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A rod of length $l$ is hinged from one end. It is brought to a horizontal position and released. The angular velocity of the rod when it is in the vertical position is
$\sqrt {\frac {g}{l}}$
$\sqrt {\frac {2g}{l}}$
$\sqrt {\frac {3g}{l}}$
$\sqrt {\frac {g}{2l}}$
Solution
Mass of the rod is concentrated at its centre of
mass. In the vertical position of the rod, the centre
of mass is at $C^{\prime} .$ Now loss in $\mathrm{PE}=$ Gain in Rotational KE
$\operatorname{mg} \frac{l}{2}=\frac{1}{2} I \omega^{2}$
Where $\quad \mathrm{I}=\frac{\mathrm{m} l^{2}}{3}$
The moment of inertia of the rod about axis passing through $\mathrm{O}$ and $1$ to the plane of paper
$\frac{m g \ell}{2}=\frac{1}{2}=\frac{m \ell^{2}}{3} \omega^{2}$
$\omega=\frac{\sqrt{3 \mathrm{g}}}{\ell}$
