A thin uniform rod (X) of mass M and length L is pivoted at a height ( L 3 ) as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top, is _ _ _ _ . ( g = gravitational acceleration)

Step-by-step solution

The rod is pivoted at a distance L 3 from one end. The centre of mass is at a distance L 6 from the pivot. Using energy conservation: mg 6 = 1 2 I ^2 Moment of inertia about the pivot (using parallel axis theorem): I = m ^2 12 + m ( 6 )^2 = m ^2 12 + m ^2 36 = m ^2 9 mg 6 = 1 2 m ^2 9 ^2 ^2 = 9g 6 2 = 3g = 3g