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1 Question with regarded to Centripical Forces

  1. Apr 20, 2006 #1
    Hey there, so we started learning centripical forces today. We learned the basic formulas:

    [tex]Fc = \frac{v^2}{r}[/tex]

    and

    [tex]Fc = m \frac{4 pi^2 r}{T^2}[/tex]

    http://elarune.net/admins/josh/q-angle.jpg

    That is the question, I had to get rid of stuff that i had written in, probably not appropriate for the forums. I am not sure how I would go about solving this question. I am supposed to derive the relationship for it. Any help would be appreciated. Thank you.
     
  2. jcsd
  3. Apr 20, 2006 #2
    You can solve this using the rotational acceleration formula and gravity. Read the question again and try to think about what will have to be equal.
     
  4. Apr 21, 2006 #3
    Actually, I think I may have it, would this be it?

    [bare with me while I try to do that tex graphic stuff]

    [tex]Fx = Fn sin(\theta) + mew Fn cos(\theta)[/tex]

    [tex]Fy = Fn cos(\theta) - mew Fn sin(\theta) - mg[/tex]

    which to find the Fc we would solve it to

    [tex]Fc = \sqrt{\frac{r g (sin(\theta) + mew cos(\theta))}{cos(\theta) - mew sin(\theta)}}[/tex]
     
    Last edited: Apr 21, 2006
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