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Interesting Mechanics problem

  1. Aug 15, 2005 #1
    One end of a rubber band with a length L is attached to a wall. The rubber band is held horizontally and the loose end is pulled away from the wall at a speed v, stretching the band. Simultaneously, an ant begins to crawl away from the wall along the rubber band with a speed u < v. Assume that the rubber band can be infinitely stretched. Will the ant ever make it to the loose end of the rubber band? If so, how long will it take?
  2. jcsd
  3. Aug 15, 2005 #2


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    Is the speed of the ant, u, relative to the rubber band (locally) or relative to the wall?
  4. Aug 15, 2005 #3
    rubber band....

    It seemed to me that it would be like a conveyor belt at an airport kind of scenario...
    Last edited by a moderator: Aug 15, 2005
  5. Aug 15, 2005 #4
    would the ant be travelling at its velocity plus the velocity of the band?
  6. Aug 16, 2005 #5


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    In that case ...

    ... all you have to do is solve the ode:

    [tex]\frac {dx}{dt} = u + v \frac {x}{L}[/tex]

    where x is the distance from the starting position and which is equivalent to

    [tex]v \frac {dx}{dL} = u + v \frac {x}{L}[/tex]

    so the ant will reach the end (x = L) when

    [tex]t = \frac {L_0}{v} (e^{v/u} -1)[/tex]
  7. Aug 16, 2005 #6


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    P.S. - I just noticed this is the K-12 section and I'm not sure whether the problem is solvable without resorting to some fairly advanced high school calculus.
  8. Aug 16, 2005 #7
    Hey Tide, I'm wondering if you can help me out, too. I read this question and I'm intrigued at your solution. Can you explain your solution in more detail for me? For example what are your position equations before you differentiate for the end of the rubber band and for the ant?

    I tried to solve it a different way using strictly position equations, but I'm pretty sure it's wrong (compared to your fancy stuff!). Here's how I did it.

    * Let L = the end of the rubber band
    * Let v = velocity of L
    * Let x = the ant
    * Let u = velocity of x
    * The wall is the left edge of this diagram and the motion of the band goes to the right, so after some time t, we have


    Position equations for the rubber band and the ant, relative to us the observers:
    s_T(t) = vt
    s_x(t) = (u+v)t [the ant's velocity relative to us is its velocity relative to the band plus the band's velocity. I think this is wrong but I'm not sure why...]

    Set the two equations equal to each other to find out the time they merge:
    vt = (u+v)t
    vt = ut + vt
    1 = ut
    t = 1/u

    According to this "solution", if the band leaves the wall at 5 m/s and the ant travels at 2 m/s, the ant will reach the end of the band after 1/2 of a second. This solution has nothing to do whatsoever with the rubber band's velocity, I assume because we determined that the ant's velocity is relative to the rubber band. Where's the rub?

    Thanks in advance, dude. I look forward to your feedback.
  9. Aug 16, 2005 #8
    Furthermore after this little slip-up, it becomes apparent that I truly am confused:

    vt = ut + vt
    1 = ut

    oops! This actually resolves to

    1 = ut/vt + 1 or
    0 = u/v

    which resolves to total nonesense.
  10. Aug 16, 2005 #9


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    Details ...

    The speed of any point on the rubber band varies linearly with position - the point connected to the wall never moves while the other end moves at constant speed v. Let's call it w(x) so that [itex]w(x) = v \frac {x}{L}[/itex]

    If the ant moves along at speed u with respect to the band then its speed is the sum of the two so that

    [tex]\frac {dx}{dt} = u + w(x)[/tex]

    which is just

    [tex]\frac {dx}{dt} = u + v \frac {x}{L}[/tex]

    Now, L depends only on time so I decided to turn it into the independent variable rather than using time so I wrote

