Calculate the kinetic energy of a uniform disk of mass M

Calculate the kinetic energy of a uniform disk of mass $M$ and radius $R$ rotating with angular velocity $\omega$ about the axis through its center, perpendicular to its plane.

Step 1: Mass element in polar coordinates

With surface density $\sigma=M/(\pi R^2)$, a mass element at polar radius $\rho$ is $dm=\sigma\,\rho\,d\rho\,d\theta$.

Step 2: Moment of inertia

\[I=\int_0^{2\pi}\!\!\int_0^R\rho^2\cdot\sigma\rho\,d\rho\,d\theta=\sigma\cdot2\pi\cdot\left[\frac{\rho^4}{4}\right]_0^R=\frac{\pi\sigma R^4}{2}=\frac{\pi R^4}{2}\cdot\frac{M}{\pi R^2}=\frac{MR^2}{2}.\]

\[\boxed{I=\frac12MR^2}\]

Step 3: Kinetic energy

\[E_k=\frac12I\omega^2=\frac12\cdot\frac12MR^2\cdot\omega^2=\frac14MR^2\omega^2.\]

\[\boxed{E_k=\frac14MR^2\omega^2}\]

Step 4: Comparison with other shapes about a central axis

For the same $M,R,\omega$: a thin ring gives $E_k=\tfrac12MR^2\omega^2$ (all mass at the rim, twice the disk's energy); a solid sphere gives $E_k=\tfrac15MR^2\omega^2$ (mass concentrated closer to the axis, less energy). The disk sits between these, consistent with its mass being spread uniformly from center to rim.