the basis ( ), , S Si j k

attached to the solid. Its expression in the

basis ( ), , i j k

will be then deduced by applying the basis change (22.1). Thus:

( ) ( ) ( )cos sin sin cosO l l l l lS L M i L M j N kψ ψ ψ ψ= − + + +

. (22.35)

22.1.6.2 Rotation with Friction and without Friction

In the case of a perfect connection, i.e. without friction, the power developed

(22.26) is zero. Thus, it results:

0lN = , (22.36)

and the equation of motion is written:


sinC mga Nψ ψ+ = . (22.37)

The expression of ψ as a function of time will depend on the component N of the

driving or braking couple.

In the case of a hinge connection with a friction of viscous type, the component

Nl of connection is opposed to the angular velocity of rotation:

lN cψ= − , (22.38)

and the equation of motion (22.37) is modified as:

sinC c mga Nψ ψ ψ+ + = . (22.39)

22.2 EXAMPLES OF MOTIONS OF ROTATION

ABOUT AN AXIS

22.2.1 Solid in Rotation Submitted only to the Gravity

In the case where the solid (S) is submitted only to the gravity, the solid in

motion of rotation is usually called simple pendulum. Among Equations (22.32),

only the equation of motion is modified and is written:

sin lC mga Nψ ψ= − + . (22.40)

Recall that the parameter C is the moment of inertia of the solid (S), with respect

to the axis of rotation, depending on the geometry and the mass of the solid.

In the case of a connection without friction, the equation of motion is reduced to:

sin 0C mgaψ ψ+ = . (22.41)

The position of equilibrium is obtained when sin 0ψ = ; what leads to the two

values of the angle eq 0ψ = and eqψ π= .

The characterization of the stability of these equilibriums can be evaluated, in

accordance with the following definitions:

The equilibrium of a solid is said to be stable if and only if this solid, slightly

disturbed from its position of equilibrium and then released, swings around this

position and moves back to this position.

The equilibrium of a solid is said to be unstable, if the solid, slightly disturbed

from its position of equilibrium, keeps moving away from this position when the

solid is released.

Thus we search for the motion ε of the solid around the position eqψ of equili-

brium. We have eqψ ψ ε= + , and Equation (22.41) is written:

( )eq eqsin cos cos sin 0C mgaε ψ ε ψ ε+ + = . (22.42)

22.2 Examples of Motions of Rotation about an Axis 355

Expanding to the first order and taking account of the fact that eqsin 0ψ = , we

obtain:

eqcos 0C mgaε ε ψ+ = , (22.43)

or

20 eqcos 0ε ω ε ψ+ = , (22.44)

introducing the natural angular frequency:

20

mga

Cω = . (22.45)

In the case where eq 0ψ = , whence eqcos 1ψ = , Equation (22.44) is written:

20 0ε ω ε+ = . (22.46)

We are brought back to the reduced form (21.24) of the vibrations of a system

with one degree of freedom. The motion of the solid is pendular around the po-

sition of equilibrium eqψ , and the equilibrium is thus stable.

In the case where eqψ π= , whence eqcos 1ψ = − , Equation (22.44) is written:

20 0ε ω ε− = . (22.47)

The motion for low amplitudes is of the form:

0tAeωε = . (22.48)

The function ε is a function which increases with time and the equilibrium is

unstable.

In the case of a connection with viscous friction (22.38), the equation of motion

(22.39) is modified as:

sin 0C c mgaψ ψ ψ+ + = . (22.49)

This equation is written in the form:

202 sin 0ψ δψ ω ψ+ + = , (22.50)

setting:

2

c

Cδ = . (22.51)

Near the position of equilibrium eq 0ψ = , we have sinψ ψ≈ and the equation of

motion (22.50) is reduced to:

202 0ψ δψ ω ψ+ + = (22.52)

We are brought back to the reduced form (21.57) of the vibrations of a system

with one degree of freedom with viscous friction. The results developed in Chap-

ter 21 can thus be applied to the present case.


22.2.2 Pendulum of Torsion

We consider (Figure 22.3) the system constituted of a disk in rotation about

horizontal axis and a spiral spring (R) exerting a couple of torsion. The disk is

homogeneous and its mass centre coincides with the centre of symmetry O of the

disk. The spring exerts a restoring couple of direction k

and component:

N Kψ= − , (22.53)

where K is the constant of torsion and the angle ψ is measured with respect to the

position of equilibrium. If a is the radius of the disk and m its mass, the moment

of inertia with respect to the axis of rotation is:

2

2

aC m= . (22.54)

Neglecting the mass of the axis of rotation, the equation of motion of the system is

written: 2

2l

am K Nψ ψ+ = . (22.55)

In the case of a viscous friction (22.38), the equation of motion is written in the

reduced form: 202 0ψ δψ ω ψ+ + = , (22.56)

setting:

202 2

2 and

c K

ma maδ ω= = . (22.57)

The equation of motion is the one (21.57) of a system with one degree of freedom,

studied in Chapter 21. Contrary to Equation (22.52) valid only for the low values

of the angle of rotation, Equation (22.56) is obtained whatever the values of the

angles of rotation.

FIGURE 22.3. Pendulum of torsion.

xS

()

x

yS

y

z

(S)

(R)

O

22.3 Problem of the Balancing of Rotors 357

22.3 PROBLEM OF THE BALANCING OF ROTORS

22.3.1 General Equations of an Unbalanced Solid in Rotation

22.3.1.1 Kinetics of the Motion

A rotor (S) is connected to a frame (T) through two bearings (P1) and (P2), of

respective centres P1 and P2 (Figure 22.4). As coordinate system attached to the

frame, we choose the system (Oxyz) such as the axis Oz

is the axis of rotation

and such as the axis Ox

is downward vertical. The rotor is supposed to be unba-

lanced, and the mass centre is located outside the axis of rotation in a position

which is not known a priori. As coordinate system attached to the rotor (S), we

choose the system ( )S SOx y z of which the orientation at a given instant is defined

by the angle ψ . The position vector of the mass centre in the basis ( ), , S Si j k

is:

S SOG a i b j c k= + +

, (22.58)

where (a, b, c) are the Cartesian coordinates of G in the reference ( )S SOx y z .

The velocity vector of the mass centre is obtained by deriving the preceding

expression: ( )

( , )TS SG t a j b iψ ψ= − . (22.59)

Deriving once again, we obtain the acceleration vector:

( ) ( ) ( )

2 2( , )TS Sa G t a b i a b jψ ψ ψ ψ= − + + −

. (22.60)

The acceleration vector can then be expressed in the basis ( ), , i j k

attached to the

frame, by using the basis change (22.1). We obtain:

( ) ( ) ( )

( ) ( )

2 2

2 2

( , ) cos sin

sin cos .

Ta G t a b a b i

a b a b j

ψ ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ ψ

= − + + −

+ − + + −

(22.61)

FIGURE 22.4. Rotor.

yS

P2G

(S)

O

xS

x

y

z

P1

d1

d2


The resultant of the dynamic torsor is thus:

( ) ( )( , )T T

SR ma G t= , (22.62)

where the acceleration vector of the mass centre is given by the preceding expres-

sion (22.61).

The matrix of inertia at the point O of the rotor (S) expressed in the basis (bS) is

arbitrary and thus expressed in the general form (22.14). The moment at the point

O of the dynamic torsor is then given by Expression (22.17). Expressing this

moment in the basis ( ), , i j k

, we obtain:

( ) ( ) ( )

( ) ( )

2 2

2 2

cos sin

sin cos .

TO S E D D E i

E D D E j C k

ψ ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ ψ ψ

= − + + +

+ − + − + +

(22.63)

22.3.1.2 Mechanical Actions Exerted on the Rotor

The mechanical actions exerted on the rotor are the action of gravity, the

actions induced by the frame at the level of the bearings and possibly a driving or

braking couple.


The action is represented by the torsor ( ) e S of which the elements of redu-

ction at the mass centre are:

( )

( )

e ,

e 0.G

R S mg i

S

= −

=

(22.64)

The moment at the point O of the action of gravity is:

( ) ( ) e eO S OG R S= ×

, (22.65)

hence:

( ) ( ) e sin cosO S mgcj mg a b kψ ψ= − + + (22.66)

2. Action of the frame exerted at the level of the bearing (P1)

The action is considered to be a force of which the line of action passes through

the centre P1 of the bearing. The action is represented by the torsor ( ) 1 S , of

elements of reduction at the point O:

( )

( )

1

1 1 1 1

1

,

0.P

R S X i Y j Z k

S

= + +

=

(22.67)

The moment at the point O is expressed by:

( ) ( ) 11 1O S OP R S= ×

. (22.68)

Hence:

( ) 1 1 1 1 1O S Y d i X d j= − + (22.69)

where d1 is the distance from the centre of the bearing P1 to the point O (Figure


22.4). According to the position of the point P1 with respect to the point O, this

distance will be taken positive or negative.

3. Action of the frame exerted at the level of the bearing (P2)

As previously, the action is considered to be a force of which the line of action

passes through the centre P2 of the bearing. By analogy with the preceding results,

the elements of reduction at the point O will be thus:

( ) 2 2 2 2R S X i Y j Z k= + +

, (22.70)

( ) 2 2 2 2 2O S Y d i X d j= − + , (22.71)

where d2 is the distance from the centre of the bearing P2 to the point O (Figure

22.4), taken positive or negative according to the position of the bearing with

respect to the point O.

4. Driving or braking couple

In order to put the rotor in rotation, to maintain it in rotation or to stop it, it will

be necessary to exert a driving or braking couple. This couple is represented by

the torsor ( )S of elements of reduction at the point O:

( ) 0,

( ) .O

R S

S N k

=

=

(22.72)

22.3.1.3 Equations of Dynamics

The equations of the dynamics are obtained by applying to the motion of the

rotor the fundamental principle of dynamics which is written as:

( ) ( ) ( ) ( ) 1 2e ( )TS S S S S= + + + . (22.73)

While expressing the moments of the actions at the point O, we obtain the six

scalar equations:

( ) ( )

( ) ( )

( ) ( )( ) ( )

2 21 2

2 21 2

1 2

2 21 1 2 2

2 21 1 2 2

cos sin ,

sin cos ,

0 ,

cos sin ,

sin cos ,

sin c

m a b a b X X mg

m a b a b Y Y

Z Z

E D D E Y d Y d

E D D E mgc X d X d

C mg a b

ψ ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ ψ

ψ ψ

− + + − = + −

− + + − = +

= +

− + + + = − −

− + − + = − + +

= +

( )os .Nψ +

(22.74)

The last equation is the equation of the motion. This equation allows us to

derive the angle of rotation ψ as a function of time. Having obtained ψ , the other

equations then enable us to deduce the components of the actions of connections

exerted by the frame at the level of the two bearings.


22.3.2 Mechanical Actions Exerted on the Shaft of Rotor

During the motion of rotation, the mechanical actions exerted on the shaft of

the rotor at the level of the bearings are represented respectively by the torsors

( ) 1 S and ( ) 2 S . Reciprocally, the rotor exerts opposed mechanical actions,

represented by the torsors ( ) 1 S− and ( ) 2 S− . The components (X1, Y1)

and (X2, Y2) are obtained by solving Relations (22.74). We obtain:

( ) ( ) ( )

( ) ( )

( ) ( ) ( )

( ) ( )

2 21 2 2 2 2 2

2 1

2 21 2 2 2 2

2 1

2 22 1 1 1 1 1

2 1

2 22 1 1 1 1

2 1

1sin cos ,

1sin cos ,

1sin cos ,

1sin cos

X mg c d E D D Ed d

Y D E E Dd d

X mg c d E D D Ed d

Y D E E Dd d

ψ ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ

= − − + − + + −

= + + − + −

= − + − + + − −

= − + + −−

,ψ

(22.75)

setting:

1 1 1 1

2 2 2 2

, ,

, .

E E mad D D mbd

E E mad D D mbd

= − = −

= − = − (22.76)

22.3.3 Principle of the Balancing

The components (22.75) of the mechanical actions exerted on the shaft of the

rotor depend on , ψ ψ and more especially on 2ψ . They can thus reach values

rapidly high when ψ increases. Moreover, the shaft is deformable and the rotor-

shaft system behaves in a way similar to that of a system with one degree of

freedom such as that studied in Chapter 21. Vibrations are thus generated which

will lead to a premature deterioration of the shafts or bearings. In order to increase

the life duration of shafts, it is necessary to reduce as well as possible the actions

exerted on the shaft. The examination of Expressions (22.75) shows that the

actions are reduced to minimal values if:

0a b= = , (22.77)

and

0D E= = . (22.78)

The first condition corresponds to the case where the mass centre is located on

the axis of rotation. It is said that the rotor-shaft system is statically balanced. The

equilibrium of the rotor is then indifferent or neutral. The second condition corres-

ponds to the case where the axis of rotation is principal axis of inertia. When these

two conditions are satisfied, it is said that the rotor-shaft system is dynamically

balanced. The components of the actions of connections are then reduced as:


Mi

(i)

O

yS

xS

x

y

z

zi

yS

xS

i

FIGURE 22.5. Planes of balancing.

21 1

2 1

12 2

2 1

, 0,

, 0.

c dX mg Y

d d

c dX mg Y

d d

−= − =

−

−= =

−

(22.79)

The conditions of dynamic balancing are as well as possible approached when

the rotor-shaft system is constructed. Then, these conditions are adjusted by

setting masses of low dimensions in two planes (1) and (2) orthogonal to the axis

of rotation. In each plane (i) (i = 1 or 2), the mass mi of low dimensions can be

considered as located at the point Mi (Figure 22.5). The position of the point Mi is

characterized by its polar coordinates ( ), , i i ir zα in the system ( )S SOx y z atta-

ched to the rotor (S). Its Cartesian coordinates in this same coordinate system are:

cos ,

sin ,

, 1, 2.

i i i

i i i

i

x r

y r

z z i

α

α

=

=

= =

(22.80)

Adding the two masses m1 and m2 modifies the position of the mass centre of

the system, in accordance to the relation:

( ) 1 21 2 1 2m m m OG m OG m OM m OM′+ + = + +

, (22.81)

where G′ is the position of the mass centre of the rotor-shaft-masses set. The

Cartesian coordinates of G′ in the reference ( )S SOx y z are:

( )

( )

( )

1 1 1 2 2 21 2

1 1 1 2 2 21 2

1 1 2 21 2

1cos cos ,

1sin sin ,

1.

a ma m r m rm m m

b mb m r m rm m m

c mc m z m zm m m

α α

α α

′ = + ++ +

′ = + ++ +

′ = + ++ +

(22.82)


Condition (22.77), applied to a′ and b′ , leads to the relations of static balancing:

1 1 1 2 2 2

1 1 1 2 2 2

cos cos 0,

sin sin 0.

ma m r m r

mb m r m r

α α

α α

+ + =

+ + = (22.83)

The matrix of inertia at the point O of the mass mi considered as located at the

point Mi is:

( )

( )( )

( )

2 2

2 2

2 2

( )S

i i i i i i i i i

bi i i i i i i i i iO

i i i i i i i i i

m y z m x y m x z

M m x y m x z m y z

m x z m y z m x y

+ − − = − + −

− − +

I . (22.84)

The matrix of inertia at the point O of the rotor-shaft-masses set is written as:

( ) ( ) ( )1 2( ) ( ) ( )S S Sb b b

O O OS M M+ +I I I . (22.85)

In particular, the moments of inertia D and E of the rotor are modified respect-

tively according to the expressions:

1 1 1 1 2 2 2 2

1 1 1 1 2 2 2 2

sin sin ,

cos cos .

D D m r z m r z

E E m r z m r z

α α

α α

′ = + +

′ = + + (22.86)

The conditions (22.78) applied to D′ and E′ lead to the relations:

1 1 1 1 2 2 2 2

1 1 1 1 2 2 2 2

sin sin 0,

cos cos 0.

D m r z m r z

E m r z m r z

α α

α α

+ + =

+ + = (22.87)

Relations (22.83) and (22.87) constitute the four relations of dynamic balancing of

a rotor.

Usually the masses are set in given planes, easily accessible. The relations of

balancing thus allow us to derive four of the parameters ( ) 1 1 1, ,m r α and

( ) 2 2 2, , ,m r α in fact two of the parameters ( ) 1 1 1, ,m r α and two of the parameters

( ) 2 2 2, , .m r α The choice of the parameters will be conditioned to facilities of

implementation. In practice, balancing is carried out by using an electronic system

of balancing. The mechanical actions exerted on the axis of rotation are measured

using accelerometers. Preliminary measurements make it possible to determine

the parameters a, b, D and E of the unbalanced rotor, and then to deduce the four

parameters of balancing by Relations (22.83) and (22.87).

EXERCISES

22.1 Analyse the motion of a parallelepiped connected to a support by a hinge

connection of horizontal axis passing through its mass centre and submitted to the

action of a spiral spring (R) exerting a couple of torsion (Figure 22.6).

22.2 Analyse the motion of a parallelepiped in the case of a hinge connection of

eccentric horizontal axis (Figure 22.7a).

Exercises 363

FIGURE 22.6. Motion of a parallelepiped around an axis passing through its centre.

22.3 Analyse the motion of a parallelepiped connected with a hinge connection

of eccentric horizontal axis and submitted to the action of a spiral spring (R)

exerting a couple of torsion (Figure 22.7b).

FIGURE 22.7. Motion of a parallelepiped around an eccentric axis: a) without spring of

torsion and b) with a spring of torsion.

G

(R)

G

Od

(R)

G

Od


COMMENTS

The motion of rotation about an axis occurs in many industrial appli-

cations, such as rotors, wheels, crankshafts, revolving machines, etc. The

motion of rotation induces actions exerted at the level of the connections

which can lead to a deterioration of the connections and shafts. This

problem is solved by implementing a balancing of the rotor and will be

considered by the reader with a great attention.

CHAPTER 23

Plane Motion of a Rigid Body

23.1 INTRODUCTION

Kinematics of a plane motion of a rigid body was studied in Subsection 9.4.5.

The motion of a solid relatively to a given reference is a plane motion, if an only

if a plane attached to the solid remains in coincidence, during the motion, with a

plane attached to the reference. In the most general case, the motion is then defi-

ned by two parameters of translation and one parameter of rotation. In this chapter,

we analyse three examples of plane motion: the motion of a parallelepiped moving

on an inclined plane, the sliding and rocking of a parallelepiped on an inclined

plane, the rolling and sliding of a cylinder on an inclined plane.

23.2 PARALLELEPIPED MOVING

ON AN INCLINED PLANE

23.2.1 Parameters of Situation and Kinematics

We consider (Figure 23.1) the motion of the parallelepiped (S) on the inclined

plane (T), of inclination α with respect to a horizontal plane. During the motion,

the plane (ABCD) of the parallelepiped remains in coincidence with the plane (T).

As coordinate system attached to the plane (T), we choose the trihedron (Oxyz) of

which the plane (Oxy) coincides with the plane (T) and such as the axis Ox

is the

direction of greater slope. As coordinate system attached to the solid (S), we

choose the trihedron ( )S SAx y z constructed on the edges of the parallelepiped.

During the motion, the z-coordinate of the mass centre is constant. The situation

of the solid (S) is then determined by:

366 Chapter 23 Plane Motion of a Rigid Body

FIGURE 23.1. Motion of a parallelepiped on an inclined plane.

— the position of the mass centre G, defined by its two Cartesian coordinates

(x, y) depending on time relatively to the reference system (Oxyz),

— the orientation of the trihedron ( )S SAx y z defined by the angle ψ between

the direction i

and the direction Si

.

The motion is a motion with 3 parameters of situation or 3 degrees of freedom.

The basis change is expressed by the usual relations:

cos sin ,

sin cos .

S

S

i i j

j i j

ψ ψ

ψ ψ

= +

= − +

(23.1)

The kinematic torsor ( ) TS associated to the motion of the parallelepiped with

respect to the inclined plane has for elements of reduction at the mass centre:

( ) ( )T TS SR kω ψ= =

, (23.2)

( ) ( ) ( , )T T

G S G t x i y j= = +

. (23.3)

The acceleration vector of the mass centre is obtained by deriving the velocity

vector (23.3) with respect to time:

( ) ( , )Ta G t x i y j= + (23.4)

23.2.2 Kinetics of the Motion

The elements of reduction of the dynamic torsor ( ) T

S relative to the motion

of (S) with respect to the plane (T) are at the mass centre:

( ) ( ) ( ) ( , )T TSR ma G t m x i y j= = +

, (23.5)

z

G

A

B

CD

(S)

x

x

xS

yS

y z

(T)

O

23.2 Parallelepiped Moving on an Inclined Plane 367

( ) ( ) ( ) ( ) ( ) ( )T T T TG S G S S G SS Sω ω ω= + ×

. (23.6)

The operator of inertia at the point G is represented in the basis ( ) ( ), , S S Sb i j k=

by the matrix of inertia:

( )0 0

( ) 0 0

0 0

SbG

A

S B

C

=

I , (23.7)

with:

( ) ( ) ( )2 2 2 2 2 2, , ,12 12 12

m m mA b c B a c C a b= + = + = + (23.8)

where m is the mass of the solid and (a, b, c) the respective lengths of the edges of

the parallelepiped. The moment at the mass centre of the dynamic torsor is thus:

( ) ( )

2 2

12T

G Sm

C k a b kψ ψ= = + (23.9)

The kinetic energy can be determined from Relation (16.27). We obtain:

( ) ( ) ( )2 2 2 2 2c

1( )2 24

T mE S m x y a b ψ= + + + . (23.10)

The first term is the kinetic energy of translation, and the second term is the

kinetic energy of rotation.

23.2.3 Mechanical Actions Exerted on the Parallelepiped

The mechanical actions exerted on the parallelepiped are reduced to the action

of gravity and the action of contact exerted by the inclined plane.


The action is represented by the torsor ( ) e S of which the elements of

reduction at the mass centre are:

( )

( )

e ,

e 0,G

R S mg u

S

=

=

(23.11)

where u

is the unit vector of the downward vertical direction:

sin cosu i kα α= −

.

Hence:

( ) ( )e sin cosR S mg i kα α= −

. (23.12)


( ) ( ) ( ) ( ) e e sinT TSP S S mgx α= =⋅ . (23.13)


2. Action of contact exerted by the inclined plane

The action of contact is represented by the torsor ( ) S , of which the ele-

ments of reduction at the mass centre are:

( )

( )

,

.

l l l

G l l l

R S X i Y j Z k

S L i M j N k

= + +

= + +

(23.14)

The components Xl, Yl, ..., Nl, of the action of contact are to be determined.

The power developed by the action of contact is:

( ) ( ) ( ) ( ) T TS l l lP S S X x Y y N ψ= = + +⋅ . (23.15)

23.2.4 Equations Deduced from the Fundamental Principle

In the case where the inclined plane is a pseudo-Galilean reference (attached to

the Earth), the fundamental principle applied to the motion of the parallelepiped is

written: ( ) ( ) ( ) eT

S S S= + . (23.16)

The equation of the resultant and that of the moment at the mass centre lead to the

six scalar equations:

sin ,

,

0 cos ,

0 ,

0 ,

.

l

l

l

l

l

l

mx mg X

my Y

mg Z

L

M

C N

α

α

ψ

= +

=

= − +

=

=

=

(23.17)

The theorem of power-energy:

( )( ) ( ) ( ) ( ) ( ) cd ed

T T TE S P S P St

= + (23.18)


( ) ( ) ( )2 2 sin12

l l lm

m xx yy a b mg X x Y y Nψψ α ψ+ + + = + + + . (23.19)

This equation is a linear combination of Equations (23.17). In the present case, the

theorem of power-energy does not lead to an equation being of an interest.

Finally, we obtain 6 equations for 9 unknowns: Xl, Yl, Zl, Ll, Ml, Nl, x, y, ψ.

Three equations are solved:

cos ,

0, 0.

l

l l

Z mg

L M

α=

= = (23.20)


The three others are the equations of motion:

sin ,

,

.

l

l

l

mx mg X

my Y

C N

α

ψ

= +

=

=

(23.21)

The physical nature of the action of contact allows us to express the components

Xl, Yl, Nl and to deduce from (23.21) the expressions of x, y and ψ as functions of

time.

23.2.5 Motion without Friction

In the case where the contact is perfect between the plane and the solid

(absence of friction), the power developed (23.15) by the action of contact is zero:

0l l lX x Y y N ψ+ + = . (23.22)

This relation is satisfied, whatever , and x y ψ , if:

0, 0, 0.l l lX Y N= = = (23.23)

Equations (23.21) of the motion are then written:

sin ,

0,

0.

x g

y

α

ψ

=

=

=

(23.24)

The integration of these equations gives first:

0

0

0

sin ,

,

.

x gt x

y y

α

ψ ψ

= +

=

=

(23.25)

Then

20 0

0 0

0 0

1sin ,

2

,

,

x gt x t x

y y t y

t

α

ψ ψ ψ

= + +

= +

= +

(23.26)

where 0 0 0 0 0 0, , , , and x y x yψ ψ are the respective values of , , , , and x y x yψ ψ

at the initial time 0t = .

The trajectory of the mass centre G is a parabola contained in the plane parallel

to the inclined plane and distant of c/2. The parabola is tangent to the axis

( )0 0,G , G0 being the position of the mass centre at 0t = :

0 0 02

cOG x i y j k= + +

, (23.27)


and 0

being the velocity vector of G at 0t = :

0 0 0 .x i y j= +

(23.28)

The motion of the mass centre G is accelerated along the axis Ox

and uniform

along the axis Oy

. Once initiated, the motion does not stop. To this motion, a

motion of rotation of the parallelepiped is superimposed about the axis ( ),G k

.

23.2.6 Motion with Dry Friction

In the case where there is dry friction, we make the assumption that the plane

exerts a force of resistance to sliding of resultant tR

and a couple of resistance to

spinning of moment n

, satisfying the laws (13.2) to (13.25) of dry friction. The

resultant and the moment are expressed by:

t l lR X i Y j= +

(23.29)

and

n lN k=

(23.30)

Moreover, the normal force of contact is expressed by:

cosn lR Z k fmg kα= =

, (23.31)

where f is the coefficient of dry friction between the parallelepiped and the plane.

In the case where there is motion, the magnitude of the force of resistance to

sliding is expressed from (13.6) by:

2 2 cosl lX Y fmg α+ = , (23.32)

and the resultant is opposed to the velocity vector of the mass centre. Hence:

( ) ( ) ( , ) 0, with ( , ) 0.T T

t tR G t R G t× = <⋅

Thus

0, with 0.l l l lX y Y x X x Y y− = + < (23.33)

Substituting the first two equations of Equations (23.21) of motion into Expres-

sions (23.30) and (23.31), we obtain the equations of the motion of the mass

centre:

( )

( )

2 2 2 2 2sin cos ,

sin 0,

0.

x g y f g

x g y yx

yy

α α

α

− + =

− − =

<

(23.34)


These equations allow us to derive x and y as functions of time and so to obtain

the trajectory of the mass centre. The motion of the mass centre is complex. The

mass centre has first a trajectory of parabolic type. The motion stops then, when

the coefficient of friction is high enough, or tends towards an accelerated recti-

linear motion, parallel to the axis Ox

, when the coefficient of friction is low.

To this motion of translation, a motion of rotation is superimposed, for which

the moment of resistance to spinning is expressed (Subsection 13.1.3.3) by the

relation:

coslN hmg α= . (23.35)

The moment is opposed to the rotation vector. Hence:

coslN hmg α= − . (23.36)

This expression, associated to the third equation (23.21) leads to the equation of

the motion of rotation:

cosC hmgψ α= − . (23.37)

The integration with respect to time leads to:

0coshmg

tC

αψ ψ= − + , (23.38)

where 0ψ is the angular velocity at the initial instant t = 0. The motion of proper

rotation is thus uniformly retarded. The angular velocity of rotation decreases and

vanishes at the date:

0

cos

Ct

hmg

ψα

=

(23.39)

23.2.7 Motion with Viscous Friction

In the case of a viscous friction between the parallelepiped and the plane, we

make the assumption that the parallelepiped is submitted to a force of friction

opposed to the velocity vector of the mass centre:

( ) ( , )T

t tR c G t= −

, (23.40)

and to a couple of resistance to spinning opposed to the rotation vector, hence:

l rN c ψ= − . (23.41)

The coefficients ct and cr are the coefficients of viscous friction, respectively in

translation and in rotation. Taking account of (23.3) and (23.29), the resultant

(23.40) of the force of friction, leads to the two relations:

, .l t l tX c x Y c y= − = − (23.42)

By substituting Relations (23.41) and (23.42) into Equations (23.21), we obtain

the equations of motion:


sin ,

0,

0.

t

t

r

mx c x mg

my c y

C c

α

ψ ψ

+ =

+ =

+ =

(23.43)

Solving these equations leads to results similar to those obtained in the case of a

dry friction (preceding subsection). The motion will be analysed in Chapter 27

(Section 27.4), where the equations of motion will be solved using a numerical

method.

23.3 ANALYSIS OF SLIDING AND

ROCKING OF A PARALLELEPIPED


23.3.1 Introduction

A parallelepiped (S), of mass m and edges a, b, c, is placed on the plane (T), so

that one of the edges of (S) remains horizontal when the plane is inclined (Figure

23.2). When we incline the plane by an angle α with respect to the horizontal

plane, we observe the following events:

1. For low enough values of the inclination α, the parallelepiped (S) stays in

equilibrium on the plane.

2. For high enough values, the equilibrium is upset and, according to the values

of the inclination, of the edges and of the friction between the parallelepiped and

the plane, we observe three possible motions:

— the parallelepiped slides on the plane without rocking,

— the parallelepiped rocks around the lower edge without sliding,

— the parallelepiped rocks and slides.

We intend to study, in the following subsections, each of these early motions.

FIGURE 23.2. Initial situation of a parallelepiped, before sliding and rocking.

G

(S)

(T)

ac

b

23.3 Sliding and Rocking of a Parallelepiped on an Inclined Plane 373

FIGURE 23.3 Equilibrium, sliding and rocking of the parallelepiped: a) equilibrium or

sliding, b) rocking or sliding and rocking.


The vertical plane passing through the mass centre G of the parallelepiped (S)

is a plane of symmetry for (S) and the motion of (S). In particular the mass centre

remains in this plane. We choose this plane as plane (Oxy) of the coordinate

reference (Oxyz) attached to the plane (T). The plane of symmetry ABCD of (S)

remains in coincidence with the plane (Oxy) and the motion is a plane motion

(Figure 23.3). The situation of the parallelepiped is determined by:

— the Cartesian coordinates (x, y, 0) of the mass centre G with respect to the

system (Oxyz),

— the angle ψ between the system ( )S SOx y z attached to the solid (S) and the

system (Oxyz).

Two descriptions have to be considered (Figure 23.3). The first description (Fi-

gure 23.3a) corresponds to the equilibrium or sliding without rocking of the paral-

lelepiped, description for which ψ = 0. The other description (Figure 23.3b)

corresponds to the rocking motion or to the motion of sliding and rocking of the

parallelepiped on the inclined plane.

The kinematic torsor ( ) TS associated to the motion of the parallelepiped with


( ) ( ),T T

S SR kω ψ= = (23.44)

( ) ( ) ( , ) .T T

G S G t x i y j= = +

(23.45)

The acceleration vector of the mass centre is:

( ) ( , ) .Ta G t x i y j= + (23.46)

Note that Equations (23.44)-(23.46) have the same form as Equations (23.2)-

(23.4), the role of the plane (Oxy) being however different in both cases.

A

B

C

D

x

xS

yS

y

O

G

x

GA

B

C

D

x

xS

yS

y

O

I

(a) (b)



The mechanical actions exerted on the parallelepiped are reduced to the action

of gravity and the action of contact exerted by the inclined plane. The action of

gravity is represented by the torsor ( ) e S of which the elements of reduction at

the mass centre are:

( ) ( )

( )

e sin cos ,

e 0.G

R S mg i k

S

α α= −

=

(23.47)

The action of contact exerted by the plane is represented by the torsor ( ) S of

elements of reduction at G:

( )

( )

,

.

l l l

G l l l

R S X i Y j Z k

S L i M j N k

= + +

= + +

(23.48)

The fundamental principle of dynamics expressed in the reference (T), consi-

dered as pseudo-Galilean reference is written as:

( ) ( ) ( ) eTS S S= + . (23.49)

The dynamic torsor associated to the motion of the solid (S) with respect to the

reference (T) has elements of reduction at the mass centre whose expressions are

identical to those found in (23.5) and (23.9). The fundamental principle leads thus

to the six scalar equations:

sin ,

cos ,

0 ,

0 ,

0 ,

.

l

l

l

l

l

l

mx mg X

my mg Y

Z

L

M

C N

α

α

ψ

= +

= − +

=

=

=

=

(23.50)

We have 6 equations to determine 9 unknowns. So as to supplement these

equations, we admit that the parallelepiped-plane contact is a dry friction contact

characterized by the coefficient of friction f and the action exerted by the plane on

the parallelepiped is a force of which the support passes through a point I of the

surface of contact. This point is located between A and B in the case where there is

equilibrium or sliding (Figure 23.3a), or coincides with the point B in the case

where there is rocking (Figure 23.3b). The validity of this description is verified

by the good agreement which is observed between the experimental facts and the

results that we will deduce from this modelling. We have thus an additional vector

equation:

( ) ( ) ( ) 0I GS S R S GI= + × =

. (23.51)


This equation leads to:

( ) ( ) G l lS X i Y j IG= + × . (23.52)

Furthermore, the laws of friction allow us to write:

0lY > , (23.53)

l lX f Y= − , (23.54)

if the solid slides on the plane, and:

l lX f Y< , (23.55)

if the solid does not slide.

23.3.4 Analysis of the Different Motions

23.3.4.1 Equilibrium of theParallelepiped

When the parallelepiped is in equilibrium, we have 0x y ψ= = = , and Equa-

tions (23.50) are written:

( )

sin 0,

cos 0,

0,

0.

l

l

l

G

X mg

Y mg

Z

S

α

α

+ =

− =

=

=

(23.56)

The components of the resultant of the action of contact exerted by the plane on

the parallelepiped are thus:

sin , cos , 0.l l lX mg Y mg Zα α= − = = (23.57)

Moreover the forth equation (23.56) associated to Relation (23.52) leads to:

( ) 0l lX i Y j IG+ × =

, (23.58)

or

( ) 02 2

l lb a

X AI Y+ − = , (23.59)

writing:

( )2 2

a bIG AI i j= − − +

. (23.60)

The position of the point I is thus given by:

( )1tan

2AI a b α= + . (23.61)

The condition (23.55) of non sliding and the condition that the point I is between

the points A and B (0 )AI a< < lead to the inequalities:

tan and tan .a

fb

α α< < (23.62)


23.3.4.2 Sliding without Rocking of the Parallelepiped

Sliding without rocking of the parallelepiped occurs when 0y ψ= = , with

0x > . Equations (23.50) are written

( )

sin ,

0 cos ,

0 ,

0.

l

l

l

G

mx mg X

mg Y

Z

S

α

α

= +

= − +

=

=

(23.63)

As previously, the last relation leads to Expression (23.61) of the position of the

point I. Equations (23.63) are associated to the conditions:

0,

0, ,

0 .

l l l

x

Y X fY

AI a

>

> = −

< <

(23.64)

Combining (23.63) and (23.64) leads to:

( )

cos ,

cos

sin cos ,

l

l

Y mg

X fmg

x g f

α

α

α α

=

= −

= −

(23.65)

with the conditions:

tan , .a

f fb

α > < (23.66)

The equation of motion is given by the last equation (23.65). The motion is

uniformly accelerated of equation:

( ) 20 0

1sin cos

2x g f t x t xα α= − + + , (23.67)

where 0x et 0x are the respective values of x and x at the initial instant 0t = .

23.3.4.3 Rocking without Sliding of the Parallelepiped

In the case of a motion with rocking of the parallelepiped (Figure 23.3.b), the

position of the mass centre G is related to that of the point B of contact by the

relation:

( )OG OB GB OB l u ψ γ= − = − +

, (23.68)

with

2 2 11 and tan

2

bl a b

aγ −= + = , (23.69)

and where ( )u ψ γ+

is the unit vector of the direction forming an angle of ψ γ+

with the direction i

. The coordinates of the mass centre are thus expressed as

follows:


( ) cos( ),

sin( ),

x x B l

y l

ψ γ

ψ γ

= − +

= − + (23.70)

where x(B) is the abscissa of the point B on the axis Ox

.

In the case where there is not sliding of the point B, the abscissa x(B) is inde-

pendent of time and we obtain by deriving (23.70) with respect to time:

sin( ),

cos( ),

x l

y l

ψ ψ γ

ψ ψ γ

= +

= − +

(23.71)

and then:

2

2

sin( ) cos( ) ,

cos( ) sin( ) .

x l

y l

ψ ψ γ ψ ψ γ

ψ ψ γ ψ ψ γ

= + + +

= − + − +

(23.72)

We examine the case of the early motion, for which 0ψ ≈ and 0ψ ≈ . Equations

(23.72) are then written:

sin ,

cos ,

x l

y l

ψ γ

ψ γ

=

= −

(23.73)

and Equations (23.50) become:

( )

sin sin ,

cos cos ,

0 ,

.

l

l

l

G

ml mg X

ml mg Y

Z

S C k

ψ γ α

ψ γ α

ψ

= +

− = − +

=

=

(23.74)

The last equation may be rewritten taking account of the fact that the action of

contact exerted by the plane is, for this motion, a force of which the line of action

passes through the point B. Thus:

( ) ( ) ( ) 0B G l lS S X i Y j GB= + + × = , (23.75)

with

( )GB l u ψ γ= +

. (23.76)

Hence finally the characteristic equations of the motion:

sin sin ,

cos cos ,

0,

sin cos ,

l

l

l

l l

ml X mg

ml Y mg

Z

C X Y

ψ γ α

ψ γ α

ψ γ γ

= +

− = −

=

= − +

(23.77)

equations for which it is necessary to add the conditions:

0, 0, .l l lY X f Yψ < < < (23.78)


Solving Equations (23.77), with respect to the unknowns ψ, Xl and Yl leads to:

( )( )

2 2

2

2

3cos( ),

2

3 31 sin sin sin cos cos ,

4 4

3 31 sin cos sin cos sin .

4 4

l

l

g

a b

X mg

Y mg

ψ α γ

γ α γ γ α

γ α γ γ α

= −+

= − − −

= − −

(23.79)

The motion of rotation about the point D is uniformly accelerated. The angle γbeing negative, the second condition (23.78) is always satisfied. The condition

0ψ < leads to:

1tan

tanα

γ> − , (23.80)

and the condition l lX f Y< is written:

( ) ( )2 23 3 3 31 sin sin cos tan 1 sin sin cos

4 4 4 4f fγ γ γ α γ γ γ − + < − +

. (23.81)

23.3.4.4 Rocking and Sliding of the Parallelepiped

In the case of the motion of rocking and sliding of the parallelepiped, the

coordinates of the mass centre are given by Relations (23.70). In the present

motion, the abscissa of the point B depends on time and the component x of the

acceleration vector is written:

2( ) sin( ) cos( )x x B l ψ ψ γ ψ ψ γ = + + + + , (23.82)

and for the early motion:

( ) sinx x B lψ γ= + . (23.83)

The equations of the early motion are thus from (23.50):

[ ]( ) sin sin ,

cos cos ,

0,

sin cos ,

l

l

l

l l

m x B l X mg

ml Y mg

Z

C X Y

ψ γ α

ψ γ α

ψ γ γ

+ = +

− = −

=

= − +

(23.84)

equations to which it is necessary to associate the conditions:

( ) 0, 0, 0, l l lx B Y X f Yψ> < > = − . (23.85)

Solving Equations (23.84) leads to:

( )22 2

6 cos sin cos1,

1 3 sin cos 3cos

fg

fa b

γ γ αψ

γ γ γ

+=

+ ++


( )2

2

3sin cos 1 3sin( ) sin cos ,

1 3 sin cos 3cos

fx B g

f

γ γ γα α

γ γ γ

+ += −

+ +

2

cos,

1 3 sin cos 3cosl

fmgX

f

α

γ γ γ= −

+ +

(23.86)

2

cos

1 3 sin cos 3cosl

mgY

f

α

γ γ γ=

+ +.

The condition 0lY > is written:

21 3cos

3sin cosf

γγ γ

+<

−. (23.87)

This condition associated to 0ψ < leads to:

1

tanf

γ> − . (23.88)

Similarly the condition ( ) 0x B > is written:

( )2

2

3sin cos 1 3sintan

1 3 sin cos 3cos

f

f

γ γ γα

γ γ γ

+ +>

+ +. (23.89)

23.3.4 Conclusions

The results derived in the preceding subsection 23.3.3 show that the different

studied motions occur according as such or such conditions are verified, depen-

ding upon the values: of the edges of the parallelepiped (introduced in the expres-

sion of γ), of the coefficient of friction and of the inclination of the plane. As an

example, we examine the case of a parallelepiped with a square section. Thus:

2 2, sin , cos .

2 2a b γ γ= = − = (23.90)

The different conditions which were obtained lead in this case to the following

results. The parallelepiped is in equilibrium if:

tan , tan 1fα α< < . (23.91)

The motion of sliding without rocking occurs if:

tan , 1f fα > < . (23.92)

The motion of rocking without sliding occurs when:

( )tan 1, 5 3 tan 5 3f fα α> − < − . (23.93)

Lastly, the parallelepiped has a motion of sliding and rocking if:

( )51 , 5 3 tan 5 3

3f f fα< < − > − . (23.94)


1

2

3

4 ( )5 3 tan 5 3f

fα−

= −

1

5/3

3/5

10Plane inclination tan

Co

effi

cien

t o

f fr

icti

on

f ta

n1

α=

1f =

tan

f

α

=

FIGURE 23.4 The different motions according the values of the plane inclination and of

the coefficient of friction: 1) equilibrium, 2) sliding without rocking, 3) rocking without

sliding, 4) rocking and sliding.

The conditions (23.91) to (23.94) can be represented graphically in a system of

axes (tan α, f ). In this system of axes, we have to plot the straight lines:

5tan , tan 1, 1, ,

3f f fα α= = = =

and the equilateral hyperbola asymptote to the line f = 5/3:

( )5 3 5 3 tanf f α− = − .

The curves obtained delimit the space (tan α, f) in four areas (Figure 23.4): the

area 1 where the parallelepiped is in equilibrium, the area 2 where the paralle-

lepiped slides without rocking, the area 3 where the parallelepiped rocks without

sliding and the area 4 where the parallelepiped rocks and slides.

23.4 MOTION OF A CYLINDER


23.4.1 Introduction

A cylinder (S) of mass m and radius a is placed on a plane (T) inclined by an

angle α with the horizontal plane (Figure 23.5a). In the following subsections, we

intend to study the plane motion of the cylinder for which the line of contact AB

remains horizontal. The different possible motions are then: equilibrium of the

23.4 Motion of a Cylinder on an Inclined Plane 381

cylinder on the inclined plane, rolling of the cylinder on the plane without sliding,

sliding of the cylinder without rolling, simultaneous sliding and rolling of the

cylinder on the plane.

23.4.2 Parameters of Situations and Kinematics

As plane (Oxy) of the coordinate reference attached to the inclined plane, we

choose the plane of symmetry of the cylinder and motions, vertical plane passing

through the mass centre G of the cylinder (Figure 23.5b). We consider the

motions for which the cylinder remains in contact with the inclined plane. The

situation of the cylinder is determined by:

— the Cartesian coordinates (x, a, 0) of the mass centre G with respect to the

system (Oxyz),

— the angle ψ between the trihedron ( )S SOx y z attached to the cylinder (S) and

the trihedron (Oxyz).

The motion of the cylinder with respect to the inclined plane is a motion with

two degrees of freedom.

FIGURE 23.5. Cylinder on an inclined plane.

(T)

(S) G

A

B

(a)

G

a

x

y

xS

yS

O

I

y

x

(S)

(b)


The kinematic torsor ( ) TS associated to the motion of the cylinder with


( ) ( )T TS SR kω ψ= =

, (23.95)

( ) ( ) ( , )T T

G S G t x i= =

. (23.96)


( )( , )Ta G t x i=

. (23.97)

The velocity vector of sliding of the point of contact I is, from (10.2), expressed

by: ( ) ( ) ( ) ( )

( , ) ( , )T T T TIgS S SI t G t GIω= = + ×

. (23.98)

Thus: ( )

( )( , )TgS I t x a iψ= +

. (23.99)

The condition of non sliding of the cylinder on the plane is thus written:

0x aψ+ = . (23.100)

When the cylinder does not slide on the plane, the motion is thus a motion with

one degree of freedom.

23.4.3 Mechanical Actions Exerted on the Cylinder

The mechanical actions exerted on the cylinder are reduced to the action of

gravity and the action of contact exerted by the inclined plane. The action of

gravity is represented by the torsor of which the elements of reduction at the mass

centre are:

( ) ( )

( )

e sin cos ,

e 0.G

R S mg i k

S

α α= −

=

(23.101)

We suppose that the action of contact exerted by the plane on the cylinder can be

reduced to a force of friction of support passing through the point of contact I and

to a couple of resistance to rolling. The force of friction is represented by the

torsor ( ) f S of which the elements of reduction at the point I are:

( ) ( )

,

0.

f l l l

I f

R S X i Y j Z k

S

= + +

=

(23.102)

The moment at the mass centre is:

( ) G f l lS aZ i a X j= − + (23.103)

The couple of resistance to rolling is represented by the torsor ( ) r S of ele-

ments of reduction:


( )

( )

0,

.

r

G tr l l l

R S

S L i M j N k

=

= = + +

(23.104)

The moment t

is independent of the point considered.

Moreover, we admit that the usual laws (Section 13.1) of contact between

solids are verified, the contact cylinder-plane being characterized by the coef-

ficient of friction f and the coefficient of resistance to rolling h.


The fundamental principle of dynamics expressed in the reference (T) consi-

dered as pseudo-Galilean reference is written:

( ) ( ) ( ) ( ) e .TS f rS S S= + + (23.105)

The dynamic torsor relative to the motion of the cylinder with respect to the

inclined plane has for elements of reduction at the mass centre:

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( , ) ,

.

T TS


R ma G t mx i

S Sω ω ω

= =

= + ×

(23.106)

The operator of inertia at the point G is represented in the basis ( ) ( ), , S S Sb i j k=

by the matrix of inertia:

( )

2

2

2

0 04

( ) 0 04

0 02

SbG

am

aS m

am

=

I . (23.107)

The moment of the dynamic torsor at the mass centre is thus written:

( ) 2

2T

G Sa

m kψ= (23.108)

Combining Relations (23.101) to (23.108) leads to the six scalar equations

deduced from the fundamental principle of dynamics:

2

sin ,

0 cos ,

0 ,

0 ,

0 ,

.2

l

l

l

l l

l

l l

mx mg X

mg Y

Z

aZ L

M

am N a X

α

α

ψ

= +

= − +

=

= − +

=

= +

(23.109)


Hence, finally:

0, 0, 0,l l lZ L M= = = (23.110)

and

2

sin ,

cos ,

.2

l

l

l l

mx mg X

Y mg

am N a X

α

α

ψ

= +

=

= +

(23.111)

Equations (23.111) have to be coupled with the kinematic conditions of the

contact and the physical laws of the contact (Relations (13.2) to (13.19)) which

introduce the force of friction tR

, the normal force of contact nR

of magnitude

Rn and the couple of resistance to rolling. We have in the present case:

, cos .t l n lR X i R Y mg α= = =

(23.112)

In the case where there is not sliding of the cylinder on the plane, the kinematic

condition is given by the condition (23.100) of non sliding:

0x aψ+ = , (23.113)

and Relation (13.9) of the law of contact leads to:

coslX fmg α< . (23.114)

In the case where there is sliding of the cylinder, the velocity vector of sliding has

a positive component (the cylinder goes down) and the kinematic condition is

written as:

0x aψ+ > . (23.115)

During sliding, the resultant is expressed by Relation (13.6) and has a sign oppo-

sed to that of the sliding velocity vector. Hence:

coslX fmg α= − . (23.116)

In the case where there is not rolling of the cylinder, the kinematic condition is:

0ψ = , (23.117)

and Relation (13.20) of the law of contact is written:

coslN hmg α< . (23.118)

Lastly, in the case where there is rolling of the cylinder, the kinematic condition is

written:

0ψ < . (23.119)

The couple of resistance to rolling is expressed by Relation (13.22) and has a sign

opposed to that of the rotation vector. Hence:

coslN hmg α= . (23.120)



23.4.5.1 Equilibrium of the Cylinder

When the cylinder is in equilibrium, we have 0x = and 0ψ = . Equations

(23.111) are written:

sin ,

cos ,

sin .

l

l

l l

X mg

Y mg

N a X mga

α

α

α

= −

=

= − =

(23.121)

These equations, associated to the conditions of non sliding (23.114) and non

rolling (23.118), lead to the conditions of equilibrium:

tan , tan .h

fa

α α< < (23.122)

23.4.5.2 Sliding without Rolling of the Cylinder

The kinematic condition (23.117) of non rolling and Equations (23.111) lead to

the equations of sliding without rolling of the cylinder:

sin ,

cos ,

.

l

l

l l

mx mg X

Y mg

N a X

α

α

= +

=

= −

(23.123)

To these equations it is necessary to add Relation (23.116) of friction. We obtain:

( )sin cos ,

cos .l

x g f

N afmg

α α

α

= −

=

(23.124)

The condition (23.118) of non rolling is written:

hf

a< , (23.125)

and the kinematic condition of sliding 0x > (the cylinder goes down) gives:

tan .fα > (23.126)

The motion of the mass centre is a motion uniformly accelerated, of which the

equation is obtained by integrating twice the first equation (23.124). Thus:

( ) 20 0

1sin cos

2x g f t x t xα α= − + + , (23.127)

where 0x and 0x are the respective values of x and x at the initial instant 0t = .


