The analogy between linear and rotational motion has been so consistent that you might expect one more. The Chapter 11 result $\vec{P}=M\vec{v}$, which we can now write as $M\vec{v}{}_{\text{cm}}$ because it is translational motion of the object as a whole, might give us reason to anticipate that angular momentum and angular velocity are related by $\vec{L}=I\vec{\omega }$. Unfortunately, the analogy breaks down here. For an arbitrarily shaped object, the angular momentum vector and the angular velocity vector don’t necessarily point in the same direction. The general relationship between $\vec{L}$ and $\vec{\omega }$ is beyond the scope of this text.

The good news is that the analogy does continue to hold for the important situation of rotation about a fixed axle. In this case,

This relationship is shown for a spinning disk in FIGURE 12.54. Equation 12.55 is particularly important for applying the law of conservation of angular momentum.

FIGURE 12.54 The angular momentum vector for rotation on a fixed axle.

If an object’s angular momentum is conserved, its angular speed is inversely proportional to its moment of inertia. The rotation of the rod in Example 12.18 slowed dramatically as it expanded because its moment of inertia increased. Similarly, the ice skater in FIGURE 12.55 uses her moment of inertia to control her spin. She spins faster if she pulls in her arms, decreasing her moment of inertia. Similarly, extending her arms increases her moment of inertia, and her angular velocity drops until she can skate out of the spin. It’s all a matter of conserving angular momentum.

FIGURE 12.55 An ice skater’s rotation speed depends on her moment of inertia.

Table 12.4 summarizes the analogies between linear and angular quantities.

EXAMPLE 12.19 Two interacting disks

A 20-cm-diameter, 2.0 kg solid disk is rotating at 200 rpm. A 20-cm-diameter, 1.0 kg circular loop is dropped straight down onto the rotating disk. Friction causes the loop to accelerate until it is “riding” on the disk. What is the final angular velocity of the combined system?

MODEL The friction between the two objects creates torques that speed up the loop and slow down the disk. But these torques are internal to the combined disk $+$ loop system, so ${\tau }_{\text{net}}=0$ and the total angular momentum of the disk $+$ loop system is conserved.

VISUALIZE FIGURE 12.56 is a before-and-after pictorial representation. Initially only the disk is rotating, at angular velocity $\vec{\omega }{}_{\text{i}}$. The rotation is about a fixed axle, so the angular momentum $\vec{L}=I\vec{\omega }$ is parallel to $\vec{\omega }$. At the end of the problem, $\vec{\omega }{}_{\text{disk}}=\vec{\omega }{}_{\text{loop}}=\vec{\omega }{}_{\text{f}}$.

FIGURE 12.56 The circular loop drops onto the rotating disk.

FIGURE 12.56 Full Alternative Text

SOLVE Both angular momentum vectors point along the rotation axis. Conservation of angular momentum tells us that the magnitude of $\vec{L}$ is unchanged. Thus

$$L_{\text{f}}=I_{\text{disk}}{\omega }_{\text{f}}+I_{\text{loop}}{\omega }_{\text{f}}=L_{\text{i}}=I_{\text{disk}}{\omega }_{\text{i}}$$

Solving for ${\omega }_{\text{f}}$ gives

$${\omega }_{\text{f}}={\displaystyle \frac{I_{\text{disk}}}{I_{\text{disk}}+I_{\text{loop}}}}{\omega }_{\text{i}}$$

The moments of inertia for a disk and a loop can be found in Table 12.2, leading to

$${\omega }_{\text{f}}={\displaystyle \frac{\frac{1}{2}{M}_{\text{disk}}{R}^2}{\frac{1}{2}{M}_{\text{disk}}{R}^2+M_{\text{loop}}{R}^2}}{\omega }_{\text{i}}=100\text{rpm}$$

REVIEW The angular velocity has been reduced to half its initial value, which seems reasonable.