Physics 430: Lecture 6
Center of Mass, Angular
Momentum
Dale E. Gary
NJIT Physics Department
3.3 Center of Mass
We are now going to discuss the notion of center of mass, with which you
are certainly already familiar. Think of a system of N particles a = 1, …, N,
with masses ma and positions ra. The center of mass (or CM) is defined to
be the position
m1r1    mN rN
1 N
R
m
r

 aa
M a 1
M
 Like any vector equation, this represents separate equations for each of the
components (X, Y, Z):
1 N
1 N
1 N
X
ma ya , Z 
ma za .
 ma xa , Y  M 

M a 1
M
a 1
a 1

You can think of the center of mass as a weighted average of the positions
of each mass element, i.e. weighted by the mass of that element, or
equivalently it is the vector sum of the ra, each multiplied by the fraction of
mass at that location.
 To get a feeling for CM, let’s look at the center of mass for a two particle
system, which might, for example, represent the Sun and Earth, or two
stars in orbit around each other.

September 17, 2008
Center of Mass and Equation of Motion


1 N
m1r1  m2r2
In this case, R 
, which can be seen in the figure.
m
r


a a
M a 1
m1  m2
CM
m1
It is easy to show that the distance of the CM from m1
and m2 is in the ratio m2/m1. The figure shows the case
r1 R
m2
where m1  4m2. In particular, if m1 >> m2, then the CM
r2
will be very close to m1.
O
Note that the time derivative of the center of mass for N particles is just the
CM velocity
1 N
1 N

R
ma ra 
pa


M a 1
M a 1
.
so the momentum of an N-particle system is related to its CM by P  MR
 Differentiating this expression, we get the very useful relation for the

equation of motion: Fext  MR
 This says that the CM of a collection of particles moves as if the external
forces on all of the individual particles were concentrated at the CM. This is
why we can treat extended objects (e.g. a baseball) as a point mass.

September 17, 2008
Calculating the Center of Mass
Although we developed the foregoing for a set of point particles, the result
obviously applies to extended objects by replacing the summation with an
integral, and treating infinitesimal parts of the object as having mass dm.
The CM expression then becomes
1
R
r dm
M
where the integral extends over the object.
 If you have a uniform extended object of total mass M, you may be given
the size or volume, from which you can determine the density, or
alternatively you may be given the density, from which you determine the
volume. In either case, the integral over the mass is replaced by an
integral over the volume
1
R
 r dV

M


Let’s do an example, the CM of a solid cone. You will have a chance to
practice this with a solid hemisphere, for homework.
September 17, 2008
Example 3.2: The CM of a Solid Cone

Statement of the problem:


z
Find the CM position for the uniform solid cone shown in the figure.
Solution:

R
You should be able to see immediately from the symmetry
of the problem that the CM lies on the z axis. This greatly
simplifies the problem, since we can now concentrate only
h
on the z component. To find the height Z of the CM,
Z

r=Rz/h
1
1

z
dV

z dx dy dz


M
V
where the density   M/V can be brought outside the
integral because the cone is uniform, and we have
replaced dV with the cartesian element of volume dx dy dz.
x
Note that we have to do an integration over x, y and z,
despite the fact that this is only the z component.
If there is any trick to this, it is that we can do the x, y integrals in our head—the
area at a given height z is a circle of radius r = Rz/h, of area pr2 = pR2z2/h2. The
integral then becomes
pR 2 3
pR 2 h 4 3
pR 2 h
Z  2 z dz  2
 h where
V
Vh

Vh 4
4
3
September 17, 2008
y
Example 3.2: Cylindrical Coords

Solution:

Although the text solves the problem as just shown, from the symmetry of the
problem it is a more natural choice to use cylindrical coordinates. The cylindrical
element of volume is dV = r dr df dz. (Convince yourself this is right.) The integral
1
is then
Z
z rdr df dz
V
2p
1
  zdz   rdr   df 

 0

V
Rz / h
 R2 z2 
1
1


  zdz  rdr 2p    zdz p 2 
 0
 V
V
 h 

which then leads to the previous result.
If you do not know the volume of a cone, the way to calculate it parallels the
above, but without the z:
V   dV   rdr df dz
2p
  dz   rdr   df 

 0

Rz / h
h
 R 2 z 2  pR 2 h


  dz  rdr 2p    dz p 2  
 0
 0  h 
3
September 17, 2008
3.4 Angular Momentum for a Single
Particle
As you know, in addition to the law of conservation of momentum, there is
an independent but obviously related law of conservation of angular
momentum.

 The angular momentum  of a single particle is defined as the vector

  rp

where I am forced to use the over-arrow because I cannot make the script 
bold. Here r × p is the vector product of the particle’s position vector r,
relative to the chosen origin O, and its momentum p, as shown in the figure.

