Kinetic Energy of Rotation — Question 8

Question 8

A variable rate stores more average energy than its mean suggests

A flywheel of constant inertia $I=0.40\,\mathrm{kg\,m^2}$ has angular velocity $\omega(t)=10+2\cos(\Omega t)\,\mathrm{rad/s}$. Average over a whole modulation cycle; use $\langle\cos\rangle=0$ and $\langle\cos^2\rangle=1/2$.

1. Find the average kinetic energy.

2. Compare with computing energy from the average angular velocity, and give a general expression for the difference.

Question 8 - Solution

A variable rate stores more average energy than its mean suggests

(a) Derivation and calculation

Square before averaging: $$\omega^2=100+40\cos(\Omega t)+4\cos^2(\Omega t).$$ $$\langle\omega^2\rangle=100+40(0)+4(1/2)=102,\qquad \boxed{\langle K\rangle=\tfrac12(0.40)(102)=20.4\ \mathrm J}.$$

(b) Result and interpretation

The energy evaluated at the mean rate is $K(\langle\omega\rangle)=\tfrac12(0.40)(10)^2=20$ J. Generally, $$\boxed{\langle K\rangle-K(\langle\omega\rangle) =\tfrac12I\left(\langle\omega^2\rangle-\langle\omega\rangle^2\right) =\tfrac12I\operatorname{Var}(\omega)\ge0}.$$ $$\boxed{\Delta K=\tfrac12(0.40)(2)=0.40\ \mathrm J}.$$

Physical check

The difference vanishes only when the angular rate has zero variance.