Phys 221
Chapter 11
Angular Momentum
adzyubenko@csub.edu
© 2012, 2016 A. Dzyubenko
http://www.csub.edu/~adzyubenko
© 2004, 2012 Brooks/Cole
1
Vector Product

Given two vectors A and B, the vector product
(cross product) A×B is a vector C
having a magnitude
C  AB sin 
Θ is the angle
between
A and B
2
Vector Product, cont
The quantity AB sin Θ is equal to the
area of the parallelogram formed by A and B
 The direction of the vector C is perpendicular to
the plane formed by A and B
 This direction is defined by the right-hand rule

AB sin 
3
Some Properties of the Cross Product

The vector product is not commutative
AB  BA
The order is important!
Non-commutative…

If A is parallel to B (Θ = 0º or 180º), then A×B = 0
 A×A= 0
Θ = 90º AB sin 
 If A is perpendicular to B, then |A × B | = AB

The vector product obeys the distributive law:
A  (B  C)  A  B  A  C
4
Some Properties of the
Cross Product, cont

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

It is important to preserve the
multiplicative order of A and B
5
Unit Vectors

ˆi , ˆj , and kˆ form a set of
mutually perpendicular unit
vectors in a right-handed
coordinate system
ĵ
y
î
x
z
k̂
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Cross Products of Unit Vectors

The cross products of the rectangular unit vectors
ˆi , ˆj, and kˆ obey the following rules:
ˆi  ˆi  ˆj  ˆj  kˆ  kˆ  0
ˆi  ˆj   ˆj  ˆi  kˆ
ˆj  kˆ   kˆ  ˆj  ˆi
kˆ  ˆi   ˆi  kˆ  ˆj

y
x
z
Signs are interchangeable in cross product:
ˆi  ( ˆj)   ˆi  ˆj
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Determinant Form of Cross Product

The cross product of any two vectors A (Ax, Ay, Az)
and B (Bx, By, Bz) can be expressed in the following
determinant:
ˆi
A  B  Ax
Bx
or
ˆj
Ay
By
kˆ
Az
Bz
Ay

By
Az
ˆi  Ax
Bz
Bx
A  B  ( Ay Bz  Az B y ) ˆi 
( Az Bx  Ax Bz ) ˆj 
( Ax B y  Ay Bx ) kˆ
Az ˆ Ax
j
Bx
Bz
Ay
kˆ
By
Trick to use:
y
x
z
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Vector Product Example
Given
A  2ˆi  3ˆj; B  ˆi  2ˆj

A B
Find
Result
A  B  (2ˆi  3ˆj)  ( ˆi  2ˆj)
 2ˆi  ( ˆi )  2ˆi  2ˆj  3ˆj  ( ˆi )  3ˆj  2ˆj
 0  4kˆ  3kˆ  0  7kˆ
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Torque Vector Example
Given
the force and location
F  (2.00 ˆi  3.00 ˆj) N
r  (4.00 ˆi  5.00 ˆj) m
Find
the torque produced
  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
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Vector Product and Torque
The torque vector τ is the cross
product of the position vector r
and force F

τ  rF

The magnitude of the torque τ is
  rF sin 
φ is the angle between r and F

Vector τ lies in a direction perpendicular to the plane
formed by the position vector r and the applied force F.
Along the axis of rotation!
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Rotational Dynamics

A particle of mass m
located at position r,
moves with linear momentum p
 The net force on the particle:
 F  dp


dt
Take the cross product on the left
side of the equation
dp
r F  τ  r 
dt
Add the term dr dt  p  0
dp dr
 τ  r  dt  dt  p ?
dr dt  v  p
d (r  p )
 τ  dt
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Angular Momentum
d (r  p )
 τ  dt

looks similar in form to
 F  dp
Define the instantaneous angular momentum L
of a particle relative to the origin O as the cross
product of the particle’s instantaneous position
vector r and its instantaneous linear momentum p
L  rp

