Rotational Mechanics
Part IV – Angular Momentum
Pre AP Physics
Angular Momentum and Its Conservation
In an analogy with linear momentum, we can define
angular momentum L:
L = I
Since  = I and since  
 f  i
t
, We can then
write the total torque as being the rate of change of angular
momentum.
  f  i 
I

t




I   f  i 
t
L

t
If the net torque on an object is zero, the total angular
momentum is constant.
Angular Momentum and Its Conservation
Another way of expressing the angular momentum L of a particle
relative to a point O is the cross product of the particle's position
r relative to O with the linear momentum p of the particle.
   

L  r  p  r  mv
Angular momentum of a particle
The value of the angular momentum depends on the choice
of the origin O, since it involves the position vector relative
to the origin
The units of angular momentum: kgm2/s
NOTE:
If  = v/r and I = mr2, then L = mr2 OR L = mvr
Change in angular momentum
τ  (t) = Iωf – Iωi
τ = torque
t = time
I = moment of inertia
ωf = final angular velocity
ωi = initial angular velocity
The angular momentum of a system remains unchanged
unless an external torque acts on it
Spinning ice skater
Arms extended, inertia is larger, velocity is smaller
Pull arms in tight, inertia decreases, velocity increases
Therefore, systems that can change their rotational inertia through
internal forces will also change their rate of rotation:
Example #1
If the ice cap at the South Pole melted and the water were
uniformly distributed over the earth's oceans, what would
happen to the earth’s angular velocity? Would it increase,
decrease or stay the same?
REASONING AND SOLUTION
Consider the earth to be an isolated system. Note that the
earth rotates about an axis that passes through the North
and South poles and is perpendicular to the plane of the
equator. If the ice cap at the South Pole melted and the
water were uniformly distributed over the earth's oceans,
the mass at the South Pole would be uniformly distributed
and, on average, be farther from the earth's rotational axis.
The moment of inertia of the earth would, therefore,
increase.
Example #1 con’t
Since the earth is an isolated system, any torques involved in
the redistribution of the water would be internal torques;
therefore, the angular momentum of the earth must
remain the same. If the moment of inertia increases, and the
angular momentum is to remain constant, the angular
velocity of the earth must decrease.
Rotational Inertia, I, INCREASES
BUT angular momentum, L, stays the same
 Since L = I
Angular velocity, , DECREASES
Example #2 – Pulling Through a Hole
A particle of mass m moves with speed vi in a circle of
radius ri on a frictionless tabletop. The particle is
attached to a massless string that passes through a hole
in the table as shown. The string is pulled slowly
downward until the particle is a distance rf from the hole
and continues to rotate in a circle of that radius. (A) Find
the final velocity vf. (B) Find the tension T in the string
when the particle moves in a circle of radius r in terms of
the angular momentum L.
(A)
vi ri
v

f
Lf = Li  mvfrf = mviri
rf
T = centripetal Force
mv 2 m  L 
 
(B) T 

r
r  mr 
2
L = mvr  v = L/mr
L2
mr 3
Example #3
A skater has a moment of inertia of 3.0 kg m2 when her
arms are outstretched and 1.0 kg m2 when her arms are brought
in close to her sides. She starts to spin at the rate if 1 rev/s when
her arms are outstretched, and then pulls her arms to her sides.
(A) What is her final angular speed? (B) How much work did she
do?
 Ii 
(A)
Lf = Li  Iff = Iii  f   I   i
 f 
 3 kg  m2 
f  
 1 rev / sec  3 rev/sec
2  
 1 kg  m 
(B)
Work (rotational) = KE (rotational)  Work-Energy Theorem
Work 
1
1
I f 2f  i i2 W = (.5)(1.0)(3 x 2π)2 – (.5)(3)(1 x 2π)2
2
2
118 J
Angular Momentum
Angular momentum depends on linear momentum and
the distance from a particular point. It is a vector quantity with
symbol L. If r and v are  then the magnitude of angular
momentum with respect to point Q is given by L = p∙r = m v r.
In the case shown below L points out of the page. If the mass
were moving in the opposite direction, L would point into the
page.
The SI unit for angular
momentum is the kg  m2 / s. (It has
no special name.) Angular momentum
v
is a conserved quantity. A torque is
r
m
needed to change L, just a force is
Q
needed to change p. Anything
spinning has angular has angular
momentum. The more it has, the
harder it is to stop it from spinning.
If r and v are not  then the angle between these two
vectors must be taken into account. The general definition of
angular momentum is given by a vector cross product:
   

L  r  p  r  mv
This formula works regardless of the angle. The magnitude
of the angular momentum of m relative to point Q is: L = r p sin
= m v r.
v

