How can I find the minimum volume of two rings in soap water with given radii?

[LMZ]

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.

 

[Ben Niehoff]

Soap films try to minimize surface area, not volume.

 

[ozymandias]

Yep, area ~ surface tension.
If I'm not mistaken, for r=R your solution should look like a hyperbolic cosine.

 

[LMZ]

Yep, area ~ surface tension.
If I'm not mistaken, for r=R your solution should look like a hyperbolic cosine.

please be more concrete

 

[ozymandias]

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:

E = C \int_{\mathrm{sample}} dA

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:

\Delta E(x) = 2 \pi y(x) ds

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

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

and so:

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

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!

--------
Assaf
http://www.physicallyincorrect.com/"

 

[LMZ]

thanks a lot!