Question

1. a merry - go - round has a moment of inertia of 300 kg·m² and an angular velocity of 15 revolutions minute. determine the merry - go - rounds angular momentum.

471 kg·m²/s
400 kg·m²/s
4,710 kg·m²/s
942 kg·m²/s

Explanation:

Step1: Convert angular - velocity to rad/s

The angular velocity is given as \(15\) revolutions per minute. We know that \(1\) revolution \( = 2\pi\) radians and \(1\) minute \(=60\) seconds. So, \(\omega=15\times\frac{2\pi}{60}\text{ rad/s}= \frac{15\times2\pi}{60}=\frac{\pi}{2}\text{ rad/s}\).

Step2: Use the formula for angular momentum

The formula for angular momentum \(L = I\omega\), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity. Given \(I = 300\text{ kg}\cdot\text{m}^2\) and \(\omega=\frac{\pi}{2}\text{ rad/s}\), then \(L=300\times\frac{\pi}{2}\text{ kg}\cdot\text{m}^2/\text{s}= 471\text{ kg}\cdot\text{m}^2/\text{s}\) (since \(300\times\frac{\pi}{2}=150\pi\approx150\times 3.14 = 471\)).

Answer:

\(471\frac{\text{kg}\cdot\text{m}^2}{\text{s}}\)