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18 October, 09:49

A coin is placed on a turntable that is rotating at 33.3 rpm. If the coefficient of static friction between the coin and the turntable is 0.3406, how far from the center of the turntable can the coin be placed without having it slip off?

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  1. 18 October, 10:02
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    Answer: 3.65 m

    Explanation:

    from the question we have

    angular speed = 33.3 rpm

    coefficient of static friction (μ) = 0.3406

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

    to get how far from the center of the turntable can 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

    we have to convert the angular speed from rpm to rads/seconds

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

    angular speed (rads/seconds) = ((2 x π) / 60) x 33.3 = 3.49 rads / seconds

    now

    velocity = 3.49 x r = 3.49r

    now substituting the value of velocity into equation 1

    v^2 / r = g x μ

    (3.49r) ^2 / r = 9.8 x 0.3406

    12.18 x r = 3.34

    r = 3.65 m
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