23.4.5.3 Rolling without Sliding of the Cylinder

During rolling, the component Nl of the couple of resistance to rolling is

expressed by Relation (23.120). Substituting this expression into Equations

(23.111), we obtain the two equations of the motion:

2

sin ,

cos .2

l

l

mx mg X

am hmg a X

α

ψ α

= +

= +

(23.128)

To these equations, it is associated the kinematic condition (23.113) of non

sliding. The association of these three equations leads to:

( )( )

( )

2sin cos ,

3

2sin cos ,

3

1sin 2 cos .

3l

hx g

ag h

a a

hX mg

a

α α

ψ α α

α α

= −

= − −

= − +

(23.129)

The kinematic condition ( 0 or 0x ψ> < ) imposes:

tanh

aα > . (23.130)

Moreover, the condition of non sliding (23.114) leads to:

( )12 tan

3

hf

aα> + . (23.131)

The motion of translation of the cylinder and its motion of rotation are uniformly

accelerated.

23.4.5.4 Rolling and Sliding of the Cylinder

The component Nl of the couple of resistance to rolling is expressed by Rela-

tion (23.120). The component Xl of the resultant of the force of friction is given by

(23.116). Reporting these two expressions into Equations (23.111), we obtain the

two equations of motion:

( )

( )sin cos ,

2 cos .

x g f

g hf

a a

α α

ψ α

= −

= − −

(23.132)

To these two equations, must be associated the kinematic conditions of sliding

(23.115) and rolling (23.119). These two conditions lead to the conditions which

must be satisfied to have sliding and rolling of the cylinder:

( )12 tan , .

3

h hf f

a aα< + > (23.133)


1

2

3

4

0Inclination of the plane tan

Co

effi

cien

t o

f fr

icti

on

f

hf

a=

tan

f

α

=

()

1 2tan

3

h

f a

α

=+

ha

2

3

h

a

tan

h aα

=

ha

23.4.6 Conclusions

The results established in the preceding subsection 23.4.5 show that the

different motions studied occur according as such or such conditions are verified,

between the inclination of the plane, the radius of the cylinder, the coefficient of

friction and the coefficient of resistance to rolling. The cylinder is in equilibrium

if:

tan , tan .h

fa

α α< < (23.134)

The cylinder slides without rolling if:

, tan .h

f fa

α< > (23.135)

The cylinder rolls without sliding if:

( )1tan , 2 tan .

3

h hf

a aα α> > + (23.136)

Lastly, the cylinder slides and rolls on the inclined plane, if:

( )12 tan , .

3

h hf f

a aα< + > (23.137)

The conditions (23.134) to (23.137) can be represented (Figure 23.6) in a system

of axes (tan α, f ). The space (tan α, f ) is thus delimited into four areas: the area 1

where there is equilibrium of the cylinder, the area 2 where the cylinder slides

without rolling, the area 3 where the cylinder rolls without sliding and the area 4

where the cylinder slides and rolls on the inclined plane.

FIGURE 23.6. The different motions of the cylinder according to the values of the

inclination of the plane and the coefficient of friction: 1) equilibrium, 2) sliding without

rolling, 3) rolling without sliding, 4) sliding and rolling.


In practice, the coefficient h of resistance to rolling is low and the inequality /f h a> is always satisfied. The motion of sliding of the cylinder without rolling is then not observed and the equilibrium of the cylinder occurs only for very low values of the inclination of the plane.

COMMENTS

The various motions studied in the present chapter make it possible to

highlight how the actions of friction occur in the equilibrium and the

motions of bodies. In the case of a parallelepiped moving on an inclined

plane (Section 23.2), the different types of friction can be considered and

the motion observed depends on the conditions of friction: absence of

friction, viscous friction or dry friction. In contrast, in the case of the

analyses of the motions developed in Sections 23.3 and 23.4, it is necessary

to consider a dry friction between solids to describe the different motions

which are observed. Also, these examples illustrate simply how the laws of

friction interact between solids. The reader will pay a great attention to the

development of the analyses implemented in this chapter.

CHAPTER 24

Other Examples of Motions of Rigid Bodies

24.1 SOLID IN TRANSLATION

24.1.1 General Equations of a Solid in Translation

24.1.1.1 Kinetics of the Motion

The kinematic study of a solid (S) in translation was developed in Subsection

9.4.2. We take again similar notations. The coordinate system attached to the

Galilean reference (g) is the axis system ( ) / , ,O i j k

(Figure 24.1). As coordinate

system attached to the solid (S), we choose the trihedron ( ) / , ,G i j k

, of which

the origin is the mass centre of the solid.

Figure 24.1. Solid in translation.

Oi

j

k

y

z

x

jk

i

G(S )

z

y

x

(g )

390 Chapter 24 Other Examples of Motions of Rigid Bodies

The situation of the solid (S) with respect to the reference (g) is entirely defined

by the knowledge of the position of the mass centre, determined for example by

its Cartesian coordinates (x, y, z). The position vector is thus written in the form:

OG x i y j z k= + +

. (24.1)

The kinematic torsor is a couple (Relations (9.59 and (9.60)) of moment:

( ) ( , )

gG t x i y j z k= + +

. (24.2)


( ) ( , )

ga G t x i y j z k= + +

(24.3)

So, it results that the elements of reduction at the mass centre of the kinetic

torsor ( ) g

S are:

( ) ( ) ( )( ) ( ) ( )

( , ) ,

0,

g gS

g TG G SS

R m G t m x i y j z k

S ω

= = + +

= =

(24.4)

where m is the mass of the solid (S).

Similarly, the elements of reduction of the dynamic torsor ( ) g

S are:

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )

( , ) ,

0.

g gS

g g ggG G GS S S S

R ma G t m x i y j z k

S Sω ω ω

= = + +

= + × =

(24.5)

Lastly, the kinetic energy is:

( ) ( ) ( )2

2 2 2c

1 1( ) ( , )2 2

g gE S m G t m x y z = = + +

. (24.6)

24.1.1.2 Mechanical Actions Exerted on the Solid

The mechanical actions exerted on the solid can be separated in known actions

and actions of connection. The whole of the known actions are represented by the

torsor ( ) S of which the elements of reduction at the mass centre are:

( )

( )

,

,G

R S X i Y j Z k

S Li M j N k

= + +

= + +

(24.7)

where the components X, Y, ..., N, are known. The actions of connection are

represented by the torsor ( ) S of which the elements of reduction at the mass

centre are:

( )

( )

,

.

l l l

G l l l

R S X i Y j Z k

S L i M j N k

= + +

= + +

(24.8)

where the components Xl, Yl, ..., Nl, are to be determined.

24.1 Solid in Translation 391

24.1.1.3 Equations Deduced from the Fundamental Principle

The fundamental principle of dynamics applied to the solid in translation is

written as: ( ) ( ) ( ) g

S S S= + . (24.9)

This equation leads to the six scalar equations:

,

,

,

0 ,

0 ,

0 .

l

l

l

l

l

l

mx X X

my Y Y

mz Z Z

L L

M M

N N

= +

= +

= +

= +

= +

= +

(24.10)

We have 6 equations to determine 9 unknowns: the six components of the actions

of connection (Xl, Yl, ..., Nl) and the three parameters of translation (x, y, z). The

physical nature of the connection will allow us to obtain three additional equa-

tions, necessary to determine entirely the problem.

24.1.2 Free Solid in Translation

The case of a free solid in translation, submitted to known actions, can be

deduced from the preceding general equations, by making null the actions of

connection. Equations (24.10) are written in this case:

,

,

,

0, 0, 0.

mx X

my Y

mz Z

L M N

=

=

=

= = =

(24.11)

The first three equations allow us to derive x, y and z as functions of time. These

coordinates determine the motion of the solid. The last three equations express the

fact that the moment of the actions exerted must be null at the mass centre. Hence

the following result: the necessary (but not sufficient) condition in order that a

free body has a motion of translation is that the moment at the mass centre of the

mechanical actions exerted on the body is zero. The mechanical actions exerted

on the body are equivalent to a force of which the support passes through the mass

centre.

An example of such a case is that of a free solid subjected to the action of

gravity and not having a motion of proper rotation: ball without spinning for

example. If the axis Oz

is the upward vertical axis, the three equations of motion

are written:

0, 0, .x y z g= = = (24.12)


Such a motion was studied in Chapter 7 (Section 7.3): the motion of the mass

centre is a motion with parabolic or rectilinear trajectory. The trajectories of the

other points of the solid are deduced from the trajectory of the mass centre by

translation (Subsection 9.4.2.1).

Lastly, note that the theorem of the kinetic energy (18.24) allows us to obtain

the expression which relates the velocity vectors and the altitudes of the mass

centre for two given positions:

( ) ( ) ( )

2 2

2 12 1( , ) ( , ) 2g g

G t G t g z z − = − − . (24.13)

24.2 MOTION OF A SOLID PLACED

ON A WAGON

24.2.1 Introduction

A parallelepiped (S1) of masse m1 and edges a, b, c is placed on a wagon (S2)

of mass m2. The wagon is animated by a motion of rectilinear translation along the

horizontal axis Ox

, guiding the wagon by rails (Figure 24.2). The parallelepiped

is set so that its mass centre G1 is located in the same vertical plane (Oxy) as the

mass centre G2 of the wagon. It is exerted on the wagon a driving action equi-

valent to a force of horizontal support passing through the mass centre of the

wagon. This action is represented by the torsor ( ) 2S such as:

( ) ( )

2

2

,

0.G

R S F i

S

=

=

(24.14)

FIGURE 24.2. Parallelepiped on a moving chariot.

A B

CD

y

O

I

G2

G1

(S2)

(S1)

y y1

x1

x

x(g)

24.2 Motion of a Solid Placed on a Wagon 393

A

B

C

y

O

G2

G1

(S2)

(S1)

yy1

x1

x

x(g)

D

1 x

According to the values of F, we observe one of the following events when the

wagon is moving: 1) the parallelepiped (S1) stays in equilibrium on the wagon; 2)

the parallelepiped slides on the wagon without rocking; 3) the parallelepiped

rocks around the edge passing through D, without sliding; 4) the parallelepiped

rocks around the edges and slides. We intend to analyse each type of early motion.


The most general motion of the parallelepiped is that of rocking and sliding

(Figure 24.3). To the wagon (S2), we associate the coordinate system (G2xyz), of

fixed orientation with respect to the reference system (Oxyz). The mass centre G2

of the wagon has for coordinates (x2, h, 0) in the system (Oxyz), where the height

h is independent of the position of the wagon. The situation of the wagon is

determined by the only parameter of translation x2. The position vector of G2 is

written:

2 2OG x i h j= +

. (24.15)

To the parallelepiped (S1), we associate the system (Gx1y1z) of axes parallel to the

edges. The situation of the parallelepiped is determined by the coordinates

( )1 1, , 0x y of the mass centre G1 in the system (Oxyz) and by the angle of rotation

ψ1 between the axes Gx

and 1Gx

. The position vector of G1 is written:

1 1 1OG x i y j= +

. (24.16)

Finally, in the most general case of rocking and sliding of the parallelepiped on

the wagon, the parameters of situation are: x1, y1, ψ1 and x2.

FIGURE 24.3. Rocking and sliding of the parallelepiped on the wagon.


24.2.3 Kinetics

24.2.3.1 Kinetics of the Motion of the Parallelepiped

The kinematic vectors of the mass centre G1 are deduced from (24.16). Thus:

( ) 1 1 1( , )

gG t x i y j= +

, (24.17)

( ) 1 1 1( , )

ga G t x i y j= +

. (24.18)

The instantaneous vector of rotation, relatively to the motion of the paralle-

lepiped with respect to the rails, is:

( )1

1g

Skω ψ=

. (24.19)

The elements of reduction at the mass centre of the dynamic torsor are thus

written:

( ) ( )

( ) ( )

1

1 1

1 1 1

2 211

,

.12

g

S

gG S

R m x i y j

ma b kψ

= +

= +

(24.20)

24.2.3.2 Kinetics of the Motion of the Wagon

The kinematic vectors of the mass centre G2 are deduced from (24.15). Thus:

( ) ( ) 2 2 2 2( , ) , ( , ) .

g gG t x i a G t x i= =

(24.21)

The wagon having a motion of translation, the rotation vector is null. So, it results

that the elements of reduction at the mass centre of the dynamic torsor are written:

( ) ( )

2

2 2

2 2 ,

0.

g

S

gG S

R m x i=

=

(24.22)

24.2.4 Analysis of the Mechanical Actions

24.2.4.1 Actions Exerted on the Parallelepiped

The actions exerted on the parallelepiped are reduced to the action of gravity

and the action of contact exerted by the wagon. The action of gravity is repre-

sented by the torsor ( ) 1e S of which the elements of reduction at the mass

centre are:

( )

( )

1

1 1

1

e ,

e 0.G

R S m g j

S

= −

=

(24.23)


As in Section 23.3 of Chapter 23, we consider a dry friction between the paralle-

lepiped and the wagon, characterized by the coefficient of friction f, and that the

action of contact exerted by the wagon can be assimilated to a force of which the

support passes through the point I of the surface of contact, point located between

D and C (Figure 24.2). The action of contact is thus represented by a torsor

( ) 2 1S of elements of reduction at the point I:

( )

( )

2 1 21 21 21

2 1

,

0,I

R S X i Y j Z k

S

= + +

=

(24.24)

where the components X21, Y21, Z21, are to be determined. In the case where there

is rocking around the edge passing through the point D (Figure 24.3), the point I

coincides with the point D.

24.2.4.2 Actions Exerted on the Wagon

The actions exerted on the wagon are: the action of contact exerted by the

parallelepiped, the action of gravity, the action of contact exerted by the rails and

the driving action. The action of contact exerted by the parallelepiped is repre-

sented by the torsor ( ) 1 2S opposed to the torsor ( ) 2 1S :

( ) ( ) 1 2 2 1 .S S= − (24.25)

The action of gravity is represented by the torsor ( ) 2e S of which the ele-

ments of reduction at the mass centre are:

( )

( )

2

2 2

2

e ,

e 0.G

R S m g j

S

= −

=

(24.26)

The action of contact exerted by the rails is represented by the torsor ( ) 2S

of which the elements of reduction at the mass centre are:

( )

( )

2

2 2 2 2

2 2 2 2

,

,G

R S X i Y j Z k

S L i M j N k

= + +

= + +

(24.27)

where the components X2, Y2, ..., N2, are to be determined. The driving force is

represented by the torsor ( ) 2S of which the elements of reduction have been

expressed in (24.14).

24.2.5 Equations of Dynamics

The fundamental principle of dynamics applied to the motion of the paralle-

lepiped (S1) with respect to the rails is written as:

( ) ( ) ( ) 1

1 2 1eg

SS S= + . (24.28)


The fundamental principle of dynamics applied to the motion of the wagon

with respect to the rails leads to: ( ) ( ) ( ) ( ) ( ) 2

2 2 1 2 2eg

SS S S S= − + + . (24.29)

Lastly, one of the two equations can be replaced by the equation obtained while

applying the fundamental principle to the set of the two solids. We obtain:

( ) ( ) ( ) ( ) ( ) ( )

1 21 2 2 2e e

g g

S SS S S S+ = + + + . (24.30)

Among these equations, we consider those which will allow us to describe the

different motions observed. Equation (24.28) leads to:

( ) ( ) 1

1 1 21

1 1 1 21

21

2 211 2 1

,

,

0 ,

.12

G

m x X

m y m g Y

Z

ma b Sψ

=

= − +

=

+ =

(24.31)

The last equation has to be expressed while taking account of Relations (24.24).

In this way, we set:

1 1 1G I a i b j= +

, (24.32)

where a1 and b1 depend on the motion. We obtain, considering the relation

21 0Z = :

( ) ( )1 2 1 1 21 1 21G S b X a Y k= − +

. (24.33)

Equation (24.28) leads finally to the equations:

( )

1 1 21

1 1 1 21

21

2 211 1 21 1 21

,

,

0 ,

.12

m x X

m y m g Y

Z

ma b b X a Yψ

=

= − +

=

+ = − +

(24.34)

Also, we shall have to use the equation between the resultants derived from

Equation (24.30). Thus:

( )1 1 2 2 2

1 1 1 2 2

2

,

,

0 .

m x m x X F

m y m m g Y

Z

+ = +

= − + +

=

(24.35)

We shall suppose that the contact between the wagon and the rails occurs without

friction, thus that X2 = 0. The preceding equations are then written:

( )1 1 2 2

1 1 1 2 2

2

,

,

0.

m x m x F

m y m m g Y

Z

+ =

= − + +

=

(24.36)

The first equation will allow us to determine the motion of the wagon, whereas

the second one expresses the vertical component of the resultant of the action of

contact exerted by the rails on the wagon.



24.2.6.1 The Parallelepiped is in Equilibrium on the Wagon

The equilibrium of the parallelepiped on the wagon is characterized by:

1 2 1 1 1, 0, 0, ,2

bx x y bψ= = = = − (24.37)

and Equations (24.34) and (24.36) are written:

( )( )

1 1 21

1 21

21 1 21

1 2 1

1 2 2

,

0 ,

0 .2

,

0 .

m x X

m g Y

bX a Y

m m x F

m m g Y

=

= − +

= +

+ =

= − + +

(24.38)

To these equations, we have to associate the conditions of friction:

21 21 210, Y X f Y> < . (24.39)

From Equations (24.38), we deduce:

( )

1 21 2

121

1 2

21 1

11 2

,

,

,

.2

Fx x

m m

mX F

m m

Y m g

b Fa

m m g

= =+

=+

=

= −+

(24.40)

The condition 21 0Y > is satisfied, and the condition 21 21X f Y< is written:

( )1 2F f m m g< + . (24.41)

Moreover, the point I must be located between the points C and D

1( /2 /2)a a a− < < . It results that, in the case where F is positive, the following

inequality must be satisfied:

( )1 2a

F m m gb

< + . (24.42)

The conditions (24.41) and (24.42) being satisfied, the parallelepiped is in equi-

librium on the wagon. The motion of the parallelepiped-wagon set is imposed by

the force F, in accordance with the first of Equations (24.40).

24.2.6.2 The Parallelepiped Slides on the Wagon without Rocking

The motion is then characterized by:

1 2 1 1 1, 0, 0, .2

bx x y bψ≠ = = = − (24.43)


Equations (24.34) and (24.36) are written:

( )

1 1 21

21 1

21 1 21

1 1 2 2

2 1 2

,

,

0.2

,

.

m x X

Y m g

bX a Y

m x m x F

Y m m g

=

=

+ =

+ =

= +

(24.44)

To these equations, it is necessary to associate the conditions of friction:

21 21 21 1 2 10, , , .2 2

a aY X f Y x x a> = < − < < (24.45)

Combining Equations (24.44) and (24.45) allows us to determine first the compo-

nents of the action of contact:

21 1 21 1, ,Y m g X fm g= = (24.46)

and then the motions of the parallelepiped and the wagon:

1 22

, .F

x fg x fgm

= = − (24.47)

Sliding of the parallelepiped occurs if the following conditions are satisfied:

( )1 2 , .a

F f m m g fb

> + < (24.48)

24.2.6.3 The Parallelepiped Rocks and Slides on the Wagon

In the case where there is rocking of the parallelepiped (Figure 24.4), this

motion occurs around the edge passing through the point D. The coordinates of

the mass centre G1 can be expressed as functions of the Cartesian coordinates

( )( ), ( ), 0x D y D of the point D as:

( )( )

1 1

1 1

1

( ) cos ,

( ) sin ,

0,

x x D l

y y D l

z

ψ γ

ψ γ

= + +

= + +

=

(24.49)

introducing the geometrical characteristics of the parallelepiped:

2 2 11, tan .

2

al a b

bγ −= + = (24.50)

From this, we deduce then:

( )( )

1 1 1

1 1 1

( ) sin ,

cos ,

x x D l

y l

ψ ψ γ

ψ ψ γ

= − +

= +

(24.51)

and

( ) ( )

( ) ( )

21 1 1 1 1

21 1 1 1 1

( ) sin cos ,

cos sin .

x x D l

y l

ψ ψ γ ψ ψ γ

ψ ψ γ ψ ψ γ

= − + + +

= + − +

(24.52)


A

y

y1

x1

x

D

G1

x(D)

y(D)

(S1)

(S2) wagon

1

FIGURE 24.4. Rocking motion of the parallelepiped.

Moreover:

( ) ( )1 1 1 1cos sinG I G D l i jψ γ ψ γ = = − + + +

(24.53)

For the early motion, we have 1 0ψ ≈ and 1 0ψ ≈ . The preceding expressions

are reduced to:

( )

1 1

1 1

1 1

( ) sin ,

cos ,

cos sin .

x x D l

y l

G I G D l i j

ψ γψ γ

γ γ

= −

=

= = − +

(24.54)

Hence:

1 1cos , sina l b lγ γ= − = − . (24.55)

In the case where the parallelepiped rocks and slides on the wagon, Equations

(24.34) and (24.36) are written for the early motion in the form:

[ ]

[ ]

1 1 21

1 1 1 21

1 1 21 21

1 1 2 2

( ) sin ,

cos ,

sin cos .3

( ) sin .

m x D l X

m l m g Y

lm X Y

m x D l m x F

ψ γ

ψ γ

ψ γ γ

ψ γ

− =

= − +

= −

− + =

(24.56)

To these equations, are associated the conditions of contact:

21 21 21 1 20, , 0, ( ) 0, ( )Y X fY x D x D xψ> = > > < . (24.57)

Solving Equations (24.56) leads to:


( )

( )

( )

1 2

121 2

21 21

2

2

12 2

2 2

3 sin cos,

1 3cos 3 sin cos

,1 3cos 3 sin cos

,

1 3sin 3sin cos( ) ,

1 3cos 3 sin cos

.1 3cos 3 sin cos

g f

l f

m gY

f

X f Y

fx D g

f

gmFx f

m m f

γ γψ

γ γ γ

γ γ γ

γ γ γ

γ γ γ

γ γ γ

−=

+ −

=+ −

=

+ −=

+ −

= −+ −

(24.58)

The condition 21 0Y > is satisfied if:

21 3cos

3sin cosf

γγ γ

+< . (24.59)

The condition 1 0ψ > leads to:

.b

fa

> (24.60)

Lastly, the conditions ( ) 0x D > and 2( )x D x< are written:

2

3sin cos

1 3sinf

γ γ

γ>

+, (24.61)

and

( )21 2 2

2

1 3sin 3 sin cos.

1 3cos 3 sin cos

f m m mF

f

γ γ γ

γ γ γ

+ + − >+ −

(24.62)

24.2.6.4 The Parallelepiped Rocks without Sliding

In the case where there is not sliding of the point D on the wagon: 2( )x D x=

and Equations (24.56) are written:

( )

( )

1 2 1 21

1 1 1 21

1 1 21 21

1 2 2 1 1

sin ,

cos ,

sin cos ,3

sin .

m x l X

m l m g Y

lm X Y

m m x m l F

ψ γψ γ

ψ γ γ

ψ γ

− =

= − +

= −

+ − =

(24.63)

To these equations, we have to associate the conditions:

21 21 21 1 20, , 0, 0Y X fY xψ> < > > . (24.64)

Solving Equations (24.63), we obtain:

( )

( )1 2

1 21 2

sin cos3,

1 3cos 4

F m m g

l m m

γ γψ

γ

− +=

+ +


( )( )

( )( )

( )

12 2

1 2

21 2

21 1 21 2

22

21 1 21 2

4 3 sin cos,

1 3cos 4

3 sin cos 4 3cos,

1 3cos 4

4 3sin 3 sin cos.

1 3cos 4

F m gx

m m

F m m gY m

m m

F m gX m

m m

γ γ

γ

γ γ γ

γ

γ γ γ

γ

−=

+ +

+ + − =+ +

− +=

+ +

(24.65)

The condition 1 0ψ > leads to:

( )1 2b

F m m ga

> + . (24.66)

The condition 2 0x > is written:

13

sin cos4

F m g γ γ> . (24.67)

The condition 21 0Y > is always satisfied. Lastly, the condition 21 21X fY< leads

to the inequality:

( )21 2 2

2

4 3cos 3 sin cos

4 3sin 3 sin cos

f m m mF g

f

γ γ γ

γ γ γ

+ − − <− −

. (24.68)

24.2.6.5 Conclusions on the Analysis of the Different Motions

The different motions studied take place according as such or such of the con-

ditions (24.41), (24.42), (24.48), (24.59) to (24.62), (24.66) to (24.68) are

satisfied. Thus, the different motions occur according to the values of the edges of

the parallelepiped, of the resultant of the driving force, of the masses of the paral-

lelepiped and wagon, of the coefficient of friction.

As an example, we examine the case of a parallelepiped with a square section,

for which:

1, sin cos ,

2a b γ γ= = = (24.69)

and with a mass equal to that of the wagon: 1 2m m= . The conditions obtained are

written as follows:

— Equilibrium of the parallelepiped on the wagon:

1 12 , 2F fm g F m g< < . (24.70)

— Sliding without rocking of the parallelepiped:

12 , 1F fm g f> < . (24.71)

— Rocking and sliding of the parallelepiped:

17 35

1 , 3 5 3

ff F m g

f

−< < >

−. (24.72)


FIGURE 24.5. The different motions of the parallelepiped according to the values of the

driving force and the coefficient of friction.

— Rocking without sliding of the parallelepiped:

1 17 3

2 , 5 3

fF m g F m g

f

−> <

−. (24.73)

The conditions (24.70) to (24.73) can be represented graphically (Figure 24.5) in a

system of axes (F, f ). The space (F, f ) is thus delimited into four areas: the area 1

where there is equilibrium of the parallelepiped on the wagon, the area 2 where

the parallelepiped slides without rocking, the area 3 where the parallelepiped

rocks and slides and the area 4 where the parallelepiped rocks without sliding.

24.3 COUPLED MOTIONS OF TWO SOLIDS

24.3.1 Introduction

We consider the simple mechanical system schematized in Figure 24.6. The

solid (S1) of mass m1 is connected to the support (T) through a prismatic conne-

ction (not represented in the figure) of horizontal axis (∆1). The solid is subjected

to the action of the spring (R) of negligible mass, of axis (∆1), stiffness k and length

l0 in the absence of action exerted on the spring. A solid (S2) of mass m2 is

connected to the solid (S1) using a hinge connection of horizontal axis (∆2),

passing through the mass centre G1 of the solid (S1) and orthogonal to the axis

(∆1). The point of the connection of the spring with (S1) is located at a distance d

from the mass centre G1. Lastly, the mass centre G2 of the solid (S2) moves in the

vertical plane passing through G1. It is located at a distance a from G1.

1

2

3

4

1

5/3

3/7

12m g0Driving force F

Co

effi

cien

t o

f fr

icti

on

f

1f =

12

Fm

g=

1

2

F

fmg

=

( ) ( ) 1

5 37 3

f F f m g

−= −

24.3 Coupled Motions of Two Solids 403

FIGURE 24.6. System of two coupled solids.


We choose the coordinate system (Oxyz) attached to the support such that the

origin O coincides with the point of connection of the spring with the support, and

such that the axis Oy

coincides with the axis of the spring and that the axis Ox

is

downward vertical.

24.3.2.1 Motion of the Solid (S1) with respect to the Support (T)

As coordinate system attached to the solid (S1), we choose the trihedron

(G1xyz) of invariable orientation with respect to the support. The motion of the

solid (S1) is a motion of rectilinear translation characterized by the abscissa y of

the mass centre G1 along the axis 1G y

.

The kinematic torsor ( ) 1

TS of the motion of the solid (S1) with respect to the

support has for elements of reduction at G1:

( ) ( )

( ) ( )

1 1

1 11

0,

( , ) .

T TS S

T TG S

R

G t y j

ω= =

= =

(24.74)

24.3.2.2 Motion of the Solid (S2) with respect to the Support (T)

As coordinate system attached to the solid (S2), we choose the trihedron

( )1 2 2G x y z , such that the axis 1 2G x

passes through the mass centre G2 of the solid

(S2). The situation of the solid (S2) with respect to the solid (S1) is characterized

by the angle of rotation ψ between the axes 1G x

and 1 2G x

. The basis change is

written:

(T) (R) (S1)

(S2)

G2

G1 (1)

(2) y

x

x x2

y2

z

z

d

O

a


2

2

cos sin ,

sin cos .

i i j

j i j

ψ ψ

ψ ψ

= +

= − +

(24.75)


TS relative to the motion of the solid (S2) with respect

to the support has for elements of reduction at the point G1:

( ) ( )

( ) ( )

2 2

1 21

,

( , ) .

T TS S

T TG S

R k

G t y j

ω ψ= =

= =

(24.76)

The velocity vector of the mass centre G2 is:

( ) ( ) ( )

222 1 1 2( , ) ( , )T T T

SG t G t G G y j a jω ψ= + × = +

. (24.77)

The acceleration vector of the mass centre is obtained by deriving the preceding

expression with respect to time. We obtain:

( )

22 2 2( , )Ta G t y j a j a iψ ψ= + −

, (24.78)

or by taking account of Relation (24.75) of basis change:

( ) ( ) ( )

2 22( , ) cos sin cos sinTa G t a i y a jψ ψ ψ ψ ψ ψ ψ ψ = − + + + −

. (24.79)

24.3.3 Kinetics

24.3.3.1 Motion of the Solid (S1)

The kinetic torsor ( ) 1

TS relative to the motion of the solid (S1) with respect to

the support has for elements of reduction at the mass centre G1:

( ) ( )

( ) ( ) ( )

1

1 11 1

1 1 1

1

( , ) ,

0.

T TS

T TG GS S

R m G t m y j

S ω

= =

= =

(24.80)

The kinetic energy is deduced easily from (24.74) and (24.80). Thus:

( ) 2c 1 1

1( )

2TE S m y= . (24.81)

Lastly, the dynamic torsor ( ) 1

TS relative to the motion of the solid (S1) with

respect to the support has for elements of reduction at the mass centre G1:

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

1

1 1 11 1 1 1

1 1 1

1 1

( , ) ,

0.

T TS

T T T TG G GS S S S

R m a G t m y j

S Sω ω ω

= =

= + × =

(24.82)





the support has for elements of reduction at the point G1:

( ) ( ) ( )( ) ( ) ( ) ( )

2

1 12 2

2 2 2 2

22 1 1 2

( , ) ,

( , ) .

T TS

T T TG GS S

R m G t m y j a j

m G G G t S

ψ

ω

= = +

= × +

(24.83)

The operator of inertia at the point G1 of the solid (S2) is represented, in the

basis (2) = ( )2 2, , i j k

attached to the solid, by the matrix of inertia:

( )

1

2 2 22

2 2 2 2

2 2 2

( )G

A F E

S F B D

E D C

− − = − − − −

I . (24.84)

The moment at the point G1 of the kinetic torsor is thus written by expressing

(24.83) in the form: ( ) ( )

1 22 2 2 2 2 2 cosT

G SE i D j C m ay kψ ψ ψ ψ= − − + +

. (24.85)

The kinetic energy is calculated by the relation:

( ) ( ) ( ) 2 2

c 21

( )2

T T TS S

E S = ⋅ . (24.86)

Taking account of Expressions (24.76), (24.83) and (24.85), we obtain:

( ) 2 2c 2 2 2 2

1 1( ) cos

2 2TE S m y C m a yψ ψ ψ= + + . (24.87)

The total kinetic energy associated to the motions of the two solids is obtained

by adding Expressions (24.81) and (24.87). Thus:

( )( ) ( ) 2 2c 1 2 1 2 2 2

1 1cos

2 2TE S S m m y C m a yψ ψ ψ∪ = + + + . (24.88)

The first term represents the kinetic energy of translation of the two solids, the

second term the kinetic energy of rotation of the solid (S2) and the third term is a

coupling kinetic energy.

The dynamic torsor ( )

2


to the support has for elements of reduction at the point G1:

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

2

1 1 12 2 2 2

22 2 2 2 2

22 1 1 2 2

( , ) ,

( , ) .

T TS

T T T T TG G GS S S S

R m a G t m y j a i a j

m G G a G t S S

ψ ψ

ω ω ω

= = − +

= × + + ×

(24.89)

The resultant will be expressed in the basis ( ), , i j k

by considering the basis

change given by Relation (24.75). Introducing the matrix of inertia (24.84) in


the expression of the moment, we obtain:

( ) ( ) ( ) ( )

1 2

2 22 2 2 2 2 2 2 2 cosT

G SE D i D E j C m ay kψ ψ ψ ψ ψ ψ= − + − + + +

(24.90)

The moment can then be expressed in the basis ( ), , i j k

using Relation (24.75)

of basis change. We obtain:

( ) ( ) ( )

( ) ( )( )

1 2

2 22 2 2 2

2 22 2 2 2

2 2

cos sin

sin cos

cos .

TG S

E D D E i

E D D E j

C m ay k

ψ ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ ψ

ψ ψ

= − + + +

+ − + − +

+ +

(24.91)

24.3.4 Analysis of the Mechanical Actions

24.3.4.1 Actions Exerted on the Solid (S1)

The mechanical actions exerted on the solid (S1) are: the action of gravity, the

action of the spring, the action of the support induced by the prismatic connection

and the action of the solid (S2) induced by the hinge connection.

The action of gravity is represented by the torsor ( ) 1e S of elements of

reduction at the mass centre:

( )

( )

1

1 1

1

e ,

e 0.G

R S m g i

S

=

=

(24.92)

The power developed by the action of gravity in the reference attached to the

support is: ( ) ( ) ( ) ( )

11 1e e 0T T

SP S S= =⋅ . (24.93)

The action exerted by the spring is a force of axis (∆1), represented by the

torsor 1( )S and of which the elements of reduction at the point G1 are:

( )

1

1 0

1

( ) ,

( ) 0.G

R S k y d l i

S

= − − −

=

(24.94)

The power developed by the action of the spring is:

( ) ( ) ( )1

1 1 0( ) ( )T TS

P S S k y d l y= = − − −⋅ . (24.95)

The action exerted by the support through the prismatic connection is repre-

sented by the torsor 1( )S . Its elements of reduction at the point G1 are:

1

1 1 1 1

1 1 1 1

( ) ,

( ) .G

R S X i Y j Z k

S L i M j N k

= + +

= + +

(24.96)


The components X1, Y1, ..., N1, are to be determined. The power developed by the

action of connection is:

( ) ( ) 1

1 1 1( ) ( )T TS

P S S Y y= =⋅ . (24.97)

The action exerted by the solid (S2) through the hinge connection is represented

by the torsor 2 1( )S , of which the elements of reduction at the point G1 are:

1

2 1 21 21 21

2 1 21 21 21

( ) ,

( ) .G

R S X i Y j Z k

S L i M j N k

= + +

= + +

(24.98)

The components X21, Y21, ..., N21, are to be determined. The power developed by

the action of connection exerted by the solid (S2), power evaluated in the refe-

rence (T), is:

( ) ( ) 1

2 1 2 1 21( ) ( )T TS

P S S Y y= =⋅ . (24.99)

24.3.4.2 Actions Exerted on the Solid (S2)

The mechanical actions exerted on the solid (S2) are reduced to the action of

gravity and the action of the solid (S1) induced by the hinge connection.

The action of gravity is represented by the torsor ( ) 2e S of elements of


( )

( )

2

2 2

2

e ,

e 0.G

R S m g i

S

=

=

(24.100)

We shall need the moment vector at the point G1. It is written:

( ) ( ) 1 12 2 2 2e e sinG S R S G G m ga kψ= × = −

. (24.101)

The power developed by the action of gravity, evaluated in the reference (T), is:

( ) ( ) ( ) ( )

22 2 2e e sinT T

SP S S m gaψ ψ= = −⋅ . (24.102)

The action exerted by the solid (S2) induced by the hinge connection is repre-

sented by the torsor 1 2( )S opposed to the torsor 2 1( )S :

1 2 2 1( ) ( )S S= − . (24.103)

The power developed by this action, evaluated in the reference (T), is:

( ) ( ) 2

1 2 1 2 21 21( ) ( )T TS

P S S Y y N ψ= = − −⋅ . (24.104)


24.3.5 Equations Deduced from the Fundamental Principle of Dynamics


In the case where the support is a pseudo-Galilean support (support attached to

the Earth), the fundamental principle of dynamics applied to the motion of the

solid (S1) with respect to the support is written as:

( ) ( ) 1

1 1 1 2 1e ( ) ( ) ( )TS

S S S S= + + + . (24.105)

The vector equations of the resultant and moment at the point G1 lead to the six

scalar equations:

( )1 1 21

1 0 1 21

1 21

1 21

1 21

1 21

0 ,

,

0 ,

0 ,

0 ,

0 .

m g X X

m y k y d l Y Y

Z Z

L L

M M

N N

= + +

= − − − + +

= +

= +

= +

= +

(24.106)


The fundamental principle of dynamics applied to the motion of the solid (S2)

with respect to the support is written:

( ) ( ) 2

2 2 1e ( )TS

S S= − . (24.107)

The vector equations of the resultant and moment at the point G1 lead to the six

scalar equations:

( )

( )

( ) ( )( ) ( )

22 2 21

22 21

21

2 22 2 2 2 21

2 22 2 2 2 21

2 2 21 2

cos sin ,

cos sin ,

0 ,

cos sin ,

sin cos ,

cos sin .

m a m g X

m y a Y

Z

E D D E L

E D D E M

C m a y N m ga

ψ ψ ψ ψ

ψ ψ ψ ψ

ψ ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ ψ

ψ ψ ψ

− + = −

+ − = −

=

− + + + = −

− + − + = −

+ = − −

(24.108)

24.3.5.3 Motion of the Set of the Two Solids

The fundamental principle of dynamics can be applied to the set constituted of

the two solids. It is then written:


( ) ( ) ( ) ( )

1 21 1 1 2e ( ) ( ) eT T

S SS S S S+ = + + + . (24.109)

This equation results from the addition of the two equations (24.105) and

(24.107). In the case where the moments of the torsors are expressed all at the

point G1 (and only in this case), the six scalar equations deduced from (24.109)

are the result of the addition of Equations (24.106) and (24.108). We obtain:

( ) ( )

( ) ( ) ( )

( ) ( )( ) ( )

22 1 2 1

21 2 2 0 1

1

2 22 2 2 2 1

2 22 2 2 2 1

2 2 1 2

cos sin ,

cos sin ,

0 ,

cos sin ,

sin cos ,

cos sin .

m a m m g X

m m y m a k y d l Y

Z

E D D E L

E D D E M

C m a y N m ga

ψ ψ ψ ψ

ψ ψ ψ ψ

ψ ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ ψ

ψ ψ ψ

− + = + +

+ + − = − − − +

=

− + + + =

− + − + =

+ = −

(24.110)

The components of the action of connection exerted between the two solids do not

appear in these equations. This property is general. The fundamental principle

applied to a set of solids does not introduce the actions exerted between the solids

of this set.

24.3.6 Analysis of the Equations Deduced from the Fundamental Principle


The fundamental principle of dynamics leads to 12 independent equations

chosen among Equations (24.106), (24.108), (24.110) or possibly among the

linear combinations of these equations. We have 14 unknowns to determine: X1,

Y1, ..., N1, X21, Y21, ..., N21, y and ψ. The physical nature of the actions of

connection will supplement to 14 the number of the equations. Solving the

selected equations will then make it possible to derive the parameters of situation

y and ψ as functions of time, then the components of connection.

24.3.6.2 Case of Connections without Friction

In the case where the connections are frictionless, the power developed by the

actions of connections are zero. For the prismatic connection, Relation (24.97)

leads to:

1 0Y = . (24.111)

An important point must be underlined for the hinge connection. Indeed, one

should not consider the powers (24.99) or (24.104) calculated with respect to the

support. The connection is made between the solids (S1) and (S2) and the power


to be considered is, for example, the power developed by the action exerted by the

solid (S1) on the solid (S2) and expressed in a reference attached to (S1). The

power developed is written:

( ) ( ) 1 1

21 2 1 2( ) ( )

S S

SP S S= ⋅ , (24.112)

where the kinematic torsor relative to the motion of the solid (S2) with respect to

the solid (S1) is expressed as:

( ) ( ) ( ) 1

2 2 1

S T TS S S

= − . (24.113)

Combining Relations (24.99), (24.103), (24.104), (24.112) and (24.113) leads to:

( ) ( ) ( ) 11 2 1 2 2 1 21( ) ( ) ( )

S T TP S P S P S N ψ= + = − . (24.114)

We find the usual expression of the power developed by the action induced by a

hinge connection. In the absence of friction, we have:

21 0N = . (24.115)

The equations of motion are among Equations (24.106), (24.108) and (24.110)

those which introduce only the components of connections Y1 and N21. Thus, the

second equation (24.110) relative to the motion of the set of the two solids

(component along j

of the resultant) and the sixth equation (24.108) relative to

the motion of the solid (S2) (component along k

of the moment at the point G1):

( ) ( ) ( )21 2 0 2

2 2 2

cos sin 0,

cos sin 0.

m m y k y d l m a

m ay C m ga

ψ ψ ψ ψ

ψ ψ ψ

+ + − − + − =

+ + =

(24.116)

Solving these equations will allow us to obtain y and ψ as functions of time. The

components of the actions of connections will be then deduced from Equations

(24.108) and (24.110).

The equilibrium of the system is obtained when 0,y ψ ψ= = = what leads to:

0 , sin 0 ( 0 ou ).y l d ψ ψ π= + = = (24.117)

The equilibrium corresponding to ψ π= is unstable. In practice, the analytical

resolution of Equations (24.116) is not possible when ψ is arbitrary. It is then

necessary to use a numerical method (Chapter 27).

In the case of low values of ψ around the position of stable equilibrium 0ψ = ,

such as cos 1ψ ≈ and sinψ ψ≈ , Equations (24.116) are written in the form:

( ) ( ) ( )21 2 0 2

2 2 2

0,

0.

m m y k y d l m a

m ay C m ga

ψ ψ ψ

ψ ψ

+ + − − + − =

+ + =

(24.118)

The equations of motion are linearized, and an analytical solution can be obtained

(Subsection 27.5.2 of Chapter 27).

Exercises 411

24.3.6.2 Case of Connections with Viscous Friction

In the case of connections with viscous friction, we have:

1 21, ,t rY c y N c ψ= − − = − (24.119)

where ct and cr are the respective coefficients of friction of the prismatic con-

nection and of the hinge connection. Equations (24.116) of motion are then mo-

dified as:

( ) ( ) ( )21 2 0 2

2 2 2

cos sin 0,

cos sin 0,

t

r

m m y c y k y d l m a

m ay C c m ga

ψ ψ ψ ψ

ψ ψ ψ ψ

+ + + − − + − =

+ + + =

(24.120)

or in linearized form:

( ) ( ) ( )21 2 0 2

2 2 2

0,

0.

t

r


m ay C c m ga

ψ ψ ψ

ψ ψ ψ

+ + + − − + − =

+ + + =

(24.121)

EXERCISES

24.1 Analysis of the motion of two solids (Figure 24.7)

The solid (S1) is a hollowed cylinder in motion of rotation with respect to a

support (T) about a horizontal axis (1). A cylinder (S2) is connected to the cylin-

der (S1) through a traction-compression and torsion spring. The solid (S2) has thus

a motion of translation along the axis of the cylinder (S1) and a motion of rotation

about this axis.

Study the motion of the two solids.

FIGURE 24.7. Motion of two solids.

(S1)

(S2)

(1)


(T)

(S1)

G1

(1)

(2)

(S2) G2

(M1

(M2

24.2 Study of the motion a radar antennaFor the analysis of its motion, a radar antenna can be considered (Figure 24.8)

as being constituted of a support (S1) and a reflector (S2).

The support (S1) is connected through a hinge connection of vertical axis (1)

with the frame (T) attached to the Earth, using an electric motor (M1). The stator

of the motor is attached to the frame and the rotor is rigidly locked with (1).

The reflector (S2) has a cylindrical symmetry. It is connected through a hinge

connection of horizontal axis (2) with the support (S1), using a motor (M2). The

stator of the motor is attached to the support (S1) and the rotor is rigidly locked

with (2). The mass centre G2 of (S2) is located at the intersection of (1) and (2).

The mass centre G1 of the set constituted of the support (S1) and the stator of

the motor (M2) is located on the axis (1) and the matrix of inertia is arbitrary.

Study the motion of the antenna.

FIGURE 24.8. Radar antenna.

COMMENTS

The motions studied in the present chapter come in complement of the

examples of motions already studied in the preceding chapters. They cons-

titute simple illustrations of the analysis of dynamics of rigid bodies and

they do not call particular comments.

The two exercises will allow the reader to apply the process of the ana-

lysis of dynamics to two simple cases.

Commentaires 413

CHAPTER 25

The Lagrange Equations


25.1.1 Free Body and Connected Body

In the case of a solid (S) free with respect to a reference (T), the situation of the

solid is determined by the knowledge of:

— the position of the mass centre G with respect to the reference (T), charac-

terized by three parameters of translation p1, p2, p3 (Cartesian, cylindrical or

spherical coordinates, etc.);

— the orientation of a coordinate system attached to the solid (S) with respect to the reference (T), characterized by the three Eulerian angles ψ, θ, ϕ which we denote by Qi ( 1 ,Q ψ= 2Q θ= and 3Q ϕ= ).

The situation of the solid is determined by the six parameters (p1, p2, p3, Q1,

Q2, Q3), which we denote by the general form qi ( 1 to 6i = ).

In the case where the solid (S) is connected in the reference (T), the situation of

the solid is determined by the position of a particular point P of the solid, cha-

racterized by the parameters of translation pj ( 3j ≤ ), and by the orientation of the

solid (S), characterized by the angles of rotation Qk ( 3k ≤ ). The set of the para-

meters of situation will be also denoted by qi, with 6i < .

25.1.2 Partial Kinematic Torsors

In the case of a solid (S) free in the reference (T), the kinematic torsor, relative

to the motion of the solid, has for elements of reduction at the mass centre:

( ) ( )

( ) ( )( ) ( ) ( )

3

1 2 31 2 3

,

( , ) ,

T TS S S

T T TT T

G S

R k i k

G t p OG p OG p OGp p p

ω ψ θ ϕ= = + +

∂ ∂ ∂= = + +

∂ ∂ ∂

(25.1)

414 Chapitre 25 The Lagrange Equations

where O is a point of reference attached to the reference (T).

The preceding expressions show that the kinematic torsor can be expressed in

the form of a linear combination of torsors as:

( ) ( ) 6

,

1i

T TS i S q

i

q

=

= , (25.2)

where the elements of reduction at the point G of the torsors ( ) , i

TS q are defined in

the following way:

( ) ( )

( )

( ) ( )

( )

( ) ( )

( )

1

2

3

,,1

, 3,2

,,3

0, , , 0 ,

0, , , 0 ,

0, , , 0 .

TT T

S GS p GGG

TT T

S GS p GGG

TT T

S S GS p GGG

OG kp

OG ip

OG kp

ψ

θ

ϕ

∂= =

∂

∂= =

∂

∂= =

∂

(25.3)

The torsors ( ) , i

TS q thus introduced are called the partial kinematic torsors rela-

tive to the respective parameters qi of situation.

