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Circular Motion Ferris Wheel

  1. Oct 11, 2006 #1
    I'm on the right track, but I'm stuck here....

    The radius of a Ferris wheel is 5 m and it makes one rev in 10 sec

    a Find the difference b/w the apparent weight of a passenger at the highest and lowest points, expressed as a fraction of his weight, W

    b What would the time for one rev be if teh apparent weight at the top were zero?
    c What would be the apparent weight at the low point??

    I have at the top that

    mv^2/r = mg - Fn

    and at the bottom

    mv^2/r = Fn -mg

    I really dont undestand what a is asking...(Fnbottom - Fntop)/ W ??
  2. jcsd
  3. Oct 11, 2006 #2


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    Yes you do. It's exactly what you said. Write your answer as Fnb - Fnt = _______W
  4. Oct 11, 2006 #3
    but im confused as if anything else would be needed...liek that blank befor e the W...would i need to expand any further or would i just keep it
    Fnbottom - Fntop = xW

    b would just be v^2/r = g? and then t = 2pir/v

    and c would be Fnb = mv^2/r + mg...but how would i eliminate m?
    Last edited: Oct 11, 2006
  5. Oct 11, 2006 #4


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    Solve your earlier top and bottom equations for Fn. Take the difference between the two. Your result will be of the form
    Fnb - Fnt = mA where A is a number that can be computed from the given information. You can do that. Once you have that form, multiply and divide by g
    mA = mgA/g = WA/g
    You can calculate A/g to express your answer as
    Fnb - Fnt = (A/g)W with A/g replaced by a number.
  6. Oct 11, 2006 #5
    i got A and B...but how would i manage part c?

    im not seein anyway to cancel out the mass...
  7. Oct 12, 2006 #6


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    What is your understanding of "apparent weight". All the forces in the problem are proportional to mass. It will cancel out.
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