Chapter 8

Rotational Equilibrium and Dynamics

• The Torque, τ

• Elements of Statics. Center of Mass

• Rotational Dynamics. Moment of Inertia, I

• Rotational Energy

• Angular Momentum, L

What causes rotations?

• To rotate an object which is initially at

rest, a force is needed

• Beside the magnitude of the force, the

rotational effect of the force depends on

the position of the force with respect to

the axis of rotation, and on its direction

• So, in order to study how one can

produce rotations, one must take into

consideration all these quantities, not

only the simple force as in the purely

translational case

• Let’s first define some geometrical

characteristics:

a) the strength (magnitude) of the force

b) the position where the force is applied with

respect to the hinges (axis of rotation)

Ex: Consider a door acted upon by a force F.

The rotational effect depends on

c) the direction of the force (or line of action)

zero effect

maximum effect

larger

effect

smaller

effect F F

F

F

Def 1: A line along the vector force is

called the line of action

Def 2: The perpendicular distance from

the axis of rotation to the line of action

is called the lever arm

Ex: A longer lever arm is very helpful in rotating objects:

this is why a wrench can be used to loosen a bolt

• Formally, this is due to the fact that a longer r increases

the torque for the same F and θ

• Also, applying the force perpendicular on the arm will

maximize the torque for the same F and r, since sinθ is

maximum when θ = 90°

• What about the direction of the torque? The vector torque is

perpendicular on the plane of the rotation that it attempts to produce

• We can describe the direction using a sign convention:

sinFr

Torque – Definition

Def: The physical quantity that models the rotational

effect of a force is called torque, τ, with magnitude

defined by taking into account all the influences upon

the rotation:

Magnitude of force F Distance r to the axis Angle between F and r

+ – τ > 0 if it tries to rotate the

object counterclockwise

τ < 0 if it tries to rotate the

object clockwise

Axis of rotation

r

F

θ

SIm N

Torque – Calculation

The magnitude of the torque can be viewed in two ways:

1. Note that that the force is perpendicular on the lever

arm ℓ, so

Torque = (Force Magnitude) × (Lever Arm)

sinrF F

sinFr rF

Axis of rotation

r

F

θ

ℓ = rsinθ

Axis of rotation

r

F

θ

θ

2. Note that that the component of the force along r does not produce

rotation, so we remain only with the perpendicular component:

Torque = (Distance to Axis) × (Perpendicular Component of Force)

sinF F

F

Problem:

1. Net torque on a wheel: Two forces, F1 = 7.50 N and F2 = 5.30 N,

are applied to a wheel with radius r = 0.330 m, as in the figure.

What is the net torque on the wheel due to these forces for an axis

perpendicular on the wheel and passing through its center?

Exercise 1: Net torque on a dumbbell: Two balls of masses m1 = m2 = m are attached to the

ends of a massless rod which pivots, at distances L1 and L2 from the pivot, as shown in the

figure. Initially the rod is held in the horizontal position and then released.

Axis of rotation

r

F1

F2 ½r

L1 L2 m1

a) Sketch the forces action on the

dumbbell.

b) Calculate the magnitude and

direction of the net torque acting

on the system relative to the pivot.

m2

Statics – Elements

• Based on our discussion about different types motion, we see that the motion is

controlled by forces in the case of translations and torques in the case of rotations

• Therefore, we can introduce the conditions for an object to be completely at rest (or

moving with constant velocity)

• A solid body that is at rest is said to be in motional equilibrium

• To be in complete equilibrium, a solid body must satisfy both translational and a

rotational conditions of equilibrium:

00

0

x

y

FF

F

0

2. if the net torque is zero with respect with any arbitrary

center of rotation (pivot), there is no rotation

Ex: Rotation without

translation: Even when the

net force is zero, the object

can still rotate. So, a zero

balance of force does not

grant complete equilibrium

1. if the net force is zero there is no translation

• Since the pivot is arbitrary, we can choose its such

that the equation won’t contain unknown forces

Calculating the position of the Center of Mass

• If the mass of an object is uniformly distributed, and the system is geometrically

symmetric, the center of mass (CM) is located either in the center of symmetry or on

the axis of symmetry

1 1 2 2CM CM CM

1 2

...1

...iMg m g i i i i

i i i

m x m xMx m x x m x x

M m m

Ex: In the center of a sphere, or along the height of a cone.

