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10 July, 18:41

A coin is placed on a large disk which rotates uniformly at a rate of 1 rot/s. The coefficient of friction between the coin and the disk is 0.1. What is the maximum distance from the center of the disk that the coin could be placed without slipping? Report your answer in meters.

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  1. 10 July, 19:37
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    r = 0.02 m

    Explanation:

    from the question we have:

    speed = 1 rps = 1x 60 = 60 rpm

    coefficient of friction (μ) = 0.1

    acceleration due to gravity (g) = 9.8 m/s^{2}

    maximum distance without falling off (r) = ?

    to get how far from the center of the disk the coin can be placed without having to slip off we equate the formula for the centrifugal force with the frictional force on the turntable force

    mv^2 / r = m x g x μ

    v^2 / r = g x μ ... equation 1

    where

    velocity (v) = angular speed (rads/seconds) x radius

    angular speed (rads/seconds) = (/frac{2π}{60}) x rpm

    angular speed (rads/seconds) = (/frac{2 x π}{60}) x 60 = 6.28 rads / seconds

    now

    velocity = 6.28 x r = 6.28 r

    now substituting the value of velocity into equation 1

    v^2 / r = g x μ

    (6.28r) ^2 / r = 9.8 x 0.1

    39.5 x r = 0.98

    r = 0.02 m
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