Chapter 11
Angular Momentum
Schedule 2+ Weeks left!
10Apr
Ch 11: Angular
Ch 11: Angular Mom. Mom.+ Chapt 12.
Ch 12: Statics
17Apr
Ch 12: Statics
Ch 15: Oscillations
24Apr
Ch 15: Oscillations
1May
FINAL EXAM
Ch 15: Oscillations
No Class
No Class
No Class
No Class
So…
 This week
– Angular Momentum
– Statics (?)
– Quiz on Friday
 Next Week
– Statics & Oscillations
 Following Week
– One Class
 Then … the final examination
– VERY CLOSE!
The Exam
 You had seen most of it.
 It is not graded yet.
 You probably did ok!
– Or not.
 Let’s just make a few comments on the
exam
As shown in the figure, a bullet of mass m and speed v
passes completely through a pendulum bob of mass M.
The bullet emerges with a speed of v/2. The pendulum
bob is suspended by a stiff rod of length and negligible
mass. What is the minimum value of v such that the
pendulum bob will barely swing through a complete
vertical circle? Have you seen this one before??
A rod of length 30.0 cm has linear mass density (mass-perlength) that varies with position. It is given by
= 50.0 g/m + 20.0 x g/m2,
where x is the distance from one end, measured in meters.
(a) What is the mass of the rod? (b) How far from the x = 0
end is its center of mass? This one is new.
L
M   dm
0
X CM
1

M
L
xdm

0
This one is in your notes
So is the other one
The Vector Product (Brief Review)

There are instances where the product
of two vectors is another vector



AXB
The vector product of two vectors is
also called the cross product
You can learn more about this in the
textbook and in the notes on the
website.
The Vector Product and Torque



The torque vector lies in
a direction
perpendicular to the
plane formed by the
position vector and the
force vector
t=rxF
The torque is the vector
(or cross) product of
the position vector and
the force vector
The Vector Product Defined




Given two vectors, A and B
The vector (cross) product of A and B is
defined as a third vector, C
C is read as “A cross B”
The magnitude of C is AB sin q

q is the angle between A and B
More About the Vector Product



The quantity AB sin q is
equal to the area of the
parallelogram formed by
A and B
The direction of C is
perpendicular to the
plane formed by A and B
The best way to
determine this direction
is to use the right-hand
rule
Properties of the Vector Product



The vector product is not commutative.
The order in which the vectors are
multiplied is important
To account for order, remember
AxB=-BxA
If A is parallel to B (q = 0o or 180o),
then A x B = 0

Therefore A x A = 0
More Properties of the Vector
Product


If A is perpendicular to B, then |A x B|
= AB
The vector product obeys the
distributive law

A x (B + C) = A x B + A x C
Final Properties of the Vector
Product

The derivative of the cross product with
respect to some variable such as t is
d
dA
dB
A  B   B  A 
dt
dt
dt
where it is important to preserve the
multiplicative order of A and B
Vector Products of Unit
Vectors
ˆi  ˆi  ˆj  ˆj  kˆ  kˆ  0
ˆi  ˆj  ˆj  ˆi  kˆ
ˆj  kˆ  kˆ  ˆj  ˆi
kˆ  ˆi  ˆi  kˆ  ˆj
Vector Products of Unit
Vectors, cont

Signs are interchangeable in cross
products


A x (-B) = - A x B
ˆi   ˆj  ˆi  ˆj
 
Using Determinants

The cross product can be expressed as
ˆi
A  B  Ax
Bx
ˆj
Ay
By
kˆ
Ay
Az 
By
Bz
Az
ˆi  Ax
Bz
Bx
Az ˆ Ax
j
Bx
Bz
Ay
By
kˆ
Expanding the determinants gives
A  B   Ay Bz  Az B y  ˆi   Ax Bz  Az Bx  ˆj   Ax B y  Ay Bx  kˆ

Torque Vector Example


Given the force
F  (2.00 ˆi  3.00 ˆj) N
r  (4.00 ˆi  5.00 ˆj) m
t=?
t  r  F  [(4.00ˆi  5.00ˆj)N]  [(2.00ˆi  3.00ˆj)m]
 [(4.00)(2.00)ˆi  ˆi  (4.00)(3.00)ˆi  ˆj
(5.00)(2.00)ˆj  ˆi  (5.00)(3.00)ˆi  ˆj
 2.0 kˆ N  m
Using the determinant …
i
j k
r  F  4 5 0  (4 x3  2 x5)k  2.0k ( Nm)
2 3 0
New Definition
Angular Momentum: L
p=mv
Angular Momentum

Consider a particle of mass m located at
the vector position r and moving with
linear momentum p
dp
r   F  t  r 
dt
dr
Adding the term
p
dt
d (r  p )
t  dt
This is equal to zero
because v=(dr/dt) is
parallel to p
New Rules



The torque applied to a mass m is equal to
the time rate of change of its angular
momentum.
The force applied to a mass is still equal to
the time rate of change of its linear
momentum.
From the definition, if there is no torque
applied to a mass, its angular momentum will
remain constant!
So .. no force, no torque, no
change in angular momentum!
r sin(q)

