10.5 Calculating moments of inertia  (Page 3/9)

Note the rotational inertia of the rod about its endpoint is larger than the rotational inertia about its center (consistent with the barbell example) by a factor of four.

The parallel-axis theorem

The similarity between the process of finding the moment of inertia of a rod about an axis through its middle and about an axis through its end is striking, and suggests that there might be a simpler method for determining the moment of inertia for a rod about any axis parallel to the axis through the center of mass. Such an axis is called a parallel axis    . There is a theorem for this, called the parallel-axis theorem    , which we state here but do not derive in this text.

Let’s apply this to the rod examples solved above:

This result agrees with our more lengthy calculation from above. This is a useful equation that we apply in some of the examples and problems.

Check Your Understanding What is the moment of inertia of a cylinder of radius R and mass m about an axis through a point on the surface, as shown below?

$I_{\text{parallel-axis}}=I_{\text{center of mass}}+m{d}^2=m{R}^2+m{R}^2=2m{R}^2$

Got questions? Get instant answers now!

A uniform thin disk about an axis through the center

Integrating to find the moment of inertia of a two-dimensional object is a little bit trickier, but one shape is commonly done at this level of study—a uniform thin disk about an axis through its center ( [link] ).

Since the disk is thin, we can take the mass as distributed entirely in the xy -plane. We again start with the relationship for the surface mass density    , which is the mass per unit surface area. Since it is uniform, the surface mass density $\sigma$ is constant:

Now we use a simplification for the area. The area can be thought of as made up of a series of thin rings, where each ring is a mass increment dm of radius r equidistanct from the axis, as shown in part (b) of the figure. The infinitesimal area of each ring dA is therefore given by the length of each ring ( $2\pi r$ ) times the infinitesimmal width of each ring dr :

The full area of the disk is then made up from adding all the thin rings with a radius range from 0 to R . This radius range then becomes our limits of integration for dr , that is, we integrate from $r=0$ to $r=R$ . Putting this all together, we have

Note that this agrees with the value given in [link] .

Calculating the moment of inertia for compound objects

Now consider a compound object such as that in [link] , which depicts a thin disk at the end of a thin rod. This cannot be easily integrated to find the moment of inertia because it is not a uniformly shaped object. However, if we go back to the initial definition of moment of inertia as a summation, we can reason that a compound object’s moment of inertia can be found from the sum of each part of the object: