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In a previous subsection we have derived the moment of inertia for a spherical shell (I = 2/3 MR2) and a solid sphere (I = 2/5 MR2). From these basic shapes and the technique we have introduced, we can find the moment of inertia for a number of symmetric forms like cylindrical shells, etc. However, what if the shape is not symmetric about the axis of rotation? Unless the shape is relatively simple, the integration needed to find the rotational inertia becomes difficult quite quickly.
To get around the difficulty of doing an integration for each new
shape or axis of rotation, we now introduce another method for finding
the moment of inertia once the moment of inertia is known for an axis
through the center of mass. The parallel-axis theorem
allows a speedy calculation of the moment of inertia about any rotation axis
for any shape provided the moment of inertia for a parallel
axis through the center-of-mass is known. The proof of the parallel-axis
theorem is not too difficult. Consider any shape, for example the one below,
for which the mass, M, and location of the center-of-mass are known. For now,
we make the mass flat and two dimensional, but the argument extends easily
to three dimensions.
Suppose also that the moment of inertia of the body about a rotation axis
that goes through the center of mass is also known. Let's call it ICM.
Now, pick a different point on the mass. The location of this point is
most conveniently specified by setting the center-of-mass (whose location is
known) as the origin for our coordinate system. The new point is then
located a horizontal distance x and vertical distance y away from the
origin as shown below.
We wish to find the moment of inertia of the mass about an axis parallel
to the one through the center-of-mass, but going through the new point.
To do that, consider the location of an infinitesimal bit of mass,
dm, whose location from the center-of-mass
is (x1, y1)
while the location relative to the new axis of rotation is
(x2, y2).
Now, if we relate the distance of dm from
the new point (x,y) in terms of the origin (center-of-mass position), then,
you should be able to spot that x2 =
x1 - x and
y2 = y1 - y.
Therefore, the moment of inertia can be calculated by integrating
over all the infinitesimal masses in the whole solid:
The first term is easy to calculate since the values of x and y are constant. Therefore, the first term is just (x2 + y2)M. If we say the distance of the new axis of rotation from the center-of-mass is D, then D2 = x2 + y2. The second term is just the formula for the rotational inertia about the center-of-mass, ICM. The third and fourth terms are, by the definition of center-of-mass, -2*x*xCM and -2*y*yCM. However, since xCM = 0 and yCM = 0, then Inew = ICM + MD2.
Solution: We can calculate the inertia in the usual way by breaking the rod into pieces of infinitesimal length, dx. The ratio of the mass of dx to the mass of the whole rod is dm/M = dx/L ==> dm = M/L dx. The moment of inertia of this piece is dI = x2dm = Mx2/L dx. Integrating this over the length 0 to L, gives M/L(L3/3) = 1/3 ML2.
Solution: We can do this problem in two ways. The first way is to integrate the same infinitesimal length as before, except now we integrate over the limits -L/2 to L/2. This yields M/L[(L/2)3/3 - (-L/2)3/3] = M/L[L3/24 + L3/24] = 1/12 ML2.
The other method is to use the parallel-axis theorem. The relationship between the inertia around axis through the center-of-mass (which we presume to be at the center of the rod) and the inertia around the axis at the end, for which we have already calculated the inertia is
so, ICM = Iend - M(L/2)2 = ML2/3 - ML2/4 = 1/12 ML2. Certainly this is a considerably faster way of doing the problem.
larryg@upenn5.hep.upenn.edu