Download Lecture 7.3 1. Angular Momentum

Lecture 7.3
1. Angular Momentum
We have already discussed many problems involving collisions. The most
powerful tool to solve such problems is the law of conservation of linear momentum. Let
us see if there is a similar concept in the case of rotational motion. If a particle of mass m



moves with velocity v then it has linear momentum p  mv . The position of this particle

can be defined by means of the radius vector r directed from the origin of coordinate
system. Then we shall define the angular momentum of this particle as
  
 
l  r  p  m r v .
b g
(7.3.1)
The angular momentum of the particle is a vector which is equal to the cross product of
the radius vector of the particle and its momentum. The SI unit for angular momentum is
d
i
kilogram times meter squared per second kg m2 s . It is easy to see that angular
momentum plays the same role with respect to momentum as torque plays with respect to
force. The direction of angular momentum-vector can be found using the right-hand rule.


It is perpendicular to both r and p vectors. The magnitude of the angular momentum is
l  rmv sin  ,
(7.3.2)


where  is the angle between the directions of r and p vectors. We can also rewrite the
last equation in terms of vector components
l  rp  rmv  r p  r mv ,
(7.3.3)


where r is the component of vector r perpendicular to vector p and p is the component


of vector p perpendicular to vector r .
We have also seen that the Newton's second law can be written as


dp
.
Fnet 
dt
(7.3.4)
This equation can be used to describe motion on of the particle-like object. Now let us
see how this equation will change if we apply it to rotational motion. Let us consider the
angular momentum of a particle, defined according to the equation 7.3.1, and find its time
derivative, which is



dl
d  
d  
dr   dv

rp m
r v  m
v r 

dt dt
dt
dt
dt

   
   
m v  v  r  a  m r  a  r  Fnet   net ,
b
b
g b g FGH
g b g
so for a single particle we have
IJ
K

 net

dl
.

dt
(7.3.5)
The vector sum of all the torques acting on a particle is equal to the time rate of change
of the angular momentum of that particle, which is the angular form of the Newton's
second law for a single particle.
Now let us consider the system of many particles. The total angular momentum of
this system with respect to the same origin is the sum of the angular momenta of
individual particles, which is
 n 
L   li .
(7.3.6)
i 1
This angular momentum is changing according to



dL d n  n dli n 
  li  
   net i   net .
dt dt i 1
i 1 dt
i 1
In this equation summation includes torques acting on individual particles. These torques
are due to the forces acting in between the particles as well as due to the external forces
acting on the system. According to the Newton's third law forces acting in between the
particles are equal in magnitude and opposite in direction, so are the torques. This is why

the only torques remaining in this summation are the external torques. So,  net is the net
external torque acting on the system and


dL
.
 net 
dt
(7.3.7)
This is the angular form of Newton's second law for rotation of the system of particles.
The net external torque acting on the system of particles is equal to the time rate of
change of the system's total angular momentum. This law has a similar form as the
Newton's second law for translational motion of the system of particles. However, the
angular momentum and the torque have to be calculated for the same origin. In the case
of the pure rotation it can be any origin. In the case if the system's center of mass is
accelerating relative to an inertial frame, this origin has to be the center of mass. This is
the case for the rolling wheel, which accelerates along the road and participates in two
types of motion translation of the center of mass and rotation around center of mass.
A special case of the system of particles is a uniform rigid body, rotating around
fixed axis. Let this body rotate around the axis with angular speed  , what is then its
angular momentum? As usual we shall divide this body into small elements of masses



mi , each having its own velocity vi and momentum pi  mi vi . So, the angular
momentum for each of these elements is
  
 
li  ri  pi  mi ri  vi .


The direction of this angular momentum is perpendicular to both vectors ri and vi , so it is
going to be in the direction of the rotational axis and it has the magnitude of
li  mi ri vi ,
where ri is the component of the radius-vector perpendicular to the rotational axis which,
at the same time, is the radius of the circular path for rotating element. The total angular
momentum of the rigid body is
L   li   mi ri vi   mi riri    mi ri2 ,
i
i
i
i
where we have used the fact that linear velocity of the rotating element is equal vi  ri .
Finally taking into account definition of the moment of inertia, we have
L  I .
(7.3.8)
In this equation I is the moment of inertia for the same axis of rotation.
2. Conservation of Angular Momentum
In the case if there is no external torque acting on the system of particles or the
rigid body, the Newton's second law 7.3.7 becomes

