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

Answer 2

This question involves the concepts of angular momentum, angular speed, and radius.

The angular momentum of the stick is "B) 0.473 kg.m²/s".

Angular Momentum

The angular momentum of the stick can be found using the following formula:

$L=mvr$

where,

• L = Angular momentum = ?

• r = radius of the circle = length of stick = 1 m
• ω = angular speed = 3.5 rad/s
• v = linear speed = rω = (1 m)(3.5 rad/s) = 3.5 m/s
• m = mass of stick = 135 g = 0.135 kg

Therefore,

L = (0.135 kg)(3.5 m/s)(1 m)

L = 0.473 kg.m²/s

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