Relation (25.2) is transposed to the case of a connected solid, of which the

situation is defined by p parameters, in the form:

( ) ( ) ,

1i

pT T

S i S q

i

q

=

= , (25.4)

where the partial kinematic torsors are characterized by their elements of reduc-

tion at the point P, where the parameters of translation have been defined. If the

parameter qi is the parameter of translation pj, the elements of reduction of the

partial kinematic torsor are:

( ) ( ) ( )

, ,0,

i j

TT T

S q S pP P j P

OPp

∂= = ∂

. (25.5)

If the parameter qi is the parameter of rotation Qk, we shall have:

( ) ( )

, ,, 0

i k

T Tk PS q S QP P

u= =

(25.6)

with

3

, si ,

, si ,

, si .

k

k k

S k

k Q

u i Q

k Q

ψ

θ

ϕ

=

= =

=

(25.7)

25.1 General Elements 415

25.1.3 Power Coefficients

The power developed in the reference (T) by the mechanical action exerted on

the solid (S) and represented by the torsor ( ) S is written from Relation (11.13):

( ) ( ) ( ) ( ) T TSP S S= ⋅ . (25.8)

Taking account of Expressions (25.2) and (25.4), we may write the power in the

form:

( ) ( ) ( ) ( ) 1

i

pT T

qi

i

P S q P S

=

= , (25.9)

setting:

( ) ( ) ( ) ( ) ,i i

T Tq S q

P S S= ⋅ . (25.10)

The coefficient ( ) ( ) i

TqP S thus introduced is called power coefficient relative to

the variable qi.

The power coefficients can be derived from Relation (25.10), requiring to

express the partial kinematic torsors as a preliminary. Usually, they can be obtain-

ned more simply while implementing a direct calculation of the power using

Relation (25.8). The expression obtained then displays the power coefficients.

25.1.4 Perfect Connections

In the case where the solid (S) is connected in the reference (T) through a

connection, the reference (T) exerts an action of connection represented by the

torsor ( ) T S . The connection is perfect if the power developed is zero:

( ) ( ) ( ) ( ) 0T TT T SP S S= =⋅ , (25.11)

what leads, taking account of (25.9), to p relations:

( ) ( ) ( ) ( ) ,

0, 1, 2, . . . , .i i

T Tq T T S q

P S S i p= = =⋅ (25.12)

In the case where the connection cannot be regarded as perfect, it is necessary

to consider hypotheses on the physical nature of the friction processes induced.

The model of viscous friction is the simplest one to implement. This model leads

to write the power coefficients in the form:

( ) ( ) i

Tq T i iP S c q= − (25.13)

where ci is the viscous damping coefficient relative to the parameter of situation qi.


25.2 LAGRANGE EQUATIONS RELATIVE TO

A RIGID BODY

25.2.1 Introduction to the Lagrange Equations

We consider the case of a solid (S) in motion relatively to the reference (T). The

kinetic energy relative to the motion of the solid (S) with respect to the reference

(T) can be put in the form (16.27):

( ) ( ) ( ) ( ) ( )2

c1 1( ) ( , )2 2

T T T TS G SE S m G t Sω ω = + ⋅

. (25.14)

We search for expressing in this subsection the expression:

( )( ) ( )( )c c

d

dT T

i i

E S E St q q

∂ ∂−

∂ ∂. (25.15)

Deriving the first term of Expression (25.14) with respect to the variable ,iq

we obtain:

( ) ( ) ( )

21 ( , ) ( , ) ( , )2

T T T

i i

m G t m G t G tq q∂ ∂ = ∂ ∂

⋅

. (25.16)

Then deriving this expression with respect to time, we have:

( )( )

( ) ( ) ( )( )

( )

21 d ( , )2 d

d( , ) ( , ) ( , ) ( , ) .d

TT

i

TT T T T

i i

m G tt q

ma G t G t m G t G tq t q

∂ = ∂

∂ ∂+∂ ∂

⋅ ⋅

(25.17)

To calculate the derivative of the second term, we represent the operator of

inertia by the matrix of inertia at the point G in the principal basis (bS) of inertia:

( )( )

0 0

0 0

0 0

SbG

A

S B

C

=

I . (25.18)

If ω1, ω2 and ω3 are the components of the rotation vector ( )TSω

in the basis (bS),

the second term of Expression (25.14) of the kinetic energy is written:

( ) ( ) ( ) ( )2 2 21 2 3

1 12 2

T TS G SS A B Cω ω ω ω ω= + +⋅ . (25.19)

Deriving this expression with respect to the variable ,iq we obtain:

( ) ( ) ( ) 31 21 2 3

12

T TS G S

i i i i

S A B Cq q q q

ωω ωω ω ω ω ω

∂∂ ∂∂ = + + ∂ ∂ ∂ ∂⋅

, (25.20)

or

25.2 Lagrange Equations Relative to a Rigid Body 417

( ) ( ) ( )( ) ( )

( ) ( )12

S TT T TS

S G S G Si i

S Sq q

ωω ω ω

∂∂ = ∂ ∂⋅ ⋅

. (25.21)

Deriving this expression with respect to time, we obtain:

( )( ) ( ) ( )

( ) ( ) ( )

( ) ( )( ) ( ) ( )

( ) ( )

1 d2 d

d d .d d

TT T

S G Si

S T S TT TT TS S

G S G Si i

St q

S St q q t

ω ω

ω ωω ω

∂ = ∂

∂ ∂ + ∂ ∂

⋅

⋅ ⋅

(25.22)

Implementing a calculation similar to the preceding one, but substituting the

variable iq for the variable qi, we obtain:

( )( ) ( )( )

( )( )

( ) ( ) ( ) c ( , ) ( , ) .

T ST T T T T

S G Si i i

E S m G t G t Sq q q

ω ω∂ ∂ ∂= +∂ ∂ ∂

⋅ ⋅

(25.23)

Combining Expressions (25.17), (25.22) and (25.23) leads, taking account of

Relations (A.25.19) and (A.25.24) of the appendix to this chapter, to the result:

( )( ) ( )( )

( ) ( )

( )( )

( ) ( )

c cd

d

( , ) ,

T T

i i

T ST T T T

GS S Si i

E S E St q q

R G tq q

ω

∂ ∂− =

∂ ∂

∂ ∂+

∂ ∂⋅ ⋅

(25.24)

while introducing the elements of reduction at the point G of the dynamic torsor.

From Relations (25.1) to (25.3), the elements of reduction at the point G of the

partial kinematic torsors are:

( ) ( )

( )

( ) ( )

( )

,

,

,

( , ) .

i

i

ST T

SS qi

TT T

G S qi

Rq

G tq

ω∂=

∂

∂=

∂

(25.25)

Relation (25.24) is thus written finally:

( )( ) ( )( ) ( ) ( ) c c ,

d

d i

T T T TS S q

i i

E S E St q q

∂ ∂− =

∂ ∂⋅

. (25.26)

25.2.2 Lagrange Equations

If the reference (T) is a Galilean reference (g), the fundamental principle of

dynamics leads to:

( ) ( ) ( ) ( ) ( ) ( ) , , ii i

g g g gqS S q S q

S P S= =⋅ ⋅ , (25.27)

where ( ) S is the torsor representing the whole of the actions exerted on the

solid (S) and expressed in the Galilean reference (g). The association of Relations


(25.26) and (25.27) leads to the equations:

( )( ) ( )( ) ( ) ( ) c cd

, 1, 2, . . . , ,d i

gg gq

i i

E S E S P S i pt q q

∂ ∂− = =

∂ ∂ (25.28)

where the power coefficients are expressed in the Galilean reference:

( ) ( ) ( ) ( ) ,i i

ggq S qP S S= ⋅ . (25.29)

Equations (25.28) are known under the name of Lagrange equations. They can be

rewritten while distinguishing between the known actions ( ) S exerted on the

solid and the actions of connection ( ) S . Thus:

( )( ) ( )( ) ( ) ( ) ( ) ( )

c cd

,d

1, 2, . . . , ,

i i

g gg gq q

i i

E S E S P S P St q q

i p

∂ ∂− = +

∂ ∂

=

(25.30)

with

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

,

,

,

.

i i

i i

ggq S q

ggq S q

P S S

P S S

=

=

⋅

⋅

(25.31)

The Lagrange equations are derived from the fundamental principle of dyna-

mics, and consequently they do not bring any new information compared to the

scalar equations deduced from the fundamental principle. Indeed, the Lagrange

equations are linear combinations of these last equations. Their interest lies in the

fact that they constitute p differential equations of the motion of the solid (S).

Indeed, once determined the physical nature of the connections, Equations (25.30)

form a system of p differential equations where the only unknowns are the p

parameters of situation qi. Solving this system allows us to determine these para-

meters as functions of time. Note however that the determination of the compo-

nents of the actions of connections, other than those appearing in (25.31), requires

to return to the equations derived from the fundamental principle of dynamics.

In the case of frictionless connections, the power coefficients relative to the

actions of connections are zero, and the Lagrange equations (25.30) are reduced

to:

( )( ) ( )( ) ( ) ( ) c cd

, 1, 2, . . . , .d i

gg gq

i i

E S E S P S i pt q q

∂ ∂− = =

∂ ∂ (25.32)

25.2.3 Case where the Mechanical Actions Admit a Potential Energy

In the case where the mechanical actions exerted on the solid, other that the

actions of connections, admit a potential energy not depending explicitly on time,

25.3 Lagrange Equations for a Set of Rigid Bodies 419

Relation (11.29) leads to:

( ) ( ) ( ) ( ) p 1 2 p 1 2

1

d ( , , . . . , ) ( , , . . . , )d

pg g g

p p ii

i

P S E q q q E q q q qt q

=

∂= − = −∂ .

Thus, it results that the power coefficient relative to the parameter qi is expressed

as follows:

( ) ( ) ( )( )pi

g gq

i

P S E Sq∂= −

∂ . (25.33)

In the case of perfect connections, the Lagrange equations (25.32) are thus written

in the form:

( )( ) ( )( ) ( )( ) c c pd

0, 1, 2, . . . , .d

g g g

i i i

E S E S E S i pt q q q

∂ ∂ ∂− + = =

∂ ∂ ∂ (25.34)

25.3 LAGRANGE EQUATIONS FOR A SET

OF RIGID BODIES

We consider in this section a set (D) constituted of n solids (S1), (S2), ..., (Sn).

The mechanical actions exerted on the solids were considered in Subsection

14.2.2. We take here again the notations already used. Moreover, the situation of

the whole of the solids with respect to the reference (g) used for the analysis is

characterized by a total of p independent parameters of situation q1, q2, ..., qp.

25.3.1 Lagrange Equations for each Solid

The actions exerted on each solid (Sj) are (14.5) and (14.6):

( ) ( ) ( ) ( ) ( ) 1

n

j j j j j k j k j

kj

S S S S S S S

=≠

= → = + + + , (25.35)

separating the internal and external actions, the known actions and the actions of

connections.

The Lagrange equations (25.30) relative to each solid (Sj) are thus written,

taking account of (25.35), as:

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

c c

1

d( ) ( )

d

,

1, 2, . . . , .

i i

i i

g gT Tq qj j j j

i in

g gq qk j k j

kj

E S E S P S P St q q

P S P S

i p

=≠

∂ ∂− = +

∂ ∂

+ +

=

(25.36)


The power coefficients introduce the partial kinematic torsors ( ) ,j i

g

S q . The

calculation of these torsors can be implemented either directly if (Sj) is located

directly with respect to the reference (g), or through the other solids (Sα), (Sβ), ...,

(Sγ), if (Sj) is located through these solids:

( ) ( ) ( ) ( ) , , , ,

. . . j i j i i i

SSg g

S q S q S q S qβα

α γ= + + + . (25.37)

Writing the Lagrange equations for the set of the n solids, as many equations

will be obtained as there are partial kinematic torsors ( ) ,j i

g

S q which are non null.

25.3.2 Lagrange Equations for the Set (D)

The kinetic energy of the set (D) of the solids is written:

( )( ) ( )c c

1

( )

ng g

j

j

E D E S

=

= . (25.38)

Taking account of this expression and adding member with member, Equations

(25.36) obtained for each solid, we deduce the Lagrange equations for the set (D):

( )( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

c c

1

1 1

d

d

,

1, 2, . . . , .

i i

i i

ng gT T

q qj ji i

jn n

g gq qk j k j

j kj

E D E D P S P St q q

P S P S

i p

=

= =≠

∂ ∂ − = + ∂ ∂

+ +

=

(25.39)

This system of equations constitutes the system of the p equations of motion.

The nature of the connections being introduced, these equations will allow us to

determine the parameters qi as functions of time.

The Lagrange equations can be also rewritten while taking account of the

mutual actions exerted between the solids:

( ) ( ) ( ) ( ) , .j k k j j k k jS S S S= − = − (25.40)

Thus, it results that:

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

1 1 1 1

1 1

.

i j

k

j k j

n n n ng g

q k j k j S

j k j kj j

n n n nSg g

k j k jS S S

j k j j k j

P S S

S S

= = = =≠ ≠

= < = <

=

= − =

⋅

⋅ ⋅

(25.41)

25.3 Lagrange Equations for a Set of Rigid Bodies 421

A similar relation exists in the case of the actions of connections. The Lagrange

equations (25.39) can thus be written in the form:

( )( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

c c

1

1 1

d

d

,

(25.42)1, 2, . . . , .

i i

k k

j j

ng gT T

q qj ji i

jn n n n

S Sk j k jS S

j k j j k j

E D E D P S P St q q

S S

i p

=

= < = <

∂ ∂ − = + ∂ ∂

+ +

=

⋅ ⋅

Lastly, in the case where the mechanical actions exerted on the solids admit a

potential energy, the Lagrange equations (25.39) are written, for the set of the

solids, in a form similar to the form (25.34).

25.3.3 Case where the Situation Parameters are Linked

Suppose that while choosing the parameters qi, we did not take account of all

the connections. The number of parameters of situation is thus m (m > p): q1, q2,

..., qp, ... qm. There exist then m – p relations of connection between the

parameters, written as:

1 2( , , . . . , ) 0, 1, 2, . . . , .r mf q q q r m p= = − (25.43)

These relations involve, between the parameters iq , the linear relations:

1

0, 1, 2, . . . , .

mr

ii

i

fq r m p

q=

∂= = −

∂ (25.44)

By introducing the torsor ( ) ,jS representing (25.35) the whole of the actions

exerted on the solid (Sj), the Lagrange equations (25.39) lead to:

( )( ) ( )( ) ( ) ( ) c c

1 1 1

d

d i

m m ngT T

qi j ii i

i i j

E D E D q P S qt q q

= = =

∂ ∂ − = ∂ ∂

. (25.45)

The association of Relations (25.44) and (25.45) leads to the following equation:

( )( ) ( )( ) ( ) ( ) c c

1 1 1

d0

d i

m pm ngT T r

qr j ii i i

i r j

fE D E D P S q

t q q qλ

−

= = =

∂∂ ∂− − − =

∂ ∂ ∂

, (25.46)

obtained by subtracting Relations (25.44) affected of the multipliers λr from Equations (25.45). It is then possible to choose the multipliers λr, so as to make zero in Equation (25.46) the coefficients of the parameters .iq The Lagrange equations are thus written:

( )( ) ( )( ) ( ) ( ) c c

1 1

d,

d

1, 2,..., .

i

m pngT T r

q j ri i i

j r

fE D E D P S

t q q q

i m

λ−

= =

∂∂ ∂− = +

∂ ∂ ∂

=

(25.47)


Adding the m – p conditions of connection (25.43), we obtain a system of 2m p−unknowns: 1 2 1 2, , . . . , , , , . . . ,m m pq q q λ λ λ − , which can be determined as func-

tions of time.

The parameters 1 2, , . . . , m pλ λ λ − are called the Lagrange multipliers. Their

interest lies in the fact that the system is implemented in an automatic way. This

will compensate in some cases the disadvantage that there is to introduce new

unknowns: 1 2, , . . . , m pλ λ λ − .

25.4 APPLICATIONS

25.4.1 Motion of a Parallelepiped Moving on an Inclined Plane

We return to the problem considered in Section 23.2 of Chapter 23, by

searching for the equations of motion using the Lagrange equations.

The three parameters of situation are: the coordinates x, y of the mass centre and the angle of rotation ψ. The kinetic energy of the parallelepiped (S) with respect to the inclined plane was expressed in (23.10):

( ) ( ) ( )2 2 2 2 2c

1( )2 24

T mE S m x y a b ψ= + + + . (25.48)

The power developed by the action of gravity was also determined. We obtained

(23.13):

( ) ( ) e sinTP S mgx α= . (25.49)

Thus, we deduce the power coefficients relative to the parameters x, y and ψ:

( ) ( ) ( ) ( ) ( ) ( ) e sin , e 0, e 0.T T T

x yP S mg P S P Sψα= = = (25.50)

The power developed by the action of contact exerted by the plane is expressed in

(23.15): ( ) ( ) T

l l lP S X x Y y N ψ= + + , (25.51)

hence the power coefficients:

( ) ( ) ( ) ( ) ( ) ( ) , , .T T Tx l y l lP S X P S Y P S Nψ= = = (25.52)

1. Lagrange equation relative to the parameter x

The Lagrange equation relative to the parameter x is written, from (25.30):

( )( ) ( )( ) ( ) ( ) ( ) ( ) c cd

ed

T T T Tx xE S E S P S P S

t x x

∂ ∂− = +

∂ ∂ . (25.53)

We have:

25.4 Applications 423

( )( ) ( )( ) ( )( )c c c

d, , 0.

dT T TE S mx E S mx E S

x t x x

∂ ∂ ∂= = =

∂ ∂ ∂

(25.54)

Substituting Expressions (25.50), (25.52) and (25.54) into Equation (25.53), we

obtain the first Lagrange equation:

sin lmx mg Xα= + (25.55)

2. Lagrange equation relative to the parameter y

The Lagrange equation relative to the parameter y is written:

( )( ) ( )( ) ( ) ( ) ( ) ( ) c cd

e ,d

T T T Ty yE S E S P S P S

t y y

∂ ∂− = +

∂ ∂ (25.56)

with

( )( ) ( )( )c c

d, 0.

dT TE S my E S

t y y

∂ ∂= =

∂ ∂

(25.57)

Hence the second Lagrange equation:

lmy Y= . (25.58)

3. Lagrange equation relative to the parameter ψ

The Lagrange equation relative to the parameter ψ is written:

( )( ) ( )( ) ( ) ( ) ( ) ( ) c cd

e ,d

T T T TE S E S P S P St

ψ ψψ ψ∂ ∂

− = +∂ ∂

(25.59)

with

( )( ) ( ) ( )( )2 2c c

d, 0.

d 12T Tm

E S a b E St

ψψ ψ∂ ∂

= + =∂ ∂

(25.60)

Hence the third Lagrange equation:

( )2 2

12l

ma b Nψ+ = . (25.61)

The Lagrange equations (25.55), (25.58) and (25.61) coincide with the equations

of motion (Equations 1, 2 and 6) of Equations (23.17) deduced from the funda-

mental principle of dynamics.

25.4.2 Coupled Motions of Two Solids

In this subsection, we return to the problem considered in Section 24.3 of

Chapter 24. The problem is that of two solids (S1) and (S2) coupled, with two

parameters of situation: the parameter y of translation of the solid (S1) and the

parameter ψ of rotation of the solid (S2) with respect to the solid (S1).

The Lagrange equations for the set (D) constituted of the two solids are written

from (25.39):


( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

c c 1 1 1

2 1 2 1 2

de

d

e ,

(25.62), .

i i i

i i i

T T T T Tq q q

i iT T T

q q q

i

E D E D P S P S P St q q

P S P S P S

q y ψ

∂ ∂− = + +

∂ ∂

+ + +

=

The kinetic energy is given by Expression (24.88):

( )( ) ( ) 2 2c 1 2 2 2

1 1cos

2 2TE D m m y C m a yψ ψ ψ= + + + . (25.63)

The powers developed were also determined in (24.93), (24.95), (24.97), (24.99),

(24.102) and (24.104). From these results, we deduce the power coefficients:

— action of gravity exerted on the solid (S1):

( ) ( ) ( ) ( ) 1 1e 0, e 0,T TyP S P Sψ= = (25.64)

— action exerted by the spring on the solid (S1):

( ) ( ) ( ) ( ) ( ) 1 0 1, 0,T TyP S k y d l P Sψ= − − − =

— action exerted by the support on the solid (S1):

( ) ( ) ( ) ( ) 1 1 1, 0,T TyP S Y P Sψ= = (25.66)

— action exerted by the solid (S2) on the solid (S1):

( ) ( ) ( ) ( ) 2 1 21 2 1, 0,T TyP S Y P Sψ= = (25.67)


( ) ( ) ( ) ( ) 2 2 2e 0, e sin ,T TyP S P S m gaψ ψ= = − (25.68)


( ) ( ) ( ) ( ) 1 2 21 1 2 21, .T TyP S Y P S Nψ= − = − (25.69)

1. Lagrange equation relative to the parameter y

We have:

( )( ) ( )c 1 2 2 cosTE D m m y m ay

ψ ψ∂= + +

∂

, (25.70)

( )( ) ( ) 2c 1 2 2 2

dcos sin

dTE D m m y m a m a

t yψ ψ ψ ψ∂

= + + −∂

, (25.71)

( )( )c 0TE D

y

∂=

∂. (25.72)

Hence the first Lagrange equation:

( ) ( ) ( )21 2 2 0 1cos sinm m y m a k y d l Yψ ψ ψ ψ+ + − = − − − + . (25.73)



We have:

( )( )c 2 2 cosTE D C m ayψ ψ

ψ∂

= +∂

, (25.74)

( )( )c 2 2 2

dcos sin

dTE D C m ay m ay

tψ ψ ψ ψ

ψ∂

= + −∂

, (25.75)

( )( )c 2 sinTE D m a yψ ψ

ψ∂

= −∂

. (25.76)


2 2 21 2cos sinC m ay N m gaψ ψ ψ+ = − − . (25.77)

The Lagrange equations (25.73) and (25.77) coincide with the equations of

motions derived from the fundamental principle of dynamics (second equation

(24.110) and sixth equation (24.108)).

25.4.3 Double Pendulum


As another example of application, we study the motion of the set of two solids

(S1) and (S2), schematized in Figure 25.1. The solid (S1) is connected to the

support (T) through a hinge connection of horizontal axis. The solid (S2) is

connected to the solid (S1) by a hinge connection of also horizontal axis. At rest,

the two mass centres G1 and G2 of the two solids are located on the same vertical.

The only mechanical actions external to the set of the two solids are the actions of

gravity. The support will be considered as a pseudo-Galilean reference.

25.4.3.2 Parameters of Situation and Kinematics

As coordinate system attached to the support, we choose the trihedron (Oxyz)

such that the axis Ox

is downward vertical and the axis Oz

coincides with the

hinge connection between the solid (S1) and the support. To the solid (S1), we

attach the system (Ox1y1z), such that the axis 1Ox

passes through the mass centre

G1. The situation of the solid (S1) with respect to the support is determined by the

angle of rotation ψ1 between the axes Ox

and 1Ox

. The relation of basis change is:

1 1 1

1 1 1

cos sin ,

sin cos .

i i j

j i j

ψ ψ

ψ ψ

= +

= − +

(25.78)

To the solid (S2), we attach the trihedron (O1x2y2z), such that the axis 21O x


(S1)

(S2)

O

(1)

(2)

y

x

1

x

x2

y1

zy2

G1

G2

O1

d

a1

a2

1

2

x1

2 z

FIGURE 25.1. Double pendulum.

passes through the mass centre G2 of the solid (S2) (the points O, G1 and O1 are

aligned). The situation of the solid (S2) with respect to the support is determined

by the angle of rotation ψ2 between the axes 11O x

and 21O x

. The relation of basis

change is:

2 2 2

2 2 2

cos sin ,

sin cos .

i i j

j i j

ψ ψ

ψ ψ

= +

= − +

(25.79)

The position vectors of the mass centres G1 and G2, and of the point O2 are:

121 1 1 1 2 2 1, , ,OG a i O G a i OO d i= = =

(25.80)

where a1 is the distance from the mass centre G1 to the axis of rotation Oz

, a2 the

distance from the mass centre G2 to the axis of rotation 1O z

and d the distance

between the two axes of rotation.

The kinematic torsor relative to the motion of the solid (S1) with respect to the

support has for elements of reduction at the point O:

( ) ( )

( ) ( )

1 1

1

1 ,

( , ) 0.

T TS S

T TO S

R k

O t

ω ψ= =

= =

(25.81)

The velocity vector of the mass centre G1 of the solid (S1) is:

( )1 1 1 1( , )T G t a jψ=

. (25.82)

The kinematic torsor relative to the motion of the solid (S2) with respect to the


support has for elements of reduction at the point O1:

( ) ( )

( ) ( )

2 2

1 2

2

1 1 1

,

( , ) .

T TS S

T TO S

R k

O t d j

ω ψ

ψ

= =

= =

(25.83)

The velocity vector of the mass centre G2 of the solid (S2) is: ( ) ( ) ( )

122 1 1( , ) ( , )T T T

SG t O t O Gω= + ×

, (25.84)

hence: ( )

2 1 1 2 2 2( , )T G t d j a jψ ψ= + . (25.85)

25.4.3.3 Kinetic Energies

The calculation of the kinetic energies needs the determination of the kinetic

torsors, which introduce the masses and the operators of inertia of the solids. The

mass of the solid (S1) is denoted by m1 and its matrix of inertia at the point G1 in

the basis (1) = ( )1 1, , i j k

is:

( )

1

1 1 11

1 1 1 1

1 1 1

( )G

A F E

S F B D

E D C

− − = − − − −

I . (25.86)

Similarly, the mass of the solid (S2) is m2 and its matrix of inertia at the point G2

in the basis (2) = ( )2 2, , i j k

is:

( )

2

2 2 22

2 2 2 2

2 2 2

( )G

A F E

S F B D

E D C

− − = − − − −

I . (25.87)

Thus, it results that the kinetic torsor ( ) 1

TS relative to the motion of the solid

(S1) with respect to the support has for elements of reduction at the point G1:

( ) ( )

( ) ( ) ( )

1

1 11 1

1 1 1 1 1 1

1 1 1 1 1 1 1 1 1

( , ) ,

.

T TS

T TG GS S

R m G t m a j

S E i D j C k

ψ

ω ψ ψ ψ

= =

= = − − +

(25.88)

Similarly, the elements of reduction at the point G2 of the kinetic torsor relative

to the motion of the solid (S2) with respect to the support are:

( ) ( ) ( )( ) ( ) ( )

2

2 22 2

2 2 2 1 1 2 2 2

2 2 2 2 2 2 2 2 2

( , ) ,

.

T TS

T TG GS S

R m G t m d j a j

S E i D j C k

ψ ψ

ω ψ ψ ψ

= = +

= = − − +

(25.89)

The kinetic energy relative to the motion of the solid (S1) with respect to the

support is:


( ) ( ) ( ) 1 1

c 11

( )2

T T TS S

E S = ⋅ , (25.90)

with: ( ) ( ) ( ) 11 1 1

1T T T

O GS S SR G O= + ×

, (25.91)

or taking account of (25.88):

( ) ( )

1

21 1 1 1 1 1 1 1 1 1

TO S

E i D j C m a kψ ψ ψ= − − + +

. (25.92)

Introducing this result into Expression (25.90) leads to:

( ) ( )2 2c 1 1 1 1 1

1( )

2TE S C m a ψ= + . (25.93)

The kinetic energy relative to the motion of the solid (S2) with respect to the

support is:

( ) ( ) ( ) 2 2

c 21

( )2

T T TS S

E S = ⋅ , (25.94)

or expanding the product of the torsors at the point O1:

( ) ( ) ( ) ( ) ( ) 12 2

c 2 2 2 11 1

( ) ( , ) ( , )2 2

T T T T TOS S

E S m G t O t ω= +⋅ ⋅ , (25.95)

with: ( ) ( ) ( ) 1 22 2 2

12T T T

O GS S SR G O= + ×

. (25.96)

Taking account of (25.89), we obtain:

( ) ( )

1 22

2 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2cos .

TO S

E i D j C m da m a kψ ψ ψ ψ ψ ψ ψ

=

− − + + − +

(25.97)

Introducing (25.83), (25.85) and (25.97) into Expression (25.95) for the kinetic

energy leads to:

( ) ( ) ( )2 2 2 2c 2 2 1 2 2 2 2 2 2 1 2 2 1

1 1( ) cos

2 2TE S m d C m a m daψ ψ ψ ψ ψ ψ= + + + − . (25.98)

From Expressions (25.93) and (25.98), we deduce that the kinetic energy of the

set (D) of the two solids is:

( )

( ) ( ) ( )c

2 2 2 2 21 1 1 2 1 2 2 2 2 2 2 1 2 2 1

( )1 1

cos .2 2

TE D

C m a m d C m a m daψ ψ ψ ψ ψ ψ

=

+ + + + + − (25.99)

25.4.3.4 Powers Developed

The mechanical actions exerted by the solid (S1) are: the action of gravity,

represented by the torsor ( ) 1e S ; the action of the support induced by the hinge

connection, represented by the torsor ( ) 1S ; the action of the solid (S2) induced

by the second hinge connection, represented by the torsor ( ) 2 1S . The power


developed by the action of gravity is:

( ) ( ) ( ) ( ) ( ) ( )

1 11 1 1e e eT T T

OS SP S S Sω= =⋅ ⋅

, (25.100)

with

( ) 1 1 1 1e sinO S m ga kψ= − . (25.101)

Hence: ( ) ( ) 1 1 1 1 1e sinTP S m gaψ ψ= − , (25.102)

and the power coefficients are:

( ) ( ) ( ) ( )

1 21 1 1 1 1e sin , e 0.T TP S m ga P Sψ ψψ= − = (25.103)

Similarly, the power developed by the action exerted by the support is:

( ) ( ) 1

1 1 1 1( ) ( )T TS

P S S N ψ= =⋅ , (25.104)

where N1 is the component along the direction k

of the moment at the point O of

the axis of the hinge connection. The power coefficients are:

( ) ( ) 1 21 1 1( ) , ( ) 0.T TP S N P Sψ ψ= = (25.105)

The power developed by the action of connection exerted by the solid (S2) is:

( ) ( ) 1

2 1 2 1 21 1 21 1( ) ( )T TS

P S S N dYψ ψ= = +⋅ , (25.106)

where N21 is the component along the direction k

of the moment at the point O1

and Y21 the component along the direction 1j

of the resultant of the action of con-

nection. The power coefficients are thus:

( ) ( ) 1 22 1 21 21 2 1( ) , ( ) 0.T TP S N dY P Sψ ψ= + = (25.107)

The mechanical actions exerted on the solid (S2) are reduced to the action of

gravity, represented by the torsor ( ) 2e S , and the action of the solid (S1)

induced by the hinge connection, represented by the torsor ( ) 1 2S opposed to

the torsor ( ) 2 1S . The power developed by the action of gravity is:

( ) ( ) ( ) ( )

22 2e eT T

SP S S= ⋅ , (25.108)

or expanding at the point O1:

( ) ( ) ( ) ( ) ( ) ( ) 122 2 1 2e e ( , ) eT T T

OSP S R S O t Sω= +⋅ ⋅

, (25.109)

with:

( ) 1 2 2 2 2e sinO S m ga kψ= − . (25.110)

Thus: ( ) ( ) 2 2 1 1 2 2 2 2e sin sinTP S m gd m gaψ ψ ψ ψ= − − . (25.111)

The power coefficients are:


( ) ( ) ( ) ( )

1 22 2 1 2 2 2 2e sin , e sin .T TP S m gd P S m gaψ ψψ ψ= − = − (25.112)

The expression of the power developed by the action of connection exerted by the

solid (S1) on the solid (S2) has the same form as Expression (25.109). By applying

this expression to the action of connection, we obtain:

( ) 1 2 21 1 21 2( )TP S dY Nψ ψ= − − , (25.113)

hence the power coefficients:

( ) ( ) 1 21 2 21 1 2 21( ) , ( ) .T TP S dY P S Nψ ψ= − = − (25.114)

25.4.3.5 Lagrange Equations

The Lagrange equations for the set (D), constituted of the two solids, are

written from (25.39) as:

( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

c c 1 1 2 1

2 1 2

1 2

de

d

e ,

, 115(25. ).

i i i

i i

T T T T Tq q q

i iT T

q q

i


P S P S

q ψ ψ

∂ ∂− = + +

∂ ∂

+ +

=

1. Lagrange Equation relative to the parameter ψ1

We have:

( )( ) ( ) ( )2 2c 1 1 1 2 1 2 2 2 2 1

1

cosTE D C m a m d m daψ ψ ψ ψψ∂

= + + + −∂

, (25.116)

( )( ) ( ) ( )

( ) ( )

2 2c 1 1 1 2 1 2 2 2 2 1

1

2 2 2 2 1 2 1

dcos

dsin ,

TE D C m a m d m dat

m da

ψ ψ ψ ψψ

ψ ψ ψ ψ ψ

∂= + + + −

∂− − −

(25.117)

( )( ) ( )c 2 2 1 2 2 11

sin .TE D m da ψ ψ ψ ψψ∂

= −∂

(25.118)


( ) ( ) ( )

( )

2 2 21 1 1 2 1 2 2 2 2 1 2 2 1

1 1 2 1 1 21

cos sin

sin .

C m a m d m da

m a m d g N N


ψ

+ + + − − −

= − + + +

(25.119)

2. Lagrange Equation relative to the parameter ψ2

We have:

( )( ) ( ) ( )2c 2 2 2 2 2 2 1 2 1

2

cosTE D C m a m daψ ψ ψ ψψ∂

= + + −∂

, (25.120)

( )( ) ( ) ( )

( ) ( )

2c 2 2 2 2 2 2 1 2 1

2

2 2 1 2 1 2 1

dcos

dsin ,

TE D C m a m dat

m da

ψ ψ ψ ψψ

ψ ψ ψ ψ ψ

∂= + + −

∂− − −

(25.121)

A.25 Appendix 431

( )( ) ( )c 2 2 1 2 2 12

sin .TE D m da ψ ψ ψ ψψ∂

= − −∂

(25.122)


( ) ( ) ( )2 22 2 2 2 2 2 1 2 1 1 2 1

2 2 1 21

cos sin

sin .

C m a m da

m ga N


ψ

+ + − + −

= − −

(25.123)

The two equations (25.119) and (25.123) constitute the two equations of the

motions of the double pendulum.

A.25 APPENDIX

A.25.1 Properties of Derivatives of Rotation Vector

The instantaneous vector of rotation relative to the motion of the solid (S) with

respect to the reference (T) is written (9.12):

( ) 3

TS Sk i kω ψ θ ϕ= + +

. (A.25.1)

The expressions in the bases ( ), , i j k

and ( ), , S S Si j k

are given respectively by

Expressions (9.16) and (9.17).

The relations between the different bases, introduced by the rotations, are

expressed by Relations (9.13) to (9.15). Thus:

— Rotation of angle ψ about the direction k

:

3 3 3

3 3 3

cos sin , cos sin ,

sin cos , sin cos ,

. .

i i j i i j

j i j j i j

k k

ψ ψ ψ ψ

ψ ψ ψ ψ

= + = −

= − + = +

(A.25.2)

— Rotation of angle θ about the direction 3i

:

3 3

4 3 3 4

3 4

, ,

cos sin , cos sin ,

sin cos . sin cos .

S

S S

i i

j j k j j k

k j k k j k

θ θ θ θ

θ θ θ θ

= + = − = − + = +

(A.25.3)

— Rotation of angle ϕ about the direction Sk

:

3 4 3

3 4 4

cos sin , cos sin ,

sin cos , sin cos ,

. .

S S S

S S S

S S

i i j i i j

j i j j i j

k k

ϕ ϕ ϕ ϕ

ϕ ϕ ϕ ϕ

= + = −

= − + = +

(A.25.4)

These basic relations allow us to express the derivatives in the reference (T)

with respect to time of the vectors k

, 3i

and Sk

. The vector k

being independent


of time in the reference (T), we have:

( )d

0d

T

kt

=

. (A.25.5)

The derivative of the vector 3i

is obtained by writing:

( )

( )3 3 4d

cos sind

T

Si j j kt

ψ ψ θ θ= = −

. (A.25.6)

In the basis ( ), , S S Si j k

its expression is thus:

( )

3d

cos sin cos cos sind

T

S S Si i j kt

ψ θ ϕ ψ θ ϕ ψ θ= + −

. (A.25.7)

From (A.25.3), the derivative of Sk

is written:

( ) ( )

3 3d d

sin cos sind d

T T

Sk j j kt t

θ θ θ θ θ= − − −

, (A.25.8)

or

( )

3 3d

sin cos sind

T

Sk i j kt

ψ θ θ θ θ θ= − − −

. (A.25.9)

Transforming the vectors 3i

, 3j

and k

in the basis ( ), , S S Si j k

, we obtain finally:

( )

( ) ( )dsin cos sin sin cos cos

d

T

S S Sk i jt

ψ θ ϕ θ ϕ ψ θ ϕ θ ϕ= − + − −

. (A.25.10)

From Expression (9.17) of the rotation vector in the basis ( ), , S S Si j k

, we

deduce easily the following relations:

( )( )

0,S

TSω

ψ∂

=∂

(A.25.11)

( )( )

sin cos cos cos sin ,S

TS S S Si j kω ψ ϕ θ ψ ϕ θ ψ θ

θ∂

= + −∂

(A.25.12)

( )( ) ( ) ( )sin cos sin sin sin cos .

ST

S S Si jω θ ϕ ψ ϕ θ ψ ϕ θ θ ϕϕ

∂= − − + − −

∂

(A.25.13)

The comparison of Relations (A.25.5) and (A.25.11), (A25.7) and (A.25.12), then

(A.25.10) and (A.25.13) leads to the relations:

( ) ( )( )

( ) ( )( )

( ) ( )( )

3d d d

, , .d d d

T S T S T ST T T

S S S Sk i kt t t

ω ω ωψ θ ϕ

∂ ∂ ∂= = =

∂ ∂ ∂

(A.25.14)

From Relation (A.25.1), we deduce:

( )( )

ST

S kωψ

∂=

∂

. (A.25.15)

A.25 Appendix 433

Deriving this result with respect to time, we obtain then:

( ) ( )( )

( ) ( )( )d d

d d

T S T ST T

S Skt t

ω ωψ ψ

∂ ∂= =

∂ ∂

. (A.25.16)

While operating in the same way with the variables θ and ϕ , we obtain:

( ) ( )( )

( )( )d

d

T S ST T

S St

ω ωθθ

∂ ∂=

∂∂

, (A.25.17)

( ) ( )( )

( )( )d

d

T S ST T

S St

ω ωϕ ϕ

∂ ∂=

∂ ∂

. (A.25.18)

These three relations can be written in the form:

( ) ( )( )

( )( )d

, , , d

T S ST T

S S iii

Qt QQ

ω ω ψ θ ϕ∂ ∂= =

∂∂

. (A.25.19)

A.25.2 Properties of the Derivatives of the Velocity Vector of the Mass Centre

In the case where the position of the mass centre is defined by the parameters

of translation pi, the velocity vector is expressed as:

( )( ) ( )

1

d( , ) , 3.

d

kT TT

jj

j

G t OG p OG kt p

=

∂= = ≤

∂

(A.25.20)

From this expression, we deduce:

— first:

( )( )

( )2

1

( , ) ,

kT TT

ji i j

j

G t p OGp p p

=

∂ ∂=

∂ ∂ ∂

(A.25.21)

— then: ( )

( )( )

( , ) .T T

T

i i

G t OGp p

∂ ∂=

∂ ∂

(A.25.22)

Thus we obtain:

( ) ( )( )

( )2

1

d( , ) ,

d

kT T TT

ji i j

j

G t p OGt p p p

=

∂ ∂=

∂ ∂ ∂

(A.25.23)

The comparison of Expression (A.25.21) and (A.25.23) leads to the relation:

( ) ( )( )

( )( )

d

( , ) ( , )d

T T TT T

i i

G t G tt p p

∂ ∂=

∂ ∂

. (A.25.24)


EXERCISES

25.1 Using Lagrange equations, derive the equations of motion of the set of the

two solids considered in Exercise 24.1.

25.2 Similarly, derive the equations of motion of the radar antenna studied in

Exercise 24.2.

COMMENTS

The Lagrange equations constitute an efficient tool to obtain the equa-

tions of motion directly. Establishing these equations is rather complex and

the reader will endeavour first to understand the development which leads

to these equations.

The Lagrange equations which constitute the equations of motion are

Equations (25.30) in the case of one body and Equations (25.39) for a set

of bodies. The reader will have to thus know these equations, as well as the

way for applying the equations. The examples given at the end of the

chapter illustrate well the way for using the Lagrange equations. The reader

will apply them to the cases considered in the two exercises.

Part VI

Numerical Methods for Solving Differential Equations.

Application to Equations of Motion

The equations of motion of bodies are generally complex, and the

most of the equations cannot be solved analytically. This part is an

introduction to the numerical resolution of the equations of motion.

The tools which will be introduced can be easily implemented using a

software of general use, such as Matlab.

CHAPTER 26

Numerical Methods for Solving First Order Differential Equations


26.1.1 Problem with Given Initial Conditions

A differential equation of the first order can be put in the general form:

d

( , ) or ( , )d

yf t y y f t y

t= = . (26.1)

The problem consists in searching for the function y of the variable t, of which the

derivative with respect to this variable is known. The variable t can be arbitrary.

When we shall apply the results obtained to the resolution of the equations of

motion, the variable t will be the time variable.

In a general way, there exists an infinity of solutions to Equation (26.1), differ-

ring by an arbitrary constant. Setting a value 0y of the solution for an initial value

t0 of the variable t: 0 0( ),y y t= the solution is then unique. The problem is said to

be a problem with given initial conditions. We shall limit thereafter the analysis to

this type of problem, which makes it possible to solve the equations of motion.

The problem with given initial conditions is thus stated in the following way:

Find the function y(t), defined on the interval [ ]0 f, t t , satisfying:

0 0d

( , ) with ( ) .d

yf t y y t y

t= = (26.2)

It is shown that this problem admits a unique solution if the function f satisfies

the Lipschitz condition:

1 2 1 2 ( , ) ( , ) f t y f t y L y y− ≤ − , (26.3)

where L is a positive constant. It is also shown that if the derivative of the

function f with respect to y is continuous and bounded, then f satisfies the

Lipschitz condition .

438 Chapter 26 Numerical Methods for Solving First Order Differential Equations

TABLE 26.1 Numerical values and exact values.

variable t 0t 1 0t t h= + 2 0 2t t h= + .... 0it t ih= + .... 0nt t nh= +

computed values for y 0y 1y 2y .... iy .... ny

exact values for y 0( )y t 1( )y t 2( )y t .... ( )iy t .... ( )ny t

26.1.2 General Method of Resolution

The general method of numerical resolution consists in increasing the variable

t, from the initial value t0, by steps equal to h ( )h t= ∆ and calculating succes-

sively the values of the function y, for 1 0 ,t t h= + 2 0 2 ,t t h= + ..., 0 ,it t ih= + ...,

fnt t= . The computed values will be denoted by yi and the exact values by y(ti)

(Table 26.1). The numerical procedure used must lead to 0( ).iy y t ih≈ + It will be

possible to use either a constant step independent of the variable t (h = cst), or a

variable step function of the variable t (h = h(t)). The use of a variable step will

make it possible to control the value of the step according to the value of the

variable t.

The numerical procedures can be divided into single-step procedures and

multiple-step procedures. In the single-step procedures, each computed value 1iy +

is only deduced as a function of the step and of the preceding value yi. At each

stage of the procedure, calculations are independent of the preceding stages, hence

the name of single-step procedures. These methods are easily implemented and

their interest lies in the possibility of easily using variable step in the course of the

computation. In contrast, the multiple-step procedures re-use former calculations:

the value of 1iy + is evaluated from the preceding values yi , 1iy − , ..., y1, y0. Such

methods make it possible to increase the speed of calculations clearly. However,

the use of a variable step in the course of computation raises difficulties.

A large variety of numerical methods was developed to solve the initial value

problems (26.2). Most of the procedures were generalized so as to be able to apply

to higher order problems. This chapter aims to present some elements on these

methods of numerical resolution, with for objective to apply the results to the

numerical resolution of the equations of motion.

26.1.3 Euler Method

The Euler method is practically not used owing to the fact that it is too slow to

reach accurate values. However, its simplicity constitutes a good introduction to

the other more powerful methods. The Euler method consists in deriving the value

of 1iy + starting from the value yi and considering only the first order term of the

Taylor series:

( ) ( ) ( )i i iy t h y t hy t+ = + . (26.4)

26.1 General Elements 439

t1 t0 t2 t3 t5 t4 t7 t6 t8 t

y

h

numerical values

exact values

FIGURE 26.1. Approximation using theEuler method.

The set of the values of the function is thus derived from the relation:

1 ( , ), 0, 1, . . . , .i i i iy y h f t y i n+ = + = (26.5)

A difference equation is then substituted for the differential equation (26.2). Thus,

the curve y(t) is replaced by its tangent at each step of the procedure (Figure 26.1).

As an example, we consider the differential equation:

1/3y t y= , (26.6)

with for initial condition (1) 1y = . This differential equation has an exact solution:

3/22 2

3

ty

+=

. (26.7)

Table 26.2 compares the results obtained by the Euler method for different values

of the step with the exact values (26.7) of the function. The values reported in

TABLE 26.2 Results derived from the Euler method.

variable t 1 2 3 4 5

exacty 1.0000 2.8284 7.0211 14.6969 27.0000

0.10h = iy 1.0000 2.7239 6.7148 14.0799 25.9595

error 0.0000 –0.1045 –0.3064 –0.6171 –1.0405

0.05h = iy 1.0000 2.7754 6.8662 14.3856 26.4759

error 0.0000 –0.0530 –0.1549 –0.3113 –0.5241

0.01h = iy 1.0000 2.8177 6.9899 14.6342 26.8946

error 0.0000 –0.0107 –0.0313 –0.0627 –0.1054


the table show an appreciable improvement of the numerical values when the step

decreases, at the price of an important increase of the computing delay. For

example the 25.9595 value for t = 5 is obtained using 50 steps for computation,

whereas the 26.8946 value needed 500 steps.

26.2 SINGLE-STEP METHODS

26.2.1 General Elements

26.2.1.1 Formulation

The single-step methods replace the function ( , )i if t y of the Euler method by a

more general function ( , , )i it y hϕ . The evaluation of 1iy + at a given stage is

derived from the relation:

1 ( , , )i i i iy y h t y hϕ+ = + . (26.8)

The differential equation (26.2) is replaced by a non-linear difference equation.

The choice of the function depends on the method used. Thus, the Euler method is

a single-step method, for which the function ϕ is reduced to the function f.

26.2.1.2 Convergence

The methods used must lead to a decrease in the error when the step of compu-

tation decreases, the error having to vanish when the step becomes infinitely

small. It is said that the method must be convergent.

The numerical procedure is convergent on the interval [ ]0 f, t t if, for all the

values yi calculated, the maximum difference with the exact solution decreases

when the step h of computation decreases.

Thus the method is convergent if:

0max ( ) 0, when 0i i

i ny y t h

≤ ≤− → → . (26.9)

26.2.1.3 Stability

Furthermore, the errors induced by the method should not be amplified. It is

said that the method must be stable.

A method is said to be stable if a perturbation on the computed value of yi

involves only a small perturbation on the value of 1iy + , and that whatever the step

value h.

An unstable procedure leads to an amplification of the errors at each step of

26.2 Single-Step Methods 441

computation. It is shown that if the function ϕ satisfies the Lipschitz condition:

( , , ) ( , , ) , 0i i i i i it y y h t y h L y Lϕ ϕ+ ∆ − ≤ ∆ < < ∞ , (26.10)

then the method is stable.

26.2.1.4 Consistence

The method of approximation (26.8) is consistent with the differential equation

if:

( )

10 0

1lim max ( , , ) 0i i i ih i n

y y t y hh

ϕ+→ ≤ ≤

− − →

. (26.11)

Consistency means that the approximation (26.8) must be a probable and well

constructed approximation. It is shown that the method is consistent if:

( , , 0) ( , )t y f t yϕ = . (26.12)

If the method is stable and consistent, then it is convergent for any value of the

initial value.

26.2.1.5 Order

A numerical method is said to be convergent of order p, if:

0max ( ) , 0p

i ii n

y y t Kh K≤ ≤

− ≤ < < ∞ . (26.13)

We observe that if the step is divided by the value λ, the error on the approxi-mation is divided by pλ . Hence the interest to have a high order method.

26.2.1.6 Example of the Euler Method

We considered that the Euler method is a single-step method, for which fϕ = .

1. The method is consistent if ( , , 0) ( , )i i i it y f t yϕ = . This result is immediate, since fϕ = does not depend on the step h. The Euler method is thus consistent.

2. The method is stable if the Lipschitz condition (26.10) on the function ϕ is satisfied. The Euler method is stable since fϕ = , and that the function f by as-sumption satisfies the Lipschitz condition (26.3). The Euler method is thus con-vergent.

