PHYS 218sec. 517-520

ReviewChap. 10

Dynamics of Rotational Motion

What you have to know

• Dynamics of rotational motion

• Torque

• Equation of motion for a rotational motion

• Rotation about a moving axis

• Work & power

• Angular momentum and its conservation

Torque

Translational motion: governed by forces

Rotational motion: governed by torques

When you define a rotational motion, you should specify the axis of rotation. Therefore, physical quantities in rotational motion must include information

about the location of the axis of rotation.This makes the definitions of those physical quantities include x or r.

For example, even if the same force (same magnitude & same direction) is applied, the rotational motion depends on where the force exerts on, i.e., the distance between the axis of ration and the point where the force exerts on.

When you have a rotational motion, determine whether the rotation is clockwise or counterclockwise.

Torque

definition

Torque is a vector:

magnitude sin , where is the angle between and

r F

Fr r F

t

t ff

= ´

= =

rrr

rr

rr

Fr

This force causes counterclockwise rotation. (So, is positive)

Fr r Ft = ^rr

Q

rr

Frf

sin 0Frt f= >

Fr

f

rr

sin 0 : clockwiseFrt f=- <

Axis of rotation

Torque

rr

Fr

f

rad cosF F f=

tan sinF F f=

f

sinr f=l

Line of action of the force F

Lever arm tan

Therefore,

sinFr F r Ft f= = = l

Only the tangential force can give a torque.

Torque for a rigid body

Translational motion:

Rotational motion:

F ma

It a

=

=

åå

r r

Note the similarity of the two equations.

( )

tan

,tan ,tan

,tan ,tan ,tan

2

Since rotational motion depends on ,

first consider Newton's second law for a mass element ;

torque on by :

Then the total torque is

i

i i i

i i i i i i i i i i i

i i i

F

m

F m a

m F F r m a r m r r

m r

t a

t t

=

= = =

= =å ( )2 ,

where is the moment of inertia.

i im r I

I

a a a= =å å

x

y

z

rr

ir

,radiF

,taniF,i zF

use a ra=

It a=åRotational analog of Newton’s 2nd

law for a rigid body

Ex 10.2

F

Same situation as in Ex 9.8

Free-body diagram

x

y

Mg

n

2

2

2

2

1In this motion, ,

2then gives

26.0 rad/s

12

then the tangential acceleration is

0.36 m/s

I MR

FR I

FR F

I MRMR

a R

t a

ta

a

=

= =

= = = =

Þ

= =+

Direction of the rotation & sing of the torque

Reproduce the result of Ex 9.8as explained in the class

2 20Use 2v v ad- =

Ex 10.3

m

h

Same situation as in Ex 9.9

Free-body diagram

cylinder

+Mg

n

T

block

x x

y

y

T

mg

2

For the block,

For the cylinder,

acceleration of the cable tangential acceleration of the surface of the cylinder

1122

1Then the first eq. becomes

2

yF mg T ma

RT I

a R

aMRI RT Ma

R R

mg Ma ma

t a

a

a

= - =

= =

=

® =

Þ = = =

- =

åå

1 2

ga

M mÞ =

+

21

2I MR=

Obtain v !

Rigid-body rotation about a moving axis

Extend the analysis to the case where the axis of rotation moves

Translation + Rotation

cm cmv

ivriv¢

i cm iv v v¢= +r r r

( ) ( )

( )

( ) ( ) ( )2 2

The kinetic energy of the -th particle

1 1

2 21

22

Then the total kinetic energy of the system is

1 1

2 21

2

i i i i i cm i cm i

i cm cm cm i i i

i i cm cm i i i i

i

K m v v m v v v v

m v v v v v v

K K m v v m v m v

M

¢ ¢= × = + × +

¢ ¢ ¢= × + × + ×

¢ ¢= = + × +

=

å å å å

r r r r r r

r r r r r r

r r

2 21

2cm cmv I w+

( )

Second term:

0 0

Therefore, the 2nd term vanishes

i cm i

i i i cm i cm i i

i i i i

r r r

m r m r r M r m r

m r m v

¢= +

¢ ¢Þ = + = +

\ = Þ =

å å åå å

r r r

r r r r r

r r Using the definition of CM,

1 cm i i i i cmr m r m r Mr

M= Þ =å å

r r r r

2 2 2 2

Third term:

: moment of inertia of the object when the axis of rotation

passes its CM

i i i i i i cm

cm

v r m v m r I

I

w w w¢ ¢ ¢ ¢= Þ = =å å

Rolling without slipping

is a general expression.

