Chapter 5 Rotation of a Rigid Body

§5-5 Angular Momentum of a rigid Body Conservation of Angular Momentum

§5-1 Motion of a Rigid body

§5-2 Torque The Law of Rotation rotational Inertia

§5-3 Application the Law of Rotation

§5-4 Kinetic Energy and Work in Rotational Motion

§§5-1 5-1 Motion of a Rigid body

1. Rigid body1. Rigid bodyThe body has a perfectly definite and unchanged The body has a perfectly definite and unchanged shape and size no matter how much external shape and size no matter how much external force acts on it.force acts on it.

2. Motion forms of rigid body2. Motion forms of rigid bodyTranslation( 平动

)Can be regarded as a particleCan be regarded as a particle

The distance between two points in a rigid body maintains constant forever

RotationRotating around axisRotating around axis

Rotating around a pointRotating around a point

Fixed axisFixed axis

Moving axisMoving axis

----Superposition of several rotations Superposition of several rotations around axisaround axis

axis

Translation + rotation

3. Rotation of a rigid body around a fixed axis

Every point of the rigid body moves in a circle

P

Q

axis

,,

They have the same angular displacement, angular speed & angular acceleration.

Select arbitrary point P.P’s rotation can represent the rotation of the rigid body.

P

Q

axis

,, Angular position

Angular displacement

Angular speed

td

d

tt d

d

d

d 2

Angular acceleration

4. Relation between angular and linear quantities

r

v s2

nt a ra

rr

z

x0

v

r P

,,,

nt aavsr ,,,,

§5-2 Torque The Law of Rotation Rotational Inertia

1. Torque

FrM

F

r φ

d

M

is located in the plane perpendicular to the axisF

F

r

No effect at all to rotate the rigid body around the axis

FrM

1Fr

)( 21 FFr

F

F2

1

is not placed in the plane perpendicular to the axisF

Only is useful for the rotation around the axis

2Fr

0

0`

imir

if

iF

i

2. The law of rotation

i

Use Newton’s second law to im

Tangential component

itiiiii amfF )(sinsin

)( iirm

The radial component of force passes through the axis and can’t cause the body to rotate.

iiiiii rfrF sinsin )( 2iirm

Multiply both side by ri ,

External torque Internal torque

dt

dIIM

--Law of rotation

For entire rigid body

iii

iiii

i rfrF sinsin )( 2ii

i

rm

resultant external torque M

Rotational inertia I

resultant internal torque =0

3. Calculation of moment of inertia

The magnitude of moment of inertia depends on the total mass and mass distribution of body, the location and orientation of the axis.

discrete particles

iiirmI 2

It describes the rotating inertia of a rigid body about an axis.

i

iirmI 2 --the rotational inertia (moment of inertia) of the rigid body about the axis

dl dm

—the mass per unit length

the mass distribution over a surface

ds dm —the mass per unit area

the mass distribution over a volume

dvdm —the mass per unit volume

the mass distribution over a line

Continuous distribution of mass

I dmr 2

Exa. A slender rod has mass m ， length l . Find its I about some axis follow as

⑴the axis through O and perpendicular with the rod.⑵ the axis through an end of the rod and perpendicular with it.

⑶ The axis at arbitrary distance h from O

m lo

dm

x

dxdm l

m

dx x

Solution

⑴

m l⑵

dmrI 2

dm

x

dxdm

l

m

dxxl

0

2

2

3

1ml

xdxo o

dmrI 20 dxx

l

l 2

2

2 2

12

1ml

m l

h

⑶

dxxIh

l

hlh

2

)2

(

2

xo o

22

12

1mhml

20 mhIIh --Parallel-axis theorem

Perpendicular-axis theorem

x y

z

O

dm

r

mz dmrI 2

m

dmyx )( 22

mm

dmydmx 22

xyz III -- -- Perpendicular-axis theorem

Suppose a thin board is Suppose a thin board is located in located in xOyxOy plane plane

Calculate the Acce. of the blocks and the tension of the rope.

Example are fastened together by a weightless rope across a fixed pulley ( )

21 mm ,

2`1 mm <

mThe pulley has mass and radius . There is a fractional torque exerting on the axis. The rope does not slip over the pulley.

r rMr

1m

2m

§5-3 Application the Law of Rotation

1T 2T

rM

a

a gm1

1T

1ma

gm2

2T

2m a

r

1m

2m

SolutionAccording to New.’s Second Law

1m

2m

amgmT 111 amTgm 222

：：

According to Rotational Law

Pulley ： IMrTrT r 12

ra 2

2

1mrI

221

12

mmm

rM

gmma

r

)(

)(

)(

22

11

agmT

agmT

We get

Example A uniform circular plate with mass m and radius R is placed on a roughly horizontal plane. At the beginning, the plate rotates with angular speed 0 around the axis across the center of it. Suppose the fractional coefficient between them is . Calculate how many times does the plate rotate before it stops?

