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 Posted: Fri Oct 06, 2006 10:01 am
This was a problem off of my last problem set. Nearly everyone I knew ended up using Mathematica, which is a cop-out (i.e. I have low self-esteem 'cos I haven't learned it yet). So now that the problem sets have been graded and whatnot, I'd like to see some other takes on it.
S'pose you have two parallel circular loops of radius R, a coaxial distance s apart. Now s'pose that there's a bubble film spanning the two loops, which will obviously be radially symmetric. The potential energy of this film is equal to the surface area times the tension of the film; assume the tension to be constant. (There's a diagram to go with this, if anyone would like to see it. But it's pretty straightforward).
a) Demonstrate that there are three configurations that are local extrema of the potential energy functional. You may assume radial symmetry. The boundary conditions are that the soap film ends on the two circular wire loops.
b) Which of the three extrema has the lowest potential energy? For which values of R/s?
c) S'pose we start with a nearly cylindrical configuration with very large R/s (loops close together). Holding R constant, gradually increase s. What happens, and at what R/s does it happen?
Parts (b) and (c) are really the same problem, just more specific. I s'pose this is really more a question of mathematics than physics, but ah well. Just wondering what you guys' take on the solution is.
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Posted: Fri Oct 06, 2006 10:57 pm
Place the rings with radius R so that their centers are at (s/2, 0, 0) and (-s/2, 0, 0) respectively. Let the soap have tension T.
Then, by symmetry arguments without real mathematical backing (um...curvature 0?), the extrema are where the film is {(x, y, z)| y^2+z^2 < R^2, x = +-s/2}, {(x, y,z)| y^2+z^2 = R^2} and {(x, y, z)| |x| < s/2, y^2+z^2 = (Rx/s)^2}. In other words, disks inside the rings, a cylinder between the rings, or a pair of cones with bases on the rings and vertices at the origin.
The disks have 2pi*R^2*T energy total. The cylinder has pi*R*s*T, while the cones have 2pi*R*sqrt(R^2+(s^2)/4)*T total.
Now we get to set things equal:
2R^2 = R*s for R*2 = s
2R^2 < 2R*sqrt(R^2+(s^2)/4) for s >0
R*s < 2R*sqrt(R^2+(s^2)/4) for R > 0
Thus for R/s < 1/2, 2R^2 is smaller. For R/s > 1/2, R*s is smaller.
As for c...I don't actually know, but I'd say that it starts as the cylinder, but when you get within some epsilon of R/s = 1/2, the soap suddenly snaps into the disk configuration.
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Posted: Fri Oct 13, 2006 6:46 pm
So, since I s'pose no one else is going to try it, I'll post what my preceptor sent us (Yay! I got most of it right...): Quote: The first configuration is obviously two "drums" separately stretched on two hoops. We will find more two solutions assuming they also allow rotational symmetry. In this case the hole shape of the soap film could be described by the single function r(x) - the dependence of the radius. The energy is proportional to the area. Hence our Lagrangian is L = r * Sqrt(1+r'^2) Notice that this is independent of x so the "Hamiltonian" H = (part L/part r')r' - L = r/Sqrt(1+r'^2) is conversed. Setting it to a constant we will obtain other solutions. Notice that if r is itself constant then H will also be. This corresponds to a solution were there is a cylindrical soap bubble between the two hoops. It turns out that the area corresponding to this solution is always greater than one of the other solutions so we will not concern ourselves with it in what follows. Assuming that r is not a constant we obtain dx = dr/Sqrt((r/c)^2-1) Integrating we obtain the solution r(x) = C * cosh((x - x_0)/C) Where x_0 is an integration constant. This is the equation for a catenary. The boundary conditions are (choosing the origin just between the hoops) R = C*cosh((ell-2x_0)/(2c)) = C*cosh((ell+2x_0)/(2c)) with an obvious solution x_0 = 0 and the equation for the smallest radius C: R = C*cosh(ell/(2C)) Rewriting, we get R/ell = (C/ell) * cosh(ell/(2C)) and think of C/ell and R/ell as the independent variables. By plotting the RHS of this equation it is clear that there is a critical value of R/l for each there is one solution to this equation. Above this value, there are always to values of C/ell that satisfy the equation and below there are none. This critical value can be found to be R/ell = 0.75444 by numerical methods. Note that the higher C the flatter the catenary becomes, so one solution is shallower - "almost cylindrical". I'll post the answers to (b) and (c) tomorrow or soon.
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