A uniform rod of length $l$ and mass $m$ is free to rotate in a vertical plane about $\mathrm{A}$. The rod…

A uniform rod of length $l$ and mass $m$ is free to rotate in a vertical plane about $\mathrm{A}$. The rod initially in horizontal position is released. The initial angular acceleration of the rod is (moment of inertia of the rod about $\mathrm{A}$ is $\frac{m l^2}{3}$ ).

1. $m g \frac{l}{2}$
2. $\frac{3 g}{2 l}$
3. $\frac{2 l}{3 g}$
4. $\frac{3 g}{2 l^2}$

Solution

Here torque $\tau=m g\left(\frac{1}{2}\right)$ M. I. of $\operatorname{rod}$ about $A$ is: $I=\frac{m l^2}{3}$ $\therefore$ Angular acceleration of the rod is $\alpha=\frac{\tau}{I}$ $\alpha=\frac{m g\left(\frac{1}{2}\right)}{\frac{m l^2}{3}}=\frac{3 g}{2 l} .$ ~

Asked in: NEET 2006

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