A uniform rod of mass M and length L is hinged at one of its ends. It is held in a horizontal position and then released from rest. The angular speed of the rod when it reaches the lowest (vertical) position is:

Step-by-step solution

When the rod moves from the horizontal to the vertical position, its center of mass falls by a distance h = L 2 . The loss in gravitational potential energy is: U = Mg ( L 2 ) This loss in potential energy is converted into rotational kinetic energy. The moment of inertia of a uniform rod about an axis through its end is I = ML^2 3 . Gain in rotational kinetic energy: K = 1 2 I ^2 = 1 2 ( ML^2 3 ) ^2 By conservation of mechanical energy: Mg ( L 2 ) = 1 2 ( ML^2 3 ) ^2 MgL = ML^2 3 ^2 ^2 = 3g L = 3g L Assuming the c