A uniform rod of mass $M$ and length $L$ is pivoted at one end and oscillates as a physical pendulum. What is the period of small oscillations?

Solution

1. The moment of inertia about the pivot is $I=\frac{1}{3}M{L}^2$.
2. The distance from the pivot to the center of mass is $d=\frac{L}{2}$.
3. The period is $$\begin{array}{rl}T& =2\pi \sqrt{\frac{I}{Mgd}}\\ & =2\pi \sqrt{\frac{M{L}^2/3}{MgL/2}}\end{array}$$.
4. Simplifying gives $T=2\pi \sqrt{\frac{2L}{3g}}$.
5. Hence the answer is (B).