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Homework Help: Tetherball - find theta and tension

  1. Sep 28, 2010 #1


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    1. The problem statement, all variables and given/known data
    A 450-g tetherball A is moving along a horizontal circular path at a constant speed of 4m/s. Determine (a) the angle θ that the cord forms with pole BC, (b) the tension in the cord.

    2. Relevant equations
    [tex]\sum F_{y}=ma_{y}[/tex]
    [tex]\sum F_{n}=ma_{n}=m\frac{v^{2}}{r}[/tex]

    3. The attempt at a solution

    Find θ and T.



    Sum of forces equations:
    (1)[tex]\sum F_{y}=-mg+Tsin\theta =0 \to Tsin\theta =mg[/tex]
    (2)[tex]\sum F_{n}=Tcos\theta =m\frac{v^{2}}{r} \to T=m\frac{v^{2}}{rcos\theta}[/tex]

    Substitute expression for T found in (2) into (1):
    [tex]\left (m\frac{v^{2}}{r} \right )tan\theta =mg[/tex]
    [tex]\theta =tan^{-1}\left (\frac{gr}{v^{2}} \right )[/tex]

    This is where I get stuck. I know that if I can find r, I can find θ and subsequently T. How do I find r?
  2. jcsd
  3. Sep 28, 2010 #2


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    You have found that

    tanθ = gr/v^2.

    Check this expression. In FBD angles indicated are different.

    In the problem, the length of cord L is given. So sinθ = r/L.

    Put it in the expression and find θ.
  4. Sep 29, 2010 #3


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    I fixed my FBD so that θ is in the correct place:

    [tex]sin\theta=\frac{r}{1.8}\rightarrow r=1.8sin\theta[/tex]

    [tex]\sum F_{y}=-mg+Tcos\theta=0[/tex]

    (1) [tex]Tcos\theta=mg[/tex]

    [tex]\sum F_{n}=Tsin\theta=m\frac{v^{2}}{r}[/tex]

    (2) [tex]T=\frac{mv^{2}}{rsin\theta}[/tex]

    Substituting (2) into (1) yields:

    [tex]\left (\frac{mv^2}{rsin\theta} \right )cos\theta=mg[/tex]


    How do I simplify the [itex]\frac{cos\theta}{sin^2\theta}[/itex] to solve for θ?
    Last edited: Sep 29, 2010
  5. Sep 29, 2010 #4

    Doc Al

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    Staff: Mentor

    Hint: Look up the Pythagorean trig identities. (I'm sure it's one you already know.)
  6. Sep 29, 2010 #5


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    I did this:

    Making the equation:

    However, this doesn't get me any closer to getting theta out on its own.
  7. Sep 29, 2010 #6

    Doc Al

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    Staff: Mentor

    Sure it does. Hint: Rearrange that equation and solve for cosθ. (You'll get a quadratic.)
  8. Sep 29, 2010 #7


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    Thanks to both of you for the persistence. With your help, the solution presented itself:

    [tex]\left (\frac{1.8g}{v^2} \right )cos\theta+cos\theta-\left (\frac{1.8g}{v^2} \right )=0[/tex]

    [tex]cos\theta=\frac{-1+\sqrt{1^2-4\left (\frac{1.8g}{v^2} \right )\left (\frac{-1.8g}{v^2} \right )}}{2\left (\frac{1.8g}{v^2} \right )}[/tex]

    [tex]\theta\approx49.9[/tex] degrees

    [tex]T=\frac{mg}{cos\theta}=\frac{\left (0.45kg \right )\left (9.81m/s^2 \right )}{cos(49.9)}\approx6.85N[/tex]

    These answers line up with the book's answers. Again, thanks to both of you for the help.
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