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A solid sphere (mass 9.14kg], radius 88 cm) is rolling without slipping on a horizontal table. What fraction of its total kinetic energy is translational kinetic energy?

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  • Whome
    Lv 7
    2 months ago
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  • 2 months ago

    Let the velocity of the solid sphere is v m/s

    =>Angular velocity (ω)  = v/r

    Now KE(translational) = 1/2mv^2 = 0.5mv^2------------(i)

    & KE (rotational) = 1/2Iω^2

    {As I = 2/5mr^2 for solid sphere}

    =>KE(rotational) = 1/2 x 2/5 x mr^2 x (v/r)^2

    =>KE(rotational) = 1/5mv^2 = 0.20mv^2--------------(ii)

    Thus KE (Total) = KE(translational) + KE (rotational)

    =>KE(Total) = 0.5mv^2 + 0.2mv^2

    =>KE(translational)= 0.7mv^2 -----------------------(iii)

    Thus fraction of its total kinetic energy is translational kinetic energy by  =>KE(translational)/KE(Total) = 0.5mv^2)/(0.7mv^2)

    =>KE(translational)/KE(Total) = 0.71

    =>KE(translational) = 0.71 x KE(Total)

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