A thin uniform rod of mass m and length L is hinged at one end and is free to rotate in the vertical plane. A small ball of mass m is fixed to the other end of the rod. A constant horizontal force F =

Question

A thin uniform rod of mass m and length L is hinged at one end and is free to rotate in the vertical plane. A small ball of mass m is fixed to the other end of the rod. A constant horizontal force $F = \frac{mg}{2}$ is also acting on the rod at the free end when the rod is released from horizontal position as shown in figure.

CT42 m11 diagram

TestHub · Accepted Answer

The initial torque about the hinge is m gL + m gL /2 = 3 m gL /2 ; the horizontal force has zero torque initially. I h ​ = m L 2 + m L 2 /3 = 4 m L 2 /3 , so α = 9 g / ( 8 L ) and ball acceleration is 9 g /8 . The rod centre acceleration is 9 g /16 downward. Vertical force balance gives the hinge reaction 5 m g /16 upward.

Physics - Rotation Question with Solution | TestHub

Options:(select one or more)

Doubts & Discussion