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. Given that the moment of inertia of the rod about $A$ is $m{l}^2/3,$ the initial angular acceleration of the rod will be:

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

Answer 1

Answer: (D)

The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is,
$l=\frac{m{l}^2}{3}$
where $m$ is the mass of the rod and $l$ is the length Torque $\left(\tau =I\alpha \right)$ acting on the centre of gravity of rod is given by $\tau =mg\frac{l}{2}$
or $I\alpha =mg\frac{l}{2}$
or $\frac{m{l}^2}{3}\alpha =mg\frac{l}{2}$
or $\alpha =\frac{3g}{2l}$