Moment of inertia of a uniform solid cylinder ($I=m{a}^2/2$)

For a uniform solid cylinder of radius $a$, length $L$ and mass density $\rho$, a coaxial shell carries mass $dm=2\pi \rho Lrdr$. The axial moment of inertia is $I={\int }_0^a{r}^{2}dm=\pi \rho L{a}^4/2=m{a}^2/2$. The axial moment of inertia is independent of the length once mass and radius are fixed. A thin cylindrical shell instead has $I=m{a}^2$, demonstrating why the uniform solid assumption matters.

For a uniform solid cylinder of length $l$, radius $r$ and mass $m$, a thin circular slice at axial coordinate $z$ has transverse moment of inertia $(z^2+r^2/4)dm$ about a diametral axis through the cylinder's midpoint. Integrating from $-l/2$ to $l/2$ gives $I_{\perp ,G}=m{l}^2/12+m{r}^2/4$. The parallel axis theorem gives $I_{\perp ,\mathrm{end}}=m{l}^2/3+m{r}^2/4$ about a parallel axis through an end-face centre. The transverse and symmetry-axis moments of inertia coincide exactly when $l=\sqrt{3}r$.