## University Physics with Modern Physics (14th Edition)

(a) At the base of the hill, it was rotating at a rate of 67.9 rad/s. (b) $K_{rot} = 8.36~J$
(a) We can use conservation of energy to solve this question. The potential energy at the top will be equal to the total kinetic energy at the bottom. $\frac{1}{2}mv^2+\frac{1}{2}I\omega^2 = mgh$ $\frac{1}{2}mv^2+\frac{1}{2}(\frac{2mR^2}{3})(\frac{v}{R})^2 = mgh$ $\frac{1}{2}v^2+\frac{1}{3}v^2 = gh$ $v^2 = \frac{6gh}{5}$ $v = \sqrt{\frac{6gh}{5}}$ $v = \sqrt{\frac{(6)(9.80~m/s^2)(5.00~m)}{5}}$ $v = 7.67~m/s$ We can use the velocity to find the angular velocity. $\omega = \frac{v}{R}$ $\omega = \frac{7.67~m/s}{0.113~m}$ $\omega = 67.9~rad/s$ At the base of the hill, it was rotating at a rate of 67.9 rad/s. (b) We can find the rotational kinetic energy. $K_{rot} = \frac{1}{2}I\omega^2$ $K_{rot} = \frac{1}{2}(\frac{2mR^2}{3})\omega^2$ $K_{rot} = \frac{1}{3}(mR^2)\omega^2$ $K_{rot} = \frac{1}{3}(0.426~kg)(0.113~m)^2(67.9~rad/s)^2$ $K_{rot} = 8.36~J$