Bug on a Rubber Band
An infinitely long 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?
Labels: funphysics
posted by Rajesh Lal at
7:23 AM
95 Comments:
The question seems contridictory. First it says the rubber band is infinitely long, then it says it's 1 meter long.
Do you mean it's 1 meter long, but can strecth to an infinte length?
If so, I think the bug will fall off, because eventually the rubber band will become so thin, he can no longer balance on it.
This is a very nice on, didn't knew it =)
Euclid's Brother. I agree that some of the problems are not well written. You have (in my opinion) correctly interpreted that the elastic band starts at 1 m and can be stretched indefinitely.
As for the band becoming so thin that the bug falls off - I suppose that is a valid out of the box answer, but obviously isn't the "correct" answer.
no, for every 0.001cm the bug travels, the end of the rubber band gets 99.999cm further away. so the bug will die before the rubber band stops at infinty and the bug catches it up
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Assume the bug is immortal, if necessary.
The mere existence of the question on this site makes me believe that the bug is going to get to the other end, because that's the opposite of what you would probably expect.
Anonymous, your numbers are wrong. Remember that the bug is on a band that is stretching, so it gains some speed from the band. The 0.001 cm/sec is the speed relative to the band. So if the bug gets to the end, it would be travelling at 100.001 cm/sec (big hint there, I think).
I'm not sure if the bug's starting point is crucial to the success/fail result. My intuition isn't that good.
Well you know Chris since your smarter than all of us why you just figure out the problem by yourself. After all it well help you suceed in life.
Wow, lets have a huge debate about a bug going across a rubber band. Chris knows all so lets listen to his theoretical genius work.
By the way the bug would not make it.
hooray for the previous "anonymous". Having only just discovered these puzzles, I noticed that Chris either has his comment removed or is furnishing us with an arrogant " I know everything" answer. I also noticed that since he has been more active, Ragnot has been quiet.
Chris, please stick to answering the puzzles and dont comment on other people's failings.
GET A LIFE!!
ps. I bet your single
Anonymous, please check my response to your post in the "proof that 1 = -1". It is not rude or silly.
I don't know why you visit this site, you don't seem to have noticed that it is for people who enjoy solving problems like these. No one here gives a rats about the bug, what we enjoy is solving problems.
New anonymous. All the removed comments were removed by me - notice it says "This comment has been removed ny the author" and not by an administrator.
I have never said "I know everything", please point at a single post of mine that implies any such claim.
If you check, I usually don't simply knock someone for the heck of it. Quite the reverse in fact, I try to help them. But I have been a bit silly a couple of times.
I think that most of my contributions are about answering the problems.
Why don't you live by the rules that you wish me to live by. Why do you think I'm single? Why do you suggest that I get a life? What's that got to do with the bug on a band? It's just childish name calling.
If the site administrator's think I'm crossing a line, then I'll abide by their wishes.
Personally I thought that Ragknot and I were geting along quite nicely.
Chris, yes the bug would be moving at 100.001 cm/sec, but the band is still extending further away from the bug at the same time. Since we do not know what the bug's original position on the band was, we do not know how much of the 1 m/sec extension is in front and how much is behind the bug. Since it states that the bug started near the wall end, we can assume that it is between 1 cm and 49 cm from the wall. However, since the band is stretching between 51 cm/sec and 99 cm/sec away from the bug, I don't see how one would go about solving the problem. In my opinion, I dont believe that the bug would in fact ever reach the end
Hi Anonymous, thanks for not being one of the silly one's round here.
Big IF the bug got to the end, then it would would be travelling 0.001 cm/sec faster than the end of the band, the bug would fall off on its next step.
The only reasons I think it would make it is because of the just mentioned point and because it sounds so unlikely to be possible that it must be possible.
I am finding it hard to write down the equations to solve as well.
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For the Anonymous trolls, this is by no means the first time that I have admitted to struggling with a problem. Please take that into consideration when you next see fit to suggest that I claim to be super smart. If you were actually the decent people that you implicitly think you are, you would check out a fair sample of my posts and see that you are very wrong about me, instead of holding some sort of kangaroo court with no material evidence to show. I'm actually a nice guy. But I am human and occasionally am a bit silly in my posts.
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ok.. lets assume that the bug is infinitely small and can stay on the band no matter how thin it gets... lol
(Had to call my brother for this part)
Let's assume the following.
