Answer
(a) At the base of the hill, it was rotating at a rate of 67.9 rad/s.
(b) $K_{rot} = 8.36~J$
Work Step by Step
(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.
$K_{rot} = \frac{1}{2}I\omega^2$
$K_{rot} = \frac{1}{2}(\frac{2mR^2}{3})\omega^2$
$K_{rot} = \frac{1}{3}(mR^2)\omega^2$
$K_{rot} = \frac{1}{3}(0.426~kg)(0.113~m)^2(67.9~rad/s)^2$
$K_{rot} = 8.36~J$
