Conservation of Angular
Momentum
8.01
W10D2
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:
r
r r
L S  rS  p
dL S d
  rS  p 
dt
dt
dL S d
drS
d
  rS  p  
 p  rS  p
dt
dt
dt
dt
Time derivative of the angular momentum about S:
Product rule
Key Fact:
Result:
drS
drS
v

 mv  v  mv  0
dt
dt
dL S
d
 rS  p  rS  F   S
dt
dt
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 .
dL S
S 
dt
Conservation of Angular
Momentum
r
dL S
r
Rotational dynamics
S 
dt
r
r r
dL S
No torques
0  S 
dt
Change in Angular momentum is zero
r
r
r
r
L S  L S , f  L S ,i  0
Angular Momentum is conserved
r
r
L S , f  L S ,i
Concept Question: Change in
Angular Momentum
A person spins a tennis ball on a string in a horizontal
circle with velocity v (so that the axis of rotation is
vertical). At the point indicated below, the ball is given
a sharp blow (force F ) in the forward direction.
This
r
causes a change in angular momentum L in the
1. r̂ direction
2. φ direction
3. k̂ direction
Concept Question: Change in
Angular Momentum
Answer
3. The torque about the center of the circle
r̂
points in the positive k̂ -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
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
L S ,i  rS ,i  pi
Angular momentum
for system about S
iN
iN
i 1
i 1
Lsys
S   L S ,i   rS ,i  p i
Conservation of Angular Momentum
for System of Particles
Assumption: all internal torques cancel in pairs
Rotational dynamics
No external torques
 S ext
dLsys
S

dt
0   S ext
dLsys
S

dt
Change in Angular momentum is zero r
sys
r sys
r sys r
L S  L S , f  L S ,i  0
Angular Momentum is conserved
r sys
r sys
L S , f  L S ,i
Angular Impulse and Change in
Angular Momentum
Angular impulse
J  ( S ext )ave tint  Lsys
S
tf
J    S ext dt
ti
r sys r sys
r sys
L S  L S , f  L S ,i
Change in angular momentum
tf
Rotational dynamics
r sys
r sys
r ext
  S dt  LS , f  LS ,i
ti
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
r
   z kφ
z  0
Angular Momentum
Kinetic energy
K rot
Lz  I z z
2
L
1
 I z z2  z
2
2I 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.
2.
3.
4.
the same.
larger.
smaller.
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
1 2 1 L2 L2
K  I  I 2 
2
2 I
2I
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
0  S
ext
dLsys
S

dt
• No external work: mechanical energy constant
0  Wext  Emechanical
• No external forces: momentum constant
F
ext
dpsys

dt
Rotational and Translational
Comparison
Quantity
Rotation
Force
Torque
Kinetic Energy
r
dL cm
r
 cm 
K rot 
Kinetic Energy
r
r
r ext
dpsys
F 
 mtotal A cm
dt
dt
1
I cm 2
2
Momentum
Angular Momentum
Translation
K trans 
1
mVcm2
2
p  mVcm
Lcm  Icm
K rot 
L2cm
2Icm
K trans
p2

2m
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.
2.
3.
4.
the same.
larger.
smaller.
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-zcomponents 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 6
A 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
iN
i  N dr
 S ,i
dLsys
dpi 
S
  L S ,i   
 pi  rS ,i 

dt
dt
dt
i 1
i 1 

iN
dLsys
dpi

S
   rS ,i 
dt
dt
i 1 
iN
 iN
total

r

F




S
  S ,i i  S ,i
 i 1
i 1

dLsys
S
  S total
dt

Internal and External Torques
Total external torque
S
iN
ext
 
i 1
iN
ext
S ,i
  rS ,i  Fiext
i 1
Total internal torque
N
iN iN
iN iN
j 1
j 1 i 1
j i
i 1 j 1
j i
int
 S int    S int


 S ,i, j   rS ,i  Fi , j
,j
Total torque about S
 S total   S ext   S int
Internal Torques
Internal forces cancel in pairs
r
r
r
r
Fi, j  Fj,i  Fint  0
Does the same statement hold about pairs of
internal torques?
int
int


S ,i , j
S , j ,i  rS ,i  Fi , j  rS , j  Fj ,i
``By the Third Law this sum becomes
int
int


S ,i , j
S , j ,i   rS ,i  rS , j )  Fi , j
The vector rS ,i  rS , j points from
the jth element to the ith element.
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
int
int


S ,i , j
S , j ,i   rS ,i  rS , j )  Fi , j  0
With this assumption, the total torque
is just due to the external forces
S
ext
dLsys
S

dt
No isolated system has been
encountered such that the angular
momentum is not constant.
Fi , j  rS ,i  rS , j )