Chapter 10

Rotational Motion About a Fixed Axis

2

Rigid body

A body with a definite shape that doesn’t change

1. Vibrating or deforming can be ignored

2. Distance between particles does not change

Motion of a rigid body can be expressed as

Translational motion of its CM

+ Rotational motion about its CM

Pure rotational motion: all move in circles

Axis of rotation Fixed axis

centers of these circles all lie on a line

3

Angular quantities

Angular coordinate / Angle

Angular velocity

d

dt

1 ds v

R dt R

y

xoR s

v

Angular accelerationd

dt

/s R

tan1 adv

R dt R

Radial acceleration 2 2/Ra v R R

Frequency ,2

f

Period1

Tf

4

Hard drive

Example1: The platter of hard disk rotates 5400rpm. a) Angular velocity; b) Speed of reading head located 3cm from the axis; c) Acceleration at that point; d) How many bits the writing head writes per second at that point if 1 bit needs 0.5µm length.

Solution: a) 5400 rev / min 90 rev / s 90Hzf

2 565 rad/sf

b) speed 17 m/sv R

angular velocity

c) acceleration 2 29580m/sRa R

d) 77

17 3.4 10 bit/s

5 10 N

4 MB/s

5

Useful Similarities

0 t 2

0

1

2t t

2 20 2

20

1

2x v t at

2 20 2v v a x

ov v at Uniformly accelerated rotational motion

Notice the similarities in different motion

, dx dv

v adt dt

, d d

dt dt

Analogous thinking is very helpful

6

Vector nature of angular quantities

Angular velocity and acceleration → vectors

d

dt

Points along the axis,

follows the right-hand rule

v r

d

dt

Angular acceleration

How does change?v r

spinning toppulley

Rotational dynamics

7

What causes acceleration of rotational motion?

Force: magnitude, direction and point of action

Push a door

Archimedes’ Lever

“Give me a fulcrum, and I shall

move the world. ” ———— Archimedes

Lever arm: the perpendicular distance from the

axis to the line along which the force acts

Torque about fixed axis

8

R F

R⊥: lever arm or moment arm

Torque = force × lever arm

F

R

R

The effect of force → angular acceleration

Net torque causes acceleration of rotational motion

Balance of rigid body

Play on a seesaw

positive rotational direction

s i nR F tanF R

9

Torque and balance

Example2: A 15kg mass locates 20cm from the axis of massless lever. Determine: a) torque on the lever; b) Force required respectively to balance the lever.

Solution: a)

b)

150N

3 0R F N m 20cm

1F

25cm1 2 5 3 0F c m N m

1 1 2 0F N

2F

5cm

4F

2 5 3 0F c m N m

3F

30 15cm

2 6 0 0F N

3 1 5 s in 3 0 3 0F c m N m 3 4 0 0F N

What about F4 ?

Rotational theorem

10

Fi

fi

mi

Ri

A particle mi in rigid body

ta n ta n ta ni i iF f m a i im R 2

ta n ta ni i i i i iF R f R m R

2ta n ta ni i i i i iF R f R m R

ext Net external torque

Rotational inertiaI

Rotational theorem about fixed axisI

0in or moment of inertia

Properties of Rotational theorem

11

2) Rotational equivalent of Newton’s second law.

I F m a

1) Only external torques are effective

Sum of the internal torques is 0 from N-3

3) Analogy of rotational and translational motion

change in motion a cause of the change F inertia of motion I m

Determining rotational inertia

12

2i iI m R

several particles

2I R dmcontinuous object

How mass is distributed with respect to the axis

Example3: Rotational inertia of 3 particles fixed on a massless rod about a, b, c axis.

cba

m 2m 3ml l

Solution:

222 3 2aI m l m l 214ml2 23bI m l m l 24ml

222 2cI m l m l 26ml

Uniform thin rod

13

Example4: Rotational inertia of uniform thin rod with mass m and length l. a) Through center;

Cxx

o dx

dm

Solution: Choose a segment dx

mdx

l2x2

2

l

lI

21

12ml

b) Through endo2

0

l mI x dx

l 21

3ml

Typical result, should be memorized

Uniform circles

14

Example5: Uniform thin hoop (mass m, radius R)

R

dm

Solution: Choose a segment dm

2I R dm 2m R

Example6: Uniform disk/cylinder (m, R)

r drSolution: Choose a hoop dm

2 r d r2

m

R2

0

RI r 21

2mR

Homework: Uniform sphere (m, R) (P246)

Parallel-axis theorem

15

If I is the rotational inertia about any axis, and IC is

the rotational inertia about an axis through the CM, and parallel to the first but a distance l, then

2CI I M l

Proof:

I Icl

CMo

r r

l( )I r dm r r dm 2

( )( )r l r l dm

2 2cI M l l r dm

2 2( 2 )r l r l dm

0CM r

IC is always less than other I of parallel axes

Perpendicular-axis theorem

16

The sum of the rotational inertia of a plane about any two perpendicular axes in the plane is equal to the rotational inertia about an axes through the point of intersection ⊥ the plane.

yxz III

xy

z

1) Only for plane figures or 2-dimensional bodies

2) x ⊥ y ⊥ z and intersect at one point

3) Try to prove it by yourself

Application of two theorems

17

dm

Solution: You can choose a dm

or apply the ⊥-axis theorem

xyxz IIII 2 21

4I mR

Example8: Thin hoop about a tangent line

Example7: Rotational inertia of uniform thin disk about the line of diameter

Solution: Perpendicular-axis theorem

Parallel-axis theorem

2 2 / 2mR mR2 2/ 2 3 / 2mR mR

Massive Pulley

18

mg

m

MR

T

Example9: A box is hanging on a pulley by massless rope, then the system starts to move without slip, determine the angular acceleration and tension.

