Conservation of Angular Momentum — Question 10

Question 10

Common rotation minimizes energy at fixed total angular momentum

Two coaxial rotors have fixed positive inertias $I_1,I_2$ and speeds $\omega_1,\omega_2$. Their total angular momentum is $L=I_1\omega_1+I_2\omega_2$.

1. Show that kinetic energy equals a common-rotation term plus a nonnegative relative-motion term.

2. Use the result to predict energy lost if an internal clutch brings them to a common speed without external torque.

Question 10 - Solution

Common rotation minimizes energy at fixed total angular momentum

(a) Derivation and calculation

Let $I_\Sigma=I_1+I_2$, $\bar\omega=L/I_\Sigma$ and $\Delta\omega=\omega_1-\omega_2$. Then $$\omega_1=\bar\omega+\frac{I_2}{I_\Sigma}\Delta\omega,\qquad \omega_2=\bar\omega-\frac{I_1}{I_\Sigma}\Delta\omega.$$ Substitute into $K=(I_1\omega_1^2+I_2\omega_2^2)/2$. The terms linear in $\Delta\omega$ cancel, leaving $$\boxed{K=\frac{L^2}{2I_\Sigma}+\frac{I_1I_2}{2I_\Sigma}(\omega_1-\omega_2)^2}.$$

(b) Result and interpretation

With negligible external torque, $L$ remains fixed. Clutching gives $$\omega_f=L/I_\Sigma,\qquad K_f=L^2/(2I_\Sigma).$$ Therefore $$\boxed{K_{\mathrm{lost}}=\frac{I_1I_2}{2(I_1+I_2)}(\omega_1-\omega_2)^2\ge0}.$$ For positive inertias, equality holds only when the initial speeds already agree. Common rotation is the minimum-energy state at fixed total angular momentum.

Physical check

The loss depends on the square of relative speed and is unchanged by adding the same angular velocity to both rotors.