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3 July, 04:07

A computer hard disk 8.0 cm in diameter is initially at rest. A small dot is painted on the edge of the disk. The disk accelerates at 610 rad/s2 for 12s, then coasts at a steady angular velocity for another 12s.

A. What is the speed of the dot at t = 1.0 s? 12m/s

B. Through how many revolutions has it turned?

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  1. 3 July, 04:19
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    a) 282.8 m/s

    b) 13812 revolutions

    Explanation:

    a) The aceleration (a) is the angular velocity (w) divide by the time. The disk had cosated for 12 s in a constant angular velocity, so it will be:

    a = w/t

    610 = w/12

    w = 7230 rad/s

    The angular velocity is related by the linear velocity:

    V = w*r, where r is the radius of the circumference (r = 0.08/2 = 0.04 m)

    V = 7230*0.04 = 292.8 m/s

    b) The number of revolutions is the frequency (f) of the moviment, which is related to the angular velocity as:

    w = 2π*f

    f = w/2π

    f = 7230/6.28

    f = 1151 revolutions/s

    So, the total revolutions (n) is 12 s was:

    n = 12*1151 = 13812 revolutions
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