Ant on a stretchy rope puzzle

[DaveC426913]

 

[DaveC426913]

Here is my take on it:

Answer: Yes.

More and more of the rope is stretching out *behind* the ant. That halves the expansion the ant needs to deal with.

After one second, the ant has covered (1cm/(2km/2)) = 1/100,000th of the distance to the far end of the rope.
After two seconds the ant has covered (2cm/(3km/2)) = 1/75,000th of the distance to the far end of the rope.
After three seconds the ant has covered (3cm/(4km/2)) = 1/66,666th of the distance to the far end of the rope.

The ant is definitely making progress.

 

[DaveC426913]

As a spreadsheet:

I guess I haven't proven that column G reaches one in a finite number iterations....

 

[jedishrfu]

I would say no. If you think of the end of the rope as the destination then for every second forward the ant is (1km - 1cm) further away.

Basically, the ant is effectively traveling away from the end of rope destination.

 

[Ibix]

Assuming I made no mistakes, here's a complete answer.

Spoiler: Complete solution

Let the band have initial length $L$, with one end at rest and the other moving at constant speed $V$. Thus at time $t$ the band has length $L+Vt$, and a point at distance $x$ from the stationary end has speed $\frac x{L+Vt}V$. The ant's speed relative to the rubber is $u$, so its speed relative to the stationary end is $$\frac{dx}{dt}=u+\frac{xV}{L+Vt}$$Maxima says that this is satisfied by $$\frac xL=\frac uV\left(1+\frac{Vt}L\right)\ln\left(1+\frac{Vt}L\right)$$meaning that the ant reaches $x=L+Vt$ when$$\begin{eqnarray*}
\frac Vu&=&\ln\left(1+\frac{Vt}L\right)\\
t&=&\frac LV\left(e^{V/u}-1\right)
\end{eqnarray*}$$Plugging in $L=10^3\,\mathrm{m}$, $V=10^3\,\mathrm{ms^{-1}}$, and $u=10^{-2}\,\mathrm{ms^{-1}}$ this becomes $t=e^{10^5}-1\approx 10^{43400}\,\mathrm{s}$.

Bear in mind that the universe is less than $10^{18}\,\mathrm{s}$ old. So there is a solution to the maths (assuming I didn't make any mistakes), but it's not realistic.

 

[DaveC426913]

The one guy who said he recalls the problem said it would take 10^^{some ungodly number} years, so that tracks.

 

[DaveC426913]

I would say no. If you think of the end of the rope as the destination then for every second forward the ant is (1km - 1cm) further away.

The rope is expanding in both directions.
After one second, the ant is already about 25% along its length.

 

[haruspex]

Edit: only just saw @Ibix' solution. Seems to be even a very similar choice of variable names.


The rope starts at length L and grows at rate u.
If at time t the ant is x from the far end of the rope then in further time dt the whole rope grows from $L+ut$ to $L+ut+udt$, and the rope ahead of the ant grows in the same ratio: from $x$ to $x\frac{L+ut+udt}{L+ut}$. That would make the ant $x(1+\frac{udt}{L+ut})$ from the far end. Against that, the ant moves on $vdt$, so
$\dot x=x\frac{u}{L+ut}-v$.

I get $x=(L+ut)(1-\frac vu\ln(\frac{L+ut}L))$, so $x=0$ at $t=\frac Lu(e^{v/u}-1)$.

 

[Ibix]

The rope is expanding in both directions.
After one second, the ant is already about 25% along its length.

I don't agree. Plugging the numbers into my formula, after 1s, $\frac xL=10^{-5}\times 2\ln 2\approx 1.4\times 10^{-5}$, which is barely 0.007% of the length of the rope at that time.

The ant is barely moving with respect to the rope, and one tiny bit of the rope is barely moving with respect to the next tiny bit, so it would be quite surprising if the ant moved very far from its end in such a short time.

 

[Hill]

If there were a bit of acceleration of the moving end of the rope, then there would've been an "event horizon", i.e., such L beyond which the ant would've never reached the other end. Is it correct?

 

[Ibix]

If there were a bit of acceleration of the moving end of the rope, then there would've been an "event horizon", i.e., such L beyond which the ant would've never reached the other end.

An evant horizon, amirite?

Is it correct?

I suspect so, but haven't checked. It shouldn't be difficult to write down a generalisation of the differential equations @haruspex and I wrote down for this case, but I don't know if there is an analytical solution.

 

[jedishrfu]

The rope is expanding in both directions.
After one second, the ant is already about 25% along its length.

But does it matter, the ant is still moving away from the destination.

 

[Ibix]

But does it matter, the ant is still moving away from the destination.

It picks up speed as it moves (slowly!) because every tiny step it takes moves it on to a bit of elastic that's moving a bit faster.

 

[Steve4Physics]

Of course, relativistic effects need to be accounted....
Edit - please ignore. misread question.

 

[A.T.]

If you think of the end of the rope as the destination then for every second forward the ant is (1km - 1cm) further away.

That separation rate only applies initially.

Basically, the ant is effectively traveling away from the end of rope destination.

At a continuously decreasing rate.

 

[A.T.]

Of course, relativistic effects need to be accounted....

Why? Nothing moves faster than 1 km/s here.

 

[martinbn]

This is quite famous it has a wiki page
https://en.wikipedia.org/wiki/Ant_on_a_rubber_rope

 

[DaveC426913]

I don't agree. Plugging the numbers into my formula, after 1s, $\frac xL=10^{-5}\times 2\ln 2\approx 1.4\times 10^{-5}$, which is barely 0.007% of the length of the rope at that time.

The ant is barely moving with respect to the rope, and one tiny bit of the rope is barely moving with respect to the next tiny bit, so it would be quite surprising if the ant moved very far from its end in such a short time.

Right. Yeah. You're right.

The ant is 1/100,000th of the way along the rope, so only 1/100,000th (~1cm) of the total 1km expansion occurs behind it.

Oops.

 

[DaveC426913]

This is quite famous it has a wiki page
https://en.wikipedia.org/wiki/Ant_on_a_rubber_rope

Ah. There it is.


"What we need to do is think about the ant's position as a fraction of the length of the rope. The above reasoning shows that this fraction is always increasing, but this is not yet enough. (The ant might asymptotically approach some fraction of the rope and never come close to reaching the target point.)"


This is how I went about intuitively resolving it (as a fraction) but I quickly realized it doesn't prove that it actually reaches the end.

 

[Steve4Physics]

Why? Nothing moves faster than 1 km/s here.

Aagh. Misread the question. Post struck-through.