## College Physics (4th Edition)

The lower end is moving with a speed of $~\sqrt{3gL}$
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 $~\sqrt{3gL}$.