Moment of inertia of a disc

The moment of inertia of a uniform disc is $\frac{1}{2}m{R}^2$. Let the mass per unit area of the disc be $\rho$.

Mass of “donut” with radius difference $\mathrm{d}r$ is $\rho \pi (r+\mathrm{d}r{)}^2-\pi r^2=2\rho \pi r\mathrm{d}r+\rho \pi \mathrm{d}{r}^2$. We ignore the $\rho \pi \mathrm{d}{r}^2$ term as it becomes negligible compared to the $2\rho \pi r\mathrm{d}r$ term as $\mathrm{d}r$ becomes small, so we have that the total moment of inertia is ${\int }_0^{R}2\rho \pi r^3\mathrm{d}r$ which is $[\frac{1}{2}\rho \pi r^4{]}_0^R=\frac{1}{2}\rho \pi R^4$. Noting that $\rho =\frac{m}{\pi R^2}$, we have that the moment of inertia is $\frac{1}{2}m{R}^2$.

List of moments of inertia of shapes