Question

A penny is placed at the outer edge of a disk (radius = 0.134 m) that rotates about an axis perpendicular to the plane of the disk at its center. The period of the rotation is 1.86 s. Find the minimum coefficient of friction necessary to allow the penny to rotate along with the disk.

Answer 1

When the penny is rotating with the disk (and not sliding relative to it), it is the static frictional force that provides the centripetal force required to keep the penny moving on a circular path.

here maximim static friction force = fsmax = mu*N

mu = fsmax/N

Since the maximum centripetal force that can act on the penny is the maximum static frictional force

Fc = fsmax

Fc = mv^2/r

mu = (mv^2/r)/N

v = 2pi*r/T

since the penny does not accelerate in the vertical direction, the upward normal force must be balanced by the downward-pointing weight so N = mg  

mu = m(2pi*r/T)^2/r /mg

mu = 4*pi^2*r/gT^2

given : T = 1.86 s

r = 0.134 m

g = 9.8 m/s^2

mu = 0.156