Solution: T = mg ?

Free-body diagram

Rotational theorem: TR I

Newton’s second law: mg T ma

No slip motion: a R 21( )

2I MR

2

, 2 2

mg MmgT

m M R m M

I

Two boxes?

Rotating rod

19

Example10: A uniform rod of mass M and length l can pivot freely about axis o, released horizontally. Determine: a) α; b) aC; c) force acted by the axis.

Solution: a) Gravity → torque

o

Mg

C

2

lMg

I

3

2

g

l

b) Acceleration of CM3

2 4C

la g

c) CM a M g F

F

3 1

4 4F M g M g M g

a of the end? when 0?

Angular momentum

20

L I

I

Linear momentum p = m v

Angular momentumrotational analog

Rotational theorem can be written in terms of L

dI

dt

d I dL

dt dt

dL

dt

The rate of change in angular momentum of a rigid body is equal to the net torque applied on it.

Comparing with Newton’s second lawdp

F ma Fdt

Conservation of angular momentum

21

1) It is valid even if I changes

dLI

dt Torque & angular momentum

2) Valid for a fixed axis or axis through its CM

The total angular momentum of a rotating body remains constant if the net external torque is zero.

Law of conservation of angular momentum:

1) It holds for inertial frames or frame of CM

2) One of fundamental laws of conservation

22

Examples in sports

Figure skating Diving

23

Helicopter

24

Falling cat

A falling cat can adjust his posture to avoid injury, how to make it?Angular momentum is conserved

1) Bend his body

2) Rotate his upper part to proper position about blue axis, meanwhile the lower part rotates a less angle

3) Rotate his lower part to proper position about the red axis

4) Get the work done

Rotating disk

25

Example11: A disk is rotating about its center axis, and two identical bullets hit into it symmetrically. The angular velocity of system will ________

.o

A. increase; B. decrease; C. remain constant

Total rotational inertia increases

Solution: Total angular momentum of the system is conserved

L I c o n s t a n t

So angular velocity will decrease

What if two opposite forces act on the disk?

Man on rotating platform

26

Example12: A platform is rotating about its center axis, and a man standing on it (treat as a particle) starts to move. How does change if he goes: a) to point o; b) along the edge with relative speed v.

I oR

m

Solution: a) Conservation of angular momentum

2( )I m R I 2I mR

I

b) choose a positive direction2 2( ) ( )I m R I m R m v R

2

mvR

I mR

Hits on a rod

27

Example13: A bullet hits into a hanged uniform rod, determine the angular velocity after collision.

Solution: Conservation of total L

A

r

mv

o

M, l

m v r 2 21[ ]

3Ml mr

2 2

3

3

mvr

Ml mr

Notice: Momentum is not conserved in general!

Only if the bullet hits on position r = 2l / 3

Require: Force acted by axis remains constant

Rotational kinetic energy

28

21

2kE mv

Kinetic energy in translational motion

21

2kE I

Total Ek is the sum of Ek of all particles

21

2k i iE m v 2 21

2 i im R 2 2 21 1

2 2i im R I

rotational motion

Work done on a rotating body:

tanW F dl F dl

tanF Rd d dW d

Pdt dt

Power

Energy in rotational motion

29

Work-energy principle:2 2

1 1

2 22 1

1 1

2 2W d I d I I

Rotational theorem:d

I Idt

d d dI I

d dt d

Comparing with

Total mechanical energy is conserved in rotational motion if only conservative forces do work.

2 22 1

1 1

2 2W F dl mv mv

Potential energy of gravity: CU mgh

Rotating rod

30

Example14: A uniform rod (M, l ) can pivot freely about axis o (Ao=l/3), and it is released horizontally. Determine ω at the vertical position.

Solution: Distance oC=l/6

2

6C

lI I M

A o. .

C Rotational inertia

22 21 1

12 6 9

lMl M Ml

Conservation of mechanical energy

21

6 2

lMg I 3 /g l

Any position?

α = ?

General motion

31

The total kinetic energy

C CI

2 21 1

2 2k kC kR C CE E E mv I (Proved in P259)

Translational motion of its CM:

+ Rotational motion about its CM:

For general motion of a rigid body

net CF ma

Examples:

Rolling motion

32

A typical motion: rolling without slipping

vC

To make sure 0Pv

Relationship between translational and rotational motion

Cv R

Valid only if no slipping

Motion of pulley, tire, …

Rolling down an incline

33

Example15: A uniform cylinder (m, R) rolls down an incline without slipping, determine its speed if the CM moves a vertical height H.

Solution: Conservation of energy

2 21 1

2 2C CmgH mv I

21,

2C CI mR v R

2 / 3Cv gH

where

Comparing with sliding: 2Cv gH

Which is faster?

34

Fmg

NForces acting on the cylinder

Dynamics in rolling

Translational motion of CM

s in Cm g F m a Rotational motion about CM

C CI 21

2FR mR

We can obtain: aC = 2gsin /3 ， F = mgsin /3

Static friction causes the rolling motion

It also rearranges the kinetic energy

( )Ca R

Challenging question

35

A stick (M, l ) stands vertically on a frictionless table, then it falls down. Describe the motion of its CM, and of each end.