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Ferris Wheel Trick Question?

  1. Nov 16, 2003 #1
    Ferris Wheel ...Trick Question???

    This is one of those that might be a typo or I am really missing something. A 250lb student is riding on a steadily rotating Ferris Wheel. If the student has an apparent weight of 250lb at the top what will be the students apparent weight at the bottom? What would be the apparent weight of the student at the top if the speed of the Ferris wheel were exactly tripled?
    If the weight is the same at the top as it normaly is then all the forces are equaling out so there wouldn't be a change in weight anywhere on the wheel. Right??? As for tripling the speed his weight would increase by a factor of 9 so it would be 2250lbs. Am I on the right track here? Thanks for the help
  2. jcsd
  3. Nov 16, 2003 #2
    Re: Ferris Wheel ...Trick Question???

    If his apparent weight (the magnitude of the normal force) at the top is equal to his true weight mg, then the Ferris wheel isn't moving: your apparent weight at the top of a rotating Ferris wheel is always less than your true weight (or negative).

    Perhaps they really intended to discuss the loop-the-loop situation where you're on the inside of a track (so the normal force is down at the top of the loop, instead of up in the Ferris wheel case).
  4. Nov 16, 2003 #3
    Are you sure with the question.Because student with original weight 250 lb will have weight less than 250lb at the top of the wheel provided the wheel move with constant speed.

    because the new weight becomes=250-m(v^2/r)
    m=mass of student
    v=speed of the wheel
    r=radius of the wheel

    it is possible to have the same weight if only when speed is zero.but the question says"steadily rotating Ferris Wheel".So I am questioning that part.
  5. Nov 16, 2003 #4
    It was a typo

    The weight of the student was 225 at the top. Having all the info makes the problem much easyer.
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