Chapter 7

Rotational Motion and the Law of Gravity

• What is a Rigid Body?

• Rotational Kinematics

• Angular Velocity ω and Acceleration α

• Uniform Rotational Motion: Kinematics

• Uniform Circular Motion: Kinematics and Dynamics

• Applications

• Newton’s Law of Universal Gravitational Attraction

• Planetary motion

Rigid Body and its Motions

• Until now we considered only the motion of point-like objects. Objects with

extended size can be considered as a collection of many point-like particles.

• When these particles do not move with respect to each other, the system is called a

rigid body: it cannot be deformed.

• The general motion of a rigid body can be split into two types:

Translational (Linear) Motion

Rotational (Angular) Motion – we need an “angular” formalism

Object and terminology:

, , , , , ...r r v a F p

, , , , , ...L

Linear velocity

Translation velocity

Center or axis of rotation

Rigid body A particle of the body

• In purely rotational motion, all points on the object move

in circles around the axis of rotation (“O”), with each point

described by a vector position r

Convention:

• angles measured counterclockwise are positive

• angle measured clockwise are negative

x

y

O

θ′

θ r′ r

• In our approach to rotations, angles will be measured in

radians: 1 radian is the angle at the center of a circle subtending

an arc equal to the radius of the circle

• When the angle at the center is expressed in radians, the

length of the arc subtended is given by:

θ

θ = 1 rad

r

r

l = r

θ

r

r

l = rθ

l r

+

‒

Def: The angle θ made by the position vector r with respect

to an arbitrary axis (say x) is called angular position

Ex: see the two points on the adjacent bicycle wheel: notice that,

as long they are not on the same radius, the points on the rigid

body will have different angular positions

Angular Kinematics – Angular Position θ

Ex: The circumference 2πr of a complete

rotation subtends an angle of 2π

How can we use angular positions to describe rotations?

• Notice that even though the angular positions of

different points of a wheel are in general different, when

the wheel rotates, all points rotate through the same angle

2 1

0lim

t t

Def: The change in angular position of all the points on

a rotating rigid body is called angular displacement:

• Then, if we want to refer to how fast the angular position changes we have to define

first the average angular velocity as the angular displacement divided by time:

• Therefore, like in the linear case, the instantaneous angular velocity is given by:

x

θ2

θ1

Δθ

Arbitrary

radius

Ex: If the bicycle wheel makes two complete rotations

every second, we say that it has a constant angular

speed of 2×(2π/1 s) = 4π rad/s

Angular Kinematics – Angular Displacement and Velocity

2 1

2 1t t t

SI

rad s

• Hence we can define the average angular acceleration as the rate at which the

angular velocity changes with time:

2 1

2 1t t t

0lim

t t

2

SIrad s

Angular Kinematics – Angular Acceleration. Angular vector directions

The direction of

angular velocity

is given by a

right hand rule

The direction of

angular acceleration

is parallel or anti-

parallel with the

angular velocity

depending on

whether ω increases

or decreases 0 0

• Hence, the instantaneous angular acceleration:

• Although it is not as intuitive as in the translational case, the angular velocity and

acceleration are vectors, perpendicular on the circle of rotation:

Ex: If the bicycle wheel spins each

second through an angle larger and

larger by π, we say that each complete

rotation its average angular velocity is

π, so its instantaneous angular velocity

increases by 2π rad/s, so it has a

constant angular acceleration 2π rad/s2

Relating Linear and Angular Kinematics

• So, the angular displacement, velocity and acceleration characterize the entire rigid

body: all points have the same Δθ, ω and α

• However, the linear distance, velocity and acceleration of points at various radii r

from the axis of rotation are different: each point has a different Δl, v and a

lr

tv r

t

• So, as long as they are not at the same distance r from

the center of rotation, the different points on a rigid body

have different linear speeds, increasing from zero in the

center of rotation to a maximum value on the outer rim

of the rotating rigid body

x Δθ

Δl = rΔθ

ω

r r

• The linear kinematics of each point on a rigid body can

be related to the overall angular characteristics based on

the relationship l = rθ

• For instance, consider a wheel rotating with constant

angular speed ω. A point at distance r from the center of

rotation will rotate with constant linear speed v traveling

an arc Δl in a time Δt: therefore, we obtain

v v

ω

r1 v1 = ωr1

v2 = ωr2

v3 = ωr3

r2

r3

How about Acceleration?

• The acceleration is a bit more complex since in general the vector linear

acceleration of a particle in circular motion is not tangent to the trajectory.