3. It is possible to show (the demonstration is not immediate) that:

( ) , 1, 2, . . . , .i iy y t Kh i n− = = (26.14)

The Euler method is thus locally (for each step) of order 1. The local error (26.14)

is induced to each stage of the procedure between the step i and the step 1i + .

These errors are propagated through the stages of the procedure and it is generally

not possible to have a precise idea, a priori, of the propagation of the errors. The

results of Table 26.1 show that the order 1 of the Euler method is maintained, in

the example considered, throughout the stages going from 0 1t = to f 5t = .


26.2.2 Methods of Runge-Kutta Type

26.2.2.1 General Formulation

In the methods of Runge-Kutta type, the function ( , )t yϕ of the approximation

relation (26.8) is formed as a linear combination of the values of the function

( , )f t y calculated at suitable points ( , )j jt y , in such a way to obtain high order

procedures. The Runge-Kutta methods lead to errors clearly lower than that

obtained by the Euler method, when the implementation remains simple. In these

methods, the approximation 1iy + is expressed as a function of the approximation

iy in the form (26.8), with:

1 1 2 2( , , ) . . . i i n nt y h k k kϕ ϕ ϕ ϕ= + + + , (26.15)

where 1 2, , . . . , nk k k are coefficients and 1 2, , . . . , nϕ ϕ ϕ are the values of the

function ( , )f t y expressed as:

1

2 1 1 1

3 2 2 2

1 1 1

( , ) ,

( , ) ,

( , ) ,

.

.

.

( , ) .

i i

i i

i i

n i n i n n

f t y

f t a h y

f t a h y

f t a h y

ϕ

ϕ α ϕ

ϕ α ϕ

ϕ α ϕ− − −

=

= + +

= + +

= + +

(26.16)

26.2.2.2 Runge-Kutta Methods of Order 2

In the case of the Runge-Kutta methods of order 2, the function (26.15) is

expressed in the form:

1 1 2 2( , , )i it y h k kϕ ϕ ϕ= + , (26.17)

with

1

2 1

( , ) ,

( , ) .

i i

i i

f t y

f t ah y

ϕ

ϕ αϕ

=

= + + (26.18)

The coefficients k1, k2, a and α are derived from the Taylor expansion to the

second order of the function y at the vicinity of ti. Thus:

2

( ) ( ) ( ) ( )2

i i i ih

y t h y t hy t y t+ = + + , (26.19)

with

( ) ( , )i i iy t f t y= , (26.20)

and

d d

d d

f f f y fy f y

t t y t y

∂ ∂ ∂= = + = +

∂ ∂ ∂ . (26.21)


Thus:

( ) ( , ) ( , ) ( , )i i i i i i if

y t f t y t y f t yy

∂= +

∂ . (26.22)

The expansion (26.19) is thus written:

2

( ) ( ) ( , ) ( , ) ( , ) ( , )2

i i i i i i i i i ifh

y t h y t h f t y f t y t y f t yy

∂ + = + + + ∂ . (26.23)

Moreover, the function 2ϕ (26.18) expanded to the first order at the vicinity of

( , )i it y is written as:

1 1( , ) ( , ) ( , ) ( , )i i i i i i i if

f t ah y f t y ah f t y t yy

αϕ αϕ∂

+ + = + +∂

, (26.24)

or

1( , ) ( , ) ( , ) ( , ) ( , )i i i i i i i i i if

f t ah y f t y ah f t y h f t y t yy

αϕ α∂

+ + = + +∂

. (26.25)

The evaluation of 1iy + at a given stage is thus written finally, from (26.8), (26.17),

(26.18) and (26.25), in the form:

( ) 2

1 1 2 2( , ) ( , ) ( , ) ( , )i i i i i i i i i if

y y k k h f t y h k f t y t y a f t yy

α+∂ = + + + + ∂

. (26.26)

Comparison between Expressions (26.23) and (26.26) leads to the relations:

1 2

12 2

12 2

1,

,

.

k k

k

ak

α

+ =

=

=

(26.27)

Hence the expressions of the parameters:

1

2

,

2 1,

2

1.

2

a

ak

a

ka

α =

−=

=

(26.28)

The choice of the value of the parameter a is arbitrary, and this choice determines

the method used.

1. In the case where we choose 1/2a α= = , the values of the parameters k1 and

k2 are: 1 0k = and 2 1k = . These values lead to the improved Euler method, where

the evaluation of 1iy + is computed in the form:

1 1( , )2 2

i i i ih h

y y h f t y ϕ+ = + + + , (26.29)

with:

1 ( , )i if t yϕ = . (26.30)


2. In the case where we choose 1a α= = , the values of the parameters k1 and

k2 are 1 2 1/2k k= = . These values lead to the method known as Euler-Cauchy,

where the evaluation of 1iy + is computed in the form:

( )1 1 22

i ih

y y ϕ ϕ+ = + + , (26.31)

with:

1

2 1

( , ) ,

( , ) .

i i

i i

f t y

f t h y h

ϕ

ϕ ϕ

=

= + + (26.32)

The Euler-Cauchy method can be considered as the cumulative result of the

Euler method applied to the two half-intervals [ ], /2i it t h+ and [ ]/2, i it h t h+ + .

Indeed, the approximation (26.5) is written for these two intervals:

12

( , )2

i i ii

hy y f t y

+= + , (26.33)

1 1 12 2 2

1 ( , )2

i i i i

hy y f t y+ + + +

= + . (26.34)

Combination of these two results leads well to Relation (26.31).

The improved Euler and Euler-Cauchy methods are methods of order 2. Table

26.3 reports the results derived from the Euler-Cauchy method to solve Equation

(26.6). The results show that the method is effectively of order 2. In the case

where the step is equal to 0.01, the maximum error observed is equal to 48 10−− × .

26.2.2.3 Runge-Kutta Methods of Order 4

In the case of the Runge-Kutta methods of order 4, the function (26.15) is

expressed in the form:

1 1 2 2 3 3 4 4( , , )i it y h k k k kϕ ϕ ϕ ϕ ϕ= + + + , (26.35)

TABLE 26.3 Results derived from the Euler-Cauchy method.


exacty 1.000000 2.828427 7.021132 14.696938 27.000000

0.10h = iy 1.000000 2.827609 7.018633 14.692080 26.992260

error 0.000000 –0.000818 –0.002499 –0.004859 –0.007740

0.05h = iy 1.000000 2.828219 7.020495 14.695701 26.998031

error 0.000000 –0.000208 –0.000637 –0.001237 –0.001969

0.01h = iy 1.000000 2.828419 7.021106 14.696889 26.999920

error 0.000000 –0.000008 –0.000026 –0.000050 –0.000080


with

1

2 1 1 1

3 2 2 2

4 3 3 3

( , ) ,

( , ) ,

( , ) ,

( , ) .

i i

i i

i i

i i

f t y

f t a h y

f t a h y

f t a h y

ϕ

ϕ α ϕ

ϕ α ϕ

ϕ α ϕ

=

= + +

= + +

= + +

(26.36)

The various coefficients intervening in the functions (26.35) and (26.36) are

derived by a method comparable to that used in the preceding subsection, by

expanding to the forth order in Taylor series the function y at the vicinity of ti(Exercise 26.1). The identification of the expansions leads at first to the equality

of the coefficients αi and ai:

, 1, 2, 3.i ia iα = = (26.37)

Then to the equations:

11 2 3 4 3 1 2 4 2 3 6

2 21 12 1 3 2 4 3 3 1 2 4 2 32 8

2 2 2 2 21 12 1 3 2 4 3 3 1 2 4 2 33 12

3 3 3 1 12 1 3 2 4 3 4 1 2 34 24

1, ,

, ,

, ,

, .

k k k k k a a k a a

k a k a k a k a a k a a

k a k a k a k a a k a a

k a k a k a k a a a

+ + + = + =

+ + = + =

+ + = + =

+ + = =

(26.38)

The system of these equations is overdetermined: eight equations for seven un-

knowns. The usual solution which is considered is:

1 1 11 2 3 1 4 2 32 6 3

, 1, , .a a a k k k k= = = = = = = (26.39)

The evaluation of 1iy + at the step i is thus computed in the form:

( )1 1 2 3 42 26

i ih

y y ϕ ϕ ϕ ϕ+ = + + + + , (26.40)

with

1

12

23

4 3

( , ) ,

( , ) ,2 2

( , ) ,2 2

( , ) .

i i

i i

i i

i i

f t y

hf t y

hf t y

f t h y

ϕ

ϕϕ

ϕϕ

ϕ ϕ

=

= + +

= + +

= + +

(26.41)

Table 26.3 reports the results obtained by this Runge-Kutta method to solve the

differential equation (26.6). These results show that this method is effectively of

order 4, making it possible to decrease clearly the number of values to compute.

The interest of this method lies thus in a high precision associated with a rather

simple implementation.


TABLE 26.4 Results obtained by the Runge-Kutta method of order 4.


exacty 1 2.82842712 7.02113212 14.69693846 27.00000000

0.10h = iy 1 2.82842678 7.02113135 14.69693723 26.99999834

error 0 –0.00000034 –0.00000078 –0.00000122 –0.00000166

0.05h = iy 1 2.82842710 7.02113207 14.69693838 26.99999989

error 0 –0.00000002 –0.00000005 –0.00000008 –0.0000001

0.01h = iy 1 2.82842712 7.02113212 14.69693846 26.99999999

error 0 100.35 10−− × 100.80 10−− × 101.26 10−− × 101.71 10−− ×

26.2.3 Romberg Method

In the preceding procedures, the error observed on the computed values is not

controlled. The single-step methods can then be improved by using an iterative

process which makes it possible to control the error obtained on the computed

value. We consider in this subsection the Romberg method.

1. In the Romberg method, the interval [ ]0 f, t t of integration is first divided into

1n − intervals characterized by the step h: 1 0 2 f, , ..., , ...,p nt t t t t t= = . Then,

each interval , p pt t h+ is divided into 21, 2, 2 , ..., 2 , ...j sub-intervals. The

successive numerical values of y are computed on these intervals for:

, 2 , ..., ,2 2

p p pj j

h ht t t h+ + + (26.42)

using one of the methods considered previously. For the simplicity of the deve-

lopment, we consider the Euler method. The step of the sub-intervals will be

denoted by:

02 j

hh = . (26.43)

In the case of one sub-interval 0, p pt t h+ 0( 0, )j h h= = , the value compu-

ted while using the Euler method is:

( )11 0 ( , )p p py y h f t y= + . (26.44)

In the case of two sub-intervals 0 0, , 2p p pt t h t h+ + 0( 1, /2)j h h= = , the

values computed by the Euler method are successively:


( )

( ) ( ) ( )

21 0

2 2 202 1 0 1

( , ),

( , ).2

p p p

p

y y h f t y

hy y h f t y

= +

= + + (26.45)

In the case of 2 j sub-intervals, the successive values computed by the Euler

method are:

( )

( ) ( ) ( )

( ) ( ) ( ) ( )

01

01 02 1

00

2 2 1 2 1

( , ),

( , ),2

.

.

.

( 2 1 , ).2

j j j

jp p p

j jjp j

j j jjp j

y y h f t y

hy y h f t y

hy y h f t y

− −

= +

= + +

= + + −

(26.46)

Finally, the procedure leads to the numerical values:

( ) ( ) ( ) ( )

1 2 31 2 4

2, , , . . . , , . . . ,j

jy y y y (26.47)

which are at each time better approximations of ( )py t h+ . These successive

approximations can be denoted by:

( ) ( ) ( ) ( )

1 2 31 1 2 2 3 4

2, , , . . . , , . . . .j

jkY y Y y Y y Y y= = = = (26.48)

2. It is possible to be contented with the approximation Yk. However, it is shown

that the convergence is accelerated by constructing the following table using a

linear extrapolation:

11

21 22

31 32 33

1,1 1,2 1, 1, 1

1 2 , 1 ,

. ..

............ ............

............ ............

. ..

k k k l k k

k k kl k k k k

y

y y

y y y

y y y y

y y y y y− − − − −

−

In this table, the first column coincides with the values calculated previously: ( ) ( ) ( )

1 1 1

11 1 21 2 12

, , . . . , , . . . ,jky y y y y y= = = (26.49)

and the following terms are deduced from the linear extrapolation:

1,

, 1

2, 2,3, . . . and 1.

2 1

lkl k l

k l l

y yy k l k

−+

−= = ≤ −

− (26.50)

In a general way, , 1k ly + is a better approximation than kly .


11y

21y 22y 22y

32y 33y

11y

21y

31y

21y

31y 32y

11y

22y

31y

41y 42y

11y

21y

33y

22y

32y

The table is constructed gradually starting from the values computed by the Euler

method: y11, y21, ..., yk1, ..., in the following way:

In the process of forming the table, we observe that we need to keep only one row

to derive the following row. This remark implies that it is not necessary to store

the 2-dimension table ykl, but simply a 1-dimension table Tk, by using the follo-

wing process for filling the table:

In forming the table, we have thus:

1k l klT y− + = , (26.51)

and Relation (26.50) is written:

12

, 2, 3, . . . , 1,2 1

lk l k l

k l l

T TT k l k− + −

−−

= = = −−

(26.52)

with for 1 111, k T y= = . (26.53)

The procedure of computation is stopped when the difference between the last

computed value k lT − and the preceding one 1k lT − + is lower in absolute value than

the desired precision. The value computed at the step p + 1 is then k lT − . Thus:

1p k ly T+ −= (26.54)

Table 26.5 reports the results obtained by the Euler-Romberg method for

solving the differential equation (26.6). The main step chosen is equal to 1: the

interval [1, 5] of integration is thus divided into 4 intervals only. The results

obtained are reported with two imposed values of precision: the one higher than

1T

2T 1T 1T

2T 1T

1T

2T

3T

2T

3T 2T

1T

1T

3T

4T 3T

1T2T

1T

2T 1T

26.3 Multiple-Step Methods 449

TABLE 26.5 Results derived from the Euler-Romberg method.


yexact 1 2.82842712 7.02113212 14.69693846 27.00000000

desired precision higher than 0.01

yp 1 2.8277 7.0194 14.6946 26.9971

error 048 10−− × 31.7 10−− × 32.3 10−− × 32.9 10−− ×

number of

iterations0 3 3 4 4

desired precision higher than 10–5

yp 1 2.82842723 7.02113238 14.69693900 27.00000081

error 071.08 10−× 72.6 10−× 75.4 10−× 78.1 10−×

number of

iterations0 5 5 5 5

0.01 and the other higher than 10–5. The number of iterations necessary at each

step to reach the desired precision is also reported in the table.

26.3 MULTIPLE-STEP METHODS

26.3.1 Introduction to the Multiple-Step Methods

In the single-step methods, the numerical computations are independent of the

computations for the preceding steps. The multiple-step methods consist in re-

using the numerical values computed in the preceding steps, thus allowing to

improve the speed of the computations.

The solution y(t) of the differential equation satisfies the relation:

[ ]0 f( ) ( ) ( , ( )) d , , ,t t

t

y t t y t f u y u u t t t t t+∆

+ ∆ = + ∀ + ∆ ∈ . (26.55)

If yp are the computed values of the solution at the point tp, we can substitute for

the function ( )f t an interpolation polynomial P(t), taking at the point tp the values

( , )p p pf f t y= . On replacing the integral of f by the integral of the polynomial P,

we shall obtain then an approximation of ( ) ( )y t t y t+ ∆ − .

If we are at the step 1 ( 0, 1, . . .)i i+ = corresponding to the value ti of the

variable t, we know the approximations 1 2, , , . . . ,i i iy y y− − evaluated at the prece-


ding steps. It is then possible to determine the values 1 2, , , . . . ,i i if f f− − which make

it possible to approximate the function ( , )y f t y′ = at the points 1 2, , , . . . ,i i it t t− −

by the interpolation polynomial P(t). If we choose 1it t t ++ ∆ = and i kt t −=

( )k i≤ , Relation (26.55) is written:

1

1( ) ( ) ( ) di

i k

t

i i kt

y t y t P t t+

−

+ −− = . (26.56)

The value 1iy + , computed at the step 1i + is written as a function of the value

computed at the step i k− by the relation:

1

1 ( ) di

i k

t

i i kt

y y P t t+

−

+ −= + . (26.57)

According to the possible choices of the value of k and of the points ti , we are

leaded to methods of two types: 1) if the point 1it + and the value 1iy + are not

included in the computation of the interpolation polynomial, therefore of its

integration, the method is said to be explicit; 2) if the point 1it + and the value 1iy +

are included, the method is said to be implicit.

Furthermore, the multiple-step methods do not have the property to be able to

start by themselves, since the first numerical values y0, y1, ..., do not exist. It is

then necessary to start computations, by using a sufficiently accurate single-step

method to compute the first values.

26.3.2 Methods Based on the Newton Interpolation

26.3.2.1 Interpolation Polynomial of Newton

In Relation (26.57), the indices being decreasing, it is interesting to appro-

ximate the function f using the regressive interpolation polynomial of Newton.

This polynomial is associated to the backward differences defined as:

( )

( )

0

11

21 2

11 2

,

,

2 ,...

. . . ( 1) ,1 2

i i

i i i i

i i i i i

n n ni i i i i i n

f f

f f f f

f f f f f

n nf f f f f f

−

− −

−− − −

∇ =

∇ = ∇ = −

∇ = ∇ ∇ = − −

∇ = ∇ ∇ = − + + + −

(26.58)

where the coefficients n

m

are expressed by the relation:


!

!( )!

n n

m m n m

= − . (26.59)

The Newton interpolation polynomial is written in the form:

1 212

1 1

1 1( ) ( ) ( )( ) . . .

2

1( )( ) . . . ( ) .

!

n i i i i i i

ni i i n in

P t f t t f t t t t fh h

t t t t t t fn h

−

− − +

= + − ∇ + − − ∇ +

+ − − − ∇

(26.60)

Note that the interpolation error of the function f by the Newton polynomial is

of order 1n + .

26.3.2.2 Explicit Methods

On introducing the Newton interpolation polynomial (26.60) into Relation

(26.57), we obtain:

11 2

1 12

1 1

1 1( ) ( )( ) . . .

2

1( )( ) . . . ( ) d .

!

i

i k

t

i i k i i i i i it

ni i i n in

y y f t t f t t t t fh h

t t t t t t f tn h

+

−

+ − −

− − +

= + + − ∇ + − − ∇ +

+ − − − ∇

(26.61)

The integration is operated by introducing the reduced variable:

it tu

h

−= , (26.62)

and setting:

i j i i i j it t t t t t t t jh− −− = − + − = − + , (26.63)

thus:

( )i jt t h u j−− = + . (26.64)

Relation (26.61) is then written:

11 2

11

( 1) . . .2

1( 1)( 2) . . . ( 1) d .

!

i i k i i ik

ni

y y h f u f u u f

u u u u n f un

+ −−

= + + ∇ + + ∇ +

+ + + + − ∇

(26.65)

The integration leads to:

( ) 1 2

1 0 1 2 . . . ni i k i i i n iy y h p f p f p f p f+ −= + + ∇ + ∇ + + ∇ , (26.66)

with

11

( 1)( 2) . . . ( 1) d!

jk

p u u u u j uj −

= + + + − ,


1 2 . . . ( 1)1 2

j ji i i i i j

j jf f f f f− − −

∇ = − + + + −

, (26.67)

!

!( )!

jj

m m j m

= − .

Relation (26.65) is finally of the form:

( ) 1 0 1 1 2 2 . . .i i k i i i n i ny y h p f p f p f p f+ − − − −= + + + + + . (26.68)

This relation thus makes it possible to compute the value of y at 1it + as a function

of the value computed at i kt − and as a function of the values of the function f

taken for the values 1, , . . . ,i i i nt t t− − of the variable t.

The error induced by the multiple-step methods depends: 1) on the error intro-

duced by evaluating (using a single-step method) the initial values necessary to

the starting of the multiple-step methods, 2) on the error introduced by approxi-

mating the integral of f in (26.57) by that of an interpolation polynomial. We

noted that the error introduced by the Newton interpolation polynomial (26.60) is

of order n + 1. If the starting procedure introduces a negligible error, the multiple-

step method (26.66) can be considered as a method of order n.

26.3.2.3 Implicit Methods

The implicit methods can be formulated by considering the Newton regressive

polynomial interpolated starting from the value 1if + . Relation (26.60) is thus writ-

ten as:

1 21 1 1 12

1 2 1

1 1( ) ( ) ( )( ) . . .

2

1( )( ) . . . ( ) .

!

n i i i i i i

ni i i n in

P t f t t f t t t t fh h

t t t t t t fn h

∗+ + + +

+ − + +

= + − ∇ + − − ∇ +

+ − − − ∇

(26.69)

On introducing this polynomial into Relation (26.57) and proceeding as pre-

viously, Expression (26.65) is modified according to:

( )

11 2

1 1 1 11

1

1( 1) . . .

2

1( 1)( 2) . . . ( 1) d .

!

i i k i i ik

ni

y y h f u f u u f

u u u u n f un

+ − + + +− +

+

= + + ∇ + + ∇ +

+ + + + − ∇

(26.70)

The procedure of computation is well of implicit type.

26.3.3 Generalization of the Multiple-Step Methods

The general form of the multiple-step methods is obtained by generalizing the

formulations (26.68) and (26.70). The value of 1iy + is formulated in the general


form:

( )1 0 1 1 1 1 0. . . . . .i i i k i k i i k i ky y y y h f f fα α α β β β+ − − − + −= + + + + + + + . (26.71)

If 1 0β− = , the value of 1iy + is deduced from the computed values 1, , . . . ,i if f −

i kf − . The method is explicit. If 1 0β− ≠ , the method is implicit.

After having use a single-step method for starting, an explicit method makes it

possible to compute the value 1iy + as a function of the values evaluated at the pre-

ceding step, using Relation (26.71) with 1 0β− = . Relation (26.71) is thus called

predictor formula. The precision of 1iy + can then be improved by using Relation

(26.71) with 1 0β− ≠ . This process is then implemented using an iterative pro-

cedure. Relation (26.71) is then called corrector formula. Coupling the two for-

mulations leads to a so-called predictor-corrector procedure.

26.3.4 Examples of Multiple-Step Methods

26.3.4.1 Milne Methods

Taking the values 3k = and 2n = , Relation (26.65) is written:

11 2

1 33

1( 1) d .

2i i i i iy y h f u f u u f u+ −

−

= + + ∇ + + ∇ (26.72)

Thus while integrating:

( )

1 21 3

84 4

3i i i i iy y h f f f+ −= + − ∇ + ∇ . (26.73)

Hence, taking account of (26.58):

( )1 3 1 24

2 23

i i i i iy y h f f f+ − − −= + − − (26.74)

This relation is known as the Milne predictor formula.

In the same way, by taking the values 1k = and 2n = , Relation (26.70) is

written:

11 2

1 1 1 1 12

1( 1) d

2i i i i iy y h f u f u u f u+ − + + +

−

= + + ∇ + + ∇ . (26.75)

Thus while integrating:

( )

1 21 1 1 1 1

3 33

2 4i i i i iy y h f f f+ − + + += + − ∇ + ∇ . (26.76)

The introduction of Relations (26.58) transposed to the value fi + 1 leads to:

( ) 1 1 1 14 .3

i i i i ih

y y f f f+ − + −= + + + (26.77)

This relation is known as the Milne corrector formula.


26.3.4.2 Adams-Moulton Procedures

The Adams-Moulton procedures are obtained by taking 0k = , in Relations

(26.65) and (26.70). Hence the predictor and corrector relations:

1p 1 2

10

1( 1) . . .

2

1( 1)( 2) . . . ( 1) d .

!

i i i ii

ni

y y h f u f u u f

u u u u n f un

+= + + ∇ + + ∇ +

+ + + + − ∇

(26.78)

and

0c 1 2

1 1 1 11

1

1( 1) . . .

2

1( 1)( 2) . . . ( 1) d .

!

i i i i i

ni

y y h f u f u u f

u u u u n f un

+ + + +−

+

= + + ∇ + + ∇ +

+ + + + − ∇

(26.79)

From these relations we deduce the following results:

1n =

( )

( )

p11

c1 1

3 ,2

.2

i i ii

i i i i

hy y f f

hy y f f

−+

+ +

= + −

= + +

(26.80)

2n =

( )

( )

p1 21

c1 1 1

23 16 5 ,12

5 8 .12

i i i ii

i i i i i

hy y f f f

hy y f f f

− −+

+ + −

= + − +

= + + −

(26.81)

3n =

( )

( )

p1 2 31

c1 1 1 2

55 59 37 9 ,24

9 19 5 .24

i i i i ii

i i i i i i

hy y f f f f

hy y f f f f

− − −+

+ + − −

= + − + −

= + + − +

(26.82)

4n =

( )

( )

p1 2 3 41

c1 1 1 2 3

1901 2984 2616 1274 251 ,720

251 646 264 106 19 . (26.83)720

i i i i i ii

i i i i i i i

hy y f f f f f

hy y f f f f f

− − − −+

+ + − − −

= + − + − +

= + + − + −

26.3.5 Results

Table 26.6 gives the results obtained by using the Adams-Moulton predictor

formula corresponding to 2n = :

( )p1 21 23 16 5

12i i i ii

hy y f f f− −+ = + − + , (26.84)


TABLE 26.6 Results derived from an Adams-Moulton predictor procedure, for three

values of the step of computation.


yexact 1 2.82842712 7.02113212 14.69693846 27.00000000

h = 0.10 yi 1 2.82835406 7.02101203 14.69677999 26.99980445

error 0 – 0.00007306 – 0.00012009 – 0.00015847 – 0.00019555

h = 0.05 yi 1 2.82841644 7.02111523 14.69691628 26.99997266

error 0 – 0.00001069 – 0.00001690 – 0.00002218 – 0.00002734

h = 0.01 yi 1 2.82842703 7.02113197 14.69693826 26.99999976

error 0 – 0.00000009 – 0.00000015 – 0.00000019 – 0.00000024

for solving the differential equation (26.6). The starting procedure, used to

compute the first three values, is the Runge-Kutta method of order 4. The results

obtained for the different values of the step ( 0.10,h = 0.05h = and 0.01h = )

show that the procedure is appreciably of the third order.

The precision can be improved by associating to the predictor formula a

corrector formula of the same order. The precision of the result could then be

possibly controlled by using an iterative corrector process until the difference

obtained between two successive corrections is acceptable.

Table 26.7 shows the results obtained by associating to the predictor formula

(26.84) the corrector formula:

( )c1 1 1 29 19 5

24i i i i i i

hy y f f f f+ + − −= + + − + . (26.85)

TABLE 26.7 Results obtained by an Adams-Moulton predictor-corrector procedure, for a

step of h = 0.2 and in the case of several successive corrections k.


yexact 1 2.82842712 7.02113212 14.69693846 27.00000000

k = 0 yi 1 2.82801553 7.02037351 14.69592343 26.99874364

error 0 – 0.00041160 – 0.000758610 – 0.001015029 – 0.001256356

k = 1 yi 1 2.82840571 7.02109451 14.69688877 26.99993877

error 0 – 0.00002141 – 0.00003762 – 0.00004969 – 0.00006123

k = 2 yi 1 2.82841553 7.02111213 14.69691211 26.99996754

error 0 – 0.00001160 – 0.00001999 – 0.00002635 – 0.00003246

k = 5 yi 1 2.82841579 7.02111259 14.69691271 26.99996827

erreur 0 - 0.00001134 - 0.00001953 - 0.00002575 - 0.00003172


Table 26.7 compares, for a step of computation 0.20,h = the results obtained

without correction ( 0k = ) to the results obtained k with successive corrections

( 1k = , 2k = and 5k = ) at each step of computation. The results show that the

precision is clearly improved when one correction is used (the error obtained for

5t = is appreciably divided by 20). However, the improvement is then limited,

when the number of successive corrections is increased.

EXERCISES

26.1 Establish Relations (26.37) and (26.38) which are obtained in the case of the

Runge-Kutta methods of order 4.

26.2 Implement the numerical procedures, using the different methods consi-

dered in this chapter, to solve the differential equation:

2d2

d

yty

t= −

in the interval [ ]0, 2t ∈ , with the initial value (0) 1y = .

COMMENTS

The numerical methods constitute an indispensable tool for solving the

differential equations. In a first approach, the reader will be interested in

the Euler-Cauchy method, then in the Runge-Kutta methods of order 4.

Then, the reader will study thoroughly the methods which make it

possible to control the precision of the computation.

The implementation of the different methods will be carried out by the

reader by deriving the results reported in the different tables of this

chapter.

CHAPTER 27

Numerical Procedures for Solving the Equations of Motions

27.1 EQUATION OF MOTION

WITH ONE DEGREE OF FREEDOM

27.1.1 Form of the Equation of Motion with One Degree of Freedom

We studied (Part IV) the case of body motions with one degree of freedom. For

a solid in translation, the equation of motion was expressed in (21.23) in the case

where there is no friction and in (21.56) in the case of viscous friction. The study

of a solid in rotation about an axis was considered in Chapter 22. The equation of

motion is given by (22.37) when there is no friction and by (22.39) in the case of

viscous friction.

In a general way, the differential equation of motion of a system with one

degree of freedom y is a differential equation of order 2 of the form:

2

2

d( ) ( , , )

dy t f t y y

t= . (27.1)

The problem of the motion is an initial value problem: for the value t0 are imposed

the values of the parameter of situation 0 0( )y y t= and of the velocity 0 0( )y y t= .

27.1.2 Principle of the Numerical Resolution

To numerically solve the differential equation of motion (27.1), we introduce

458 Chapter 27 Numerical Procedures for Solving the Equations of Motions

the velocity variable:

y y= . (27.2)

The resolution of Equation (27.1) is then led to the simultaneous resolution of two

differential equations of first order:

( , , )y yf t y= , (27.3)

yy = , (27.4)

with for initial conditions at t0:

0 0 0 0( ) , ( )y t y y y t= = .

The procedure of numerical resolution thus consists in computing at each step

the value of y and the value of y, using one of the methods considered in Chap-

ter 26.

27.1.3 Application to the case of the Motion of a Simple Pendulum

27.1.3.1 Analysis in the Absence of Friction

The motion of a simple pendulum was studied in Section 22.2.1 of Chapter 22.

In the absence of friction, the equation of motion (22.41) is written in the form:

20 sin 0ψ ω ψ+ = , (27.5)

where 0ω is the natural angular frequency expressed by (22.45).

In the case of low values of the angle of rotation, its sine can be replaced by the

angle and the equation of motion (27.5) is reduced to:

20 0ψ ω ψ+ = . (27.6)

The motion is sinusoidal, of period:

00

2T

πω

= , (27.7)

and the angle of rotation is given by:

00 0 0

0

( ) cos sint t tψ

ψ ψ ω ωω

= +

, (27.8)

where 0ψ and 0ψ are the initial conditions at 0t = :

0 0(0), (0).ψ ψ ψ ψ= = (27.9)

The Euler-Romberg method (Section 26.2.3) was applied to the resolution of

the equation of motion (27.5), in the case of a period 0 2 sT = and an zero initial

angular velocity: 0 0ψ = . A precision of 10–6 on the value of the angle was

27.1 Equation of Motion with One Degree of Freedom 459

TABLE 27.1 Resolution of the equation of motion of a simple pendulum.

variable t(s) 0 0.1 0.2 0.3 0.4 0.5

0 5ψ = °

ψ (num.) (°/s) 0.000000 –4.848166 –9.223304 –12.69774 –14.9308 –15.70297

ψ (anal.) (°/s) 0.000000 –4.854028 –9.232909 –12.70800 –14.93916 –15.70796

ψ (num.) (°) 5.000000 4.755586 4.046185 2.941046 1.548151 0.003739

ψ (anal.) (°) 5.000000 4.755283 4.045085 2.938926 1.545085 0.000000

0 10ψ = °

ψ (num.) (°/s) 0.000000 –9.661204 –18.38900 –25.33385 –29.81162 –31.37593

ψ (anal.) (°/s) 0.000000 –9.708055 –18.46582 –25.41602 –29.87833 –31.41593

ψ (num.) (°) 10.000000 9.512986 8.098966 5.894803 3.114699 0.029920

ψ (anal.) (°) 10.000000 9.510565 8.090170 5.877853 3.090170 0.000000

0 20ψ = °

ψ (num.) (°/s) 0.000000 –19.04267 –36.31789 –50.17341 –61.62167 –62.50881

ψ (anal.) (°/s) 0.000000 –19.41611 –36.93164 –50.83204 –62.05829 –62.83185

ψ (num.) (°) 20.000000 19.040421 16.250503 11.891120 3.349357 0.239702

ψ (anal.) (°) 20.000000 19.021130 16.180340 11.755705 3.128689 0.000000

0 40ψ = °

ψ (num.) (°/s) 0.000000 –35.88890 –68.98037 –96.35467 –115.1266 –122.9781

ψ (anal.) (°/s) 0.000000 –38.83222 –73.86327 –101.6641 –119.5133 –125.6637

ψ (num.) (°) 40.000000 38.194043 32.915291 24.588522 13.931709 1.928657

ψ (anal.) (°) 40.000000 38.042261 32.360680 23.511410 12.360680 0.000000

prescribed, and the numerical resolution was implemented on a quarter of the

natural period. Table 27.1 give the results obtained for various values of the initial

angle: 0 5 , 10 , 20 and 40ψ = ° ° ° ° . The table compares the values ( ψ (anal.)

and ψ (anal.)) deduced from Relation (27.8) to the values ( ψ (num.) and ψ

(num.)) derived from the numerical computation. The results obtained show that

the solution (27.8) describes rather well the exact solution of the equation of

motion (27.5), up to amplitudes of the angles of the order of 10°. Figure 27.1

gives the variations of ψ and ψ over two natural periods T0, with the initial

conditions 0 40ψ = ° and 10 60 sψ −= ° . We observe that the actual period of the

motion ( 2.08 sT = ) differs slightly from the natural period.


temps ( s )

0 1 2 3 4 5 6 7 8 9 10

ang

le d

e ro

tati

on (

° )

-50

-40

-30

-20

-10

0

10

20

30

40

50

temps ( s )

0 1 2 3 4 5 6 7 8 9 10

ang

le d

e ro

tati

on

(

° )

-50

-40

-30

-20

-10

0

10

20

30

40

50

FIGURE 27.1 Motion of a simple pendulum, for the initial conditions 0 40ψ = ° and

0 60 /sψ = ° .

FIGURE 27.2. Motion of a simple pendulum as a function of time, for four values of the

damping.

temps ( s )0 1 2 3 4

angle

de

rota

tio

n

ψ

( °

)

-40

-20

0

20

40

-160

vitesse d

e rotatio

n ψ

( °/s )

-80

160

80

0

angle of rotation

rotation velocity ang

le o

f ro

tati

on

(°

)

rotatio

n v

elocity

ψ

( °/s )

time ( s )

temps ( s )

0 1 2 3 4 5 6 7 8 9 10

angle

de

rota

tio

n (

° )

-50

-40

-30

-20

-10

0

10

20

30

40

50

temps ( s )

0 1 2 3 4 5 6 7 8 9 10

angle

de

rota

tion

(

° )

-50

-40

-30

-20

-10

0

10

20

30

40

50

= 0.25 s−1

= 1 s−1

= 3.14 s−1

= 5 s−1

time ( s ) time ( s )

time ( s ) time ( s )

ang

le o

f ro

tati

on

( °

)

ang

le o

f ro

tati

on

( °

)

ang

le o

f ro

tati

on

( °

)

ang

le o

f ro

tati

on

( °

)

27.2 Equations of Motions with Several Degrees of Freedom 461

27.1.3.2 Analysis in the Presence of Friction

In the case of a friction of viscous type, the equation of motion (22.50) is:

202 sin 0ψ δψ ω ψ+ + = . (27.10)

For low values of the angle of rotation, the equation of motion is reduced to

Equation (22.52), which is the reduced form for a system with one degree of

freedom with viscous friction. The free vibrations were studied in Section 21.3.2

of Chapter 21. For a natural period 0 2 sT = , the critical damping is (21.91): 13.1416 scδ −= .

Figures 27.2 report the results derived from a numerical resolution of Equation

(27.10) using the Runge-Kutta method of order 4. Four values for the damping

are considered: 0.25 s–1, 1 s–1, 3.1416 s–1 and 5 s–1, with the initial conditions:

0 40ψ = ° and 10 60 sψ −= ° . We find well the different types of motions studied in

Section 21.3.2.

27.2 EQUATIONS OF MOTIONS WITH SEVERAL

DEGREES OF FREEDOM

27.2.1 Form of the Equations of Motions with Several Degrees of Freedom

In the general case, the motion of a solid or a system of solids is a motion with

several degrees of freedom. Various motions were studied in Chapters 23 to 25.

The equations of motions are either deduced from the fundamental principle of

dynamics, or obtained automatically using the Lagrange equations. For a mecha-

nical system having p degrees of freedom de: y1, y2, ..., yp, the equations of

motions are written in the general form:

1 1 1 1 2 2

2 2 1 1 2 2

1 1 2 2

( , , , , , . . . , , ),

( , , , , , . . . , , ),...

( , , , , , . . . , , ),

p p

p p

p p p p

y f t y y y y y y

y f t y y y y y y

y f t y y y y y y

=

=

=

(27.11)

with the initial conditions:

0 0

0 0( ) , ( ) , 1, 2, . . . , .i i i iy t y y t y i p= = = (27.12)


27.2.2 Principle of the Numerical Resolution

We have to solve a system (27.11) of p differential equations of the second

order. To solve it numerically, we generalize the method used in Section 27.1.2

for the numerical resolution of the equation of the motion of a system with one

degree of freedom. For that, we introduce the variables:

, 1, 2, . . . , ,iy iy i p= = (27.13)

and we lead the system of the p equations of the second order to a system of 2p

differential equations of the first order:

1

1 1 2

2

2 1 2

1 2

1

1 1 2

2

2 1 2

1 2

,

( , , , , , . . . , , ),

,

( , , , , , . . . , , ),...

,

( , , , , , . . . , , ),

p

p

p

p p

y

y y y p y

y

y y y p y

p y

y p y y p y

y

f t y y y

y

f t y y y

y

f t y y y

=

=

=

=

=

=

(27.14)

with the initial conditions:

0 0

0 0( ) , ( ) , 1, 2, . . . , .i ii i y yy t y t i p= = = (27.15)

The procedure of numerical resolution consists in computing at each step, using

one of the methods considered in Chapter 26, the values of 1 2, ,..., ,py y y and

the values of y1, y2, ..., yp.

27.2.3 Trajectories and Kinematic Vectors

Once solved the equations of motion, the trajectories and the kinematic vectors

of an arbitrary point of a solid are obtained as functions of the parameters of

situation, in accordance to the results of the kinematics of the solid (Chapter 9).

The trajectory of an arbitrary point M of the solid (S) in motion with respect to the

reference (T) is obtained by Relation (9.5):

t

( , ) ( , ) ( )

( , ) ( , ) ( ) ( )

( , ) ( , ) ( )

S

S

S

x M t x P t x M

y M t y P t t y M

z M t z P t z M

= +

A . (27.16)

trajectory of deduced from coordinates of the point M

the point M in the parameters of in a system attached to (S)

the reference (T) situation y1, y2, ... yp and of origin P

27.3 Motions of Planets and Satellites 463

The point P is the point of the solid (S) at which the parameters of translation

were chosen.

The velocity vector of the point M is calculated while using the law (9.11) of

composition of the velocities: ( ) ( ) ( )

( , ) ( , )T T TSM t P t PMω= + ×

. (27.17)

derived from the

velocities 1 2, ,...,

py y y

In the same way, the acceleration vector is derived from Relation (9.24) for the

composition of the accelerations:

( ) ( ) ( ) ( ) ( )( ) ( , ) ( , )T T T T TS S Sa M t a P t PM PMω ω ω= + × + × ×

. (27.18)

The acceleration vector of the point P: ( )

( , )Ta P t

at every time and the rotation

acceleration vector ( )TSω are deduced from the velocity vectors obtained at the

various times of the numerical resolution.

27.3 MOTIONS OF PLANETS

AND SATELLITES

27.3.1 Motion of a Planet about the Sun

27.3.1.1 Equation of the Motion

The mechanical actions exerted on a planet are reduced to the action of gravi-

tation exerted by the Sun. So, it results that the motions of the planets about the

Sun are plane motions with central accelerations (Section 18.5.2). If O is the

centre of the Sun of mass mSo, and G the mass centre of the planet of mass m, the

fundamental relation of dynamics is written for the resultant as:

( )So 3

( , )g OG

m a G t Km mOG

= −

, (27.19)

where (g) is a Galilean reference (Section 18.5.1) and K the universal constant of

gravitation of value 6.67 × 10–11 m3 kg–1 s–2. If x and y are the Cartesian coor-

dinates of the mass centre G in the plane of the trajectory of the planet, the

equations of motion deduced from (27.19) are written:

So 3

So 3

,

,

xx A

r

yy A

r

= −

= −

(27.20)


FIGURE 27.3. Plane of the trajectory of a planet.

where r is the distance between the points O and G (Figure 27.3):

2 2r x y= + , (27.21)

and ASo is a constant expressed by:

So SoA Km= . (27.22)

With a mass of the Sun appreciably equal to 302 10 kg× , the value of this cons-

tant is: 19 3 2So 13.34 10 m sA −= × .

27.3.1.2 Numerical Resolution of the Equations of Motion

To numerically solve Equations (27.20) of motion, it is interesting to use a

constant step of length of the trajectory, rather than a constant step of time. A

constant step of trajectory makes it possible to have an appreciably constant

computation time throughout the trajectory. As input of the numerical procedure,

we introduce the step hs of curvilinear abscissa corresponding to a given length of

the trajectory. The step in time is then calculated at every moment by the relation:

shh =

, (27.23)

where is the instantaneous velocity at the moment under consideration.

27.3.1.3 Examples of Results

As an example, we consider the case of a planet which would have at 0,t = a

position G0 of Cartesian coordinates 80 0 0( 1.5 10 km, 0, 0)x y z= − × = = and a

velocity 0

of components 1

0 0 0( 0, 20 km s , 0)x y z−= = = .

For solving the equations of the motion, we use the Euler method with a

computation step equal to 100,000 km. The computation is stopped when the

planet reaches its initial position G0. Figure 27.4a shows the trajectory obtained.

The step chosen for the computation is in fact relatively low, since the total

y

x

G

r

O


-1,5 -1,0 -0,5 0,0 0,5-1,0

-0,5

0,0

0,5

1,0

( × 108 km)

( × 108 km)

Sun

-1.5 -1.0 -0.5 0.0 0.5 -1.0

-0.5

0.0

0.5

1.0

-1,5 -1,0 -0,5 0,0 0,5-1,0

-0,5

0,0

0,5

1,0

( × 108 km)

( × 108 km)

Sun

-1.5 -1.0 -0.5 0.0 0.5 -1.0

-0.5

0.0

0.5

1.0

number of computations implemented is high: 5,823cn = . We observe a notable

error however, since the trajectory does not loop at the initial position. Decreasing

the step of the computation makes it possible to improve the precision, however

with a high time of computation. For example, for a step of 50,000 km, the total

number of computations necessary to reach the initial point is 12,000cn = . The

distance remains still notable (Figure 27.4b). These results illustrate well the

limits associated to the Euler method, easy to implement but whose precision is

limited.

On implementing the numerical resolution of the equations of motion with the

Runge-Kutta method of order 4, the number of computed values decreases rapidly

with the computation step, while keeping a good precision. Figure 27.5 shows the

trajectory obtained while taking a computation step equal to 107 km. The total

FIgure 27.4. Trajectory of a planet derived by the Euler method for two values of the

computation step: a) step of 100,000 km, b) step of 50,000 km.


-1,5 -1,0 -0,5 0,0 0,5 1,0 1,5

-1,5

-1,0

-0,5

0,0

0,5

1,0

1,5( × 10

8 km)

( × 108 km)

Sun

-1.0

-0.5

0.0

0.5

1.0

1.5

-1.5

-1.5 -1.0 -0.5 0.0 0.5 1.5 1.0

FIGURE 27.5. Trajectory of a planet derived by the Runge-Kutta method of order 4, with

a computation step of 107 km.

number of computations to reach the initial point is 56cn = . The trajectory loops

well at its initial point in this case. This result shows thus the interest to use a

resolution method of high order.

By modifying the initial velocity of the preceding example according to

0( 0,x = 10 030 km s , 0)y z−= = , the trajectory obtained by the Runge-Kutta

method is the one of Figure 27.6. This trajectory is fairly the one of the Earth

about the Sun (elliptic trajectory of semi-major axis equal to 81.496 10× km and

eccentricity equal to 0.017).

FIGURE 27.6 Trajectory obtained by the Runge-Kutta method of order 4, with for initial

conditions 80 ( 1.5 10 km, 0, 0)G − × and 1

0 (0, 30 km s , 0).−

-1,5 -1,0 -0,5 0,0 0,5-1,0

-0,5

0,0

0,5

1,0

( × 108 km)

Sun

( × 108 km)

-1.0

-0.5

0.0

0.5

1.0

-1.5 -1.0 -0.5 0.0 0.5


-1,0 -0,5 0,0 0,5 1,0 1,5

-1,0

-0,5

0,0

0,5

1,0

Ea

h t r

( × 104 km )

( × 104 km )

(a)

(b) (c)

(d)

(a): 6 km/h

(b): 7 km/h

(c): 7.4 km/h

(d): 7.8 km/h

-1.0 -0.5 0.0 0.5 1.5 1.0

-1.0

-0.5

0.0

0.5

1.0

27.3.2 Motion of a Satellite around the Earth

In the case of the motion of a satellite in the vicinity of the Earth, the mecha-

nical actions exerted on the satellite are reduced to the action of gravitation

exerted by the Earth. The fundamental relation of dynamics for the resultant is

transposed from Relation (27.19), thus:

( )TeTe 3

( , )OG

m a G t Km mOG

= −

, (27.24)

where (Te) is a reference attached to the plane of the ecliptic, mTe is the mass of

the Earth, O its centre and G the mass centre of the satellite of mass m. If x and y

are the Cartesian coordinates of the mass centre G in the plane of the trajectory of

the satellite, the equations of the motion have the same form as Equation (27.20),

thus:

Te 3

Te 3

,

,

xx A

r

yy A

r

= −

= −

(27.25)

where the constant ATe is expressed by:

Te TeA Km= . (27.26)

With a mass of the Earth fairly equal to 246 10× kg, the value of this constant is:

13 3 2

Te 40.02 10 m sA −= × .

Figure 27.7 shows the trajectories computed by the Runge-Kutta method of

FIGURE 27.7. Trajectories computed for a satellite launched at the point

8

0 ( 1.5 10 km, 0, 0)G − × with an initial velocity 0(0, , 0)y , for four values of 0y .


order 4 with a step of 100 km, for a satellite launched at the point G0 with an

initial velocity of components 0(0, , 0)y . Four values of the component 0y are

considered: 16, 7, 7.4 and 7.8 km s− . The results obtained show the influence of

the launching velocity. In the case where the component is equal to 6 km s– 1, the

launching velocity is not high enough, and the satellite is crashed on the surface

of the Earth after its launching. For a value of the component of 7 km s– 1, the

orbit is slightly eccentric, the satellite keeping a rather constant altitude. The

increase of the value of the component enhances then the eccentricity of the

ellipse.

27.3.3 Launching and Motion of a Moon Probe

In this subsection, we consider the motion of a Moon probe launched from a

earthly orbit. Launched from this orbit, the objective is that the probe turns around

the Moon, then returns in the vicinity of the Earth to be recovered. Furthermore,

its recovery needs a return in the atmosphere under a favourable angle.

The equation of motion of the mass centre G of the probe is written:

( ) Te LTe L3 3

Te L

( , )T O G O Ga G t K m m

r r

= − +

, (27.27)

where OTe and OL are the respective centres of the Earth and of the Moon, mTe

and mL (mL = 227.4 10 kg× ) the respective masses of the Earth and of the Moon,

and rTe and rL the distances from the probe respectively to the Earth and to the

Moon. If (x, y) are the Cartesian coordinates of the probe in the plane of its tra-

jectory (figure 27.8), Equation (27.27) leads to the equations of motion:

LTe L3 3

Te L

Te L3 3Te L

,

,

x xxx K m m

r r

y yy K m m

r r

− = − +

= − +

(27.28)

with

( )22 2 2Te L L, ,r x y r x x y= + = − +

and where xL is the distance from the Moon to the Earth (xL = 384,000 km). For

reasons of simplicity, the equations do not take into account the relative motion of

the Moon with respect to the Earth. This motion can be taken into account

without difficulty, in the case of a numerical procedure.

The examples considered (Figure 27.9) use the same initial position for the

launching ,( 19 000 km, 0, 0)− with initial velocities directed along the direction

Oy

of component 0y having various values. The trajectories were computed,

using the Runge-Kutta method of order 4 and choosing a step of 1,000 km.

