1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Tetherball - find theta and tension

  1. Sep 28, 2010 #1


    User Avatar
    Gold Member

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Gold Member

    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

    User Avatar

    Staff: Mentor

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


    User Avatar
    Gold Member

    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

    User Avatar

    Staff: Mentor

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


    User Avatar
    Gold Member

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Tetherball find theta Date
Bead on a string, find y(x) if horizontal velocity is const. Thursday at 10:18 PM
Lagrange's Equations for a Tetherball Mar 17, 2014
Tetherball problem Jan 29, 2014
Sin^2 tetherball Mar 19, 2013
Centripetal Motion: Tetherball Jan 10, 2010