A disk, with a radius of 0.25 m, is to be rotated like a merry-go-round through 800 rad, starting from rest, gaining angular speed at the co

Question

A disk, with a radius of 0.25 m, is to be rotated like a merry-go-round through 800 rad, starting from rest, gaining angular speed at the constant rate through the first 400 rad and then losing angular speed at the constant rate – until it is again at rest. The magnitude of the centripetal acceleration of any portion of the disk is not to exceed 100 m/s2. (a) What is the least time required for the rotation

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Dulcie 6 days 2021-07-22T20:13:48+00:00 1 Answers 0 views 0

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    2021-07-22T20:15:34+00:00

    Answer:

    a) T_{min} = 80\,s

    Explanation:

    a) Let consider that disk accelerates and decelerates at constant rate. The expression for angular acceleration and deceleration are, respectively:

    Acceleration

    \alpha_{1} = \frac{\omega_{max}^{2}}{2\cdot (400\,rad)}

    Deceleration

    \alpha_{2} = -\frac{\omega_{max}^{2}}{2\cdot (400\,rad)}

    Since angular acceleration and deceleration have same magnitude but opposite sign. Let is find the maximum allowed angular speed from maximum allowed centripetal acceleration:

    a_{r,max} = \omega_{max}^{2}\cdot r

    \omega_{max} = \sqrt{\frac{a_{r,max}}{r} }

    \omega_{max} = \sqrt{\frac{100\,\frac{m}{s^{2}} }{0.25\,m} }

    \omega_{max} = 20\,\frac{rad}{s}

    Maximum magnitude of acceleration/deceleration is:

    \alpha = 0.5\,\frac{rad}{s^{2}}

    The least time require for rotation is:

    T_{min} = 2\cdot \left(\frac{20\,\frac{rad}{s} }{0.5\,\frac{rad}{s^{2}} }  \right)

    T_{min} = 80\,s

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