27.4 Motion of a Solid on an Inclined Plane 469

FIGURE 27.8. Coordinates of the probe in the plane of its trajectory.

We observe (Figure 27.9a) that for the lowest velocities of launching (6 to 6.25

km s–1), the trajectories stay in the vicinity of the Earth and the influence of the

Moon is negligible. It appears only for 10 6.30 km sy −= .

Figure 27.9b shows two trajectories: the one where the probe escapes the Moon

attraction ( 10 6.40 km sy −= ) and the other ( 1

0 6.3155 km sy −= ) corresponding

to an ideal launching, with passing round the Moon and return in the vicinity of

the Earth

Figure 27.9c shows the notable influence on the trajectories of low variations of

the launching velocity at the vicinity of the ideal trajectory: for 0y = 16.30 km s−

the probe does not reach the vicinity of the Moon, for 0y = 6.33 km s–1 the probe

is crashed on the Moon, for 10 6.35 km sy −= the probe passes behind the Moon

and returns at the vicinity of the Earth.

Lastly, Figure 27.9d shows a complex trajectory of the probe, obtained for 1

0 6.31 km sy −= . Arrived in the vicinity of the Moon, the probe starts to pass

around the Moon in the direct direction, then it continues to turn around in the

inverse direction, to return then on the Earth.

In practice, the satellites are provided with auxiliary engines which make it

possible to correct the variations of the trajectory at every moment.

27.4 MOTION OF A SOLID ON AN INCLINED PLANE

The equations of motion on a solid on an inclined plane have been expressed in

(23.43) in the case of viscous friction:

sin ,

0,

0,

t

t

r

mx c x mg

my c y

C c

α

ψ ψ

+ =

+ =

+ =

(27.29)

where ct and cr are the coefficients of viscous friction in translation and rotation,

y

x

G

rTe

OTe

rL

xL

L


0,0 0,5 1,0 1,5 2,0 2,5 3,0 3,5 4,0

-1,0

-0,5

0,0

0,5

1,0

Earth Moon

( × 105 km )

( × 105 km )

0 6.31 km/sy =

(d) 2.0 1.5 2.5 3.0 3.5 4.0 1.0 0.5 0.0

0.0

-0.5

-1.0

1.0

0.5

0,0 0,5 1,0 1,5 2,0 2,5 3,0 3,5 4,0

-1,0

-0,5

0,0

0,5

1,0

Earth Moon

( × 105 km )

( × 105 km )

(a) (b) (c)

(a): 0 6.30 km/sy =

(b): 0 6.33 km/sy =

(c): 0 6.35 km/sy =

2.0 1.5 2.5 3.0 3.5 4.0 1.0 0.5 0.0

0.0

-0.5

-1.0

1.0

0.5

0,0 0,5 1,0 1,5 2,0 2,5 3,0 3,5 4,0

-1,0

-0,5

0,0

0,5

1,0

Earth Moon

( × 105 km )

( × 105 km )

(a)

(b)

(a): 0 6.3155 km/sy =

(b): 0 6.5000 km/sy =

0.0

-0.5

-1.0

1.0

0.5

2.0 1.5 2.5 3.0 3.5 4.0 1.0 0.5 0.0

0,0 0,5 1,0 1,5 2,0 2,5 3,0 3,5 4,0

-1,0

-0,5

0,0

0,5

1,0

Earth Moon

( × 105 km )

( × 105 km )

(a) (b) (c) (d) (e)

(a): 0 6.00 km/sy =

(b): 0 6.10 km/sy =

(c): 0 6.20 km/sy =

(d): 0 6.25 km/sy =

(e): 0 6.30 km/sy =

2.0 1.5 2.5 3.0 3.5 4.0 1.0 0.5 0.0

0.0

-0.5

-1.0

1.0

0.5

FIGURE 27.9 Trajectories of a Moon probe launched from the position (–19,000 km, 0,

0) with various initial velocities.

(a)

(b)

(c)


respectively. In reduced forms these equations are written:

sin ,

,

,

t

t

r

x g f x

y f y

f

α

ψ ψ

= −

= −

= −

(27.30)

introducing the viscous damping coefficients in translation and rotation:

, .t rt r

c cf f

m m= = (27.31)

Note that the x-axis is the direction of greater slope of the inclined plane.

The equations (27.30) are solved numerically using for example the Runge-

Kutta method of order 4. The resolution makes it possible to derive the para-

meters of situation x, y, ψ, and their derivatives , , x y ψ as functions of time. The

trajectory of the mass centre is deduced from the parameters (x, y). Then, it is

possible to obtain the trajectory of an arbitrary point of the solid by using Relation

(27.16).

Figures 27.10 show the results obtained for various values of the inclination of

the plane and of the damping coefficients ft and fr. At the initial moment ( 0),t =the mass centre is at the origin point of the coordinate system and has a velocity

vector 10 (0, 5 m s , 0)− . At 0t = , the orientation of the coordinate system

attached to the solid coincides with the one of the system attached to the inclined

plane (Section 23.2.1) and the angular velocity is taken as 10 360 sψ −= ° . Com-

putations were implemented in the interval [0, 5 s], with a computation step equal

to 0.05 s. In Figures 27.10, are reported the trajectories of the mass centre and of a

point attached to the solid of coordinates (4 m, 0, 0) relatively to the system

attached to the solid. The results obtained show the influence of the angle of

inclination of the plane, and the influence of the frictions in translation and

rotation.

27.5 COUPLED MOTIONS OF TWO SOLIDS


In this section we return to the motion of the two solids considered in Section

24.3 of Chapter 24. In the case of a viscous friction, the equations of motion have

been written in (24.120):

( ) ( ) ( )21 2 0 2

2 2 2

cos sin 0,

cos sin 0.

t

r


m ay C c m ga

ψ ψ ψ ψ

ψ ψ ψ ψ

+ + + − − + − =

+ + + =

(27.32)

The motion of the solid (S1) is induced around the position of equilibrium

0y l d= + . Hereafter, we study the motion around this equilibrium introducing


-5 0 5 10 15 20 25 30 35 40

0

5

10

15

20

25

0 10 20 30 40 50 60 70

0

5

10

15

20

25

30

0 10 20 30 40 50 60

0

5

10

15

20

25

30

-5 0 5 10 15 20 25 30 35 40 45

0

5

10

15

20

25

30

FIGURE 27.10. Trajectories of the mass centre and of a point attached to a solid in mo-

tion on an inclined plane, for various values of translation and rotation frictions.

( m )

( m )

( m )

( m )

( m )

( m )

( m )

( m )

20 ,

0,

0.

t

r

f

f

α = °

=

=

30 ,

0,

0.

t

r

f

f

α = °

=

=

45 ,

0,

0.

t

r

f

f

α = °

=

=

1

20 ,

0.1 s ,

0.

t

r

f

f

α−

= °

=

=


-5 0 5 10 15 20 25

0

2

4

6

8

10

12

14

-5 0 5 10 15 20 25

0

2

4

6

8

10

12

14

-5 0 5 10 15 20 25 30 35 40

0

5

10

15

20

25

-5 0 5 10 15 20 25 30 35 40

0

5

10

15

20

25

FIGURE 27.10. (continued). Trajectories of the mass centre and of a point attached to a

solid in motion on an inclined plane, for various values of translation and rotation

frictions.

( m )

( m )

1

1

20 ,

0.1 s ,

0.1 s .

t

r

f

f

α−

−

= °

=

=

( m )

( m )

( m )

( m )

( m )

( m )

1

1

20 ,

0,1 s ,

0,5 s .

t

r

f

f

α−

−

= °

=

=

1

1

20 ,

0.5 s ,

0.5 s .

t

r

f

f

α−

−

= °

=

=

1

1

20 ,

0.5 s ,

0.8 s .

t

r

f

f

α−

−

= °

=

=


variable change : 0y l d y− + → . We obtain the equations of motion:

( ) ( )21 2 2

2 2 2

cos sin 0,

cos sin 0.

t

r

m m y c y ky m a

m ay C c m ga

ψ ψ ψ ψ

ψ ψ ψ ψ

+ + + + − =

+ + + =

(27.33)

In the absence of the solid (S2), the spring-mass system constituted of the solid

(S1) and the spring has a natural period and damping expressed (Chapter 21) by:

11

1 1

22 , .

2t

tcm

Tk m

π π δω

= = = (27.34)

In the same way (Section 22.2.1), the solid (S2) in rotation has a natural period

and a damping given by:

22

2 2 2

22 , .

2r

rC c

Tm ga C

π π δω

= = = (27.35)

27.5.2 Analytical Solving in the case of Low Amplitudes and in the Absence of Friction

In the case of low amplitudes of the angle of rotation for which:

sin , cos 1,ψ ψ ψ≈ ≈ (27.36)

the equations of motion (27.33) are reduced to:

( )1 2 2

2 2 2

0,

0.

m m y ky m a

m ay C m ga

ψ

ψ ψ

+ + + =

+ + =

(27.37)

This system of equations is solved by searching for harmonic solutions of angular

frequency Ω , written in the complex form as:

1 2, .i t i ty A e A eΩ Ωψ= = (27.38)

Introducing these expressions in Equations (27.37), we obtain the system of equa-

tions of the amplitudes A1 and A2:

( )

( )

2 21 2 1 2 2

2 22 1 2 2 2

0,

0.

m m k A m a A

m a A m ga C A

Ω Ω

Ω Ω

− + + − =

− + − = (27.39)

A solution different from zero ( 1 20 and 0A A≠ ≠ ) is obtained in the case where

the determinant of the system is zero, what provides the characteristic equation:

( ) ( )[ ]2 2 4 21 2 2 2 1 2 2 2 2 0m m C m a m m m ga kC m gakΩ Ω − + − − − + + + = .

(27.40)

This equation can be rewritten in the form:

27.5 Coupled Motion of Two Solids 475

( )4 2 2 2 2 212 2 1 1 1 1 2 0α Ω ω α ω Ω α ω ω− + + = , (27.41)

introducing the natural pulsations of the two systems and the coefficients:

( )2

1 21 2 12 1 2

1 2 2

, , 1 1 .m m a

m m Cα α α α α= = = − −

+ (27.42)

The characteristic equation (27.41) generally has two real and positive roots 21Ω

and 22Ω . The general solution of the system (27.37) is then a linear combination

of the complex solutions:

1 1 2 2, , , .i t i t i t i te e e eΩ Ω Ω Ω− − (27.43)

This linear combination may be expressed in the form:

( ) ( )

( ) ( )11 1 1 12 2 2

21 1 1 22 2 2

cos cos ,

cos cos ,

y A t A t

A t A t

Ω φ Ω φ

ψ Ω φ Ω φ

= + + +

= + + + (27.44)

where the first index of the amplitudes Aij is relative to y ( 1)i = or toψ ( 2)i = ,

and the second one is relative to the pulsation 1Ω or 2Ω . The amplitudes A1j and

A2j are not however independent, since according to the value of Ω , we have by

reporting 1Ω or 2Ω in the system (27.39):

( )

( )

21 2 1

21 1 11 1 22 1

21 2 2

22 2 12 2 22 2

, with ,

, with .

m m kA a A a

m a

m m kA a A a

m a

Ω

Ω

Ω

Ω

− + += =

− + += =

(27.45)

Finally, y and ψ have for respective expressions:

( ) ( )

( ) ( )11 1 1 12 2 2

1 11 1 1 2 12 2 2

cos cos ,

cos cos .

y A t A t

a A t a A t

Ω φ Ω φ

ψ Ω φ Ω φ

= + + +

= + + + (27.46)

The values of the amplitudes A11, A12 and of the phases 1 2, φ φ are determined

from the initial conditions: 0 0 0 0, , and y y ψ ψ .

Suppose that initially at 0t = , we have, for example:

0 0 0 00, 0, 0, 0.y y ψ ψ= = ≠ = (27.47)

The initial conditions are then expressed from (27.46) in the form:

11 1 12 2

11 1 1 12 2 2

0 1 11 1 2 12 2

1 11 1 1 2 12 2 2

0 cos cos ,

0 sin sin ,

cos cos ,

0 sin sin .

A A

A A

a A a A

a A a A

φ φ

Ω φ Ω φ

ψ φ φ

Ω φ Ω φ

= +

= − −

= +

= − −

(27.48)


Solving this system leads to:

0 01 2 11 12

1 2 1 2

0, 0, , .A Aa a a a

ψ ψφ φ= = = = −

− − (27.49)

Expressions (27.46) of y and ψ are then given in this case as:

( )01 2

1 2

1 20 1 2

1 2 1

cos cos ,

cos cos .

y t ta a

a at t

a a a

ψΩ Ω

ψ ψ Ω Ω

= −−

= − −

(27.50)

27.5.3 Numerical Computation of the Equations of Motion

The numerical resolution of Equations (27.33) of motion requires at first to

separate terms in y and ψ . We obtain:

( )

( ) ( )

( ) ]

22 2 2 2

22 2

2 22 1 2 2

2 1 2 2

1sin cos

( )

sin cos ,

1sin cos cos

( )

cos sin ,

t r

r t

y m aC c C y m acD

kC y m a g

m a m m c m ac yD

km ay m m m ga

ψ ψ ψ ψψ

ψ ψ

ψ ψ ψ ψ ψ ψψ

ψ ψ

= − +

− +

= − + + −

− + +

(27.51)

setting

( ) ( )21 2 2 2( ) cosD m m C m aψ ψ= + − . (27.52)

Equations (27.51) are then led (Section 27.2) to a linear system of differential

equations which can be solved numerically using one of the methods considered

in Chapter (26). As input parameters, we will have on the one hand the para-

meters relative to the solid (S1): m1, k, ct, and the parameters relative to the solid

(S2): m2, C2, a, cr. Starting from these data, the numerical procedure will compute

the values of the natural periods T1 and T2. On the other hand, we shall have to

introduce the initial conditions (for 0t = ): y0 and 0y for the solid (S1), ψ0 and 0ψfor the solid (S2). Lastly, it would be necessary to give the duration of the compu-

tation tf and the step of the computation.

The results derived from the numerical computation are reported in Figures

27.11, which plot y and ψ as a function of time. The values of the parameters

corresponding to the different figures are reported in Table 27.2. The whole of the

results was obtained with for the initial conditions at 0t = : 0 0,y = 0 0,y =

0 020 and 0.ψ ψ= ° = These numerical results could be compared with the

results deduced from the relations developed in Section 27.5.2. The results


TABLE 27.2 Values of the parameters used for deriving the results of Figures 27.11.

Figure 27.11 a b c d e

m1 (kg) 40 20 5 40 5

k (N m–1) 160 80 10 160 10

ct (N m s–1) 0 0 0 0.05 0.05

m2 (kg) 1.5 1.5 1.5 1.5 1.5

C2 (kg m–2) 1.8 1.8 1.8 1.8 1.8

a (m) 0.48 0.48 0.48 0.48 0.48

cr (N m s–1) 0 0 0 0.1 0.1

ω1 (s–1) 2 2 1.4142 2 1.4142

T1 (s) 3.1416 3.1416 4.4429 3.1416 4.4429

ω2 (s–1) 1.9808 1.9808 1.9808 1.9808 1.9808

T2 (s) 3.1720 3.1720 3.1720 3.1720 3.1720

α1 0.9639 0.9302 0.7692 0.9639 0.7692

α2 0.1920 0.1920 0.1920 0.1920 0.1920

α12 0.9931 0.9931 0.9557 0.9931 0.9557

Ω1 (s–1) 2.061 2.0817 2.0535 2.0601 2.0535

Ω2 (s–1) 1.8943 1.8477 1.2237 1.8943 1.2237

a1 – 5.2797 – 4.2202 – 5.7341 – 5.2797 – 5.7341

a2 4.2920 2.6847 0.2470 4.2920 0.2470

tf (s) 80 80 80 160 160

h (s) 0.08 0.08 0.08 0.08 0.08

obtained strongly depend: 1) on the natural periods of each of the two systems, 2)

on the mechanical energies induced in each motion of the solids (S1) and (S2).

According to the values of these energies, there is exchange or not of the energies

induced during the motions of each solid. Beat processes are observed in the case

where the natural frequencies are close (Figures 27.11a, 27.11b and 27.11d),

caused by alternating constructive and destructive interferences.


FIGURE 27.11. Coupled motions of two solids for various values of the parameters.

temps ( s )

0 10 20 30 40 50 60 70 80

rota

tio

n ψ

(

° )

-20

-10

0

10

20

time ( s )

temps ( s )

0 10 20 30 40 50 60 70 80

dép

lace

men

t y

(

m )

-10

-5

0

5

10

dis

pla

cem

ent

y

( m

)

temps ( s )0 10 20 30 40 50 60 70 80

rota

tion

ψ

(

° )

-20

-10

0

10

20

time ( s )

temps ( s )

0 10 20 30 40 50 60 70 80

dép

lace

men

t y

(

m )

-10

-5

0

5

10

time ( s )

dis

pla

cem

ent

y

( m

)

temps ( s )

0 10 20 30 40 50 60 70 80

rota

tion

ψ

(

° )

-20

-10

0

10

20

time ( s )

1

22

2

20 kg,

80 N/m,

0,

1.5 kg,

1.8 kg/m ,

0.48 m,

0.

t

r

m

k

c

m

C

a

c

====

===

1

22

2

5 kg,

10 N/m,

0,

1.5 kg,

1.8 kg/m ,

0.48 m,

0.

t

r

m

k

c

m

C

a

c

====

===

(a)

(b)

(c)

temps ( s )

0 10 20 30 40 50 60 70 80

dép

lace

men

t y

(

m )

-8

-4

0

4

8

time ( s )

dis

pla

cem

ent

y

( m

)

time ( s )

1

22

2

40 kg,

160 N/m,

0,

1.5 kg,

1.8 kg/m ,

0.48 m,

0.

t

r

m

k

c

m

C

a

c

====

===


FIGURE 27.11 (continued). Coupled motions of two solids for various values of the

parameters .

EXERCISES 27.1 Implement numerical procedures making it possible to study the various

motions considered in this chapter.

temps ( s )

0 20 40 60 80 100 120 140 160

rota

tion

ψ

(

° )

-20

-10

0

10

20

time ( s )

temps ( s )

0 20 40 60 80 100 120 140 160

dép

lace

men

t y

(

m )

-10

-5

0

5

10

time ( s )

dis

pla

cem

ent

y

( m

)

temps ( s )

0 20 40 60 80 100 120 140 160

rota

tio

n ψ

(

° )

-20

-10

0

10

20

time ( s )

temps ( s )0 20 40 60 80 100 120 140 160

dép

lace

men

t y

(m

)

-6

-4

-2

0

2

4

6d

isp

lace

men

t y

(

m )

time ( s )

1

22

2

40 kg,

160 N/m,

0.05 Nm/s,

1.5 kg,

1.8 kg/m ,

0.48 m,

0.1 Nm/s.

t

r

m

k

c

m

C

a

c

====

===

(d)

1

22

2

5 kg,

10 N/m,

0.05 Nm/s,

1.5 kg,

1.8 kg/m ,

0.48 m,

0.1 Nm/s.

t

r

m

k

c

m

C

a

c

====

===

(e)


COMMENTS

The numerical resolution of the equations of motion of a solid or a

system of solids is particularly important owing to the fact that it is gene-

rally necessary to use it. The reader will pay an extreme attention to the

development of the concepts introduced in the preceding chapter then

applied to the cases of the resolution of the equations of the motions

considered in the present chapter.

So as to implement the procedures for numerically solving the equa-

tions of motion, the reader will endeavour to set up the numerical

procedures which make it possible to analyze the various motions studied

in this chapter. The procedures thus implemented could be then trans-

posed easily to the study of any motion of solids.

Part VII

Solutions of the Exercises

This last part of the textbook reports the solutions of the exercises

proposed all along the chapters of the textbook. The writing has been

developed extensively and structured in such a way to underline the

development of the analyses, and so to improve the capacity of the

comprehension of the reader. The author estimates that a well

structured development of the solutions is indispensable for a good

understanding and a good application of the concepts introduced in

the textbook.

Chapter 1

Vector Space 3

1.1 Let V

be a given vector and u

a vector collinear to this vector.

u

is collinear to V

is expressed as:

, where u Vα α= ∈

.

u

is a unit vector is expressed as:

1u =

.

The two preceding relations lead thus to:

1 ou 1V Vα α= =

.

Hence:

1

Vα = .

Thus, we deduce:

1

Vα = ± .

Hence the expression of the unit vectors:

Vu

V= ±

.

There exist thus two unit vectors collinear to a given vector. They are opposed

and obtained by dividing the vector by its norm.

Numerical application

(2, 5, 3) in the canonical basis.V = −

We have: 38V =

.

Hence: ( )

12 5 3

38u i j k= ± − +

,

or

2 5 3 2 5 3, , and , , .

38 38 38 38 38 38u u

− − − = =

1.2 The necessary and sufficient condition so that the vectors 1V

and 2V

are

orthogonal is that the scalar product is zero. Thus:

1 2 0V V =⋅

or 5 5 0α − = .

Hence: 1α = .

The vector 1V

is thus expressed by 1 (1, 2, 1)V = −

.


1.3 Let 1V

and 2V

be the two given vectors. Their vector product 1 2V V V= ×

is

orthogonal to 1V

and to 2 .V

We are brought back to Exercise 1.1. The unit

vectors u

orthogonal to 1V

and to 2V

are thus:

1 2

1 2

V Vu

V V

×= ±

×

.


1 2(2, 5, 3) and ( 2, 1, 3)V V= − = − −

.

Their vector product is: 1 2 12 8V V i k× = −

.

So, the unit vectors are:

( )

112 8

208u i k= ± −

.

1.4 Expansions

1. By applying the distributivity of the scalar product:

( ) ( ) 2 21 2 1 2 1 1 2 2 1 2 .V V V V V V V V V V+ − = − + −⋅ ⋅ ⋅

The scalar product is commutative. Hence:

( ) ( ) 2 21 2 1 2 1 2 .V V V V V V+ − = −⋅

2. Similarly, by applying the distributivity of the vector product:

( ) ( )1 2 1 2 1 1 1 2 2 1 2 2 .V V V V V V V V V V V V+ × − = × − × + × − ×

The vector product of a vector by itself is the null vector and the vector product is

antisymmetric. Hence:

( ) ( ) ( )1 2 1 2 1 22 .V V V V V V+ × − = − ×

1.5 The vector V

has for components (4, –9, 3) in the basis ( )1 1 1(1) , , i j k=

.

Thus:

1 1 14 9 3V i j k= − +

.

The basis (2) is deduced from the basis (1) by the relations:

2 1 2 1 2 12 , 2 , .i i j j k k= = = −

Hence by substituting into the expression of V

:

2 2 29

2 32

V i j k= − −

.

In the basis (2), the components of V

are thus (2, –3.5, –3).

1.6 We have to derive the vectors V

such that: 1 2 1V V V V× = ×

, where 1V

and

2V

are two given vectors.

Solution of Exercise 1.6 485

We have to proceed by equivalences starting from the given relation. While

passing the first member of the equation into the second one, we have:

1 1 2 0V V V V× − × =

.

Considering the distributivity, it comes:

( )1 2 0V V V× − =

.

The necessary and sufficient condition so that the vector product of the vectors 1V

and 2V V−

is null is that these vectors are collinear. For example:

2 1, V V Vα α− = ∀ ∈

.

Hence the expression of the vector:

2 1, V V Vα α= + ∀ ∈

.


1 24 and 5 6 2V i j V i j k= − = + −

.

The result is:

( ) ( ) 5 6 4 2 , V i j kα α α= + + − − ∀ ∈

.

Chapter 2

The Geometric Space

2.1 We express that the point H is the orthogonal projection of the point M on

the line (D) (Figure 2.16).

The point H is the orthogonal projection of the point M. Hence MH is

orthogonal to the line (D). Thus:

0HM V =⋅

.

The point M is a point of (D):

, OH Vα α= ∈

.

To derive the position of the point H, we have to obtain the expression of its

position vector as a function of the data of the problem: direction vector V

and

position vector OM

of the point M.

The first relation is written:

( ) 0 or 0OM OH V OM V OH V− = − =⋅ ⋅ ⋅

.

Hence by introducing the expression of OH

, we have:

20OM V Vα− =⋅

.

Hence the expression of α :

2

OM V

Vα =

⋅

,

and the expression of the position vector OH

:

2

OM VOH V

V=

⋅

.

If the vector V

is the unit direction vector u

of the line (D): 2 1u =

, the

expression of OH

is reduced to:

( )OH OM u u= ⋅

.


The vector V

has for components (1, −2, 3) and the position vector OM

has

for components (x, y, z), the coordinates of the point M. Hence:

214 and 2 3V OM V x y z= = − +⋅

.

The coordinates ( , , )H H Hx y z of the point H are thus:

1 1 3( 2 3 ), ( 2 3 ), ( 2 3 ).

14 7 14H H Hx x y z y x y z z x y z= − + = − + − = − +

Solution of Exercise 2.2 487

2.2 Straight line passing through the point A and orthogonal to the plane passing

through the points A, B and C.

The line may be defined by ( ), A u

where u

is the unit vector of the direction

orthogonal to the plane (A, B, C).

Two direction vectors of the plane are given by AB

and AC

image vectors of

the respective bipoints (A, B) and (A, C). An orthogonal vector is given by

AB AC×

. The vector u

is a unit vector collinear to the vector product AB AC×

.

We are thus brought back to Exercice 1.1. Hence:

AB ACu

AB AC

×=

×

,

taking the + determination of the vector.


A (–1, 2, 1), B (2, 3, –1), C (–3, 4, –2).

Hence:

AB =

(3, 1, –2), AC =

(–2, 2, –3),

and

( ) + 1, 13, 8 , + 234AB AC AB AC= =

.

The vector u

is thus:

1 13 8

234 234 234u i j k= + +

.

It is then easy to find the Cartesian equations of the line.

2.3 So that the triangle ABC is isosceles and right-angled at A, we have to show

that the edges AB and AC are equal and are orthogonal. Or:

? ? .and 0AB ACAB AC == ⋅

Derivation of the vectors AB

and AC

:

( ) ( )1 2, 2, 1 2 , 1 2, 2, 1 2 .AB AC= − + = − − − − +

Orthogonality:

( )( ) ( )( )1 2 1 2 2 1 2 1 2 .AB AC = − − − − + + − +⋅

Hence: 0.AB AC =⋅

The triangle is thus right-angled at A.

Equality of the edges:

( ) ( )

( ) ( )

2 2

2 2

1 2 2 1 2 ,

1 2 2 1 2 .

AB AB

AC AC

= = − + + +

= = − − + + − +

Thus: AB AC=

.

The two edges of the triangle are equal. So it is isosceles and right-angled at A.


Figure Exercise 2.4.

2.4 Area of the triangle ABC

The area of the triangle is: 1

2S AC BH= , where BH is the height of the

triangle (Figure Exercise 2.4).

The height is expressed as: sinBH AB α= , where α is the angle at A.

From Expression (2.14) of the vector product, we have:

sin ,AB AC u AB AC α× =

where u

is the unit vector orthogonal to the vectors AB

and AC

. It results that:

1

2S AB AC= ×

.


A (–1, –2, –1), B (2, 2, –1), C (3, 2, 1).

Hence:

AB =

(3, 4, 0), AC =

(4, 4, 2), AB AC∧ =

(8, –6, –4).

Thus:

116S = .

2.5 The volume V of the parallelepiped (Figure Exercise 2.5) is expressed as:

area of basis height .V DH= ×


A

B

C H

α

A

B

C

D

H

β

Solution Exercise 2.5 489

From the preceding exercise, the area of the basis is: S AB AC= ×

.

The height is expressed by cosDH AD β= , where β is the angle between the

bipoints (D, A) and (D, H). Hence the expression of the volume:

cosV AB AC AD β= ×

.

Thus, from the expression (2.13) of the scalar product:

( ) V AB AC AD= × ⋅

.

The volume of the parallelepiped is thus equal to the mixed product of the

vectors images of the bipoints on which the parallelepiped is constructed.


A (0, 0, 0), B (3, 2, 1), C (1, 1, 2), D (−1, −1, 2).

Hence:

AB =

(3, 2, 1), AC =

(1, 1, 2), AD =

(−1, −1, 2),

AB AC× =

(3, −5, 1), ( ) 4AB AC AD× =⋅

.

Hence the volume: 4V = .

2.6 The distance d from the point D to the plane passing through the three points

A, B and C is the height of the parallelepiped constructed on the bipoints (A, B),

(A, C) and (A, D). The results of the preceding exercises lead to:

( )

volume

area of basis

AB AC ADd

AB AC

×= =

×

⋅

.


A (0, 0, 0), B (1, 2, 3), C (2, 1, 1), D (−2, −1, −3).

Hence:

AB =

(1, 2, 3), AC =

(2, 1, 1), AD =

(−2, −1, −3),

AB AC× =

(−1, 5, −3), 35AB AC× =

, ( ) 6AB AC AD× =⋅

.

The distance is thus: 6

35d = .

2.7 The necessary and sufficient condition so that the four points A, B, C and D

are contained in the same plane is that the point D is contained in the plane

passing trough the points A, B and C. Thus, the distance from the point D to the

plane is zero. From the preceding exercise, the condition is:

( ) 0AB AC AD× =⋅

.

2.8 The two rotations considered are reported in Figure Exercise 2.8.

2.8.1. First rotation

( ) 1 1 1/ , ,O i j k

( ) 1 3 2/ , ,O i j k

. ( )1, 30i °




1

3 1 1

2 1 1

,

cos30 sin 30 ,

sin 30 cos30 ,

i

j j k

k j k

= ° + ° = − ° + °

or

1

3 1 1

2 1 1

,

3 1 ,2 2

31 .2 2

i

j j k

k j k

= + = − +

Second rotation

( ) 1 3 2/ , ,O i j k

( ) 2 2 2/ , ,O i j k

.


2 1 3

2 1 3

2

cos 45 sin 45 ,

sin 45 cos 45 ,

,

i i j

j i j

k

= ° + °

= − ° + °

or

2 1 3

2 1 3

2

2 2 ,2 2

2 2 ,2 2

.

i i j

j i j

k

= +

= − +

z1

x1

O

1i

y1

y3

2k

x2

z2y2

2i 1j3j

2j

1k

45°

45°

30°

30°

45°

30°

( )2 , 45k °


Thus, by expressing the vectors 3j

and 2k

:

2 1 1 1

2 1 1 1

2 1 1

62 2 ,2 4 4

62 2 ,2 4 4

31 .2 2

i i j k

j i j k

k j k

= + +

= − + +

= − +

The matrix of the basis change is thus:

2 6 2

2 4 4

2 6 2

2 4 4

1 30

2 2

= − −

A .

2.8.2. The point M has for Cartesian coordinates (−1, 2, 4) relatively to the system

( )1 1 1(1) Ox y z= . Its Cartesian coordinates relatively to the system ( )2 2 2(2) Ox y z=

are given from Relation (2.51) by:

( )

( )

( )

( )

( )

( )

(2) (1)

(2) (1)

(2) (1)

x M x M

y M y M

z M z M

=

A ,

or

( )

( )

( )

(2)

(2)

(2)

62 22 4 4 1

62 2 22 4 4

4310

2 2

x M

y M

z M

− = − −

.

The coordinates of the point M in the system (2) are thus: (1.32, 3.346, 2.464).

2.8.3. The point N has for Cartesian coordinates (3, −4, 8) relatively to the system

( )2 2 2(2) Ox y z= . From Relation (2.51), its coordinates in the system (1) are:

( )

( )

( )

(1)

(1)

(1)

2 2 02 2 36 6 1 4

4 4 28

32 24 4 2

x M

y M

z M

− = − −

.

The coordinates of the point N in the system (1) are thus: (4.950, −4.612, 6.575).

Chapter 4

Elementary Concepts on Curves

4.1 The position vector of a point M of the curve (C) is expressed by:

( )3 3sin cos cos 2OM a i q j q k q= + −

,

with 0a > and 02

qπ

< < .

4.1.1. Unit direction vector of the tangent

A direction vector V

of the tangent is obtained by deriving the position vector

OM

with respect to the parameter q. Thus:

( )2 2d3 sin cos 3 cos sin 2 sin 2

d

OMV a i q q j q q k q

q= = − −

,

or

( )dsin cos 3 sin 3 cos 4

d

OMV a q q i q j q k

q= = − −

.

From Exercise 1.1, the unit vector te

of the tangent is given by:

tV

eV

= ±

,

with

2 2sin cos 9sin 9cos 16 5 sin cosV a q q q q a q q= + + =

.

Hence:

( )13 sin 3 cos 4

5te i q j q k= ± − −

.

Hereafter, we shall take the + determination which orientates the curve in the

direction of increasing q.

4.1.2. Curvilinear abscissa

The curvilinear abscissa s can be introduced while writing:

d d d

d d d

OM OM s

q s q=

.

Thus from (4.9):

d d

d dt

OM se

q q=

.

The expressions of d

d

OM

q

and te

obtained previously lead then:

d5 sin cos

d

sa q q

q= ,

or integrating:


0

0( ) ( ) 5 sin cos dq

q

s q s q a q q q− = .


( )2 20 0

5( ) ( ) sin sin

2s q s q a q q− = − .

By taking as the origin of the curvilinear abscissae the point M0 where the point

point M is located when 0( ) 0,s q = the curvilinear abscissa is given by:

25( ) sin

2s q a q= .

4.1.3. Unit vector of the principal normal direction and radius of curvature

They are expressed by Relation (4.11), with here:

( )d d d 1 13 cos 3 sin

d d d 5 5 sin cost te e q

i q j qs q s a q q

= = +

.

Hence:

( )3cos sin

25 sin cosne

i q j qa q q

= +

.

Thus, we deduce:

cos sin ,

25sin cos .

3

ne i q j q

a q q

= +

=

The vector ne

and the radius of curvature were already separated.

4.1.4. Frenet basis The third vector of the Frenet basis is obtained by Relation (4.12). Thus:

( ) ( )13 sin 3 cos 4 cos sin

5be i q j q k i q j q= − − × +

.

That leads to:

( )14 sin 4 cos 3

5be i q j q k= − +

.

Chapter 5

Torsors

5.1 Let ( ) D be the torsor associated to the field of sliders defined on the

domain (D) constituted of the four points M1, M2, M3 and M4.

5.1.1. Resultant of the torsor ( ) R D

( ) 4

1

i

i

R D R

=

= .

Thus:

( ) 8 2 4R D i j k= + +

.

5.1.2. Moment of the torsor at the point OGenerally the calculation of the moment is easier at the origin of the coor-

dinates. Thus:

( ) 4

1

iiO

i

D OM R

=

= × ,

with

11

22

33

44

15 10 ,

2 8 ,

6 15 ,

2 6 .

OM R j k

OM R i k

OM R i j

OM R i k

× = +

× = − +

× = − −

× = −

Hence:

( ) 6 12O D i k= − +

.

5.1.3. Caracterization of the torsorThe scalar invariant of the torsor is:

( ) ( ) ( ) OI D R D D= ⋅

.

Calculation leads to ( ) 0I D = . The resultant being different from the null

vector, the torsor is a slider.

5.1.4. Moment of the torsor at a point PThe point P has for coordinates (x, y, z) relatively to the system (Oxyz). Hence:

OP x i y j z k= + +

.

The moment at the point P is given by:

( ) ( ) ( ) P OD D R D OP= + × ,

with

( ) ( ) ( ) ( )2 4 4 8 8 2R D OP z y i x z j y x k× = − + − + −

.

Hence:

( ) ( ) ( ) ( )2 2 3 4 2 2 4 6P D y z i x z j x y k= − + − + − + − + −

.


5.1.5. Axis of the torsorThe torsor being a slider, it has an axis of null moments, set of the points at

which the moment of the torsor is null. Thus, the set of the points P such as:

( ) 0P D = .

That leads to

2 3 0,

2 0,

4 6 0.

y z

x z

x y

− + − =

− =

− + − =

A straight line is defined by two equations. Among the three preceding equations,

one of the equations is a linear combination of the two others. For example, the

equations of the axis will be:

2 3 0,

2 0.

y z

x z

− + − =

− =

The axis can be determined by two points A and B. For example the point A for

which 0x = and the point B for which 0y = . We obtain:

A (0, 32

− , 0) and B (6, 0, 3).

The axis can also be defined by a point (A for example) and a direction vector:

the resultant of the torsor 8 2 4i j k+ +

.

5.2 On the same domain (D) as the preceding exercise, it is defined a new field

of sliders. And let ( ) D be the torsor associated to this new field.

5.2.1. Resultant of the torsor ( ) R D

( ) 4

1

i

i

R D R

=

= .

Thus:

( ) 0R D = .

5.2.2. Moment of the torsor at the point O

( ) 4

1

iiO

i

D OM R

=

= × ,

with

11

22

33

44

300 200 ,

100 200 800 ,

100 50 200 ,

0.

OM R j k

OM R i j k

OM R i j k

OM R

× = +

× = − + −

× = − −

× =

Thus:

( ) 450 800O D j k= −

.


5.2.3. Caracterization of the torsor The torsor having a null resultant and a non null moment is a couple-torsor. Its

moment is independent of the point:

( ) 450 800 , P D j k P= − ∀

.

5.2.4. Resolution of the couple-torsor at the point O

The couple-torsor ( ) D can be resolved into two sliders ( ) 1 D and

( ) 2 D . Thus:

( ) ( ) ( ) 1 2D D D= + .

The resolution being implemented at the point O, we have:

( ) ( ) ( )

( ) ( ) ( ) 1 2

1 2

,

.O O O

R D R D R D

D D D

= +

= +

We choose the slider ( ) 1 D so that its axis passes through the point O. Thus:

( ) 1 0O D = .

Hence the elements of reduction at the point O of the slider ( ) 2 D :

( ) ( )

( ) ( )

2 1

2

,

450 800 .O O

R D R D

D D j k

= −

= = −

The sliders are then completely determined by choosing a resultant for one of the

sliders.

5.3 Let ( ) D be the new torsor associated to the field of sliders defined on

the domain (D).

5.3.1. Resultant

( ) 8 2 5R D i j k= − +

.

5.3.2. Moment at the point O

11

22

33

44

10 13 2 ,

2 8 ,

6 15 ,

2 6 .

OM R i j k

OM R i k

OM R i j

OM R i k

× = + +

× = − +

× = − −

× = −

Hence:

( ) 4 2 4O D i j k= − +

.

5.3.3. Caracterization of the torsor The scalar invariant of the torsor is:

( ) ( ) ( ) 56OI D R D D= =⋅

.


The scalar invariant is different from zero. Thus it results that the torsor is arbi-

trary.

5.3.4. Moment at a point P of coordinates ( ), , x y z

( ) ( ) ( ) P OD D R D OP= + × ,

with

( ) ( ) ( ) ( )2 5 5 8 8 2R D OP z y i x z j y x k× = − − + − + +

.

Hence:

( ) ( ) ( ) ( )5 2 4 5 8 2 2 8 4P D y z i x z j x y k= − − + + − − + + +

.

5.3.5. Central axis of the torsor It is the set of the points P such as the moment of the torsor at the point P is

collinear to the resultant:

( ) ( ) , P D R Dα α= ∀ ∈

.

That leads to:

5 2 4 8 ,

5 8 2 2 ,

2 8 4 4 .

y z

x z

x y

α

α

α

− − + =

− − = −

+ + =

The equations of the axis are obtained by eliminating α. Thus:

20 5 34 4 0,

12 8 16 2 0.

x y z

x y z

− − − =

+ − + =

The axis can be eventually defined by two points of coordinates deduced from the

preceding equations.

5.3.6. Resolution of the torsor at the point O

The arbitrary torsor ( ) D is resolved into a slider ( ) 1 D and a couple-

torsor ( ) 2 D . Thus:

( ) ( ) ( ) 1 2D D D= + .

The resolution being implemented at the point O, we have:

( ) ( ) ( )

( ) ( ) ( ) 1 2

1 2

,

.O O O

R D R D R D

D D D

= +

= +

The couple-torsor is such as ( ) 2 0R D = and the slider is chosen so that its axis

of null moments passes through the point O. It results that the slider has for

elements of reduction:

( ) ( )

( )

1

1

8 2 5 ,

0,O

R D R D i j k

D

= = − +

=

and the couple-torsor is such as:


( )

( ) ( )

2

2

0 ,

4 2 4 , .P O

R D

D D i j k P

=

= = − + ∀


5.4 We consider then the domain (D) constituted of a rectangular area. To every

area element surrounding the point M it is associated a slider of vector density

( )p M i

:

( )M D∀ ∈ slider of resultant d ( ) ( ) d ( )R M p M i S M=

and of axis ( ), M i

.

Let ( ) D be the torsor associated to this field of sliders.

5.4.1. Resultant The resultant is expressed by:

( ) ( )

d ( )D

R D R M= .

Considering the Cartesian coordinates, the area element surrounding the point M

is an elementary rectangle of edges d and dx y : d ( ) d dS M x y= . The resultant is

then expressed as:

( )

0 0

( )d da b

x y

R D i p M x y= =

= ,

where a and b are the respective edges of the rectangle. The integral depends on

the expression of ( )p M .

In the case of a constant function for ( )p M : 0( )p M p= , the resultant is

simply expressed as:

( ) 0R D p S i= ,

where S is the area of the rectangle: S ab= .

5.4.2. Moment of the torsorThe moment of the torsor is determined while expressing this moment at a

point. It is also possible to search for a point where the moment is known. We are

here in the important particular case considered in Subsection 5.3.3: the resultant

of the slider associated to the element surrounding the point M is a vector whose

the direction i

is independent of the point M. In this case there exists a measure

centre H and the moment at this point is null. By searching this point H, we

answer in search for the moment of the torsor since:

( ) 0H D = .

Furthermore in this case, the resultant is expressed (5.66) as:

( ) ( )R D D iµ= ,

where ( )Dµ is the measure of the domain (D), associated to the field of sliders

under consideration:

0 0

( ) ( ) d da b

x y

D p M x yµ= =

= .

If ( )p M is constant over the domain (D), we have:

0( )D p Sµ = .


The position of the point H is given by Expression (5.72). That leads here to:

( )

0 0

1( ) d d

( )

a b

x y

OH x i y j p M x yDµ = =

= +

.

The coordinates ( ), , 0H Hx y of the point H are then deduced from this expres-

sion and are expressed as:

0 0

0 0

1( ) d d ,

( )

1( ) d d .

( )

a b

Hx y

a b

Hx y

x x p M x yD

y y p M x yD

µ

µ

= =

= =

=

=

If p(M) is constant over the domain (D):

0 0

0 0

1d d ,

1d d .

a b

Hx y

a b

Hx y

x x x yS

y y x yS

= =

= =

=

=

The point H coincides with the centre of the rectangle.

Having obtained the position of the measure centre H, it is then possible to

derive the moment at an arbitrary point using the expression:

( ) ( ) P D R D HP= × .

Chapter 6

Kinematics of Point

6.1 The Cartesian coordinates of the point M are given by:

( )2 2 3 33 , 3 , 0,x a t y a t t zω ω ω= = − =

where a and ω are positive constants.

6.1.1. Plot of the trajectory of the point M for 0t ≥ .

Some general characteristics of the curve can be first derived.

The trajectory is plane, contained in the plane 0z = .

At 0t = , the point M is at the origin ( )0, 0, 0x y z= = = . The trajectory

intersects the x-axis for 0y = , hence for 2 2 3tω = . The abscissa along this axis is

then: 9x a= .

The tangent at the point M to the curve has a slope given by:

2 2d

d 1ddd 2

d

yy tt

xx t

t

ωω

−= = .

For 0tω = , the tangent is parallel to the y-axis. The tangent is parallel to the x-

axis for 1tω = .

Some particular points can then be obtained. Hence the table:

tω 0 1/2 1 3 2

/x a 0 3/4 3 9 12

/y a 0 11/8 2 0 –2

The curve can then be plotted using a general-purpose software package

(Figure Exercise 6.1).

6.1.2. The position vector of the point M is given by:

( )2 2 3 33 3OM a t i t t jω ω ω = + −

.

1. Velocity vector It is expressed by:

( )( )

( )2 3 2d( , ) 6 3 3

d

TT OM

M t a t i t jt

ω ω ω = = + −

.

Hence: ( ) ( )2 2( , ) 3 2 1T M t a t i t jω ω ω = + −

.

2. Instantaneous algebraic velocity The velocity is given by:

( )( , )T

tM t e= ,

where te

is the unit vector of the tangent to the trajectory at the point M.


0 2 4 6 8 10 12 14-4

-3

-2

-1

0

1

2

/x a

/y

a


By taking the norm of the velocity vector, we obtain: ( )

( , )T M t=

,

with

( ) ( ) ( )22 2 2 2 2 2( , ) 3 4 1 3 1T M t a t t a tω ω ω ω ω= + − = + .

Wence the instantaneous algebraic velocity:

( )2 23 1a tω ω= ± + .

A plus determination corresponds to the case where the trajectory is orientated in

the sense of the motion of the point M. A minus sign to the contrary orientation.

3. Acceleration vector It is expressed by:

( )( )

( )

d( , ) ( , )

d

TT Ta M t M t

t=

.

Hence: ( ) ( )2( , ) 6Ta M t a i t jω ω= −

.

4. Tangential and normal components of the acceleration vector We have:

( ) ( , )T

t t n na M t a e a e= +

,

while introducing the tangential component:

d

dta

t=

,

and the normal component:


2

na =

,

where is the radius of curvature of the trajectory at the point M.

The tangential component is then:

36ta a tω= .

The radius of curvature being not known, it is not possible to deduce the normal

component from the previous expression. So, we use the expression of the accele-

ration vector while expressing its norm:

( )

2 2( , )Tt na M t a a= +

.

Thus:

( )2 4 2 2 2 6 2 236 1 36 na t a t aω ω ω+ = + or 2 2 436na a ω= .

Hence:

26na aω= ± .

The component an being positive, we deduce then:

26na aω= .

5. Radius of curvature

It is expressed by: 2

na=

.

Hence: ( )22 231

2a tω= + .

6. Curvilinear abscissa The curvilinear abscissa at time t can be derived from the expression of the

instantaneous algebraic velocity:

d

d

s

t= .

Thus integrating between the instants t0 and t:

( )

0 0

2 20( ) ( ) d 3 1 d

t t

t t

s t s t t a t tω ω− = = + .

We obtain:

( ) ( )2

3 30 0 0( ) ( ) 3

3s t s t a t t t t

ωω

− = − + − .

Taking the origin of the curvilinear abscissae for 0t = ( 0( 0) 0s t = = ), we have:

2 2

( ) 3 13

ts t a t

ωω

= +

.

6.2 A cyclist, a car and a truck move between the cities A and B distant of 160

km (Figure Exercise 6.2).



6.2.1. Equations of the motionsThe motions of the cyclist, car and truck are considered as being averagely

uniform between the two cities. Thus:

dcst , cyclist 1, car 2, truck 3,

di

is

i i it

= = = = =

where is is the curvilinear abscissa of the moving body i and i its average alge-

braic velocity. Integrating between the instants t and 0it , we obtain:

( )0 0( ) ( )i i i i is t s t t t− = − .

Hence:

( )0 0( ) ( )i i i i is t t t s t= − + .

1. The cyclist leaves the city A at 8 h and moves towards the city B at the average

speed of 30 km/h.

We orientate the trajectory from A towards B. Thus, it follows that:

1 30 km/h= .

Taking the origin of the abscissae at the point A, we have:

1 01 01( ) 0 with 8 h.s t t= =

The equation of the motion of the cyclist is then:

( )1 1 01 1 01( ) , 30 km/h, 8 hs t t t t= − = = .

2. The car leaves the city A at 9 h and moves towards the city B at the average

speed of 85 km/h. By analogy with the preceding motion, the equation of the

motion is:

( )2 2 02 2 02 02( ) , 85 km/h, 9 h, s t t t t t t= − = = > .

3. The truck leaves the city B at 9h30 and moves towards the city B at the average

speed of 60 km/h. We obtain easily:

( )3 3 03

3 03 03

( ) ,

with 160 km, 60 km/h, 9h30, .

B

B

s t s t t

s t t t

= + −

= = − = >

6.2.2.

1. The car draws ahead of the cyclist when 1 2( ) ( )s t s t= . Thus:

( ) ( )1 01 2 02t t t t− = − .

A

B


Hence:

2 02 1 01

2 1

t tt

−=

−

.

We obtain: 9h32min 44st = .

The place where the car passes the cyclist is given by:

2 02 1 011 1 01

2 1

( )t t

s s t t− = = − −

.

Thus: 46.364 kms = .

2. The truck meats the cyclist when 3 1( ) ( )s t s t= . Hence:

( ) ( )3 03 1 01Bs t t t t+ − = − .

