Answer
The minimum starting height is $~3 r$
Work Step by Step
To find the minimum height $h$ where the cylinder 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 cylinder moving in a circle. We can find the speed at the top of the loop:
$\frac{Mv^2}{r} = Mg$
$v = \sqrt{gr}$
At the top of the loop, the cylinder also has rotational kinetic energy. We can use conservation of energy to find the minimum starting height $h$:
$Mgh = Mgh_2+\frac{1}{2}Mv^2+ \frac{1}{2}I~\omega^2$
$Mgh = Mgh_2+\frac{1}{2}Mv^2+ \frac{1}{2}(MR^2)(\frac{v}{R})^2$
$Mgh = Mgh_2+\frac{1}{2}Mv^2+ \frac{1}{2}Mv^2$
$Mgh = Mgh_2+Mv^2$
$h = h_2+\frac{v^2}{g}$
$h = 2r+\frac{(\sqrt{gr})^2}{g}$
$h = 2r+\frac{gr}{g}$
$h = 2r+r$
$h = 3 r$
The minimum starting height is $~3 r$.
