A coin is placed on a record that is rotating at 33.3 rpm. If the coefficient of static friction between the c

A coin is placed on a record that is rotating at 33.3 rpm. If the coefficient of static friction between the coin and the record is 0.1, how far from the center of the record can the coin be placed without having it slip off

A coin is placed on a record that is rotating at 33.3 rpm. If the coefficient of static friction between the c
Firstly as soon as coin gets centripetal force it will move inwards. Therefore mv^2 / r must be zero. However there is also frictional force acting.


Thus making the equation like following:





mv^2 / r - Frictional Force = 0


mv^2 / r - static friction*Normal Force = 0


mv^2 / r - static friction*mgcos@ = 0





Remember: v = Angular Velocity*r and also since the object is at 0 degress, cos@ = cos 0=1


If we make r (distance from center) the subject it will look like following:


r = Static Friction / (Angular Velocity)^2


r = Static Friction/ (2pi*frequency)^2


r = 0.1/ (2*pi*33.3)^2





Hence: r= 2.28 * 10^-6 meters
Reply:You want the REAL answer to that question done with real coins?