Hence:

3 03 1 01

1 3

Bs t tt

− +=

−

.

We obtain: 10h46min 40st = .

The place of the meeting is given by:

3 03 1 011 1 01

1 3

( ) Bs t ts s t t

− + = = − −

.

Hence: 83.333 kms = .

3. The truck meets the car when 3 2( ) ( )s t s t= . By analogy with the preceding

results, we have:

3 03 2 02

2 3

Bs t tt

− +=

−

,

3 03 2 022 2 02

2 3

( ) Bs t ts s t t

− + = = − −

.

Hence at the time 10h18min 37st = , at the distance 111.379 km from the city A.

Chapter 7

Study of Particular Motions

7.1 Performances of a car7.1.1. The performances are established on a car-track of high radius of curvature

and the different stages of the motions are considered as being, on average,

uniformly accelerated. From Section 7.1.3, the equation of the different stages of

the motions can be written in the form:

( ) ( )200 0 0 0

2

as t t t t s= − + − + ,

The car starting at 0t from the position 0s with the velocity 0 , and where 0a is

the average acceleration. The velocity is given by:

( )0 0 0a t t= − + .

Initial accelerations The car starts at time 0 0t = , from the position 0 0s = with a velocity 0 0= .

The equation of the motion is:

20

1

2s a t= ,

and the velocity is:

0a t= .

This latest relation allows us to determine the average acceleration 0a , then to

obtain the distance s covered during the acceleration:

01

, 2

a s t= =

,

where t is the duration of the acceleration and the velocity reached at the end of

the acceleration.

Acceleration stages The car passes at the time 0 0t = at the position 0 0s = while accelerating from

the velocity 0 . The equation of the motion is thus:

20 0

1

2s a t t= + ,

and the velocity is:

0 0a t= + .

This expression allows us to obtain the average acceleration 0a , then the distance

s covered during the acceleration:

( )00 0

1,

2a s t

−= = +

,

where t is the duration of the acceleration and the velocities 0 and at the

beginning and at the end of the acceleration are known.


The results obtained for the initial accelerations and the acceleration stages are

reported in the following table.

Table of the accelerations

time

(s)

average

acceleration a0

(m/s2)

covered

distance s

(m)

Initial accelerations

0 to 60 km/h 6.4 2.60 53.3

0 to 80 km/h 10.5 2.12 116.7

Acceleration stages

30 to 100 km/h in 4th 21.6 0.900 390.0

in 5th 30.0 0.648 541.7

40 to 100 km/h in 4th 18.7 0.891 363.6

in 5th 26.4 0.631 513.3

80 to 100 km/h in 3rd 5.7 0.974 142.5

in 4th 6.9 0.805 172.5

in 5th 9.5 0.585 237.5

80 to 120 km/h in 4th 14.6 0.761 405.6

in 5th 18.4 0.604 511.1

7.1.2. Succession of stages:

a. acceleration from 0 to 80 km/h: time 10.5 s and covered distance 116.7 m;

b. 80 to 100 km/h in 3rd: time 5.7 s and covered distance 142.5 m;

c. 100 to 120 km/h in 4th: average acceleration 20 0.761 m/sa = ;

d. beyond 120 km/h in 5th: average acceleration 20 0.604 m/sa = .

7.1.2.1. Time and distance necessary to reach the speed of 100 km/h

This acceleration phase corresponds to the stage a followed by the stage b.

Thus, the time is 12.6 s and the distance is 259.2 m.

Time and distance necessary to reach the speed of 120 km/h: stages a, b and c.

For the stage c, we have: 0 0a t= + , and the duration of the stage c is:

0

0

ta

−=

,

with 20 0.761 m/sa = , 0 100 km/h= and 120 km/h= . Hence:

stage c 7.3 st = ,

and the covered distance is given by:

( ) ( ) 00 0

0

1 1

2 2s t

a

−= + = +

.

Hence:

stage c 223.1 mx = .

Thus, the time and the distance necessary to reach 120 km/h are respectively:

23.5 s and 482.3 m.


7.1.2.2. Time and speed reached after 1 000 m

To the stages a, b and c, we have to add the stage with an average acceleration

0a = 20.604 m/s . Taking the origin of times and distances at the beginning of

the stage d, the equation of the motion is:

20 0

1

2s a t t= + .

The distance which remains to be covered is 517.7 m. The duration t of the stage d

is thus solution of:

2 1200.302 517.7 0

3.6t t+ − = .

Solving this equation leads to 13.8 st = .

The time necessary to cover the distance of 1,000 m is thus 37.3 s.

The velocity reached is given by 0 0a t= + , that leads to 150.007 km/h.

7.2 The coordinates of the point M are expressed (7.51) by:

200 00, cos , sin .

2

ax y t z t tϕ ϕ= = = −

7.2.1. If 2

πϕ = , the trajectory is a straight line which reaches a maximum along z:

20

max02

za

=

.

If 2

πϕ ≠ , the trajectory of the point M is:

( )2 2020

1 tan tan2

az y yϕ ϕ= − + +

.

For a given value of the initial velocity 0 , the trajectories depend on the parameter

tanϕ .

So that a point P of the plane (Oxyz) can be a position of the point M, it is

necessary that its coordinates yP and zP satisfy the inequality:

20 02

0022

P Pa

z ya

≤ − +

.

Thus, the point P has to be located inside the parabola of equation:

2 420 0 0 0

2 2 200 0 0

22 2

a az y y

a a

= − + = − −

.

This parabola is envelope of the parabolic trajectories when tanϕ varies. This

envelope has for axis the axis Oz

, its maximum along this axis is 20

02a

and this


envelope is reached by the trajectory at the point of coordinate20

0 tanz

a ϕ=

.

7.2.2. Any point Q inside this envelope has its coordinates related by:

( )2

0 2

20

1 tan tan2

QQ Q

a yz yϕ ϕ= − + +

,

where the coordinates yQ and yQ are given. This equation has two roots and these

roots determine the angles 1ϕ and 2ϕ for which it is possible to reach the point Q.

If the point Q is located along the axis Oy

( 0)Qz = , the preceding equation is

written:

( )2

0 2

20

1 tan tan 02

QQ

a yyϕ ϕ− + + =

,

or 2 2

0 02

2 20 0

tan tan 02 2

Q QQ

a y a yyϕ ϕ− + =

.

This equation has two roots 1tanϕ and 2tanϕ , such as:

1 2tan tan 1ϕ ϕ = .

The angles 1ϕ and 2ϕ are thus related by:

1 22

πϕ ϕ+ = .

To reach a given horizontal point, there exist thus two shoot angles, a low angle

and a high angle. These two angles differ from 90°.

Chapter 9

Kinematics of Rigid Body

9.1 Motion of a parallelepiped on a plane (Figure 9.10)

The analysis of the kinematics of a solid is always implemented according the

same process: 1) determination of the parameters of situation, 2) evaluation of the

kinematic torsor, 3) derivation of the kinematic vectors of particular points.

1. Parameters of situation The determination of the parameters of situation is also always carried out

using the same process. Here we associate the Cartesian system (Oxyz) to the

plane (T), such that the plane (Oxy) coincides with the plane (T) (Figure Exercise

9.1). Next, we search for the parameters of translation, and then for the parameters

of rotation.

1.1. Parameters of translation

To find the parameters of translation, we have to choose a particular point of the solid. This particular point must have the smallest number ( 3≤ ) as possible of coordinates depending of time. For example it is possible to consider if there exists a fixed point. If so, the particular point is the fixed point. If not, we search for a point which has only one coordinate depending of time. Etc.

Here, all the points of the solid (S) have two coordinates depending of time.

We choose the point A of the solid (S). Its Cartesian coordinates relatively to the

system (Oxyz) are (x, y, 0). Hence the position vector of the point A:

OA x i y j= +

.

The motion has two parameters of translation: x, y.

1.2. Parameters of rotation

To find the parameters of rotation, we associate a trihedron attached to the solid

(S). Thus the system ( )S SAx y z such as the axes SAx

and Ax

pass respectively


(T)

(S)

A

B

C

D A'

B'

C'

D'

O

x

x

xS

yS

y z z

ijk

Si

Sj

511 Chapter 9 Kinematics of Rigid Body

through the vertices B and D. The orientation of the solid (S) is then defined by

the rotation ψ about the direction ,k

where ψ is the angle between the axes SAx

and Ax

.

Next, we write the relation of basis change:

cos sin ,

sin cos ,

.

S

S

i i j

j i j

k

ψ ψ

ψ ψ

= +

= − +

Finally, the motion of the parallelepiped is characterized by two parameters of

translation x and y, and one parameter of rotation , thus three parameters of

situation in total.

2. Kinematic torsor

Let ( ) TS be the kinematic torsor associated to the motion of the parallele-

piped (S) on the plane (T). It is defined by its elements of reduction at the point A:

( ) ( )

( ) ( )

, instantaneous rotation vector relative to the motion

of the solid ( ) with respect to the reference ( ),

( , ) , velocity vector of the particular point .

T TS S

T TA S

R

S T

A t A

ω =

=

Note that we must consider the moment of the kinematic torsor at the particular

point for which the parameters of translation have been defined.

We have a rotation about the direction k

. Hence:

( )TS kω ψ=

.

The velocity vector of the point A is given by:

( )( )

d( , )

d

TT A t OA

t=

.

Thus, considering the expression of the position vector of A: ( )

( , )T A t x i y j= + .

Hence the elements of reduction of the kinematic torsor:

( ) ( )

( ) ( )

,

( , ) .

T TS S

T TA S

R k

A t x i y j

ω ψ = =

= = +

The resultant (the rotation vector) of the kinematic torsor does not depend upon

the parameter of rotation, when the moment (the velocity vector of the point A)

depends only on the parameters of translation.

3. Kinematic vectors of a point of the solid, for example the point ′C

3.1. Velocity vector

It is derived from: ( ) ( ) ( )

( , ) ( , )T T TSC t A t ACω′ ′= + ×

.

This expression is deduced from the expression of the moment at C′ of the kine-

matic torsor. If a, b and c are the respective lengths of the edges of the solid:


S SAC a i b j c k′ = + +

,

We obtain: ( )

TS S SAC b i a jω ψ ψ′× = − +

.

Hence: ( )

( , )TS SC t x i y j b i a jψ ψ′ = + − +

.

We have to transform the vectors Si

and Sj

using the relations of basis change.

We obtain: ( ) ( )[ ] ( )[ ] ( , ) sin cos cos sinT C t x a b i y a b jψ ψ ψ ψ ψ ψ′ = − + + + +

.

3.2. Acceleration vector

It is possible either to differentiate the preceding expression, or to use the

composition (9.24) of the acceleration vectors. Deriving, we have:

( )

( )( )

d( , ) ( , )

d

TT Ta C t C t

t′ ′=

.

Hence: ( )

( ) ( )[ ]

( ) ( )

2

2


cos sin sin cos .

Ta C t x a b a b i

y a b a b j

ψ ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ ψ

′ = − + − −

+ + − − +

9.2 Motion of a cylinder on a plane (Figure 9.11)

The cylinder remains in contact with the plane (T) during its motion. Thus, it

can move on the plane (T) while sliding, rolling and spinning (the motions of

sliding, rolling and spinning will be considered in the next chapter).

The process of analysis is similar to the process used in the preceding exercise.

1. Parameters of situation 1.1. Parameters of translation

We choose first a coordinate system attached to the plane (T), hence (Oxyz)

such as the plane (Oxyz) coincides with the plane (T) (Figure Exercise 9.2). Next,

we choose a particular point of the cylinder with the smallest number of coordi-

nates depending on time. We may choose one of the axis of the cylinder. For

example, the centre A of one of the sides. The coordinates of the point A relatively

to the system (Oxyz) are (x, y, a), where a is the radius of the cylinder. The

position vector is:

OA x i y j a k= + +

.

The motion has thus two parameters of translation: x, y.


To find the parameters of rotation, we associate a trihedron attached to the

cylinder. This trihedron is obtained considering first the motion of spinning of the

cylinder, then the motion of rolling on the plane.

We consider first the motion of spinning of the cylinder. The cylinder is sub-

mitted to a rotation about the direction k

. To describe this motion, we associate

to the cylinder the trihedron ( )3SAx y z for which the axis SAx

coincides with the

Solution exercice 9.1 513


axis of the cylinder (Figure Exercise 9.2). The cylinder is submitted to a rotation

of angle about the direction k

. The basis change is:

3

cos sin ,

sin cos ,

.

Si i j

j i j

k

ψ ψ

ψ ψ

= +

= − +

Next, we consider the motion of rolling on the plane. We associate then to the

cylinder the trihedron ( )S S SAx y z obtained by the rotation of angle about the

direction Si

(Figure Exercise 9.2). The basis change is:

3

3

,

cos sin ,

sin cos ,

S

S

S

i

j j k

k j k

θ θ

θ θ

= + = − +

We have thus four parameters of situation: two parameters of translation x, y,

and two parameters of rotation , .

2. Kinematic torsor

Let ( ) TS be the kinematic torsor associated to the motion of the cylinder (S)

on the plane (T). Its elements of reduction at the point A are:

( ) ( )

( ) ( )

, instantaneous rotation vector relatively to the motion

of the solid ( ) with respect to the reference ( ),

( , ) , velocity vector of the particular point .

T TS S

T TA S

R

S T

A t A

ω =

=

Thus: ( ) ( )

( ) ( )

, a rotation about

and a rotation about ,

( , ) .

T TS S S

S

T TA S

R k i k

i

A t x i y j

ω ψ θ ψθ

= = +

= = +

(T)

(S)

A

B

x

x

xS

yS y z

O

z

ijk

Si

Sj

3j

Sk

y3

zS

I


The instantaneous rotation vector depends only on the parameters of rotation and

the velocity vector depends only on the parameters of translation.

3. Kinematic vectors of a point of the solid As an example, we consider the case of the point B of the cylinder (Figure

Exercise 9.2) contained in the plane ( )S SAx y . Its coordinates relatively to the

system ( )S S SAx y z are: l, a, 0 where l is the length of the cylinder. Thus:

S SAB l i a j= +

.


The expression of the moment at the point B of the kinematic torsor leads to: ( ) ( ) ( )

( , ) ( , )T T TSB t A t ABω= + ×

.

The calculation of the vector product ( )TS ABω ×

is easier in the basis ( ), , S Si j k

.

However, the vector ( )

( , )T A t is expressed in the basis ( ), , i j k

. There exist

then several possibilities to implement the calculation. For example, we may carry

out all the calculations in the basis ( ), , i j k

. In this way, we transform the com-

ponents of the vectors in this basis:

( ) cos sinT

S Sk i i j kω ψ θ θ ψ θ ψ ψ= + = + + ,

( ) ( ) cos sin cos sin cos cos sinS SAB l i a j l a i l a j a kψ ψ θ ψ ψ θ θ= + = − + + +

.

After derivation of the vector product ( )TS ABω ×

, we obtain:

( ) ( )

( )

( , ) sin cos cos sin sin

cos sin cos cos sin cos .

T B t x l a a i

y l a a j a k

ψ ψ ψ θ θ ψ θ

ψ ψ ψ θ θ ψ θ θ θ

= − + +

+ + − − +


The acceleration vector can be obtained either by derivation of the preceding

expression, or using Relation (9.24).

9.3 Motion of two solidsThe process of analysis is again the same.

1. Parameters of situation

1.1. Motion of the solid (S1) with respect to the support (T)

We associate a coordinate system (Oxyz) to the support (T), such as the axis Oy

coincides with the axis (1) and the axis Ox

is downward vertical (Figure Exercise

9.3). The axis Oz

has then the same direction as the axis of rotation (2).

1.1.1. Parameters of translation

We choose a particular point of the solid (S1): the point A1. The coordinates of

the point are (0, y, 0). The solid (S1) has one parameter of translation y.

1.1.2. Parameters of rotation

We associate a trihedron attached to the solid (S1): (A1xyz). This system keeps

the same directions during the motion. There is thus no parameter of rotation.

Solution exercice 9.1 515


1.2. Motion of the solid (S2) with respect to the solid (S1)

A coordinate system is already associated to the solid (S1).


We have to choose a particular point of the solid (S2): the point A1. It is fixed

relatively to the solid (S1). There is thus no parameter of translation.


We associate a coordinate system attached to the solid (S2): the system (A1x2y2z)

such as the axis 21A x

passes through the point A2. The system is obtained from

the system (A1xyz) through a rotation of angle about the direction .k

We have

thus one parameter of rotation . The basis change is:

2

2

cos sin ,

sin cos ,

.

i i j

j i j

k

ψ ψ

ψ ψ

= +

= − +

2. Kinematic torsors


Let ( ) 1

TS be the kinematic torsor relative to the motion of the solid (S1) with

respect to the support (T). Its elements of reduction at the point A1 are:

( ) ( )

( ) ( )

1 1

1 11 1

0, instantaneous rotation vector,

( , ) , velocity vector of point with respect to ( ).

T TS S

T TA S

R

A t y j A T

ω = =

= =

2.2. Motion of the solid (S2) with respect to the solid (S1)

Let ( ) 1

2

S

S be the kinematic torsor relative to the motion of the solid (S2) with

respect to the solid (S1). Its elements of reduction at the point A1 are:

(T) (R) (S1)

(S2)

A2

A1 (1)

(2) y

x

x x2

y2

z

z

d

O


( ) ( )

( ) ( )

1 1

2 2

1 11 2

1 1 1

, instantaneous rotation vector,

( , ) 0, velocity vector of point with respect to ( ).

S S

S S

S SA S

R k

A t A S

ω ψ = = = =


Let ( ) 2

TS be the kinematic torsor relative to the motion of the solid (S2) with

respect to the support (T). Its elements of reduction can be obtained either directly,

or using the relation of combination of motions:

( ) ( ) ( ) 1

2 2 1

ST TS S S

= + .

Thus, we deduce the elements of reduction at the point A1:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

1

2 2 2 1

11 1 12 2 1

1

,

( , ) .

ST T TS S S S

ST T TA A AS S S

R R R k

A t y j

ω ψ = = + = = = + =

3. Kinematic vectors of the point A2

We have:

21 2A A a i=

.


From the expression of the moment at the point A2 of the kinematic torsor, we

obtain:

( ) ( ) ( )

222 1 1( , ) ( , )T T T

SA t A t A Aω= + ×

.

Hence: ( )

2 2( , )T A t y j a jψ= +

,

or considering the basis change:

( ) ( )2( , ) sin cosT A t a i y a jψ ψ ψ ψ= − + +

.


The vector is obtained by deriving the preceding expression. Thus:

( ) ( ) ( )2 22( , ) sin cos cos sinTa A t a i y a jψ ψ ψ ψ ψ ψ ψ ψ = − + + + −

.

Chapter 10

Kinematics of Rigid Bodies in Contact

10.1 Motion of a wheel on a straight line (Figure 10.14)

This motion can also be applied to the motion of a cylinder on a plane, when

there is no spinning of the cylinder.

The kinematic analysis of solids in contact is first implemented by the same

process as the one used in the exercises of the preceding chapter. The conditions

of sliding, spinning and rolling are next considered.

1. Parameters of situation We choose a coordinate system attached to the reference (T) containing the line

(D). Thus the system (Oxyz) such as the axis Ox

coincides with the line (D) and

the plane (Oxy) contains the plane of the wheel (Figure Exercise 10.1).


We choose a particular point of the wheel (S), defined with the smallest

number of parameters depending on time: the centre A of the wheel. The coordi-

nates of the point A relatively to (Oxyz) are (x, a, 0). The position vector is:

OA x i a j= +

,

where a is the radius of the wheel.

We have thus one parameter of translation: x.


We associate a trihedron attached to the wheel: the system ( )S SAx y z (Figure

Exercise 10.1). The orientation of the wheel is then given by the rotation of angle

ψ and direction k

. The basis change is written:


I

A(S)

(D) MO

y

x

z

z

y

x

xS

yS

(T)

518 Chapter10 Kinematics of Rigid Bodies in Contact

cos sin ,

sin cos ,

.

S

S

i i j

j i j

k

ψ ψ

ψ ψ

= +

= − +

There is thus one parameter of rotation: ψ .

In the general case, the motion of the wheel along the line is a motion with two

parameters of situation: x and ψ . In this case, there is thus sliding and rolling of

the wheel on the straight line.

2. Kinematic study

2.1. Kinematic torsor

Let ( ) TS be the kinematic torsor relative to the motion of the wheel (S) with

respect to reference (T). It is defined by its elements of reduction at the centre A:

( ) ( )

( ) ( )

, instantaneous rotation vector relative to the motion,

( , ) , velocity vector of the centre of the wheel .

T TS S

T TA S

R k

A t x i A

ω ψ = =

= =

2.2. Velocity vector of sliding

The velocity vector of sliding ( )

( , )TgS I t of the point of contact I in the motion

of the wheel (S) with respect to (T) is given by:

( ) ( ) ( , )T TIgS SI t =

,

with ( ) ( ) ( ) T T T

I AS S SR AI= + × .

Thus: ( ) ( ) ( )

( , ) ( , )T T TgS SI t A t AIω= + ×

,

with

AI a j= −

.

Hence: ( ) ( )( , )TgS I t x a iψ= +

.

2.3. Condition of non sliding

The condition of non sliding at the point of contact I is: ( )

( , ) 0TgS I t =

.

Thus:

0x aψ+ = .

In this case the wheel rolls on the line (D) without sliding.

The integration of the preceding equation leads to:

cstx aψ+ = .

We may take the origin of the angles of rotation ψ so as, when the centre A of the

wheel is on the axis Oy

( 0x = ), the axis SAy

of the trihedron attached to the


wheel coincides with the axis Oy

. In this case, we have 0ψ = for 0x = and the

preceding equation is written:

0x aψ+ = .

The parameters x and have opposed signs. When the wheel moves in the direc-

tion of 0x < , is positive: the wheel rolls in the direct sense. When the wheel

moves in the direction 0x > , is negative: the wheel rolls in the inverse sense.

The motion has only one parameter of situation: x or .

3. Motion of a point of the wheel

3.1. Kinematic vectors

We consider the point M located on the circumference of the wheel and on the

axis SAy

(Figure Exercise 10.1) of Cartesian coordinates ( )0, , 0a− relatively

to the system ( )S SAx y z . Thus: SAM a j= −

.

The velocity vector of the point M is given by:

( ) ( ) ( ) ( , ) ( , )T T T

SM t A t AMω= + ×

,

thus: ( )

( , )TS SM t x i k a j x i a iψ ψ= + × − = +

.

Hence: ( ) ( )( , ) cos sinT M t x a i a jψ ψ ψ ψ= + +

.

The acceleration vector is then deduced using the relation:

( )( )

( )

d( , ) ( , )d

TT Ta M t M t

t=

.

We obtain: ( ) ( ) ( )2 2( , ) cos sin sin cosTa M t x a i a jψ ψ ψ ψ ψ ψ ψ ψ = + − + +

.

When the wheel rolls without sliding, the kinematic vectors are written: ( ) ( )[ ]( , ) 1 cos sinT M t a i jψ ψ ψ= − + +

,

( ) ( ) ( ) 2 2( , ) 1 cos sin sin cosTa M t a i jψ ψ ψ ψ ψ ψ ψ ψ = − + − + +

.

These expressions can also be expressed as a function of the parameter x.

3.2. Trajectory

The trajectory of the point M relatively to the system (Oxyz) is deduced from

the expression of its position vector:

OM OA AM= +

.

Thus:

( ) sin cosSOM x i a j x a i a jψ ψ= − = + −

.

This trajectory depends on the variations of the parameters x and ψ as functions

of time.

In the case where the wheel rolls without sliding, the position vector is written:

( ) ( )sin 1 cosOM a i a jψ ψ ψ= − + + −

.


This expression is identical to Expression (7.76) with qψ = − .

10.2 Motion of a cylinder (or a sphere) inside a cylinder (Figure 10.15)

We study the motion of a cylinder or a sphere, in such a way that the centre of

the cylinder or the sphere moves in the same vertical plane during the motion.

1. Parameters of situation We associate a trihedron attached to the cylinder (T): (Oxyz), so as the axis

Oz

coincides with the axis of the cylinder and the axis Ox

is upwards vertical

(Figure Exercise 10.2).


We choose a particular point of the cylinder or of the sphere (S), so as this

point is defined with the smallest number of parameters depending of time. Thus

the point OS centre of the cylinder or of the sphere. The position of the point OS is

defined by the angle α between the axes SOO

and Ox

. We have thus one para-

meter of translation: α , which is in fact one of the cylindrical coordinates of the

point OS.

The position vector of the point OS is given by:

( )1SOO b a i= −

,

where b and a are the respective radii of the cylinders (T) and (S), and where 1i

is

the unit direction vector of SOO

:

1 cos sini i jα α= +

.


We associate a coordinate system attached to the cylinder or to the sphere: the

system ( )S S SO x y z (Figure Exercise 10.2). The orientation is then given by the

angle of rotation ψ about the direction k

. The basis change is written:


(S)

I

(T)

M

O

y

z

y

xxS

yS

OS

z

x

i

j

1i

Si

Sj


cos sin ,

sin cos ,

.

S

S

i i j

j i j

k

ψ ψ

ψ ψ

= +

= − +

Finally the motion has two parameters of situation: , .α ψ

2. Kinematic study

2.1. Kinematic torsor

Let ( ) TS be the kinematic torsor associated to the motion of the cylinder or

of the sphere (S) with respect to the cylinder (T). Its elements of reduction at the

point SO are:

( ) ( )

( ) ( )


( , ) , velocity vector of the point

with respect to the cylinder ( ) .S

T TS S

T TO S S S

R k

O t O

T

ω ψ = =

=

With:

( )( )

( )1

d( , )

d

TT

SSO t OO b a jt

α= = −

.

2.2. Velocity vector of sliding

The velocity vector of sliding ( )

( , )TgS I t of the point of contact I in the motion

of the cylinder or of the sphere (S) with respect to the cylinder (T) is given by:

( ) ( ) ( , )T TIgS SI t =

.

The expression of the moment leads to:

( ) ( ) ( )( , ) ( , )T T T

gS S S SI t O t O Iω= + ×

,

with

1SO I a i=

.

Hence: ( ) ( )[ ] 1( , )TgS I t b a a jα ψ= − +

.

The velocity vector of sliding is collinear to the direction 1j

of the plane tangent

to the two solids at the point of contact I.

2.3. Condition of non sliding

The condition of non sliding at the point of contact I is written: ( )

( , ) 0TgS I t =

.

Thus: ( ) 0b a aα ψ− + = .

In this case, the cylinder or the sphere rolls on the cylinder (T) without sliding.

The motion is then a motion with only one parameter of situation or .α ψ

The integration of the preceding equation leads to:


( ) cstb a aα ψ− + = .

Taking the origin of the angle ψ so as, when the centre SO is on the axis Ox

, the

axis S SO x

of the trihedron attached to the solid (S) coincides with the axis Ox

,

we have 0ψ = for 0α = . The preceding equation is then written as:

( ) 0b a aα ψ− + = .

The angle of rotation is thus expressed as:

b a

aψ α−

= − .

The signs of the angles and ψ α are opposed. When the cylinder or the sphere

(S) moves in the direction of 0α > , (S) rolls in the inverse sense. And conversely.

3. Motion of a point We consider the point M located on the axis S SO x

on the periphery of the

solid (S) (Figure Exercise 10.2) of Cartesian coordinates ( ), 0, 0a relatively to

the system ( )S S SO x y z . Thus: S SO M a i=

.

3.1. Kinematic vectors

The velocity vector of the point M at the instant t is expressed by:

( ) ( ) ( ) ( , ) ( , )T T T


,

Hence: ( ) ( ) ( )

1 1( , )TS SM t b a i k a i b a i a jα ψ α ψ= − + × = − +

.

Thus: ( ) ( )[ ]

( )[ ]

( , ) cos sin

sin cos .

T M t b a a i

b a a j

α α ψ ψ

α α ψ ψ

= − −

+ − +

The acceleration vector is obtained by deriving the velocity vector. Hence:

( ) ( ) ( ) ( )( )( ) ( )

2 2

2 2

( , ) cos sin sin cos

sin cos cos sin .

Ta M t b a a i

b a a j

α α α α ψ ψ ψ ψ

α α α α ψ ψ ψ ψ

= − − − +

+ − + + −

3.2. Trajectory

The trajectory of the point M relatively to the system (Oxyz) is deduced from

the expression of its position vector:

S SOM OO O M= +

.

Thus:

( )1 SOM b a i a i= − +

,

or

( )[ ] ( )[ ]cos cos sin sinOM b a a i b a a jα ψ α ψ= − + + − +

.

This trajectory depends of the variations of the parameters α and ψ as functions

of time, according to the conditions of sliding and rolling.

In the case where there is rolling without sliding, the position vector is given by:


( ) ( )cos cos sin sinb a b aOM b a a i b a a ja a

α α α α− − = − + + − −

.

The trajectory is a hypocycloïde.

10.3 Cylinder on a planeThe case of a cylinder in contact with a plane was studied in Exercice 9.2. We

consider (Figure Exercise 9.2) a point I of contact between the cylinder and the

plane distant of 1x from the side of the cylinder. The coordinates of this point

relatively to the system ( )S SAx y z are: ( )1, 0, x a− . The position vector is thus

written:

1 SAI x i a k= −

.

The velocity vector of sliding of the point of contact I in the motion of the

cylinder (S) with respect to the plane (T) is: ( ) ( ) ( )

( , ) ( , )T T TgS SI t A t AIω= + ×

.

The vector product is:

( ) ( )3

1 3

1

0

0

ST

S

i j k

AI x a j

x a

ω θ ψ ψ θ× = = +

.

We obtain thus: ( ) ( ) ( )1 1( , ) sin cosTgS I t x x a i y x a jψ θ ψ ψ θ ψ = − + + + +

.

There is non sliding at the points I such as ( )

( , ) 0.TgS I t =

Thus at the points I

such as:

( )

( )1

1

sin 0,

cos 0.

x x a

y x a

ψ θ ψ

ψ θ ψ

− + =

+ + =

Case where the cylinder does not spin We consider the case where there is non spinning of the cylinder. Thus 0ψ = .

The velocity vector of sliding of the point I of contact is thus written as:

( ) ( ) ( )( , ) sin cosTgS I t x a i y a jθ ψ θ ψ= − + +

.

The velocity vector of sliding is then independent of the abscissa x1, thus the same

for all the points of the generator of contact.

The case of rolling without sliding nor spinning leads thus to:

sin 0,

cos 0.

x a

y a

θ ψ

θ ψ

− =

+ =

By excluding θ , we obtain: tan 0x y ψ+ = . Then integrating:

tan cstx y ψ+ = .

The constant depends on the position of the cylinder at the initial instant. The

trajectory of the point A is thus the straight line of orientation .ψ

Chapter 11

General Elements on the Mechanical Actions

11.1 Mechanical action exerted on a frame (Figure 11.5)

11.1.1. Field of the forces exerted

We have first to characterize the field of the forces i. At each point Mi it is

exerted a force of resultant iR

, of which the characteristics are deduced from the

data and Figure 11.5. Note that to express the components of a force of magnitude

Ri and direction iα (Figure Exercise 11.1), it is advised to use the relations:

( ) cos , sin , 0i i i i iR R Rα α

.

Thus, the field of forces is:

M1 (4, 0, 0) 1R

(0, –2000, 0)

M2 (8, 0, 0) 2R

(0, –1000, 0)

M3 (12, 0, 0) 3R

(0, –1500, 0)

M4 (2, 1,5, 0) 4R

(–900, 1200, 0)

M5 (6, 4, 0) 5R

(–447, 894, 0)

M6 (14, 4, 0) 6R

(–671, –1342, 0)

M7 (18, 1,5, 0) 7R

(–1200, –1600, 0)

11.1.2. Action exerted on the frame

Let ( ) S be the torsor representing the resultant of the mechanical action

exerted on the frame (S).

1.Resultant of the action

The resultant is:

( ) 7

1

i

i

R S R

=

=

.

Thus: ( ) 3218 5348R S i j= − − .


x

y

iα

iR

iR

524 Chapter 11 General Elements on the Mechanical Actions

2. Moment of the action at the point O

The moment at the point O is expressed by:

( ) 7

1

iiO

i

S OM R

=

= ×

.

We have:

1 21 2

3 43 4

5 65 6

77

8000 , 8000 ,

18000 , 3750 ,

7152 , 16104 ,

27000 .

OM R k OM R k

OM R k OM R k

OM R k OM R k

OM R k

× = − × = −

× = − × =

× = × = −

× = −

Hence:

( ) 66202O S k= − .

3. Type of action

The scalar invariant of the torsor is:

( ) ( ) ( ) 0OI S R S S= =⋅

.

The mechanical action exerted is thus a force.

4. Support of the force

It is the set of the points P such as ( ) 0P S = . We evaluate the moment

at the point P of coordinates (x, y, z):

( ) ( ) ( ) P OS S R S OP= + ∧

.

Thus:

( ) ( ) 5348 3218 5348 3218 66202P S z i z j x y k= − + + − − .

The equation of the support of the force is then:

0,

5348 3218 66202 0.

z

x y

=

− − =

The support is contained in the plane (Oxy). We may determine two points of the

support:

A (0, 20.57, 0) and B (12.38, 0, 0).

11.2 Mechanical action exerted on a barrage11.2.1. The action exerted by the water on an elementary surface dS(M) of the

barrage surrounding the point M is a force of resultant:

d ( ) ( ) d ( )R M p M S M i=

,

and support ( ), M i

.

The pressure p(M) exerted at the point M results from the atmospheric pressure

0p and the height h z− of the water which is above the point M. Thus:


( )0( )p M p g h zρ= + − .

This expression can be written in the form:

( )p M zλ µ= + ,

introducing the coefficients:

0 , .p gh gλ ρ µ ρ= + = −

11.2.2. The action exerted by the water of the dam on the side (D) of the barrage

in contact with the water is represented by the torsor ( ) .D

1. Resultant

The resultant is:

( ) ( ) ( )

d ( ) ( )d ( )D D

R D R M i p M S M= = .

Thus, introducing the Cartesian coordinates (0, y, z) of the point M, it comes:

( ) ( )

0 0

d da h

y z

R D i z y zλ µ= =

= + .


( ) ( ) ( )

2 2h hR D ah i S iλ µ λ µ= + = +

,

where S is the area of the face of the barrage.


3 3 2 5010 kg m , 9.81 m s , 1.013 10 Pa, 50 m, 30 m.g p a hρ − − −= = = × = =

Hence: 5 3 13.956 10 Pa, 9.81 10 Pa m .λ µ −= × = − ×

Thus: 87.45 10 (en N).R i= ×

2. Moment

It is needed to determine the moment at a point or to find a point which is asso-

ciated to a particular property for the moment.

Here the field of forces is such as:

( )M D∀ ∈ d ( ) ( ) d ( )R M p M S M i=

.

At any point M of the face of the barrage it is associated a force (a slider) of

support (of axis) i

which is independent of the point M. We are in the case

studied in Section 5.3.3. From the results established in this section, there exists a

measure centre H, here a centre of thrust. This centre is such as the moment at the

point H of the torsor ( ) D is null. Thus, the determination of the measure

centre stands in for the determination of the moment at a point.

The position of the centre of thrust H is given, from Expression (5.72), by:

( ) ( )

1 ( ) d ( )D

OH OM p M S MDµ

=

,


introducing:

( ) ( )2hD Sµ λ µ= + ,

which is the measure associated to the field of the forces exerted on the face (D)

of the barrage. The global resultant (thrust on the face) is thus expressed by:

( ) ( )R D D iµ= .

Introducing the Cartesian coordinates, the expression of the centre of thrust is

written:

( )( )( )

0 0

1 d da h

y z

OH y j z k z y zD

λ µµ = =

= + +

.

Separating the different integrals, the integration leads to:

( ) ( ) ( )1 22 2 3h aOH j h k ah

Dλ µ λ µ

µ = + + +

,

or

2 2a hOH j kα= +

,

setting:

23

2

h

h

λ µα

λ µ

+=

+.

The coefficient µ being negative, it follows that the centre of thrust is located on

the vertical line passing through the centre of the face and under this point.


50 12 (in m).OH j k= +

The centre of thrust is located 3 m under the centre of the face of the barrage.

In conclusion, the action exerted by the dam on the face of the barrage is a force

of resultant ( )D iµ

and support ( ), H i

, H being the centre of thrust.

3. Moment at an arbitray point of the face of the barrage

Let P be a point of the face of coordinates (0, y, z). The moment at the point P

is expressed by:

( ) ( ) ( ) P HD D R D HP= + ×

.

Thus:

( ) ( ) P D R D HP= ×

.

The position vector HP

is expressed by:

( ) ( )2 2a hHP OP OH y j z kα= − = − + −

.

Hence the moment at P:

( ) ( ) ( ) ( )2 2P

h aD D z j y kµ α = − − + −

.


11.2.3. Action exerted on a sluice We consider a circular sluice (D) completely immersed: the centre of the sluice

is at the depth d, with /2d D≥ (Figure Exercise 11.2). We choose the coordinate

system (Oxyz) such as O is the centre of the sluice and Oz

is upward vertical.

1. As previously, the action exerted on an element of surface of the sluice

surrounding the point M is a force of resultant expressed in 11.2.1 and of support

( ), M i

. Compared to 11.2.1, the height h is substituted by the depth d and the

expression of the pressure at the point M becomes:

( ) dp M zλ µ= + ,

introducing the coefficient:

0d p gdλ ρ= + .

2. The action exerted by the water on the face (D) of the sluice in contact with the

water is represented by the torsor ( ) .D

2.1. Resultant

As previously, the resultant is expressed by:

( ) ( ) ( )

d ( ) ( )d ( )D D

R D R M i p M S M= = .

To evaluate the integral, we have to use the polar coordinates of the point M:

( ), r α in the plane 0x = (Figure Exercise 11.2). The elementary surface is

obtained while icreasing by d and by dr r α α (Figure Exercise 11.2). Thus

d ( ) d dS M r rα= . The resultant is thus written as:

( ) ( )

/2 2

0 0

sin d dD

dr

R D i r r rπ

αλ µ α α

= == +

.

Thus:

( ) 2

4d d

DR D i S iλ π λ= =

,

where S

is the area of the sluice.


z

x

y

M

d ( )S M/2D

O

d

(D)y

z

O

d

d S(M) = rddr

rdr


2.2. Moment

As previously, there exists a measure centre H

, the centre of thrust on the

sluice. Its position is given by the same relation as previously while introducing

the measure:

( )dD Sµ λ=

.

The global resultant of the action exerted on the sluice has the same form as

previously in the case of the face of the barrage.

The preceding expression of the centre of thrust leads to:

( )( )( )

/2 2

0 0

1 cos sin sin d dD

dr

OH r j k r r rD

π

αα α λ µ α α

µ = == + +

.

The integrations of sin , cos and sin cosα α α α between the values 0 and 2πlead to values which are zero. So, we have simply:

( )

/2 23 2

0 0

1 sin d dD

r

OH k r rD

π

αµ α α

µ = ==

.

Thus:

2

16 d

OH D kµλ

=

,

or

016 1

DdOH D k

p

gdρ

= − +

.

The centre of thrust is located under the centre of the sluice.

Finally the action exerted by the water on the face of the sluice is a force of

resultant d S iλ

and support ( ), H i

.

11.3 Action exerted by a liquid on an immersed sphereThe sphere (D) of radius a is completely immersed in a liquid (Figure Exercise

11.3): the centre of the sphere is at the depth h ( h a≥ ). We choose the coordinate

system (Oxyz) such as O is the centre of the sphere and the axis Oz

is upward

vertical.

1. The action exerted by the liquid on the element of surface d ( )S M of the sphere

surrounding the point M is a force of resultant:

d ( ) ( ) d ( ) ( )R M p M S M n M= −

,

and support ( ), ( )M n M−

where ( )n M

is the unit vector at the point M of the

normal to the sphere orientated towards the exterior of the sphere.

The pressure ( )p M exerted at the point M is expressed by the same relation as

in Exercise 11.2.1, where ρ is the mass per unit of volume of the liquid.



2. Action exerted by the liquid on the sphere

The action exerted on the sphere (D) is represented by the torsor ( ) .D

2.1. Resultant

The resultant is expressed by:

( ) ( ) ( )

d ( ) ( ) ( )d ( )D D

R D R M p M n M S M= = − .

The evaluation of the integral is implemented while using the spherical coor-

dinates of the point M (Figure Exercise 11.3). The point M is located on the

sphere by its longitude α and its latitude β . The elementary surface is obtained

by increasing the longitude and the latitude by dα and dβ , respectively. Thus:

2d ( ) cos d dS M a β α β= .

The unit vector ( )n M

at the point M is expressed by:

( ) ( , ) ( ) cos sinn M n u kα β α β β= = +

,

where ( )u α

is the unit vector of the longitude α . Hence:

( ) cos cos sin cos sinn M i j kα β α β β= + +

.

The resultant is thus expressed by:

( )

( ) ( )

22 2

02

sin cos cos sin cos sin cos d d .

R D

a i j k a

ππ

πα βλ µ β α β α β β β α β

= =−

=

− + + +

The integrations of sin and cosα α between 0 and 2π give null values. Hence:

( ) ( )

222

02

sin sin cos d dR D k a a

ππ

πα βλ µ β β β α β

= =−= − +

,

z

x

y

M

d ( )S M

O

h

(D)

( )n M

x

yO

d

d S(M) = a

2cos dd

z

a

d

a cos a cos d

i

j

k

( )u α


or

( )

223 2

02

sin cos d dR D a k

ππ

πα βµ β β α β

= =−= −

.

Thus:

( ) 343

R D a g kπ ρ= ,

or

( ) R D mg k= ,

where m is the mass of liquid having the same volume as the sphere.

2.2. Moment

In the present case, the normal ( )n M

at every element of surface of the sphere

depends on the point M. Thus, there does not exist a measure centre.

We determine the moment at the centre O of the sphere expressed by:

( ) ( )

d ( )OD

D OM R M= ×

,

or

( ) ( )

( ) ( )d ( )OD

D OM n M p M S M= − × .

The vectors OM

and ( )n M

are collinear. So, it follows that their vector pro-

duct is null. Hence:

( ) 0O D = .

In conclusion, the mechanical action exerted by the liquid on the sphere is a

force of which the resultant has a magnitude mg and the support is the downward

vertical axis passing through the centre of the sphere.

Hence the result: the mechanical action exerted by the liquid on the sphere is

opposed to the action of gravity (Chapter 12) exerted on the mass of liquid

occupying the same volume as the sphere. This result is described by the Principle

of Archimedes which is applied to a solid with an arbitrary shape immersed in a

liquid.

Chapter 12

Gravitation. Gravity

Mass Centre

12.1 Mass centre of an arc of circle (figure 12.13)

We take (Figure Exercise 12.1) the coordinate system (Oxyz) such as the point

O is the centre of the circle on which the arc is constructed, the plane (Oxy) con-

tains the arc of circle and the axis Ox

coincides with the axis of symmetry.

The solid being homogeneous, the mass centre coincides with the centroid

given by Expression (12.47). A point M of the arc of circle can be located by its

polar angle θ (Figure Exercise 12.1). The length of an element of arc, obtained

while increasing θ by dθ , is da θ . The total length of the arc is 2aα . Expres-

sion (12.47) is thus written:

( )1 cos sin d2

OG a i j aa

α

αα α θ

α −= +

.

Hence:

sinOG a iαα

=

.

12.2 Mass centre of a circular sector (Figure 12.14)

The mass centre coincides with the centroid. A point M of the sector (Figure

Exercise 12.2) is characterized by its polar coordinates ( ), r θ and the element of

surface of the sector is obtained while increasing r by dr and θ by dθ . Its area is

d d .r rθ The surface of the sector is 2.aα Expression (12.47) is thus written:

( )

20

1 cos sin d da

r

OG r i j r ra

α

θ αα α θ

α = =−= +

.

Hence: 2 sin3

OG a iαα

=

.

Figure Exercise 12.1. Figure Exercise 12.2.

a

M

O

y

x

z

a

M

O

y

x

z

r

532 Chapter 12 Gravitation. Gravity. Mass centre


12.3 Mass centre of a circular segment (Figure 12.15)

To determine the mass centre (coinciding with the centroid), we consider the

circular sector (S) of the preceding exercise, constituted (Figure Exercise 12.3) of

the triangle (1) and the circular segment (2).

The circular sector has for surface and centroid:

2 2 sin, .3

S a OG a iααα

= =

The triangle has for surface and centroid:

211

2sin cos , cos .3

S a OG a iα α α= =

The circular segment has for surface ( )22 1 sin cosS S S a α α α= − = − and the

centroid 2G must be determined. We have:

1 21 2S OG S OG S OG= +

.

We deduce: 3

22 sin3 sin cos

aOG iαα α α

=−

.

12.4 Mass centre of a cone (Figure 12.16)

The mass centre coincides with the centroid which is located on the axis Oz

(Figure Exercise 12.4). It follows that we can take as an element of volume of the

cone a slice of thickness dz located at the height z (Figure Exercise 12.4). The

radius of this slice is a zh

and its volume2

2

2d .a z z

hπ Expression (12.47) leads to:

22

22 0

1 d

3

haOG k z z zha h

ππ

=

,

where 2

3a hπ is the volume of the cone. Thus, we obtain:

34

OG h k=

.

a

G1O

y

x

z

G2

(1) (2)



12.5 Mass centre of a spherical segment (Figure 12.17)

We choose (Figure Exercise 12.5) a coordinate system (Oxyz) such as O is the

centre of the base of the segment and Oz

is the axis of symmetry. The mass

centre coincides with the centroid of the segment which is located on the axis Oz

.

As in the preceding exercise, we can take for an element of volume a slice of

thickness dz located at the height z. The radius r of this slice is such as:

( ) ( ) ( )22 2 2 2 2r a z a h z a h z a h h= − + − = − − − + − .

Expression (12.47) leads to:

( ) ( )[ ]

2

0

2 2 dh

OG k z z a h z a h h zVπ= − − − + −

,

where V is the volume of the spherical segment expressed by:

( )2 33

V h a hπ= − .


( )414 3

a h hOG ka h−=−

.


h

a

z

dz

z

y

x

O

h

aOz

dz

z

y

x

r



12.6 Mass centre of a hollowed cylinder (Figure 12.18)

The solid cylinder (S) can be considered as the union of the hollowed cylinder

(S1) and of the cylinder (S2) which is hollowed out.

The solid cylinder (S) has a volume 2V a hπ= and its mass centre G has for

coordinatees (0, 0, 2h ).

The cylinder (S2) which is hollowed out has a volume 2

24

aV hπ= and its mass

centre G2 has for coordinates (0, , )2 2a h .

Te hollowed cylinder (S1) has a volume 21 2

34

V V V a hπ= − = and a mass

centre G1 which is to be determined.

From Relation (12.41), we have:

22 2

1 23 4 4

aa h OG a h OG h OGπ π π= +

.

Thus:

16 2a hOG j k= − +

.

The coordinates of G are (0, , )6 2a h− .

12.7 Action of gravitation exerted by a sphereThe action of gravitation exerted by a sphere at a point M external to the sphere

is defined in Section 12.1.3. Its moment at the point M is null and its resultant

h

a

z

y

x

/2a

O



is expressed by Relation (12.7). To determine this resultant, it is necessary to use

the spherical coordinates of the point M ′ of the sphere. For that, we choose the

coordinate system (Oxyz) such as the axis Oz

passes through the point M (Figure

Exercise 12.7). The spherical coordinates and the element of volume of the sphere

are considered in Subsection 15.4.3.2 of Chapter 15. Using the results of this

subsection, Expression (12.7) of the resultant is written:

( )

22 2

30 0

2

( ) cos d d da

R

MMR Km M R RS MMM

ππ

πα βρ β α β

= = =−

′ ′=→′

.

The vector MM ′

is expressed as:

( )MM OM OM R n M r k′ ′ ′= − = −

,

where r is the distance from the point M to the centre O of the sphere and ( )n M ′

is the unit vector of ,OM ′

already introduced in Exercise 11.3. This vector is

expressed as:

( ) ( , ) ( ) cos sinn M n u kα β α β β′ = = +

,

where the vector ( )u α

is the unit vector of the direction α :

( ) cos sinu i jα α α= +

.

We have thus:

( ) ( )cos sinMM R u R r kβ α β′ = + −

.

Hence:

( ) ( )

1 122 2 2 22 2cos sin 2 sinMM MM R R r R r rRβ β β ′ ′= = + − = + −

.

Moreover, the mass per unit volume is a function of , , and R α β :

( ) ( , , )M Rρ ρ α β′ = .

z

x

y

M ′d ( )V M ′

O

( )n M ′

M

a

r


The resultant is thus written:

( ) ( )

( )

22 2

30 0 2 2 22

cos sin( , , ) cos d d d .

