Answer
$h = 38~cm$
Work Step by Step
We can write an expression for the rotational inertia of a solid sphere:
$I = \frac{2}{5}Mr^2$
We can find an expression for the total kinetic energy of the ball when it has a speed $v$:
$K = \frac{1}{2}Mv^2+\frac{1}{2}I\omega^2$
$K = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{2}{5}Mr^2)(\frac{v}{r})^2$
$K = \frac{1}{2}Mv^2+\frac{1}{5}Mv^2$
$K = \frac{7}{10}Mv^2$
If the ball is on the verge of leaving the track when it reaches the top of the loop, then the normal force on the ball from the track must be zero. The centripetal force on the ball at the top of the track is provided by the gravitational force:
$F_N+Mg = \frac{Mv^2}{R}$
$0+Mg = \frac{Mv^2}{R}$
$v^2 = gR,$ where $R$ is the radius of the loop
We can write an expression for the total kinetic energy at the top of the loop:
$K = \frac{7}{10}Mv^2$
$K = \frac{7}{10}MgR$
We can use conservation of energy to find $h$:
$U_1+K_1 = U_2+K_2$
$U_1+0 = U_2+K_2$
$Mgh = Mg(2R)+\frac{7}{10}MgR$
$h = 2R+\frac{7}{10}R$
$h = (2)(0.14~m)+\frac{7}{10}(0.14~m)$
$h = 0.38~m$
$h = 38~cm$
