Conservation of Angular Momentum

8.01W10D2

Young and Freedman: 10.5-10.6

Announcements

Math Review Night Next Tuesday from 9-11 pm

Pset 10 Due Nov 15 at 9 pm

No Class Friday Nov 11

Exam 3 Tuesday Nov 22 7:30-9:30 pm

W011D1 Rolling without Slipping, Translation and Rotation Reading Assignment Young and Freedman: 10.3-4

Time Derivative of Angular Momentum for a Point Particle

Angular momentum of a point particle about S:

Time derivative of the angular momentum about S:

Product rule

Key Fact:

Result:

( )SS

d d

dt dt= ×

Lr p

rr r

( )S SS S

d dd d

dt dt dt dt= × = × + ×

L rr p p r p

r rr r r r r

SS S S

d d

dt dt= × = × =

Lr p r F

rrr r r r

τ

S Sd dm m

dt dt= ⇒ × = × =

r rv v v v 0

r r rr r r r

rL

S=

rrS ×

rp

Torque and the Time Derivative of Angular Momentum: Point

Particle

Torque about a point S is equal to the time derivative of the angular momentum about S .

SS

d

dt=

Lr

rτ

Conservation of Angular Momentum

Rotational dynamics

No torques

Change in Angular momentum is zero

Angular Momentum is conserved

rτS =

drL S

dt

Δ

rL S ≡

rL S, f −

rL S,i =

r0

r0 =

rτS =

drL S

dt

rL

S , f=

rL S,i

Concept Question: Change in Angular Momentum

A person spins a tennis ball on a string in a horizontal circle with velocity (so that the axis of rotation is vertical). At the point indicated below, the ball is given a sharp blow (force ) in the forward direction. This causes a change in angular momentum in the

1. direction

2. direction

3. direction

r̂

θ̂

k̂

F

ΔrL

v

Concept Question: Change in Angular Momentum

Answer 3. The torque about the center of the circle points in the positive -direction. The change in the angular momentum about the center of the circle is proportional to the angular impulse about the center of the circle which is in the direction of the torque

r̂k̂

Table Problem: Cross-section for Meteor Shower

A meteor of mass m is approaching earth as shown on the sketch. The distance h on the sketch below is called the impact parameter. The radius of the earth is Re. The mass of the earth is me. Suppose the meteor has an initial speed of v0. Assume that the meteor started very far away from the earth. Suppose the meteor just grazes the earth. You may ignore all other gravitational forces except the earth. Find the impact parameter h and the cross section πh2 .

Angular Momentum for System of Particles

Treat each particle separately

Angular momentum for system about S

, ,S i S i i= ×L r pr r r

sys, ,

1 1

i N i N

S S i S i ii i

= =

= =

= = ×∑ ∑L L r pr r r r

Conservation of Angular Momentum for System of Particles

Assumption: all internal torques cancel in pairs

Rotational dynamics

No external torques

Change in Angular momentum is zero

Angular Momentum is conserved

sysext S

SS

d

dt=

Lr

rτ

sysext S

SS

d

dt= =

L0

rr r

τ

rL

S , fsys =

rL S,i

sys

Angular Impulse and Change in Angular Momentum

Angular impulse

Change in angular momentum

Rotational dynamics

Δ

rL S

sys ≡rL S, f

sys −rL S,i

sys

rτSS

ext dtti

tf

∫ =rL S, f

sys −rL S,i

sys

ext sysave int( )SS St= Δ =ΔJ L

r rrτ

extf

i

t

SS

t

dt=∫Jr r

τ

Kinetic Energy of Cylindrically Symmetric Body

A cylindrically symmetric rigid body with moment of inertia Iz rotating about its symmetry axis at a constant angular velocity

Angular Momentum

Kinetic energy

K

rot=

12

I zω z2 =

Lz2

2I z

rω =ω z k̂

ω z > 0

Lz=I zω z

Concept Question: Figure Skater

A figure skater stands on one spot on the ice (assumed frictionless) and spins around with her arms extended. When she pulls in her arms, she reduces her rotational moment of inertia and her angular speed increases. Assume that her angular momentum is constant. Compared to her initial rotational kinetic energy, her rotational kinetic energy after she has pulled in her arms must be

1. the same. 2. larger.3. smaller. 4. not enough information is given to decide.

Concept Question: Figure Skater

Answer 2. Call the rotation axis the z-axis. When she pulls her arms in there are no external torques in the z-direction so her z-component of her angular momentum about the axis of rotation is constant. The magnitude of her angular momentum about the rotation axis passing through the center of the stool is L= Iω. Her kinetic energy is

Since L is constant and her moment of inertia decreases when she pulls her arms, her kinetic energy must increase.

Constants of the Motion

When are the quantities, angular momentum about a point S, energy, and momentum constant for a system?

