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)

Question

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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Gia Bảo 5 years 2021-07-14T04:58:01+00:00 1 Answers 17 views 0

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    2021-07-14T04:59:30+00:00

    Answer:

    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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