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26 August, 13:43

A machinist turns the power on to a grinding wheel, at rest, at time t = 0 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 56 rad/s. The wheel is run at that angular velocity for 21 s and then power is shut off. The wheel slows down uniformly at 1.7 rad/s2 until the wheel stops. In this situation, the time interval of deceleration is closest to:

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  1. 26 August, 13:45
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    t = 32.94 sec

    Explanation:

    From 1st equation of motion, we know that:

    Vf = Vi + at

    So, for angular motion it will become:

    ωf = ωi + at

    ωf - ωi = αt

    t = (ωf - ωi) / α

    where,

    ωi = initial angular velocity = 56 rad/s

    ωf = final angular velocity = 0 rad/s (since, wheel finally stops)

    α = deceleration = - 1.7 rad/s²

    t = time interval of deceleration = ?

    Therefore,

    t = (0 rad/s - 56 rad/s) / (-1.7 rad/s²)

    t = 32.94 sec
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