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Weightlessness; Ferris Wheel

  1. Nov 20, 2007 #1
    1. The problem statement, all variables and given/known data
    The apparent weight a person feels is the normal force that acts on the person. Suppose a Ferris Wheel has a diameter of 28.om and makes one revolution every 13.3 seconds. What is the ratio of a person's apparent weight to real weight a) at the top and b) at the bottom.

    2. Relevant equations

    3. The attempt at a solution
    First I found the circumference. C=[tex]\pi[/tex]196
    I can get the velocity by dividing that by 13.3.
    v=46.2971549 m/s

    So, I can then make my ratio comparison.
    At top: Fn/m=2143.426552/14
    At bottom: Fn/m=2143.426552/14+9.8

    1) Am I working the solution out correctly?
    2) How should I make my comparison? Should I leave it all in variables Fn/m=v^2/r? Leave it like I have it above? Or completely work it out Fn/m=153.1018966
  2. jcsd
  3. Nov 20, 2007 #2
    The person's weight will always be [tex]W=mg[/tex], top or bottom.

    However, the person will always be accelerating towards the center of the wheel, and therefore there will be a force counteracting that in the opposite direction. At the top of the wheel, it will be up. At the bottom, down.

    That force is [tex]ma=m\frac{v^2}{r}=m4\pi^2 \nu^2 r[/tex]
  4. Nov 20, 2007 #3
    Is your diameter 196m or 28.0m?
  5. Nov 20, 2007 #4
    28.0m. But, C=pir^2. So it'd be pi14^2.
  6. Nov 20, 2007 #5
    [tex]v=\frac{2\pi r}{t}[/tex]
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