But in the case of rolling without slipping,

CM

v R

v v R

w

w

=

= =2 2

Kinetic energy

1 1

2 2CM CMK M v I w= +

distance traveled by CM

l

=

s

In this case,

CM

ds dls l v v

dt dt= Þ = Þ =

Translation + Rotation

ext

Equations describing this motion

CM

CM

F M a

It a

=

=

åå

r r For translation, concentrate the mass of the object on its CM

same as the motion of a point-like particle

For rotation, consider the rotation about CM same as the rotational motion when the CM

is fixed.

The second equation is valid, if1. The axis through the CM must be an axis of symmetry

2. The axis must not change direction

Ex 10.4 Speed of a primitive yo-yo

initial 2

The yo-yo has mass and radius

since it is a cylinder

and the axis of rotation is passing its CM

1

2

M R

I MR=

0

0CMv

w

=

=

?CMv =

final

h1 1

22 2 2 2

2

2

2

Since this asks the speed of the yo-yo,

use energy conservation.

0,

1 1 1 1 1

2 2 2 2 2

3

40

CMCM CM CM

CM

K U Mgh

vK M v I M v MR

R

M v

U

w

= =

æ öæ ö ÷÷çç= + = + ÷÷çç ÷÷ç çè øè ø

=

=

2

Therefore,

3 3

4 4CM CMMgh Mv v gh= Þ =

Ex 10.5 Race of the rolling bodies

h

Rigid bodies are released from rest at the top of an inclined plane

speed at the bottom of the incline?

2

2

The moment of insertia is proprotional to .

So we can write

,

where depends on the shape of the rigid body.CM

MR

I cMR

c

=

MR

( )

( )

1 1

22 2 2 2

2

2

2

To obtain the speed of the rigid body, use energy conservation.

At the top,

0 (initially it is at rest),

At the bottom,

1 1 1 1

2 2 2 2

11 ( )

2

CMCM CM CM

CM CM

K U Mgh

vK M v I M v c MR

R

c M v v R

U

w

w

= =

æ ö÷ç= + = + ÷ç ÷çè ø

= + =

=

Q

0

0y =

1 2Energy conservation means ,

which gives

2

1CM

E E

ghv

c

=

=+

Ex 10.6 Acceleration of a primitive yo-yo (cf. Ex 10.4)

This is asking the acceleration of the body. So we draw free-body diagram.

O

Rr

y+

x+

Mg

T

21

2

Eq. for the translational motion of CM

Eq. for the rotational motion about CM

1 1 1

2 2 2

Then the first eq. becomes

1

22 1

,3

y

C

y

yCMy

y y

y

M

ma

I M

F Mg T

TR T R

aIT MR MR Ma

R R

Mg Ma Ma

a g

R

T

a at

aa

== -

= \ ^

æ ö÷ç ÷Þ = = = =ç ÷ç ÷çè ø

- =

Þ = =

= =

å

år r

3Mg Rolling without slipping

Try to reproduce the result of Ex 10.4

Ex 10.7 Solid bowling ball (rolling without slipping)

Free-body diagramacceleration & friction force?

draw free-body diagram®

22solid sphere:

5: its radius

CMI MR

R

=

Mg

x+

+

n

y+

R

bcosMg b

sinMg b

Since there is no motion in the -direction,

we do not consider the eq. for the -component.

y

y

2

for CM: sin

2for rotation:

5

2

5

x s x

xs CM

s x

F Mg f Ma

af R I MR

R

f Ma

b

t a

= - =

æ öæ ö ÷÷çç= = = ÷÷çç ÷÷ç çè øè ø

Þ =

å

åsf

2Then the eq. for CM becomes sin

55 2

sin and sin7 7

x x

x s

Mg Ma Ma

a g f Mg

b

b b

- =

Þ = =

If is small, the friction force which is required to have the rolling is small.

If is large, the ball easily slips, then we need a large friction force to make it

roll.

