Accounting for Angular
Momentum
Chapter 21
Objectives


Understand the basic fundamentals behind
angular momentum
Be able to define measures of rotary
motion and measures of rotation
Accounting for Angular Momentum


Linear momentum deals with objects
moving in a straight line.
Angular momentum deals with objects that
rotate or orbit.
Angular Motion
Consider a person on a
spinning carnival ride.
Position, velocity, and
acceleration are a
function of angular,
rather than linear
measurements.
r
q1
q2
Time
passes
Angular Position and Speed

Position can be described by q
q=
arc length s
=
radius
r

Angular displacement, Dq = q2 - q1

Average angular speed is

Instantaneous angular speed is
q 2 - q1
Dq
=
=
t2 - t1
Dt
D q dq
 = lim
=
Dt  0 D t
dt
Angular Acceleration

Average angular acceleration is
 =

 2 - 1
t 2 - t1
D
=
Dt
Instantaneous angular acceleration is
D d
 = lim
=
Dt 0 Dt
dt
Analogy to Linear Motion

Linear
x = xo + vot + ½aot2
dx/dt = v = vo + aot
dv/dt = a = ao

Angular
q = qo + ot + ½ot2
dq/dt =  = o + ot
d/dt =  = o
Pair Exercise #1

A carnival ride is rotating at 2.0 rad/s. An
external torque is applied that slows the ride
down at a rate of –0.05 rad/s2.
 How long does it take the ride to come to
rest?
 How many revolutions does it make while
coming to rest?
Measures of Rotation
Frequency, f is the number of cycles (revolutions) per second
Unit: Herzt or Hz = 1 cycle/second
1 revolution
n
q
= n=
2π radians q
2π


n revolutions sweep out q radians. There are 2p radians per
cycle (revolution). Therefore
Frequency is given by
dn d  q  1  dq 
1
f =
= 

=
 =
dt dt  2p  2p  dt  2p
 = 2pf
Angular frequency, ω, is the amount of radians per second.
Measures of Rotation
Period, T, is the time it takes to complete one
cycle (seconds per revolution)
1 2p
T= =
f

Pairs Exercise #2
A carnival ride completes 2 revolutions per
second. What is its frequency, period, and
angular velocity?
More Basic Equations


Speed (magnitude of velocity) is
ds
Speed = v = V =
= r
dt
Using some basic calculus and algebra you can
find (Section 20.1.3):
2
dv v
2
a=
=
= r
dt
r
Centripetal Forces
Applying Newton’s
second law:
2
v
2
F = ma = m = mr
r
This is the inward force
required to keep a mass
in a circular orbit. If the
force stops being applied,
the mass will fly off
tangentially.
Centrifugal Force
From the Inertial Frame The ball
is seen to rotate. Centripetal force
keep the ball rotating
From a co-rotating frame: a
rotating object appears to have an
outward force acting on it. This
fictitious centrifugal force has
the same magnitude as the
centripetal force, but acts in the
opposite direction.
The centripetal (not the
centrifugal) force can be
used to design a
centrifuge, which
separates materials
based upon density
differences.
Angular momentum (particles)


Angular momentum (L) is a vector quantity
The direction is perpendicular to the plane of the
orbit and follows the right hand rule
L = rmv = rmr  = mr 
2
Moment of Inertia
The mass moment of inertia is define as:
I  mr
2
Thus angular momentum can be expressed as:
L = I
Which is analogous to linear momentum:
p = mv
Angular Momentum for Bodies
Finding the angular momentum of a rotating body
requires you to integrate over the volume of the
body. (see Figure 21.7)
dL = rv dm
dL = r r rdq dr dh 
L
 dL =  
H
0
0
2p
R
0
0
 
r 3 dr dq dh
Angular Momentum for Bodies



This gets messy for most shapes (even simple ones), so
tables with moments of inertia have be developed for
standard shapes (see Table 21.2).
From AutoCAD you can use the moment of inertia
from Mass Properties times the density of the material.
Once you have the moment of inertia, the angular
momentum is found from the formula:
L = I
Parallel Axis Theorem
Because we want properties about an axis
other than the center of mass, we must use
the Parallel Axis Theorem, which states that
Iaxis=Icg+MD2
where Iaxis and ICG are parallel and D is the
distance between them. M is the mass of the
part.
Using the Theorem



With the mass and CG known, and the desired
axis known, we can find the correct moment of
inertia.
The axis we want is through the point 55, 25 mm.
Because the CG’s X and Y are 35.54, 25 mm, the
distance between them is 19.46 mm.
So the the correct moment is
275.46 kg•mm2 + 0.401 kg • (19.46 mm)2
= 427.4 kg•mm2
Solution
rev 2p rad min
rad
 = 20


= 2.09
min
rev
60 s
s
2
rad
kg

mm
L = I = 427.4 kg  mm 2  2.09
= 893.3
s
s
Torque

Torque is a twisting force. It is created by
applying a force to an object in an attempt to
make the object rotate. (see Figure 21.8)
T = rF

Like momentum, it is a vector quantity.
Conservation of Angular Momentum


Angular momentum is a conserved
quantity.
Changing angular momentum of a system


Changing mass
(see Figure 21.9)
Applying an external torque
Torque and Newton

When Torque is applied to an object, it changes
the angular momentum of the object in a manner
similar to a force applied to a body changes its
linear momentum.
dL d I 
d
T=
=
=I
= I
dt
dt
dt
dp d mv
dv
F=
=
= m = ma
dt
dt
dt
Pair Exercise #3


The previous AutoCAD part is to be spun
by applying a torque of 0.001N·m to the
hole for 10 seconds.
What is its angular velocity (in rpm’s) at
the end of the 10-s period?
Systems without Net Momentum Input


Many times you do not have unbalanced
torque or mass transfer. In this case the
UAE simplifies to:
Final Amount=Initial Amount
This is how ice skaters spin faster.
Pair Exercise #4

A space satellite has an electric motor in it
with a flywheel attached (see Figure
21.11). The motor causes the flywheel to
rotate at 10 rpm. If the moment of inertia
of the flywheel is 10 kg·m2 and the satellite
is 10000 kg·m2, how long must the motor
run to twist the satellite by 10 degrees?