It is important to understand the implications of the
statement that the angular momentum is about the
origin O. In the figure at left, we can make the
angular momentum of the particle disappear by simply
shifting our origin. How can we define away a
conserved quantity like this?
 Just consider that angular momentum has little
meaning for a single particle, but when a second body
is included, shifting the origin affects that one, too.

r
r
O
p=mv

   rp  0
  rp
into page
September 17, 2008
Angular Momentum and Torque


The time derivative of angular momentum is
 d
  r  p  r  p  r  p
dt

p

m
r
but because
, the first term is identically zero (the vector product of a
vector with itself is zero). In addition, we can replace the p in the second
term with the net force F, and we
 then recognize the torque.
  rF  
The text using the greek capital gamma for torque, and I will, too. Other
popular symbols are t and N.
In many two-body problems one should choose the origin O so that the net
torque is zero. For example, a planet orbiting the Sun feels a gravitational
force F = GmM/r2 from the Sun. A hallmark of such motion is that the force
is central, i.e. is directed along the line between the two centers. Choosing
the origin at the Sun greatly simplifies the problem because this ensures
that there is no torque (r × F = 0), so the angular momentum r × p is
constant, from which we can immediately deduce that r and p must remain
in a fixed plane through the Sun. Let’s take a closer look at that problem.
September 17, 2008
Kepler’s Second Law

Kepler’s second law states that
As each planet moves around the Sun, a line drawn from the
planet to the Sun sweeps out equal areas in equal times.

The situation is shown in the figure below, where we show two segments of
the orbit that I will approximate as triangles (the approximation becomes
exact in the limit as the width of the triangles goes to zero). Kepler’s 2nd law
is equivalent to saying that so long as the elapsed time dt for the planet to
go from P to Q is the same as for it to go from P’ to Q’, then the areas of
these two triangles must be equal. Equivalently, dA/dt = constant.
 A well-known property of the vector product is that
Q
P
two sides of a triangle are given by vectors a and b,
dA
then the area is A  12 a b (see problem 3.24—this is
related to area = ½ base × height). Thus, the area of
triangle OPQ is dA  12 r  vdt .
dA
1


r

p

r
 This can be rearranged to get:
dA
Q
dt 2m
2m
which,
since
the
angular
P
dr = vdt
momentum   constant implies that Kepler’s law holds.
September 17, 2008
3.5 Angular Momentum for N Particles


We can extend these ideas to N particles, a = 1, 2, …, N, following very much
the same procedure as we did for
 momentum in lecture 1. Each particle has
a  all
ra rapameasured from the same origin
angular momentum
(with
O), so the total angular momentum is
L   ra  pa
a
The time derivative of the total angular momentum is
  r  F
L
a
a
a
where, exactly as before, the net force on particle a is
Fa   Fab  Faext
b a
consisting of the inter-particle forces Fab, and the external force Faext.
Substituting into the L-dot equation, we have
    r  F   r  F ext
L
a
ab
a
a

a b a
a
As before, we use the fact that Fab = -Fba to replace the sum over ba with
one over b>a containing matching pairs, to get
    r - r  F   r  F ext
L
a
b
ab
a
a
a b >a
a
September 17, 2008
Conservation of Angular Momentum

In lecture 1, the paired terms canceled directly. This time, the first sum is
again zero, but for a different reason. Clearly ra – rb  0, but rather, we
assume that the inter-particle forces are central forces, so that the vector ra
– rb is aligned with the force Fab, so the cross-product is zero. So finally we
are left with the time derivative of total angular moment equal to the
externally applied torque
   r  F ext   ext
L
a
a
a

In particular, if there are no applied torques, then L = constant, which leads
to
Principle of Conservation of Angular Momentum
If the net external torque ext on an N-particle system is zero,
the system’s total angular momentum L = S ra× pa is constant.

The validity depends on two assumptions: the inter-particle forces are
central, and that they obey Newton’s third law. You can imagine a system
where the first of these is not true, but for nearly all cases the law holds.
September 17, 2008
Moment of Inertia
We will deal with (and extend) the idea of moment of inertia in Chapter 10.
However, you should be familiar with the basics from your Introductory
Physics course. In particular, you should recall that the angular momentum
about an axis of rotation (say the z axis) is
Lz  Iw
where I is the moment of inertia about the axis of rotation, and w is the
angular velocity.
 In your earlier course, you learned that the moment of inertia is known for a
few standard bodies (i.e. for a uniform sphere of mass M, radius R, the
moment of inertia through the center is I = 2/5 MR2).
 In general, for any multiparticle system, I = Smara2, where ra is the distance
of mass ma from the axis of rotation. The moment of inertia for an
extended object, can be calculated by replacing the sum with an integral.

