Ask Question
Yesterday, 22:38

A man stands on a platform that is rotating (without friction) with an angular speed of 1.2 rev/s; his arms are outstretched and he holds a brick in each hand. The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is 6.0 k g times m squared. If by moving the bricks the man decreases the rotational inertia of the system to 2.0 k g times m squared, what is the resulting angular speed of the platform in rad/s

+3
Answers (1)
  1. Today, 01:04
    0
    resulting angular speed = 3.6 rev/s

    Explanation:

    We are given;

    Initial angular speed; ω_i = 1.2 rev/s

    Initial moment of inertia; I_i = 6 kg/m²

    Final moment of inertia; I_f = 2 kg/m²

    From conservation of angular momentum;

    Initial angular momentum = Final angular momentum

    Thus;

    I_i * ω_i = I_f * ω_f

    Making ω_f the subject, we have;

    ω_f = (I_i * ω_i) / I_f

    Plugging in the relevant values;

    ω_f = (6 * 1.2) / 2

    ω_f = 3.6 rev/s
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A man stands on a platform that is rotating (without friction) with an angular speed of 1.2 rev/s; his arms are outstretched and he holds a ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers