Angular Momentum & Torque
for Systems of Particles
Lecturer:
Professor Stephen T. Thornton
Reading Quiz
A particle is located in the xy-plane at a
location x = 1 and y = 1 and is moving
parallel to the +y axis. A force is exerted
on the particle along the +x axis. L and t
are in what directions about the origin?
A)
B)
C)
D)
E)
O
L and t are along the +z axis.
L and t are along the -z axis.
L is along the +z axis; t is along the –z axis.
L is along the -z axis; t is along the +z axis.
L is along the +y axis; t is along the +x axis.
·
y
v
F
x
z
Answer: C
L = r´ p
y
r
x
p
·
·
F
t
Last Time
Angular momentum
Vector (cross) products
Torque again with vectors
Today
Angular momentum and torque
system of particles
rigid objects
Unbalanced torque
Kepler’s 2nd law
Atwood Machine. An Atwood
machine consists of two masses,
mA  7.0 kg and mB  8.2 kg
connected by a cord that passes
over a pulley free to rotate about a
fixed axis. The pulley is a solid
cylinder of radius R0  0.40 m and
mass 0.80 kg. (a) Determine the
acceleration a of each mass. (b)
What percentage of error in a
would be made if the moment of
inertia of the pulley were ignored?
Ignore friction in the pulley
bearings.
System of Particles
The angular momentum of a system of
particles can change only if there is an
external torque—torques due to internal
forces cancel.
This equation is valid in any inertial
reference frame. It is also valid about a
point uniformly moving in an inertial
frame of reference. We are starting to get
very technical!
System of Particles
The equation above is not valid in general about a
point accelerating in an inertial frame of reference.
But the center of mass is special! The equation is
true even for an accelerating center of mass of a
system of particles or for a rigid object:
Angular Momentum
for a Rigid Object
For a rigid object, we can show that its
angular momentum when rotating
around a particular axis is given by:
Add up all the particles. If L is
along a symmetry axis (z here)
through CM, particles on one side
of symmetry axis cancel L on the
other side.
rotating
Li

L  Iω
dL



dt
 axis  I
So we finally have these equations for a rigid object.
The values must be calculated about
1) Origin or axis fixed in an inertial frame.
or
2) An origin at the CM or about an axis passing
through the CM.
If we do not have this, then things get real
complicated! We have reached our limit here!!
Torque and Angular
Momentum Vectors
  dL
dt
L  τt
r
ω
Torque
Gravity and Extended Objects
Gravitational torque acts at the center of mass, as
if all mass were concentrated there:
Torque
Gravity and Extended Objects
Gravitational torque acts at the center of mass, as
if all mass were concentrated there.
Do the Falling Rigid Body demo again.
Conceptual Quiz
You are looking at a bicycle wheel along
its axis. The wheel rotates CCW and is
supported by a string attached to the
rear of the handle. When the wheel is
released, the end of the handle closest to
you will
A)
B)
C)
D)
move up
move to the left
move to the right
move down
Do bicycle wheel demo.
Answer: C
Move to the right. The picture below
is looking from above.
Li
L
Lf
L
 
t
Torque
r
Lf
Li
L

Lf
Top view

Li
Conceptual Quiz
A man sits at rest on a frictionless rotating stool.
He holds a rotating bicycle wheel that has an
angular momentum L directed up. When he flips
the wheel over, so that it has L directed down, the
angular momentum of the system (man + stool +
wheel) is
A)
B)
C)
D)
E)
zero.
L, up.
L, down.
2L, up.
2L, down.
Answer: B
Angular momentum has to be
conserved. There is no torque to
change it.
Do experiment.
Angular Momentum
and Torque for a
Rigid Object
A system that is
rotationally imbalanced
will not have its angular
momentum and angular
velocity vectors in the
same direction. A torque
is required to keep an
unbalanced system
rotating.
L and  in
p in
An unbalanced car wheel will cause
problems on your wheel bearings. We
need to keep our wheels well balanced,
dynamically not just statically.
Kepler’s 2nd Law
1
dA  r (v dt sin  )
2
dA 1
 rv sin  but L  r  mv  mrv sin 
dt 2
dA L

 constant
dt 2m
There is no torque so L is
constant, and Kepler’s second
law states that each planet
moves so that a line from the
Sun to the planet sweeps out
equal areas in equal times.
Conceptual Quiz
You are holding a spinning bicycle
wheel while standing on a
stationary turntable. If you
suddenly flip the wheel over so
that it is spinning in the opposite
direction, the turntable will:
A) remain stationary
B) start to spin in the same
direction as before flipping
C) to spin in the same direction
as after flipping
Conceptual Quiz
You are holding a spinning bicycle
wheel while standing on a
stationary turntable. If you
suddenly flip the wheel over so
that it is spinning in the opposite
direction, the turntable will:
The total angular momentum of the
system is L upward, and it is
conserved. So if the wheel has
−L downward, you and the table
must have +2L upward.
A) remain stationary
B) start to spin in the same
direction as before flipping
C) start to spin in the same
direction as after flipping
Conceptual Quiz
Two different spinning disks have
the same angular momentum, but
disk 1 has more kinetic energy than
disk 2.
A) disk 1
B) disk 2
C) not enough info
Which one has the bigger moment of
inertia?
See hint on
next slide.
Disk 1
Disk 2
Conceptual Quiz
Two different spinning disks have
the same angular momentum, but
disk 1 has more kinetic energy than
disk 2.
A) disk 1
B) disk 2
C) not enough info
Which one has the bigger moment of
inertia?
1
KE = 2 I 2 = L2 / (2 I)
(used L = I ).
Disk 1
Disk 2
Conceptual Quiz
Two different spinning disks have
the same angular momentum, but
disk 1 has more kinetic energy than
disk 2.
A) disk 1
B) disk 2
C) not enough info
Which one has the bigger moment of
inertia?
1
KE = 2 I 2 = L2 / (2 I)
(used L = I ).
Because L is the same,
bigger I means smaller KE.
Disk 1
Disk 2