A uniform rod of length L and mass M is pivoted at one end and is free to rotate in a vertical plane. The rod is released from rest in a horizontal position. What is the angular acceleration of the rod just after it is released?

Core Concept

The moment of inertia of a rod about one end is I = (1/3)ML². The torque about the pivot point is τ = Mg(L/2) since the center of mass is at L/2. Using τ = Iα, we get Mg(L/2) = (1/3)ML²α. Solving for α gives α = 3g/(2L).