Consider an object that follows a determined path, with it's position given by $f(t)$, parameterized by time. Then consider a second object that is connected to the first by a rigid rod of length $l$, so that it is free to rotate around the first object, but remains a constant distance from it. Given an initial position of the second object, $x_0$, (and perhaps initial velocity $v_0$), what is the differential equation that describes the position of the second object?

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    $\begingroup$ Welcome to Physics SE :) We are unfortunately not a homework help site, so it would be mostly appreciated if you showed some attempt at solving the question yourself. $\endgroup$ – Sanya Aug 6 '16 at 20:47
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    $\begingroup$ It's a neat question though. I think the answer would have to look fairly complicated, as this system has nontrivial holonomy: if you wiggle in a loop, you can make the rod come back to itself except rotated. So the solution is path-dependent. $\endgroup$ – knzhou Aug 6 '16 at 21:27
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    $\begingroup$ The object that follows a predetermined path is a kind of 'irresistible force', if I understand the question. So, it's not a real physical situation. $\endgroup$ – Whit3rd Aug 7 '16 at 2:08
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    $\begingroup$ @Sanya Are you sure it is a homework-like question? To me it seems much more like a personal curiosity one. $\endgroup$ – user259412 Aug 7 '16 at 2:26
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    $\begingroup$ @peterh if I understand the application of the guidelines correctly, the motivation does not matter for that, simply the fact that the question is asking for a specific solution. It is a very nice question and I don't want to judge its motivation. $\endgroup$ – Sanya Aug 7 '16 at 7:21

This is a broad and complex problem, so start with some simplifying assumptions. For example :
* motion is confined to a plane;
* the 1st object A moves along a straight line, eg the x axis;
* the 2nd object B is a point-mass $m$, the rod has no mass, and A has mass $M >>m$ so its motion is not affected by that of B;
* there is no gravity.

In the frame which is moving with object A, the motion of B is affected by that of A only if A accelerates. Otherwise B generally moves in a circle with constant speed.

If A accelerates in the +x direction at $a(t)$ then in the A-frame a pseudo-force acts on B, of magnitude $ma(t)$ in the -x direction. If AB currently makes angle $\theta$ clockwise from the -x direction then the pseudo-force exerts an anti-clockwise torque on B of $ma(t)l\sin\theta$. The moment of inertia of B about A is $ml^2$. So the eqn of motion is
$\ddot \theta=-\frac{a(t)}{l}\sin\theta$.

In the simplest case of $a(t)=0$, the solution is uniform circular motion (superposed on the uniform velocity of A) :
$\theta(t) = \theta_0 - \omega t$.

If $a(t)=g$, a constant, then the equation of motion resembles that of a rigid-rod pendulum which swings under gravity :
$\ddot \theta = -\frac{g}{l}\sin\theta$.
Except for restricting oscillations to small amplitude, as with the simple pendulum, the solution is already difficult, requiring an elliptic integral.

If A performs SHM with amplitude A and angular frequency $\Omega$ such that $x=-A\cos(\Omega t)$ then $a(t)=\ddot x=\Omega^2 A\cos(\Omega t)$. Setting $A\Omega^2=g$ for convenience, the eqn of motion is
$\ddot \theta = -\frac{g}{l}\cos(\Omega t)\sin\theta$.
Except for special cases, this would have to be solved numerically. One special case is that B moves with SHM along the y axis.

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    $\begingroup$ This gives a neat hint towards the general solution. We can think of the system as a pendulum, but with variable gravity! The solution is to solve the equation you have, but with $g$ a vector and varying in time. $\endgroup$ – knzhou Aug 7 '16 at 22:36
  • $\begingroup$ @knzhou : Please feel free to post a more general solution along these lines. In view of my last observation I am thinking of revising my approach. $\endgroup$ – sammy gerbil Aug 7 '16 at 22:43

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