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Coefficient of static friction problem

  1. Sep 21, 2004 #1
    Hey guys,

    I was wondering if you could help me out with the following problem...

    In a "Rotor-ride" at a carnival, people pay money to be rotated in a vertical cylindrically walled "room". If the room radius is 5.0 m and the rotation frequency is .60 revoloutions per second when the floor drops out, what is the minimum coefficient of static friction so that people will not slip down.

    I keep getting stuck in the same place...I have calculated the velocity using the formula for circular motion that states v=(2*pi*r)/T, getting (10*pi)/2 and could easily get the acceleration of the ride by squaring the velocity and dividing by the radius..but I'm not 100% sure what to do after that. I know I need an equation in which the mass will cancel out on both sides, but I'm not sure how to tie the friction coefficient into the problem. :frown:

    Any help would be greatly appreciated

    Thanks alot,

  2. jcsd
  3. Sep 21, 2004 #2


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    Homework Helper

    Remember in the cyclone the cetripetal force will be the normal force, (forces on x-axis) and on the y-axis we got the friction force pointin up and the weight pointing down.

    Also a hint [tex] F_{f} = \mu N [/tex]
  4. Oct 19, 2004 #3
    perhaps the equation you are looking for is

    u (coeffi of static friction)= [(4pi^2)(f^2)(r)]/g
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