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6 August, 11:43

cylinder of mass 6.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed of 15.0 m/s. (a) Determine the translational kinetic energy of its center of mass. J (b) Determine the rotational kinetic energy about its center of mass. J (c) Determine its total energy.

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  1. 6 August, 15:37
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    a). 675J

    b). 337.5J

    c). 1012.5J

    Explanation:

    M = 6.0kg

    V = 15.0m/s

    a). Translational energy

    E = ½ * mv²

    E = ½ * 6 * 15²

    E = 675J

    b). Rotational kinetic energy K. E (rot) = Iw²

    But moment of inertia of a cylinder (I) = ½Mr²

    I = ½mr²

    V = wr, r = v / w

    K. E (rot) = ¼ mv²

    K. E (rot) = ¼ * 6 * 15²

    K. E (rot) = 337.5J

    Total energy of the system = K. E (rot) + Translational energy = 337.5 + 675

    T. E = 1012.5J
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