A uniform meter stick is pivoted to rotate about a horizontal axis through the 25-cm mark on the stick. The stick is released from rest in a horizontal position. The rotational inertia of a uniform rod about an axis perpendicular to the rod and through the center of mass of the rod is given by (1/12)ML². Determine the magnitude of the initial angular acceleration of the stick. A)17 rad/s² B)13 rad/s² C)15 rad/s² D)19 rad/s² E)23 rad/s²

Question

Quizplus · Accepted Answer

The torque due to gravity about the pivot is τ = Mg(0.25 m), where M is the mass of the stick and g is the acceleration due to gravity. The rotational inertia about the pivot is I = (1/12)ML^2 + M(0.25 m)^2. Using τ = Iα, solve for angular acceleration α.

A Uniform Meter Stick Is Pivoted to Rotate About a Horizontal