dL
 0,
dt

L  const ,


Li  L f ,
(7.3.9)
which is known as the law of conservation of the angular momentum. If the net external
torque acting on the system is zero, the angular momentum of the system remains
constant. Equation 7.3.9 is a vector equation, so it can be represented as three equations
for the components. In some cases this law can only work in one particular direction, if
the net torque does not have a component in this direction.
We have already seen that conservation laws are powerful tools to solve problems
which can hardly be resolved by means of Newton's equations. Let us consider several
examples of this kind for the case of rotation. If we have a rotating body, which changes
its shape and distribution of mass during its rotation but no external torques are acting on
it, it still will change its angular speed with accordance to the law of conservation of the
angular momentum, which is
I i i  I f  f .
(7.3.10)
Change of mass distribution changes moment of inertia and, as a result, the angular speed
changes as well. Let us now discuss several examples of this conservation law.
The spinning volunteer. A person is seating on the stool that can rotate freely about
vertical axis. He/she and the stool are rotating with some initial angular speed  i . At the
same time the person holds dumbbells in his/her outstretched hands. His/her angular
momentum is directed along vertical rotational axis of the stool. The external forces
acting on the stool-person system are force of gravity and normal force. They both are
directed vertically, so they can not affect the angular momentum of this system, since
they have zero torques. If the person pulls his/her arms in, he/she will reduce the moment
of inertia. Indeed the same mass is now distributed closer to the rotational axis. Since the
angular momentum of the system should stay the same this will result in the increase of
the angular speed of rotation, thus if I f  Ii ,  f   i .
The springboard diver. Let us consider a diver, who is doing one-and-a-half somersault dive. The diver's center of mass follows a parabolic path. She/he also has
some angular momentum around this center of mass, when she/he leaves the board.
When the diver is in the air there is no external torque acting on her/him with respect to
the center of mass. The only force acting is gravitational force applied at the center of
mass, so it does not produce any torque. This is why the angular momentum about the
center of mass should be conserved. By pulling her/his legs and arms into closed tuck
position, she/he can reduce the moment of inertia and increase the angular speed of
her/his rotation. Pulling out to the open layout position, she/he increases her/his moment
of inertia and decreases her/his angular speed, so she/he can enter water without splash.
The similar idea can be used to construct a device allowing the maintaining of the
spacecraft orientation. If the spacecraft has a rotating flywheel, it can make a turn rotating
this wheel with respect to itself. Since the angular momentum of the system should stay
the same, the spacecraft will rotate too and will be able to make a turn.
Example 7.3.1 (from the book) Two m=2.00 kg balls are attached to the ends of the
thin rod of negligible mass, l=50.0 cm long. The rod is free to rotate in a vertical plane
without friction about horizontal axis through its center. With the rod initially horizontal
a M=50.0g wad of wet putty drops onto one of the balls, hitting it with a speed of v=3.00
m/s and then sticking to it. (a) What is the angular speed of the system just after the putty
wad hits? (b) What is the ratio of the kinetic energy of the entire system after the collision
to that of the putty wad just before? (c) Through what angle will the system rotate until it
momentarily stops?
The picture below shows three snapshots of this rod: before the collision, right after
the collision and when it momentarily stops.
(a) To answer this question let us use the law of the conservation of the angular
momentum for this system before and after the collision with respect to rotational axis of
the rod. Let us note that the net torque due to gravitational forces of both balls at the ends
of the rod is zero, when the rod is in horizontal position. And we shall also neglect the
torque of gravitational force acting on the wad, since its mass is very small.
The angular momentum of the system before the collision is just due to the wad. This
angular momentum is Li  rMv , where r 
l 50.0cm

 25.0cm is the distance from the
2
2
rotational axis to the point where the wad is attached. The final angular momentum of the
system after the impact is
i b
d
g
L f  I f   mr 2  mr 2  Mr 2   2m  M r 2 ,
where  is the angular velocity, so
b
g
rMv  2m  M r 2 ,

b
rMv
Mv
0.0500kg  3.00 m s
rad .


 0148
.
2
2m  M r
2  2.00kg  0.0500kg 0.250m
s
2m  M r
g
b
g b
g
(b) To answer this question let us find kinetic energy of the system before and after
the collision. The original kinetic energy was
Mv 2
.
Ki 
2
Final kinetic energy is
I
b2m  M gr FGH b2mMv
 M gr JK

2
2
Kf 
I f 2
2
2
M 2v 2

.
2 2m  M
b
g
So the ratio of the kinetic energies is going to be
Kf
Ki

M 2v 2
2
M
1
1



 0.0123 .
2
2 2m  M Mv
2m  M 1  2 m M 1  2 2.0kg 0.0500kg
b
g
(c) To answer this question let us use the law of energy conservation for this system after
the collision. Indeed we shall consider horizontal position of the rod to be the ground
level for gravitational potential energy. In this case total energy of the system right after
the collision is the kinetic energy K f . When the rod stops momentarily its kinetic energy
is all gone into the potential energy of the rod's center of mass. So we have
b
g
K f  2m  M gh,
h
Kf
M 2v 2

b2m  M gg 2b2m  M g g ,
2
which is the height for which the center of mass will raise. To find the angle,  , shown in
the picture, we need to know the position of the center of mass with respect to the axis of
rotation. This distance is shown in the second snapshot and can be found as
Rcom 
sin  
b
b
g
r M  m  rm
Mr
, so

M  2m
M  2m
b
g
2m  M
h
M 2v 2
Mv 2



Rcom 2 2m  M 2 g
Mr
2 2m  M gr
b
g
b
0.0500kg 3.00 m s
g
g
b
g
2
2 2  2.00kg  0.0500kg 9.8 m s2 0.250m
 0.023
Taking into account that angle of rotation, we are looking for is 180 degrees larger, we
b
g
have 180o  sin 1 0.023  181o .