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Revolving Ball on String Problem

  1. Apr 22, 2005 #1
    Another physics question for all you smart people out there

    A 0.50-kg ball that is tied to the end of a 1.5-m light cord is revolved in a horizontal plane with the cord making a certain degree with the vertical. If the ball is moving at 4.0 m/s, how can I find the angle that the cord makes with the vertical?

    This one is a toughie!

    I worked on this a couple of hours, but I keep on running into the same dead ends, could anyone give us some pointers?

    (This is the direction I've been trying to go with no luck)

    Fx = Tsin(theta) = m((v^2)/r)
    Fy = Tcos(theta) - Fg = 0
     
  2. jcsd
  3. Apr 22, 2005 #2
    What does [itex]r[/itex] stand for in your equation?
     
  4. Apr 22, 2005 #3
    r stands for the radius
     
  5. Apr 22, 2005 #4
    Well, of course. :tongue2: I meant what value does [itex]r[/itex] take in your equation.

    EDIT: Because, you know, it's not 1.5 m.
     
    Last edited: Apr 22, 2005
  6. Apr 23, 2005 #5

    Doc Al

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    Staff: Mentor

    So far, so good. As I think Berislav was trying to point out, r depends on [itex]\theta[/itex]: rewrite r in terms of the length and the angle. Also, realize that [itex]F_g = mg[/itex].

    Then just eliminate T and solve for the angle. Hint: You may need to use a trig identity.

    (Note: We can't tell where you got stuck, since you only showed the first step in your solution.)
     
  7. Apr 27, 2005 #6
    Well,

    I got up to where

    tan(theta)sin(theta) = 1.088

    I just need to find (theta) now.

    Any help????

    Thanks
     
  8. Apr 27, 2005 #7

    Doc Al

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    Maybe this will help:
    [tex]\sin^2\theta / \cos\theta = (1 - \cos^2\theta)/ \cos\theta[/tex]
     
  9. Apr 27, 2005 #8
    Yea, that was my next step.

    **New info**

    I was informed by my friend that I should use the half-angle identity for this!

    I don't know how to answer this problem yet, but I'm getting there
     
  10. Apr 28, 2005 #9

    Doc Al

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    Use what I gave you in post #7 and solve for [itex]\cos\theta[/itex]. (Rewrite the expression as a quadratic equation in [itex]\cos\theta[/itex], then solve it.)
     
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