Answer
The bowling ball would have to spin at 91 rpm
Work Step by Step
We can find the moment of inertia of the bowling ball.
$I = \frac{2}{5}MR^2$
$I = \frac{2}{5}(5.0~kg)(0.11~m)^2$
$I = 0.0242~kg~m^2$
We can use the angular momentum to find the angular velocity of the bowling ball.
$I~\omega = L$
$I~\omega = 0.23~kg~m^2/s$
$\omega = \frac{0.23~kg~m^2/s}{I}$
$\omega = \frac{0.23~kg~m^2/s}{0.0242~kg~m^2}$
$\omega = 9.5~rad/s$
We can convert the angular velocity to units of rpm.
$\omega = (9.5~rad/s)(\frac{1~rev}{2\pi~rad})(\frac{60~s}{1~min})$
$\omega = 91~rpm$
The bowling ball would have to spin at 91 rpm
