A uniform rod of length $L$ and mass $M$ is pivoted at one end and allowed to swing in a vertical plane. What is the angular frequency of small oscillations?

Solution

1. The angular frequency of a physical pendulum is given by $\omega =\sqrt{\frac{Mgd}{I}}$, where $d$ is the distance from the pivot to the center of mass and $I$ is the moment of inertia about the pivot.
2. For a uniform rod pivoted at one end, the distance to the center of mass is $d=\frac{L}{2}$ and the moment of inertia is $I=\frac{1}{3}M{L}^2$.
3. Substituting these values into the formula, we get $\omega =\sqrt{\frac{Mg(L/2)}{\frac{1}{3}M{L}^2}}$.
4. Simplifying the expression yields $$\begin{array}{rl}\omega & =\sqrt{\frac{MgL/2}{M{L}^2/3}}\\ & =\sqrt{\frac{3g}{2L}}\end{array}$$.
5. Hence the answer is (A).