A uniform meter stick with a mass of 135 g rotates about an axis perpendicular to the stick passing through the center of the stick with an angular speed of 3.50 rad/s². What is the magnitude of the angular momentum of the stick? A) 0.0394 kg∙m²/s B) 0.473 kg∙m²/s C) 0.0739 kg∙m²/s D) 0.158 kg∙m²/s E) 0.0236 kg∙m²/s

Question

Quizplus · Accepted Answer

The angular momentum of the stick is given by Iω, where I is the moment of inertia of the stick and ω is the angular speed. The moment of inertia of a uniform meter stick rotating about an axis perpendicular to the stick passing through the center of the stick is (1/12)ML^2, where M is the mass of the stick and L is the length of the stick. Therefore, the angular momentum of the stick is:Iω = (1/12)ML^2ω = (1/12)(0.135 kg)(1 m)^2(3.50 rad/s) = 0.0394 kg∙m^2/s

A Uniform Meter Stick with a Mass of 135 G