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For the stated square side length $2a$, the volume is $4a^{2}h$ and the uniform mass density is ${\rho }_1=m_1/(4a^{2}h)$. Direct integration about its central vertical axis gives the moment of inertia of a uniform square plate:Thus the PDF's coefficient $1/6$ is inconsistent with its stated side length. The coefficient $1/6$ would apply to a square of side $a$.

For the disc, use cylindrical coordinates and $m_2={\rho }_2\pi a^{2}h$:For the sphere, use spherical coordinates and $m_3=4\pi {\rho }_3{a}^3/3$:These agree with the moment of inertia of a uniform disc and the moment of inertia of a uniform solid sphere. The square's thickness does not enter its axial moment of inertia once the mass is fixed.

Use Cartesian coordinates along the edges from the pivot. With surface mass density $M/L^2$, the moment of inertia isEquivalently, the moment of inertia of a uniform square plate about its central perpendicular axis is $M{L}^2/6$, and the parallel axis theorem adds $M(L/\sqrt{2}{)}^2$.

The source figure measures $\theta$ from the upward vertical to $OC$, with $OC=L/\sqrt{2}$. The centre's height above the pivot is $(L/\sqrt{2})\cos \theta$. The smooth fixed pivot does no work. Conservation of energy, with the initially upright configuration as reference, givesFor rotation in the positive $\theta$ direction on the first descent, take the positive root. Its equivalent form is $\dot{\theta }=\sqrt{3\sqrt{2}g/L}\sin (\theta /2)$ for $0<\theta <\pi$.

Strictly, release from rest at a small angle ${\theta }_0$ replaces $1-\cos \theta$ by $\cos {\theta }_0-\cos \theta$. The displayed result is the ${\theta }_0\to 0$ limiting energy relation. Exactly upright and exactly at rest is an unstable equilibrium; without a perturbation it stays there. This distinction prevents the infinitesimal displacement convention from being mistaken for spontaneous motion.