A uniform rod of length (L = 2.0 m) and mass (M = 1.5 kg) is pivoted about a horizontal frictionless pin through one end. The rod is released from rest at an angle of 30° below the horizontal. What is the angular speed of the rod when it passes through the vertical position? (The rotational inertia of the rod about the pin is 2.0 kg⋅m².) A)3.5 rad/s B)2.7 rad/s C)3.1 rad/s D)2.3 rad/s E)1.6 rad/s

Question

Quizplus · Accepted Answer

The potential energy lost by the rod as it moves from 30° below the horizontal to the vertical position is converted into rotational kinetic energy. Using conservation of energy: mgh = (1/2)Iω², where h = L/2 * (1 - cos(30°)), I = 2.0 kg⋅m², and solving for ω gives approximately 2.7 rad/s.

A Uniform Rod of Length (L = 2