PHYS 218sec. 517-520

ReviewChap. 9

Rotation of Rigid Bodies

What you have to know

• Rotational kinematics (polar coordinate system)• Relationship & analogy between translational

and angular motions• Moment of inertia• Rotational kinetic energy

• Section 9.6 is not in the curriculum.

Analog between translation and rotation motion

Translation Rotation

position angle

velocity angular velocity

acceleration angular accele

x

v

a

q

w

2 2

ration

mass moment of inertia

1 1kinetic energy rotational kinetic energy

2 2force torque

momentum

m I

mv I

F

p

a

w

t

2 1 2 1

2 2 2

angular momentum

1 1 1

2 2 2

L

F ma I

W E E W E E

E mv U E mv I U

t w

w

= =

= - = -

= + = + +

å å

Angular velocity and acceleration

q

s length of the arc : , where is in radianss s rq q=

r Angular velocity

ratio of the angular displacement to t

d

dt

q

qw

D D

=

Angular velocity

ratio of the angular velocity to t

d

dt

w

wa

D D

=

The angular velocity and angular acceleration are vectors.

Follow the right hand rule.

0

counterclockwise

w> 0

clockwise

w<

Rotation with constant angular acceleration

All the formulas obtained for constant linear acceleration are valid for the analog quantities to translational motion

( )

( )

0

20 0

2 20 0

0 0

angular motion (fixed-axis rotation)

constant

1

2

2

1

2

t

t t

t

a

w w a

q q w a

w w a q q

q q w w

=

= +

= + +

- = -

- = +

( )

( )

0

20 0

2 20 0

0 0

linear motion (straight-line motion)

constant

1

2

2

1

2

a

v v at

x v t at

v v a x x

x x v v t

q

=

= +

= + +

- = -

- = +

Polar coordinate system

rr

x

yrq

i

j

ˆˆunit vectors in polar coordinate system: ,

ˆ ˆunit vectors in Cartesian coordinate system: ,

ˆPosition vector of :

r

i j

P r r r

q

=rP

( )

ˆ ˆˆ cos sin ,

ˆ ˆˆ ˆsin cos ,

ˆ ˆ ˆˆ ˆsin cos

r i j

dri j

ddr dr d

i jdt d dt

q q

q q qq

qq q w wq

q

= +

=- + =

= = - + = ( )

ˆ ˆ ˆsin cos ,

ˆˆ ˆ ˆcos sin ,

ˆ ˆˆ ˆ ˆcos sin

i j

di j r

d

d d di j r

dt d dt

q q q

qq q

q

q q qq q w w

q

=- +

=- - =-

= = - - =-

( ) tan

Then

ˆ ˆˆ ˆ ˆ .dr d dr dr dr

v r r r r r rdt dt dt dt dt

v rwq w= = = + = + \ =r

rTherefore, this is valid

in general.

Polar coordinate system

22

2

2

2

2

2 22

tan

ˆ ˆˆ ˆ 2

For a circular motion, 0 and we get

ˆˆ

,rad

dv d dr d r dra r r r r r

dt dt dt dt dt

dr d r

dt dt

a r r r

v va r r a r

r r

wq w w a q

w aq

w a

æ öæ ö æ ö÷ç÷ ÷ç ç÷= = + = - + +÷ ÷çç ç÷÷ ÷ç ç ç÷çè ø è øè ø

Þ = =

=- +

æö÷çÞ =- =- =- =÷ç ÷çè ø

rr

r

Energy in rotational motion

Rotational motion of a rigid body

irw

( )

( ) ( )

2 2 2 2 2

2 2 2 2 2

Kinetic energy of the -th particle

1 1 1

2 2 2Then the total kinetic energy of a rigid body reads

1 1 1

2 2 2

i i i i i i i

i i i i i

i

K m v m r m r

K K m r m r I

w w

w w w

= = =

= = = =å å å

2

This defines the moment of inertia;

i iI m r=å

Depends on1. How the body’s mass is

distributed in space,2. The axis of rotation

Moment of inertia

2

If the mass distribution is continuous,

I r dm=ò

Moments of inertia for various rigid bodies are given in section 9.6

Rotational kinetic energy is obtained by summing kinetic

energies of each particles.

Each particle satisfies Work-Energy theorem

Work-Energy theorem holds true for rotational kinetic energy2 1W K K= -

includes rotational kinetic energy

Parallel-axis theorem

Moments of inertia depends on the axis of rotation.There is a simple relationship between Icm and IP if the two axes are parallel to each other.

x

y

CM

P

Two axes of rotation

( )

( )

2 2

2 2

Position of , and

; choose the CM as the origin 0CM i i i cm cm

P a b d a b

I m x y x y

= = +

= + = =å

( ) ( ) ( ){ }( )

2 2 2

2 2

Then

2

P i i i i i

i i i i i

I m r P m x a y b

m x y a m x

= - = - + -

= + -

å å

å å

rr

2 i ib m y- å ( )2 2

2

i

CM

m a b

I Md

+ +

= +

å

2Parallel-axis theorem: P CMI I Md= +

1. If you know ICM, you can easily calculate IP.2. IP is always larger than ICM. Therefore, ICM is smaller than any IP, and it is natural for a

rigid body to rotate around an axis through its CM.

Ex 9.8 Unwinding cable I

F

F

initial final

2m

mass of the cylinder: 50 kg,

radius of the cylinder: 0.06 m

9.0 N,

final angular speed and the final speed of the cable?

m

R

F

=

=

=

2

21 1 2 2

1 2

2

speed

1Moment of inertia of the cylinder in this motion:

2Use Work-Energy theorem to obtain the

10, , 0

2Then

, where 18 J

10

2

220 rad/s,

other other

other

other

I mR

K U K I U

E W E W Fd

W I

Wv R

I

w

w

w

=

= = = =

+ = = =

+ =

Þ = = = 1.2 m/sw=

Ex 9.9 Unwinding cable IIcylinder with mass and radius

and block of mass

what is ?

M R

m

v

m

h

initial final

v

1 1

2 22 2

0,

1 1, 0

2 2

K U mgh

K mv I Uw

= =

= + =

Kinetic energy of mRotational kinetic

energy of M;I=MR2/2, =v/R

22 2 2

By energy conservation,

1 1 1 1 10

2 2 2 2 2

22 : speed when we neglect the rotational motion of the cylinder

1 2

vmgh mv MR m M v

R

ghv gh

M m

æ öæ ö æ ö÷ ÷ ÷ç ç ç= + + = +÷ ÷ ÷ç ç ç÷ ÷ ÷ç ç çè øè ø è ø

Þ = <+