Angular Momentum of a Particle
Angular Momentum (l) is the vector cross
product of translational momentum and the
displacement of that momentum relative to a
fixed point :
l=rxp
z
Since angular
momentum is a vector
l
cross product its
y direction follows the
r
v
right hand rule!
m
x
In order to produce a change in linear
momentum, a force must act :
∑F = m dv
In order to produce a change
in the angular momentum, a
net torque must act : ∑  = d l
dt
dt
A particle of mass m is held a horizontal
distance x from point P and dropped. A) Find
the torque acting at any time t with respect to P:
A)
x
At any given time:
P
 = rFsinø
rsinø = x and F = mg
r
therefore torque is constant:
 = mgx
into
ø
mg
out of
By using the right hand
rule the direction would
be perpendicular to the
plane of the page and
into the page (negative)!
B) Find the angular momentum of m at any
given time t with respect to P:
l = rpsinø
l = mgxt
rsinø = x and p = mv = m(gt)
Perpendicular into the plane
C) Show that ∑ = dl / dt
 = dl
dt
mgx = d(mgxt)
dt
mgx = mgx
Systems of Particles
The sum of the net forces acting on an object will
equal the timed rate change of linear momentum:
∑F = m∆v = dp
dt
dt
The sum of the net torque acting on a system of
particles is equal to the time rate change of total
angular momentum:
∑ = dL
dt
For a rigid body that is rotating with axial
symmetry (the body is symmetric to the axis of
rotation), total angular momentum can be found:
L = I
units: kg•m2/s
A sanding disk with a rotational inertia of 1.22 X
10-3 kg-m2 is attached to an electric drill whose
motor delivers a torque of 15.8 N-m. Find A) the
angular momentum and B) the angular speed of
the disk 33.0 ms after the motor is turned on.
I = 1.22 x 10-3 kg•m2
 = 15.8 N•m
t = .033 s
=/I
 = I
= 13,000 rad/s2
 = o + t =
427 rad/s
L = I = .521 kg•m2/s
A wheel of radius 24.7 cm, moving initially at 43.3
m/s, rolls to a stop in 225 m. Calculate its linear
and angular acceleration. If the moment of inertia
of the wheel is .155 kg-m2, what torque does
friction exert on the wheel?
r = .247 m
vo = 43.3 m/s
v=0
s = 225 m
a = v2 - vo2
= - 4.17 m/s2
2s
 = a / r = - 16.9 rad/s2
 = I = - 2.62 N•m
I = .155 kg•m2
A uniform stick with a mass of 4.42 kg and a length
of 1.23 m is initially lying flat and motionless on a
frictionless surface. It is then struck perpendicularly
by a puck imparting an impulse of 12.8 N-s at a
distance of 46.4 cm from the center. Determine the
subsequent motion of the stick. 2.90 m/s & 10.7 rad/s
The angular momentum of a flywheel having a
rotational inertia of .142 kgm2 decreases from
3.07 to .788 kgm2/s in 1.53 s. A)Find the average
torque acting on the flywheel during this time. B)
Assuming a uniform angular acceleration, through
what angle will the flywheel have turned during
this time? C) How much work was done on the
wheel? D) How much average power was supplied
by the flywheel?
11.27.2 A 0.16 meter long, 0.15 kg thin rigid rod
has a small 0.22 kg mass stuck on one of its ends
and a small 0.080 kg mass stuck on the other end.
The rod rotates at 1.7 rad/s through its physical
center without friction. What is the magnitude of
the angular momentum of the system taking the
center of the rod as the origin? Treat the masses on
the ends as point masses.
11.27.3 A solid uniform cylinder rolls down a ramp,
starting from a stationary position at height 1.4 m. At
the bottom of the ramp, the cylinder has no potential
energy. If the mass of the cylinder is 2.9 kg and its
radius is 0.076 m, what is the magnitude of angular
momentum of the cylinder at the bottom of the ramp
with respect to the cylinder's center of mass?
11.31.3 A string is wound around the edge of a solid
1.60 kg disk with a 0.130 m radius. The disk is initially
at rest when the string is pulled, applying a force of
6.50 N in the plane of the disk and tangent to its edge.
If the force is applied for 1.90 seconds, what is the
magnitude of its final angular velocity?
Conservation of Angular Momentum
If not net external torque acts upon a system, the
total angular momentum of the system will
remain constant.
• a spinning skater will spin faster when
pulling in her arms because she reduces
her rotational inertia in doing so.
Li = Lf
Iii = If f
A disk of 125 g and radius 7.2 cm is spinning with and
angular speed of .84 rev/s about a vertical axis through
its center and perpendicular to the plane of the disk.
Two additional identical disks are then dropped onto
the first disk and friction between the disks cause them
to all end up rotating at the same angular speed. What
is that angular speed?
A man stands on a frictionless platform that is rotating
at 1.22 rev/s. His arms are outstretched and in each
hand he holds a weight. In this position, the man, the
platform and the weights have a total moment of inertia
of 6.13 kg-m2. By moving his hands in, the man
reduces the rotational inertia by 4.16 kg-m2. A) What is
the angular speed of the platform now?
B) What is the ratio of total kinetic energy of the
man with outstretched arms to the man with the
arms pulled in?
Li = Lf
1 = 1.22 rev/s
I1 = 6.13 kg•m2
A) 2 = I1 1
I2
I2 = 1.97 kg•m2
= 3.80 rev/s
B) K1
= .5I1 12
K2
.5I2 22
= 3.21
A wheel with a rotational inertia of 1.27 kg-m2 is
rotating with an angular speed of 824 rpm when a
second wheel with a rotational inertia of 4.85 kgm2 is suddenly clamped onto it. What is the
angular speed of both wheels and what is the
percent change of kinetic energy of the system?
I1 = 1.27 kg•m2
1 =824 rev/min
I2 = 4.85 kg•m2
A)
Li = Lf
I1 1 = (I1 + I2) 2
2 = I1 1
I1 + I2
= 171 rpm
B) %∆K = ?
% = ∆K
Ki = .5I1 12 = 4.31 x 105
Ki
Kf = .5(I1 + I2) 22 = 8.95 x 104
∆K = Kf - Ki = -3.42 x 105
= .792 = 79.2%
A 45.0 kg cube is resting 0.750 m from the center
of a solid 125 kg disk that is rotating at 2.80 rad/s.
The radius of the disk is 2.00 m. The cube slides
radially to a location 1.60 m from the center of the
disk, changing the moment of inertia of the system.
What is the new angular velocity of the system?
Treat the cube as a point mass in your calculation.
Gyroscopic Precession
F
N
L
O
Precession- the
movement of
the axis of
rotation
mg
There is no net torque here, so the axis of
rotation remains constantly fixed!
No Precession!
When the end of the axle is released:
N
∆L

r
O
The weight will
produce an
unbalanced torque
directed into the
page.
mg
L
The torque changes the
direction of the axis of
rotation and the shaft
moves horizontally in
that direction.
The Spinning Top
The spinning top is a common example of when
the axis of rotation will move in its own circle of
rotation.
The torque is provided by the weight of the top
itself.
The speed of precession can be found in terms
of the weight, the radius, and the angular
momentum:
p = Mgr
L