Chapter 11
Angular Momentum
Angular Momentum
1. Angular momentum
- Definition
2. Newton’s second law in angular form
3. Angular momentum
- System of particles
- Rigid body
- Conservation
Angular momentum
- Vector quantity:
   

 
L  r  p  r  mv  m(r  v )
Units: kg m2/s
Magnitude:
L  r  p sin   r  m  v sin 
Direction: right hand rule.
L positive  counterclockwise
L negative  clockwise
Direction of L is always perpendicular to plane formed by r
and p.
IV. Newton’s second law in angular form
Linear

Fnet

dp

dt
Angular

 net

dL

dt
Single particle
The vector sum of all torques acting on a particle is equal to the time
rate of change of the angular momentum of that particle.



 
dL
  dv
L  m(r  v ) 
 m r 

dt
dt

Proof:

  
 
dL 
 r  m a  r  Fnet   r  F
dt


   
 
dr  
 v   m r  a  v  v   m ( r  a )
dt

  
net
V. Angular momentum
- System of particles:
n 
  

L  l1  l 2  l3  ...  l n   li
i 1



n
n


d li
dL
dL

   net ,i   net 
dt i 1 dt i 1
dt
Includes internal torques (due to forces between particles within
system) and external torques (due to forces on the particles from
bodies outside system).
Forces inside system  third law force pairs  torqueint sum =0 
The only torques that can change the angular momentum of a system
are the external torques acting on a system.
The net external torque acting on a system of particles is equal to
the time rate of change of the system’s total angular momentum L.
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
•
  Fr
• The torque is the vector (or
cross) product of the
position vector and the
force vector
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
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  [( 4.00iˆ  5.00 ˆj )m]  [( 2.00iˆ  3.00 ˆj ) N ]
 [( 4.00)( 2.00)iˆ  iˆ  (4.00)(3.00)iˆ  ˆj
 (5.00)( 2.00) ˆj  iˆ  (5.00)(3.00) ˆj  ˆj
 (2.0kˆ) N  m
Angular Momentum
• The instantaneous angular
momentum Lof a particle
relative to the origin O is defined
as the cross product of the
particle’s instantaneous position
vector
and its rinstantaneous
linear
momentum
•
p
L  r p
Torque and Angular Momentum
• The torque is related to the angular momentum
– Similar to the way force is related to linear momentum
dL
   dt
• The torque acting on a particle is equal to the time
rate of change of the particle’s angular momentum
• This is the rotational analog of Newton’s Second Law
–
must be measured about the same origin
  and L
Angular Momentum
• The SI units of angular momentum are
(kg.m2)/ s
• Both the magnitude and direction of the
angular momentum depend on the choice of
origin
• The magnitude is L = mvr sinf
– f is the angle between p and r
• The direction of L is perpendicular to the
plane formed by and p
r
Angular Momentum of a Particle, Example
• The vector L = r  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
–
L tot  L 1  L 2 
 Ln 
L
i
i
• Differentiating with respect to time
d L tot
dL i

  i
dt
dt
i
i
Angular Momentum of a System of Particles
• Any torques associated with the internal forces
acting in a system of particles are zero
• Therefore,

