question_answer

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\[\frac{m{{l}^{2}}}{3}.\] the Initial angular acceleration of the rod will be:

A) \[\frac{2g}{3l}\]

B) \[mg\frac{l}{2}\]

C) \[\frac{3}{2}gl\]

D) \[\frac{3g}{2l}\]  

Correct Answer: D

Solution :

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