12.4 Calculating Moment of Inertia

The equation for rotational energy is easy to write, but we can’t make use of it without knowing an object’s moment of inertia. Unlike mass, we can’t measure moment of inertia by putting an object on a scale. And while we can guess that the center of mass of a symmetrical object is at the physical center of the object, we can not guess the moment of inertia of even a simple object. To find I, we really must carry through the calculation.

Equation 12.12 defines the moment of inertia as a sum over all the particles in the system. As we did for the center of mass, we can replace the individual particles with cells 1, 2, 3, … of mass $Δ m$ . Then the moment of inertia summation can be converted to an integration:

I = \sum_{i} r_{i}^{2} Δ m \underset{Δ m \to 0}{\to} \int r^{2} d m

(12.15)

where r is the distance from the rotation axis. If we let the rotation axis be the z-axis, then we can write the moment of inertia as

I = \int_{}^{} (x^{2} + y^{2}) d m (rotation about the z -axis)

(12.16)

NOTE You must replace $d m$ by an equivalent expression involving a coordinate differential such as $d x$ or $d y$ before you can carry out the integration.

You can use any coordinate system to calculate the coordinates $x_{cm}$ and $y_{cm}$ of the center of mass. But the moment of inertia is defined for rotation about a particular axis, and r is measured from that axis. Thus the coordinate system used for moment-of-inertia calculations must have its origin at the pivot point.

EXAMPLE 12.5 Moment of inertia of a rod about a pivot at one end

Find the moment of inertia of a thin, uniform rod of length L and mass M that rotates about a pivot at one end.

MODEL An object’s moment of inertia depends on the axis of rotation. In this case, the rotation axis is at the end of the rod.

VISUALIZE FIGURE 12.13 defines an x-axis with the origin at the pivot point.

The figure illustrates a long rod with a horizontal axis displayed below the rod.

FIGURE 12.13 Full Alternative Text

SOLVE Because the rod is thin, we can assume that $y \approx 0$ for all points on the rod. Thus

I = \int_{}^{} x^{2} d m

The small amount of mass $d m$ in the small length $d x$ is $d m =$ $(M / L) d x$ , as we found in Example 12.2. The rod extends from $x = 0$ to $x = L$ , so the moment of inertia about one end is

I_{end} = \frac{M}{L} \int_{0}^{L} x^{2} d x = \frac{M}{L} {[\frac{x^{3}}{3}]}_{0}^{L} = \frac{1}{3} M L^{2}

REVIEW The moment of inertia involves a product of the total mass M with the square of a length, in this case L. All moments of inertia have a similar form, although the fraction in front will vary. This is the result shown earlier in Table 12.2.

EXAMPLE 12.6 Moment of inertia of a circular disk about an axis through the center

Find the moment of inertia of a circular disk of radius R and mass M that rotates on an axis passing through its center.

VISUALIZE FIGURE 12.14 shows the disk and defines distance r from the axis.

The figure illustrates the method used to set up the integral to find the moment of inertia of a disk.

FIGURE 12.14 Full Alternative Text

SOLVE This is a situation of great practical importance. To solve this problem, we need to use a two-dimensional integration scheme that you learned in calculus. Rather than dividing the disk into little boxes, let’s divide it into narrow rings of mass $d m$ . Figure 12.14 shows one such ring, of radius r and width $d r$ . Let $d A$ represent the area of this ring. The mass $d m$ in this ring is the same fraction of the total mass M as $d A$ is of the total area A. That is,

\frac{d m}{M} = \frac{d A}{A}

Thus the mass in the small area $d A$ is

d m = \frac{M}{A} d A

This is the reasoning we used to find the center of mass of the rod in Example 12.2, only now we’re using it in two dimensions.

The total area of the disk is $A = π R^{2}$ , but what is $d A$ ? If we imagine unrolling the little ring, it would form a long, thin rectangle of length $2 π r$ and height $d r$ . Thus the area of this little ring is $d A = 2 π r d r$ . With this information we can write

d m = \frac{M}{π R^{2}} (2 π r d r) = \frac{2 M}{R^{2}} r d r

Now we have an expression for $d m$ in terms of a coordinate differential $d r$ , so we can proceed to carry out the integration for I. Using Equation 12.15, we find

I_{disk} = \int^{​} r^{2} d m = \int^{​} r^{2} (\frac{2 M}{R^{2}} r d r) = \frac{2 M}{R^{2}} {\int^{​}}_{0}^{R} r^{3} d r

where in the last step we have used the fact that the disk extends from $r = 0$ to $r = R$ . Performing the integration gives

I_{disk} = \frac{2 M}{R^{2}} {[\frac{r^{4}}{4}]}_{0}^{R} = \frac{1}{2} M R^{2}

REVIEW Once again, the moment of inertia involves a product of the total mass M with the square of a length, in this case R.

If a complex object can be divided into simpler pieces $1, 2, 3, \dots$ whose moments of inertia $I_{1}, I_{2}, I_{3}, \dots$ are already known, the moment of inertia of the entire object is

I_{object} = I_{1} + I_{2} + I_{3} + \dots

(12.17)

12.4 Calculating Moment of Inertia

EXAMPLE 12.5 Moment of inertia of a rod about a pivot at one end

FIGURE 12.13 Setting up the integral to find the moment of inertia of a rod.

EXAMPLE 12.6 Moment of inertia of a circular disk about an axis through the center

FIGURE 12.14 Setting up the integral to find the moment of inertia of a disk.