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 :
Answer & explanation
Correct answer: option 3
Weight of the rod will produce torque,
\(\tau = mg × \frac{l}{2}\)
Also, \(\tau = I\alpha\)
and I = \(\frac{ml^2}{3}\)
\(\frac{ml^2}{3} \alpha = mg × \frac{l}{2}\)
\(\alpha = \frac{3g}{2l}\)