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Circular Motion, Velocity at top/bottom question

  1. Oct 4, 2008 #1
    1. The problem statement, all variables and given/known data

    In an amusement park ride, passengers stand inside a 16-m-diameter roating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane.
    Suppose the ring rotates once every 4.5 seconds. If a riders mass is 55 kg, with how much force does the ring push on her at the top of the ride? At the bottom?

    2. Relevant equations

    3. The attempt at a solution

    First I solved for velocity:

    8 m radius * 2pi = 16pi m circumference
    16pi m / 4.5 s = 11.17 m/s

    Then acceleration:
    a=(v^2)/r = (124.77)/8= 15.6 m/s^2

    Sum of forces in radial direction at top = n + mg = ma = 55 kg * 15.6 m/s^2 = 857.8 N
    n + mg = 857.8 N

    If I follow this logic, I would get the same answer for the sum of forces at the bottom.
    My book gives the equation n=mg + [m(v_top)^2]/r and n= [m(v_bottom)^2] - mg .
    I understand this, because at the top, the normal and gravitational force point in the same direction, and at the bottom they point in different directions. The problem is, I don't know how to computer the velocity at the top or at the bottom. My book also gives the equation v_c = sqrt(rg) which is the slowest speed at which the object can complete rotation, but I don't think this ride is travelling near that speed.

    Can anyone tell me how to find the velocity at the top or bottom of circular motion? I'm pretty sure this is uniform, but I'm not sure.
  2. jcsd
  3. Oct 4, 2008 #2


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    Science Advisor
    Homework Helper

    Hi swooshfactory! :smile:

    Yes it is uniform … the question makes that clear.

    It's a powered ride. :smile:
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