L = the length of the band at any given time.
B = the length of the band behind the bug (Before)
A = the length of the band in front of the bug (After)
As L increases towards infinity, A+B also proportionaly increases to infinity.
If the bug was standing still, the the ratio between A and B would always be the same. The bug would get further from the wall because the band is being stretched.
But, the bug is moving too, which increases A and decreases B, so the bug is making headway relitive to the ratio between A and B..
so eventually, he'll reach his goal. We can't tell exactly how long it will take w/o knowing the exact position of the bug when evertying started. We only know it was near the wall.
if all of the parts of the rubber band are increasingy its length which is very important because u can also pull it by holding the beggining and then the first 1m doenst really move then:
t=1m/0.001cm/s=100000s
if thats not the case then the first has to travel to the 1m from the begining and then move together with the rubber.then:
t1=1m/0.001cm/s=100000s
s=100000s*1m/s=100000m and
t2=s/0.001cm/s=100000000s and just t=t1+t2
Euclid's Brother. Lucky you having Euclid as a brother. That was a beautiful argument, very clearly expressed.
Thank you for confirming my intuition so quickly because now I can already deal with one of my self appointed prosecutors/persecutors at one level:
So Anonymous 11:21, did you understand EB's argument? Either way it is evident that you're not the super duper brain that you thought you were. You got that emphatically wrong AS WELL.
OK my system's cleared (for now). BTW I will try to avoid responding to the trolls as it isn't what this site is for.
Hey! Please quit making this web site a place to grip, complain, and try to demean anyone.
We need thoughtful comments, but non-thoughtful comments should be lighthearted. if you can't add thought, just add comedy.
The rubber band has to break sometimes... when it does, does the bug fly forward, backward, or just fall down?
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Ragknot. Well said. I'm not sure if that was mainly addressed to me or the Anonymous's who have caused me to deviate from my normal desire to be a helpful member of this site?
Whichever, I hate the fact that I've been silly enough to continue with my response to their jibes. My initial responses were trying to be positive, but I have do seem to have descended to their level.
I will not respond in such a fashion to them again. Thank you for the wake up call.
My first thought was that the bug is too slow, he would never reach the end. But I now know that his speed needs to include the "stretch". But the stretch is dependent on his place on the rubber band. I think I could write the equations to this, and figure it out, but I don't think the bug has a chance. In reality, he should be able to move faster, and the band has to break sometime. There's just no truth here and I don't like the unreal assumptions. Why solve for impossible solutions? I think I'll make a guess.
The bug will reach the end when he catches up to infinity.
Hi Ragknot. I've usually assumed that problems like these were essentially mathematical in nature. They're usually much simpler to solve than real world problems, but hint at the theoretical limits at what's possible. They are not completely without real worth. With a bit of luck, the idealised solution is often good enough for real world use or to act as a good starter for that. e.g. you recently posted a problem to do with measuring dam radii. You gained a useful starter for an iterative solution. The starter came about by simplifying the problem a bit.
The result that the bug can do it is surprising, and I think that's wonderful in it's own right.
The proof that EB gave was beautiful. I was profoundly impressed by it.
The question does say that the band is infinitely long (OK another poorly phrased question). That obviously includes the idea that it won't break.
I'm now very confident that the bug will get there in a finite time. I'm not sure just how much the starting position affects the time it takes and am pretty sure that the worst case time is finite (and that's theone when the bug starts off at the wall). I've not had time to look at it, as I've been dealing with the unpleasantries of earlier.
harmonic series diverge ==> it reaches the end. q.e.d.
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Let u = 0.001 cm/s be the speed of the bug relative to the band.
Then using Euclid brother's ratio, we see that the total speed of
the bug at time t is: x' = u + x/(1+t)
where x is the position of the bug relative to the wall and
the ' denotes the time derivative.
This differential equation cannot be solved directly.
Let x = y(1+t). Then the equation can be written: (y(1+t))' = u+y
=> (1+t)y'+y = u+y => y' = u/(1+t)
Using [] to denote function brackets,
Integrate over time from 0 to T => y[T]-y[0] = u*ln[1+T]
But y[t] = x[t]/(1+t). At start (bug at wall => x=t=0), so y[0]=0
and x[T]=1+T, so y[T]=1. Therefore u*ln[1+T] = 1 => T = e^(1/u)-1
As u = 0.001 cm/s = 10^(-5) m/s, we get
T = e^(10^5)-1 = 2.8*10^43429 seconds (approx). This time is
imcomprehensibly greater than the age of the universe.
