Two objects connected by a rigid rod, one of which follows a determined path 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? 
 A: 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.
