Answer
Initially, there are $~~8~J~~$ of rotational kinetic energy.
Work Step by Step
The rotational inertia of a hollow sphere is $I = \frac{2}{3}MR^2$
We can find an expression for the total kinetic energy:
$K = \frac{1}{2}Mv^2+\frac{1}{2}I\omega^2$
$K = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{2}{3}MR^2)(\frac{v}{R})^2$
$K = \frac{1}{2}Mv^2+\frac{1}{3}Mv^2$
$K = \frac{5}{6}Mv^2$
Note that the expression for the rotational kinetic energy is $\frac{1}{3}Mv^2$
We can find the fraction of the total kinetic energy that is rotational:
$\frac{\frac{1}{3}Mv^2}{\frac{5}{6}Mv^2} = \frac{2}{5}$
We can find the amount of the initial kinetic energy that is rotational:
$K_{rot} = (\frac{2}{5})(20~J) = 8~J$
Initially, there are $~~8~J~~$ of rotational kinetic energy.
