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Circular Motion and carnival ride

  1. Nov 16, 2009 #1
    1. The problem statement, all variables and given/known data
    An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that a person inside is stuck to the wall and does not slide down when the floor drops away. The acceleration of gravity is 9.8m/s^2.
    Given g=9.8 m/s^2, the coefficient of static friction between the person and the wall = 0.337. The radius of the cylinder, R=5.8m. For simplicity, neglect the person's depth and assume he or she is just a physical point on the wall. The person's speed is v= (2{pi}R)/T where T is the rotation period of the cylinder (the time to complete a full circle).
    Find the maximum rotation period T of the cylinder which would prevent a 47kg person from falling down. Answer in units of seconds.

    *For clarity this ride is one that spins and the person is pressed against the wall*
    -attached is an image of the ride-



    2. Relevant equations

    v= (2{pi}R)/T
    Ff=(coefficient of friction)(Fn)
    Ff=m*((v^2)/R)


    3. The attempt at a solution
    M=coefficient of Friction
    m=mass

    Ff=M(Fn)
    Ff=(0.337)(460.6)
    Ff=155.22

    Ff=m*((v^2)/R)
    155.22=47*v^2/5.8
    v=4.3766m/s^2

    v= (2{pi}R)/T
    4.3766=2{pi}(5.8)/T
    T=8.326 seconds

    -this answer is wrong-
     

    Attached Files:

    Last edited: Nov 16, 2009
  2. jcsd
  3. Nov 16, 2009 #2

    mgb_phys

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    Almost - you are trying to find what normal force is needed so that the frictional force is equal to their weight.
     
  4. Nov 16, 2009 #3
    Thanks for the reply.

    However, I am still confused as to what I need to do.

    Does Ff=47?
     
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