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Calculate the coefficient of friction on a rotating carnival ride

  1. Oct 15, 2008 #1
    1. The problem statement, all variables and given/known data
    In a "rotor-ride" at a carnival, people rotate in a vertical cylindrically walled room. If the room radius was 5.5m, and the rotation frequency 0.50 revolutions per second when the floor drops out, what minimum coefficient of static friction keeps the people from slipping down?


    2. Relevant equations
    F(friction)</= usF(normal)
    F(normal)=mg
    us=g/(T*r)

    3. The attempt at a solution
    us=g/(T*r)
    us=9.8/(2*5.5)
    us=.891
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 15, 2008 #2
    Draw a free body Diagram of a person on the ride. The normal force between the person and the wall of the ride is what is providing the centripetal acceleration (directed towards the center of the ride).
    Force of Gravity acts down on the person and so friction acts up on the person to keep them from falling.

    Looking at the vertical components (of the forces) we know that Fg = Ff in order to keep the person from falling.
    Solve for normal force by finding the centripetal force required at the given speed.

    Once you know normal force and Force of Friction you can then use the following equation to solve for the coefficient of friction.
    Ff(Friction) = N(normal force) x Coefficient of Friction
     
  4. Oct 15, 2008 #3

    LowlyPion

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    Homework Helper

    Welcome to PF.

    I'm not sure where you are getting these equations. But that said I'm not seeing your answer in terms of π .

    w = 2*π*f = v/r
     
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