Chapter 11

General Rotation

Vector cross product

2

x y zA A i A j A k

x y zB B i B j B k x y z

x y z

i j k

A B A A A

B B B

a b c

b

a

0,a a

sinc ab where

Right-hand rule and c a c b

c

a b b a

The torque vector

3

r F

sinRF r

OF

P

Torque vector about point O

Expressed by cross product

Example1: A particle is at position , Calculate the torque about origin if .

10 12r i j

3 5F i k

Solution: We use the determinant form

10 12 0

3 0 5

i j k

r F

60 50 36i j k

Angular momentum of a particle

4

L r p

rO p

L

magnitude

Angular momentum about point O

sinL rp

Angular momentum theorem

dL dr dpp r

dt dt dt

v p r F

dL

dt

Rotational equivalent of Newton’s second law

Angular momentum in circle

5

L rmvAgreed with the rigid body case

Example2: Determine the angular momentum of a particle in uniform circular motion (m, v, r).

Solution: It depends on the choice of point O!

First calculate it about the center of circle2mr I

Direction?

What about another O’ ?

O

l

r

O’

Component along axis OO’

Conservation of angular momentum

6

The angular momentum of a particle remains constant if there is no net torque acting on it.

Kepler’s 2nd law: A line from the sun to a planet sweeps out equal area in equal time.

0 0

dL

dt

Area:1

sin2

A r v t

1sin

2

Arv

t

consant2

L

m

Typical case: Acted by a central force

Move in a spiral line

7

Example3: A mass m connected by a rope moves on frictionless table circularly with uniform and r. Then one pulls the rope slowly through the center, determine the work done when r changes to r/2.

Solution: No torque, L about o is conserved

22 / 2mr m r 4

F

r

om

2 21 1

2 2W mv mv Work

2 2 2 21 1( )

2 2 2

rm mr

2 23 / 2m r

Homework

8

A massless spring (l0=0.2m, k=100N/m) connects mass m=1kg to point o on a horizontal frictionless table. Mass m moves with v0=5m/s ⊥ the spring, and the length of spring becomes l=0.5m after rotating 90°, determine the final velocity v’ and θ.

90°

90°

Angular quantities for a system

9

Consider a system of particles (rigid body or not)

The total angular momentum:iL L

The net torque:net i ext

Angular momentum theorem for a system:

ext

dL

dt

(about the same origin O)

Valid in a inertial frame or frame of the CM

CC

dL

dt

(about the CM of system)

Rigid body & fixed axis

10

For a rigid body rotates about a fixed axis

Consider the component along axis2

iz i i i i iL m v r m r

Total angular momentum2

z i iL m r I

(rotational theorem)zz

dL dI I

dt dt

Angular momentum theorem for a rigid body:

1) Vector & Component 2) 1-dimensional case

Conservation of L for system

11

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

Example4: A uniform thin rod (m, l) rotates about fixed axis o with on frictionless horizontal table, and collides elastically with a resting mass m at its end. Determine the final angular velocity.

Solution: Elastic collision:

2 21 1

3 3ml ml mvl

2 2 2 2 21 1 1 1 1

2 3 2 3 2ml ml mv

2

o

m, l

m

.

Rotating about varying axis

12

L I

For a body rotating about a symmetry axis

(symmetry axis, through CM)

Rigid body rotates about a fixed axis zL I

/ext dL dt

Angular momentum theorem:

where and ext L may have different directions!

How angular momentum changes?

1) No initial rotation

2) With initial rotation

*The spinning top

13

W

N

Motion of a rapidly spinning top, or a gyroscope

L

/ext dL dt

External torques are acted by a pair of forces:

We always have ext L

Only change the direction of L

It is called precession

Precession of Earth

Bullet, bicycle, …

Noninertial reference frame

14

Inertial force: a type of fictitious force

Newton’s first law does not hold in such frames

To use Newton’s laws, we have to use a trick

iF ma where S Sa a

S’

S

mS mS S Sa a a

mS mS S Sm a m a m a

iF F ma

mS S S mSm a ma m a

Dynamics in noninertial frame

15

With considering inertial forces, N-2 is still valid

iF F ma

1) It is not a real force: no object exerts it

2) For rotating frame, also called centrifugal force2

iF m r

(opposite to the centripetal force)

Weight loss

Ring form spaceship

16

Example5: A ring-form spaceship with radius r is rotating to obtain a virtual gravity of g, determine the period.

r=r=50m50mOO

Solution: Centrifugal force → virtual gravity2

2 vm r m mg

r

22.1 m/sv gr

2 14 s

rT

v

*The Coriolis effect

17

If a body is moving relative to a rotating frame

There is another inertial force: Coriolis force

2CF mv

where is the relative velocityv

Coriolis effect

River, wind, whirlpool

Falling objects

Foucault pendulum

*Foucault pendulum

18

Earth

cFcF

1

1

2

23