Physics I
Class 16
Conservation of
Angular Momentum
Rev. 23-Oct-03 GB
16-1
Angular Momentum of a Particle
Review
center of rotation (defined)

r


p  mv
Angular momentum of a particle
once a center is defined:
  
l  r p
(What is the direction of angular
momentum here?)
Once we define a center (or axis) of rotation, any object with a
linear momentum that does not move directly through that point
has an angular momentum defined relative to the chosen center.
16-2
Angular Momentum of a Particle
Angular Momentum of an Object
For a solid object, each atom has its own angular momentum:
  

l i  ri  p i  ri  ( m i v i )
The direction is the same as the direction of angular velocity.
The magnitude is

 
 
2
| l i |  | ri | | p i | sin( )  m i | ri | | v i |  m i ri  ri   m i ri
so
 
2
l i   m i ri
The total angular momentum, summing all atoms, is

 

2
L   l i   m i ri  I 
16-3
How Does Angular Momentum of
a Particle Change with Time?
Take the time
 derivative of angular momentum:


dl
d   d r   dp
 ( r  p) 
p  r 
dt dt
dt
dt
Find each term separately:
so

dr   
 p  v  p  0 (Why?)
dt

 dp  

r
 r  Fnet  net (Why?)
dt

dl 
 net (Newton’s 2nd Law for angular momentum.)
dt
16-4
Angular Momentum of a Particle:
Does It Change if  = 0?
Y
(0,0)

r

r
(blue)
(0,–3)
X
(red)
(4,–3)


p  m v = 1 kg m/s (+X dir.)
The figure at the left shows the same
particle at two different times. No forces
(or torques) act on the particle.
Is its angular momentum constant?
(Check magnitudes at the two times.)
Blue angle:  = 90º
l = r p sin() = (3) (1) sin(90º) = 3 kg m2/s
Red angle:  = arctan(3/4) = 36.87º
l = r p sin() = (5) (1) sin(36.87º) = 3 kg m2/s


[r sin()] is the component of r at a right angle to p . It is constant.
It is also the distance at closest approach to the center.
16-5
Conservation of
Angular Momentum
Take (for example)
two rotating objects that interact.

dl1 

 on 1 from 2  ext on 1
dt
dl2 

 on 2 from 1  ext on 2
dt
The total angular
momentum


 is the sum of 1 and 2:
dL d l 1 d l 2 



 ext on 1  ext on 2 (Why?)
dt
dt
dt
If there are no external torques, then
dL
0
dt
16-6
Example 1
An ice skater spins at 6 rad/sec with out-stretched hands.
Her rotational inertia is 1.5 kg m2. She then pulls her arms
in, thereby changing her rotational inertial to 1.2 kg m2.
What is her angular speed now?
No external torque, so L remains constant
I before before  L  I after after
after
I before before 1.5  6


 7.5 rad/sec
I after
1.2
16-7
Example 2
A wheel is rotating freely with an angular speed of 30 rad/sec on
a shaft whose rotational inertia is negligible. A second wheel,
initially at rest and with twice the rotational inertia of the first is
suddenly coupled to the same shaft. What is the angular speed of
the resultant combination of the shaft and two wheels?
No external torque, so L remains constant
I1 before  L  I1 after  I 2 after
after 
I1 before I1 before 1 30


 10 rad/sec
I1  I 2
I1  2 I1
3
16-8
Class #16
Take-Away Concepts
  
1. Angular momentum of a particle (review): l  r  p .
2.
Newton’s 2nd Law for angular momentum:
3.
Conservation of angular momentum (no ext. torque):

dl 
 net
dt

dL
0
dt
16-9
Class #16
Problems of the Day
___1. When a woman on a frictionless rotating turntable extends
her arms out horizontally, her angular momentum:
A. must increase
B. must decrease
C. must remain the same
D. may increase or decrease depending on her initial
angular velocity
E. changes into kinetic energy
16-10
Answer to Problem 1 for Class #16
The answer is C, angular momentum stays the same.
There is no external torque, so there is no way to change the
angular momentum. The woman’s rotational inertia increases, but
her angular speed decreases so that the angular momentum
remains the same.
There is no way to change momentum into energy!
16-11
Class #16
Problems of the Day
2. Two ice skaters of equal mass perform the following trick:
Skater A is at rest on the ice while skater B approaches. As skater
B passes by at 10 m/s, his center of mass is 1.8 m from skater A’s
center of mass at the instant of closest approach. At that instant,
the skaters reach out and clasp each other’s hands.
Find the rotational speed of the skaters, find the speed of their
center of mass, and describe the subsequent path of the center of
mass in terms of geometric shape.
Treat the skaters as point masses and ignore the friction of the
skates on the ice.
16-12
Answer to Problem 2 for Class #16
The system consists of the two skaters. There are no external
forces in this system, so we can use conservation of linear and
angular momentum. Let the mass of each skater be m.
The magnitude of the initial linear momentum is 10 m. The total
mass of the system is m+m. Therefore, the speed of the center of
mass = (10 m) / (m+m) = 5 meters/s. (before = after)
The magnitude of the initial angular momentum about the center
of mass at the instant the skaters clasp hands is 0.9 x 10 m = 9 m.
After clasping hands, the total rotational inertia of the skaters is
2 m (0.9)2 = 1.62 m. Therefore,  = 9 m / 1.62 m = 5.56 rad/s.
The path of the center of mass after clasping hands is a straight
line, not some kind of cycloid or other curve.
16-13
Activity #16 - Conservation of
Angular Momentum
Objective of the Activity:
1.
2.
3.
Think about conservation of angular momentum.
Use conservation of momentum to predict the
change in rotational speed in a simple system.
Compare measurements with predictions.
16-14