Physics 430: Lecture 22
Rotational Motion of Rigid Bodies
Dale E. Gary
NJIT Physics Department
10.1 Properties of the CM
The text develops a number of properties of the center of mass that we will
need, but the derivations are rather obvious and the results are certainly
not surprising or controversial, so here I will simply state them and refer
you to the text for the derivations.
 In all of these equations, R and its derivatives refer to the coordinate of the
center of mass, and r’ and its derivatives refer to coordinates relative to the
center of mass.
and
Fext  MR. Object acts like a point mass
 Total Momentum P  MR;

Total Angular Momentum L  R  P   r  m r . L of CM + L relative to CM
e.g. Lplanet = Lorb + Lspin
2
2
 Kinetic Energy T  12 MR  12  m r .
T of CM + T of motion relative to CM


Potential Energy U  U ext  U int  U ext .
Ignore Uint
(Uint is due to internal arrangement of masses,
but since this is fixed for a rigid body, Uint = const)
R
r
O
November 19, 2009
r
m
10.2 Rotation about a Fixed Axis
Let’s look at the special case of a rigid body rotating about a fixed axis, such
as a piece of wood on a fixed rod. (The rod need not be through the CM)
 For this case, there is a fixed rotation axis, which we will call the z axis, about
which the body is rotating with angular velocity w. Let’s calculate the vector
angular momentum L.
 Imagine the wood divided into many small pieces of mass m, so that the total
angular momentum is
L   l   r  m v ,
where v  ω  r .
 Since the rotation is about the z axis, the components of w and r are:
ω  (0, 0, w ) and r  ( x , y , z ).

so v  ω  r  (w y , w x , 0)
Thus, the angular momentum for mass element m is
l  m r  v  m r  ω  r   m w( z x ,  z y , x2  y2 ).
 The z component of L is
I z   m 2
Lz   m w(x2  y2 )   m w2  I z w.
Moment of inertia
where   x 2  y 2 is the distance of any point from the axis.
about the z axis

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Rotation about a Fixed Axis-2






So far, we haven’t seen anything new. The angular momentum about the z
axis (which we have called Lz) is just the moment of inertia about that axis, Iz,
times the angular frequency of rotation.
Likewise, the kinetic energy of rotation is T  12  m v2  12  m 2w 2  12 I zw 2 .
What is new, however, is that l  m w ( z x ,  z y , x2  y2 ),
so there are two other components of L:
Lx   m x z w
and
Ly   m y z w.
We shall see in a moment that these two sums are in general not zero, so
even though w points in the z direction, there may be orthogonal components
of L so that L points in some other direction, i.e. L = Iw is generally not true.
As you might guess, this is going to have something to do with the symmetry
of the rotating body relative to the rotation axis.
w
Here is a simple example: A point mass on a massless rod,
m
rotating at an angle q relative to the rod.
q r
At the moment shown, v is into the page, and L = r × p is in the L
direction shown, perpendicular to r. Note that L precesses.
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Rotation about a Fixed Axis-3
Because L is changing direction, clearly L  0. That means there is a non-zero
torque, which should be obvious to you, since the centrifugal force on the
mass is going to try to bend or straighten the rod. In order to keep the mass
in the position shown, the rod must exert a sideways force, presumably due
to the stiffness of the rod, which amounts to a torque G = r × F.
 Usually one wants to design rotating mechanisms so that no torque is needed
(i.e. the wheels on your car), which amounts to designing them so that their
angular momentum is aligned with the rotation axis.
 You know that if your wheels are not balanced, they generate a wobble or
vibration, which is just this precessing angular momentum vector. If that
happens, you can take your car to the shop and have the wheels balanced
(small weights attached to the rims).
 Whether L and w are aligned will lead us to the idea of
w F
m
principle axes. The rotation axis shown for this mass is
NOT a principle axis.
q r
L

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Products of Inertia

Notice that we can factor w out of the sums

These are called the products of inertia, and basically carry the subscript z
to tell us that these are the x and y products when the rotation axis is in the z
direction. To conform with this notation, we need to write the z product of
inertia as
I zz   m 2   m (x2  y2 ).
Lx    m x z  w
and
Ly    m y z  w.
 This leads us to identify the terms in parentheses as moments of inertia, and
because they depend on two coordinates we will write them with two
subscripts:
I xz   m x z
and
I yz   m y z .
Putting these together, we see that
L   I xz w , I yz w , I zz w  .
 To use these ideas, we will need to know how to calculate the products of
inertia for bodies of different shapes, so we will do an example.
 Before doing that, note that we can also have Ixy, Izy, Iyx, etc. => matrices!

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Example 10.1:
Calculating Products of Inertia

Statement of the problem:

Calculate the moment and products of inertia for rotation about the z axis of the
following rigid bodies: (a) A single mass m located at the position (0, yo, zo) as
shown in the figure below left. (b) The same as in part (a), but with a second equal
mass placed symmetrically below the xy plane (below middle). (c) A uniform ring of
mass M and radius o centered on the z axis and parallel to the xy plane (below
right)
z
(0, yo, zo)
m
z
y
x
z
(0, yo, zo)
m
o
y
x
M
y
x
m
(0, yo, zo)
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Example 10.1: Solution

Solution:

(a) The products we have to calculate are:
I xz   m x z ;
Since x = 0, these reduce to
I xz  0;

I yz  myo zo ;
I zz   m (x2  y2 ).
I zz  myo2
This confirms that L is not along the rotation axis (this one is going to wobble).
(b) For the two mass system, we have to calculate sums:
I xz  0;

I yz   m y z ;
I yz  m[ yo zo  yo ( zo )  0;
I zz  2myo2
This illustrates that when symmetries occur (like the one about the y axis in this
case), the product of inertia for that axis can become zero. In this case, there is
reflection symmetry about the y axis. Because both Ixz and Iyz are zero, L is along
the rotation axis. Note that both rods experience a torque, but they balance.
(c) For the ring, we would normally have to do an integration, but due to the
symmetry we can do this without a formal mathematical integration. You can
easily see that Ixz = 0, since the x coordinate for each element of mass can be paired
with another with negative x coordinate. Likewise for Iyz. Since (x2 + y2) = o2 =
constant, Izz is just
I zz    m  o2  M o2 .
November 19, 2009

Problem 10.10: I for Rotating Rod
Statement of Problem:


(a) A thin uniform rod of mass M and length L lies on the x axis with one end at the
origin. Find its moment of inertia for rotation about the z axis. (b) What if the rod’s
center is at the origin?
Solution

(a) Because the rod is uniform, it has a uniform density that we can characterize as
a linear mass density m = M/L.: We want to find the products of inertia Ixz, Iyz and
Izz. Because this is a thin rod (i.e. we are not supposed to worry about the
thickness), we need to take y = z = 0, so Ixz and Iyz are identically zero. The
remaining sum,
2
2
I zz   m (x  y ),
is to be converted to an integral over the length by considering an element of mass
dm = m dx:
L
I zz   m x 2 dx.
0


Because m = constant, we have the solution I zz 

L
0
L3 1
m x dx  m  3 ML2 .
3
2
(b) To calculate it for rotation about the rod center, we simply integrate from L/2
to L/2:
  L / 2 3   L / 2 3 
L /2
2
2
1
I zz  
 L /2
m x dx  m 

3

3
  12 ML .

November 19, 2009