Work & Power in rotational motion

Work 2

1W F dr= ×ò

r r

To rotate a body, a force should be applied. Then this force do work on it.

tan tan

Since only the tangential force causes rotational motion,

the work done by the tangential force for the rotation of

a body through an infinitesimal angle is

( )

Therefore,

dW

d

dW F Rd d F R

W dq

q

q t q t

t q

= = =

=

Q

2 2 2 2

1 1 1 1

2

1

2 22 1

1 1

2 2

d dI d I d I d

dt dt

I d I I

q q q q

q q q

w

w

w qa q q w

w w w w

= = =

= = -

ò ò ò ò

ò

R

R

ds

dqtanFr

Work –energy theorem

2 22 1

2 22 1

Translational motion

1 1

2 2Rotational motion

1 1

2 2

W K mv mv

W K I Iw w

=D = -

=D = -

, , ,m I v F aw t a« ® ® «

Angular momentum

(linear) momentum p p mv=r r

Similarly, we define angular momentum L L r p= ´r r r

.cf r Ft = ´rrr

This is defined with r, so it depends on the choice of the origin.

This is a vector product. Therefore, the angular momentum is perpendicular to the plane spanned by r and p. If r and p lie on the xy-plane, L is in the z-

direction.

0L r L p× = × =r rr r

This is always true.

x

y

rr

f

p mv=r r

sinmv f

sinl r f=

Therefore, the magnitude of angular momentum is

sin sinL rp mvr mvlff= = =

perpendicular distance from the line of to .v Or

O

Newton's 2nd law of motion

dpF

dt=

rr

( ) ( )dL d dr dpr p p r v mv r F

dt dt dt dt

dL

dtt

æ ö æ ö÷ ÷ç ç= ´ = ´ + ´ = ´ + ´÷ ÷ç ç÷ ÷ç çè ø è ø

Þ =

r r r rr r r r r r r

rr

0=

Note the similarity.

F

p L

t«

«

Linear momentum is .

Similar expression for , i.e., in terms of and ?

F ma

L I a

=r r

r

Angular momentum of a rigid-body

w

iLr

O irr

im

i iv rw=( )

( )

2

2 2

angular momentum of mass

Then the total angular momentum of the rigid-body is

i

i i i i i i

i i i i i

m

L mvl m r r m r

L L m r m r I

w w

w w w

= = =

= = = =å å å

This is valid if the rigid-body lies on the xy-plane.

Angular momentum of a rigid-bodyz

w

-planexy

Consider a slice of the ridig-body, which is parallel to the -plane,

to calculate the angular momentum of each slice.

The total angular momentum of the rigid-body is the sum of

the angular momentum of

xy

each slice.

2

the perpendicular distance

from the particle to the axis of ro

The moment of inertia is defined as

,

where is

.

The angular momentum of the particle i

tation

the distanc

s

,

where is

i i

i

Note

I m r

r

L r p

r

=

= ´

å

r r r

re from the particle

to the origin.

They are different

in general

Angular momentum of a rigid-body

If the body rotates about an axis of symmetry, then we get the simple relation,

L Iw=r r

is valid for any system of particlesdL

dtt =å

rr w

This is a case where the axis of rotation is

an axis of symmetry of the rigid body,

then we have L Iw=-planexy

z

1m2m

( )

1 2 1 2

1 2

1 2

1, 2,

2

-axis is the axis of symmetry, therefore,

,

therefore,

The -components of and are canceled out.

Their -compnents add up,

sin sin

sin

z z

z

m m m r r r

L L L rmv

x L L

z

L L L rmv

mvr mvl m l l

ml I

q q

q w

w w

= = = =

= = =

= = =

= = =

= =

z w

xO

1Lr

2Lr

1rr

2rr q

sinl r q=

Note that the moment of inertia is defined with NOT with I l r

Conservation of angular momentum

Since , if 0,

0

constant

ext ext

dL

dt

dL

dt

L

t t= =

=

Þ =

å år

rr r

rr

r

.

If 0, constantext

cf

F p= =årr r

Ex 10.14

d

l

bvr

w

m

M

2

A door is rotating. Its moment of inertia is

1

3I Md=

Collision problem including rotational motion.

use angular momentum conservation

beforeafter

( ) 2

Initial angular momentum

, 0

Final angular momentum

, where

bullet door

door bullet bullet

L mvl L

L I I I ml

mvl

I

w

w

= =

= + =

Þ =

Use the numbers given in the textbook to verify the result.

Also check the change in kinetic energies.

Gyroscopes & precession

The motion of gyroscope was discussed in the class.

Here I do not repeat it.

But you should understand that the gyroscope has precession and this is due to the vector nature of angular momentum.