Imagine a membrane is attached in a ring.
Then a weight is hanged exactly in the middle
of the membrane.
How is the resulting form of the membrane called ?
Possibly some variant on a catenary curve.
Catenary curves at Google
Mathworld on Catenarys.
I’m pretty sure it’s a catenary.
I don’t think it’s a catenary as such. A catenary is the solution for a uniform mass distribution; what the OP describes seems to have a uniform mass distribution + a delta function in the middle. Shouldn’t that screw up the answer? I assume that the solution for a uniform 2D mass distribution constrained so that a particular ring is at a constant height (i.e. what the OP describes, without the weight in the middle) is a catenary of revolution, but I’m not even 100% sure that it’s true there.
If the membrane has finite strength, then the weight can’t hang from a single point without breaking the membrane; instead I’ll specify that the weight is hung from a small ring in the center of the membrane. (Now the unstretched membrane occupies the annulus between the two coaxial rings.) If the membrane is assumed to be massless and isotropic, then the membrane’s surface tension T acts to minimize its total surface area between the two rings. The membrane’s surface in this case is a catenoid (the surface of revolution of a catenary); this can be found with variational calculus as shown here (one of Squink’s MathWorld links). (The radial function z = f®, giving the height of the membrane as a function of its radius, is actually the inverse catenary function, f® = d acosh(r/d), where d = M g / 2 pi T.)
If the membrane is not massless, the equation is more complicated. For a membrane with constant mass per unit area m, variational calculus gives the differential equation r z’’ = c r [1 + (z’)[sup]2[/sup]][sup]3/2[/sup] - z’ - (z’)[sup]3[/sup], where c = m g / T. Solving for z’ is not too difficult (make the substitution z’ = tan sin[sup]-1[/sup] v) but the resulting equation is too ugly for me to want to typeset it in HTML. I’m not sure if an elementary closed-form solution exists for z in this case.
We need to know the potential energy function for the membrane. Omphaloskeptic’s result assumes that the energy is monotonic in the surface area, which is a reasonable assumption, but certainly not the only one possible. Further, he’s using the distance pulled down as a given, rather than the weight: We would need a more detailed form of the potential energy function, and the size of the outer ring, to determine how far it’s pulled down.
Yeah, I assumed that the membrane’s potential energy E = T A, where T is the surface tension. Other types of membranes exist, but I think this is a pretty common assumption for the energy of a thin elastic membrane. (But I haven’t done much with membranes, so I could easily be wrong.)
I wasn’t actually using the distance pulled down as a given; I was using the object’s weight M g and the membrane’s surface tension T (and the size of the two rings) as parameters. Variational calculus gives the solution in this case; the membrane height as a function of radius is, if I haven’t made any mistakes, z = - d acosh (r/d), where d = M g / 2 pi T.