ext
d L 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
• 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  and L are evaluated
relative to the center of mass
System of Objects
• The masses are
connected by a light
cord that passes over a
pulley; find the linear
acceleration
– The sphere falls, the
pulley rotates and the
block slides
– Use angular momentum
approach
Find an expressions for the linear acceleration
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
• L and w are directed along
the z axis
Angular Momentum of a Rotating Rigid
Object
• To find the angular momentum of the entire
object, add the angular momenta of all the
individual particles
Lz 
 L    m r  w  Iw
2
i
i
i i
i
• This also gives the rotational form of Newton’s
Second Law
dLz
dw
  ext  dt  I dt  I
Angular Momentum of a Rotating Rigid
Object
• The rotational form of Newton’s Second Law is also
valid for a rigid object rotating about a moving axis
provided the moving axis:
(1) passes through the center of mass
(2) is a symmetry axis
• 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
L tot = constant or L i = L f
• For a system of particles,
L tot =
L
n
= constant
Conservation of Angular Momentum
• If the mass of an isolated system undergoes
redistribution, the moment of inertia changes
– The conservation of angular momentum requires
a compensating change in the angular velocity
– Ii wi = If wf = constant
• This holds for rotation about a fixed axis and for
rotation about an axis through the center of mass of a
moving system
• The net torque must be zero in any case
Conservation Law Summary
• For an isolated system (1) Conservation of Energy:
– Ei = Ef
(2) Conservation of Linear Momentum:
–p
i
 pf
(3) Conservation of Angular Momentum:
–L
i
 Lf
A solid cylinder of radius 15 cm and mass 3.0 kg rolls
without slipping at a constant speed of 1.6 m/s. (a) What
is its angular momentum about its symmetry axis? (b)
What is its rotational kinetic energy? (c) What is its total
kinetic energy?
1
( I cylinder= MR 2)
2
A light rigid rod 1.00 m in length
joins two particles, with masses
4.00 kg and 3.00 kg, at its ends.
The combination rotates in the xy
plane about a pivot through the
center of the rod. Determine the
angular momentum of the system
about the origin when the speed
of each particle is 5.00 m/s.
• A 4.00 kg counterweight is attached to a light cord,
which is wound around a spool. The spool is a uniform
solid cylinder of radius 8.00 cm and mass 2.00 kg. (a)
What is the net
torque on the system? (b) What
is the angular momentum function?
(c) What is the acceleration
of the counterweight?
A ball having mass m is fastened at the end of a flagpole
connected to the side of a tall building at point P,
shown in the figure. The length of the flagpole is ℓ, and
θ is the angle the flagpole makes with the horizontal.
The ball becomes loose and starts to fall with
acceleration –g .
ĵ
(a) Determine the angular momentum (as
a function of time) of the ball about point
P. Neglect air resistance.
(b) For what physical reason does the
angular momentum change?
(c) What, if any, is its rate of change?
A wooden block of mass M resting on a frictionless, horizontal
surface is attached to a rigid rod of length ℓ and of negligible
mass. The rod is pivoted at the other end. A bullet of mass m
traveling parallel to the horizontal surface and perpendicular to
the rod with speed v hits the block and becomes embedded in it.
A conical pendulum consists of a bob of
mass m in motion in a circular path in a
horizontal plane as shown. During the
motion, the supporting wire of length ℓ
maintains the constant angle with the
vertical. Show that the magnitude of
the angular momentum of the bob
about the center of the circle is
 m g  s in f 
L  

c o sf


2
3
4
1/2
The position vector of a particle of mass 2.00 kg is
given as a function of time by r  6 . 0 0 ˆi  5 . 0 0 t ˆj m
Determine the angular momentum of the particle
about the origin, as a function of time.


A particle of mass m is shot with an initial velocity vi making an angle
with the horizontal as shown. The particle moves in the gravitational
field of the Earth. Find the angular momentum of the particle about
the origin when the particle is (a) at the origin, (b) at the highest
point of its trajectory, and (c) just before it hits the ground. (d) What
torque causes its angular momentum to change?
A wad of sticky clay with mass m and velocity vi is fired at a solid
cylinder of mass M and radius R. The cylinder is initially at rest, and is
mounted on a fixed horizontal axle that runs through its center of
mass. The line of motion of the projectile is perpendicular to the axle
and at a distance d < R from the center. (a) Find the angular speed of
the system just after the clay strikes and sticks to the surface of the
cylinder. (b) Is mechanical energy of the clay-cylinder system
conserved in this process? Explain your answer.
Two astronauts, each having a mass of 75.0 kg, are connected by a
10.0-m rope of negligible mass. They are isolated in space, orbiting
their center of mass at speeds of 5.00 m/s. Treating the astronauts as
particles, calculate (a) the magnitude of the angular momentum of
the system and (b) the rotational energy of the system. By pulling on
the rope, one of the astronauts shortens the distance between them
to 5.00 m. (c) What is the new angular momentum of the system? (d)
What are the astronauts’ new speeds? (e) What is the new rotational
energy of the system? (f) How much work does the astronaut do in
shortening the rope?
A uniform solid disk is set into rotation with an angular speed ωi
about an axis through its center. While still rotating at this speed,
the disk is placed into contact with a horizontal surface and
released as in Figure. (a) What is the angular speed of the disk
once pure rolling takes place? (b) Find the fractional loss in kinetic
energy from the time the disk is released until pure rolling occurs.
(Hint: Consider torques about the center of mass.)
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 the angular
momentum constant