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2 January, 13:08

A solid uniform cylinder is rolling without slipping. What fraction of its kinetic energy is rotational?

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  1. 2 January, 14:47
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    Let m be the mass of cylinder and r be the radius. It is moving with velocity v and angular velocity is ω. Let I be the moment of inertia of the cylinder.

    I = 0.5 mr²

    Total kinetic energy, T = 0.5 mv² + 0.5 Iω²

    T = 0.5 (mv² + 0.5 mr²ω²)

    v = rω

    So, T = 0.5 (mv² + 0.5 mv²) = 0.75 mv²

    Rotational kinetic energy is

    R = 0.5 Iω² = 0.5 x 0.5 mr²ω²

    R = 0.25 mv²

    So, R / T = 0.25 / 0.75 = 1/3
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