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Weightlessness on a Ferris Wheel

  1. Oct 8, 2009 #1
    1. The problem statement, all variables and given/known data
    How many revolutions per minute would a 15.1 m diameter Ferris wheel need to make for the passengers to feel "weightless" at the topmost point of the trip?


    2. Relevant equations



    3. The attempt at a solution
    I assume this problem has something to do with finding acceleration and converting from there. I have been having a lot of trouble with this problem.
     
  2. jcsd
  3. Oct 8, 2009 #2

    rl.bhat

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    What are the things you know about vertical circular motion?
     
  4. Oct 8, 2009 #3
    I know that Centripetal acceleration = (velocity)^2/radius
     
  5. Oct 8, 2009 #4
    If you're weightless at the top, what does that mean? Draw your free body diagram, then figure out which forces are acting where.
     
  6. Oct 8, 2009 #5

    rl.bhat

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    Can write this in terms of angular velocity?
    What happens at the top of the wheel?
     
  7. Oct 8, 2009 #6
    If you are weightless at the top you should have mass * gravity acting down and a centripetal force equal to gravity acting up. But how does this information help to discover revolutions per minute?
     
  8. Oct 8, 2009 #7
    To set up the equation would I use the following:

    9.80=V^2/7.55 and solve for velocity? But how do you convert velocity to revolutions per minute?
     
  9. Oct 8, 2009 #8
    I found the answer! Thank you for all the help.
     
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