The torque acting on a particle
is equal to the time rate of change
of the particle’s angular momentum
dL
 τ  dt
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dt
Angular Momentum, cont
dL
 τ  dt
 F  dp dt

Is the rotational analog of Newton’s second law
for translational motion

Torque causes the angular momentum L to change
just as force causes linear momentum p to change

Is valid only if Σ τ and L are measured
about the same origin
The expression is valid for any origin fixed
in an inertial frame

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More About Angular Momentum

The SI unit of angular momentum is
kg·m2/s
 L is perpendicular to the plane
formed by r and p
 The magnitude and the direction of
L depend on the choice of origin
 The magnitude of L is
L  mvr sin 
φ is the angle between r and p

L is zero when r is parallel to p (φ = 0º or 180º)

L = mvr when r is perpendicular to p (φ = 90º)
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Angular Momentum of a System
of Particles: Motivation

The Newton’s second law for a system of particles

The net external force on a system of particles
is equal to the time rate of change of the total
linear momentum of the system
dp tot
 Fext  dt

Is there a similar statement
that can be made for
rotational motion?
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Angular Momentum of a System, cont.

The total angular momentum of a system of
particles is the vector sum of the angular
momenta of the individual particles
L tot  L1  L 2  ...  L n   L i
i
dL tot
dL i

  i
dt
dt
i
i

Differentiate with
respect to time:

The total angular momentum
varies in time according
to the net external torque:
τ
ext
d L tot

dt
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Angular Momentum of a System
Relative to the System’s
Center of Mass

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 theorem applies even if the center of mass is
accelerating, provided τ and L are evaluated relative to
the center of mass
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Angular Momentum of a Rotating
Rigid Object L  I
z
 Each particle rotates in the xy plane
about the z axis with an angular
speed 
 The magnitude of the angular
momentum of a particle of mass mi
about z axis is
L i  mi vi ri  mi ri 
2
The angular momentum
of the whole object is

2
Lz   Li   mi ri     mi ri 
i
i
 i

2
I is the moment of inertia of the object
19
Angular Momentum of a
Rotating Rigid Object, cont

Differentiate with respect to time,
noting that I is constant for a rigid
object
dL
d
z
dt
I
dt
 I
 is the angular acceleration relative to the axis of rotation


ext
 I
If a symmetrical object rotates about a fixed axis passing
through its center of mass, you can write in vector form
L  Iω
L is the total angular momentum measured
with respect to the axis of rotation
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QQ

A solid sphere and a hollow sphere have the
same mass and radius. They are rotating with
the same angular speed.
The one with the higher angular momentum is
(a) the solid sphere
(b) the hollow sphere
(c) they both have the same angular speed
(d) impossible to determine
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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. That is, if the system is isolated


ext
d L tot

0
dt
L tot  constant
Li  L f
For an isolated system consisting of N particles
N
L tot   L n  constant
n 1
22
Conservation of Angular Momentum, cont

If the mass of an isolated system
undergoes redistribution in some
way, the system’s moment of
inertia I changes

A change in I for an isolated
system requires a change in ω
Li  L f
I ii  I f  f
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Conservation Laws for an
Isolated System
 Ei  E f


p i  p f


L

L
i
f


Energy,
linear momentum,
and angular momentum
of an isolated system
all remain constant

Manifestations of some certain symmetries of space
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Angular Momentum as a
Fundamental Quantity

The concept of angular momentum is also valid
on a submicroscopic scale
 Angular momentum is an intrinsic property of
atoms, molecules, and their constituents
 Fundamental unit of momentum is   h 2

h is called Planck’s constant
  1.054  10  34 kg  m 2 s
s, p, d, f, …
electronic orbitals: L=0, 1, 2, 3, … in terms of

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Reading assignment: Gyroscopes
http://aesp.nasa.okstate.edu/fieldguide/pages/skylab/skylabhu1.html
http://www.pbs.org/wgbh/nova/lostsub/torpworks.html
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