Q.
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$\frac{{ml}^{2}}{3}$, the initial angular acceleration of the rod will be
A) $\frac{2g}{3l}$
B) $mg\frac{l}{2}\ $
C) $\frac{3}{2}gl$
D) $\frac{3g}{2l}$
Explanation
(d) The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is
$I = \frac{ml^{2}}{3}$
Where $m$ is mass of rod and $l$ is length.
Torque $(\tau = I\alpha)$ acting on centre of gravity of rod is given by
$\tau = mg\ \frac{1}{2}$
or $I\alpha = mg\ \frac{1}{2}$
or $\frac{ml^{2}}{3}\alpha = mg\frac{1}{2}$
or $\alpha = \frac{3g}{2l}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.