September 17, 2008
Example 3.3: Collision of a Lump of
Putty with a Turntable

Statement of the problem:


A uniform circular turntable (mass M, radius R, center O, moment of inertia about O,
½ MR2) is at rest in the x, y plane and is mounted on a frictionless axle, which lies
along the vertical z axis. I throw a lump of putty (mass m) with speed v toward the
edge of the turntable so it approaches along a line that passes within a distance b of
O, as shown in the figure. When the putty hits the turntable, it sticks to the edge
and the two rotate together with angular velocity w. Find w.
Solution:

We use conservation of angular momentum. Since the turntable is not moving
initially, the initial angular momentum is that of the putty about O,
Lz ,ini  r  mv  mvr sin   mvb

After the putty sticks, the turntable starts to turn at some
unknown angular velocity, so that the angular momentum is
Lz , fin  I tbl puttyw 


1
2

MR 2  mR2 w
Equating these two equations and solving for w, we have
w
m
vb
m  M / 2 R2
r 
v
b
Location of
stuck putty
September 17, 2008
Special Case: Angular Momentum
About CM
   ext was made under the (unstated) assumption
The foregoing derivation of L
that Newton’s second law F = ma holds, but recall that this is only valid in an
   ext holds for
inertial (non-accelerating) reference frame. We can state that L
any origin O only for inertial reference frames.
 We will see in Chapter 10, but state it now without proof, that the law holds
even in accelerating (non-inertial) frames as long as the origin is about the CM
of the system, even when the CM is undergoing acceleration.
d
L(about CM)   ext (about CM)
dt

Stated another way, if ext(about CM) = 0 then L(about CM) is conserved. You
can show this yourself with the guidance of Prob. 3.37 of the text, if you are
curious.
 This result shows the special nature of the CM.

September 17, 2008
Example 3.4: A Sliding and Spinning
Barbell

Statement of the problem:


A barbell consisting of two equal masses m mounted on the ends of a rigid massless
rod of length 2b is at rest on a frictionless horizontal table, lying on the x axis and
centered on the origin, as shown in the figure. At time t = 0, the left mass is given a
sharp tap, in the shape of a horizontal force F in the y direction, lasting for a short
time Dt. Describe the subsequent motion.
Solution:



The type of force described is called an impulse. When dealing with such a force, we
want to focus on the change in momentum due to the force, rather than the force
  F ext, the momentum after the force acts is
itself. Since P
y
ext
P  F Dt (this is actually the change in momentum, DP, but
the momentum before the impulse is zero).
x
It is important to recognize that this impulse does two things.
It provides momentum to the CM, but at the same time, since
2b
F
it is off-center, it also provides a torque. We need to calculate
both.
The CM relation is easy: vCM = FextDt / 2m. Since Fext is in the +y direction, vCM is too.
September 17, 2008
Example 3.4, cont’d

Solution, cont’d:

The rotational motion due to the torque is found using the methods of the chapter.
The torque is going to cause a change in angular momentum,
   ext
L
with magnitude ext = Fb. Using the same approach as before to deal with the fact
that the force is an impulse, we find the angular momentum after the impulse is
L  Lz  ext Dt  FbDt  Iw

The moment of inertia of the dumbbell is easily calculated (since the rod is massless)
as I = 2mb2, since each mass can be considered a point mass, and each contributes
y
mb2 to the moment of inertia. Solving for the angular velocity:
w


FbDt FDt

2
2mb
2mb
wb
vcm
Note that just after the impulse, the velocity of the left mass
is vleft = vcm + wb = FDt/m, while the velocity of the right hand
2b
mass is vright = vcm – wb = 0.
The subsequent motion is straightforward. The CM moves straight up the y axis
while the barbell continues to rotate at angular velocity w.
September 17, 2008
x
Example 3.4, Further Remarks

Remarks:




It always bothered me that if the same impulsive force were applied on the bar
between the two masses, i.e. at the CM, the barbell would move at the SAME speed
we just calculated for the CM, but without rotation.
From energy considerations, it seems that applying the same force in two locations
imparts DIFFERENT amounts of energy to the barbell. It always seemed to me that
the energy should be the same in the two situations, if the force is the same.
Hopefully this gives you the same sense of unease.
After more thought, however, one can understand the difference. Recall that work
(energy) is force through a distance. We are given the force, but what about the
distance over which it acts? In the case of hitting the barbell at the CM, the force
acts for a time Dt on the barbell moving at speed vCM, so the distance is s = vCM Dt. In
the case of this problem, where we hit the left mass, at a location a distance b from
the CM, the left mass moves at speed 2vCM (check it), so the force acts through a
distance s = 2vCM Dt. It is this difference that accounts for the difference in energy.
If you really want to be confused, however, think about the case where the impulse
force is due to a lump of putty that sticks to the barbell. The putty has the same
energy in both cases, but the putty plus barbell energy is different depending on
where the putty lands. What the… (There is an explanation—can you find it?)
September 17, 2008