Kinetic Energy of Rotation — Question 7

Question 7

A rim-speed limit bounds flywheel energy per mass

An ideal thin hoop of mass $M$ and radius $R$ is restricted to rim speed $v_{\max}=100\,\mathrm{m/s}$. Use $I=MR^2$. Ignore material engineering beyond this supplied speed limit.

1. Find the maximum energy per unit mass.

2. If its radius doubles while mass and rim-speed limit stay fixed, does its maximum stored energy increase?

Question 7 - Solution

A rim-speed limit bounds flywheel energy per mass

(a) Derivation and calculation

Eliminate angular speed using the rim constraint: $$I=MR^2,\qquad\omega=\frac vR,\qquad K=\tfrac12MR^2\left(\frac vR\right)^2=\tfrac12Mv^2.$$ $$\boxed{\frac{K_{\max}}M=\frac12(100)^2=5000\ \mathrm{J/kg}}.$$

(b) Result and interpretation

Doubling radius at fixed mass gives $$I^{\prime}=4I,\qquad\omega_{\max}^{\prime}=\frac{\omega_{\max}}2,\qquad \boxed{K_{\max}^{\prime}=\tfrac12(4I)(\omega_{\max}/2)^2=K_{\max}}.$$ Within the supplied speed-limit model, radius does not independently increase maximum energy.

Physical check

The energy per mass has units of speed squared, equivalent to J/kg.