Torsional waves in a solid circular rod (${\varphi }_{tt}=(G/\rho ){\varphi }_{xx}$)

For small torsional displacement $\varphi (x,t)$, the transmitted torque is $C=G{J}_p{\varphi }_x$. A slice of length $dx$ receives net torque $G{J}_p{\varphi }_{xx}dx$ and has axial moment of inertia $\rho J_{p}dx$. Its angular equation therefore gives the displayed wave equation, with wave speed $\sqrt{G/\rho }$. The same geometric factor cancels from inertia and stiffness, making this ideal torsional speed independent of rod radius. This long-wave circular-rod model neglects additional elastic modes and dissipation.

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.

For a slender homogeneous circular shaft in small-amplitude linear elasticity, torque is $G{J}_p{\theta }_x$ and axial angular inertia per length is $\rho J_p$. Their common geometric factor cancels in the angular equation of motion, leaving the displayed wave equation. Here $J_p=\int r^{2}dA$ is the polar second moment of area, rather than the mass moment of inertia itself.