• However, it can be considered as having two components – one tangent to the

trajectory (parallel or anti-parallel with the velocity) and one perpendicular on the

velocity:

at: tangent, describes how the magnitude of the linear velocity varies

ar: radial (or centripetal), describes how the direction of the velocity varies

at • The component at of a particle at distance r from the

axis of rotation can be easily related to the angular

acceleration α of the rigid body:

• The component ar (also called centripetal since it

always point toward the center of rotation) makes

necessary a separate discussion – a few slides further…

t

vr

ta r

t

x

ω

r

a

ar

Exercise : Components of acceleration

A particle moves as shown in the figure. Between points B and D, the path is a straight line.

Let’s figure out the net acceleration vectors in points A, C and E along the path represented

below, for each of the following cases:

a) the particle moves with steadily constant speed

b) the particle moves with steadily increasing speed

c) the particle moves with steadily decreasing speed

x

ω0

r

v0

x Δθ

Δl

r v

t0 = 0 At time t ω

• The linear-angular relationship provides an easy way to describe circular motion

with constant angular acceleration α, by simply noticing that each point on the

rotating rigid body accelerates uniformly along the respective arc of circle

• Assume that the motion starts at t0 = 0 when the rotation is characterized by θ0, ω0

210 0 2

0

2 2

0

10 02

2

t tt

t t

t

210 2

0

2 2

0

102

2

t

t

t

l v t a t

v v a t

v v a l

l v v t

The linear motion

of one particle of

the rigid body at

distance r from

the center of

rotation

if each linear

quantity is

divided by r,

we obtain…

The rotational

motion of the

entire rigid body

(valid for any of

its parts)

• Then, if the angular acceleration α is constant, at a later instant t,

Angular Kinematics – Uniformly accelerated rotation

Rolling Motion (Without Slipping)

Observation useful in problems:

• In figure (a), if the wheel of radius r is rolling

without slipping, the point P on the rim is at rest with

respect to the floor when it touches it, while the center

C moves with velocity v to the right

• In figure (b), the same wheel is seen from a reference

frame where C is at rest. Now point P is moving with

velocity –v. Since P is a point at distance r from the

center of rotation, we have:

• Therefore, when a wheel, or a sphere, or a cylinder

rolls, its translational speed (that is, the speed vcm of

its center of mass) is related to its angular speed ω by

• Caution: even though this has the same form as the

relation between linear speed of a point and the

angular speed, it is not the same thing.

cmv r

v

v

a) Wheel seen by someone on

the ground:

C

P

C

P

b) Wheel seen by someone on

the bike:

v r

ω

ω

r

srad 83srad s 60

22500rpm 2500

srad 150srad s 60

24500rpm 4500

1

0

Problems:

1. Uniformly accelerated rotation: An automobile engine slows down from 4500 rpm to

2500 rpm in 2.5 s. Calculate

a) its angular acceleration (assumed constant)

b) the total number of revolutions the engine makes in this time.

2. Rotational and linear motion: You are to design a rotating cylindrical axle to lift buckets

of cement from the ground to a rooftop. The buckets will be attached to a hook on the free end

of a cable that wraps around the rim of the axle; as the axle turns, the buckets will rise.

a) What should the diameter of the axle be in order to raise the buckets at a steady 2.00 cm/s

when it is turning at 7.5 rpm?

b) If instead the axle must give the buckets an upward acceleration of 0.400 m/s2, what should

be the angular acceleration of the axle be?

7.5 27.5 rpm rad s rad s

60 s 4

• This expression is valid for any particle

moving along a curved trajectory: if the

curvature of the path can be fitted locally by a

circle of radius r and the instantaneous speed

is v, the expression above gives the radial

component of the acceleration in the

respective point

• So, the radial (or centripetal) acceleration describes how the direction of the velocity

changes: that is, if the object moves in a straight line, ar = 0

• To see how ar is related to the speed v, we’ll consider a particle moving in a circle of

radius r with α = 0, that is, a constant speed (caution! the velocity is not constant)

2

r

va

r

Centripetal Acceleration

v

1v

Two similar triangles form, such that: 2

r

v v v v l vv l

l r r

va

rt t r

2v

~Δl

Δθ

1v

r

r

v

v • Assume that the particle travels an arc-distance Δl

subtending an angular displacement Δθ in a time Δt

• Then, looking at the corresponding change in

velocity of the revolving particle and using the

definition of acceleration, we find that the centripetal

acceleration is given by

path

path

1v

2v

ra ra

r1

r2

2

1

1

r

va

r

2

2

2

r

va

r

• Summarizing list of correspondences between linear and rotational quantities:

Linear Type Rotational Relation

x displacement θ x = rθ

v velocity ω v = rω

at acceleration α at = rα

ar acceleration - ar = rω2

1f T

2

rv r

T

• We can introduce a new set of parameters describing periodic motion:

1. Frequency f : the number of revolutions per time

2. Period T: time required to complete a revolution

• Frequency and period are the inverse of each other:

