Answer
The speed of the ball at the lowest point of the track will be $v = (3.74~\sqrt{R-r}~)~m/s$
Work Step by Step
The kinetic energy at the bottom will be equal to the magnitude of the change in potential energy. Note that the change in height is $R-r$. So,
$PE = K$
$Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I_{CM}\omega^2$
$Mg(R-r) = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{2}{5}Mr^2)(\frac{v^2}{r^2})$
$Mg(R-r) = \frac{1}{2}Mv^2 + \frac{1}{5}Mv^2$
$Mg(R-r) = \frac{7}{10}Mv^2$
$v = \sqrt{\frac{10}{7}(9.80~m/s^2)(R-r)}$
$v = (3.74~\sqrt{R-r})~m/s$
Therefore, the speed of the ball at the lowest point of the track will be $v = (3.74~\sqrt{R-r}~)~m/s$
