A spherical shell of radius 3.84 cm and a sphere of radius 7.72 cm are rolling without slipping along the same floor. The two objects have t

Question

A spherical shell of radius 3.84 cm and a sphere of radius 7.72 cm are rolling without slipping along the same floor. The two objects have the same mass. If they are to have the same total kinetic energy, what should the ratio of the spherical shell’s angular speed to the sphere’s angular speed be?

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Tài Đức 3 years 2021-07-29T19:43:34+00:00 1 Answers 31 views 0

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  1. Edana
    0
    2021-07-29T19:45:12+00:00
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    Answer:

    \frac{\omega_{SS}}{\omega_{S}} \approx 1.698

    Explanation:

    Let assume that both the spherical shell and the sphere are rigid bodies. The kinetic energy due to rolling is:

    K = \frac{1}{2}\cdot I_{g}\cdot \omega^{2}

    The moments of inertial for the spherical shell and the sphere are, respectively:

    Spherical Shell

    I_{g, SS} = \frac{2}{3}\cdot m \cdot r_{SS}^{2}

    I_{g,SS} = 9.830\cdot m \times 10^{-4}

    Sphere

    I_{g,S} = \frac{2}{5}\cdot m \cdot r_{S}^{2}

    I_{g,S} = 2.384\cdot m \times 10^{-3}

    Given that both rigid bodies have the same kinetic energy and the same mass, then:

    \frac{1}{2}\cdot (9.830\cdot m\times 10^{-4})\cdot \omega_{SS}^{2} = \frac{1}{2}\cdot (2.834\cdot m\times 10^{-3})\cdot \omega_{S}^{2}

    (9.830\cdot m\times 10^{-4})\cdot \omega_{SS}^{2} = (2.834\cdot m\times 10^{-3})\cdot \omega_{S}^{2}

    \frac{\omega_{SS}^{2}}{\omega_{S}^{2}} = 2.883

    The ratio of the spherical shell’s angular speed to the sphere’s angular spped is:

    \frac{\omega_{SS}}{\omega_{S}} \approx 1.698

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