2 sin

a

R

R S M

R u R r kKm R R R

R r rR

ππ

πα β

β α βρ α β β α β

β= = =−

=→

+ −

+ −

To continue the integration, it is necessary to introduce assumptions on the

mass per unit volume.

First assumption. The mass per unit volume does not depend on the longitude

α :

( ) ( , )M Rρ ρ β′ = .

In the expression of the resultant, the vector ( )u α

introduces terms in sinα and

cosα . Their integration with respect to α between the limits 0 and 2π leads to

null terms. Thus, integrating with respect to α , we obtain:

( )

( )

22

30 2 2 22

sin cos2 ( , )d d .

2 sin

a

R

R r RR Kmk R RS M

R r rR

π

πβ

β βπ ρ β β

β= =−

−=→

+ −

At this stage, we find that the resultant of the action of gravitation is collinear

to k

. The action is thus a force of support OM

.

Second assumption. The mass per unit volume depends only on R:

( ) ( )M Rρ ρ′ = .

The sphere is said homogeneous by concentric layers.

The resultant is then written as:

( )

( )

222

30 2 2 22

sin cos2 ( ) d d .

2 sin

a

R

R r RR Kmk R R RS M

R r rR

π

πβ

β βπ ρ β

β= =−

− =→

+ −

We evaluate the integral with respect to β :

( )

( )

22

32 2 22

sin cosd

2 sin

R r RI

R r rR

π

βπ

β ββ

β−

−=

+ − .

To integrate, we introduce the variable u such as:

( )1

2 2 22 sinu R r rR β= + − ,

thus:

2 2 2 2 sin et d cos du R r rR u u rRβ β β= + − = − .

Furthermore, for , 2

u r Rπβ = − = + and for , 2

u r Rπβ = = − . Hence:

2 2 2

3

1 d2

r R

r R

R r u uI r ur rRu

β

−

+

+ −= − − .


The rearrangment of the integral, then its integration lead to: 2

2Ir

β = − . Hence

the expression of the resultant of the action of gravitation:

2

20

4 ( ) da

KmR k R R RS Mr

π ρ= −→

.

We determine the mass of the sphere. Since the mass per unit volume depends

only on R, it is possible to take as element of volume the volume included between

the spheres of radii R and dR R+ , of mass: 2d 4 ( ) dm R R Rπ ρ= ,

and the mass of the sphere is:

2

0 0

d 4 ( ) da a

Sm m R R Rπ ρ= = .

The resultant of the action of gravitation is thus simply expressed as:

2

SKmmR kS M

r= −→

,

or

3S

MOR KmmS MOM

=→

.

This result was introduced in Relation (12.9) of Chapter 12.

Chapter 14

Statics of Rigid Bodies

14.1 Equilibrium of a system of two beamsWe choose (Figure Exercise 14.1) a coordinate system (Oxyz) associated to the

set of the two beams, such as the axis Ox

is horizontal and passes through the

point C and such as the axis Oy

is upward vertical and passes through the point

A. The plane (Oxy) contains the plane (ABC) of the two beams.

The angles and α γ of the triangle ABC are expressed as:

2 2 2 2 2 2 2 21 11 2 2 1

2 2 2 21 2

cos , cos ,2 2

h l l l l h l l

l h l l h lα γ− −+ + − + + −

= =+ +

and the angle ε is given by:

1tan hl

ε −= .

We introduce then the cosines and sines:

( )

( )

( )

( )21

21

cos , cos ,

sin .sin ,

cc

ss

γ εα ε

γ εα ε

= += −

= += −

So, coordinates of the different points are determined by:

( ) ( ) ( )1 1 1 1 0, , 0 , , , 0 , , 0, 0 ,A h B c l h s l C l+

( ) ( )1 1 1 1 1 1 1 2 2 2 2 2 2 2 , , 0 , , , 0 .M c l h s l M l c l s lα α α α+ −


M1

A

C

1l1

M2

B

2l2

l

hhorizontal

O

y

x

z

(S1) (S2)


14.1.1. Mechanical actions exerted on the beams

1. Actions exerted on the beam (S1)

1.1. Action induced by the mass m1, represented by the torsor ( ) 1S :

( )

( ) 1

1 1

1

,

0.M

R S m g j

S

= −

=

1.2. Action of the frame induced by the connection at A, represented by the torsor

( ) 1S :

( )

( )

1 1 1 1

1 1 1 1

,

.A

R S X i Y j Z k

S L i M j N k

= + +

= + +

The components X1, Y1, . . . , N1 are to be determined.

1.3. Action of the beam (S2) induced by the connection at B, represented by the

torsor ( ) 2 1S :

( )

( )

2 1 21 21 21

2 1 21 21 21

,

.B

R S X i Y j Z k

S L i M j N k

= + +

= + +


2. Actions exerted on the beam (S2)

2.1. Action induced by the mass m2, represented by the torsor ( ) 2S :

( )

( ) 2

2 2

2

,

0.M

R S m g j

S

= −

=

2.2. Action of the frame induced by the connection at C, represented by the torsor

( ) 2S :

( )

( )

2 2 2 2

2 2 2 2

,

.C

R S X i Y j Z k

S L i M j N k

= + +

= + +


2.3. Action of the beam (S1) induced by the connection at B, represented by the

torsor ( ) 1 2 .S We have:

( ) ( ) 1 2 2 1S S= − .

14.1.2. Equations of equilibrium

1. Equilibrium of the beam (S1)

The equilibrium of the beam (S1) is expressed by the equation:

( ) ( ) ( ) 1 1 2 1 0S S S+ + = . (1)

1.1. Equation of the resultant

This equation is deduced from the preceding equation. Thus:


( ) ( ) ( ) 1 1 2 1 0R S R S R S+ + = . (2)

Hence:

1 21

1 1 21

1 21

0,

0,

0.

X X

m g Y Y

Z Z

+ =

− + + =

+ =

(3)

1.2. Equation of the moment

The equation is deduced from the equation between torsors, while choosing a

point. For example the point A. Hence:

( ) ( ) ( ) 1 1 2 1 0A A AS S S+ + = . (4)

We have to express the moments at the point A:

( ) ( ) ( ) ( ) 1 11 1 1 1 1A MS S R S M A AM R S= + × = ×

.

Thus:

( ) ( ) ( ) 1 1 1 1 1 1 1 1 1 1A S l c i s j m g j c m gl kα α= + × − = − .

( ) ( ) ( ) 2 1 2 1 2 1A BS S R S BA= + ×

.

It follows:

( ) ( ) ( )

( )

2 1 21 1 1 21 21 1 1 21

21 1 1 21 1 1 21 .

A S L s l Z i M c l Z j

N s l X c l Y k

= + + −

+ − +

Hence the scalar equations of the moment at A:

1 21 1 1 21

1 21 1 1 21

1 1 1 1 1 21 1 1 21 1 1 21

0,

0,

0.

L L s l Z

M M c l Z

c m gl N N s l X c l Yα

+ + =

+ − =

− + + − + =

(5)

Finally, the equilibrium of the beam (S1) leads to six scalar equations:

1 21

1 1 21

1 21

1 21 1 1 21

1 21 1 1 21

1 1 1 1 1 21 1 1 21 1 1 21

0,

0,

0,

0,

0,

0.

X X

m g Y Y

Z Z

L L s l Z

M M c l Z

c m gl N N s l X c l Yα

+ =

− + + =

+ =

+ + =

+ − =

− + + − + =

(6)


The equilibrium of the beam (S2) leads to the equation:

( ) ( ) ( ) 2 2 2 1 0S S S+ − = . (7)


The equation is:

( ) ( ) ( ) 2 2 2 1 0R S R S R S+ − = . (8)

Hence:


2 21

2 2 21

2 21

0,

0,

0.

X X

m g Y Y

Z Z

− =

− + − =

− =

(9)

2.2. Equation of the moment, for example at the point C

( ) ( ) ( ) 2 2 2 1 0C C CS S S+ − = . (10)

We have to express the moments at the point C:

( ) ( ) ( ) ( ) 2 22 2 2 2 2C MS S R S M C CM R S= + × = ×

.

Thus:

( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2C S l c i s j m g j c m gl kα α= − + × − = .

( ) ( ) ( ) 2 1 2 1 2 1C BS S R S BC= + ×

.

It comes:

( ) ( )[ ] ( )[ ]( ) ( )[ ]

2 1 21 1 1 21 21 1 1 21

21 1 1 21 1 1 21 .

C S L h s l Z i M l c l Z j

N h s l X l c l Y k

= + + + + −

+ − + − −

Hence the scalar equations of the moment at C:

( )

( )

( ) ( )

2 21 1 1 21

2 21 1 1 21

2 2 2 2 2 21 1 1 21 1 1 21

0,

0,

0.

L L h s l Z

M M l c l Z

c m gl N N h s l X l c l Yα

− − + =

− − − =

+ − + + + − =

(11)

Finally, the equilibrium of the beam (S2) leads to the six scalar equations:

( )

( )

( ) ( )

2 21

2 2 21

2 21

2 21 1 1 21

2 21 1 1 21

2 2 2 2 2 21 1 1 21 1 1 21

0,

0,

0,

0,

0,

0.

X X

m g Y Y

Z Z

L L h s l Z

M M l c l Z

c m gl N N h s l X l c l Yα

− =

− + − =

− =

− − + =

− − − =

+ − + + + − =

(12)

The equilibrium of the beam (S1) and of the beam (S2) leads in total to 12

scalar equations: Equations (6) and Equations (12).

3. Equilibrium of the set of the two beams

Considering the equilibrium of the set of the two beams, we obtain 6 scalar

equations which are linear combinations of Equations (6) ans (12). So, we do not

obtain new information, but another form of these equations.

The equilibrium of the set of the two beams leads to:

( ) ( ) ( ) ( ) 1 1 2 2 0S S S S+ + + = . (13)

This equation is the superimposition of the preceding equations (1) and (7). This

equation removes the actions internal to the set of the two beams.



The equation is:

( ) ( ) ( ) ( ) 1 1 2 2 0R S R S R S R S+ + + = . (14)

It is the superimposition of the equations of the resultants (2) and (8) of the

equilibriums of the beam (S1) and of the beam (S2). Hence:

( )1 2

1 2 1 2

1 2

0,

0,

0.

X X

m m g Y Y

Z Z

+ =

− + + + =

+ =

(15)

These equations are the superimpositions of Equations (3) and (9).

3.2. Equation of the moment

We have to write the equation of the moment at a point. For the equilibrium of

the beam (S1), the equation of the moment (4) was written at A. For the

equilibrium of the beam (S1), the equation of the moment (10) was written at C.

So, the equation of the moment of the set of the two beams cannot be obtained by

superimposition of these two equations. We ought to have chosen the same point

for the two equations of the moment.

We express the equation of the moment at the point A:

( ) ( ) ( ) ( ) 1 1 2 2 0A A A AS S S S+ + + = . (16)

The firs two moments are known. It remains to be determined:

( ) ( ) 22 2A S AM R S= ×

.

Thus:

( ) ( ) ( )[ ] ( ) ( )2 2 2 2 2 2 2 2 2 2 2 2A S l c l i s l h j m g j m g l c l kα α α= − + − × − = − − .

( ) ( ) ( ) 2 2 2A CS S R S CA= + ×

.

Thus:

( ) ( ) ( ) ( )2 2 2 2 2 2 2 2A S L hZ i M lZ j N hX lY k= − + − + + + .

Hence the scalar equations of the moment at A:

( )

1 2 2

1 2 2

1 1 1 1 1 2 2 2 2 2 2 2

0,

0,

0.

L L hZ

M M lZ

c m gl N N hX lY m g l c lα α

+ − =

+ − =

− + + + + − − =

(17)

The equilibrium of the two beams leads thus to the six scalar equations:

( )

( )

1 2

1 2 1 2

1 2

1 2 2

1 2 2

1 1 1 1 1 2 2 2 2 2 2 2

0,

0,

0,

0,

0,

0.

X X

m m g Y Y

Z Z

L L hZ

M M lZ

c m gl N N hX lY m g l c lα α

+ =

− + + + =

+ =

+ − =

+ − =

− + + + + − − =

(18)

Deriving the equations of the moments shows the difficulty in choosing the


points for expressing the moments of the actions. For the equilibrium of the beam

(S1), we chose the point A. For the equilibrium of the beam (S2), the point C.

These choices led us to rewrite the moments for the equilibrium of the set of the

two beams.

The development that we carried out shows that the best choice would have

been to choose the intermediate point B to express the equations of the moments

for the beam (S1) and for the beam (S2).

14.1.3. Choice of the connections

We have 12 independent scalar equations among Equations (6), (12) and (18),

for 18 unknowns of connection: X1, Y1, …, N1; X2, Y2, …, N2; X21, Y21, …, N21.

So that the system of equations can be solved (it is said that the system of the two

beams is isostatic), we must find 6 other equations. These 6 equations are deduced

from the nature of the connections, assuming that these connections are perfect.

1. We want that the beams AB and CB are in the same plane. Hence the

necessity to put at B a hinge connection of axis normal to the plane (ABC). If the

connection is perfect, we have:

21 0N = . (19)

2. The plane (ABC) must be vertical. It is possible to put a cylindrical connec-

tion at C of horizontal axis along the direction k

. If the connection is perfect, we

have:

2 20, 0.Z N= = (20)

3. It remains to put at the point A a connection with three degrees of freedom.

The point A must have a fixed location. So, it is necessary to put at A a spherical

connection of centre A. If the connection is perfect, we have:

1 1 10, 0, 0.L M N= = = (21)

We have thus 6 equations of connection. If the connections are not perfect,

Equations (19)-(21) will be modified to take account of friction induced by the

connections.

14.1.4. Determination of the actions induced by the connections Considering the equations of connection (19)-(21), Equations (6), (12) and (18)

lead to the follwing equations.


1 21

1 1 21

1 21

21 1 1 21

21 1 1 21

1 1 1 1 21 1 21

0,

0,

0,

0,

0,

0.

X X

m g Y Y

Z Z

L s l Z

M c l Z

c m g s X c Yα

+ =

− + + =

+ =

+ =

− =

− − + =

(22)



( ) ( )

2 21

2 2 21

21

2 21

2 21

2 2 2 2 1 1 21 1 1 21

0,

0,

0,

0,

0,

0.

X X

m g Y Y

Z

L L

M M

c m gl h s l X l c l Yα

− =

− + − =

=

− =

− =

+ + + − =

(23)

3. Equilibrium of the set of the two beams

( )

( )

1 2

1 2 1 2

1

1 2

1 2

1 1 1 1 2 2 2 2 2 2

0,

0,

0,

0,

0,

0.

X X

m m g Y Y

Z

L L

M M

c m gl hX lY m g l c lα α

+ =

− + + + =

=

+ =

+ =

− + + − − =

(24)

4. Solving the equations

Solving the equations allows us to obtain the components of the actions of

connection on which no assumption was formulated.

From equations (22)-(24), we deduce first:

1 2 2

21 21 21

0, 0, 0,

0, 0, 0,

Z L M

Z L M

= = =

= = = (25)

then

21 2

21 2 2

,

,

X X

Y Y m g

=

= − (26)

and

( )1 2

1 1 2 2

,

.

X X

Y m m g Y

= −

= + − (27)

It remains to find two equations for the determination of X2 and Y2. For example,

the sixth equation of (24) and the sixth equation of (22) associated to (26). We

obtain:

( )[ ]( )

2 2 1 1 1 1 2 2 2 2

1 2 1 2 1 2 1 1

,

.

hX lY c l m l c l m g

s X c Y c m m g

α α

α

+ = + −

− + = +

Solving these two equations leads to:

( )

( ) ( )[ ]

1 1 1 1 1 2 2 1 2 22

1 1

1 1 1 1 1 1 1 2 2 2 22

1 1

,

.

c c l l m c c l mX g

s l c h

c h s l m c h s l c l mY g

s l c h

α α

α α

− −=

+

+ + + −=

+

The components of the actions of connection are thus all determined.


14.2 Equilibrium of a ladderThe analysis of the equilibrium of the person-ladder set is rather long and we

develop here only some elements.

To implement the analysis we choose a coordinate system (Oxyz) such as the

plane (Oxy) contains the plane of symmetry of the person-ladder set and such the

axis Oy

is upward vertical (Figure Exercise 14.2).

14.2.1. Mechanical actions

1. Actions exerted on the ladder (S)

1.1. Action of gravity, represented by the torsor ( ) e S :

( )

( )

e ,

e 0,Ge

R S Mg j

S

= −

=

where M is the mass of the ladder and eG its mass centre. If l is the length of the

ladder and α its inclination, the coordinates of eG are ( )cos , sin , 02 2l lα α .

1.2. Action exerted by the ground at A, represented by the torsor ( ) A S

We may describe this action as a force of which the support passes through the

point A. Thus:

( )

( )

,

0.

A A A A

A A

R S X i Y j Z k

S

= + +

=


B

A

C

D

ground

wa

ll

G

y

O x

z

C

legD

arm

u


The components XA, YA and ZA are to be determined. Moreover, considering the

Coulomb’s law for dry friction (Chapter 13), equilibrium at A will occur only if:

sA AX f Y< ,

where fs is the coefficient of friction between the ladder and the ground.

1.3. Action exerted by the wall at B, represented by the torsor ( ) B S :

This action can also be described as a force of which the support passes

through the point B. Thus:

( )

( )

,

0.

B B B B

B B

R S X i Y j Z k

S

= + +

=

The components XB, YB and ZB are to be determined. As previously the law for the

dry friction implies that equilibrium at B occurs only if:

mB BX f Y< ,

where fm is the coefficient of dry friction between the ladder and the wall.

1.4. Action exerted by the person (P)

The problem for describing this action is complex, since this action depends on

the position of the person: support on the feet, assistance with the arms, etc.

We describe this action as the superimposition (Figure Exercise 14.2) of two

actions: one exerted at C, represented by the torsor ,C S→ and the other exerted

at D, represented by the torsor .D S→

1.4.1. Action exerted at C, represented by the torsor C S→

The action may be modelled as the action of a plane on a cylinder (Figure

Exercise 14.2). So:

,

.

C C C

C C C C

R C S X i Y j Z k

C S L i M j N k

→ = + +

→ = + +

Neglecting friction of spinning about the direction ,k

we shall have:

0CN = .

Such as to have equilibrium at C the components XC and YC they will have to

satisfy the conditions of friction between feet and ladder.

1.4.2. Action exerted at D, represented by the torsor D S→In the case where the person does not grip the rung at D, the action may be

modelled as a force of which the support is the axis of the arms (Figure Exercise

14.2) of direction u

. We have:

,

0.

D

D

R D S F u

D S

→ =

→ =

The direction u

is known.

2. Actions exerted on the person (P)

2.1. Action of gravity, represented by the torsor ( ) e P :


( )

( )

e ,

e 0,G

R P mg j

P

= −

=

where m is the mass of the person.

2.2. Action exerted by the ladder at the contact C, represented by the torsor

( ) C P :

( ) .C P C S= − →

2.3. Action exerted by the ladder at the point D, represented by the torsor

( ) D P :

( ) .D P D S= − →

14.2.2. Equilibrium person-ladder

The equilibrium of the ladder is written:

( ) ( ) ( ) e .0A BS S S C S D S+ + + =→ + →

The equilibrium of the person is written:

( ) e 0 .P C S D S− − =→ →

One of the two equations can be replaced by expressing the equilibrium of the

person-ladder set:

( ) ( ) ( ) ( ) e e .0A BS S S P+ + + =

This latter equation removes the actions internal to the person-ladder set.

We have thus 12 scalar equations to derive 12 components of the actions of

connection: XA, YA, ZA; XB, YB, ZB; XC, YC, ..., MC; FD.

The reader will solve easily these equations and will take account of the

conditions of friction at the points of contact A, B and D.

Chapter 15

The Operator of Inertia

15.1 Matrix of inertia of a rectangular plate (Figure 15.22)

The axes of the trihedron (Oxyz) being axes of symmetry of the plate, the

matrix of inertia at the centre O of the plate in the basis ( ) ( ), , b i j k=

is of the

form:

( )( )

0 0

0 0

0 0

Oxb

O Oy

Oz

I

S I

I

=

I ,

with from (15.41) and (15.42):

( )

( )

2

2

d ( ),

d ( ),

.

OxS

OyS

Oz Ox Oy

I y m M

I x m M

I I I

=

=

= +

Using the Cartesian coordinates, the mass of the element of plate is d ( )m M =

d ds x yρ , where sρ is the mass per unit surface of the plate. Hence:

2 2 2

2 2

d d

a b

Ox sa bx y

I y x yρ=− =−

= .

Thus: 3

2

12 12Ox s

b mI a bρ= = ,

where m is the mass of the plate. The moment OyI is obtained by changing the

roles of a and b. Hence the matrix of inertia:

( )( )

( )

2

2

2 2

0 012

0 012

0 012

bO

m b

mS a

m a b

= +

I .

It is the result (15.86).

15.2 Matrix of inertia of a quarter of disc

We choose the coordinate system (Oxyz) such as the axes Ox

and Oy

are along

the edges of the disc (Figure Exercise 15.2).

The matrix of inertia is of the form:



( )( )

0

0

0 0

Ox Oxyb

O Oxy Oy

Oz

I P

S P I

I

− = −

I ,

with

( )

( )

( )( )

( )

2

2

2 2

d ( ),

d ( ),

d ( ),

d ( ).

OxS

OyS

Oz Ox OyS

OxyS

I y m M

I x m M

I I I x y m M

P xy m M

=

=

= + = +

=

The axes Ox

and Oy

play the same part. It follows that:

12

Ox Oy OzI I I= = .

To derive the moment of inertia OzI with respect to the axis Oz

, we use the polar

coordinates introduced in Figure 15.7. Hence:

2 2

0 0

d da

Oz sr

I r r r

π

αρ α

= == ,

where sρ is the mass per unit surface of the quarter of disc. Hence:

4 2

4 2 2Oz s

r aI mπρ= = ,

where m is the mass of the quarter of disc.

The product of inertia is:

y

O x

z

M

r

O x

z

y

()


2

0 0

sin cos d da

Oxy sr

P r r r r

π

αρ α α α

= == .

Thus: 4 21

4 2 2Oxy s

a aP mρπ

= = .

Hence the matrix of inertia:

( )( )

2 2

2 2

2

04 2

02 4

0 02

bO

a am m

a aS m m

am

π

π

−

= −

I .

Variation of the moment of inertia Let () be an axis of direction θ contained in the plane of the disc (Figure

Exercise 15.2). Its direction cosines are: cos , sin , 0.α θ β θ γ= = = From Rela-

tion (15.46), the moment of inertia with respect to the axis () is:

2 2cos sin 2 sin cosOx Oy OxyI I I P∆ θ θ θ θ= + − .

The moments of inertia and Ox OyI I being equal, we obtain:

sin 2Ox OxyI I P∆ θ= − .

Thus:

( )2 21 sin 2

4aI m∆ θ

π= − .

Hence the table of the variations of I∆ :

0 4π

2π 3

4π π

I∆2

4am ( )

2 214

amπ

−2

4am ( )

2 214

amπ

+2

4am

The moment passes through a minimum for the direction /4π and a maximum

for the direction 3 /4π . These directions correspond to the principal directions. In

these axes of basis ( )b′ , the matrix of inertia is of the form:

( )( )

( )( )

2

2

2

21 0 04

20 1 04

0 02

bO

am

aS m

am

π

π′

−

= +

I .

The property of the moment to pass through extreme values for the principal

directions of inertia is a general property.


15.3 Matrix of inertia of a hollowed cylinder We consider the hollowed cylinder (S) (Figure Exercise 15.3) of inner radius

a1, of outer radius a2 and height h.

The solid cylinder (S2) of radius a2 can be considered as the union of the hollo-

wed cylinder (S) and of the cylinder (S1) of radius a1 which was removed. The

property of associativity of the matrices of inertia is written:

( )( ) ( )( ) ( )( )2 1b b b

O O OS S S=I I + I .

Hence the matrix of inertia of the hollowed cylinder:

( )( ) ( )( ) ( )( )2 1b b b

O O OS S S= −I I I .

The matrix of inertia of the cylinder (S1) is, from (15.101):

( )( )

2 21

1

2 21

1 1

21

1

0 04 3

0 04 3

0 02

bO

a hm

a hS m

am

+

= +

I ,

where the mass m1 of the cylinder (S1) is: 2

1 1m a hπ ρ= ,

introducing the mass per unit volume of the cylinder.


y

x

z

a2 a1

O

h

(S)


The matrix of inertia of the cylinder (S2) is similarly:

( )( )

2 22

2

2 22

2 2

22

2

0 04 3

0 04 3

0 02

bO

a hm

a hS m

am

+

= +

I ,

where the mass m2 of the cylinder (S2) is: 2

2 2m a hπ ρ= .

The mass of the hollowed cylinder is:

( )2 22 1 2 1m m m a a hπ ρ= − = − .

From the expressions of the masses m, m1 and m2, we deduce:

2 21 2

1 22 2 2 22 1 2 1

, .a a

m m m ma a a a

= =− −

The application of the relation between the matrices of inertia leads to:

( )( )0 0

0 0

0 0

Oxb

O Oy

Oz

I

S I

I

=

I ,

with:

( )

( )

2 22 2

22 2

1 ,4 3

1 ,2

Ox Oy a

Oz a

a hI I m r

aI m r

= = + +

= +

introducing the radius ratio: 1

2a

ar

a= .

In the case of a hollowed cylinder of low thickness, we have 1 2a a a≈ ≈ and

the moments of inertia are written: 2 2

2

,2 3

.

Ox Oy

Oz

a hI I m

I ma

= = +

=

15.4 Matrix of inertia of a solid (Figure 15.23)

The solid is constituted of a cylinder (S1) of radius a and height h and of a half-

sphere (S2) of radius a.

The mass of the cylinder is:

21m a hπ ρ= ,

where ρ is the mass per unit volume of the solid. Its matrix of inertia in the

basis ( )b = ( ), , i j k

is:


( )( )

2 2

1

2 2

1 1

2

1

0 04 3

0 04 3

0 02

bO

a hm

a hS m

am

+

= +

I .

The mass of the half-sphere is 32

23

m aπ ρ= and its matrix of inertia at the

point O in the basis (b) is:

( )( )

22

22 2

22

20 0

52

0 05

20 0

5

bO

m a

S m a

m a

=

I .

The mass of the set cylinder and half-sphere is:

( )21 2

23

m m m a a hπ ρ= + = + .

From the expressions of the masses m1, m2 and m, we deduce:

1 2

23, .

2 23 3

ahm m m m

a h a h= =

+ +

The matrix of inertia of the set is diagonal and the diagonal terms are expressed

as:

( )2

2

1 42 2 15 ,

23

415 ,23

Ox Oy

Oz

rrI I ma

r

rI ma

r

+ += =

+

+=

+

Introducing the ratio: /r h a= .

15.5 Matrix of inertia of a non homogeneous parallelepiped (Figure

15.24)

The parallelepiped (S) is constituted (Figure Exercise 15.5) of four parallele-

pipeds (S1), (S2), (S3) and (S4).

The parallelepiped (S1) has a mass m1 and its mass centre G1 has for coordi-

nates ( )0, , 2 2b c− . Its matrix of inertia at G1 in the basis ( ) ( ), , b i j k=

is from

(15.103):


( )( )

( )

( )

( )

1

2 21

2 211

2 21

0 012

0 4 012

0 0 412

bG

mb c

mS a c

ma b

+ = +

+

I .


nates ( )0, , 2 2b c . Its matrix of inertia at G2 is similarly:

( )( )

( )

( )

( )

2

2 22

2 222

2 22

0 012

0 4 012

0 0 412

bG

mb c

mS a c

ma b

+ = +

+

I .


nates ( )0, , 2 2b c− . Its matrix of inertia is equal to the one of (S1):

( )( ) ( )( )3 1

3 1b b

G GS S=I I .


nates ( )0, , 2 2b c− − . Its matrix of inertia is equal to the one of (S2):

( )( ) ( )( )4 2

4 2b b

G GS S=I I .

The parallelepiped (S) constituted of the four parallelepipeds has a matrix of

inertia at O expressed by:

( )( ) ( )( ) ( )( ) ( )( ) ( )( )1 2 3 4b b b b b

O O O O OS S S S S+ +I = I I I + I .


2c

y

x

z

2a

2b

Om1

m1 m2

m2

(S1)

(S2)

(S3) (S4)

()

A


We have thus to express the matrices of inertia of each parallelepiped at point O.

For the parallelepiped (S1), we have:

( )( ) ( )( ) ( ) ( )1 1

1 1 1b b b

O G OGS S S=I I + D ,

with from Relation (15.24):

( ) ( )

( )( )

( )

1 1 1 11 1

1 1 1 11 1 1

1 1 1 1 1 1

2 21 1 1

2 21 1 1 1

2 21 1 1

G G G GG G

bG G G GOG G G

G G G G G G

m y z m x y m x z

S m x y m x z m y z

m x z m y z m x y

+ − − = − + −

− − +

D ,

where 1 1 1, and G G Gx y z are the coordinates of the mass centre G1. Hence:

( ) ( )

( )

1

2 21

21 11

21 1

0 04

04 4

04 4

bOG

mb c

m mS c bc

m mbc b

+ =

D .

The matrix of inertia at O is thus:

( )( )

( )

( )

( )

2 21

2 21 11

2 21 1

0 03

03 4

04 3

bO

mb c

m mS a c bc

m mbc a b

+ = +

+

I .

For the parallelepiped (S2), we have:

( )( ) ( )( ) ( ) ( )2 2

2 2 2b b b

O G OGS S S=I I + D ,

where the coordinates of the mass centre are ( )0, , 2 2b c . With the same process

as previously, we obtain:

( )( )

( )

( )

( )

2 22

2 22 22

2 22 2

0 03

03 4

04 3

bO

mb c

m mS a c bc

m mbc a b

+ = + −

− +

I .

Applying the preceding relations to the parallelepipeds (S3) and (S4), we find that: ( )( ) ( )( ) ( )( ) ( )( )3 1 4 2 et b b b bO O O OS S S S= =I I I I .

Hence the matrix of inertia:


( )( )

( )( )

( )( ) ( )

( ) ( )( )

2 21 2

2 21 2 1 2

2 21 2 1 2

20 0

32 1

03 2

1 20

2 3

bO

m m b c

S m m a c m m bc

m m bc m m a b

+ +

= + + − − − − + +

I

Moment of inertia with respect to a diagonal of the parallelepipedLet () be the diagonal passing through the vertex A of the parallelepiped. The

vector OA

is a direction vector of the axis ():

OA a i b j c k= + +

.

The direction cosines of the axis are thus:

2 2 2 2 2 2 2 2 2, , .a b c

a b c a b c a b cα β γ= = =

+ + + + + +

The moment of inertia with respect to () is deduced from Relation (15.46). Thus:

2 2 2 2Ox Oy Oz OyzI I I I P∆ α β γ βγ= + + − .

That leads to:

( )( ) ( )2 2 2 2 2 2 2 21 2 1 22 2 2

1 43

I m m a b b c a c m m b ca b c

∆ = + + + − − + +

.

15.6 Matrix of inertia of a solid sphere with a spherical holeA solid sphere (S) of radius a contains a spherical hole of radius /2a , passing

through the centre of the solid sphere (Figure Exercise 15.6). The solid sphere (S1)

can be considered as constituted of the hollowed sphere (S) and of the solid sphere

(S2) which was removed. The property of associativity of the matrices of inertia is

written: ( )( ) ( )( ) ( )( )1 2b b b

O O OS S S=I I + I


y

z

x

O G2

(S2)

(S1)


Hence the matrix of inertia of the hollowed sphere:

( )( ) ( )( ) ( )( )1 2b b b

O O OS S S= −I I I .

The matrix of inertia of the solid sphere (S1) is:

( )( )

21

21 1

21

20 0

52

0 05

20 0

5

bO

m a

S m a

m a

=

I ,

where m1 is the mass of the solid sphere (S1).

The matrix of inertia of the solid sphere (S2) at the mass centre G2 is:

( )( )2

22

222

22

0 010

0 010

0 010

bG

ma

mS a

ma

=

I ,

where m2 is the mass of the solid sphere (S2). This matrix has to be evaluated at

the point O. Thus: ( )( ) ( )( ) ( ) ( )

2 22 2 2

b b bO G OG

S S S=I I + D ,

where the coordinates of the point G2 are ( )0, , 02a . Hence:

( ) ( )2

22

2

22

0 04

0 0 0

0 04

bOG

ma

S

ma

=

D .

Thus:

( )( )

22

222

22

70 0

20

0 010

70 0

20

bO

m a

mS a

m a

=

I .

So, we deduce the expression of the matrix of inertia of the hollowed sphere:

( )( )

( )( )

( )

21 2

221

21 2

2 70 0

5 202

0 05 10

2 70 0

5 20

bO

m m a

mS m a

m m a

−

= − −

I .


The radius of the removed sphere (S2) is half the radius of the sphere (S1). It

follows that 2 1/8m m= . That leads to:

1 28 , .7 7

mm m m= =

where m is the mass of the hollowed sphere (S).

Finally the matrix of inertia is written as:

( )( )

2

2

2

570 0

14031

0 070

570 0

140

bO

ma

S ma

ma

=

I .

15.7 Matrix of inertia of a plate with a hole (Figure Exercise 15.7)

The reader will find easily that the matrix of inertia of the plate (S) with a

circular hole is:

( )( )

2 2

2 2

2 2 2

0 01 12 4

0 01 12 4

0 01 12 2

cc

bO c

c

cc

m b cr

r

m a cS r

r

m a b cr

r

− −

= − −

+ − −

I ,

introducing the mass m of the plate with the hole and the ratio of surfaces: 2

ccr

abπ= .


y

xb

a

O

c

z

Chapter 16

Kinetic and Dynamic Torsors

Kinetic Energy

16.1 Motion of rotation of a parallelepiped about an axis passing through its centre (Figure 16.2a)

The study of the motion will be always implemented according to the same

process: 1) determination of the parameters of situation, 2) kinematic study, 3)

kinetic study.

1. Parameters of situation We associate a coordinate system attached to the reference (T) of the motion:

the system (Oxyz) (Figure Exercise 16.1) such as the axis Oz

coincides with the

axis of rotation and such as the point O coincides with the centre of the paralle-

lepiped. The axis Ox

is chosen along a given direction of the plane (Oxy).


We choose a particular point of the parallelepiped (S): the point O. This point

is fixed during the motion. So, there is no parameter of translation.


We associate a trihedron attached to the parallelepiped: the trihedron ( )S SOx y z

such as the axes SOx

and SOy

are parallel to the edges. We have a rotation of


y

z

x

(T)

O

yS

xS

(S)

()


angle about the direction k

. Thus, one parameter of rotation .

The basis change is:

cos sin ,

sin cos ,

.

S

S

i i j

j i j

k

ψ ψ

ψ ψ

= +

= − +

2. Kinematic study

The kinematic torsor( ) TS relative to the motion of the parallelepiped (S) with


( ) ( )

( ) ( )


( , ) 0, velocity vector of the particular point .

T TS S

T TO S

R k

O t O

ω ψ = =

= =

3. Kinetic study 3.1. Kinetic torsor


S relative to the motion of the parallelepiped (S) with

respect to the reference (T) has for elements of reduction at the point point O,

which is the mass centre of the parallelepiped:

( ) ( )

( ) ( ) ( )

( , ) 0,

, since is the mass centre.

T TS

T TO S O S

R m O t

S Oω

= =

=

The operator of inertia ( )O S of the parallelepiped at the point O is represented

in the basis ( ) ( ), , S S Sb i j k=

by the matrix of inertia expressed in the preceding

chapter:

( )( )

( )

( )

( )

2 2

2 2

2 2

0 012

0 012

0 012

SbO

m b c

mS a c

m a b

+

= + +

I ,

where m is the mass, and a, b, c are the edges of the parallelepiped. We have to

note that the matrices of inertia were evaluated in the preceding chapter in a basis

attached to the solids. Hence the use here of the basis ( )Sb .

It results that: ( )

( ) ( )

2 2

0,

,12

TS

TO S

R

ma b k C kψ ψ

=

= + =

setting:

( )2 2

12

mC a b= + .

3.2. Kinetic energy

The expression of the kinetic energy is:


( ) ( ) ( ) c1( )2

T T TS SE S = ⋅ .

To derive the product of the torsors, we implement the sum of the crossed scalar

products of the resultants and moments of the torsors expressed at the same point.

Here:

( ) ( ) ( ) ( ) ( ) c1( )2

T T T T TO OS S S SE S R R = + ⋅ ⋅

.

Hence:

( ) ( )2 2 2 2c

1( )24 2

T mE S a b Cψ ψ= + = .

3.3. Dynamic torsor

The dynamic torsor ( ) TS relative to the motion of the parallelepiped (S) with


( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( , ) ,

.

T TS

T T T TO S O S S O S

R ma O t

S Sω ω ω

=

= + ×

The vector ( )TSω is expressed by kψ

. The vector product is null, since the two

vectors are collinear. Hence:

( ) ( ) ( )

2 2

0,

.12

TS

TO S

R

ma b k C kψ ψ

=

= + =

We observe that the expression of the moment of the dynamic torsor could

have been got by deriving the expression of the moment at O of the kinetic torsor

(Property (16.24)). If this property is interesting in the present case, it is not

recommended to use this property in a general way.

16.2 Motion of rotation of a parallelepiped about an eccentric axis (Figure 16.2b)

The analysis is implemented according the same process as the one used in the

preceding exercise.

1. Parameters of situation We associate a coordinate system attached to the reference (T) of the motion:

the system (Oxyz) (Figure Exercise 16.2) such as the axis Oz

coincides with the

axis of rotation and such as the point O is located in the medium plane of the

parallelepiped. The axis Ox

is chosen along a given direction of the medium plane.


We choose a particular point of the parallelepiped (S): the point O. This point

is fixed during the motion. So, there is no parameter of translation.


We associate a coordinate system attached to the parallelepiped: the system

( )S SOx y z such as the axes SOx

and SOy

are parallel to the edges. We have a

rotation of angle about the direction k

. Hence one parameter of rotation .




cos sin ,

sin cos ,

.

S

S

i i j

j i j

k

ψ ψ

ψ ψ

= +

= − +

2. Kinematic study 2.1. Kinematic torsor


respect to the reference (T) has for elements of reduction at O:

( ) ( )

( ) ( )


( , ) 0, velocity vector of the particular point .

T TS S

T TO S

R k

O t O

ω ψ = =

= =

2.2. Kinematic vectors of the mass centre

The velocity and acceleration vectors of the mass centre are needed for deri-

ving the resultants of the kinetic and dynamic torsors.

The velocity vector can be deduced while writing the expression of the moment

at the point G of the kinematic torsor:

( ) ( ) ( ) T T TG OS S SR OG= + ×

.

That leads to the relation between the velocity vectors:

( ) ( ) ( ) ( , ) ( , )T T T

SG t O t OGω= + ×

,

with:

y

z

x

(T)

O

yS

xS

(S)

()

d

G


( )2S S

aOG d i l i= − =

,

where a is the length of the edge of the parallelepiped and setting:

2al d= − .

Hence: ( )

( , )TSG t l jψ= .

The velocity vector can also be obtained from the relation of the definition of

the velocity vector:

( )( )

d( , )

d

TT G t OG

t=

,

where OG

was expressed previously. We thus obtain directly the preceding

expression for the velocity vector.

The acceleration vector is obtained by deriving the velocity vector with respect

to time and considering that ψ and Sj

are functions of time. We thus obtain:

( ) ( )2( , )TS Sa G t l i jψ ψ= − +

.

The kinematic vectors can be eventually expressed in the basis ( ), , i j k

intro-

ducing the relation of basis change. Hence: ( ) ( )( , ) sin cosT G t l i jψ ψ ψ= − +

,

( ) ( ) ( )2 2( , ) cos sin sin cosTa G t l i jψ ψ ψ ψ ψ ψ ψ ψ = − + + − +

.



S relative to the motion of the parallelepiped (S) with


( ) ( )

( ) ( ) ( ) ( )

( , ) ,

( , ) .

T TS S

T T TO S O S

R m G t ml j

m OG O t S

ψ

ω

= =

= × +

The velocity vector of the point O being null, the first term of the moment is null.

The operator of inertia ( )O S of the parallelepiped at the point O is represented in

the basis ( ) ( ), , S S Sb i j k=

by the matrix of inertia at G as:

( )( ) ( )( ) ( )( )S S Sb b bO G OGS S S=I I + D .

The coordinates of the point G relatively to the system ( )S SOx y z being (l, 0, 0),

we have:

( )( ) 2

2

0 0 0

0 0

0 0

SbOG S ml

ml

=

D .

Hence the matrix of inertia at O:


( )( )

( )2 2

2 22

2 22

0 012

0 012

0 012

SbO

m b c

a cS m l

a bm l

+ += +

+ +

I .

Finally we obtain: ( )

( ) ( )2

,

.

TS S

TO S

R ml j

C ml k

ψ

ψ

=

= +

3.2. Kinetic energy

The kinetic energy is expressed as:

( ) ( ) ( ) c1( )2

T T TS SE S = ⋅ .

We have to implement the crossed scalar products of the resultants and moments

of the torsors expressed at the point O. we obtain:

( ) ( )2 2c

1( )2

TE S C ml ψ= + .

3.3. Dynamic torsor

The dynamic torsor ( ) TS relative to the motion of the parallelepiped (S) with


( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

2 ( , ) ,

( , ) .

T TS S S

T T T T TO S O S S O S

R ma G t ml i j

m OG a O t S S

ψ ψ

ω ω ω

= = − +

= × + + ×

The first term of the moment is null since the point O is fixed. In the third term,

the two vectors are collinear to k

. Their vector product is thus null. Hence:

( ) ( )( ) ( )

2

2

,

.

TS S S

TO S

R ml i j

C ml k

ψ ψ

ψ

= − +

= +

16.3 Motion of a parallelepiped on a plane (Figure 16.3)

This motion was studied before in Exercise 9.1. We consider again the ele-

ments introduced.


We associate (Figure Exercise 16.3) the coordinate system (Oxyz) to the plane

(T) on which the parallelepiped moves, such as the plane (Oxy) coincides with the

plane (T). The axis Ox

is chosen along a given direction of the plane.


We choose a particular point of the parallelepiped (S). In Exercise 9.1, we

chose the point A vertex of the parallelepiped. However, it is possible to choose



the mass centre G, that will simplify the kinetic analysis. The coordinates of G are

( ), , /2x y c where c is the height of the parallelepiped. We have two parameters

of translation: x and y.


We associate the coordinate system ( )S SGx y z attached to the parallelepiped,

such as the axes SGx

and SGy

are parallel to the edges. The orientation is given

by the rotation about the direction k

. Hence:

cos sin ,

sin cos ,

.

S

S

i i j

j i j

k

ψ ψ

ψ ψ

= +

= − +

2. Kinematic study 2.1. Kinematic torsor


respect to the plane (T) has for elements of reduction at G:

( ) ( )

( ) ( )


( , ) , velocity vector of the point .

T TS S

T TG S

R k

G t x i y j G

ω ψ = =

= = +

2.2. Acceleration vector of the mass centre

The vector is:

( )( )

( )

d( , ) ( , )

d

TT Ta G t G t x i y j

t= = +

.



S relative to the motion has for elements of reduction

at the point G:

(T) (S) A

B

C

D A'

B'

C'

D'

O

x

x

xS

yS

y z z

ijk

Si

Sj G


( ) ( ) ( )( ) ( ) ( ) ( )

2 2

( , ) ,

.12

T TS

T TG S G S

R m G t m x i y j

mS a b k C kω ψ ψ

= = +

= = + =

3.2. Kinetic energy

The kinetic energy is expressed by:

( ) ( ) ( ) c1( )2

T T TS SE S = ⋅ .

Thus, expressing the crossed scalar products of the resultants and moments at the

point G, we obtain:

( ) ( )2 2 2c

1 1( )2 2

TE S m x y Cψ= + + .

The first term is the kinetic energy of translation and the second term is the kinetic

energy of rotation.

3.3. Dynamic torsor

The dynamic torsor ( ) TS relative to the motion has for elements of reduction

at the point G:

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )

( , ) ,

.

T TS


R ma G t m x i y j

S S C kω ω ω ψ

= = +

= + × =

The three exercises 16.1, 16.2 and 16.3 allow us to underline the differences

between three motions of a same solid.

Chapter 21

Dynamics of Systems with One Degree of Freedom

Analysis of Vibrations

21.1 Mass drived by a wheel (Figure 21.15)

1. Equation of motion of the mass when the wheel is motionless We consider first the case where the point A of the axle-tree is fixed (Figure

Exercise 21.1). At the equilibrium the mass centre of the mass m coincides with

O. The actions exerted on the mass are: the gravity, the action of the spring and

the viscous friction. Hence the equation of motion:

my mg ky cy= − − − .

The position of equilibrium of the mass ( 0, 0y y= = ) is:

stmg

yk

= − .

With the given values, we obtain: st 5.2 mm.y = −

Around this position of equilibrium, the equation of motion is:

0my cy ky+ + = ,

or

202 0y y yδ ω+ + = ,

where δ is the damping coefficient and the natural angular frequency is

0 /k mω = , thus 0 43.3 rad/sω = .

2. Equation of motion when the wheel moves On account of the undulated surface, the centre A of the wheel has a motion

defined by the equation sin /Ay d x lπ= . Considering that at the origin instant

0t = , the point of contact of the wheel is at 0x = , and introducing the distance

covered at the instant t: x t= , the displacement of the centre of the wheel is:

sinAy d tω= ,

where the angular frequency ω is given by /lω π= , being the speed of the

wheel.


k

m

A

y

OG

568 Chapter 21 Dynamics of Systems with One Degree of Freedom. Analysis of vibrations

The total displacement y of the mass is: A ry y y= + , where ry is the relative

displacement along y relatively to the system attached to the wheel. In the absence

of friction, the equation of motion is written as:

rmy mg ky= − − .

The position of equilibrium was obtained previously and around this position of

equilibrium, the equation of motion is:

Amy ky k y+ = − .

In the case of a viscous friction, it would be necessary to implement an analysis

similar to the one of Section 21.3.36. So as to simplify, we introduce simply the

term cy in the preceding equation. Hence:

sinmy cy ky kd tω+ + = − .

The equation of motion is thus written in the form:

20 m2 siny y y q tδ ω ω+ + = ,

setting except to the sign:

mkd

qm

= .

The equation obtained is similar to Equation (21.97) and the amplitude of the

response is given from (21.106) by:

m st( )y K yω= ,

where the magnification factor ( )K ω is expressed in (21.107). Its variation is

reported in Figure 21.8.

The amplitude my of the motion passes through a maximum, from (21.124), for:

2m 0 01 2 0.990ω ξ ω ω= − = .

This frequency corresponds to the velocity m of the displacement of the wheel

given by:

0mm 0.990

ll ωωπ π

= = .

Thus a velocity of about 49 km/h. For this velocity, the amplitude of the vibrations

is from (21.125):

m st2

1

2 1y y

ξ ξ=

−.

Thus:

m st5.025 26.1 mmy y= = .

The amplitude of the vibrations decreases when the velocity decreases or in-

creases from the value of 49 km/h.

21.2 Forced vibrations in the case of an excitation in triangle form The forced vibrations in the case of a periodic force imposed are considered in

Section 21.3.4. The force variation as a function of time is expanded in the form


of Fourier series (21.134) of which the coefficients are given by Expressions

(21.135)-(21.137).

In the case of the variation in triangle form of Figure 21.16, the function ( )f t

is antisymmmetric with respect to /2t T= . Hence:

/2

0 /2

( ) d ( ) dT T

T

f t t f t t= − .

It follows that the term a0 given by (21.135) is zero.

The term an is expressed by (21.136). The function cos n tω is symmetric with

respect to /2t T= . Associated to the antisymmetry of ( )f t , this symmetry leads to:

/2

0 /2

( ) cos d ( )cos dT T

T

f t n t t f t n t tω ω= − .

The terms an are thus zero.

The term bn is given by Expression (21.137). In the interval from 0 to /2T , the

function ( )f t is symmetric with respect to /4T , and in the interval from /2T to T,

the function is symmetric with respect to 3 /4T . Furthermore, when n is even, the

corresponding parts of the function sin n tω are antisymmetric. It follows that the

terms bn are zero for 2, 4, etc.n =

When n is odd, the function ( )f t and sin n tω are antisymmetric with respect to /2T . It results that Expression (21.137) leads to:

/4

0 0

2 8( )sin d ( )sin d

T T

nb f t n t t f t n t tT T

ω ω= = .

