A thin uniform rod AB of mass ' m ' and length ' l ' is hinged at one end A to the ground level. Initially the rod stands vertically and is allowed to fall freely to the ground in the vertical plane. The angular velocity of the rod when its B end strikes the ground is ( g = acceleration due to gravity)

Step-by-step solution

By the principle of conservation of mechanical energy, the loss in potential energy of the rod is equal to the gain in its rotational kinetic energy. The initial height of the center of mass of the rod is h_i = l 2 and the final height is h_f = 0 . Loss in potential energy = mg ( l 2 ) Gain in rotational kinetic energy = 1 2 I ^2 The moment of inertia of the rod about the hinged end A is I = ml^2 3 . Equating the two energies: mg ( l 2 ) = 1 2 ( ml^2 3 ) ^2 mgl = ml^2 3 ^2 ^2 = 3g l = 3g l Answer: 3g l