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Difficult circular motion problem

  1. Dec 7, 2011 #1
    1. The problem statement, all variables and given/known data

    http://imageshack.us/photo/my-images/266/problembd.jpg/

    2. Relevant equations

    F=mv^2/c

    3. The attempt at a solution

    I drew the free body diagram and found

    TSin(theta) = mv^2/r

    r = LSin(theta)

    TSin(theta) = mv^2/LSin(theta)

    In the y direction Fy = mg - TCos(theta) = 0
    So mg = TCos(theta)

    I tried to turn T = mg/cos(theta) into T=mg/(1-Sin(theta)) and sub it in but it doesn't seem to work.

    Any help would be appreciated!
     
  2. jcsd
  3. Dec 7, 2011 #2
    Eliminate T, then you can get the trig functions on one side as a function of all the given information.
     
  4. Dec 7, 2011 #3
    Hello. You're on the right track. Rather than trying to get cosine in terms of sine, go the other way around. For example, you have:
    [tex]T=\frac{mv^2}{Lsin^2\theta}[/tex]
    and
    [tex]T=\frac{mg}{cos\theta}[/tex]
    Put sin^2 in terms of cosine, and equate both the formulas that you have for tension. You should be able to come up with a quadratic equation in terms of cosine.
     
  5. Dec 8, 2011 #4
    How did you get [tex]T=\frac{mv^2}{Lsin^2\theta}[/tex]

    I solved for T in the y direction and got

    [tex]T=\frac{mg}{cos\theta}[/tex]

    [tex]\frac{mg}{cos\theta}*{sin\theta}[/tex]=[tex]=\frac{mv^2}{Lsin\theta}[/tex]

    I'm stuck after this I can't seem to get the sin into a quadratic equation
     
  6. Dec 8, 2011 #5
    So you have,

    mv^2/Ls^2 = mg/c (where s = sin and c = cosine)

    so,

    cv^2 = Lgs^2 = Lg(1 - c^2)

    now you can use the quadratic equation
     
  7. Dec 9, 2011 #6
    oh ok! Thank you guys!!
     
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