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I have a weight moving between two poles on a cable which is always under tension. The fact that the cable has constant length means that it moves along an elliptic arc from periapsis to apoapsis. The initial condition is 0. The length is $L$, and $D$ is the distance between poles.

I wonder if there is any special way to solve the problem. The only thing I have come up is to use the expressions
$r = a(1-e^2)/(1+e\cos\theta)$
$b = \frac12 \sqrt(L^2-D^2)$
$e = \sqrt(1-(b/a)^2)$ Then if one consider the momentum
$\frac{d^2\theta}{dt^2} = \frac{g}{r}\cos\theta$.

I also tried different parametrization in cartesian component, where
$r = norm([a*cos(E) b*sin(E)])$; and use E instead of theta for the momentum with respect to the centre of the ellipse (more intuitive since the torque by gravity is 0 when it is in -b).

Is there a way to solve the problem and find the equation of motion and maybe the time to get to the lowest point (basically on b) analytically?

  • $\begingroup$ Why do you say that the shape of the cable is an ellipse? For a cable of uniform density it is a catenary. If the weight is comparable with the weight of the hanging cable, then the shape of the cable will change as the weight moves. This is (I think) a very difficult problem to solve. $\endgroup$ – sammy gerbil Sep 13 '17 at 19:13
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    $\begingroup$ Think of a diabolo! $\endgroup$ – user154997 Sep 13 '17 at 19:45
  • $\begingroup$ The initial condition is 0? What is set equal to 0, $\theta$? a? $\endgroup$ – JMLCarter Sep 13 '17 at 19:55
  • $\begingroup$ I did express the thing incorrectly - the length of the chord is L, so while it is moving the sum of the distance from the two poles maintains constant, that is L - that is the definition of ellipse - the initial condition is theta = 0 at rest - d(theta)/dt = 0. It is not a catenary, the chord is zero mass. $\endgroup$ – Maximus1984 Sep 14 '17 at 7:34
  • $\begingroup$ If the cable is massless and frictionless and never becomes slack, isn't this problem the same as a particle sliding on an elliptical hoop? $\endgroup$ – sammy gerbil Sep 18 '17 at 17:13