A uniform metre 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 moment of inertia of a uniform rod about an axis perpendicular to the rod and through the centre 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 moment of inertia about the pivot is I = (1/12)ML^2 + M(0.25 m)^2. Using τ = Iα, solve for α to find the angular acceleration.

A Uniform Metre Stick Is Pivoted to Rotate About a Horizontal