If the bug had started at position z (between 0 and 1), then the
bug would have taken time T = e^((1-z)/u)-1. If the bug started a
mere one millimetre from the initial end, it would take about
2.7*10^43 seconds to finish the journey. That's about 434 times
the present age of the universe.
My brain hurts.
Oooops, ln[x] is the natural logarithm, base e, of x.
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Anon, AUG 25, 12:43
Wow, you think your brain hurts??? That was like trying to read Dr. Hawking for the first time
I'm sorry that it made your brain hurt. I know the first one was more mathematical than usual, but it's inherent in the problem, I couldn't see a better way to be able to calculate the time.
Try the second one again, I thought it was pretty simple. I did it because my brain just kept on hurting.
Chris take a look at my solution...
Hi milos, as I implied to Anonymous Aug 26,12:46, to the best of my knowledge, calculus is required to solve this problem. It's possible that someone's thought of a wonderful quick way of doing it, but I doubt it.
Anon AUG 25, 12:43 and Aug 26 12:46 PM.
You can sign in with a name if you want to be easy to recognise. It doesn't have to your name.
Dr. Hawking could do that stuff in his sleep, I can't, I have to work at it. I also wanted to say "thank you" for your response, it was nice after the madness above.
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milos, just to let you see why your type of maths can be used if you make a little modification to it.
If the bug had started with only 0.001 cm from the end of the band, it would take about 1.71828 secs to finish the journey.
Try to calculate that using your technique, but take a snapshot of the system at every 0.1 secs (say).
If you can program, then see what happens when you change the time resolution to 0.01 secs. As the time interval goes down, the calculated time get closer to 1.71828.
chris how can u get a solution of 1.71828...that is quite impossible...explain or this is not a logical solution.
and its also obvious that the bug starts from the very begging because its dimensions are to small comapring to the ruber...
Notice I said "if the bug had started 0.001 cm from the end" that's the same as 0.99999 m from the wall. Use z = 0.99999, u = 0.00001 m/s in the equation T = e^((1-z)/u)-1. I didn't show the trivial details of how I obtained that formula.
My main answer was for z=0, i.e. bug at the wall.
I am completely sure that my results are correct. Now try it using the way I suggested and see that for yourself.
milos, I suggested that you look at it when the bug starts very near the end of the band because doing it numerically with the bug staring at the wall is practically impossible. A computer program would take an incomprehensibly long time to run using the 0.1 second intervals analysis.
what is e???
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e is Euler's number. It has a value of approx 2.718281828459045
Check this Wiki for more:
http://en.wikipedia.org/wiki/E_(mathematical_constant)
Leonhard Euler was a mathematician of very high order. He is said to have been responsible for about 40%of the entire European output of maths in his time. What we take for granted today as how you write maths, is largely due to him.
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I finally found references to the problem. Try Googling "ant on a rubber rope".
I'm rather pleased that my solution is nicer and more complete than any other I've seen. They say things like "moderately advanced calculus required" and don't do the x = y(1+t) trick. They do it another way. That way does have some advantages in understanding the problem.
But they confirm that I've got it right. Phew!!!
A very famous result that Euler proved is known as the "Basel problem".
He determined that the infinite series:
1 + 1/2^2 + 1/3^2 + 1/ 4^2 ... = (pi^2)/6.
Dear old pi, popping up in the most unexpected places. To obtain this result, Euler used infinite series forms of the sin trigonmetric function, and so the link to the circle is clear (in this case).
We are but a short step to the celebrated Riemann zeta function which has a conjecture, made by Riemann, that is probably the most important unsolved mathematical problem there is today. Thousands of proofs of mathematical theorems rest on the truth of the Riemann conjecture/hypothesis. There's a million bucks and a guaranteed place in history for the person who solves it.
I just can't stop!
"Euler's identity" e^(pi*i) + 1 = 0 is probably the most beautiful equation in the whole of mathematics. It links the 5 most important mathematical constants together in a stunning way.
Unfortunately you have to understand what i = sqrt(-1) means to fully appreciate it.
... correction. You have to understand everything about the identity to fully appreciate it.
