a 5.67g quarter sits on a record player a distance of 5cm from the axis of rotation. if the angular speed of the record player is 33 revolutions per minute, what should the coefficient of friction be to keep it from sliding?

Answer 1

33 *2 pi radians / 60 seconds = 3.46 radians/second = omega
Ac = omega^2 R = 3.46^2 radians^2/seecond^2 * 0.05 meters
= 0.597 meters/s^2
mu m g = m (.597)
mu = .597/g = about .06