 Solution

Let $J_p=\int r^{2}dA=\pi a^4/2$ be the polar second moment of area. A twist gradient ${\partial }_x\theta$ gives shear strain $r{\partial }_x\theta$ and shear stress $Gr{\partial }_x\theta$. Integrating its moment over a cross section yieldsThe difference in torque on an element $dx$ is $G{J}_p{\partial }_x^2\theta dx$, while its angular inertia is $\rho J_{p}dx$. Hence the torsional wave equation in a circular shaft isFor uniform rotation at $\Omega$, transmitted power is $P=C\Omega$, so the given values give $C=100\mathrm{N}\mathrm{m}$. The elastic strain energy and rotational kinetic energy areThe torsional transit time is $l/c_T=6.32\times 10^{-4}\mathrm{s}$. If the requested maximum rates mean stored-energy-over-transit-time estimates, they areThese estimates require a physical qualification. They are not separate upper bounds on transmitted mechanical power. The exact local energy flux of a torsional wave is the stress-work rateFor the steadily rotating twisted shaft its magnitude is $C\Omega =30\mathrm{kW}$, even though this exceeds $c_{T}U/l$. Writing local energy densities $u_s=G{J}_p{\theta }_x^2/2$ and $u_k=\rho J_p{\theta }_t^2/2$ gives $|\mathcal{P}|=2c_T\sqrt{u_s{u}_k}\le c_T(u_s+u_k)$. A single travelling torsional wave has equal strain and kinetic energies; the steady shaft can instead be represented by oppositely travelling components whose powers largely cancel. Its stored strain pattern is not a fuel consumed once per transit. This distinguishes an energy-content estimate from the actual stress-mediated transport.