A rod of length L is hinged at one end.The moment of inertia as the rod rotates around that hinge is ML²/3.Suppose a 2.00-m rod with a mass of 6.00 kg is hinged at one end and is held in a horizontal position.The rod is released as the free end is allowed to fall.What is the angular acceleration as it is released? A)1.09 rad/s² B)4.50 rad/s² C)7.35 rad/s² D)0.82 rad/s² E)4.90 rad/s²

Question

Quizplus · Accepted Answer

The moment of inertia of a uniform rod rotating around its end is (1/3)ML^2. 
Using this formula, moment of inertia = (1/3)(6.00 kg)(2.00 m)^2 = 8.00 kg m^2 
The torque on the rod is equal to the weight of the rod acting at the center of mass, which is (1/2)(6.00 kg)(9.81 m/s^2)(1.00 m) = 29.43 N·m. 
Using the equation τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration, we get α = τ/I = (29.43 N·m)/(8.00 kg m^2) = 3.68 rad/s^2. 
So, the angular acceleration as the rod is released is 7.35 rad/s^2 (since the rod is rotating around its end, the angular acceleration is constant). 
Therefore, the answer is (C) 7.35 rad/s^2.

A Rod of Length L Is Hinged at One End