Rotational Mechanics

Let $y$ be the vertical position of the rod's center of mass. Since there are only vertical forces, the center of mass only moves in the vertical direction.

$y = \frac{l}{2} sin\theta \\ \dot{y} = \frac{l}{2} cos\theta \, \dot{\theta}$

Equate the net change in gravitational potential energy to the combined kinetic energy associated with translation and rotation:

$mg \frac{l}{2} (1- sin\theta) = \frac{1}{2} m \dot{y}^2 + \frac{1}{2} \frac{m l^2}{12} \dot{\theta}^2 = \dot{y}^2 \Big(\frac{m}{2} + \frac{m l^2}{24} \frac{4}{l^2 cos^2{\theta}}\Big) \\ g \frac{l}{2} (1- sin\theta) = \dot{y}^2 \Big(\frac{1}{2} + \frac{1}{6 cos^2{\theta}}\Big) \\ g l (1- sin\theta) = \dot{y}^2 \Big(1 + \frac{1}{3 cos^2{\theta}}\Big) \\ 3g l \, cos^2{\theta} \, (1- sin\theta) = \dot{y}^2 \Big(1 + 3 cos^2{\theta}\Big) \\ \dot{y} = \sqrt{\frac{3g l \, cos^2{\theta} (1- sin\theta)}{1 + 3 cos^2{\theta}}}$