A wheel rotates about a fixed axis with an initial angular velocity of 13 rad/s. During a 8-s interval the angular velocity increases to 57

Question

A wheel rotates about a fixed axis with an initial angular velocity of 13 rad/s. During a 8-s interval the angular velocity increases to 57 rad/s. Assume that the angular acceleration was constant during this time interval. How many revolutions does the wheel turn through during this time interval

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Acacia 2 weeks 2021-07-16T06:43:07+00:00 1 Answers 0 views 0

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    2021-07-16T06:44:46+00:00

    Answer:

    The number of revolutions is 44.6.

    Explanation:

    We can find the revolutions of the wheel with the following equation:

    \theta = \omega_{0}t + \frac{1}{2}\alpha t^{2}

    Where:

    \omega_{0}: is the initial angular velocity = 13 rad/s              

    t: is the time = 8 s

    α: is the angular acceleration

    We can find the angular acceleration with the initial and final angular velocities:

     \omega_{f} = \omega_{0} + \alpha t

    Where:

     \omega_{f} : is the final angular velocity = 57 rad/s

     \alpha = \frac{\omega_{f} - \omega_{0}}{t} = \frac{57 rad/s - 13 rad/s}{8 s} = 5.5 rad/s^{2}

    Hence, the number of revolutions is:

     \theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 13 rad/s*8 s + \frac{1}{2}*5.5 rad/s^{2}*(8 s)^{2} = 280 rad*\frac{1 rev}{2\pi rad} = 44.6 rev

    Therefore, the number of revolutions is 44.6.

           

    I hope it helps you!

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