Transcript Lecture 22.AngularMo..

Angular Momentum
Lecturer:
Professor Stephen T. Thornton
Reading Quiz:
Can an object
moving in a straight
line ever have a
nonzero angular
momentum?
A) Always
B) Never
C) Sometimes
Sometimes, because it depends upon the axis of rotation
around which you want to find the angular momentum.
There is no angular momentum when the object passes
through the rotation axis, because the moment arm is zero.
There is angular momentum when the moment arm is
nonzero (see left sketch).
Answer: C
Last Time
Rotational kinetic energy
Objects rolling – energy, speed
Rotational free-body diagram
Rotational work
Today
Angular momentum
Vector (cross) products
Torque again with vectors
Unbalanced torque
Angular Momentum of Circular Motion
This particle has linear
momentum. We can
also say it has an
angular momentum
with respect to a given
point, in this case the
center of the circle.
L  I
unit: kg  m / s
2
In the case of the particle moving
around the circle, let’s look more
carefully at the angular momentum.
v
L  I   ( m r )    m vr
r
2
B ut note that m v  p ,
so w e have
L  pr
This is another way to
determine angular
momentum.
L = m v sin q r = pr sin q
m
The Angular
Momentum of
Non-Tangential
Motion
m
L  pr sin   pr

A particle moving in any direction can
have angular momentum about any point.
O
O
O
p
O
Angular Momentum in Linear
and Circular Motion
The L in each view is constant. If v , r are the
same, then L is the same.
L = m v r^
Change in angular momentum
L = I
L = I, divide by t
L
t
 I

t
  I 
 I   torque
dL
dt
looks like F  m a 
dp
dt
This equation looks similar to Newton’s
2nd law. It is sometimes called Newton’s
2nd law for rotation.
Conservation of angular momentum
 
L
t

dL
dt
Note what happens when there is no torque.
L = 0, and angular momentum is constant.
Li  L f
if  net, ext  0
Note similarity to conservation of
linear momentum when Fnet,ext = 0.
pi  p f
if Fnet, ext  0
Conceptual Quiz:
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, how
do her rotational inertia, her angular momentum
and her rotational kinetic energy change?
They all increase.
They all remain the same.
They all decrease.
Rot inertia decreases, angular momentum
remains constant, and her KE increases.
E) Rotational inertia and angular momentum
decrease, KE decreases.
A)
B)
C)
D)
Answer: D
Angular momentum must be
conserved. No torque. Rotational
inertia decreases, because radius
decreases. Only D is possible.
How does KE increase?
K  1 I  2  1 ( I  )  1 L
2
2
2
I goes down,  goes up, L constant.
But K will increase because of .
http://www.youtube.com/watch?
v=AQLtcEAG9v0
Vector Cross Product;
Torque as a Vector
The vector cross product is defined as:
C = A´ B
C = A ´ B = AB sin q
The direction of the cross product is
defined by a right-hand rule:
The vector (cross) product can also be
written in determinant form:
Some properties of the cross product:
Conceptual Quiz
The direction of the vector cross
product j  i is along the direction
A) i
B)  i
C) j
D) k
E)  k
Answer: E
For a particle, the torque can be
defined around a point O:
t = r´ F
Here, r is the position vector to the point
of application of force relative to O.
·
Torque can be defined as the vector
product of the position vector from the
axis of rotation to the point of action of
the force with the force itself:
t = r´ F
Torque
t = r´ F
A right-hand rule
gives the direction of
the torque.
The Right-Hand
Rule for Torque
r
t = r´ F
r

F
r
Yo-yo demo
F
torque in
R otate
R otate
F
v
r
O
v
torque out
O
N o m otion
r
Angular Momentum of a Particle
The angular momentum of a particle about
a specified axis (or point) is given by:
L  rp
The Right-Hand Rule for Angular Momentum
p
L  I
L r p
Let’s do this demo!
L  I
Let’s do this demo!
L  I
No torque: L is
conserved. I
decreases, therefore
ω must increase.
A Rotational Collision – Angular
momentum will be conserved here.
Angular Momentum of a Particle
If we take the derivative of L , we find:
0
dL
dt

d
(r  p ) 
dt
Since
we have:
r´
å
dr
 p  r
dp
dt
dt
å
dp
F = r´
t =
dt
dL
dt
r
dp
dt
=
dL
dt
Opposite Particles. Two identical
particles have equal but opposite
momenta, p and  p , but they are not
traveling along the same line. Show that
the total angular momentum of this system
does not depend on the choice of origin.
Conceptual Quiz:
When a large star burns up its fuel, the
gravitational force contracts it to a small size,
even a few km. This is called a neutron star.
When neutron stars rotate at high speed, even
100 rev/sec, they are called pulsars. They have
more mass than our sun. What causes the high
rotational angular velocity?
A) Friction of gas particles
B) Conservation of angular
momentum
C) The dark force
D) Conservation of energy
Answer: B
Just like our own sun, these stars
rotate about their own axis. As
gravity contracts the particles closer
and closer, the density becomes huge.
There are no torques, so angular
momentum must be conserved.
L=I, so as I decreases,  must
increase.