text: Q:
  
  Is it really not possible to obtain an analytical expression for when (if at all) a point $Q$ on a rotating stick will hit a point $P$?
  
  Say we have a stick on length $L$ that travels toward a point $P$ with linear velocity $v$ in the direction of $P$ and rotational velocity $\omega$. The point and the stick are initially $l$ apart. Will a point $Q$ on the stick, a distance $r$ from its center, hit $P$?
  
  Let the starting position of the stick be such that it points along the $x$-axis and that the center of the stick is at the origin. Looking from the point of view of the stick, the point $P$ will have a velocity $-v$ towards the $x$-axis (and the stick). The initial position of $P$ is $(d,l)$.
  The equations of motion are thus 
  \begin{align}
  \overline{P}&=\begin{pmatrix} d \\ l-vt \end{pmatrix} \\
  \overline{Q}&=r\begin{pmatrix}\cos(\omega t) \\ \sin(\omega t) \end{pmatrix} 
  \end{align}
  and the question becomes "when is there a $t$ such that the two are equal?"
  Introducing $\gamma\equiv\arccos\frac{d}{r}$ (which clearly shows the constraint $r\geq \vert d\vert$), we have 
  $$\frac{l-\gamma \frac{v}{\omega}}{r}=\sin \gamma$$
  However, this looks transcendental. 
  Is there really no way of analytically figuring out whether or not $Q$ hits $P$? Can we determine whether or not the stick hits at all (for any $Q$)?
  Edit: Just an idea: Perhaps one can use the Intermediate value theorem to show that there must exist a solution given some initial conditions?
  Edit: Perhaps the question should be "For which $r$s (or some other, single parameter) is $\frac{l-\gamma \frac{v}{\omega}}{r}=\sin \gamma$ true (given $d,\ell,v,\omega$)?  Can we say anything sensible about this?
  
  A:
  
  $$
  \frac{\ell - \gamma \frac{v}{\omega}}{r} = \sin\gamma.
  \tag{*}
  $$
  Perhaps the problem is: Given $\ell,r,v,\omega$, solve for $\gamma$.  
  This is a transcendental equation.  In fact, it is "Kepler's equation", and it is known that the solution (although unique) cannot be written in terms of "elementary functions".
  LINK gives Kepler's equation as
  $$
  M=E-e\sin E
  \tag{K}$$
  where we should solve for $E$, given $M$ and $e$.  We may substitute
  $$
  E=\gamma
  \\
  e=\frac{-r\omega}{v}
  \\
  M=\frac{\ell \omega}{v}
  $$
  to convert (K) into (*).  
  That link also discusses algorithms for numerical and series solution of (K).
  
  