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Edited as original was flaky: The 1 m extension (each second) has to
be shared over the entire band. Put N uniformly spaced marks on the band, then the speed of separation of an adjacent pair of marks is 1/N m/s. Choose N so that 1/N m/s is less than bugs personal speed (0.001 cm/s). The bug, obviously, is able to walk between each pair of marks. There are a fixed number of marks (N), so the bug must (eventually) be able to walk past all of them.
... in this case we must choose N > 100000.
I gonna use some diff. and int. calcul. in solution. Define quantities:
1) x is a coordinate in the band direction;
2)L is the length of the band. L=L(t);
3)v is bug velocity alon x coordinate, i.e v=dx_Bug/dt;
4)u is band free end speed in x coordinate.
Band dilation law:
L=L_0+ut.
Dilation of band leads us to scaling factor, which sets correspondence between values of dispacement in x coordinate and self band coordinate. It is
a(t)=(L_0+ut)/L_0.
Correspondense itself:
dx=a(t)dl, where dl is band-coordinate distanse.
For example, if you fix at t=0 half-meter point at the band, its x-coordinate will change in time and it'll be x=1m at t=1s. In my definitions fixed band points have CONSTANT coordinates in time.
Then infinitesimal dispacement of bug is
dx=a(t)vdt=v((L_0+ut)/L_0)dt.
Integration gives
x(t)=vt+vut^2/2L_0.
The Bug will be at the end of the band if
vt+vut^2/2L_0=L_0+ut.
Solve this sq. eq. The sole positive solution is 100001 sec.
Sorry, last step was wrong. The answer is 199999sec
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Hi quantense. Even if the band wasn't being stretched, then it would
take, 100000 secs for the bug to walk the metre. After 1 second the
bug will have moved a whole 0.001 cm, but now the band is 2 m long,
so already looking like 200000 secs. Another second later, the bug
now done 0.002 cm (he's hardly picked up anything from the band's
expansion), the band is now 3 m long, so looking like 300000 secs
to go, and he's barely started.
Hi Chris.
Thank you very much for comment.
I've just read your solution. Actually I think it's very well. I also had an idea to treate the problem in the way you have done, but decided to stop at scaling factor idea.
About your 100000 secs to travel not streching band. The thing is that when band is being streched, it'll take less time to travel some x-distanse than when it's at rest. Look at my notice about FIXED point at the band, which travels in x-direction fast enough.
I'll think about correspondense between our solutions.
P.S. Strange, there's no problems for posting here. I still can't do it on rt2.
I'm finding posting here is a nightmare.
If you look back to my Aug 29, 8:20 PM post, you'll see that I've got a very good grasp of what's happening. I lived and breathed this one for 48 hours.
Soz if I'm sound harsh, I'm pee'd off with the blogger not taking my posts. It's taken a dozen attempts. Greetz.
I noticed the comment count go up, that's why I noticed your post. I loved this problem. Little Two, was my previous favourite.
All the external sites I found, with this problem, used a scaling factor - I don't understand why - it only took me two guesses to get the integrating factor type of trick.
At 0.001 cm, during the few instants, the expansion of the band is only 0.001 cm/sec (with respect to the wall), but the bug is still at 0 (as near as makes no odds). SSO he isn't getting a ride yet.
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quantese, oh I do hope that we continue this in a friendly manner. If you think I'm being to dogmatic or whatever, PLEASE say so, I won't take offence ;)
I was so annoyed with the blogger not taking posts. It made me irritable. But I did want you to be realise that I've completely undersood the problem and I couldn't be bothered to rewrite the comment. So I apologised in advance for safety.
I'm pretty sure that you line after "Then infinitesimal dispacement of bug is" is wrong. But I'm struggling a bit. It's always harder to analyse someone else's work than your own. I'm not so comfortable thinking ain terms of scaling factors. I like my safe SI system. :)
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Chris, if it's not hard I wanna ask you to prove your
x'=u+x/(t+1).
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Give me 30 mins (max). I gasping for a coffee and can't carry it to my PC (I'm on crutches, busted leg). Then I'll need to start from scratch using general variables.
... but it's related to your scaling factor.
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Hi quantense. Let Lo be the initial length of the band = 1 m,
v be the speed of the end of the band = 1 m/s, L be the current
length of the band = Lo + vt, where t is the current time and
u = bug's local speed = 0.001 cm/s. (I used u because of bUg).
A point of the band at position x from the wall will have a
speed which is the a fractition x/L of v i.e. xv/L by ratios.
