A uniform 135-g meter stick rotates about an axis perpendicular to the stick passing through its center with an angular speed of 3.50 rad/s.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:$$L = Iω$$where:* I is the moment of inertia of the stick* ω is the angular speed of the stickThe moment of inertia of a uniform rod rotating about an axis perpendicular to the rod passing through its center is:$$I = \frac{1}{12}ml^2$$where:* m is the mass of the stick* l is the length of the stickSubstituting the given values into the equations, we get:$$I = \frac{1}{12}(0.135 kg)(0.500 m)^2 = 0.00281 kg m^2$$$$L = (0.00281 kg m^2)(3.50 rad/s) = 0.00984 kg m^2/s$$Therefore, the magnitude of the angular momentum of the stick is 0.00984 kg m^2/s, which is closest to choice A, 0.0394 kg ∙ m²/s.

A Uniform 135-G Meter Stick Rotates About an Axis Perpendicular