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28 July, 07:35

A CD-ROM drive in a computer spins the 12-cm-diameter disks at 9500 rpm. Find acceleration in units of g that a speck of dust on the outside edge of the disk experiences

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  1. 28 July, 10:23
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    The net force on an object is:

    F = ma

    The only force acting on the speck of dust as it is lays on the CD is centripetal force given by:

    F = (mv²) / r

    Equating the two

    ma = (mv²) / r

    We get:

    a = v²/r

    v is the linear velocity in this case; however, we can calculate only the angular velocity with the given data. Therefore, we must use:

    v = ωr; where

    ω = 2πf; f is the rotations per second

    f = 9500 / 60

    f = 950/6

    ω = 2π (950/6)

    ω = (950π) / 3

    v = (950π) / 3 * 0.12

    v = 38π

    a = (38π) ²/0.12

    a = 1.2 * 10⁵ m/s²

    To express this in terms of gravitational acceleration, we divide by the value of gravitational acceleration, 9.81

    a = 1.2 * 10⁴ g
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