第10章固定軸まわりの回転

Now consider the same uniform thin rod of mass M and length L, but this time we move the axis of rotation to the end of the rod. We wish to find the moment of inertia about this new axis (Figure 10.26). The quantity dm is again defined to be a small element of mass making up the rod. Just as before, we obtain

However, this time we have different limits of integration. The rod extends from $x=0$ to $x=L$, since the axis is at the end of the rod at $x=0$. Therefore we find

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 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.

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 (Figure 10.27).

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 equidistant 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 infinitesimal 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 Figure 10.20.

Now consider a compound object such as that in Figure 10.28, 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:

It is important to note that the moments of inertia of the objects in Equation 10.21 are about a common axis. In the case of this object, that would be a rod of length L rotating about its end, and a thin disk of radius R rotating about an axis shifted off of the center by a distance $L+R$, where R is the radius of the disk. Let’s define the mass of the rod to be $m_{\text{r}}$ and the mass of the disk to be $m_{\text{d}}.$

The moment of inertia of the rod is simply $\frac{1}{3}{m}_{\text{r}}{L}^2$, but we have to use the parallel-axis theorem to find the moment of inertia of the disk about the axis shown. The moment of inertia of the disk about its center is $\frac{1}{2}{m}_{\text{d}}{R}^2$ and we apply the parallel-axis theorem $I_{\text{parallel-axis}}=I_{\text{center of mass}}+m{d}^2$ to find

Adding the moment of inertia of the rod plus the moment of inertia of the disk with a shifted axis of rotation, we find the moment of inertia for the compound object to be