= Solution
The <moment of inertia> of a uniform length-$2l$ rod about its transverse central axis is
$$
\boxed{I=\frac M{2l}\int_{-l}^ls^2\,ds=\frac13Ml^2}.
$$
The <kinetic energy decomposition about the center of mass> gives $\boxed{T=MV^2/2+Ml^2\dot\theta^2/6}$ for center <speed> $V$ and rod angle $\theta$.
For a falling arch with $0<\alpha<\pi/2$, the symmetric <two hinged rods with both ends sliding on a floor> have the following motion. The hinge remains vertically over the arch center. Their centers are $\mathbf R_\pm=(\pm l\cos\theta,l\sin\theta)$, so each has squared <speed> $l^2\dot\theta^2$. Total <kinetic energy> is $T=4Ml^2\dot\theta^2/3$, and total gravitational <potential energy> is $U=2Mgl\sin\theta$. <Conservation of energy> from rest at angle $\alpha$ gives $\dot\theta^2=3g(\sin\alpha-\sin\theta)/(2l)$. At the hinge impact $\theta=0$, its height derivative is $2l\cos\theta\dot\theta$, giving
$$
\boxed{v_{\rm hinge}=\sqrt{6gl\sin\alpha}}.
$$
The floor-contact assumption is consistent: differentiating the energy relation gives $\ddot\theta=-3g\cos\theta/(4l)$, and the normal reaction on each outer endpoint is $N=Mg[1+9\sin^2\theta-6\sin\alpha\sin\theta]/4$. Completing the square shows $N\ge Mg\cos^2\alpha/4\ge0$ throughout the fall.
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