physics430_lecture06.Engineering Physics

Physics 430: Lecture 6
Center of Mass, Angular
Momentum
Dale E. Gary
NJIT Physics Department

September 17, 2008
 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  = 1, …, N,
with masses m and positions r. The center of mass (or CM) is defined to
be the position
 Like any vector equation, this represents separate equations for each of
the components (X, Y, Z):
 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 r, 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.
3.3 Center of Mass
M
m
m
m
M
R N
N
N
r
r
r



 


1
1
1
1



.
1
,
1
,
1
1
1
1


 





N
N
N
z
m
M
Z
y
m
M
Y
x
m
M
X










September 17, 2008
 In this case, , which can be seen in the figure.
Center of Mass and Equation of Motion
2
1
2
2
1
1
1
1
m
m
m
m
m
M
R
N



 

r
r
r



CM
m1
m2
O
r1
r2
R
 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 where m1
 4m2. In particular, if m1 >> m2, then the CM will be very
close to m1.
 Note that the time derivative of the center of mass for N particles is just
the CM velocity
so the momentum of an N-particle system is related to its CM by .
 Differentiating this expression, we get the very useful relation for the
equation of motion:
 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.

 



N
N
M
m
M 1
1
1
1




 p
r
R 

R
P 
M

R
F 

M

ext

September 17, 2008
 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
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
 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.
Calculating the Center of Mass

 dm
M
R r
1

 dV
M
R r

1

September 17, 2008
 Solution:
 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
on the z component. To find the height Z of the CM,
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.
Note that we have to do an integration over x, y and z,
despite the fact that this is only the z component.
 Statement of the problem:
 Find the CM position for the uniform solid cone shown in the figure.
Example 3.2: The CM of a Solid Cone

 
 dz
dy
dx
z
V
dV
z
M
Z
1
1

x
z
y
h
r=Rz/h
R
 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 r2
= R2
z2
/h2
. The integral then becomes
where
h
h
Vh
R
dz
z
Vh
R
Z
4
3
4
4
2
2
3
2
2


 


3
2
h
R
V



September 17, 2008
 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 d dz. (Convince yourself this is
right.) The integral is then
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:
Example 3.2: Cylindrical Coords
  
 
  





























2
2
2
/
0
2
0
1
2
1
1
1
h
z
R
zdz
V
rdr
zdz
V
d
rdr
zdz
V
dz
d
rdr
z
V
Z
h
Rz





 
3
2
2
0 2
2
2
/
0
2
0
h
R
h
z
R
dz
rdr
dz
d
rdr
dz
dz
d
rdr
dV
V
h
h
Rz 




































 
  



September 17, 2008
 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
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.
3.4 Angular Momentum for a Single
Particle
p
r







r
p=mv
p
r



into page
O
r 0


 p
r


 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.

September 17, 2008
 The time derivative of angular momentum is
but because , the first term is identically zero (the vector product of a
vector with itself is zero). In addition, we can replace the in the second
term with the net force F, and we then recognize the torque.
The text using the greek capital gamma for torque, and I will, too. Other
popular symbols are  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.
Angular Momentum and Torque
p
r
p
r
p
r 










dt
d
r
p 
m

p




 F
r




September 17, 2008
 A well-known property of the vector product is that two
sides of a triangle are given by vectors a and b, then the
area is (see problem 3.24—this is related to area
= ½ base × height). Thus, the area of triangle OPQ is
.
 This can be rearranged to get:
which, since the angular
momentum implies that Kepler’s law holds.
 Kepler’s second law states that
 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.
Kepler’s Second Law
b
a
2
1
A
P
Q
P
Q
dA
dA
r
dr = vdt
As each planet moves around the Sun, a line drawn from the
planet to the Sun sweeps out equal areas in equal times.
dt
dA v
r
2
1
m
m
dt
dA
2
2
1 


 p
r
constant



September 17, 2008
 We can extend these ideas to N particles,  = 1, 2, …, N, following very
much the same procedure as we did for momentum in lecture 1. Each
particle has angular momentum (with all r measured from the
same origin O), so the total angular momentum is
 The time derivative of the total angular momentum is
where, exactly as before, the net force on particle  is
consisting of the inter-particle forces F, and the external force F
ext
.
Substituting into the L-dot equation, we have
 As before, we use the fact that F = F to replace the sum over 
with one over  containing matching pairs, to get
3.5 Angular Momentum for N
Particles


 p
r 



 



 F
r
L

 



 p
r
L









ext
F
F
F

  



 


 



ext
F
r
F
r
L

  
 




 


 




ext
F
r
F
r
r
L


September 17, 2008
 In lecture 1, the paired terms canceled directly. This time, the first sum
is again zero, but for a different reason. Clearly r – r  0, but rather,
we assume that the inter-particle forces are central forces, so that the
vector r – r is aligned with the force F, 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
 In particular, if there are no applied torques, then L = constant, which
leads to
 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.
Conservation of Angular Momentum
ext
ext






 F
r
L

If the net external torque ext
on an N-particle system is zero,
the system’s total angular momentum L = r× p is constant.
Principle of Conservation of Angular Momentum

September 17, 2008
 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
where I is the moment of inertia about the axis of rotation, and  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 = mr
2
, where r is the
distance of mass m from the axis of rotation. The moment of inertia for
an extended object, can be calculated by replacing the sum with an
integral.
Moment of Inertia

I
Lz 

September 17, 2008
 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 . Find .
 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,
 After the putty sticks, the turntable starts to turn at some
unknown angular velocity, so that the angular momentum is
 Equating these two equations and solving for , we have
Example 3.3: Collision of a Lump of
Putty with a Turntable
mvb
mvr
m
L ini
z 


 
sin
, v
r
 
 2
2
2
1
, mR
MR
I
L putty
tbl
fin
z 

 
b
r 
v
Location of
stuck putty
2
2
/ R
vb
M
m
m




September 17, 2008
 The foregoing derivation of was made under the (unstated)
assumption that Newton’s second law F = ma holds, but recall that this is
only valid in an inertial (non-accelerating) reference frame. We can state
that holds for 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.
 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.
Special Case: Angular Momentum
About CM
ext


L

ext


L

CM)
(about
CM)
(about ext


L
dt
d

September 17, 2008
 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 t. 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 itself. Since , the momentum after the force acts is
(this is actually the change in momentum, P, but
the momentum before the impulse is zero).
 It is important to recognize that this impulse does two things.
It provides momentum to the CM, but at the same time, since
it is off-center, it also provides a torque. We need to calculate
both.
 The CM relation is easy: vCM = Fext
t / 2m. Since Fext
is in the +y direction, vCM is too.
Example 3.4: A Sliding and Spinning
Barbell
ext
F
P 

2b
F
y
x
t

 ext
F
P

September 17, 2008
 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,
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
 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 mb2
to the moment of inertia. Solving for the angular velocity:
 Note that just after the impulse, the velocity of the left mass
is vleft = vcm + b = Ft/m, while the velocity of the right hand
mass is vright = vcm – b = 0.
 The subsequent motion is straightforward. The CM moves straight up the y axis
while the barbell continues to rotate at angular velocity .
Example 3.4, cont’d
ext


L

2b
vcm
y
x

I
t
Fb
t
Lz 





 ext
L
mb
t
F
mb
t
Fb
2
2 2





b

September 17, 2008
 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 t on the barbell moving at speed vCM, so the distance is s = vCM t. 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 t. 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?)
Example 3.4, Further Remarks

physics430_lecture06.Engineering Physics

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