Again ….. the definition of L

The instantaneous
angular momentum L of
a particle relative to the
origin O is defined as
the cross product of the
particle’s instantaneous
position vector r and its
instantaneous linear
momentum p

L=rxp
Torque and Angular
Momentum

The torque is related to the angular
momentum

Similar to the way force is related to linear
momentum
dL
t  dt

This is the rotational analog of Newton’s
Second Law


St and L must be measured about the same origin
This is valid for any origin fixed in an inertial frame
New Rules … Add the following
associations to our previous list
F ( force)  t (torque)
p (momentum)  L(angular .momentum)
More About Angular
Momentum



The SI units of angular momentum are
(kg.m2)/ s
Both the magnitude and direction of L
depend on the choice of origin
The magnitude of L = mvr sin f


f is the angle between p and r
The direction of L is perpendicular to
the plane formed by r and p
Angular Momentum of a Particle
Executing Circular Motion about a
point.


The vector L = r x p is
pointed out of the
diagram
The magnitude is
L = mvr sin 90o = mvr


sin 90o is used since v is
perpendicular to r
A particle in uniform
circular motion has a
constant angular
momentum about an
axis through the center
of its path
Angular Momentum of a
System of Particles

The total angular momentum of a
system of particles is defined as the
vector sum of the angular momenta of
the individual particles


Ltot = L1 + L2 + …+ Ln = SLi
Differentiating with respect to time
dL tot
dLi

 t i
dt
dt
i
i
Angular Momentum of a
System of Particles, cont


Any torques associated with the internal
forces acting in a system of particles are zero
Therefore,

t
ext
dL tot

dt
The net external torque acting on a system about
some axis passing through an origin in an inertial
frame equals the time rate of change of the total
angular momentum of the system about that
origin
Angular Momentum of a
System of Particles, final

The resultant torque acting on a system
about an axis through the center of
mass equals the time rate of change of
angular momentum of the system
regardless of the motion of the center
of mass

This applies even if the center of mass is
accelerating, provided t and L are
evaluated relative to the center of mass
Angular Momentum of a
Rotating Rigid Object



Each particle of the
object rotates in the xy
plane about the z axis
with an angular speed
of w
The angular momentum
of an individual particle
is Li = mi ri2 w =Iiw
L and w are directed
along the z axis
Angular Momentum of a
Rotating Rigid Object, cont

To find the angular momentum of the
entire object, add the angular momenta
of all the individual particles
Lz   Li   mi ri 2w  Iw
i

i
This also gives the rotational form of
Newton’s Second Law
t
ext
dLz
dw

I
 I
dt
dt
Angular Momentum of a
Rotating Rigid Object, final

If a symmetrical object rotates about a
fixed axis passing through its center of
mass, the vector form holds: L = Iw

where L is the total angular momentum
measured with respect to the axis of
rotation
Angular Momentum of a
Bowling Ball



The momentum of
inertia of the ball is
2/5MR 2
The angular
momentum of the
ball is Lz = Iw
The direction of the
angular momentum
is in the positive z
direction
Conservation of Angular
Momentum

The total angular momentum of a system is
constant in both magnitude and direction if
the resultant external torque acting on the
system is zero



Net torque = 0 -> means that the system is
isolated
Ltot = constant or Li = Lf
For a system of particles, Ltot = SLn =
constant
Conservation Law Summary
For an isolated system (1) Conservation of Energy:


Ei = Ef
(2) Conservation of Linear Momentum:

pi = pf
(3) Conservation of Angular Momentum:

Li = Lf
Conservation of Angular Momentum:
The Merry-Go-Round

The moment of inertia
of the system is the
moment of inertia of
the platform plus the
moment of inertia of
the person


Assume the person can
be treated as a particle
As the person moves
toward the center of the
rotating platform, the
angular speed will
increase

To keep L constant
Motion of a Top



The only external forces
acting on the top are
the normal force n and
the gravitational force M
g
The direction of the
angular momentum L is
along the axis of
symmetry
The right-hand rule
indicates that t = r  F
= r  M g is in the xy
plane
Motion of a Top, cont

The direction of d L is parallel to that of
t in part. The fact that Lf = d L + Li
indicates that the top precesses about
the z axis.


The precessional motion is the motion of
the symmetry axis about the vertical
The precession is usually slow relative to
the spinning motion of the top
Gyroscope



A gyroscope can be used
to illustrate precessional
motion
The gravitational force Mg
produces a torque about
the pivot, and this torque
is perpendicular to the
axle
The normal force
produces no torque
Gyroscope, cont



The torque results in a
change in angular
momentum d L in a
direction perpendicular to
the axle. The axle sweeps
out an angle df in a time
interval dt.
The direction, not the
magnitude, of L is
changing
The gyroscope experiences
precessional motion
Precessional Frequency

Analyzing the previous vector triangle,
the rate at which the axle rotates about
the vertical axis can be found
df Mgh
wp 

dt
Iw

wp is the precessional frequency