Question

A metal rod of length $1.25m$ and mass ' $m$ ' is pivoted at one end. A solid sphere of same mass and radius $0.25m$ is attached at its center to the free end of the rod and the sphere is free to rotate about its center. The roddisc system performs SHM in vertical plane after being released from the same displaced position. Angular frequency (in $rad/s$ ) of small oscillations is $n$. Find the value of $n^2$.

Answer 1

Answer: 9

$\tau =-\left(mg\frac{l}{2}+mgl\right)\sin \theta$
$I=\frac{m{l}^2}{3}+m{l}^2$
$\therefore \alpha =\frac{\tau }{I}=\frac{-3mgl\theta }{2\left[\frac{4m{l}^2}{3}\right]}=\frac{-9}{8}\frac{g\theta }{l};$
${\omega }^2=\frac{9g}{8l}=9$