A disk revolves in a horizontal plane at a steady speed of 3rev/s. A coin will remain on the disk if it is kept a distance of 2cm from the axis of rotation. What is the coefficient of friction between the coin and the disk?

The force of friction is holding the coin on the disk while the centripetal force is opposing it, trying to push the coin to its edge. The forces can be viewed to be in , that is, balanced so that the coin remains at rest relative to the center of the disk.

From the angular speed given (3 revolutions per second) we can determine the linear speed for the coin at that radius from the center of the disk, using the equation ##v=rw## , where ##v## is the linear speed, ##r## is the radius, and ##w## is the angular speed. Just make sure you convert the angular speed to rad/s (standard unit) before you calculate it. Since one revolution would be ##2pi## radians, the angular speed is ##6pi## rad/s, which simplifies to ##18.85## rad/s. Using the equation, the product of this angular speed and the radius of the disk (in meters, again standard unit) is ##0.377## m/s, which is the linear speed of the coin.

The reason you need to know the linear speed of the coin is so that you can calculate the centripetal force acting it. Centripetal force can be calculated using the equation ##F=(mv^2)/r## , where ##m## is the mass of the coin, ##v## is the linear speed of the coin (which we just found above) and ##r## is the radius at which the coin is located on the disk (given). Don't worry about the mass, it's going to cancel in the end.

The equation for friction is ##mu * F_n##. ##mu## is the static coefficient of friction, (which you are trying to find!) and ##F_n## is the normal force, which on a level surface like this is just equal to the force of gravity. Using Newton's 2nd Law of Motion (##F=ma##), the normal force is just the product of mass of the coin and the of gravity (##9.81## m/s/s near the Earth's surface), or mg. Again, mass is going to cancel in the end, so don't sweat that it wasn't given. But overall, the equation for friction, after subsituting, is ##F_f=mu * mg##.

Because friction and centripetal force are equal and opposite (Newton's 3rd Law), you can set these forces equal to each other with the equations we've developed. ##F_f = F_c -> mu * mg = (mv^2)/r##. See, the masses cancel and you're left with ##mu * g = v^2/r##. A little algebra from there, and ##mu = v^2/gr##.

So, the coefficient of friction turns out to be the linear velocity of the coin (which we calculated to be ##0.377## m/s) squared, divided by ##9.81## m/s/s (for gravity) and ##0.02## m (the radius). I get ##0.724## for the coefficient which, remember, is a unit-less constant. You can think of it as a number on a scale that tells you have "sticky" that particular surface is. Anything between 0 and 1 is pretty common, so our answer makes good physical sense.