Finding the center of mass in a one dimensional case: consider a number of

particles with masses mi (adding to a total mass M) and positions along x-axis xi (or

larger objects with the centers of mass located at xi) then the position of the center

of mass is given by:

• However, if an object is not symmetrically shaped, the position of the CM must be

calculated by using the fact that its weight is acted in the CM

• Hence, the torque of the net weight is equal to the sum of the torques of the weights

of each part of the system, and we can derive the following procedure:

Ex: Five particles of equal masses m are aligned at positions

given on the figure below. Find their center of mass

–2 –1 0 1 2 3 4 x (m)

1 2 3 4 5CM

15

5

2 0 1 3 4 m 1.2 m

mx mx mx mx mxx

m

CM

3. Translational equilibrium. A chandelier of mass m = 200 kg

is suspended from the ceiling by a rope making an angle θ = 60°

with the ceiling, and from a lateral wall by a horizontal rope, as

in the figure. Calculate the tension forces T1 and T2 necessary to

hold it in equilibrium.

4. Rotational equilibrium: A uniform, 256 kg beam is

supported using a cable connected to the ceiling, as shown in

the figure. The lower end of the beam rests on the floor. What

is the tension in the cable?

Problems:

2. CM of a two objects with size: A 2-kg sphere and a 1-kg

square plate are aligned as on the figure. Knowing that the

distance between their CMs is d = 2 m, find the center of mass

of this combination with respect to the center of the sphere.

256 kg

m1 m2

d

m

θ

Rotational Dynamics – Torque and Rotational Inertia

• If we consider the particle as a rigid body, what is

the corresponding torque?

2 tFr ma r mr I

2 i im r

• Let’s first consider a particle performing circular

motion of radius r

• Now we know that, if its speed varies, there must

be a tangential force accelerating it, such that, by

Newton’s 2nd law

t tF ma mr

vr path

tF

• This is for a single point mass; what about a rigid body seen as an extended object

containing many particles of mass mi at positions ri from the center of rotation?

• Then, since the angular acceleration is the same for the whole object, we can write:

2I mrDef: moment of inertia of a particle

2

i iI m rDef: moment of inertia of a system of particles

2

SI kg mI

• Notice that the same object can have different

moments of inertia with respect to different

axes of rotation

• The rotational inertia increases the further the

mass is distributed from the axis of rotation

• In this class, the moments of inertia of rigid

bodies with continuously distributed mass will

be given (see the adjacent table)

• The net moments of inertia of a system of

objects with known individual moments Ii is

2

i iI m r

1 2 3 ...netI I I I

• So, the rotational inertia of an object made

of many particles is given by its moment of

inertia

Quiz: The two cylinders

shown have the same

mass. Which has a

larger rotational inertia?

Moment of Inertia – Concept and formulas

Newton’s 2nd Law for Rotations

Comments:

• This law works only for rigid bodies (such that α is the same for all parts)

• Like forces that can modify the motion of an object only they are external forces,

the angular acceleration is determined only by external torques: the internal torques

cannot modify the rotation of the system

• Notice that the angular acceleration has the same vector direction as the net torque

(perpendicular on the circle of rotation)

I

A net torque Στ acting on a rigid body with moment of inertia I will determine an

angular acceleration α proportional to the torque as given by:

Ex: The weight of a rigid body will play a role in the

rotation in most cases. Since the weight can be

considered as acting in CM, the weight will have a

torque and rotate the body only if the if the center of

rotation doesn’t pass through the CM.

CM

mgPivot

m1 m2

Problems:

5. Net moment of inertia: A helicopter rotor blade can be

considered a long thin rod, as shown in the figure.

a) If each of the three rotor helicopter blades has length L

= 3.75 m and has a mass m =160 kg, calculate the moment

of inertia of the three rotor blades about the axis of

rotation.

b) How much torque must be applied to bring the blades

up to a speed ω in a time t?

