Question

A uniform rod of length 'l' and mass 'm' is free to rotate in a vertical plane about 'a'. The rod, initially in horizontal position, is released. The initial angular acceleration of the rod is (moment of inertia of rod about a is .)

Answer 1

Torque = Moment of inertia × Angular acc.

Force × Radius = ml^2/3 × Angular acc.

mgl/2 × 3/ml^2 = Angular acc.

Angular acceleration = 3g/2l.

Answer 2

Answer:

the initial angular acceleration of the rod is

$\frac{3g}{2l}$

Explanation:

Torque on the rod= Moment of wright of the rod about P,

$mg \frac{l}{2} = i \alpha$

$\frac{mgl}{2} = \frac{m {l}^{2} }{3} \alpha$

$\alpha = \frac{3g}{2l}$