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?