In the interval from 0 to /4T , we have:

4( )

Af t t

T= ,

and the expression of bn becomes: /4

20

32sin d

T

nA

b t n t tT

ω= .

For the integration, we introduce the change of variable: 2

u n t n tT

πω= = . Hence:

2

2 20

8sin d

n

nA

b u u un

π

π= .

Integrating by part, we obtain:

1

22 2 2 2

8 8sin ( 1) , 1, 3, 5, . . .

2

n

nA A

b n nn n

π

π π

−

= = − =

Hence:

( )2

8 1 1( ) sin sin 3 sin 5 . . .

9 25

Af t t t tω ω ω

π= − + − .

We observe that the series converges rapidly. It results that the force in triangle

570 Chapter 21 Dynamics of Systems with One Degree of Freedom. Analysis of vibrations

form produces approximatively the same effect as a sinusodal force of period T

expressed by:

2

8( ) sin

Af t tω

π= .

So as to consider the importance of the second term of the series, we can

evaluate the magnification factor ( )K ω given by Expression (21.107). At the

resonance ( 0ω ω= ), the magnification factor for the first term is:

1 01

( )2

K ωξ

= ,

And for the second term, we have:

3 02

1 1( )

9 4 36K ω

ξ=

+.

If 0.10ξ = , we obtain:

1 0 3 0( ) 5, ( ) 0.053,K Kω ω= =

thus an error of about 1 % while considering only the first term.

Chapitre 22

Motion of Rotation of a Solid

about a Fixed Axis

22.1 Motion of a parallelepiped about an axis passing through its centre (Figure 22.6)

The systems of reference used are reported in Figure Exercise 22.1. The para-

meters of situation, the kinematic and kinetic analyses of the motion were consi-

dered in Exercise 16.1 (Paragraphs 1, 2 and 3). The following stage (Stage 4)

consists in analysing the mechanical actions exerted.

4. Mechanical actions exerted on the parallelepiped

4.1. Action of gravity

This action is represented by the torsor ( ) e S of elements of reduction at the

mass centre:

( )

( )

e ,

e 0,G

R S mg i

S

=

=

where m is the mass of the parallelepiped.

4.2. Action of the spiral spring

The spring exerts a couple-action represented by the torsor ( ) S of ele-

ments of reduction:

( )

( )

0,

,G

R S

S K kψ

=

= −


G

(R)

xxS

yS

y

z

572 Chapter 22 Motion of Rotation of a Solid about a Fixed Axis

wher K is the constant of torsion and the angle ψ is measured from the position

where the spring does not exert any action on the solid.

4.3. Action of the support induced by the hinge connection

This action is represented by the torsor ( ) S of elements of reduction at the

point G:

( )

( )

,

.

l l l

G l l l

R S X i Y j Z k

S L i M j N k

= + +

= + +

The components Xl, Yl, . . . , Nl are to be determined.

The power developed by the action is: ( ) ( ) ( ) ( ) T T

SP S S= ⋅ ,

where the kinematic torsor was expressed in Paragraph 2 of the correct version of

Exercice 16.1. We then obtain: ( ) ( ) T

lP S N ψ= .

5. Fundamental relation of dynamics

The relation is: ( ) ( ) ( ) ( ) e ,TS S S S= + +

where the dynamic torsor was expressed in Paragraph 3 of the correct version of

Exercice 16.1. We obtain the six scalar equations:

resultant:

0 ,

0 ,

0 ,

l

l

l

mg X

Y

Z

= +

=

=

moment at G:

0 ,

0 ,

.

l

l

l

L

M

C N Kψ ψ

=

=

= −

The first five equations give the components of the actions of connection:

, 0, 0, 0, 0.l l l l lX mg Y Z L M= − = = = =

The last equation is the equation of motion:

.lC N Kψ ψ= −

For solving the equation, it is necessary to introduce an assumption on the physic-

cal nature of the hinge connection: with friction or without friction.

In the case where there is no friction, the power developed is zero. Thus:

( ) ( ) 0TlP S N ψ= = .

That leads to 0.lN = The equation of motion is thus written as:

20 0ψ ω ψ+ = ,


setting:

( )20 2 2

12K KC m a b

ω = =+

.

In the case of a hinge connection with friction of viscous type, the component

lN is opposed to the angular velocity of rotation:

lN cψ= − ,

and the equation of motion is written in the form:

202 0ψ δψ ω ψ+ + = ,

setting:

( )2 2

62c cC m a b

δ = =+

.

The equations of motion obtained are the ones of a motion with one degree of

freedom (Chapter 21) with friction or without friction.

22.2 Motion of a parallelepiped about an eccentric axis (Figure 22.7a)

The reference systems used are reported in Figure Exercise 22.2. The para-

meters of situation, the kinematic and kinetic analyses of the motion were

considered in Exercise 16.1 (Paragraphs 1, 2 and 3). The following stage (Stage 4)

consists in analysing the mechanical actions exerted.

4. Mechanical actions exerted on the parallelepiped

4.1. Action of gravity

This action is represented by the torsor ( ) e S of elements of reduction at


G

Od

x xS

yS

y

z


the mass centre:

( )

( )

e ,

e 0.G

R S mg i

S

=

=

4.2. Action of the support induced by the hinge connection

This action is represented by the torsor ( ) S of elements of reduction at the

point O:

( )

( )

,

.

l l l

O l l l

R S X i Y j Z k

S L i M j N k

= + +

= + +

The power developed by the action is:

( ) ( ) ( ) ( ) T TSP S S= ⋅ ,

where the kinematic torsor was expressed in Paragraph 2 of the correct version of

Exercise 16.2. We thus obtained: ( ) ( ) T

lP S N ψ= .

In the present case, we observe that it is necessary to express the moment of the

action of connection at a point of the axis of rotation (the point O) so as to obtain

a simple relation for the power and easy to apply for expressing the conditions of

friction or non friction.

4.3. Motor couple

It is possible to exert on the parallelepiped a motor couple represented by the

torsor ( ) S , of which the elements of reduction at the point O are:

( )

( )

0,

,O

R S

S N k

=

=

where N is known.

5. Fundamental relation of dynamics

The relation is written:

( ) ( ) ( ) ( ) e ,TS S S S= + +

where the dynamic torsor was expressed in Paragraph 3 of the correct version of

Exercise 16.2. This equation leads to the equation of the resultant and the one of

the moment at the point O:

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

e ,

e .

TS

TO O O OS

R R S R S R S

S S S

= + +

= + +

So as to develop the equation of the moment at O, it is necessary to express the

moment at O of the action of gravity:

( ) ( ) e e sinO S R S GO mgl kψ= ∧ = − .

Hence the six scalar equations:


( )( )

( )

2

2

2 2

cos sin ,

sin cos ,

0 ,

0 ,

0 ,

sin .12

l

l

l

l

l

l

ml mg X

ml Y

Z

L

M

m a b mgl N N

ψ ψ ψ ψ

ψ ψ ψ ψ

ψ ψ

− + = +

− + =

=

=

=

+ = − + +

We obtain 6 equations for 7 unknowns: , , . . . , , .l l lX Y N ψ One additional equa-

tion about lN will be given by the physical nature of the connection. The last

equation is the equation of motion which allows us to derive the motion ψ as a

function of time. The other equations allow us to obtain the components of the

action of connection.

In the case of a connection without friction, 0lN = and the equation of motion,

in the absence of motor couple, is written:

20 sin 0ψ ω ψ+ = ,

setting:

20 2 2

12gl

a bω =

+.

In the case of a connection with viscous friction: ,lN cψ= − the equation of

motion is written: 202 sin 0ψ δψ ω ψ+ + = ,

introducing the coefficient expressed previously.

The equations of motion are reduced to the usual equations of the motion with

one degre of freedom in the case of low values of the angle of rotation for which

sin .ψ ψ≈

22.3 Motion of a parallelepiped about an eccentric axis with the action of a spiral spring (Figure 22.7b)

The analysis differs from the one of the preceding exercise only by the intro-

duction of the mechanical action exerted by the spiral spring (action considered

previously in Exercise 22.1). The action exerted is a couple-action represented by

the torseur ( ) S of which the elements of reduction at the point O are:

( )

( )

0,

,O

R S

S K kψ

=

= −

where K is the constant of torsion of the spring. The relation of the moment

assumes that the spring is set up so as no couple of torsion is exerted when the

mass centre G of the parallelepiped is along the vertical passing through the point

O ( 0ψ = ).


Thus, it follows that only the equation of motion is modified and is written, in the

absence of motor couple, as:

( )2 2 sin .12

lm a b mgl K Nψ ψ ψ+ = − − +

In the case of low values of the angle of rotation and considering a viscous

friction, the equation of motion is written in the usual reduced form of a motion

with one degree of freedom, introducing:

( )20 2 2

12 mgl K

a bω

+=

+.

The effects of the actions of gravity and spring are superimposed.

Chapter 24

Other Examples of Motions of Rigid Bodies

24.1 Motion of two solids (Figure 24.7)

1. Parameters of situation 1.1. Motion of the solid (S1) with respect to the support (T)

We associate a coordinate system ( Oxyz ) attached to the support ( )T such as

the point O is the centre of the upper side of the solid 1( ),S the axis Oz

coincides

with the axis of rotation ( )1∆ and the axis Ox

is downward vertical (Figure

Exercise 24.1).


We choose a particular point of the solid 1( )S : the point O. This point is fixed.

Thus, there is not any parameter of translation.


We associate a coordinate system attached to the solid 1( ) :S the system

( )1 1Ox y z , such as the axis 1Ox

coincides with the axis of the cylinder 1( ).S The

orientation is defined by the angle ψ of rotation about the direction k

. We have

thus one parameter of rotation ψ .

The relation of basis change is written:


(S1)

(S2)

G2

G1

y1

y

z

z

z2

y1

y2

x1

x

O


1

1

cos sin ,

sin cos ,

.

i i j

j i j

k

ψ ψ

ψ ψ

= +

= − +

1.2. Motion of the solid (S2) with respect of the solid (S1)


We choose a particular point of the solid 2( )S : the mass centre 2G . The coor-

dinates of this point in the system ( )1 1Ox y z are ( ) , 0, 0 .x We have then one para-

meter of translation x. Thus:

2 1OG x i=

.


We associate the system ( )2 1 2 2G x y z to the solid 2( )S (Figure Exercise 24.1).

With respect to the system ( )2 1 1G x y z we have a rotation of angle θ about the

direction 1.i

Hence one parameter of rotation .θ The basis change is:

1

2 1

2 1

,

cos sin ,

sin cos .

i

j j k

k j k

θ θ

θ θ

= + = − +

2. Kinematic study 2.1. Motion of the solid (S1) with respect to the support (T)

2.1.1. Kinematic torsor


TS relative to the motion of the solid 1( )S with res-

pect to the support ( )T has for elements of reduction at the point O:

( ) ( )

( ) ( )

1 1

1


( , ) 0, velocity vector of the point .

T TS S

T TO S

R k

O t O

ω ψ = =

= =

2.1.2. Kinematic vectors of the mass centre

If 1h is the height of the cylinder 1( ),S the position vector of the mass centre

1G is:

11 1

2

hOG i=

.

The velocity vector of the mass centre can be obtained either by using the

expression of the composition of the velocity vectors, deduced from the relation

between the moments of the kinematic torsor:

( ) ( ) ( ) ( )

1 11 11( , ) ( , )T T T T

S SG t O t OG OGω ω= + × = ×

,

or by using the relation of definition of the velocity vector:

( )( )

11d( , )d

TT G t OG

t=

.


We obtain:

( ) 11 1( , )

2T h

G t jψ= .

The acceleration vector is then deduced by deriving the velocity vector:

( )

21 11 1 1( , )

2 2T h h

a G t i jψ ψ= − +

.

The kinematic vectors can eventually be expresssed in the basis ( ), , i j k

by

using the basis change. We obtain:

( ) ( )11( , ) sin cos

2T h

G t i jψ ψ ψ= − +

,

( ) ( ) ( )

2 21 11( , ) cos sin sin cos

2 2T h h

a G t i jψ ψ ψ ψ ψ ψ ψ ψ= − + + − +

.


The motion is defined by three parameters of situation: , , .x θ ψ The kinematic

torsor ( ) 2

TS relative to the motion of the solid 2( )S with respect to the support

( )T has for elements of reduction at the point 2G :

( ) ( )

( ) ( )

2 2

2 22 2


( , ) , velocity vector of the point .

T TS S

T TG S

R

G t G

ω =

=

The instantaneous rotation vector ( )

2

TS

ω

is equal to the sum of the rotation

vector ( )

1

TS

ω

and rotation vector ( )1

2

SS

ω

relative to the motion of 2( )S with respect

to 1( )S : ( )1

21

SS

iω θ= . Hence:

( )

21

TS

k iω ψ θ= + .

The velocity vector of 2G can be obtained by deriving the position vector

2OG

expressed in Paragraph 1.2.1. We obtain:

( )( )

22 1 1d( , )d

TT G t OG x i x j

tψ= = +

.

Next, the acceleration vector is obtained by deriving the velocity vector:

( )( )

( ) ( ) ( ) 2

2 2 1 1d( , ) ( , ) 2d

TT Ta G t G t x x i x x j

tψ ψ ψ= = − + +

.

The kinematic vectors can eventually be expressed in the basis ( ), , i j k

. Thus:

( ) ( ) ( ) 2( , ) cos sin sin cosT G t x x i x x jψ ψ ψ ψ ψ ψ= − + +

,

( ) ( ) ( )

( ) ( )

22

2

( , ) cos 2 sin

sin 2 cos .

Ta G t x x x x i

x x x x j

ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ

= − − + + − + +


3. Kinetic study 3.1. Motion of the solid (S1) with respect to the support (T)

3.1.1. Kinetic torsor



the support has for elements of reduction at the point O:

( ) ( )

( ) ( ) ( ) ( )

1

1 1

11 1 1 1

11 1

( , ) ,2

( , ) ,

T TS

T T TO OS S

hR m G t m j

m OG O t S

ψ

ω

= = = × +

where 1m is the mass of the solid 1( ).S

The first term of the moment is null, since the point O is fixed. The operator

( )1O S of inertia at O is represented in the basis ( ) ( )1 1 1, , b i j k=

by the matrix:

( ) ( )1

1

1 1

1

0 0

0 0 .

0 0

bO

A

S B

C

=

I

If the mass of the hollow cylinder 1( )S is preponderante compared to the mass of

the axis of rotation, we have (Exercise 15.3):

( )

( )

22 1

1 1

2 22 1 1

1 1 1

1 ,2

1 ,4 4

aA m r

a hB C m r

= +

= = + +

where r is the ratio 2 1/a a of the iner and outer radii of the cylinder. Hence:

( ) 1

1T

O SC kψ= .

3.1.2. Kinetic energy

The kinetic energy is given by:

( ) ( ) ( ) 1 1

c 11

( )2

T T TS S

E S = ⋅ .

Thus:

( ) 2c 1 1

1( )

2TE S Cψ= .

3.1.3. Dynamic torsor

The dynamic torsor dynamique ( ) 1

TS relative to the motion of the solid (S1)

with respect to the support has for elements of reduction at the point O:

( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )

1

1 1 1 1

1 1

11 1 1

( , ),

( , ) .

T TS

T T T T TO O OS S S S

R m a G t

m OG a O t S Sω ω ω

=

= × + + ×

The expression of the acceleration vector at the point 1G was expressed pre-


viously in the basis ( )1 1, , i j k

and in the basis ( ), , i j k

. The first and third terms

of the moment are null. Hence:

( ) 1

1T

O SC kψ= .





the support has for elements of reduction at the mass centre 2G :

( ) ( ) ( )( ) ( ) ( )

2

2 22 2

2 2 2 1 1

2

( , ) ,

,

T TS

T TG GS S

R m G t m x i x j

S

ψ

ω

= = +

=

where 2m is the mass of the solid 2( ).S

The operator ( )2 2G S of inertia at 2G is represented in the basis ( )2b =

( )1 2 2, , i j k

by the matrix:

( ) ( )2

2

2

2 2

2

0 0

0 0 ,

0 0

b

G

A

S B

C

=

I

with from (15.101):

2 222 2 2

2 2 2 2 2, ,2 4 3

a m hA m B C a

= = = +

where 2h is the height of the cylinder 2( ).S

To evaluate the moment at 2G , it is needed to express rotation vector in the

basis ( )2b :

( )

21 1 2 2sin cos .T

Sk i i j kω ψ θ θ ψ θ ψ θ= + = + +

On account of 2 2B C= , we obtain:

( ) 2 2

2 2 1T

G SC k A iψ θ= +

.

The result obtained is the same as if the matrix of inertia was expressed directely

in the basis ( )1b .



( ) ( ) ( ) 2 2

c 21

( )2

T T TS S

E S = ⋅ .

Expressing the crossed scalar products of the elements of reduction at the point

2G , it comes:


( ) ( )2 2 2 2 2c 2 2 2 2

1 1 1( )

2 2 2TE S m x x C Aψ ψ θ= + + + .

This expression can be put in the form:

( ) ( )2 2 2 2c 2 2 2 2 2

1 1 1( )

2 2 2TE S m x C m x Aψ θ= + + + .

The first term is the kinetic energy of translation. The second term appears to be

the sum of the usual kinetic energy 22

1

2C ψ associated to a motion of rotation

about an axis passing through the mass centre and of the kinetic energy induced

by the shift of a distance x of the axis of rotation. The third term is the energy of

rotation about the direction 1i

.


The dynamic torsor ( ) 2


to the support has for elements of reduction at the mass centre 2G :

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( )

2

2 2 22 2 2 2

22 2 2 1 2 1

2 2

2 1 2 2 1 2

( , ) 2 ,

.

T TS

T T T TG G GS S S S

R m a G t m x x i m x x j

S S

A i A C j C k

ψ ψ ψ

ω ω ω

θ ψθ ψ

= = − + +

= + ×

= + − +

4. Mechanical actions

4.1. Mechanical actions exerted on the cylinder (S1)

4.1.1. Action of gravity

This action is represented by the torsor ( ) 1e S which has for elements of


( )

( )

1

1 1

1

e ,

e 0.G

R S m g i

S

=

=

So as to apply the fundamental principle of dynamics, we shall need the

moment at the point O:

( ) ( )

11 1 1 1e e sin .

2O

hS R S G O m g kψ= × = −

The power developed by the action of gravity in the system associated to the

support is:

( ) ( ) ( ) ( )

1

11 1 1e e sin

2T T

S

hP S S m g ψ ψ= = −⋅ .

4.1.2. Action of the support induced by the hinge connection

The action of connection is represented by the torsor 1( )S which has for

elements of reduction at the point O of the axis of the connection:

1 1 1 1 1 1

1 1 1 1 1 1

( ) ,

( ) ,O

R S X i Y j Z k

S L i M j N k

= + +

= + +


where the components X1, Y1, ..., N1, are to be determined. While applying the

fundamental principle, we shall observe that it is more interesting to express the

components in the basis ( ), , i j k

. The elements or reduction are then given by:

( ) ( )

( ) ( )

1 1 1 1 1 1

1 1 1 1 1 1

( ) cos sin sin cos ,

( ) cos sin sin cos .O

R S X Y i X Y j Z k

S L M i L M j N k

ψ ψ ψ ψ

ψ ψ ψ ψ

= − + + +

= − + + +


( ) ( ) 1

1 1 1( ) ( )T TS

P S S N ψ= =⋅ .

In the case where the hinge connection is without friction, the power is zero. That

leads to:

1 0N = .

4.1.3. Action of the solid ( )2S induced by the connection between ( )1S and ( )2S

This action can be assimilated with a cylindrical connection. It is represented by

the torsor 2 1( ) .S This action is opposed to the action of connection exerted by

the cylinder ( )1S on the cylinder ( )2S , represented by the torsor 1 2( )S :

2 1 1 2( ) ( ) .S S= −

We shall express the torsor 1 2( )S in the following paragraph.

4.1.4. Action of the spring

The action exerted by the spring on the cylinder ( )1S is represented by the

torsor 1( ) .S This action is opposed to the one exerted by the spring on the

cylinder ( )2 ,S represented by the torsor 2( )S :

1 2( ) ( )S S= − .

We shall express the torsor 2( )S in the following paragraph.

4.1.5. Motor action

The solid ( )1S is eventually submitted to a motor couple represented by the

torsor ( ) 1S , of which the elements of reduction at the point O are:

( )

( ) 1

1

0,

,O

R S

S N k

=

=

where the imposed component N is known.

The power developed by this action is:

( ) ( ) 1

1 1( ) ( )T TS

P S S Nψ= =⋅ .

4.2. Mechanical actions exerted on the cylinder (S2)





( )

( )

2

2 2

2

e ,

e 0.G

R S m g i

S

=

=

The power developped by the action of gravity relatively to the support is:

( ) ( ) ( ) ( )

22 2e eT T

SP S S= ⋅ .

We obtain: ( ) ( ) 2 2 2e cos sinTP S m gx m gxψ ψ ψ= − .

4.2.2. Action of the spring

The action of the spring can be resolved into the sum of a traction-compression

force of support 1Ox

, represented by the torsor 1 2( ) ,S and a couple of torsion

of axis 1Ox

, represented by the torsor 2 2( ) .S The elements of reduction at the

mass centre 2G are:

( ) 2

21 2 0 1

1 2

( ) ,2

( ) 0,G

hR S k x l i

S

= − − −

=

where k is the traction-compression stiffness of the spring, and:

2

2 2

2 2 1

( ) 0,

( ) ,G

R S

S K iθ

=

= −

where K is the torsion stiffness, the spring being set up so as the couple of torsion

is null when 0.θ =The resultant action is:

2 1 2 2 2( ) ( ) ( ) .S S S= +

4.2.3. Action of the solid ( )1S induced by the connection between ( )1S and ( )2S

This action is represented by the torsor 1 2( )S of elements of reduction at the

point 2G :

2

1 2 12 1 12 1 12

1 2 12 1 12 1 12

( ) ,

( ) .G

R S X i Y j Z k

S L i M j N k

= + +

= + +

The components X12, Y12, ..., N12, are to be determined. The elements of reduction

are expressed in the basis ( )1b . In the basis ( ), , i j k

, they are expressed as:

( ) ( )

( ) ( )

2

1 2 12 12 12 12 12

1 2 12 12 12 12 12


( ) cos sin sin cos .G

R S X Y i X Y j Z k

S L M i L M j N k

ψ ψ ψ ψ

ψ ψ ψ ψ

= − + + +

= − + + +

The power developed in the system (T) attached to the support, is:

( ) ( ) 2

1 2 1 2( ) ( )T TS

P S S= ⋅ .


Thus: ( ) ( )1 2 12 12 12 12( )TP S X x N xY Lψ θ= + + + .

To express the conditions of connection according to the physical nature of the

connection, it is necessary to evaluate the power developed in the system attached

to ( )1S . Thus:

( ) ( ) 1 1

21 2 1 2( ) ( )

S S

SP S S= ⋅ ,

where ( ) 1

2

S

S is the kinematic torsor relative to the motion of the solid ( )2S with

respect to the solid ( )1S . Its elements of reduction are:

( ) ( ) ( )

( ) ( )

( )

1 1

2 2

1 12 2

1 1

2 1

2 1

, instantaneous rotation vector with respect to ,

( , ) ,

velocity vector of the point with respect to the solid .

S SS S

S SG S

R i S

G t x i

G S

ω θ = =

= =

Hence the expression of the power:

( ) 11 2 12 12( )

SP S X x L θ= + .

In the case of a connection without friction, the power developed with respect to

the cylinder ( )1S is zero. That leads to the usual conditions:

12 120, 0.X L= =

5. Application of the fundamental principle of dynamics


The support ( )T is attached to the Earth and the fundamental relation relative

to the motion is written as:

( ) ( ) ( ) 1

1 1 1 2 1 2 2 2 1e ( ) ( ) ( ) ( ) .TS

S S S S S S= + − − − +

The equation of the resultant leads to:

along 1i

: ( )21 21 1 1 12 0cos ,

2 2

h hm m g X X k x lψ ψ− = + − + − −

along 1j

: 11 1 1 12sin ,

2

hm m g Y Yψ ψ= − + −

along k

: 1 120 .Z Z= −

We express the equation of the moment at the point O. In this way we have to

express the moments at the point O of the torsors 1 2( ) ,S 1 2( )S et 2 2( ) .S

( ) ( )

21 2 1 2 1 2 2

12 1 12 12 1 12 12

( ) ( ) ( )

.

O GS S R S G O

L i M xZ j N xY k

= + ×

= + − + +

And:

1 2 2 2 1( ) 0, ( ) .O OS S K iθ= =


Hence the three scalar equations of the moment at the point O:

along 1i

: 1 120 ,L L Kθ= − +

along 1j

: 1 12 120 ,M M xZ= − +

along k

: 11 1 1 12 12sin .

2

hC m g N N xY Nψ ψ= − + − − +


The fundamental relation relative to the motion of the solid 2( )S with respect

to the support is written:

( ) ( ) 2

2 1 2 2 2 1 2e ( ) ( ) ( ) .TS

S S S S= + + +

The equation of the resultant leads to the three scalar equations:

along 1i

: ( ) ( )2 22 2 0 12cos ,

2

hm x x m g k x l Xψ ψ− = − − − +

along 1j

: ( )2 2 122 sin ,m x x m g Yψ ψ ψ+ = − +

along k

: 120 .Z=

We express the equation of the moment at the mass centre 2G , that leads to the

three scalar equations:

along 1i

: 2 12 ,A L Kθ θ= −

along 1j

: ( )2 2 12,A C Mψ θ− =

along k

: 2 12.C Nψ =

5.3. Motion of the set of the two solids

The fundamental principle can be applyied to the set constituted of the two

solids. The equation obtained results from the addition of the two fundamental

relations written for the motion of 1( )S and for the motion of 2( )S :

( ) ( ) ( ) ( )

1 21 1 1 2e ( ) ( ) eT T

S SS S S S+ = + + + .

This equation removes the internal actions exerted between the two solids.

The equation of the resultant leads to three scalar equations which are obtained

by superimposition of the preceding equations of the resultants:

along 1i

: ( ) ( )2 211 2 1 2 1cos ,

2

hm m x x m m g Xψ ψ ψ− + − = + +

along 1j

: ( ) ( )11 2 1 2 12 sin ,

2

hm m x x m m g Yψ ψ ψ ψ+ + = − + +

along k

: 10 .Z=

The equations of the moments were expressed at the point O for the solid 1( )S

and at the mass centre 2G for the solid 2( )S . So, it is necessary to consider a


change for the points of the moments. The easiest way is to write the equation of

the moments at the point O. We have:

( ) ( ) ( ) 22 2 22 .T T T

O GS S SR G O= + ×

Thus: ( ) ( ) ( )

2

22 1 2 2 1 2 2 22T

O SA i A C j C m x m xx kθ ψθ ψ ψ = + − + + +

.

For the action of gravity, we have:

( ) ( ) 2 2 2 2e e sinO S R S G O m gx kψ= × = −

.

The equation of the moment at the point O leads thus to the three scalar equations:

along 1i

: 2 1,A Lθ =

along 1j

: ( )2 2 1,A C Mψθ− =

along k

:

( )2 11 2 2 2 1 1 22 sin sin .

2

hC C m x m x x m g N N m gxψ ψ ψ ψ+ + + = − + + −

6. Analysis of the equations derived from the fundamental principle

The fundamental principle leads to 12 independent equations for 15 unknowns:

1 1 1 12 12 12, , ... , ; , , ... , ; , and .X Y N X Y N x ψ θ The physical nature of the actions

of connection allows us to obtain 3 additional equations to complement the 12

equations.

In the case of a hinge connection without friction with the support and in the

case of a connection without friction between the solids 1( )S ans 2( ),S we have:

1 12 120, 0, 0.N X L= = =

The equations of motion are thus, among the preceding equations derived from

the fundamental principle, the ones which introduce only the components of

connection 1 12 12, , .N X L Thus:

( ) ( )2 22 2 0 12cos ,

2

hm x x m g k x l Xψ ψ− = − − − +

component along 1i

of the resultant of the fundamental relation obtained for the

motion of 2( ),S

2 12 ,A L Kθ θ= −

component along 1i

of the moment at 2G of the fundamental relation obtained for

the motion of 2( ),S and

( ) ( )2 11 2 2 2 1 2 12 sin ,

2

hC C m x m x x m m x g N Nψ ψ ψ+ + + = − + + +

component along k

of the moment at the point O of the fundamental relation

obtained for the set of the two solids.

In the case of viscous frictions, the components of connections are expressed

as:


12 1 1 12 2, , ,t r rX c x N c L cψ θ= − = − = −

where 1 2, and t r rc c c are the coefficients of viscous friction.

The three preceding equations of motion will allow us to obtain , and x θ ψas functions of time. Next, the components of the connections will be derived from the other equations.

24.2 Motion of a radar antenna (Figure 24.8)


1.1. Motion of the support (S1) with respect to the frame (T)

We associate the coordinate systems ( )Oxyz and ( )1 1Ox y z (Figure Exercise

24.2), respectively attached to the frame ( )T and to the support ( )1S . The upward

vertical axis Oz

coincides with the axis of rotation ( )1∆ .

We have one parameter of rotation ψ about the axis Oz

. The basis change is:

1

1

cos sin ,

sin cos ,

.

i i j

j i j

k

ψ ψ

ψ ψ

= +

= − +

1.2. Motion of the reflector (S2) with respect to the support (S1)

We associate (Figure Exercise 24.2) the systems ( )2 1 1G x y z and ( )2 1 2 2G x y z

respectively to the support ( )1S and to the reflector ( )2 .S The plane ( )2 1 2G x y is

contained in the plane of symmetry of the reflector.

We have one parameter of rotation θ about the axis 12 .G x


1

2 1

2 1

,

cos sin ,

sin cos .

i

j j k

k j k

θ θ

θ θ

= + = − +

2. Kinematic study



TS relative to the motion of the support 1( )S with

respect to the frame ( )T has for elements of reduction at the mass centre 1G ,

located on the axis ( )1∆ of rotation:

( ) ( )

( ) ( )

1 1

1 11 1

, instantaneous rotation ivector,

( , ) 0, velocity vector of the point .

T TS S

T TG S

R k

G t G

ω ψ = =

= =

The acceleration vector of the mass centre is null: ( )

1( , ) 0.Ta G t =


x

y

x1

y1

z

( )T

( )1∆

O

( )1S

x1

y1

z

( )1∆

O ( )1S

( )2S

( )2∆

2G

y2z2


2.2. Motion of the reflector (S2) with respect to the frame (T)


TS relative to the motion of the reflector 2( )S with

respect to the frame ( )T has for elements of reduction at the mass centre 2G ,

located to the intersection of the axes ( )1∆ and ( )2∆ :

( ) ( )

( ) ( )

2 2

2 2

1

2 2


( , ) 0, velocity vector at the point .

T TS S

T TG S

R k i

G t G

ω ψ θ = = +

= =

The rotation vector can also be expressed in the basis ( )2b = ( )1 2 2, , i j k

:


( )

21 2 2sin cos .T

Si j kω θ ψ θ ψ θ= + +

The acceleration vector of the mass centre is null: ( )

2( , ) 0.Ta G t =

3. Kinetic study




TS relative to the motion of the support (S1) with respect

to the frame has for elements of reduction at the mass centre:

( ) ( )

( ) ( ) ( )

1

1 11 1

1 1

1

( , ) 0,

,

T TS

T TG GS S

R m G t

S ω

= =

=

where 1m is the mass of the support 1( ).S

The operator of inertia at 1G is represented in the basis ( ) ( )1 1 1, , b i j k=

by the

matrix:

( ) ( )1

1

1 1 1

1 1 1 1

1 1 1

.b

G

A F E

S F B D

E D C

− − = − − − −

I

Hence:

( ) 1 1

1 1 1 1 1T

G SE i D j C kψ ψ ψ= − − +

.


The kinetic energy is given by:

( ) ( ) ( ) 1 1

c 11

( )2

T T TS S

E S = ⋅ .

Thus:

( ) 2c 1 1

1( )

2TE S Cψ= .



TS relative to the motion of the support (S1) with

respect to the frame has for elements of reduction at the point G1:

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

1

1 1 11 1 1 1

1 1

1 1

( , ) 0,

.

T TS

T T T TG G GS S S S

R m a G t

S Sω ω ω

= =

= + ×

The calculation of the moment leads to:

( ) ( ) ( )

1 1

2 21 1 1 1 1 1 1

TG S

D E i E D j C kψ ψ ψ ψ ψ= − − + +

.





TS relative to the motion of the reflector (S2) with

respect to the frame has for elements of reduction at the mass centre 2G :

( ) ( )

( ) ( ) ( )

2

2 22 2

2 2

2

( , ) 0,

,

T TS

T TG GS S

R m G t

S ω

= =

=

where 2m is the mass of the reflector.

The reflector has a symmetry of cylindrical type. It follows that the operator of

inertia at 2G is represented in the basis ( ) ( )2 1 2 2, , b i j k=

by the matrix:

( ) ( )2

2

2

2 2

2

0 0

0 0 .

0 0

b

G

A

S A

C

=

I

The rotation vector ( )

2

TS

ω

was expressed previously in the basis ( )2b . It follows

that:

( ) 2 2

2 1 2 2 2 2sin cosTG S

A i A j C kθ ψ θ ψ θ= + +

.



( ) ( ) ( ) 2 2

c 21

( )2

T T TS S

E S = ⋅ .

Thus:

( ) ( )2 2 2 2c 2 2 2 2

1 1( ) sin cos

2 2TE S A A Cθ θ θ ψ= + + .



TS relative to the motion of the reflector (S2) with

respect to the frame ( )T has for elements of reduction at the mass centre 2G :

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

2

2 2 22 2 2 2

2 2

2 2

( , ) 0,

.

T TS

T T T TG G GS S S S

R m a G t

S Sω ω ω

= =

= + ×

The rotation vector ( )

2

TS

ω is obtained by deriving the expression of the rotation

vector ( )

2

TS

ω

in the basis ( )2b . Thus:

( ) ( ) ( )

21 2 2sin cos cos sin .T

Si j kω θ ψ θ ψθ θ ψ θ ψθ θ= + + + −


Hence:

( ) ( )

( ) ( )

2 2

22 2 2 1

2 2 2 2 2 2

sin cos

sin 2 cos cos sin .

TG S

A C A i

A A C j C k

θ ψ θ θ

ψ θ ψθ θ ψ θ ψθ θ

= + −

+ + − + −

4. Mechanical actions

4.1. Mechanical actions exerted on the support (S1)




( )

( )

1

1 1

1

e ,

e 0.G

R S m g k

S

= −

=

4.1.2. Action of the motor ( )1M

The action of the stator on the rotor of the motor ( )1M is a couple-action,

represented by the torsor ( ) 1S , of which the elements of reduction at the point

O are:

( )

( ) 1

1

1 1

0,

,G

R S

S F k

=

=

where the component 1F is known.

4.1.2. Action of the frame induced by the hinge connection

The action of connection is represented by the torsor 1( )S . Its elements of

reduction have to be expressed at a point of the connection axis. While applying

the fundamental principle to the reflector-support set, we shall observe that it is

more interesting to choose the mass centre 2G and to express the elements of

reduction in the basis ( )1b :

2

1 1 1 1 1 1

1 1 1 1 1 1

( ) ,

( ) ,G

R S X i Y j Z k

S L i M j N k

= + +

= + +

where the components X1, Y1, ..., N1, are to be determined. The elements of

reduction expressed in the basis ( ), , i j k

are obtained as:

( ) ( )

( ) ( )

1 1 1 1 1 1

1 1 1 1 1 1


( ) cos sin sin cos .O

R S X Y i X Y j Z k

S L M i L M j N k

ψ ψ ψ ψ

ψ ψ ψ ψ

= − + + +

= − + + +


( ) ( ) 1

1 1 1( ) ( )T TS

P S S N ψ= =⋅ .

In the case of a hinge connection without friction, the power developed is zero.

That leads to: 1 0N = .



The action of the rotor on the stator of the motor ( )2M is represented by the

torsor ( ) 2 1 .S This action is opposed to the action exerted by the stator on the

rotor, then to the action exerted by the motor on the reflector ( )2 .S This action is

represented by the torsor ( ) 1 2 .S We have:

( ) ( ) 2 1 1 2 .S S= −

The torsor ( ) 1 2S will be expressed afterwards.

4.1.5. Action of the reflector induced by the hinge connection between ( )1S and

( )2S

This action is represented by the torsor ( ) 2 1 .S It is opposed to the action of

connection exerted by the support ( )1S on the reflector ( )2S and represented by

the torsor ( ) 1 2S :

( ) ( ) 2 1 1 2 .S S= −

We shall express the torsor ( ) 1 2S in the following paragraph.

4.2. Mechanical actions exerted on the reflector (S2) 4.2.1. Action of gravity


reduction at the mass centre 2G :

( )

( )

2

2 2

2

e ,

e 0.G

R S m g k

S

= −

=


The action of the stator on the rotor of the motor ( )2M is a couple-action,

represented by the torsor ( ) 1 2S , of which the elements of reduction at the

point G2 are:

( )

( ) 2

1 2

1 2 2 1

0,

,G

R S

S F i

=

=

where the component 2F is known.

4.2.3. Action of the support ( )1S induced by the hinge connection with the reflector

This action is represented by the torsor 1 2( )S of elements of reduction at the

point 2G , which we express in the basis ( )1b :

2

1 2 12 1 12 1 12

1 2 12 1 12 1 12

( ) ,

( ) .G

R S X i Y j Z k

S L i M j N k

= + +

= + +

The components X12, Y12, ..., N12, are to be determined. In the case of a

connection without friction, we have: 12 0.L =


5. Application of the fundamental principle of dynamics


The fundamental relation relative to the motion is written:

( ) ( ) ( ) 1

1 1 1 2 1 2 1e ( ) ( ) ( ) .TS

S S S S S= + + + +

The equation of the resultant leads to:

along 1i

: 1 120 ,X X= −

along 1j

: 1 120 ,Y Y= −

along k

: 1 1 120 .m g Z Z= − + −

The equation of the moment can be written at the point 2G :

along 1i

: 21 1 1 2 12,D E L F Lψ ψ− = − −

along 1j

: ( )21 1 1 12,E D M Mψ ψ− + = −

along k

: 1 1 1 12.C F N Nψ = + −


The fundamental relation relative to the motion is written:

( ) ( ) 2

2 1 2 1 2e ( ) ( ) .TS

S S S= + +

The equation of the resultant leads to the three equations:

along 1i

: 120 ,X=

along 1j

: 120 ,Y=

along k

: 2 120 .m g Z= − +

The equation of the moment can be written at the point 2G , while expressing

the components in the basis ( )2b :

along 1i

: ( ) 22 2 2 2 12sin cos ,A C A F Lθ ψ θ θ+ − = +

along 2j

: ( )2 2 2 12 12sin 2 cos cos sin ,A A C M Nψ θ ψθ θ θ θ+ − = +

along 2k

: ( )2 12 12cos sin sin cos .C M Nψ θ ψθ θ θ θ− = − +

5.3. Motion of the set of the two solids

The equation obtained results from the addition of the two fundamental rela-

tions written for the motion of the support and for the motion of the reflector: ( ) ( ) ( ) ( )

1 21 1 1 2e ( ) ( ) eT T

S SS S S S+ = + + + .

The equation of the resultant leads to the three scalar equations which are the

superimposition of the preceding equations of the resultants.

To derive the equations of the moments, it is necessary to express the moments

in the same basis, for example the basis ( )1 1, , i j k

.


6. Analysis of the equations derived from the fundamental principle

The fundamental principle leads to 12 independent equations for 14 unknowns:

1 1 1 12 12 12, , ... , ; , , ... , ; and .X Y N X Y N ψ θ The physical nature of the actions of

connections allows us to obtain 2 additional equations on the components 1N and

12L .

The equations of motion are then, among the preceding equations deduced

from the fundamental principle, the ones which introduce only the components of

connections 1N and 12.L One equation of motion is thus given by the first equa-

tion of the moment for the motion of the reflector:

( ) 22 2 2 2 12sin cos .A C A F Lθ ψ θ θ+ − = +

The second equation of motion is derived from the linear combination of the

second and third equations of the moment obtained for the motion of the reflector

( )2 ,S associated to the third equation of the moment obtained for the motion of

the support ( )1 .S We obtain:

( ) ( )2 21 2 2 2 2 1 1sin cos 2 sin cosC A C A C F Nθ θ ψ ψθ θ θ+ + + + − = + .

Chapter 25

The Lagrange Equations

25.1 Lagrange equations relative to the motion of the set of the two solids studied in Exercise 24.1

In fact, the elements necessary to establish the Lagrange equations have been

expressed in the correct version of Exercise 24.1.

The Lagrange equations for the set ( )D constituted of the two solids ( )1S and

( )2S are written from (25.39):

( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

c c 1 1 2 1

1 1 2 2 1 2

de

d

e .

i i i

i i i i i

T T T T Tq q q

i iT T T T T

q q q q q


P S P S P S P S P S

∂ ∂− = + +

∂ ∂

+ + + + +

This equation introduces the power coefficients in the reference ( )T of all the

actions exerted on the two solids ( )1S and ( )2 .S The three equations which are

deduced from this equation are relative to the three parameters of situation:

, , .iq x ψ θ=The kinetic energy of the set is:

( ) ( ) ( ) ( ) ( ) ( )c c 1 c 2T T TE D E S E S= + ,

where the kinetic energies relative to the motions of the solids ( )1S and ( )2S have

been expressed in the correct version of Exercise 24.1. Hence:

( ) ( ) ( ) 2 2 2 2 2c 1 2 2 2 2

1 1 1 1.

2 2 2 2TE D C C m x m x Aψ ψ θ== + + + +

The power coefficients are deduced from the expressions of the powers deve-

loped by the different actions exerted. Some ones were expressed in the correct

version of Exercise 24.4. It remains to evaluate:

the power developed by the action of connection of ( )2S on ( )1S :

( ) ( ) ( ) 1 1

2 1 2 1 1 2( ) ( ) ( )T T TS S

P S S S= = −⋅ ⋅ .

( ) ( )2 1 12 12( )TP S N xY ψ= − + .

the power developed by the action of the spring on ( )1S :

( ) ( ) ( ) 1 1

1 1 2( ) ( ) ( )T T TS S

P S S S= = − ⋅⋅ .

( ) 1( ) 0TP S = .

the power developed by the action of the spring on ( )2S :

( ) ( ) ( ) ( )

2 2 22 2 1 2 2 2( ) ( ) ( ) ( )T T T T

S S SP S S S S= = +⋅ ⋅ ⋅ .

( ) ( )22 0( ) .

2T h

P S k x l x Kθθ= − − − −


Hence the power coefficients of the actions exerted on the solids ( )1S and ( )2S :


( ) ( ) ( ) ( ) ( ) ( ) 11 1 1 1e 0, e sin , e 0

2T T T

xh

P S P S m g P Sψ θψ= = − = ;

— action exerted by the frame on the solid (S1):

( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 10, , 0T T TxP S P S N P Sψ θ= = = ;


( ) ( ) ( ) ( ) ( ) ( ) 2 1 2 1 12 12 2 10, , 0T T TxP S P S N xY P Sψ θ= = − − = ;


( ) ( ) ( ) ( ) ( ) ( ) 1 1 10, 0, 0T T TxP S P S P Sψ θ= = = ;

— action exerted by the motor couple on the solid (S1):

( ) ( ) ( ) ( ) ( ) ( ) 1 1 10, , 0T T TxP S P S N P Sψ θ= = = ;


( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2e cos , e sin , e 0T T TxP S m g P S m gx P Sψ θψ ψ= = − = ;


( ) ( ) ( ) ( ) ( ) ( ) ( ) 22 0 2 2, 0,

2T T T

xh

P S k x l P S P S Kψ θ θ= − − − = = − ;


( ) ( ) ( ) ( ) ( ) ( ) 1 2 12 1 2 12 12 1 2 12, , T T TxP S X P S N xY P S Lψ θ= = + = .

1. Lagrange equation relative to the parameter x

We have:

( )( )c 2TE D m x

x

∂=

∂

,

( )( )c 2

d

dTE D m x

t x

∂=

∂

,

( )( ) 2c 2TE D m x

xψ∂

=∂

.


( )2 22 2 2 0 12cos .

2

hm x m x m g k x l Xψ ψ− = − − − +


We have:

( )( ) ( ) 2c 1 2 2TE D C C m xψ ψ

ψ∂

= + +∂

,

598 Chapter 25 The Lagrange Equations

( )( ) ( )2c 1 2 2 2

d2

dTE D C C m x m xx

tψ ψ

ψ∂

= + + +∂

,

( )( )c 0TE D

ψ∂

=∂

.


( )2 11 2 2 2 1 1 22 sin sin

2

hC C m x m xx m g N N m gxψ ψ ψ ψ+ + + = − + + − .

3. Lagrange equation relative to the parameter

We have:

( )( )c 2TE D A θ

θ∂

=∂

,

( )( )c 2

d

dTE D A

t xθ∂

=∂

,

( )( )c 0TE D

x

∂=

∂.

Hence the third Lagrange equation:

2 12A K Lθ θ= − + .

We find again the third equations of motion derived from the fundamental prin-

ciple of dynamics (correct version of Exercise 24.1).

25.2 Lagrange equations relative to the motion of the radar antenna

The Lagrange equations relative to the motion of the set ( )D constituted of the

support ( )1S of the antenna and of its reflector ( )2S are written:

( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

c c 1 1 1

2 1 2 1 2 1 2 1 2

de

d

e ,

i i i

i i i i i

T T T T Tq q q

i iT T T T T

q q q q q


P S P S P S P S P S

∂ ∂− = + +

∂ ∂

+ + + + +

, .iq ψ θ=

The kinetic energy of the set is the sum of the kinetic energies expressed in the

correct version of Exercise 24.2. Thus:

( ) ( )2 2 2 2c 1 2 2 2

1 1( ) sin cos

2 2TE D C A C Aθ θ ψ θ= + + + .

So as to obtain the power coefficients, we express the powers developed by the

actions exerted on the two solids:

( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 1e 0, , ,T T TP S P S F P S Nψ ψ= = =

( ) ( ) ( ) ( ) 2 1 12 2 1 12, ,T TP S F P S Nψ ψ= − = −

( ) ( ) ( ) ( ) ( ) ( ) 2 1 2 2 1 2 12 12e 0, , .T T TP S P S F P S L Nθ θ ψ= = = +


1. Lagrange equation relative to the parameter ψWe have:

( )( ) ( )2 2c 1 2 2sin cosTE D C A Cθ θ ψ

ψ∂

= + +∂

,

( )( ) ( ) ( )2 2c 1 2 2 2 2

dsin cos 2 sin cos

dTE D C A C A C

tθ θ ψ ψθ θ θ

ψ∂

= + + + −∂

,

( )( )c 0TE D

ψ∂

=∂

.


( ) ( )2 21 2 2 2 2 1 1sin cos 2 sin cosC A C A C F Nθ θ ψ ψθ θ θ+ + + + − = +

2. Lagrange equation relative to the parameter

We have:

( )( ) ( )( )c 2 c 2

d, ,

dT TE D A E D A

t xθ θ

θ∂ ∂

= =∂∂

,

( )( ) ( ) 2c 2 2 sin cosTE D A C

xψ θ θ∂

= −∂

.


( ) 22 2 2 2 12sin cos .A C A F Lθ ψ θ θ+ − = +

We find again the two equations of motion derived from the fundamental

principle of dynamics (correct version of Exercise 24.2).

Printed in France by Jean-Marie Berthelot

Les Clousures, Chemin des Horts

05290 Vallouise

France

The book "Mechanics of Rigid Bodies" develops a unified approach to the

problems of the Mechanics of Solids.

The development is based on a generalized use of the concept of “torseur” (in

French). We think that this concept is not used in the English textbooks. We will

call this concept as “torsor”.

After a first part where the necessary elements of mathematics are recalled, the

book is structured through four parts of increasing difficulties in such a way to

have a good assimilation of the concepts.

A sixth part analyses the numerical methods used to solve the motion

equations. Comments reported at the end of each chapter summarize the

fundamental notions which must which must be acquired.

Examples and simple exercises are associated to each chapter with the object to

familiarize the reader with the fundamental tools needed to solve the problems of

Mechanics of Solids.

A last part develops the solutions of the exercises which are proposed all along

the book.

Jean-Marie Berthelot is an Honorary Professor, Maine University, Le Mans France. He took part at the installation of the Institute for Advanced Materials and Mechanics (ISMANS), Le Mans, France. His current research is on the mechanical behaviour of composite materials and structures. He has published extensively in the area of composite materials and is the author of different textbooks, in particular of the textbook entitled Composite

Materials, Mechanical Behavior and Structural Analysis published by Springer, New York, in 1999.

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