= Solution
Let $J_p=\int r^2dA=\pi a^4/2$ 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 yields
$$
\boxed{C=GJ_p\frac\phi l=\frac{\pi a^4G}{2l}\phi.}
$$
The difference in torque on an element $dx$ is $GJ_p\partial_x^2\theta\,dx$, while its angular inertia is $\rho J_pdx$. Hence the <torsional wave equation in a circular shaft> is
$$
\boxed{\partial_t^2\theta=\frac G\rho\partial_x^2\theta,\qquad c_T=\sqrt{G/\rho}.}
$$
For uniform rotation at $\Omega$, transmitted power is $P=C\Omega$, so the given values give $C=100\,\mathrm{N\,m}$. The <elastic strain energy> and <rotational kinetic energy> are
$$
\phi=\frac{Cl}{GJ_p}=\boxed{9.95\times10^{-3}\,\mathrm{rad}},\qquad
U=\frac{C^2l}{2GJ_p}=\boxed{0.497\,\mathrm J},\qquad
K=\frac12\rho J_pl\Omega^2=\boxed{181\,\mathrm J}.
$$
The torsional transit time is $l/c_T=6.32\times10^{-4}\,\mathrm s$. If the requested maximum rates mean stored-energy-over-transit-time estimates, they are
$$
\boxed{\frac{c_TU}{l}\simeq786\,\mathrm W,\qquad\frac{c_TK}{l}\simeq286\,\mathrm{kW}.}
$$
These 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 rate
$$
\mathcal P=-GJ_p(\partial_x\theta)(\partial_t\theta).
$$
For the steadily rotating twisted shaft its magnitude is $C\Omega=30\,\mathrm{kW}$, even though this exceeds $c_TU/l$. Writing local energy densities $u_s=GJ_p\theta_x^2/2$ and $u_k=\rho J_p\theta_t^2/2$ gives $|\mathcal P|=2c_T\sqrt{u_su_k}\le c_T(u_s+u_k)$. 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.
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