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 $\degree$below the horizontal.What is the angular speed of the rod when it passes through the vertical position? (The moment of 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 change in potential energy is converted into rotational kinetic energy. Using conservation of energy: mgh = 0.5 * I * ω², where h = L/2 * (1 - cos(30°)). Solving for ω gives approximately 2.7 rad/s.

A Uniform Rod of Length (L = 2 °\degree° Below the Horizontal

A Uniform Rod of Length (L = 2 $\degree$ Below the Horizontal