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The local elastic strain energy and rotational kinetic energy per length are $u_s=G{J}_p{\theta }_x^2/2$ and $u_k=\rho J_p{\theta }_t^2/2$. Multiplying the torsional wave equation in a circular shaft by $\rho J_p{\theta }_t$ and integrating the product derivative gives ${\partial }_t(u_s+u_k)+{\partial }_x\mathcal{P}=0$. Thus the energy flux is torque times angular speed, with the sign set by the positive direction. The inequality $|\mathcal{P}|=2\sqrt{G/\rho }\sqrt{u_s{u}_k}\le \sqrt{G/\rho }(u_s+u_k)$ follows from a square. A one-way travelling wave has $u_s=u_k$; a steadily rotating twisted shaft need not. Stored energy divided by transit time is not a separate bound for each energy component in a general superposition.

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.

The torsional wave equation in a circular shaft has speed $\sqrt{G/\rho }$. The given shaft has torque $100\mathrm{N}\mathrm{m}$, twist $9.95\times 10^{-3}\mathrm{rad}$, strain energy $0.497\mathrm{J}$ and kinetic energy $181\mathrm{J}$. The respective transit-time estimates are $786\mathrm{W}$ and $286\mathrm{kW}$; these must not be confused with independent bounds on the actual $30\mathrm{kW}$ stress-work flux.

Full derivation: solution.