• No external torques about point S : angular momentum about S is constant

• No external work: mechanical energy constant

• No external forces: momentum constant

sysext S

SS

d

dt= =

L0

rr r

τ

0 =Wext =ΔEmechanical

sysext d

dt=

pF

rr

Rotational and Translational Comparison

Quantity Rotation Translation

Force

Torque

Kinetic Energy

Momentum

Angular Momentum

Kinetic Energy

K

rot=

12

Icmω2

K

trans=

12

mVcm2

K

rot=

Lcm2

2Icm

L

cm=Icmω

K

trans=

p2

2m

p =mVcm

rFext =

drpsys

dt=mtotal

rAcm

rτcm =

drL cm

dt

Demo: Rotating on a Chair

A person holding dumbbells in his/her arms spins in a rotating stool. When he/she pulls the dumbbells inward, the moment of inertia changes and he/she spins faster.

Concept Question: Twirling Person

A woman, holding dumbbells in her arms, spins on a rotating stool. When she pulls the dumbbells inward, her moment of inertia about the vertical axis passing through her center of mass changes and she spins faster. The magnitude of the angular momentum about that axis is

1. the same. 2. larger.3. smaller. 4. not enough information is given to decide.

Concept Question: Twirling Person

Answer 1. Call the rotation axis the z-axis. Because we assumed that there are no external torques in the z-direction acting on the woman then the z-component of her angular momentum about the axis of rotation is constant. (Note if we approximate the figure skater as a symmetric body about the z-axis than there are no non-z-components of the angular momentum about a point lying along the rotation axis.)

Demo: Train B134(1) At first the train is started without the

track moving. The train and the track both move, one opposite the other.

(2) Then the track is held fixed and the train is started. When the track is let go it does not revolve until the train is stopped. Then the track moves in the direction the train was moving.

(3) Next try holding the train in place until the track comes up to normal speed (Its being driven by the train). When you let go the train remains in its stationary position while the track revolves. You shut the power off to the train and the train goes backwards. Put the power on and the train remains stationary.

A small gauge HO train is placed on a circular track that is free to rotate.

Table Problem: Experiment 6A steel washer, is mounted on the shaft of a small motor. The moment of inertia of the motor and washer is Ir. Assume that the frictional torque on the axle is independent of angular speed. The washer is set into motion. When it reaches an initial angular velocity ω0, at t = 0, the power to the motor is shut off, and the washer slows down during the time interval Δt1 = ta until it reaches an angular velocity of ωa at time ta. At that instant, a second steel washer with a moment of inertia Iw is dropped on top of the first washer. The collision takes place over a time Δtcol = tb - ta.

a) What is the angular deceleration α1 during the interval Δt1 = ta?

b) What is the angular impulse due to the frictional torque on the axle during the collision?

c) What is the angular velocity of the two washers immediately after the collision is finished?

Experiment 6: Conservation of Angular Momentum

Appendix:Angular Momentum and Torque

for a System of Particles

Angular Momentum and Torque for a System of Particles

Change in total angular momentum about a point S equals the total torque about the point S

sys,

, ,1 1

i N i NS iS i

S i i S ii i

dd d

dt dt dt

= =

= =

⎛ ⎞= = × + ×⎜ ⎟

⎝ ⎠∑ ∑

rL pL p r

r r rr r r

systotalS

SS

d

dt=

Lr

rτ

( )sys

total, , ,

1 1 1

i N i N i NS i

S i S i i SS i SSi i i

d d

dt dt

= = =

= = =

⎛ ⎞= × = × = =⎜ ⎟

⎝ ⎠∑ ∑ ∑

L pr r F

r r rr r r rτ τ

Internal and External Torques

Total external torque

Total internal torque

Total torque about S

total ext intSS SS SS= +

r r rτ τ τ

ext ext ext, ,

1 1

i N i N

SS S i S i ii i

= =

= =

= = ×∑ ∑r Frrr r

τ τ

int int int, , , , ,

1 1 1 1 1

N i N i N i N i N

SS SS j S i j S i i jj j i i j

j i j i

= = = =

= = = = =≠ ≠

= = = ×∑ ∑∑ ∑∑r Frrr r r

τ τ τ

Internal TorquesInternal forces cancel in pairs

Does the same statement hold about pairs of internal torques?

``By the Third Law this sum becomes

int int, , , , , , , ,S i j S j i S i i j S j j i× ×r F r F

r rr rr rτ +τ = +

rF

i, j=−

rF j ,i ⇒

rFint =

r0

int int, , , , , , ,)S i j S j i S i S j i j− ×r r F

rr rr rτ +τ =(

The vector points from the jth element to the ith element.

, ,S i S j−r rr r

Central Forces: Internal Torques Cancel in Pairs

Assume all internal forces between a pair of particles are directed along the line joining the two particles then

With this assumption, the total torque is just due to the external forces

No isolated system has been encountered such that the angular momentum is not constant.

int int, , , , , , ,)S i j S j i S i S j i j− × =r r F 0

rrr rr rτ +τ =(

sysext S

SS

d

dt=

Lr

rτ

, , , )i j S i S j−F r rr r r

P(