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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?

Bobson Dugnutt
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  • There is certainly a complicated region of space that $P$ can be in where the stick entirely misses it: http://i.stack.imgur.com/B5owf.png –  Sep 11 '16 at 21:53
  • @Rahul That is a beautiful illustration! Where did you find it/how did you make it? And does the space that the stick traces out have a name, i.e. is it some kind of well-known function? – Bobson Dugnutt Sep 11 '16 at 22:12
  • It's just a plot of $Q-P$ for fixed $d,l,v,\omega$ and variable $t,r$. I made it using the ParametricPlot command in Mathematica. –  Sep 11 '16 at 22:18

1 Answers1

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$$ \frac{\ell - \gamma \frac{v}{\omega}}{r} = \sin\gamma. \tag{*} $$ 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).

GEdgar
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  • Thank you for your answer. Note that $\gamma(r)$. Also, please have a look at my latest edit - I'm sorry my question was not clear. – Bobson Dugnutt Sep 11 '16 at 22:12
  • see https://math.stackexchange.com/questions/1053472/how-to-solve-keplers-equation-m-e-varepsilon-sin-e-for-e/4732665#4732665 – IV_ Jul 08 '23 at 11:44