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Homework Help: Circular motion

  1. Jun 6, 2008 #1
    [SOLVED] Circular motion

    1. The problem statement, all variables and given/known data
    two spheres of equal mass are attached to a light wire of negligible mass of length 3 m ,as shown in the diagram attatched, the spheres and the rod are whirled about point O at a constant rate.

    2. Relevant equations
    a= 4pi^2r/T^2

    (the tension in the wire between spheres X and Y)/(Tension in the wire between sphere Y and O)

    3. The attempt at a solution
    Finding the "Tension in the wire between sphere Y and O":

    acceleration = (4pi^2(1))/(T^2) since r=1m

    From f=ma
    there are two masses flying around so ill let the mass equal to 2m.

    therefore F= 2m x (4pi^2)/t^2

    I don't know how to approach the rest of the problem. I don't even know if I've done it correctly so far.

    Attached Files:

  2. jcsd
  3. Jun 6, 2008 #2


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    You know the acceleration of each mass from your formula. The force acting on the mass X is the tension in the wire between X and Y. The force acting on Y is the difference in the tension between X and Y and that between Y and O (since they pull in opposite ways on Y). Now use F=ma. What must the ratio of the tensions be to give you the ratio of the accelerations you need?
  4. Jun 6, 2008 #3
    Im confused. How can they pull opposite ways on Y. I thought there was no such thing as centrifugal force?
  5. Jun 6, 2008 #4


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    Who said there was a centrifugal force? I was talking about the tension forces. They are real forces, and the accelerations are real accelerations. Use F=ma. No fictitious forces here. What's the net force acting on X and what's the net force acting on Y?
    Last edited: Jun 6, 2008
  6. Jun 6, 2008 #5
    Ok But for f=ma should i use 2m for Y and just 1m for X? Why are the pulling opposite ways?
  7. Jun 6, 2008 #6
    Ok But for f=ma should i use 2m for Y and just 1m for X? Why are Y being pulled opposite ways?
  8. Jun 6, 2008 #7


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    The sphere are of equal mass, use m for both. A tension along a wire pulls the body it's attached to in the direction of the wire. Y has two wires attached to it. One connects it to the center, the other connects it to the outer mass. They are pulling it in different directions. Call TXY the tension in the XY part and TOY the tension in the OY part. In terms of those, what's the force on X and what's the force on Y?
  9. Jun 6, 2008 #8
    I really need help visualising it. Is it like this?



    TXY+TOY=ma + m*3a
    therefore TOY/(TXY+TOY)=.25?

    but that's not the answer in the back. the answer is .75?
    Is this question supposed to be so difficult?
    i have an year 12 phys exam on wednesday- I'm totally screwed.
  10. Jun 7, 2008 #9


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    Don't panic. You have some of this right. Besides Wednesday is a long ways away. Yes, that's the picture. You have to consider the two masses separately. aX=TXY/m. aY=(TOY-TXY)/m. As you said aX=3*aY. Can you put that all together to find TXY/TOY=3/4?
  11. Jun 7, 2008 #10
    Im sorry i still dont understand the direction of these tension forces

    why is it like this


    ANd not also like this


    why is there forces acting this way <---- its this a action/reaction pair or something?
  12. Jun 7, 2008 #11
    TXY = m*3a
    TOY = ma+m*3a

    TXY/TOY = 3/4
    IS this right?
    Even if it is still dont understand how or why. the tension forces are confusing me. Thnks a lot.
  13. Jun 7, 2008 #12
    Okay could you please check if my reasoning is right:

    FXY=Fnet on XY

    FNet on FXY= m*3a


    There is a reaction force on Mass Y so it suffers FXY in the opposite direction:

    Fnet on Mass Y is F=ma

    FOY is the tension

    FOY +(-FXY) = Fnet
    so we have
    FOY +-m*3a = ma
    FOY + <--- = ->

    FOY = 4ma
    FOY = -(<---) + ->
    = ---->

    Hence FXY/FOY = 3/4
  14. Jun 7, 2008 #13


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    I think that's basically ok, though you are sort of mixing up forces and accelerations. The ONLY forces are the tensions TXY and TOY. The force on X is TXY inward. The force on Y is the sum of TOY inward and TXY outward - for a total of TOY-TXY inward. The 'reaction' force isn't exactly a reaction force, if you have a massless wire under tension it will create equal and opposite forces at either end. That's just because the wire is massless, so the net force on it must be zero.
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