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Effective weight Problem

  1. Oct 16, 2012 #1
    1. The problem statement, all variables and given/known data
    On a carnival ride, a girl of mass m stands in a cylindrical cage of radius R. The cage is spun about its cylindrical axis so that her speed is v. What is her effective weight?


    2. Relevant equations
    we= mg-ma

    wxe = mv/R2, wey = -mg


    3. The attempt at a solution
    This is an example problem in my book and I'm not sure how they put everything together to solve it. Upon presenting the above forumulas, it says to next use the Pythagorean theorem for it to get:

    we = m √g2 + (v2/R)2

    Then goes on to say that the girl's effective weight is greater than mg.
    I'm not sure how they set up their formula into the Pythagorean formula from the formulas above. Any clarification would be appreciated.
     
  2. jcsd
  3. Oct 16, 2012 #2

    BruceW

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    mg-ma is only the effective weight when the acceleration and gravity are in the same direction (or with sign of 'a' reversed for opposite directions). But in this problem, gravity and acceleration are not parallel or anti-parallel.

    Think about the centripetal and gravitational forces. Which directions will they be in? From this, you can see why they use Pythagoras' formula.
     
  4. Oct 17, 2012 #3
    Oh! Ok, that makes more sense, I didn't understand in reading my textbook, thanks for clarifying! So then, they each form their own vectors in coming together to produce a triangle, which is why Pythagorean Thorem is needed. Thank you!
     
  5. Oct 17, 2012 #4

    BruceW

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    yep, that's it :)
     
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