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23 September, 05:00

A puck rests on a horizontal frictionless plane. A string is wound around the puck and pulled on with constant force. What fraction of the disk's total kinetic energy is due to the rotation? KErot/KEtot

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  1. 23 September, 07:22
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    Let v be the linear velocity, ω be the angular velocity and I be the moment of inertia of the the puck.

    Kinetic energy (linear) = 1/2 mv²

    Rotational kinetic energy = 1/2 I ω²

    I = 1/2 m r² (m and r be the mass and radius of the puck)

    Rotational kinetic energy = 1/2 x1/2 m r² ω²

    = 1/4 m v² (v = r ω)

    Total energy

    = Kinetic energy (linear) + Rotational kinetic energy

    = 1/2 mv² + 1/4 m v²

    = 3/4 mv²

    rotational K E / Total K E = 1/4 m v² / 3/4 mv²

    = 1 / 3

    So 1 / 3 rd of total energy is rotational K E.
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