The flywheel of an engine has I of 1.60kg.m2 about its rotation axis. What constant torque is required to bring it up to an angular speed of

Question

The flywheel of an engine has I of 1.60kg.m2 about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 rpm in 8.00s, starting from rest?

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Orla Orla 4 years 2021-08-12T22:38:40+00:00 1 Answers 34 views 0

Answers ( )

  1. Euphemia
    0
    2021-08-12T22:40:29+00:00
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    Answer:

    Torque = 8.38Nm

    Explanation:

    Time= 8.00s

    angular speed (w) =400 rpm

    Moment of inertia (I)= 1.60kg.m2 about its rotation axis

    We need to convert the angular speed from rpm to rad/ sec for consistency

    2PI/60*n = 0.1047*409 = 41.8876 rad/sec

    What constant torque is required to bring it up to an angular speed of 40rev/min in a time of 8s , starting from rest?

    Then we need to use the formula below for our torque calculation

    from basic equation T = J*dω/dt …we get

    Where : t= time in seconds

    W= angular velocity

    T = J*Δω/Δt = 1.60*41.8876/8.0 = 8.38 Nm

    Therefore, constant torque that is required is 8.38 Nm

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