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A tether ball on pole

  1. Sep 18, 2004 #1
    A tether ball of mass 0.15 kg is attached to a vertical pole by a cord 1.1 m long. Assume the cord attaches to the center of the ball. If the cord makes an angle of 20° with the vertical, then
    a) What is the tension in the cord?
    b) What is the speed of the ball?

    Ok well i started it. On my free body diagram i only have two forces, the force of gravity (mg sin (theta)) and the force of tension (T). I calculated the radius of the circle the ball travels to be r=.3762 m. My prediction is that mg sin (theta) - T = ma. centripital acceleration is a=v^2/r. However i have one equation with 2 unknowns here. I somehow need to figure out v, or T. Is there another force that I forgot about in the free body diagram?
    Thanks
     
  2. jcsd
  3. Sep 18, 2004 #2

    arildno

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    Note that the tension works along the direction of the rope.
    The centripetal acceleration, however, lies in the plane with the vertical as normal.
    Hence, you must decompose the tension into its vertical component, and its planar component.
    The vertical component of the tension must balance the force of gravity; otherwise, the ball would start descend vertically.
    The planar component of the tension must provide the centripetal acceleration.
     
  4. Sep 18, 2004 #3
    Ok. I get the wrong answer with these components. can you check to see if these are right.....for vertical component I have T=mg cos (theta) and for planar I have mg sin (theta) - T = ma
     
  5. Sep 18, 2004 #4

    arildno

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    1.equation is correct (hopefully, you have interpreted this as balance of forces in the vertical!)

    2.equation is incorrect.
    In the horizontal, you should have:
    Tsin(theta)=ma (a=v^2/r)
     
  6. Sep 18, 2004 #5

    arildno

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    EDIT;
    Oops sorry, first equation is also wrong!!
    You should have in the vertical:
    Tcos(theta)=mg
    Sorry.. :redface:
     
  7. Sep 18, 2004 #6
    Hmm, I'm still getting the wrong answers. I calculated T = 1.383 N. and v=1.089 m/s. I'm running out of attemps...noooo. lol I dont know why they are wrong. I double checked my work and it was the same answer both times.
     
  8. Sep 18, 2004 #7

    arildno

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    Did you see my last edit?
     
  9. Sep 18, 2004 #8
    Ok, fixed it. Thanks. the right answers are T = 1.5659 N and v = 1.1589 m/s.

    but I still don't understand why the equations are the way they are.
     
  10. Sep 18, 2004 #9
    am i right when I set up a right triangle and the hypotnuse is = T, while the adjacent leg is = mg. Then cos (theta) = mg/T for vertical and then the opposite leg = ma so when you do the planar component you get sin (theta) = ma/T?
     
  11. Sep 18, 2004 #10

    arildno

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    Allright!
    Let's call the unit vector in the direction of the rope (measured from the ball) [tex]\vec{i}_{T}[/tex]
    Let's call the unit vector anti-parallell to the direction of the force of gravity [tex]\vec{k}[/tex] (i.e "upwards")
    Let the radial vector pointing inwards in the horizontal plane [tex]\vec{i}_{a}[/tex]
    Clearly, we must have:
    [tex]\vec{i}_{T}=\cos\theta\vec{k}+\sin\theta\vec{i}_{a}[/tex]

    The tensile force, [tex]\vec{T}[/tex] is parallell to [tex]\vec{i}_{T}[/tex] that is:
    [tex]\vec{T}=T\vec{i}_{T}=T(\cos\theta\vec{k}+\sin\theta\vec{i}_{a})[/tex]
    Does this help?
     
  12. Sep 18, 2004 #11

    arildno

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    Yeah, your approach is correct as well.
     
  13. Sep 18, 2004 #12
    thanks for your help i appreciate it.
     
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