Answer
The translational speed of the boulder is 29.0 m/s.
Work Step by Step
We can use conservation of energy to find the rotational kinetic energy immediately after crossing the rough patch. The kinetic energy will be equal in magnitude to the change in potential energy as the boulder drops from 50.0 meters down to 25.0 meters.
$K = PE$
$\frac{1}{2}mv^2+\frac{1}{2}I\omega^2 = mg\Delta h$
$\frac{1}{2}mv^2+\frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mg\Delta h$
$\frac{1}{2}mv^2 + \frac{1}{5}mv^2= (25.0)~mg$
We can see that $\frac{5}{7}$ of the kinetic energy is translational kinetic energy and $\frac{2}{7}$ of the kinetic energy is rotational kinetic energy. Therefore, the rotational kinetic energy after crossing the rough patch is $\frac{2}{7}\times (25.0)~mg$.
Since the bottom half of the hill has no friction, the rotational kinetic friction will not change, and the boulder will have this rotational kinetic friction at the bottom of the hill.
We can use conservation of energy to find the translational speed at the bottom of the hill. Let $PE$ be the change in potential energy for a change in height of 50.0 meters.
$K_{tran}+K_{rot} = PE$
$\frac{1}{2}mv^2+\frac{2}{7}\times (25.0)~mg = mg\Delta h$
$\frac{1}{2}mv^2 = (50.0)~mg -\frac{2}{7}\times (25.0)~mg$
$\frac{1}{2}v^2 = (50.0)~g -\frac{2}{7}\times (25.0)~g$
$\frac{1}{2}v^2 = (42.86)~g$
$v = \sqrt{(2)(42.86)~g}$
$v = \sqrt{(2)(42.86~m)(9.80~m/s^2)}$
$v = 29.0~m/s$
The translational speed of the boulder is 29.0 m/s.
