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AP Physics C: Mechanics
Chapter 11
Angular Momentum
The Vector Product
There
are instances where the product of
two vectors is another vector.

Earlier we saw where the product of two
vectors was a scalar.

This was called the dot product.
The
vector product of two vectors is called
the cross product.
Section 11.1
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 = F´r
The torque is the vector
(or cross) product of the
position vector and the force
vector.
Section 11.1
The Vector Product Defined
two vectors, A and B
The vector (cross) product of A and
defined as a third vector, C = A ´ B
.
Given

C is read as “A cross B”.
The

B is
magnitude of vector C is AB sin q
q is the angle between A and. B
Section 11.1
More About the Vector Product
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
The
best way to determine this
direction is to use the right-hand
rule.
Section 11.1
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 A ´ B = -B ´ A
If A is parallel to B (q = 0o or 180o), then A ´ B = 0
 Therefore A ´ A = 0
If A is perpendicular to B , then A ´ B = AB
The vector product obeys the distributive law.
A x (B + C) = A x B + A x C
Section 11.1
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 the vectors.
Section 11.1
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
Section 11.1
Signs in Cross Products
Signs
are interchangeable in cross
products


( )
A ´ -B = - A ´ B
and ˆi ´ (- ˆj)= -ˆi ´ ˆj
Section 11.1
Using Determinants
The
cross product can be expressed as
ˆi
A ´ B = Ax
ˆj
Ay
Bx
By
kˆ
Ay
Az =
By
Bz
Expanding
Az
ˆi + Ax
Bz
Bx
Az ˆ Ax
j+
Bx
Bz
Ay
By
kˆ
the determinants gives
A ´ B = (Ay Bz - AzBy ) ˆi + (Ax Bz - AzBx )ˆj + (Ax By - Ay Bx )kˆ
Section 11.1
Vector Product Example
Given A = 2ˆi + 3ˆj; B = -ˆi + 2ˆj
Find A ´ B
Result
A ´ B = (2ˆi + 3ˆj) ´ ( -ˆi + 2ˆj)
= 2ˆi ´ ( -ˆi ) + 2ˆi ´ 2ˆj + 3ˆj ´ ( -ˆi ) + 3ˆj ´ 2ˆj
= 0 + 4kˆ + 3kˆ + 0 = 7kˆ
Section 11.1
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
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
Section 11.1
Angular Momentum
Consider
a particle of mass m located at the vector
position r and moving with linear momentum p .
Find the net torque.
r ´ å F = åt = r ´
dp
dt
dr
´ p (sinceit = 0 )
dt
d (r ´ p )
t
=
å
dt
Add the term
This
looks very similar to the equation for the net force
in terms of the linear momentum since the torque plays
the same role in rotational motion that force plays in
translational motion.
Section 11.2
Angular Momentum

Same basic techniques that were used in linear
motion can be applied to rotational motion.





F becomes 
m becomes I
a becomes 
v becomes ω
x becomes θ
Linear momentum defined as p  mv
 What if mass of center of object is not moving,
but it is rotating?
 Angular momentum
L  I

May 3, 2017
Angular Momentum I

Angular momentum of a rotating rigid object


L  I




L


L has the same direction as 
L is positive when object rotates in CCW
L is negative when object rotates in CW

Angular momentum SI unit: kg m2/s

Calculate L of a 10 kg disc when  = 320 rad/s, R = 9 cm = 0.09 m
L = I and I = MR2/2 for disc
L = 1/2MR2 = ½(10)(0.09)2(320) = 12.96 kgm2/s


May 3, 2017
Angular Momentum II

Angular momentum of a particle
L  I  mr 2  mv r  mvr sin   rp sin 

Angular momentum of a particle
  
 
L  r  p  m( r  v )




r is the particle’s instantaneous position vector
p is its instantaneous linear momentum
Only tangential momentum component contribute
r and p tail to tail form a plane, L is perpendicular to
this plane
May 3, 2017
Angular Momentum of a Particle in
Uniform Circular Motion
Example: A particle moves in the xy plane in a circular path of
radius r. Find the magnitude and direction of its angular
momentum relative to an axis through O when its velocity is v.



The angular momentum vector
points out of the diagram
The magnitude is
L = rp sinq = mvr sin (90o) = mvr
A particle in uniform circular motion
has a constant angular momentum
about an axis through the center of
its path
O
May 3, 2017
Angular momentum III

Angular momentum of a system of particles

 

Lnet  L 1  L 2  ...  L n 


Li 
all i




ri  p i
all i
angular momenta add as vectors
be careful of sign of each angular momentum
for this case:

     
Lnet  L1  L2  r1  p1  r2  p2

| Lnet |   r1 p1 - r2 p2
May 3, 2017
Example: calculating angular momentum for particles
Two objects are moving as shown in the figure . What
is their total angular momentum about point O?

