Question

A uniform rod of length $l$ is released from rest such that it rotates about a smooth pivot. The angular speed of the rod when it becomes vertical is

• A. $2 \sqrt{\frac{6 g}{7 l}}$
• B. $\sqrt{\frac{6 g}{7 l}}$
• C. $2 \sqrt{\frac{3 g}{7 l}}$
• D. $\sqrt{\frac{3 g}{7 l}}$

Answer 1

Answer: (A)

$\Delta K+\Delta U=0$

$\left(\frac{1}{2} I_{0} \omega^{2}-0\right)+\left[-m g \frac{\ell}{4}\right]=0\ldots$ (i)
$I_{0}=\frac{M \ell^{2}}{12}+m\left(\frac{\ell}{4}\right)^{2}=\frac{7}{48} m \ell^{2}\ldots$ (ii)
From (i) and (ii), $\omega=2 \sqrt{\frac{6 g}{7 \ell}}$