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Amusement park

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known data

    In an amusement park ride called The Roundup, passengers stand inside a 17.0m -diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane.

    Suppose the ring rotates once every 4.80s . If a rider's mass is 52.0kg , with how much force does the ring push on her at the top of the ride?

    2. Relevant equations
    1.v=2*pi*r/T
    2.Fr=mg+n=Fnet=m*Ac=m*v^2/r

    3. The attempt at a solution

    What I do is T=4.80 m=52.0kg

    v=formula#1=11.126m/s
    Ac= formula#2=14.56m/s^2
    Fr=m*Ac= 757,12N

    The answer is wrong but I don't know why....
     
  2. jcsd
  3. Mar 23, 2010 #2

    rock.freak667

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    The centripetal force, you are finding, is the resultant force on the rider. At the top of the ride, what are the forces acting and their directions?
     
  4. Mar 23, 2010 #3
    n and Fg pointing downwards?
     
  5. Mar 23, 2010 #4

    rock.freak667

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    Right, I think the question wants you to find n. Your second equation n+Fg=Fr, find n.
     
  6. Mar 23, 2010 #5
    OO.. yea thx a lot :D
     
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