     
Lnet  L1  L2  r1  p1  r2  p2
m2
Lnet  r1mv1 sin q1  r2 mv2 sin q 2
 r1mv1  r2 mv2
 2.8  3.1 3.6  1.5  6.5  2.2
 31.25  21.45  9.8 kgm2 / s
m1
May 3, 2017
Linear Momentum and Force



Linear motion: apply force to a mass
The force causes the linear momentum to change
The net force acting on a body is the time rate of
change of its linear momentum





dv dp
Fnet  F  ma  m

dt dt


p  mv
 

IL
 Fnet t  p
 
t
May 3, 2017
Angular Momentum and Torque



Rotational motion: apply torque to a rigid body
The torque causes the angular momentum to change
The net torque acting on a body is the time rate of
change of its angular momentum



 dp

 dL
Fnet  F 
 net   
dt
dt

  and L


to be measured about the same origin
The origin should not be accelerating, should be an
inertial frame
May 3, 2017
Demonstration





 dp

 dL
Fnet  F 
 net   
dt
dt

Start from dL  d (r  p )  m d (r  v )
dt dt
dt
Expand using derivative chain rule



   
dL
d  
 dr   dv 
 m (r  v )  m  v  r    mv  v  r  a 
dt
dt
dt 
 dt

   
  
  

dL
 mv  v  r  a   mr  a  r  (ma )  r  Fnet   net
dt
May 3, 2017
What about SYSTEMS of Rigid
Bodies?

 dL i
Rotational 2 law for a single body : i 
dt
 • individual
Total angular momentum 
angular momenta Li
Lsys   L i • all about same
of a system of bodies:
origin


dLsys
• i = net torque on particle “i”

dLi


  i
• internal torque pairs are
dt
dt
included in sum
i
nd
BUT… internal torques in the sum cancel in Newton 3rd law
pairs. Only External Torques contribute to Lsys

dLsys
dt


  i ,ext   net
net external torque on the system
i
Nonisolated System: If a system interacts with its environment in the
sense that there is an external torque on the system, the net external
torque acting on a system is equal to the time rate of change of its
angular momentum.
May 3, 2017
Example: A Non-isolated System

A sphere mass m1 and a
block of mass m2 are
connected by a light cord
that passes over a pulley.
The radius of the pulley is R,
and the mass of the thin rim
is M. The spokes of the
pulley have negligible mass.
The block slides on a
frictionless, horizontal
surface. Find an expression
for the linear acceleration of
the two objects.
a

a
 ext  m1 gR
May 3, 2017

a
Masses are connected by a light cord
Find the linear acceleration a.
• Use angular momentum approach
• No friction between m2 and table
• Treat block, pulley and sphere as a nonisolated system rotating about pulley axis. As
sphere falls, pulley rotates, block slides
• Constraints: Equal v' s and a' s for block and sphere
v  ωR
for pulley
I

a
α  d / dt
a  αR  dv/dt
• Ignore internal forces, consider external forces only
• Net external torque on system:
  m gR
net
• Angular momentum of system:
(not constant)
1
about center of wheel
Lsys  m1vR  m2vR  Iω  m1vR  m2vR  MR 2ω
dLsys
 m1aR  m2 aR  MR 2α  (m1 R  m2 R  MR)a  τ net  m1 gR
dt
m1 g
a 
M  m1  m2
May 3, 2017
Isolated System

Isolated system: net external torque acting on
a system is ZERO


no external forces
net external force acting on a system is ZERO


dLtot
 ext 
0
dt

Ltot  constant
or
 
Li  L f
May 3, 2017
Angular Momentum Conservation

Ltot  constant
or
 
Li  L f
where i denotes initial state, f is final state
 L is conserved separately for x, y, z direction
 For an isolated system consisting of particles,


  
Ltot   Ln  L1  L2  L3    constant


For an isolated system is deformable
I ii  I f  f  constant
May 3, 2017
First Example



A puck of mass m = 0.5 kg is
attached to a taut cord passing
through a small hole in a
frictionless, horizontal surface. The
puck is initially orbiting with speed
vi = 2 m/s in a circle of radius ri =
0.2 m. The cord is then slowly
pulled from below, decreasing the
radius of the circle to r = 0.1 m.
What is the puck’s speed at the
smaller radius?
Find the tension in the cord at the
smaller radius.
May 3, 2017
Angular Momentum Conservation
m = 0.5 kg, vi = 2 m/s, ri = 0.2 m,
rf = 0.1 m, vf = ?
 Isolated system?
 Tension force on m exert zero
torque about hole, why?

 
Li  L f
   

L  r  p  r  (mv )
Li  mri vi sin 90  mri vi L f  mrf v f sin 90  mrf v f
ri
0.2
v f  vi 
2  4m/s
rf
0.1
v 2f
42
T  mac  m  0.5
 80 N
rf
0.1
May 3, 2017
Isolated
System

τ net  0

L
about z - axis

 L  constant
I ω  I
i
initial
i
final
f
ωf
Moment of inertia
changes
May 3, 2017
Controlling spin () by changing I (moment of inertia)
In the air, net = 0
L is constant
L  I ii  I f  f
Change I by curling up or stretching out
- spin rate  must adjust
Moment of inertia changes
May 3, 2017
Example: A merry-go-round problem
A 40-kg child running at 4.0
m/s jumps tangentially onto a
stationary circular merry-goround platform whose radius is
2.0 m and whose moment of
inertia is 20 kg-m2. There is
no friction.
Find the angular velocity of
the platform after the child
has jumped on.
May 3, 2017
The Merry-Go-Round
The moment of inertia of the
system = 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 moment of inertia
decreases.
 The angular speed must
increase since the angular
momentum is constant.

May 3, 2017
Solution: A merry-go-round problem

Ltot   I i ωi 
I
f
f
Li  I ii  I0  mc vT r  mc vT r
L f  I f ω f  ( I  mc r 2 )ω f
I = 20 kg.m2
VT = 4.0 m/s
mc = 40 kg
r = 2.0 m
0 = 0
( I  mc r 2 )ω f  mc vT r
ωf 
mc vT r
40  4  2

 1.78 rad/s
2
2
I  mc r
10  40  2
May 3, 2017