Physics - Rotation Question with Solution | TestHub

The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is

$\text{I}=\frac{{\text{ml}}^2}{3}$

Where m is mass of rod and l is length.

Torque$\left(\tau =\text{I}\alpha \right)$acting on centre of gravity of rod is given by

$\tau =\text{mg}\frac{\mathrm{l}}{2}$

or$\text{I}\alpha =\text{mg}\frac{\mathrm{l}}{2}$

or$\frac{{\text{ml}}^2}{3}\alpha =\text{mg}\frac{\mathrm{l}}{2}$

or$\alpha =\frac{3\text{g}}{2\text{l}}$