= Solution
Use Cartesian coordinates along the edges from the pivot. With <surface mass density> $M/L^2$, the <moment of inertia> is
$$
I_O=\frac M{L^2}\int_0^L\!\int_0^L(x^2+y^2)\,dx\,dy
=\boxed{\frac23ML^2}.
$$
Equivalently, the <moment of inertia of a uniform square plate> about its central perpendicular axis is $ML^2/6$, and the <parallel axis theorem> adds $M(L/\sqrt2)^2$.
The source figure measures $\theta$ from the upward vertical to $OC$, with $OC=L/\sqrt2$. The centre's height above the pivot is $(L/\sqrt2)\cos\theta$. The smooth fixed pivot does no work. <Conservation of energy>, with the initially upright configuration as reference, gives
$$
\frac12I_O\dot\theta^2
=\frac{MgL}{\sqrt2}(1-\cos\theta),\qquad
\boxed{|\dot\theta|=\sqrt{\frac{3g}{\sqrt2L}(1-\cos\theta)}.}
$$
For rotation in the positive $\theta$ direction on the first descent, take the positive root. Its equivalent form is $\dot\theta=\sqrt{3\sqrt2g/L}\sin(\theta/2)$ 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 $\theta_0\to0$ 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.
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