## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

(a) $\omega = \sqrt{\frac{3g}{L}}$ (b) $v = \sqrt{3gL}$
(a) The final rotational kinetic energy of the rod will be equal to the initial potential energy of the rod. Note that the center of mass of the rod is initially at a height of $\frac{L}{2}$. $KE_{rot} = PE$ $\frac{1}{2}I~\omega^2 = \frac{MgL}{2}$ $\frac{1}{2}(\frac{1}{3}ML^2)~\omega^2 = \frac{MgL}{2}$ $\omega^2 = \frac{3g}{L}$ $\omega = \sqrt{\frac{3g}{L}}$ (b) We can find the speed of the tip of the rod. $v = \omega~L$ $v = \sqrt{\frac{3g}{L}}~L$ $v = \sqrt{3gL}$