A uniform rod of length l, hinged at one end is free to rotate in the vertical plane. If the rod is held horizontal in the beginning and released, then angular velocity of the rod when it makes an angle of 60 degree with the vertical for the first time is

When it makes 60 degree with the horizontal then at this position torque acting on rod is =mgl/2 cos60                                                                                                                             =mgl/4
We know that
Torque=moment of inertia * angular acceleration
mgl/4=ml2/3 * angular acceleration, where m is the mass of rod
angular acceleration =3g/4 l