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Phy word Problem

  1. Nov 11, 2011 #1
    I added two pictures.. One of the word problem and one of my diagram I made..

    My work:

    1.Wnet = Kf - Ki
    -Ugf = Kf.... -mgy = mv^2/2 .... - gy = v^2/2

    2. x^2 + y^2 = 1780^2
    dy/dx = -x/y = velocity (v) at point (x,y)

    3. gy = x^2/(2y^2)

    4. gy = (1780^2 - y^2)/(2y^2)

    .... anyways... I know this must be totally wrong..

    Can someone please help me answer this problem?
     

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  2. jcsd
  3. Nov 11, 2011 #2

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    Hi Miike012! :smile:

    I have difficulty making sense of what you did.
    What do the dots for instance mean in (1)?

    And if x and y are coordinates in (2), how did you relate them to a force of 1780 N.
    These things have different units and can not be compared....

    Anyway, let's start with the relevant equations.
    1. Conservation of kinetic and potential energy
    2. Centripetal acceleration

    Can you write those in formula form?

    Next is the decomposition of the forces into aligned forces.
    Which forces are there to start with?
     
  4. Nov 11, 2011 #3
    1.Wnet = delta K
    Ugi - Ugf = Kf - Ki
    -Ugf = Kf

    -gy = v^2/2

    2.a = v^2/r

    3. SumFy = Ty - w = (m)(ay)
    4. SumFx = Tx = (m)(ax)

    Are these correct equations?
     
  5. Nov 11, 2011 #4
    "And if x and y are coordinates in (2), how did you relate them to a force of 1780 N."

    I related 1780 to (x,y) by saying the end point of tension started at the origin, and because the motion is a circle I gave the following equation...
    x^2 + y^2 = 1780^2... which relates it to x and y... would this not be correct????
     
  6. Nov 11, 2011 #5

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    Right!

    Can you write y in terms of theta?


    Yes.
    You need to relate this to your forces.

    Yes, these are correct equations.

    However, I suggest you try to find the forces aligned with the rope.
    The tension is already aligned.
    If you find the component of the weight aligned with the rope, you can find the resulting force along the rope.
    The resulting force along the rope must match the centripetal acceleration.
     
  7. Nov 11, 2011 #6

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    Can it be that you intended Tx^2 + Ty^2 = 1780^2?

    x and y would be coordinates that have "meter" as the unit.
    1780 is the critical tensional forces that has "newton" as the unit.

    Tx and Ty would be the cartesian force components of the tensional force.
    They also have "newton" as the unit.
     
  8. Nov 11, 2011 #7
    3. SumFx = T - wx = (m)(ax)
    4. SumFy = -wy = (m)(ay)

    like that?
     
  9. Nov 11, 2011 #8

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    Yes.

    Can you relate those quantities to the angle theta?
     
  10. Nov 11, 2011 #9
    While I relate the equations the theta.. I was wondering, how come in some situations in circular motion sometimes the SumFy or SumFx has zero acceleration??
     
  11. Nov 11, 2011 #10

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    In circular motion the radial force is not zero, or it won't be circular motion.
    The tangential force can be zero, and indeed is zero if we're talking about "uniform" circular motion.
     
  12. Nov 11, 2011 #11
    Tan(theta) = wx/wy

    cos(theta) = (...)/T
     
    Last edited: Nov 11, 2011
  13. Nov 11, 2011 #12
    Here was my reasoning

    Tx should be T
     

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  14. Nov 11, 2011 #13

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    Right.

    But I'm getting confused with what x and y stand for exactly.
    So I'll use "radial" and "tangential" to keep things clear.

    What I was looking for is:
    wradial = w sin(θ)

    And in your earlier equation for the conservation of energy:
    g (L sin(θ)) = v^2/2

    These need to be combined with:
    Fradial = m acentripetal

    Can you work that out?
     
  15. Nov 11, 2011 #14
    T- (wy)Tan(theta = (m)(ax)
    -(wx)Cot(theta) = (m)(ay)
     
  16. Nov 11, 2011 #15
    ok ill work on that
     
  17. Nov 11, 2011 #16
    F(radial) = m(2*g*L*(wradial/w))/r
     
  18. Nov 11, 2011 #17

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    How did you get that?
    And what happened to the tensional force T?
     
  19. Nov 11, 2011 #18
    im saying x and y because i labeled it with the following axis..
     

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  20. Nov 11, 2011 #19

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    Ah, okay!
     
  21. Nov 11, 2011 #20
    sin(θ) = wradial/w

    g (L sin(θ)) = v^2/2
    g(L*wradial/w) = v^2/2
    v^2 = g(L*wradial/w)(2)

    Fradial = m acentripetal = m(v^2/r)
    Fradial = m (g(L*wradial/w)(2)/r)
     
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