Question

a uniform thin rod of length L and mass M is hinged at a distance L/4 from one of the end and released from horizontal position as shown in the figure. angular velocity of the rod as it passes the vertical position is​

Answer 1

Answer:

w = (24g/7L)1/2

Explanation:

By parallel axes theorem the moment of inertia of the rod about the hinge is mL2/12 + m(L/4)2 = 7mL2/48.

when the rod reaches the vertical position the center of mass of the rod falls by L/4.

by conservation of energy we can write:

mgL/4 = (1/2)*I*w2    

  where I = moment of inertia = 7mL2/48 and w = angular velocity of the rod at the vertical position

this gives

w = (24g/7L)1/2