A thin uniform rod (length = 1.2 m, mass = 2.0 kg) is pivoted about a horizontal, frictionless pin through one end of the rod. (The rotational inertia of the rod about this axis is ML2/3.) The rod is released when it makes an angle of 37° with the horizontal. What is the angular acceleration of the rod at the instant it is released? A)9.8 rad/s² B)7.4 rad/s² C)8.4 rad/s² D)5.9 rad/s² E)6.5 rad/s²

Question

Quizplus · Accepted Answer

The correct answer is A because the angular acceleration of the rod at the instant it is released is given by:α = (3g/2L)sinθ = (3 * 9.8 m/s^2 / 2 * 1.2 m) * sin 37° = 9.8 rad/s^2

A Thin Uniform Rod (Length = 1