Let x be the position of the bug at instant t, then the bug's
total speed = u + xv/L = u + xv/(Lo+vt) = x'
Lo = v = 1 => x' = u + x/(1+t)
I don't normally put things in like Lo = v = 1 so early in a problem. I decided that it greatly improved the visual appearance by doing it early. Also saved defining extra variables ;)
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Just for safety. If a mark was made on the band (where the bug currently is) the the speed of that mark is constant for all time. The bug then moves on past that mark at local speed u.
Oh, now I see where I was wrong. Scaling factor is not uniform along the band. And it's not therefore
a(t)=(Lo+t)/Lo. Thank you, Chris, for discussion. Good night!
I must get my head round these scaled co-ordinates better though. I can't fully understand the Schwarzchild black hole metric because of it.
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I'm having a hard time posting to my own blogs! It's been getting progressively worse over the last few days.
Good night quantense. Thanks for letting me re-live some of this wonderful problem.
it can't be done...the bug, as the rubber band is first being pulled, will have a greater speed than the end of the band itself because of its extra 0.001cm/s.
but even at the beginning of when the band is being pulled, the speed of any part of the band beside the exact point on the end that is being pulled will be less than 1m/s.
if it was 1m/s at every point on the band, than you would have to assume that there was no elasticity to it, as each point would not be able to stretch out as a rubber band normally does.
so, this bug will never reach the end because the distance between the bug and the end of the band is always increasing more than what the bug can make up in any given second
same poster as above..
i don't see any numbers or variables for elasticity in the equations above...you would need that type of equation to figure this out
but i don't even think this problem can be figured out mathematically...there would need to be other information included
Hi Anonymous. You don't need to know anything about the elasticity of the band. The problem is purely kinematic; not interested in energy, force, momentum etc. You do not need any other information.
See my explanation of how the bug can do it (AUG 29, 8:20 PM). I was a little careless in that; I should have said N+1 marks, to give N sections on the rubber band, including the end points. I think it goes a step further than EBs original argument, but its not as elegant as EBs.
The full blown answer (AUG 26, 9:29) is completely correct. But it does require slightly advanced calculus to understand it.
It is a well known problem and has been scrutinised and accepted by probably tens of thousands of people - not that that guarantees anything. I was initially skeptical. I promise you that the solutions I gave are correct.
...with a problem like this, which hurts the head, it's best not to introduce unecessary IF's. The whole band stretches uniformly. e.g. the mid point always moves at 0.5 m/s.
...another point. In my AUG 29, 8:20 PM explanation, I didn't emphasixe that the distance between the marks is getting bigger all the time. So as the bug progresses, it takes him longer to go from one mark to the next. The very last section takes him the longest time, but the end of the band is moving away more slowly than the bug is walking from the previous mark.
okay thank you, i think i get it now...is this right? The bug is getting separated from the end of the wire faster than it can catch up at first, but after a very long time he eventually catches up to a point where he overtakes it and eventually gets to the end?....
i knew i was wrong somewhere in what i was saying but the only way i could figure it out was to assume i was correct. i didn't mean to try and correct you in any way...thanks for explaining it
okay, i think i get it now
the bug is going slower than the band is growing in front of him at first...but eventually he overtakes the threshold and catches up to the end of the rope?...
haha i didn't mean to put both of those in there
Your 6:44 post is right.
Your 7:01 is a little off. I assume you have got it, but didn't quite say what you meant to.
Read EBs then my 6:20 PM one again. You'll see it more clearly each time. I think my one has the edge in the long run.
If it's any consolation, it took me quite a lot of effort to get it as clearly as I have. I'm not happy that I've understood something until I can explain it to someone else, without having to wave my arms about etc.
Despite all that, I'm still gobsmacked at the solution.
...oh and don't worry, I knew you were genuinely puzzled and not really trying to say I'm a twit :)
I'm not surprised you were puzzled. Most people would have agreed with you - initially. That's the thing with some of these problems, the answer isn't what you'd expect. Take a look at the "Happy Birthday" one for a similar "I don't believe it" experience.
I would have responded differently if I thought you were a troll or just plain rude.
If the rubber band had an initial volume of 1 m^3, then it would consist of approximately a mere 10^27 atoms. By the time the bug gets to the end, the gap between the atoms would be incomprehensibly greater than the size of the known universe :)
Atoms /m^3 might be as high as 10^30.
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