12.11.2 Conservation of Angular Momentum

A net torque on a rigid body causes its angular momentum to change. Conversely, the angular momentum does not change—it is conserved—for a system with no net torque. This is the basis of the law of conservation of angular momentum.

EXAMPLE 12.18 An expanding rod

Two equal masses are at the ends of a massless 50-cm-long rod. The rod spins at $2.0 rev / s$ about an axis through its midpoint. Suddenly, a compressed gas expands the rod out to a length of 160 cm. What is the rotation frequency after the expansion?

MODEL The forces push outward from the pivot and exert no torques. Thus the system’s angular momentum is conserved.

VISUALIZE FIGURE 12.53 is a before-and-after pictorial representation. The angular momentum vectors $\vec{L}_{i}$ and $\vec{L}_{f}$ are perpendicular to the plane of motion.

The figure consists of two parts labeled Before and After that illustrate a system of two spherical masses connected by a rotating rod.

FIGURE 12.53 Full Alternative Text

SOLVE The particles are moving in circles, so each has angular momentum $L = m r v_{t} = m r^{2} ω = \frac{1}{4} m l^{2} ω$ , where we used $r = \frac{1}{2} l$ . Thus the initial angular momentum of the system is

L_{i} = \frac{1}{4} m l_{i}^{2} ω_{i} + \frac{1}{4} m l_{i}^{2} ω_{i} = \frac{1}{2} m l_{i}^{2} ω_{i}

Similarly, the angular momentum after the expansion is $L_{f} = \frac{1}{2} m l_{f}^{2} ω_{f}$ . Angular momentum is conserved as the rod expands, thus

\frac{1}{2} m l_{f}^{2} ω_{f} = \frac{1}{2} m l_{i}^{2} ω_{i}

Solving for $ω_{f}$ , we find

ω_{f} = {(\frac{l_{i}}{l_{f}})}^{2} ω_{i} = {(\frac{50 cm}{160 cm})}^{2} (2.0 rev/s) = 0.20 rev / s

REVIEW The values of the masses weren’t needed. All that matters is the ratio of the lengths.