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Finding the force on an amusement park ride

  1. Oct 20, 2008 #1
    1. The problem statement, all variables and given/known data
    In an amusement park ride called The Roundup, passengers stand inside a 16-m diameter ring. After the ring has acquired sufficient speed, it tilts into a vertical plane (figure shown in book).
    A Suppose the ring rotates once every 4.5 s. If a rider's mass is 55 kg, with how much force does the ring push on her at the top of the ride? At the bottom?
    B. What is the longest rotation period of the wheel that will prevent the riders from falling off at the top?

    2. Relevant equations
    F = ma = mv^2/r
    v = 2pi*r/t
    d = 16, therefore r = 8
    t = 4.5 s
    m = 55 kg

    3. The attempt at a solution
    v = 2([tex]\Pi[/tex](8)/4.5 = 11.17 m/s
    I think that in itself is correct.

    F = 55(11.17^2)/8 = 857.8 N, but this is the wrong answer.

    I'm just wondering where I am going wrong.
  2. jcsd
  3. Oct 20, 2008 #2


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    Homework Helper

    For a) you calculated the radial acceleration OK but didn't subtract the weight m*g to get your total force.

    For b) if you subtract at the top, then you add at the bottom.

    Finally, at what speed will the force at the top balance to 0? When is the speed such that m*g = mv2/r
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