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27 May, 07:53

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 had. The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is 6.0 kg m2. If by moving the bricks the man decreases the rotational inertia of the system to 2.0 kg m2, what are (a) the resulting angular speed of the platform and (b) the ratio of the new kinetic energy of the system to the original kinetic energy?

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  1. 27 May, 08:22
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    A. ω₂ = 3.6rev/s

    B. K₂ / K₁ = 3

    Explanation:

    Data;

    ω₁ = 1.2rev/s

    I₁ = 6.0kgm²

    ω₂ = ?

    I₂ = 2.0kgm²

    for conservation of angular momentum,

    Initial angular momentum = final angular momentum

    L₁ = L₂

    I₁ω₁ = I₂ω₂

    solving for ω₂,

    ω₂ = (I₁ * ω₁) / I₂

    ω₂ = (6.0 * 1.2) / 2.0

    ω₂ = 3.6rev/s

    the resulting angular speed of the platform is 3.6rev/s.

    b. the ratio of the new kinetic energy to the system kinetic energy

    let the new kinetic energy = K₂

    original kinetic energy = K₁

    Kinetic energy of a rotational body = ½*Iω²

    K₂ / K₁ = ½ I₁ω₁² = ½I₂ω²

    K₂ / K₁ = I₁ω₁² / I₂ω₂²

    K₂ / K₁ = (2.0 * 3.6²) / (6.0 * 1.2²)

    K₂ / K₁ = 25.92 / 8.64

    K₂ / K₁ = 3

    The ratio of the new kinetic energy to the old kinetic energy is 3.
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