A well known result in variational calculus & Lagrangian Mechanics is the solution to the "brachistochrone" problem, where it is found the path connecting two points, A & B such that the time traversed by a sliding object in a uniform gravitational field, is a cycloid.
This got me wondering: what if the 'track' is not fixed but rather is a wire of fixed length which at all instances deforms to support a massive object traversing the cable from A to B -- a "cable car" if you will? What is the optimal path then....and even more basic than that: what IS any path satisfying these constraints? In the following figure, the shape of the 'cable' at a particular instant is represented in blue, and the hypothetical path for the 'pulley' supporting the mass is shown in red.
Away from the endpoints, one would expect the path to resemble an ellipse (which is the locus of points for which the sum of distances r1 & r2 from two fixed points is the same), however the path taken by the cable car should pass through A & B. Imagine that the mass of the cable can be neglected compared to that of the car, and that the car moves on a frictionless, massless pulley. And let's also assume that the cable can't 'stretch'.
In the diagram I'm providing, the path in red with the question mark (?) is what I'm asking. The length of the cable is L = r1 + r2, where r1 & r2 are the lengths of the segments of cable moving from the 'pulley' to each fixed point. These segments may be catenaries, or perhaps are just straight lines (as drawn) if we assume a massless cable...or perhaps they have some other shape?
If you can shed some light on this, or even point me to a solution I'd appreciate it. Google searches haven't found anything.
PS: Extra part of question: So, if we figure out what the path is in general, then can you 'go the extra mile' and determine what is the optimal length of cable (or what's the optimal ratio of cable length to the "straight line" distance between A & B) such that the car traverses its path in the minimum time? Otherwise I'll do that. ;-)