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Soap bubbles sticking to each other

  1. Apr 14, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    Two soap bubbles of radii 3cm and 2cm come in contact and sticks to each other. Calculate the radius of curvature of common surface.

    3. The attempt at a solution

    Excess pressure inside soap bubble = 4T/R

    Let the radius of common surface be R'

    [itex]\dfrac{4T}{R'} = \dfrac{4T}{r_1} + \dfrac{4T}{r_2} [/itex]

    But this equation does not gives the correct answer.
     
  2. jcsd
  3. Apr 14, 2014 #2

    maajdl

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    As far as I understand you notations, ... your equation must wrong, as can be seen from a simple case.
    When two bubbles of same radius coalesce, the common surface that separate them is flat.
    This is so because the pressure on both side will be the same.
    Your equation does not predict that, as far as I understand you notations ...
    What are the meaning of r1 and r2?
    Are these numbers supposed to be always positive, or do they have an orientation (sign)?

    The curvature of the common surface is related to the pressure difference across that surface.
     
  4. Apr 14, 2014 #3

    utkarshakash

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    OK. Let me make it more clear.
    r1=3 r2=2

    Since the radius of curvature of common surface is related to pressure difference

    [itex]4T \left( | \frac{1}{r_1} - \frac{!}{r_2} | \right) = \frac{4T}{R'} [/itex]

    Am I correct this time?
     
  5. Apr 14, 2014 #4
    No that's still wrong because the radii of the bubbles change when they coalesce. The total amount of air inside each bubble is preserved and if one side of it becomes flat due to the contact with another bubble than the radius of the remainder must expand in order to preserve the air content of the bubble.
     
    Last edited: Apr 14, 2014
  6. Apr 15, 2014 #5

    utkarshakash

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    So how do I find the new radius?
     
  7. Apr 15, 2014 #6

    haruspex

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    The original equation is fine - you just have to remember that from the perspective of the common boundary one of r1 and r2 is negative. In fact, the sign of the result tells you which way the common boundary will bulge.

    Technically, yes, but I think you're expected to ignore that here.
     
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