A ball of mass m and radius R is both sliding and spinning on a horizontal surface so that its rotational kinetic energy equals its translat

Question

A ball of mass m and radius R is both sliding and spinning on a horizontal surface so that its rotational kinetic energy equals its translational kinetic energy. What is the ratio of the ball’s center-ofmass speed to the speed due to rotation only of a point on the ball’s surface? The moment of inertia of the ball is 0.63 m R2 .

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Dâu 4 years 2021-08-17T07:24:43+00:00 1 Answers 47 views 0

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  1. vankhanh
    0
    2021-08-17T07:26:33+00:00
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    Answer:

    \frac{v_{cm}}{\omega} = 1.122\cdot R

    Explanation:

    According to the statement of the problems, the following identity exists:

    K_{t} = K_{r}

    \frac{1}{2}\cdot m \cdot v_{cm}^{2} = 0.63\cdot m \cdot R^{2} \cdot \omega^{2}

    After some algebraic handling, the ratio is obtained:

    \frac{v_{cm}^{2}}{\omega^{2}}=1.26\cdot R^{2}

    \frac{v_{cm}}{\omega} = 1.122\cdot R

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