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Two rings and soap water

  1. Nov 18, 2007 #1


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    Two rings with radius R and r are let down in the soap water. Between them pellicle appeared (like in image).
    Image: here all are symmetric ;)

    problem: need to find y=f(x)

    what i think:volume of figure is minim. maybe express volume via needed function, then get minimum of volume, and find this function.

    Attached Files:

  2. jcsd
  3. Nov 18, 2007 #2

    Ben Niehoff

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    Soap films try to minimize surface area, not volume.
  4. Nov 18, 2007 #3
    Yep, area ~ surface tension.
    If I'm not mistaken, for r=R your solution should look like a hyperbolic cosine.
  5. Nov 19, 2007 #4


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    please be more concrete
  6. Nov 19, 2007 #5
    Sure, but about which part?

    Nature "seeks" to minimize the energy of a system. In your case, your system's only energy (that changes as the surface changes shape) is the surface tension. I'm not an expert on "surface"-y things, but in this problem the surface tension energy is simply proportional to the area, A:

    [tex]E = C \int_{\mathrm{sample}} dA[/tex]

    Since your problem has rotational symmetry about (say) the x-axis, we can simplify this as follows: let y=y(x) be the function describing a cross section of your sample (by a plane passing through the axis of rotational symmetry, the x-axis). Then, given x, the total area contributed to the energy term is simply:

    [tex]\Delta E(x) = 2 \pi y(x) ds[/tex]

    where ds is the "length element" of the curve y(x). It is given by:

    [tex]ds = \sqrt{dx^2 + dy^2} = \left( 1+ \left( \frac{dy}{dx} \right)^2\right)^{1/2} dx[/tex]

    and so:

    [tex]E \propto \int_{x_1}^{x_2} y(x) \left( 1+ (y')^2\right)^{1/2} dx [/tex]

    Now all that is left to do is apply Euler-Lagrange's equations using the appropriate boundary conditions (y(start) = r, y(end)=R) and you're done!

    http://www.physicallyincorrect.com/" [Broken]
    Last edited by a moderator: May 3, 2017
  7. Nov 21, 2007 #6


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    thanks a lot!
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