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, i $s_______.$
(g = gravitational acceleration)

Question

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, i $s_______.$

(g = gravitational acceleration)

G1 Physics · Accepted Answer

The pivot divides the rod so that the upper part has length 2L/3 and the lower part has length L/3. The centre of mass of the uniform rod is at its midpoint. Therefore the centre of mass is at a distance L/2 - L/3 = L/6 above the pivot initially. When the rod becomes horizontal, its centre of mass is at the level of the pivot. Hence the loss of gravitational potential energy is ΔU = Mg(L/6). The moment of inertia about the pivot is obtained by the parallel-axis theorem: I = ML²/12 + M(L/6)² = M…

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, i s_______. (g = gravitational acceleration)