## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$f = \frac{1}{2\pi}~\sqrt{\frac{5g}{2R}}$
The moment of inertia is $I = \frac{2}{5}MR^2$. The distance $d$ from the center of mass to the pivot is $R$. We can find an expression for the frequency of small-angle oscillations as: $f = \frac{1}{2\pi}~\sqrt{\frac{Mgd}{I}}$ $f = \frac{1}{2\pi}~\sqrt{\frac{MgR}{\frac{2}{5}MR^2}}$ $f = \frac{1}{2\pi}~\sqrt{\frac{5g}{2R}}$