# An astronaut and a vengeful pole

Imagine an astronaut floating in free-space with no significant nearby gravitational influences. The astronaut takes an arbitrarily thin pole of uniform density with length $l$ and mass $m$, orients it vertically from his perspective, and then positions it some distance $h$ in front of him.

Finally, once the pole is in place, he kicks it with force $F$ at some coordinate in space $p$, at distance $r$ from the pole's center of mass. Under what conditions will another coordinate on the pole's contour pass through coordinate $p$ to collide with the astronaut's foot?

-
Does it have 0 radius or 0 density? – user1778 Aug 18 '11 at 7:45
@Tim, I've edited the question to specify that the pole has uniform density and is 'thin' rather than 'infinitely thin', which should hopefully avoid some conceptual issues. – TheSheepMan Aug 18 '11 at 8:00
Observation: the problem can be reduced to two dimensions by noticing that pole will always stay in the subspace generated by $F$ and the line the pole lines on. After this, the motion of the pole is a simple combination of linear translation and uniform rotation. Further, by boosting to the CoM frame, the problem reduces to one where the pole just rotates in place while the foot is moving in a linear fashion. So now the question is whether the foot is fast enough so that it gets out of the circle generated by pole's end before one revolution. This is a simple algebra now. – Marek Aug 18 '11 at 8:33

The $h$ serves no purpose and could as well be 0. The 'kick' is best described with impulse $I$, not with 'force', albeit the actual impulse is irrelevant. It is important though that astronaut is not moving back recoiling from the kick.

The answer is: Where-ever he kicks, pole doesn't intersect that point again, meaning, it won't hit the astronaut's foot (but it may still hit astronaut on the head; perhaps a revised problem could be made about astronaut's head, or about a rod that has extra point mass attached in the middle). To cut down on writing $r/l$ and the like everywhere, I normalize it by defining my own units so that $l=2 , m=1, I=1$ , my $r$ is your $2r/l$ , and my $r$ is in range 0..1.

then

$$v = I/m = 1$$

$$w = \frac{Ir}{\text{moment of inertia}} = \frac{Ir}{m(l^2)/12} = 3r$$

let coordinate system be centered on the centre of the pole before kick; $p_x=0, p_y=-r$ and equation of motion of other end of pole is:

$$x=t-\sin(t3r)$$

$$y=\cos(t3r)$$

Some time before the intersection, the $y$ has to be equal $p.y$ and $x$ has to be be <0

let's find $x$ when $y=p.y$:

$$t3r=\arccos(-r)$$

$$t=\arccos(-r)/(3r)$$

$$x=\arccos(-r)/(3r)-\sin(\arccos(-r))$$

and the rod would subsequently intersect $p$ if $x<0$ , but if we plot it, it never goes below 0 for $r$ in range 0..1 .

Plot of the graph on wolfram alpha

edit: sorry for excessive use of paragraphs, the form deletes newlines and the javascript is bugging out. edit2: ohh, that was noscript blocking some stuff, got that solved now. Forgot to define I, improved clarity a bit.

-
Welcome to Physics.SE, and thank you for this analysis. Nic and I have added LaTeX formatting to your mathematics. The site uses the MathJax engine to support this. You'll find a bare minimum of hints about ow this works in the FAQ and a link on the sidebar of the editing page. – dmckee Aug 18 '11 at 13:28
To summarize, is this answer claiming that it would never come back and hit the astronaut? – Alan Rominger Aug 18 '11 at 14:23
I read it as claiming (correctly) that the pole doesn't return to the astronaut's toe. The astronaut is another story. – Carl Brannen Aug 18 '11 at 15:02