Kinetic Energy of Rotation — Question 10

Question 10

Revolutions per minute require two conversions before squaring

A rotor of inertia $0.50\,\mathrm{kg\,m^2}$ runs at $120\,\mathrm{rev/min}$.

1. Find its kinetic energy and the energy at $240\,\mathrm{rev/min}$.

2. A calculation inserts $120$ directly as angular speed in SI units. By what factor does it overestimate the initial energy?

Question 10 - Solution

Revolutions per minute require two conversions before squaring

(a) Derivation and calculation

Convert both angle and time units before squaring: $$\omega=120\frac{\mathrm{rev}}{\mathrm{min}} \frac{2\pi\ \mathrm{rad}}{\mathrm{rev}}\frac{\mathrm{min}}{60\ \mathrm s}=4\pi\ \mathrm{rad/s}.$$ $$\boxed{K_{120}=\tfrac12(0.50)(4\pi)^2=4\pi^2=39.48\ \mathrm J},\qquad \boxed{K_{240}=4K_{120}=16\pi^2=157.91\ \mathrm J}.$$

(b) Result and interpretation

The unconverted calculation gives $K_{\mathrm{wrong}}=\tfrac12(0.50)(120)^2=3600$ J. Its multiplicative error is $$\boxed{\frac{K_{\mathrm{wrong}}}{K_{120}}=\left(\frac{120}{4\pi}\right)^2 =\frac{900}{\pi^2}=91.19}.$$ The speed conversion factor is squared in the energy error.

Physical check

Doubling the correctly converted angular speed quadruples kinetic energy.