Consider a basketball player spinning a ball on the tip of a finger. If a player performs 2.01J of work to set the ball spinning from rest,

Question

Consider a basketball player spinning a ball on the tip of a finger. If a player performs 2.01J of work to set the ball spinning from rest, at what angular speed ω will the ball rotate? Model a basketball as a thin-walled hollow sphere. For a men’s basketball, the ball has a circumference of 0.749 m and a mass of 0.624 kg.

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Verity 4 years 2021-07-20T12:52:58+00:00 1 Answers 16 views 0

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  1. Neala
    0
    2021-07-20T12:53:58+00:00
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    Answer:

    The angular speed of ball is 25.9 rad/s .

    Explanation:

    Given :

    Work done by the player , W = 2.01 J .

    Mass of hollow spherical ball , m = 0.624 kg .

    Circumference of hollow spherical ball , C = 0.749 m .

    Therefore , its radius is ,

    r=\dfrac{C}{2\pi}\\\\r=\dfrac{0.749 }{2\pi}\\\\r=0.12\ m

    Now , this work done must be equal to the rotational energy of the ball .

    We know ,

    U=\dfrac{I \omega^2}{2}

    Therefore ,

    \omega=\sqrt{\dfrac{2U}{ I }}\\\\\omega=\sqrt{\dfrac{2U}{ \dfrac{2MR^2 }{3} }}\\\\\omega=\sqrt{\dfrac{3U}{ MR^2}}\\\\\omega=\sqrt{\dfrac{3\times 2.01}{0.624 \times 0.12^2}}\\\\\omega=25.9\ rad/s

    Hence , this is the required solution .

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