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30 January, 16:56

A grinding wheel is in the form of a uniform solid disk of radius 7.00 cm and mass 2.00 kg. it starts from rest and accelerates uniformly under the action of the constant torque of 0.600 n? m that the motor exerts on the wheel. how long does the wheel take to reach its final operating speed of 1200 rev / min? (note that moment of inertia of a disk is given by i = 0.5*mr2)

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  1. 30 January, 17:13
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    It will take 1.03 sec for the wheel to reach its final operating speed.

    Solution:

    First we calculate for the angular acceleration a from the torque formula which is the product of the moment of inertia i and the angular acceleration a:

    torque = ia

    a = torque / i

    We know that the moment of inertia i of a solid disk is i = 0.5*mr^2, where radius r is in meters:

    a = torque / 0.5 * mr^2

    Now, we substitute the values to the angular acceleration a equation:

    a = 0.600 Nm / (0.5 * 2.00 kg * (0.07m) ^2)

    = 122.45 radians/s^2

    We convert the final angular velocity Wf, which is the final operating speed 1200 rev/min, to rad/s:

    Wf = 1200 rev/min * 2π/60 = 125.66 rad/s

    For rotational motion,

    Wf = Wi + at

    125.66 rad/s = 0 + (122.45 rad/s^2) t

    The time t is therefore

    t = 125.66 / 122.45t

    = 1.03 sec
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