A horizontal rod of mass m and length L is free to rotate about a vertical axis passing through its centre. A constant horizontal force of magnitude F acts on the rod at a distance of L/4 from the centre. The force remains perpendicular to the rod at all times. Determine the angle rotated by the rod in time t after the motion begins.

Step-by-step solution

The moment of inertia of the rod about the vertical axis passing through its centre is I = mL^2 12 . The torque acting on the rod due to the force F applied at a distance of L/4 from the centre and always perpendicular to the rod is = F L 4 . Using the relation = I , the angular acceleration is given by: = I = FL 4 mL^2 12 = 3F mL Since the rod starts from rest, the angle rotated in time t is: = 1 2 t^2 = 1 2 ( 3F mL )t^2 = 3Ft^2 2mL