## Essential University Physics: Volume 1 (3rd Edition)

$h=2.7R$
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: $g(h-2R) = .7gR$ $h=2.7R$