Answer
(a) The minimum starting height is $\frac{5r}{2}$
(b) The minimum starting height is $~2.7 r$
Work Step by Step
(a) To find the minimum height $h$ where the sphere will not lose contact with the track, we can assume that the normal force is zero and the gravitational force provides the centripetal force to keep the sphere moving in a circle. We can find the speed at the top of the loop:
$\frac{mv^2}{r} = mg$
$v = \sqrt{gr}$
We can use conservation of energy to find the minimum starting height $h$:
$mgh = mgh_2+\frac{1}{2}mv^2$
$h = h_2+\frac{v^2}{2g}$
$h = 2r+\frac{(\sqrt{gr})^2}{2g}$
$h = 2r+\frac{gr}{2g}$
$h = 2r+\frac{r}{2}$
$h = \frac{5r}{2}$
The minimum starting height is $\frac{5r}{2}$
(b) At the top of the loop, the sphere also has rotational kinetic energy. Note that the speed at the top of the loop is still $v = \sqrt{gr}$. We can use conservation of energy to find the minimum starting height $h$:
$mgh = mgh_2+\frac{1}{2}mv^2+ \frac{1}{2}I~\omega^2$
$mgh = mgh_2+\frac{1}{2}mv^2+ \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2$
$mgh = mgh_2+\frac{1}{2}mv^2+ \frac{1}{5}mv^2$
$mgh = mgh_2+\frac{7}{10}mv^2$
$h = h_2+\frac{7v^2}{10g}$
$h = 2r+\frac{7(\sqrt{gr})^2}{10g}$
$h = 2r+\frac{7gr}{10g}$
$h = 2r+\frac{7r}{10}$
$h = \frac{27r}{10}$
$h = 2.7 r$
The minimum starting height is $~2.7 r$
