
In the above case the torque due to weight of the rod, here use the concept of translation work done and rotational kinetic energy,
⇒W=K.E.⇒mgl=21Iω2, now the moment of inertia of the rod when the axis of rotation passes through one end of the rod, I=31×ml2,
⇒mgl=21×3ml2ω2
⇒ω=l6g, now use the relation between linear velocity and angular velocity,
⇒v=lω=3.6g=6ms−1