## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

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. $I~\omega = 0.23~kg~m^2/s$ $\omega = \frac{0.23~kg~m^2/s}{I}$ $\omega = \frac{0.23~kg~m^2/s}{0.0242~kg~m^2}$ $\omega = 9.5~rad/s$ 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.