A thin uniform rod of mass ' $\mathrm{m}$ ' and length ' $l$ ' is suspended from one end which can oscillate…
A thin uniform rod of mass ' $\mathrm{m}$ ' and length ' $l$ ' is suspended from one end which can oscillate in a vertical plane about the point of intersection. It is pulled to one side and then released. It passes through the equilibrium position with angular speed ' $\omega$ '. The kinetic energy while passing through mean position is
$\mathrm{m} l^2 \omega^2$
$\frac{\mathrm{m} l^2 \omega^2}{4}$
$\frac{\mathrm{m} l^2 \omega^2}{6}$
$\frac{\mathrm{m} l^2 \omega^2}{12}$
Solution
The kinetic energy of the rod while passing through the mean position will be,
$\begin{aligned}
\text { K.E. } & =\frac{1}{2} \mathrm{I} \omega^2 \\
& =\frac{1}{2} \frac{\mathrm{m} l^2}{3} \times \omega^2=\frac{\mathrm{m} l^2 \omega^2}{6}
\end{aligned}$