A 0.0223 m diameter coin rolls up a 12.0◦ inclined plane. The coin starts with an initial angular speed of 49.3 rad/s and rolls in a straigh

Question

A 0.0223 m diameter coin rolls up a 12.0◦ inclined plane. The coin starts with an initial angular speed of 49.3 rad/s and rolls in a straight line without slipping. How much vertical height does it gain before it stops rolling? The acceleration due to gravity is 9.81 m/s 2 . Answer in units of m.

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Nguyệt Ánh 5 years 2021-08-31T09:16:34+00:00 1 Answers 80 views 0

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  1. minhkhang
    0
    2021-08-31T09:18:17+00:00
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    Answer:

    h = 0.0231 m

    Explanation:

    The movement of the coin is modelled after the Principle of Energy Conservation. The kinetic energy of the coin is the sum of the components associated with translation and rotation and there are no non-conservative forces. In addition, the coin starts moving at height of zero

    K_{rot} + K_{tr} = U_{g}

    \frac{1}{2} \cdot m_{coin} \cdot \omega^{2} \cdot R^{2} + \frac{1}{4} \cdot m_{coin} \cdot R^{2} \cdot \omega^{2} = m_{coin} \cdot g \cdot h\\

    \frac{3}{4} \cdot R^{2} \cdot \omega^{2} = g \cdot h

    The maximum vertical height is isolated in the previous equation:

    h = \frac{3 \cdot R^{2}\cdot \omega^{2}}{4\cdot g}

    h = \frac{3 \cdot (0.01115 m)^{2} \cdot (49.3 \frac{rad}{s} )^{2}}{4 \cdot (9.807 \frac{m}{s^{2}} )} \\h = 0.0231 m

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