Answer
The lower end is moving with a speed of $~\sqrt{3gL}$
Work Step by Step
We can use conservation of energy to find the angular speed at the lowest point:
$KE_{rot} = U_g$
$\frac{1}{2}I~\omega^2 = Mgh$
$\frac{1}{2}(\frac{1}{3}ML^2)~\omega^2 = Mg~(\frac{L}{2})$
$\omega^2 = \frac{3g}{L}$
$\omega = \sqrt{\frac{3g}{L}}$
We can find the speed that the lower end is moving:
$v = \omega~L$
$v = \sqrt{\frac{3g}{L}}~(L)$
$v = \sqrt{3gL}$
The lower end is moving with a speed of $~\sqrt{3gL}$.
