Question

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/

Answer 1

Answer: A) 0.0394 kg · m2/s

The angular momentum is equal to M = 0.0394kgm^2/s

Explanation:

Given;

Mass of stick = 135g = 0.135kg

Length L = 1m

Angular speed w = 3.50 rad/s

The angular momentum is given as;

M = I×w

Where I = moment of inertia

In this case I is given as;

I = 1/12(m×L^2)

I = 1/12( 0.135 × 1^2)

I = 0.01125kgm^2

The angular momentum can then be calculated as

M = 0.01125kgm^2×3.50rads/s

M = 0.039375kgm^2/s

M = 0.0394kgm^2/s

The angular momentum is equal to M = 0.0394kgm^2/s