Solution
= Solution
For the uniform solid cylinder, the axial <moment of inertia> is
$$
I_z=\int r^2\,dm=\rho LJ_p
=(8000)(3.0)(4.021\times10^{-6})
\simeq0.0965\ \mathrm{kg\,m^2}.
$$
The rotational <kinetic energy> at $\Omega=300\ \mathrm{rad\,s^{-1}}$ is
$$
\boxed{K=\frac12I_z\Omega^2
\simeq4.34\times10^3\ \mathrm J.}
$$
This includes the shaft's bulk steady rotation; the small twist does not appreciably alter its mass moment at this order.