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Surface tension of a liquid layer within frame with elastic wire

  1. Nov 26, 2012 #1
    Hi,

    1. The problem statement, all variables and given/known data

    A square frame with a thin crust/layer of liquid with surface tension γ is given. Into the plane of the frame an elastic wire of circular shape with radius r_0 is inserted. The thin crust/layer of liquid is then burst (by pricking) and as a result of the force of the surface tension exerted on the wire, the elastic wire extends and rapidly attains a new radius, r_1.
    It is also given that the tension in the wire is proportional to the lengthening of the wire wrt its flaccid length, with a constant k.
    I was asked to find an expression for the surface tension (γ) of the liquid, using k, r_1, and r_0.

    2. Relevant equations



    3. The attempt at a solution

    The question advises to use the diagram in the attachment.
    So I wrote down the following equation: 2Tsine(phi) = 2γdl
    And T = k(l_1-l_0) = 2pi*k(r_1-r_0)
    Hence, pi*k(r_1-r_0) = γ
    I am not sure these equations are correct. May you kindly advise?
     

    Attached Files:

  2. jcsd
  3. Nov 26, 2012 #2

    haruspex

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    It's not completely clear what that means. Is it proportional to the increase in the length (T=k(r1-r0)) or to the fractional increase in length (T=k(r1/r0-1))?
    That's true, in the limit, for small phi and dl. Breaks down for large angles because the surface tension forces are not all acting in the same direction.
    The RHS is wrong. You had 2γdl. What happened to the 2dl?
     
  4. Nov 26, 2012 #3
    Hi,
    It is proportional to the increase in length.
    The LHS should have been 2pi*k(r_1-r_0)dl (while the RHS is still 2*gamma*dl.
    Supposing this is correct (is it?), will that indeed be the correct expression for gamma?
     
  5. Nov 27, 2012 #4

    haruspex

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    No, I still think that's wrong. You had, correctly, T = 2πk(r1-r0) and T sin(dθ) = γdl.
    Now, what is the relationship between dl and dθ?
     
  6. Nov 27, 2012 #5
    Did you mean (dl/r_1) = sin (phi)?
     
  7. Nov 27, 2012 #6
    I meant to write (dl/2r_1)=sin(phi). Would you agree?
     
  8. Nov 27, 2012 #7

    haruspex

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    Since the force equation is only valid for small angles (as I mentioned) I'd prefer to make this clear by using dθ for the angle. Also, the equation is really dl = r1 dθ (no sine involved), but of course there's no difference in the limit as dθ→0.
    So you have T sin(dθ) = γdl = γ r1 dθ. Since sin(dθ) ≈ dθ, T = γ r1. You already had T = 2πk(r1-r0). So what do you get for γ?
     
  9. Nov 27, 2012 #8
    dl is the length of the entire wire segment, so I believe it should be dl=2r_1*d(theta), should it not? Which yields gamma=pi*k(r_1-r_0)/r_1
    Is that incorrect?
     
  10. Nov 27, 2012 #9

    haruspex

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    In the OP you had 2Tsine(phi) = 2γdl, implying dl is the half-length. Either way, T = γr1, and T = 2πk(r1-r0).
     
  11. Nov 27, 2012 #10
    This is a bubble, i.e. two layers, one from each side, so the surface tension had to be doubled. And the angle in the diagram was indicated as 2phi. Why is dl then the half length? If you consider it to be the full length, gamma will then be equal to pi*k(r_1-r_0)/r_1 and not to 2pi*k(r_1-r_0)/r_1.
    So which is correct?
     
  12. Nov 27, 2012 #11

    haruspex

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    OK, I missed that. So T = 2γr1, and γ = πk(r1-r0)/r1. But not πk(r1-r0), as you had a few posts back.
     
  13. Nov 27, 2012 #12
    The second part of that question asks for the change in energy of the crust between the initial and final states.
    As gamma=dE/dS, I wrote down the following equation:
    dE=[pi*k(r_1-r_0)/r_1][pi*(r_1)^2-pi*(r_0)^2]
    Is that correct?
     
  14. Nov 27, 2012 #13

    haruspex

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    Almost. dS is the increase in area, right? What would that be?
     
  15. Nov 27, 2012 #14
    Why wouldn't that be [pi*(r_1)^2-pi*(r_0)^2]?
     
  16. Nov 27, 2012 #15

    haruspex

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    Two reasons. The area of the film decreases; it's a bubble, remember?:wink:
     
  17. Nov 27, 2012 #16
    So should it be 2(pi*(r_0)^2-pi*(r_1)^2)?
     
  18. Nov 28, 2012 #17

    haruspex

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    Yes.
     
  19. Nov 28, 2012 #18
    Couldn't I have derived that by calculating the change in elastic energy in the wire? Actually, I have tried that but the two expressions are not the same.
    For the elastic energy (1/2*kx^2) I get: 1/2k(2pi*r_1-2pi*r_0)^2
    Why aren't they the same?
     
  20. Nov 28, 2012 #19
    Could it be because of the tension in the wire that the two expressions are not identical?
     
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