Solution

Select mass unit dsdm

2R

m

S

m

rdrds 2

r drR

The friction exerts on dmgdmdfr

Frictional torque correspondingly

rr rdfdM

rdrg2

drrg 22

rr dMM

drrgR 2

02 mgR

3

2

dt

dIM

According to

We have

tdtg

03

2

dt

dmR

2

2

1mgR

3

2

0

02

1

dR

We get04

3 g

Rt

QuestionHow many revolutions does the plate rotate before it stops?

§5-4 §5-4 Kinetic Energy and Work in Rotational Motion

1. The work done by torque

0

0r

F

rd

d

As rdFdA�

dFr sin

MdThe rigid rotating from 21

The work done by the torque 2

1

MdA

2. Kinetic Energy of rotation

ir

iv

im

s KE of rotation im2

2

1iiki vmE

22

2

1 iirm

For the entire rigid body

i

ikk EE 22 )(2

1 i

iirm

i

iirmI 2

Then 2

2

1 JEk

3. The theorem of KE of a RB rotating about a fixed axis

asdt

dIM

MddA d

dt

dI dI

-- KE of rotation

21

22 2

1

2

1 IIA

2

1

2

1

dIMd

-- The theorem of KE of a RB rotating about a fixed axis

4. Gravitational potential energy of RB

…can be seen as the GPE of a particle located on the center of mass with the same mass of m.

Example A uniform thin rod m, l ， One end is fixed. Find its of point A as it rotate angular from horizontal line?

?Aa?Av

m 、 l

O

A

A

Av

Solution

mgO

l

21

22 2

1

2

1 IIA as

02

1cos

22

0

Id

lmg

2

3

1mlI lvA

sin3glvA

cos2

3g

dt

vda A

t

sin32

gl

va A

n

We get

Another solutionUse conservation law of energy of rigid body, we can get first.av

Then use dt

vda A

t ,2

l

va A

n

Or use IM We can get

then lat

We can get the Acce. of A finally

im the angular momentum of :

iiii rvmL

Direction

ir

iv

imiL

§5-4 §5-4 Angular Momentum of a rigid Body Conservation of Angular Momentum

1. Angular momentum of a rigid body rotating about a fixed axis

2iirm

Angular momentum of a rigid body rotating about a fixed axis ：

I

2. Angular momentum theorem

)( 2ii

i

rmL

Asdt

dLM

Direction: Same as iL

， we have dLMdt

integration 2

1

2

1

L

L

dLdtMt

t

)( Id

22

11

)(

I

IId

1121

2

1

IIMdtt

t

=Constant when M=0IL 3. Conservation Law of Angular Momentum

2

1

t

tMdt ---impulse torque

—Angular momentum theorem

of a rigid body

Example The rotational inertia of a person and round slab is I0. The mass of dumbbell is m . Their rotating angular speed is 0 and the rotating radius of m is r1 at the beginning.

Calculate:The angular speed and the increment of the mechanical energy when the arms of the person contracts from r1 to r2

r

r1

2mm

I0

Person + round slab + dumbbells = system

21 LL

02

12

10 )( mrmrI

)( 22

220 mrmrI

0220

210

2

2 mrI

mrI

Resultant external torque is zero. So its angularmomentum is conservative.

The increment of the mechanical energy

kE

20

210 )2(

2

1 mrI 2220 )2(

2

1 mrI

0kk EE

Example A round platform has mass of M and radius of R. It can rotates around a vertical axis through its center. Suppose all resistant force can be neglected. A girl with mass of m stands on the edge of the platform. At the beginning, the platform and the girl are at rest. If the girl runs one revolution, how much degree does the girl and the platform rounds relative to the ground, respectively?

R

M

m

0 MII

Let ： MII , be rotational inertia of the girl and the platform.

,

Resultant external torque of system = 0

0LL

be angular speed of the girl and the platform relative to the ground.

22

2

1mRIMRIM

Solution

then ：

Angular momentum is conservative ：

And

We get ：

M

m2

The girl relative to the platform ：

M

mM 2

Let t refer to the time that the girl runs one revolution on the platform, then ：

tt

dtM

mMdt

00

2 2

The girl rounds relative to the ground ：

mM

Mdt

t

2

20

The platform rounds relative to the ground ：

mM

mdt

t

2

20

Example A uniformly thin rod has M,2l . It can rotate in vertical plane around the horizontal axis through its mass center O. At the beginning, the rod is placed along the horizontal position. A small ball with mass of m and speed of u falls to one end of the rod. If the collision between the ball and the rod is elastic. Find the speed of the ball and the angular speed of the rod after they collide each other. m

ou

l2M

v

m

ou

l2MSolution

As mg of the ball << the impulse force between the ball and the rod

Then the resultant external torque of the system with respect to O = 0.

So we can neglect mg during colliding.

22

3

12

12

1mllmI

As the collision is elastic, then the mechanical energy of the system is conservative

222

2

1

2

1

2

1 Imvmu

Jmvlmul

The angular momentum of the system is conservative.

And

mM

mMuv

3

)3(

lmM

mu

)3(

6

Solve above equations, we get

Rod’s

Ball’s