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Chapter 10
More on angular momentum and torque
In chapter 9 we described the rotational motion of a rigid body
and, based on that, we defined the vector of angular
momentum as: L = I
In chapter 10 we will give a more general definition of L and
introduce the principle of the conservation of angular
momentum. This principle is:
(a) A fundamental principle of Physics, and
(b) A handy tool to solve a certain class of problems on
rotational motion
(10-1)
The cross product of two vectors
C=AB
We combine A and B to form a new vector C
1. Magnitude of C
2. Direction of C
C = ABsin
C is perpendicular on the plane
defined by A and B
3. Sense of C
It is given by the right hand rule
(RHR)

(10-2)
Right Hand Rule (RHR)
I. Choose the smallest angle  by which you have to rotate
vector A (in the plane defined by A and B ) so that it coincides
with vector B
II. Curl the fingers of the right hand in that direction
III. The thumb of
the right hand
points along C
(10-3)
In terms of components:
z
C  A B
^
k
^
i
O
x
C x  Ay Bz  Az By
C y  Az Bx  Ax Bz
C z  Ax By  Ay Bx
A  iAx  jAy  k Az
^
j
B  iBx  jBy  kBz
y
C  iC x  jC y  kC z
Recipe for those who can handle determinants:
i
j
k
A  B  Ax
Bx
Ay
By
Az
Bz
(10-4)
z
x
r
P

F

rod
O
y
Example: Find the torque 
produced by the force F acting at
point P on the rod.
  rF
Indeed
 = rFsin
The cross product
gives the correct
magnitude and
direction for 
(10-5)

New definition of angular momentum L
Consider an object (mass m, position vector r, linear
momentum p = mv). According the the definition of
chapter 9 its angular momentum L = I
New definition: L  r  p

m
We will show that the
old and the new
definition give the same
result
(10-6)
Old definition: L  I , I  md
v
 mvd
d
New definition: L  r  p
2
,
v
 
d

L  md 2

L  rp sin 
From triangle OAP we have that: r sin   d

L  pd  mvd
m
Indeed the two definitions
give the same result from L
and thus are equivalent
A
(10-7)
dL
(10-8)
In chapter 9 we saw that

dt
We will derive the same result using the new definition of L
L  r p

d L d (r  p ) d r
dp


 pr
dt
dt
dt
dt
dr
dp
v ,
 F (Newton's second law)
dt
dt
dL
= v p  r F
dt
The first term is zero.
The second term is 
dL

dt

System of particles: here we deviate again from the notion of a rigid
body introduced in chapter 9. A system of particles is a collection of N
particles of mass m1, m2, … mN but it is not necessarily rigid. For
example a group of meteorites traveling in space, although not rigid, is a
system of particles
Total angular momentum is the vector sum
of the angular momenta of all the particles
in the system
L = L1 + L2 + … + LN
Total torque is the vector sum of the
torques of all the particles in the system
 = 1 + 2 + … + N
(10-9)
The total angular momentum L and total torque  of a system
obey the equation:
dL

dt
which we have shown to be true for a single particle
This equation states that the rate of change of the total
angular momentum L of a system is equal to the total
torque  on the system
Note: When we calculate  we only have to take into
account the torques generated by external forces. The torques
generated by internal forces can be ignored! This is an
important simplification
(10-10)
When we calculate  we only have to take into account the torques
generated by external forces. This can easily be shown in the simple
case of a system that consists of two particles.
Total force on particle 1:
Total force on particle 2:
F1 = F1ext + F12
F2 = F2ext + F21
F12 + F21 = 0 (Newton’s third law) 
F21 = - F12
1 = r1F1ext + r1F12
2 = r2F2ext + r2F21
2 = r2F2ext - r2F12
 = 1 + 2 
 = r1F1ext + r2F2ext + (r1 -r2)F12
the last term vanishes 
(10-11)
 = r1F1ext + r2F2ext
Conservation of angular momentum
dL
dL
In a system of particles

If   0 
0
dt
dt
 L is a constant vector. If the angular momentum at t1
is L1 and at time t 2 is is L 2 
L1  L2
More concisely:
L  0
This principle is true for any
system of particles (including
rigid bodies)
(10-12)
Example (10-6) page 280 A student sits on a spinning stool
with weights in his hands and rotates with an initial angular
velocity o The student pulls his hands so that they are close
to his torso. Find the final angular velocity 
o = 2/To = 2.1 rad/s
(10-13)
Li  I i o , L f  I f  , Li  L f
Ii
 I io  I f      o
If
2
MR 2
MR
Ii 
 2m( L  R ) 2 , I f 
 2mR 2
2
2
MR 2
 2 m( L  R ) 2
  o 2
 9 rad/s
2
MR
 2mR 2
2
Lf
Li
R
m


m
m
L+R
m
R = 0.25 m
M = 50 kg
m = 2 kg
d = 0.5 m
L=1m
To = 3s
(10-14)
Central forces
A central force is one that acts along the line joining the
source of the force and the object on which the force acts. For
example the force F exerted by the sun on a comet
The torque of the force is:  = rF  = rFcos180º = 0
The rate of change in the angular momentum of the comet is:
dL

If   0 
dt
 L is a constant vector
dL
0
dt
F
r
(10-15)
As a result of the fact that the angular momentum of the comet is
constant, the comet sweeps equal areas in equal times (shaded areas in
the figure). This is known in astronomy as “Kepler’s second law” .
For that reason the comet has to move faster when it is closer to the sun
(between points C and D) and slow down when it is further away
(between points A and B)
(10-16)
Conservation of rotational energy
Consider a rolling object of radius R and
mass m. In chapter 9 we saw that the
velocity of the center of mass and the
angular velocity  are connected via the
equation:
vcm = R
The kinetic energy of the rolling object is given by:
2
I CM  2 mvCM
K

2
2
The first term in the expression
for K is the rotational energy about the center of mass.
The second term is the kinetic energy due to the translation
of the center of mass. We can use energy conservation in
problems that involve rotational motion
(10-17)
Example (10-7), page 285. A spool of mass m, moment of
inertia about its axis I , and radius R unwinds under the force
of gravity. Use conservation of energy to find the speed of the
spool’s center of mass after it has fallen by a length h
(10-18)
before
U=0
.
y
after
Ei  0 , E f  K  U , U  mgh
CM
h
vcm
.
2
mvcm
I 2
K


2
2
2
mvcm
I 2
E f  mgh 

2
2

CM

2
mvcm
vcm
I 2
Ei  E f  0  mgh 

, 
2
2
R
2
2
mvcm
Ivcm
0  mgh 

Solve for v cm 
2
2
2R
2mgh
vcm 
m  I / R2
Substitute  
(10-19)