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Sin^2 tetherball

  1. Mar 19, 2013 #1

    tony873004

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    I'm working on the exact same problem described in this 3 year old thread:
    https://www.physicsforums.com/showthread.php?t=433104
    except the length of my rope is 3.8 m and my velocity is 5.6 m/s.
    A tetherball is on a rope 3.8 m long and moves in a circle with velocity 5.6 m/s. Solve for the angle the rope makes with the pole.

    I follow the explanation in the old thread up to this point.

    [tex]
    \begin{array}{l}
    \left( {\frac{{mv^2 }}{{r\sin \theta }}} \right)\cos \theta = mg \\
    \\
    \frac{{\cos \theta }}{{\sin ^2 \theta }} = \frac{{rg}}{{v^2 }} \\
    \end{array}
    [/tex]

    When I rearrange the top expression into the bottom expression, I don't get sin^2. I just get sin. It becomes very simple to solve for θ at that point. How did sin end up getting squared?

    My full attempt is this:
    [tex]
    \begin{array}{l}
    T\cos \theta = mg \\
    T\sin \theta = \frac{{mv^2 }}{r}\,\,\,\, \Rightarrow \,\,\,\,T = \frac{{mv^2 }}{{r\sin \theta }} \\
    \frac{{mv^2 }}{{r\sin \theta }}\cos \theta = mg \\
    \frac{{\sin \theta }}{{\cos \theta }} = \frac{{v^2 }}{{gr}} \\
    \tan \theta = \frac{{v^2 }}{{gr}}\,\,\,\, \Rightarrow \,\,\,\,\theta = \tan ^{ - 1} \frac{{\left( {5.6\,{\rm{m/s}}} \right)^2 }}{{\left( {9.8\,{\rm{m/s}}^{\rm{2}} } \right)\left( {3.8\,{\rm{m}}} \right)}} = 40^\circ \\
    \end{array}
    [/tex]

    The back of the book says 48.4 degrees.
    When I continue the problem with the sin^2, I get the correct answer. The logic makes sense. But where did that extra sin come from?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 19, 2013 #2

    TSny

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    You used the length of the rope for r. Can you find the relationship between r and the length of the rope?
     
    Last edited: Mar 19, 2013
  4. Mar 19, 2013 #3

    tony873004

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    Thanks. I missed that! The earlier explanation used r for both rope length and radius. r = r sin theta -->> r(circle) = r(rope)sin theta.
     
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