The flywheel of an engine has moment of inertia 2.50 kg m2 about its rotation axis. What constant torque is required to bring it up to an an

Question

The flywheel of an engine has moment of inertia 2.50 kg m2 about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 rev/min in 8.00s, starting from rest?

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Thu Thảo 5 years 2021-08-06T04:29:48+00:00 1 Answers 106 views 0

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  1. Tryphena
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    2021-08-06T04:31:02+00:00
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    Answer:

    Explanation:

    From the question we are told that

       The moment of inertia is  I = 2.50 \ kg \cdot m^2

        The final  angular speed is w_f = 400 rev/min = \frac{400 * 2\pi}{60} = 41.89 \ rad/s

         The time taken is  t = 8.0 s

          The initial angular speed is  w_i = 0\ rad/s

    Generally the average angular acceleration is mathematically represented as

            \alpha = \frac{w_f - w_i }{t}

    =>     \alpha = \frac{41.89}{8}

    =>      \alpha = 5.24 \ rad/s^2

    Generally the torque is mathematically represented as

       \tau = I * \alpha

    =>    \tau = 5.24 * 2.50

    =>     \tau = 13.09 \ N \cdot m

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