Question

A uniform rod $AB$ of length $l$ and mass $m$ is free to rotate about point $A$ . The rod is released from rest in the horizontal position. The initial angular acceleration of the rod will be

• A. $\frac{2\text{g}}{3l}$
• B. $\text{g}\frac{l}{2}$
• C. $\frac{3}{2}\text{g}l$
• D. $\frac{3\text{g}}{2l}$

Answer 1

Answer: (D)

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
$\tau =\text{mg}\frac{l}{2}$
or $\text{I}\alpha =\text{mg}\frac{l}{2}$
or $\frac{{\text{ml}}^2}{3}\alpha =\text{mg}\frac{l}{2}$
or $\alpha =\frac{3\text{g}}{2\text{l}}$