A rotating rigid body—whether it’s rotating freely about its center of mass or constrained to rotate on an axle—has kinetic energy because all atoms in the object are in motion. The kinetic energy due to rotation is called rotational kinetic energy.

FIGURE 12.9 shows a few of the particles making up a solid object that rotates with angular velocity $\omega$. Particle i, which rotates in a circle of radius $r_i$, moves with speed $v_i=r_i\omega$. The object’s rotational kinetic energy is the sum of the kinetic energies of each of the particles:

FIGURE 12.9 Rotational kinetic energy is due to the motion of the particles.

The quantity ${\sum }^{\text{}}{m}_i{r}_i{}^2$ is called the object’s moment of inertia I:

The units of moment of inertia are $\text{kg}{\text{m}}^2$. An object’s moment of inertia depends on the axis of rotation. Once the axis is specified, allowing the values of $r_i$ to be determined, the moment of inertia about that axis can be calculated from Equation 12.12.

Written using the moment of inertia I, the rotational kinetic energy is

EXAMPLE 12.3 A rotating widget

Students participating in an engineering project design the triangular widget seen in FIGURE 12.10. The three masses, held together by lightweight plastic rods, rotate in the plane of the page about an axle passing through the right-angle corner. At what angular velocity does the widget have 100 mJ of rotational energy?

FIGURE 12.10 The rotating widget.

FIGURE 12.10 Full Alternative Text

MODEL Model the widget as a rigid body consisting of three particles connected by massless rods.

SOLVE Rotational energy is $K=\frac{1}{2}I{\omega }^2$. The moment of inertia is measured about the rotation axis, thus

$$\begin{array}{ll}I\hfill & ={\displaystyle \sum _i{m}_i{r}_i{}^2=(0.25\text{kg}){(0.080\text{m})}^2+(0.15\text{kg}){(0.060\text{m})}^2+(0.30\text{kg}){(0\text{m})}^2}\hfill \\ \hfill & =2.14\times {10}^{-3}\text{kg}{\text{m}}^2\hfill \end{array}$$

The largest mass makes no contribution to I because it is on the rotation axis with $r=0$. With I known, the angular velocity is

$$\begin{array}{ll}\omega \hfill & =\sqrt{{\displaystyle \frac{2K}{I}}}=\sqrt{{\displaystyle \frac{2(0.10\text{ J})}{2.14\times {10}^{-3}{\text{ kg m}}^2}}}\hfill \\ \hfill & =9.67\text{ rad/s}\times {\displaystyle \frac{1\text{ rev}}{2\pi \text{ rad}}}=1.54\text{ rev/s}=92\text{ rpm}\hfill \end{array}$$

REVIEW The moment of inertia depends on the distance of each mass from the rotation axis. The moment of inertia would be different for an axle passing through either of the other two masses, and thus the required angular velocity would be different.

Before rushing to calculate moments of inertia, let’s get a better understanding of the meaning. First, notice that moment of inertia is the rotational equivalent of mass. It plays the same role in Equation 12.13 as mass m in the now-familiar $K=\frac{1}{2}m{v}^2$. Recall that the quantity we call mass was actually defined as the inertial mass. Objects with larger mass have a larger inertia, meaning that they’re harder to accelerate. Similarly, an object with a larger moment of inertia is harder to rotate. The fact that moment of inertia retains the word “inertia” reminds us of this.

Consider the two wheels shown in FIGURE 12.11. They have the same total mass M and the same radius R. As you probably know from experience, it’s much easier to spin the wheel whose mass is concentrated at the center than to spin the one whose mass is concentrated around the rim. This is because having the mass near the center (smaller values of $r_i$) lowers the moment of inertia.

FIGURE 12.11 Moment of inertia depends on both the mass and how the mass is distributed.

Moments of inertia for many solid objects are tabulated and found online. You would need to compute I yourself only for an object of unusual shape. TABLE 12.2 is a short list of common moments of inertia. We’ll see in the next section where these come from, but do notice how I depends on the rotation axis.

TABLE 12.2 Moments of inertia of objects with uniform density

If the rotation axis is not through the center of mass, then rotation may cause the center of mass to move up or down. In that case, the object’s gravitational potential energy $U_{\text{G}}=Mg{y}_{\text{cm}}$ will change. If there are no dissipative forces (i.e., if the axle is frictionless) and if no work is done by external forces, then the mechanical energy

is a conserved quantity.

EXAMPLE 12.4 The speed of a rotating rod

A 1.0-m-long, 200 g rod is hinged at one end and connected to a wall. It is held out horizontally, then released. What is the speed of the tip of the rod as it hits the wall?

MODEL The mechanical energy is conserved if we assume the hinge is frictionless. The rod’s gravitational potential energy is transformed into rotational kinetic energy as it “falls.”

VISUALIZE FIGURE 12.12 is a familiar before-and-after pictorial representation of the rod.

FIGURE 12.12 A before-and-after pictorial representation of the rod.

FIGURE 12.12 Full Alternative Text

SOLVE Mechanical energy is conserved, so we can equate the rod’s final mechanical energy to its initial mechanical energy:

$$\frac{1}{2}I{\omega }_1{}^2+Mg{y}_{\text{cm}1}=\frac{1}{2}I{\omega }_0{}^2+Mg{y}_{\text{cm}0}$$

The initial conditions are ${\omega }_0=0$ and $y_{\text{cm}0}=0$. The center of mass moves to $y_{\text{cm}1}=-\frac{1}{2}L$ as the rod hits the wall. From Table 12.2 we find $I=\frac{1}{3}M{L}^2$ for a rod rotating about one end. Thus

$$\frac{1}{2}I{\omega }_1{}^2+Mg{y}_{\text{cm}1}=\frac{1}{6}M{L}^2{\omega }_1{}^2-\frac{1}{2}MgL=0$$

We can solve this for the rod’s angular velocity as it hits the wall:

$${\omega }_1=\sqrt{{\displaystyle \frac{3g}{L}}}$$

The tip of the rod is moving in a circle with radius $r=L$. Its final speed is

$$v_{\text{tip}}={\omega }_{1}L=\sqrt{3gL}=5.4\text{m}\text{/}\text{s}$$

REVIEW 5.4 $\text{m}/\text{s}=12$ mph, which seems plausible for a meter-long rod swinging through $90\text{°}$.