= Solution
Use the arch geometry of <two hinged rods sliding on a floor>, with $A=O=(0,0)$ and $C$ on the other side of $B$. As long as $C$ remains in contact with the floor,
$$
B=(2l\cos\theta,2l\sin\theta),\qquad
C=(4l\cos\theta,0).
$$
The <centres of mass> of the two rods are $(l\cos\theta,l\sin\theta)$ and $(3l\cos\theta,l\sin\theta)$. Their squared <speeds> are respectively $l^2\dot\theta^2$ and $l^2(9\sin^2\theta+\cos^2\theta)\dot\theta^2$. Each uniform rod has <moment of inertia> $Ml^2/3$ about its midpoint, and their angular <velocities> are $\dot\theta$ and $-\dot\theta$. Applying the <kinetic energy decomposition about the center of mass> to each rod gives
$$
T=\frac M2l^2\dot\theta^2
+\frac M2l^2(9\sin^2\theta+\cos^2\theta)\dot\theta^2
+2\left(\frac12\frac{Ml^2}{3}\dot\theta^2\right)
=\frac43Ml^2(1+3\sin^2\theta)\dot\theta^2.
$$
The gravitational <potential energy> is $U=2Mgl\sin\theta$. The fixed pivot does no work; neither does the frictionless floor reaction while $C$ slides horizontally. <Conservation of energy> from rest therefore gives
$$
\boxed{\dot\theta^2=\frac{3g}{2l}
\frac{\sin\alpha-\sin\theta}{1+3\sin^2\theta}.}
$$
On the falling branch $\dot\theta$ is the negative square root. Just before the hinge reaches the floor, $\theta=0$, and its <speed> is $|\dot B|=2l|\dot\theta|$. Thus, under the maintained-contact interpretation,
$$
\boxed{v_B=\sqrt{6gl\sin\alpha}.}
$$
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/ia/paper-4-hinged-rods.png]
{title=Geometry and centers of mass of two hinged rods sliding on a floor}
There is a contact hypothesis behind this calculation. A floor can push upward but cannot pull downward. For $s=\sin\theta$ and $a=\sin\alpha$, the required normal reaction at $C$, found from the second rod's linear and angular equations, is
$$
\frac{N_C}{Mg}
=\frac{\tfrac14+\tfrac{33}4s^2+9s^4-\tfrac92as}
{(1+3s^2)^2}.
$$
For the usual acute arch, this stays nonnegative throughout $0\leq\theta\leq\alpha$ exactly when $\sin\alpha\leq35/54$. Indeed, for $s>0$, its nonnegativity is $a\leq 1/(18s)+11s/6+2s^3$, whose minimum is $35/54$ at $s=1/6$. For larger initial angles, the endpoint lifts before the hinge hits the floor, and the one-coordinate formula no longer governs the whole motion. \b[The displayed impact <speed> is the intended floor-constrained answer; for unrestricted initial angles, maintained contact is an additional assumption.] Subsequent free motion and possible reimpact would require a different model and contact/impact conditions.
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