- JEE Mains 2019, Physics, Question 23
A uniform rod of mass m and length L is hinged at one end and is free to rotate in a vertical plane. A particle of mass m is attached to the other end of the rod. The particle is released from rest when the rod is horizontal. The maximum angle that the rod will make with the vertical is
(A) $\cos^{-1} \left(\frac{1}{2}\right)$ (B) $\cos^{-1} \left(\frac{3}{4}\right)$ (C) $\cos^{-1} \left(\frac{1}{4}\right)$ (D) $\cos^{-1} \left(\frac{3}{2}\right)$
Let’s solve this question step-by-step.
The rod is in equilibrium at theta = 0, so the torque about the hinge is zero. The torque about the hinge is T = mgLsin(theta). As the rod rotates, the torque increases, until at theta = theta_max, the torque is equal to the moment of inertia of the rod about the
