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A penny is placed at the outer edge of a disk (radius = 0.147 m) that rotates about an axis perpendicular to the plane o...

A penny is placed at the outer edge of a disk (radius = 0.147 m) that rotates about an axis perpendicular to the plane of the disk at its center. The period of the rotation is 1.70 s. Find the minimum coefficient of friction necessary to allow the penny to rotate along with the disk.
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Answer #1

    Radius, R = 0.147 m

    Time-period of rotation, T = 1.7 s

    Angular velocity, ω = 2 π / T

                                 = ( 2 * 3.14 / 1.7 )

                                 = 3.69 rad/s

    Coefficient of friction, μ = ?

    Frictional force = Centripetal force

    μ M g = M ω 2R       ( where M = Mass of the penny)

    μ g = ω 2R    

    Coefficient of friction, μ = ω2 R / g

                                        = 3.69 2 * 0.147 / 9.8

                                        = 0.205

                                        = 0.21

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