= Solution
\b[The torque difference.] Let $\phi(x,t)$ now be the local angular displacement, and take $G$ and the section constant. The local twist per unit length is $\partial_x\phi$, so the transmitted torque is $C(x,t)=GJ_p\partial_x\phi$. Across a short slice,
$$
\Delta C=C(x+\Delta x)-C(x)
=GJ_p\frac{\partial^2\phi}{\partial x^2}\Delta x+O(\Delta x^2).
$$
Thus
$$
\boxed{\Delta C=\frac{\pi a^4G}{2}
\frac{\partial^2\phi}{\partial x^2}\Delta x}
$$
to first order in slice length.
\b[The slice's angular acceleration.] The slice's axial <moment of inertia> is $\Delta I=\rho J_p\Delta x$. Its net torque is the difference of the end torques, so
$$
\rho J_p\Delta x\,\partial_t^2\phi
=GJ_p\Delta x\,\partial_x^2\phi.
$$
Canceling the common geometry gives the <torsional waves in a solid circular rod>,
$$
\boxed{\frac{\partial^2\phi}{\partial t^2}
=\frac G\rho\frac{\partial^2\phi}{\partial x^2},\qquad
c_t=\sqrt{\frac G\rho}.}
$$
For the given steel, $c_t\simeq3.16\times10^3\ \mathrm{m\,s^{-1}}$. The radius cancels because both torsional stiffness and slice inertia contain the same $J_p$. This ideal one-dimensional wave model neglects dissipation and additional elastic modes.
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