r
Q
m
Now if a net force acts on an object, it must accelerate,
which means its momentum must change. Similarly, if a net
torque acts on a body, it undergoes angular acceleration, which
means its angular momentum changes. Recall, angular
momentum’s magnitude is given by
L = mvr
if v and r are
mutually
perpendicular
So, if a net torque is applied, angular velocity must
change, which changes angular momentum.
So net torque is the rate of change of angular
momentum, just as net force is the rate of change of linear
momentum.
Here is yet another pair of similar equations, one linear,
one rotational. From the formula v = r , we get
L = mvr = m r (r ) = m r 2  = I 
This is very much like p = m v, and this is one reason I is
defined the way it is.
In terms of magnitudes, linear momentum is inertia
times speed, and angular momentum is rotational
inertia times angular speed.
NOTE:
If  = v/r and I = mr2, then L = mr2 OR L = mvr
Suppose Mr. Stickman is sitting on a stool that swivels holding a
pair of dumbbells. His axis of rotation is vertical. With the weights
far from that axis, his moment of inertia is large. When he pulls his
arms in as he’s spinning, the weights are closer to the
axis, so his moment of inertia gets much
smaller. Since L = I  and L is
conserved, the product of I and  is a
constant. So, when he pulls his arms in,
I goes down,  goes up, and he starts
spinning much faster.
I  = L = I
Since an external net force is required to change
the linear momentum of an object.
An external net torque is required to change the
angular momentum of an object.
It is easier to balance on a moving bicycle than on
one at rest.
• The spinning wheels have angular momentum.
• When our center of gravity is not above a point of
support, a slight torque is produced.
• When the wheels are at rest, we fall over.
• When the bicycle is moving, the wheels have
angular momentum, and a greater torque is required
to change the direction of the angular momentum.
Angular momentum is conserved when no external
torque acts on an object.
Angular momentum is conserved for systems in
rotation.
The law of conservation of angular momentum
states that if no unbalanced external torque acts on a
rotating system, the angular momentum of that system is
constant.
With no external torque, the product of rotational
inertia and rotational velocity at one time will be the same
as at any other time.
Rotational speed is controlled by variations in the
body’s rotational inertia as angular momentum is conserved
during a forward somersault. This is done by moving some
part of the body toward or away from the axis of rotation.
Example #4
What is the angular momentum of a 0.210-kg
ball rotating on the end of a thin string in a circle of
radius 1.10 m at an tangential speed of 5.25 m/s?
m = 0.210 kg; r = 1.10 m; v = 5.25 m/s
L = mvr
L = (0.210 kg)(5.25 m/s)(1.10 m)
L = 1.21 kg∙m2/s
Example #5
What is the angular momentum of a 0.210-kg
ball rotating on the end of a thin string in a circle of
radius 1.10 m at an angular speed of 10.4 rads/sec?
m = 0.210 kg; r = 1.10 m;  = 10.4 rads/sec
L = I = mr2
L = (0.210 kg)(1.10 m)2(10.4 rads/s)
L = 2.64 kg-m2/s
Example #6
What is the angular momentum of a 2.8-kg
uniform cylindrical grinding wheel of radius 18 cm
when rotating at 1500 rpm?
m = 2.8 kg; r = 0.18 m;  = 1500 rpm
L = I
I cylinder  12 mr 2
 2 
1500 rpm 
 = 50 rads/sec
 60 
L
1
2
  2.8 kg  0.18 m  50 rads / sec 
L = 7.1 kg-m2/s
2
Example #7
An artificial satellite is placed into an elliptical orbit about the
earth. Telemetry data indicate that its point of closest
approach (called the perigee) is rP = 8.37 × 106 m from the
center of the earth, and its point of greatest distance (called
the apogee) is rA = 25.1 × 106 m from the center of the earth.
The speed of the satellite at the perigee is vP = 8450 m/s.
Find its speed vA at the apogee.
LApogee = LPerigee  IAA = IPP
 vA 
2  vP 
 mr   r    mrP   r 
 A
 P
2
A
rP vP 
vA
vA 
rA
8.37 10 m  8450 m / s 


2820 m/s
6
25.1106 m
Example #8
A horizontal disk of rotational inertia 4.25kg ⋅ m2 with
respect to its axis of symmetry is spinning counterclockwise
about its axis of symmetry, as viewed from above, at 15.5
revolutions per second on a frictionless massless bearing. A
second disk, of rotational inertia 1.80 kg ⋅ m2 with respect to its
axis of symmetry, spinning clockwise as viewed from above
about the same axis (which is also its axis of symmetry) at 14.2
revolutions per second, is dropped on top of the first disk. If the
two disks stick together
and rotate as one about
their common axis of
symmetry, at what new
angular velocity would the
combined disks move in
rads/sec?
Example #8 Continued
15.5 revs  2 rads 
i 

 = +31 rads/s (counterclockwise)
sec  rev 
14.2 revs  2 rads  = −28.4 rads/s (clockwise)
i 


sec  rev 
L Before = L After
I11 + (−I22) = I1Final + I2Final
Example #8 Continued
I11 + (−I22) = I1Final + I2Final
Final
I11  I 2

I1  I 2
4.25 kg  m   31 rads / s   1.80 kg  m   28.2 rads / s 


2
Final
2
4.25 kg  m2  1.80 kg  m2
Final = 41.9 rads/s
The Vector Nature of Rotational Motion
The vector directions of
the angular velocity vector
 and the angular

momentum
vector L

are along the axis of
rotation. Applying the
right-hand rule gives the
direction.
Right-Hand Rule Grasp the axis of
rotation with your right hand, so that
your fingers circle the axis in the same
sense as the rotation. Your extended
thumb points along the axis in the
direction of the angular velocity vector.
Angular acceleration arises when the
angular velocity changes, and the
acceleration vector also points along the
axis of rotation. The acceleration vector
has the same direction as the change in
the angular velocity.
Angular acceleration and angular
momentum vectors also point along
the axis of rotation.
Example #9 Tipping the Wheel
A student sitting on a stool on a frictionless turntable is
holding a rapidly spinning bicycle wheel. Initially, the axis of
the wheel is horizontal, with the angular momentum vector L
pointing to the right.
What happens when the student tips the wheel so that
the spin axis is vertical, with the wheel spinning
counterclockwise?
Answer
The system is free to rotate about the vertical axis (no
vertical torques) and initially the angular momentum is zero
along that axis. Therefore, vertical angular momentum is
conserved and the final angular momentum must also be zero.
In the final state, the wheel has a large angular momentum
pointing vertically upward, so the stool and student must
rotate clockwise to have an equal and opposite downward
angular momentum.