• Using the idea of period we can see once more how the relationship between the

linear velocity and the angular velocity makes sense and another relationship for ar:

22

r

va r

r

SI1 s Hertz, Hzf

SIsT

Some useful quantities and a summary…

Uniform Circular Motion – Kinematics

• The uniform circular motion is the motion of a particle in a circle of constant radius

at constant speed, such that α is zero

• Being always tangent to the circular path

the vector instantaneous velocity changes

direction, albeit its magnitude stays constant

• Hence, the tangential acceleration at is

zero, and the net acceleration is given only

by the centripetal acceleration ar pointing

everywhere perpendicular on the velocity

vr

path

Physical situation: a particle moving in a

circle:

ra

Comments:

• The magnitude of the centripetal acceleration is large if the speed is large

• The magnitude of the centripetal acceleration is large if the radius of rotation is small

Constant speed

Ex: if a car takes a turn at high speed, it will have a larger centripetal acceleration than

when taking it slowly

Ex: if a car takes a turn sharp turn (small radius), it will have a larger centripetal

acceleration than when taking a wide turn

2

r

va

r

• By Newton’s 2nd Law, since in the circular motion the acceleration is necessarily

not zero (since the velocity must change in direction), we see that for an object to

be in uniform circular motion there must be a net force acting on it.

• We already know the acceleration, so can immediately write

the force:

• This “centripetal force” is not a new force: any net force

pointing radially inward the circular trajectory (perpendicular

on the velocity) can play the role of centripetal force, since it

has as a result a change in direction of velocity

2

rrr

maF mv

v

Comments:

• A common misconception is to assume that an object on a

curved trajectory is thrown out of it by an outward “centrifugal”

force. We now see that the force must be actually inward

• The objects taking turns are apparently pushed outward by their

inertia, while the centripetal force keeps it on the trajectory

• If the centripetal force vanishes, the object flies off tangent to

the circle, not outward as if a centrifugal force were present

Uniform Circular Motion – Dynamics

vr

path rF

Exercise 1: How Angelina Succumbed to Bad Physics…

In the movie Wanted, bullets are “curved” by skilled tattooed assassins. For instance, Angelina

kills herself by firing a bullet in a circle passing through the skulls of some bald dudes before

hitting her. Say that the bullet has a mass m = 4.2 g and a muzzle speed of 600 m/s. Also, say

that the dudes arranged themselves conveniently in a circle with radius r = 5.0 m.

a) How big should be the centripetal force keeping the bullet on the circular trajectory?

Meditate about the possible origin of such a force and how realistic is such a scenario…

b) How fast should Angelina toss the gun to the sweaty guy in the middle for the scene to make

sense?

Exercise 2: Ball revolving as held by a string

A ball of mass m is connected by a string of length r

and moved into a circular trajectory with constant

speed v. Let’s estimate the force a person must exert on

the string to make the ball revolve in a horizontal

circle.

a) What is the nature of the centripetal force exerted on the

ball?

T

v

r

b) Based on Newton’s 2nd Law, what is this force in terms of given quantities?

Problem

3. Tension as a centripetal force: Now let’s assume that the ball from the exercise above is

swung in a vertical circle, still with a constant speed.

a) Determine the tension in an arbitrary point of the circle where the radius makes an angle θ

with respect to the horizontal

b) Use the result from part (a) to determine the tension in the string when the string is

horizontal, on top of the circle, and at the bottom of the circle.

Problems:

4. Normal as a centripetal force: A small remote-control car

with mass m = 1.60 kg moves at a constant speed of v = 12.0 m/s

in a vertical circle inside a hollow metal cylinder that has a

radius of r = 5.00 m. What is the magnitude of the normal force

exerted on the car by the walls of the cylinder at

a) an arbitrary point on the loop

b) point A (bottom of the vertical circle)

c) point B (top of the vertical circle)

5. Conical pendulum: A bob of mass m is suspended from a

fixed point with a massless string of length L (i.e., it is a

pendulum). What tangential speed v must the bob have so that it

moves in a horizontal circle with the string always making an

angle θ from the vertical?

ta r

Nonuniform Circular Motion – Elementary observations

r

path

Physical situation: a particle moving in a

circle:

ra

Changing speed

• If the velocity of a revolving particle changes in magnitude as well, the objects is

said to be in a nonuniform circular motion

ta

a

2 2 4 2 2 2

r ta a a r r

v

• Accordingly, the net force acting on the object will contribute to the change in

velocity direction with a centripetal component, and to the change in speed with a

tangential component

• As any revolving object, the object has

a centripetal acceleration:

• However, in this case, the particle also

has a tangential acceleration:

22

r

va r

r

• Hence, at a certain moment when the instantaneous angular speed is ω, the

instantaneous net acceleration of the particle has a magnitude given by:

Newton’s Law of Universal Gravitation – The idea

• We learned that Earth exerts a force on any mass – the weight. We studied how the

weight contributes to the motion of the object, and we quantified its strength through

its effect when it acts alone: the gravitational acceleration g

• However, what is this force? For instance, is it acted only by Earth?

