10.5 Calculating Moments of Inertia

A uniform thin rod with an axis through the center

Consider a uniform (density and shape) thin rod of mass M and length L as shown in Figure 10.25. We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. In this example, the axis of rotation is perpendicular to the rod and passes through the midpoint for simplicity. Our task is to calculate the moment of inertia about this axis. We orient the axes so that the z-axis is the axis of rotation and the x-axis passes through the length of the rod, as shown in the figure. This is a convenient choice because we can then integrate along the x-axis.

We define dm to be a small element of mass making up the rod. The moment of inertia integral is an integral over the mass distribution. However, we know how to integrate over space, not over mass. We therefore need to find a way to relate mass to spatial variables. We do this using the linear mass density $\lambda$ of the object, which is the mass per unit length. Since the mass density of this object is uniform, we can write

$$\lambda =\frac{m}{L}\text{or}m=\lambda L.$$

If we take the differential of each side of this equation, we find

$$dm=d(\lambda L)=\lambda (dL)$$

since $\lambda$ is constant. We chose to orient the rod along the x-axis for convenience—this is where that choice becomes very helpful. Note that a piece of the rod dl lies completely along the x-axis and has a length dx; in fact, $dL=dx$ in this situation. We can therefore write $dm=\lambda (dx)$, giving us an integration variable that we know how to deal with. The distance of each piece of mass dm from the axis is given by the variable x, as shown in the figure. Putting this all together, we obtain

$$I={\displaystyle \int r^{2}dm={\displaystyle \int x^{2}dm={\displaystyle \int x^2\lambda dx}}}.$$

The last step is to be careful about our limits of integration. The rod extends from $x=\text{−}L\text{/}2$ to $x=L\text{/}2$, since the axis is in the middle of the rod at $x=0$. This gives us

$$\begin{array}{cc}\hfill I& ={\displaystyle \underset{\text{−}L\text{/}2}{\overset{L\text{/}2}{\int }}{x}^2\lambda dx=\lambda }\frac{x^3}{3}{|}_{\text{−}L\text{/}2}^{L\text{/}2}=\lambda \left(\frac{1}{3}\right)\left[{\left(\frac{L}{2}\right)}^3-{\left(\frac{\text{−}L}{2}\right)}^3\right]\hfill \\ & =\lambda \left(\frac{1}{3}\right)\frac{L^3}{8}(2)=\frac{M}{L}\left(\frac{1}{3}\right)\frac{L^3}{8}(2)=\frac{1}{12}M{L}^2.\hfill \end{array}$$

Next, we calculate the moment of inertia for the same uniform thin rod but with a different axis choice so we can compare the results. We would expect the moment of inertia to be smaller about an axis through the center of mass than the endpoint axis, just as it was for the barbell example at the start of this section. This happens because more mass is distributed farther from the axis of rotation.

A uniform thin rod with axis at the end

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

$$I={\displaystyle \int r^{2}dm={\displaystyle \int x^{2}dm={\displaystyle \int x^2\lambda dx}}}.$$

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

$$\begin{array}{cc}\hfill I& ={\displaystyle \underset{0}{\overset{L}{\int }}{x}^2\lambda dx=\lambda }\frac{x^3}{3}{|}_0^L=\lambda \left(\frac{1}{3}\right)[{(L)}^3-{(0)}^3]\hfill \\ & =\lambda \left(\frac{1}{3}\right)L^3=\frac{M}{L}\left(\frac{1}{3}\right)L^3=\frac{1}{3}M{L}^2.\hfill \end{array}$$

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.

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 (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:

$$\sigma =\frac{m}{A}\text{or}\sigma A=m,\text{so}dm=\sigma (dA).$$

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:

$$A=\pi r^2,dA=d(\pi r^2)=\pi d{r}^2=2\pi rdr.$$

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

$$\begin{array}{cc}\hfill I& ={\displaystyle \underset{0}{\overset{R}{\int }}{r}^2\sigma (2\pi r)dr=2\pi \sigma {\displaystyle \underset{0}{\overset{R}{\int }}{r}^3}dr=}2\pi \sigma \frac{r^4}{4}{|}_0^R=2\pi \sigma \left(\frac{R^4}{4}-0\right)\hfill \\ & =2\pi \frac{m}{A}\left(\frac{R^4}{4}\right)=2\pi \frac{m}{\pi R^2}\left(\frac{R^4}{4}\right)=\frac{1}{2}m{R}^2.\hfill \end{array}$$

Note that this agrees with the value given in Figure 10.20.

Calculating the moment of inertia for compound objects

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

$$I_{\text{parallel-axis}}=\frac{1}{2}{m}_{\text{d}}{R}^2+m_{\text{d}}{(L+R)}^2.$$

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

$$I_{\text{total}}=\frac{1}{3}{m}_{\text{r}}{L}^2+\frac{1}{2}{m}_{\text{d}}{R}^2+m_{\text{d}}{(L+R)}^2.$$

Applying moment of inertia calculations to solve problems

Now let’s examine some practical applications of moment of inertia calculations.