Answer
$h=2.7R$
Work Step by Step
We use conservation of energy to find:
$mg(h-2R)=\frac{1}{2}mv^2 + \frac{1}{2} I \omega^2$
$mg(h-2R)=\frac{1}{2}mv^2 + \frac{1}{2} \frac{2}{5}mr^2\frac{v^2}{r^2}$
$g(h-2R)=\frac{1}{2}v^2 + \frac{1}{5} v^2$
$g(h-2R) = .7v^2 $
We know the following:
$F_n+mg=\frac{mv^2}{R}$
At the minimum velocity, $F_n$ will equal 0. Thus, we find:
$v=\sqrt{gR}$
We plug this into the above equation to obtain:
$g(h-2R) = .7gR $
$h=2.7R$
