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1 June, 10:02

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 38 rad/s. The wheel is run at that angular velocity for 30 s and then power is shut off. The wheel slows down uniformly at 2.1 rad/s2 until the wheel stops. In this situation, the angular acceleration of the wheel between t = 0 s and t = 10 s is

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  1. 1 June, 10:20
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    The angular acceleration of the grinding wheel at the given time interval is 3.8 rad/s^2

    Explanation:

    Given:

    Initial time of the grinding wheel, t1 = 0

    final time of the grinding wheel, t2 = 10 s

    Initial angular velocity of the grinding wheel, u = 0

    final angular velocity of the grinding wheel, v = 38 rad/s

    The angular acceleration of the wheel between t = 0 s and t = 10 s, is calculated as;

    a = dv/dt

    Where;

    a is angular acceleration

    dv is change in angular velocity = v - u = 38 - 0 = 38 rad/s

    dt is change in time = t2 - t1 = 10 - 0 = 10

    a = 38 / 10

    a = 3.8 rad/s^2

    Therefore, the angular acceleration of the grinding wheel at the given time interval is 3.8 rad/s^2
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