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Centripetal Motion!

  1. Jul 30, 2011 #1
    1. The problem statement, all variables and given/known data

    At amusement parks, there is a popular ride where the floor of a rotating cylindrical room falls away, leaving the backs of the riders "plastered against the wall. Suppose the radius of the room is 3.30 m and the speed of the wall is 10.0 m/s when the floor falls away. What is the minimum coefficient of static friction that must exist between a rider's back and the wall, if the rider is to remain in place when the floor drops away if the rider has a mass of 55 kg?

    2. Relevant equations

    Centripetal Force = mv^2/r = 1667

    friction = u*Fn

    3. The attempt at a solution

    centripetal force = friction = coefficient of friction * Normal Force

    Normal force = mg (weight)

    coefficient of friction = centripetal force/Normal Force = 51.5???

    Is this correct?
  2. jcsd
  3. Jul 30, 2011 #2
    You're missing a force. Friction will be resisting the motion of the person from falling, but Gravity is the one doing the pulling, you need to consider that in your force calculations
  4. Jul 30, 2011 #3
    That is incorrect. The centripetal force is equal to the normal force, not the frictional force.
    This too is incorrect. The weight of the person should be equal to the frictional force.
    Now try finding the coefficient of static friction.
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