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The moment of inertia of a uniform length-$2l$ rod about its transverse central axis isThe kinetic energy decomposition about the center of mass gives $\boxed{T=M{V}^2/2+M{l}^2{\dot{\theta }}^2/6}$ 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 $T=4M{l}^2{\dot{\theta }}^2/3$, 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 $2l\cos \theta \dot{\theta }$, givingThe 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 $N\ge Mg{\cos }^2\alpha /4\ge 0$ throughout the fall.