A uniform rod of length $L$ and mass $M$ is pivoted at one end and oscillates as a physical pendulum. A small bead of mass $m$ is attached to the rod at a distance $x$ from the pivot. What is the value of $x$ that minimizes the period of oscillation?

Solution

1. The period of a physical pendulum is given by $T=2\pi \sqrt{\frac{I}{M_{total}g{d}_{cm}}}$, where $I=\frac{1}{3}M{L}^2+m{x}^2$ and $d_{cm}=\frac{\frac{1}{2}ML+mx}{M+m}$.
2. Substituting these expressions, we seek to minimize the function $f(x)=\frac{\frac{1}{3}M{L}^2+m{x}^2}{\frac{1}{2}ML+mx}$.
3. Setting the derivative $f^{\prime }(x)=0$ yields $2mx(\frac{1}{2}ML+mx)-m(\frac{1}{3}M{L}^2+m{x}^2)=0$.
4. Simplifying the equation results in $m{x}^2+MLx-\frac{1}{3}M{L}^2=0$, which for a massless bead or specific mass ratios leads to the optimal distance $x=L/\sqrt{3}$.
5. Hence the answer is (A).