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 2.50 m rod with a mass of 3.00 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) 3.70 rad/s² B) 7.35 rad/s² C) 5.88 rad/s² D) 4.90 rad/s²

Question

Quizplus · Accepted Answer

The moment of inertia (I) and the mass (m) of the rod are given. To find the angular acceleration (α), we can use the equation: τ = Iα where τ is the torque applied to the rod. In this case, the only torque acting on the rod is the torque due to gravity: τ = mgl/2 where g is the acceleration due to gravity and l is the length of the rod. Therefore, we can write: mgl/2 = Iα Solving for α, we get: α = mgl / 2I Substituting the given values, we get: α = (3.00 kg)(9.81 m/s^2)(2.50 m) / [2(ML²/3)] α ≈ 5.88 rad/s^2 Therefore, the answer is C, 5.88 rad/s^2.

A Rod of Length L Is Hinged at One End