6. Dynamics of Atwood machine revisited : Two boxes

with masses m1 and m2 are connected by a massless cord

passing over a pulley, as in the figure. Consider the pulley

as a disk of mass M and radius R. Use the dynamics of the

system to calculate

a) the acceleration a of the masses and

b) the angular acceleration α of the pulley vv

M

R

• The rotational kinetic energy of a rigid body

formed of particles indexed by i (as the wheel

on the right) is the sum of the kinetic energies

of all its parts

2 2 21 1 1rot 1 1 2 22 2 2

... i iKE m v m v m v

Physical situation:

• Consider some rigidly connected

particles rotating about their center of

mass with angular speed ω, and in the

same time translating with speed vcm

• Then, an arbitrary particle of mass mi

located at distance ri from the center of

rotation, moves with speed vi

2 2 2 21 1rot 2 2

212i i i i IKE m r m r

1 212

2

2 cne cmmtKE m Iv

Rotation Translation

• By the argument on the left, the net

kinetic energy is the sum of the kinetic

energies associated with its two motions

• By substituting the rotational quantities, we

find that the rotational kinetic energy can be

written

• Therefore, a rigid body that has both

translational motion (motion of its cm) and

rotational motion (about its cm) has both

translational and rotational kinetic energies:

Net moment of inertia I of the

system of particles

Rotational

velocity, ω

ri

vi

Translational velocity

vcm

Energy in Rotational Motion – Kinetic energy

Energy in Rotational Motion – Conservation

• When evaluating the conservation of energy for rotating rigid bodies, the only

change from our previous approach is that, beside the kinetic energy associated with

the translation of the CM, the total kinetic energy contains a rotational term

• So the expression for the mechanical energy becomes

• The conservation of energy can still be written

21rot CM

2 21 1transl M 2C2 2

E KE PE mKE v kx gI m y

nonconservativeE W

Exercise 2: Rolling motion. Assume that all the

objects on the figure have the same mass m and

are all released down the frictionless incline from

rest, from the same initial height. If the radius r

of each rolling object is the same, which object

will move faster at the bottom of the incline?

m1 m2

m1

m2

h

h

Problem:

7. Atwood machine treated using energy: Two boxes with masses m1 and m2 are

connected by a massless cord passing over a pulley, as in the figure. Consider the pulley as

a disk of mass M and radius R. If the system is released from rest and the masses travel a

distance h, use conservation of energy to calculate:

a) the final speed of the masses

b) the final angular velocity of the pulley

c) the corresponding angular displacement of the pulley

A B

M R

Angular Momentum – Conservation

• In analogy with linear momentum, we can define angular momentum:

L I2 1

SIkg m sL

after before i i i iafter beforeL L I I

L

t

pF

t

Def: A rigid body rotating with an angular speed ω about an axis of rotation with

respect to which its moment of inertia is I, carries a “quantity of rational motion”

given by the angular momentum:

• Hence, the net torque is the rate of change of angular momentum as the force is the

rate of change in linear momentum

• This expression leads us to a condition for the conservation of momentum: if the net

external torque acting on an object is zero, the total angular momentum is constant

• Thus, if the parts of an isolated rotating system redistribute due to some internal

torques, the net angular momentum stays constant, even though its structure changes:

…compare to

Therefore, systems that can change their rotational inertia through

internal torques will also have to change their rate of rotation.

Ex 1: No net rotations but rotating body parts

• A falling cat twisting its body

• An astronaut in space trying to rotate his body

• A helicopter must have two rotors

Ex 3: A ballerina redistributing the mass of her body

The total angular momentum stays the same, but the angular

velocity changes when the mass is redistributed changing the

rotational inertia:

large small small largeI I

Ex 2: Kepler’s 2nd law: planets move faster when they are closer to the sun

Angular Momentum – Examples

Problems:

8. Conservation of angular momentum: A person of mass m stands at the center of a

rotating merry-go-round platform of radius r, moment of inertia I and rotating without

friction with angular speed ω. The person walks radially to the edge of the platform.

Calculate the angular speed when the person reaches the edge.

9. Conservation of angular momentum of a turntable: Consider a turntable to be a

circular disk of moment of inertia It rotating freely at a constant angular velocity ωi. The

axis of the disk is vertical and the disk is supported by frictionless bearings. A record of

moment of inertia Ir is dropped onto the turntable. There is friction between the two disks.

After this "rotational collision," the disks will eventually rotate with the same angular

velocity. What is the final angular velocity, ωf, of the two disks?