The center of mass of a pitched baseball or radius 2.42 cm moves at 23.3 m/s. The ball spins about an axis through its center of mass with a

Question

The center of mass of a pitched baseball or radius 2.42 cm moves at 23.3 m/s. The ball spins about an axis through its center of mass with an angular speed of 158 rad/s. Calculate the ratio of the rotational energy to the translational kinetic energy. Treat the ball as a uniform sphere.

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Doris 3 months 2021-07-22T00:33:59+00:00 1 Answers 2 views 0

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    2021-07-22T00:35:49+00:00

    To solve the problem, it will be necessary to define the rotational and translational kinetic energy in order to determine the relationship between the two. Rotational energy is defined as,

    KE_{Rotational} = \frac{1}{2} I\omega^2

    Here,

    I = Moment of Inertia

    \omega = Angular velocity

    Now the translational energy will be,

    KE_{Translational} = \frac{1}{2} mv^2

    Here,

    m = Mass

    v = Velocity

    Therefore the relation between them will be,

    \frac{KE_{Rotational} }{KE_{Translational}} = \frac{\frac{1}{2} I\omega^2 }{\frac{1}{2} mv^2 }

    Applying the moment of inertia of a sphere we have,

    \frac{KE_{Rotational} }{KE_{Translational}} = \frac{\frac{1}{2} (\frac{2}{5}mr^2)\omega^2 }{\frac{1}{2} mv^2 }

    \frac{KE_{Rotational} }{KE_{Translational}} = \frac{2}{5} \frac{r^2\omega^2}{v^2}

    \frac{KE_{Rotational} }{KE_{Translational}} = \frac{2}{5} \frac{(2.42*10^{-2})^2(158)^2}{23.3^2}

    \frac{KE_{Rotational} }{KE_{Translational}} =  0.01077

    Therefore the ratio will be 0.01077

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