Moment of inertia of a uniform square plate ($I=M{s}^2/6$)

A uniform square of side $s$ has the displayed moment of inertia about the perpendicular axis through its centre. Integrating $x^2+y^2$ over $[-s/2,s/2{]}^2$ gives $s^4/6$; multiplying by the surface mass density $M/s^2$ gives the result. A square prism of any uniform thickness has the same axial formula when expressed in terms of its total mass. If the side is denoted $2a$ instead, the result is $2M{a}^2/3$, so the side-length convention must be specified before substituting numerical data into a normal mode calculation.

For a uniform square of side $a$, the principal inertia tensor is $\mathrm{diag}(m{a}^2/12,m{a}^2/12,m{a}^2/6)$. A rotation axis tilted $45^{\circ }$ to the surface has equal in-plane and normal angular-velocity components. Multiplying each by its own principal moment gives the stated magnitude. In general $I_{\mathrm{axis}}\omega$ is only the axial projection of angular momentum.

A uniform thin square of side $l$ and mass $M$ has $I_z=M{l}^2/6$ about the perpendicular central axis. Any central in-plane axis and its perpendicular in-plane partner have equal moments by $90^{\circ }$ rotational symmetry. The perpendicular axis theorem says their sum is $I_z$, proving $I_{\varphi }=M{l}^2/12$ for every angle $\varphi$. A non-square rectangle generally lacks this angle independence.