• The formulation of a coherent theory of gravity (consistent

with his mechanics) is another example of Newton’s many

fundamental contributions to Physics

• He realized that gravity acts between any masses: it is just

logical to assume that the downward force that pulls onto an

apple or a person must be have the same nature as the attraction

exerted by Earth on the Moon in order to keep it on its orbit

• Moreover, by studying the motion of the planets, he deduced

that the magnitude of the gravitational attraction must vary

inversely proportional to the square of the distance between the

interacting masses

Quiz: Based on what law can we infer that the gravitational force is not acted only by Earth?

weight

Every two particles in the universe attract each other with gravitational forces

directly proportional to the product of their masses m and M, and inversely

proportional to the square of the distance r between them.

2gF GmM

r

r m M

Fg

Comments:

• The force of attraction is along the line

connecting the centers of gravity of the objects

(which in the cases we are interested in coincides

with their centers of mass or symmetry)

• Hence, the distance r between objects is the

distance between their centers of gravity

• Note the presence of Newton’s 3rd law in this

law: mass m attracts mass M and vice-versa with

action-reaction forces

• The gravitational constant G is very small,

meaning that the gravitational attraction will be

hardly observable between small masses

Newton’s Universal Law of Gravitation – Quantitatively

Gravitational constant 11 2 26.673 10 N m kgG

Fg

Ex: Two particles:

Two spheres:

r

m M

Two square plates:

r

m M

• The first determination of the value of the gravitational constant G is sometimes

attributed (wrongly) to Sir Henry Cavendish

• Cavendish actually tried to find experimentally the density of Earth. His data was

only much later used to calculate G

• Cavendish’s

apparatus was based on

the observation of the

gravitational attraction

between relatively

small objects using a

torsion balance

• If the force of

attraction is known, G

can be calculated easily

Modern version of Cavendish’s experimental arrangement:

Newton’s Universal Law of Gravitation – Determining G

m

m m

a

3

1

2 a

Problems:

6. Principle of superposition: Three identical particles of mass m are positioned at three

corners of a isosceles right triangle of equal sides a, as in the figure. Calculate the total

gravitational attraction on mass 1.

a√2

Gravitational Acceleration – In a more generic context

rE

mE

h

r

m h

mg G

2 22 2

1

1

E E Eh

EE E

m m mg G G

r rr h h r

• Until now, we assumed the local gravitational acceleration

constant. However, wee see that this is just an approximation

sing the weight does depend on the distance r to the center of

attraction (Earth):

• On the surface of the Earth h → 0 so

• Therefore, the gravitational acceleration at altitude h above

the surface of the Earth can be written:

2

2 2

9.8 m s

1 1h

E E

gg

h r h r

2

0 2 2~ 9.8 m sE E

h

E E

mm mmg G g G

r r

• So, the acceleration due to gravity varies over the Earth’s

surface due to altitude, local geology, and the shape of the

Earth, which is not quite spherical.

m

mg

mgh

• The motion of a satellite can be associated with a

projectile trajectory missing the surface of the Earth

due to the high speed.

• If the speed is too small, the satellite falls on the

surface (paths 1-2)

• In order to put a satellite on the orbit (paths 3-5), it

must have a minimum speed called satellite speed,

which can be estimated assuming a circular orbital

trajectory of radius r. Then

21initial 22

0 0E

Er r r

E

mmKE PE PE mv G

r 2 2 E

E

mv G

r

2

E 1

2r r

mm vmg ma G m

rr

Applications of the Law of Gravity – Satellites

• However, if the initial speed is too great, the satellite flies out into space and

becomes a spacecraft as it flies along open (paths 6-7)

• The minimum speed necessary for the object to escape the Earth’s gravity is called

escape speed and can be estimated by noticing that very far away from earth the

gravitational potential energy vanishes. Therefore

E1

mv G

r

mE

rE

Launching tower

For any two planets,

Applications of the Law of Gravity – Kepler’s Laws

1. The path of each planet about the Sun is an ellipse,

with the Sun at one focus.

Comment: This behavior is due to the specific 1/r2

dependence of the force on the distance.

3 2

1 1

2 2

T a

T a

2. The position of each planet about the Sun sweeps

out equal areas in equal times.

Comment: This behavior is explained by the

conservation of angular momentum (we’ll learn

about it in the next chapter)

3. The periods of the planets are proportional to 3/2

powers of the semi-major axis lengths of their orbits.

Comment: This is explained by Newton’s 2nd Law

applied to the orbiting body.