# Bug on a Band

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elfmotat    322

Came across this interesting problem on the Feynman Lectures site. The answer is there, so no peeking!

An infinitely stretchable rubber band has one end nailed to a wall, while the other end is pulled away from the wall at the rate of 1 m/s; initially the band is 1 meter long. A bug on the rubber band, initially near the wall end, is crawling toward the other end at the rate of 0.001 cm/s. Will the bug ever reach the other end? If so, when?

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michel123456    483

No peeking?

As a remark to the problem: if the bug is not traveling ON the rubber band but NEXT to the rubber band (on the ground for example), it will never catch the end of the band.

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Remark#2

If I understand correctly, the bug on the band, as observed by an observer at rest, is accelerating. That is not obvious from the (provided) maths.

Edited by michel123456

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let length be L, velocity be U , acceleration be A and time be T

Li=1m, Lf=?

Ui=1m/s, Uf=?

A=?

Uf=Ui+AT

Uf=Ui, T=0

A=Uf/1= 1m/Square sec

Lf=1/1=1m

Lt= Li+Lf

Lt=2m

t=2/0.00001

t=200000s

conclusion: the bug will reach the end in 200000 seconds by crawling a distance of 2 meters.

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Daedalus    294

This is from a previous post of mine that not only solves the problem, but also provides an explanation.

The ant on a rubber rope

Wikipedia actually provides a decent description of problem.

Ant on a rubber rope is a mathematical puzzle with a solution that appears counterintuitive or paradoxical. It is sometimes given as a worm, or inchworm, on a rubber or elastic band, but the principles of the puzzle remain the same. The details of the puzzle can vary,[1][2][3] but a typical form is as follows:

An ant starts to crawl along a taut rubber rope 1 km long at a speed of 1 cm per second (relative to the rubber it is crawling on). At the same time, the rope starts to stretch by 1 km per second (so that after 1 second it is 2 km long, after 2 seconds it is 3 km long, etc). Will the ant ever reach the end of the rope? At first consideration it seems that the ant will never reach the end of the rope, but in fact it does (although in the form stated above the time taken is colossal). In fact, whatever the length of the rope and the relative speeds of the ant and the stretching, providing the ant's speed and the stretching remain steady the ant will always be able to reach the end given sufficient time.

I modified the variable names used in Wikipedia to make the problem easier to read.

Consider a thin and infinitely stretchable rubber rope held taut along the $x$-axis with a starting-point marked at $x_0$ and a target-point marked at $x_1$.

At time $t=0$ the rope starts to stretch uniformly and smoothly in such a way that the starting-point $x_0$ remains stationary at $x=0$ while the target-point $x_1$ moves away from the starting-point with constant speed $v_r > 0$.

A small ant leaves the starting-point at time $t=0$ and walks steadily and smoothly along the rope towards the target-point at a constant speed $v_a > 0$ relative to the point on the rope where the ant is at each moment.

Will the ant reach the target-point?

According to the definition for the problem, the equation for position of the ant on a rubber rope is a first order linear differential equation:

A key observation is that the speed of the ant at a given time $t > 0$ is its speed relative to the rope, i.e. $v_a$, plus the speed of the rope at the point where the ant is. The target-point moves with speed $v_r$, so at time $t$ it is at $x = v_r \, t + x_1$. Other points along the rope move with proportional speed, so at time $t$ the point on the rope at $x$ is moving with speed $\left(v_r\,x\right) / \left(v_r\,t + x_1\right)$. So if we write the position of the ant at time $t$ as $x\left(t\right)$, and the speed of the ant as $x'\left(t\right)$, we can write:

$x' \left( t \right) = v_a + \frac{v_r \, x \left( t \right) }{ v_r \, t + x_1 }$

We can solve this nonhomogeneous linear differential equation using Integrating Factors. We begin by rewriting the differential equation in the form:

$\frac{dx}{dt} = \alpha (t) \, x + \beta (t)$

where

$\alpha (t) = \frac{v_r}{v_r \, t + x_1}$

$\beta (t) = v_a$

and the integrating factor is

$\mu (t) \, = \, e^{\left(-\int_{} \alpha (t) \, dt\right)} \, = \, \frac{1}{v_r \, t + x_1}$

After working the method, we arrive at:

$x(t) \ = \ \frac{1}{\mu (t)} \int_{} \, \beta (t) \mu (t) \, dt \ = \ \left(v_r \, t + x_1\right) \left(\frac{v_a}{v_r} \, \text{ln} \left(v_r \, t + x_1\right)+C\right)$

Now we have to find the value for $C$ that satisfies our initial condition $x(0) = 0$:

$x(0) \ = \ x_1 \left(\frac{v_a}{v_r} \, \text{ln} \left(x_1\right)\right)+x_1 \, C \ = \ 0$

$x_1 \, C \ = \ - x_1 \left(\frac{v_a}{v_r} \, \text{ln} \left(x_1\right)\right)$

$C \ = \ -\frac{v_a}{v_r} \, \text{ln} \left(x_1\right)$

After substituting $C$ back into our general solution and simplifying the result, we arrive at the following equation for the position of the ant at time $t$:

$x(t) \ = \ \left(v_r \, t + x_1\right) \frac{v_a}{v_r} \, \text{ln} \left(\frac{v_r \, t \, + \, x_1}{x_1}\right)$

However, it is extremely difficult using the above equation to determine the value of $t$ that will tell us how much time it took the ant to complete the journey. If you read the ant on a rubber rope problem in Wikipedia, they offer a different method that will allow us to solve for $t$.

A much simpler approach considers the ant's position as a proportion of the distance from the starting-point to the target-point.[3] Consider coordinates $\psi$ measured along the rope with the starting-point at $\psi = 0$ and the target-point at $\psi = 1$. In these coordinates, all points on the rope remain at a fixed position (in terms of $\psi$) as the rope stretches. At time $t \ge 0$, a point at $x = x_i$ is at $\psi = x_i / \left( v_r \, t \, + \, x_1 \right)$, and a speed $v_a$ relative to the rope in terms of $x$ is equivalent to a speed $v_a / \left( v_r \, t + x_1 \right)$ in terms of $\psi$. So if we write the position of the ant in terms of $\psi$ at time $t$ as $\psi \left( t \right)$, and the speed of the ant in terms of $\psi$ at time $t$ as $\psi' \left( t \right)$, we can write:

$\psi ' \left( t \right) = \frac{ v_a }{ v_r \, t \, + \, x_1 }$

$\therefore \psi \left( t \right) = \int \frac{v_a}{v_r \, t \, + \, x_1} \ dt = \frac{v_a}{v_r} \, \text{ln} \left( v_r \, t \, + \, x_1 \right) + k$ where $k$ is a constant of integration.

Now, $\psi \left( 0 \right) = 0$ which gives:

$k = - \frac{v_a}{v_r} \text{ln} \left( x_1 \right)$, so $\psi \left( t \right) = \frac{v_a}{v_r} \, \text{ln} \left( \frac{v_r \, t + x_1}{x_1} \right)$

If the ant reaches the target-point (which is at $\psi = 1$) at time $t$, we must have $\psi \left( t \right) = 1$ which gives us:

$\frac{v_a}{v_r} \, \text{ln} \left( \frac{v_r \, t \, + \, x_1}{x_1} \right) = 1$

$\therefore t=\frac{x_1}{v_r}\left(e^{v_r/v_a}-1\right)$

As this gives a finite value $t$ for all finite $x_1 \ge 0$, $v_r > 0$, and $v_a > 0$. This means that, given sufficient time, the ant will complete the journey to the target-point. This formula can be used to find out how much time is required.

For the problem as stated, $x_1 = 1 \, \text{m}$, $v_r = 1 \, \text{m/s}$, and $v_a=0.001\,\text{cm}=0.00001\,\text{m/s}$, which gives

$x_1/v_r=\frac{1\,\text{m}}{1\,\text{m/s}}=1\,\text{s}$

$v_r/v_a=\frac{1\,\text{m/s}}{0.00001\,\text{m/s}}=100000$.

Therefore, the time it takes the ant to reach the end of the rope is:

$t=\left(1\,\text{s}\right)\left(e^{100000}-1\right)\approx 2.8\times 10^{43429}\, \text{s}$

Substituting this time into the function for the position, we find the distance the ant traveled is:

$d=x_1\,e^{v_r / v_a}=(1\,\text{m})\,e^{100000}\approx 2.8\times 10^{43429}\,\text{m}$

If I understand correctly, the bug on the band, as observed by an observer at rest, is accelerating. That is not obvious from the (provided) maths.

You are correct michel123456, the bug on the rubber rope is observed to be accelerating by an observer at rest relative to the rope. Given the position function

$x(t) \ = \ \left(v_r \, t + x_1\right) \frac{v_a}{v_r} \, \text{ln} \left(\frac{v_r \, t \, + \, x_1}{x_1}\right)$

the acceleration can be found by taking the second derivative

$\frac{d^2x}{dt^2}\ x(t) \ = \ \frac{v_a \ v_r}{v_r \, t+x_1}$

We can see that the acceleration approaches zero as time approaches infinity. However, the acceleration would be very different if a nonzero net force were to act upon the bug. In that case, the bug would be accelerating relative to the rope instead of just having a constant velocity.

Edited by Daedalus

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ydoaPs    1582

Came across this interesting problem on the Feynman Lectures site. The answer is there, so no peeking!

Assuming uniform expansion of the band (it's stretching everywhere at once), I'd say yes, but I can't be bothered to Faf about with the math atm. Otherwise, probably not.

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Daedalus    294

Assuming uniform expansion of the band (it's stretching everywhere at once), I'd say yes, but I can't be bothered to Faf about with the math atm. Otherwise, probably not.

I already showed the math. Check the spoiler tag in my previous post.

Edited by Daedalus

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It will never reach the end

let length be L, velocity be U , acceleration be A and time be T

Li=1m, Lf=?

Ui=1m/s, Uf=?

A=?

Uf=Ui+AT

Uf=Ui, T=0

A=Uf/1= 1m/Square sec

Lf=1/1=1m

Lt= Li+Lf

Lt=2m

t=2/0.00001

t=200000s

conclusion: the bug will reach the end in 200000 seconds by crawling a distance of 2 meters.

Uf=Ui=1m/s for the band's end because we assume constant rates