    [tex]\frac {dx}{dt} = \frac {dL}{dt} \frac {dx}{dL}[/tex]

    but [itex]\frac {dL}{dt} = v[/itex] so

    [tex]v \frac {dx}{dL} = u + v \frac {x}{L}[/tex]

    One way to solve this is to define a new variable y such that [itex]y = x / L[/itex] so that [itex]dx = L dy + y dL[/itex] from which

    [tex]v L \frac {dy}{dL} + v y = u + v y[/tex]


    [tex]v L \frac {dy}{dL} = u[/tex]

    Integration gives [itex]\frac {v}{u} (y - y_0) = ln \frac {L}{L_0}[/itex]. Since the ant starts at the wall, [itex]y_0 = 0[/itex] and it reaches the end of the rubber band when [itex]y = 1[/itex]. Therefore,

    [tex]\frac {L}{L_0} = e^{v/u}[/tex]

    Finally, write L as [itex]L = L_0 + vt[/itex] and solve for t.
  11. Aug 16, 2005 #10
    Thank you very much for your reply, Tide. I have graduated from college and took calc in both high school and college (pursuing a CS minor). I can't even follow your logic much less develop a solution to this problem myself. But I want to, so I appreciate your extensive explanation. I feel like I'm close, and I have a lot of the tools and a lot of the concepts floating around in my head, but they're not combining to make understanding if you know what I mean. I WANT to understand this. I think I had a really lousey calculus teacher in high school, and I know I wasn't as intent on learning the material then as I am now...

    [tex]w(x) = v \frac {x}{L}[/tex]

    This was a key concept that I totally overlooked. Each point on the rubber band moves with a velocity proportionate to its position on the rubber band. The point at [itex]L[/itex] moves at [itex]v[/itex]. The point at [itex]\frac {L}{2}[/itex] moves at [itex]\frac {v}{2}[/itex]. Excellent! Furthermore, this is not a "position" equation because it's not a function of time. Instead it reveals the velocity of any given point as a function of that point's position on the band.

    [tex]\frac {dx}{dt} = u + w(x)[/tex]

    This is the first place I have trouble. Why is the ant's velocity [itex]\frac {dx}{dt}[/itex]? Thinking out loud, I think, "Okay, we know the velocity of any point x on the rubber band is w(x). And we know that the ant's speed on the rubber band at any given point x is the ant's velocity, u, plus that point's velocity, w(x). So therefore the ant's velocity is a function of x, so let's say [itex]a(x) = u + w(x)[/itex]." But where does taking the differential ([itex]\frac {dx}{dt}[/itex]) come in to this line of thinking? You can determine the ant's velocity at any point on the band with a(x). Why differentiate with respect to t? Essentially I'm saying that I'd be stuck at this point. I can understand up to this point and I can understand creating a(x) to tell you the ant's velocity at a given point, but I wouldn't know where to go from there. It looks like you knew and you introduced [itex]\frac {dx}{dt}[/itex], but what prompted you to do that?

    My guess is that [itex]\frac {dx}{dt} = u + w(x)[/itex] is another way to say, "Over an infinitismal distance, during an infinitismal period of time, the distance along the band that the ant walks equals u + w(x)." You could also write [itex]\frac {dx}{dt} = w(x)[/itex] without the term u. But this is only another way to say, "Over an infinitismal distance, during an infinitismal period of time, the distance in space that a point on the band travels equals w(x)." But we knew that. And trying to find the instantaneous velocity of a point on the band is pointless because we know that the velocity of any point on the band is constant, right? So differentiating w(x) with respect to t gives you [itex]\frac {v}{L}[/itex].

    Is this correct? If so then I've just learned something new! If not back to the drawing board.

    [tex]\frac {dx}{dt} = u + v \frac {x}{L}[/tex]

    Marching on and taking for granted that you're still reading this post (if so THANK YOU), and taking for granted that I haven't thought myself into a corner, this simply substitutes w(x). Thank god, I don't have trouble with this step.

    This is a tough leap and it seems very intuitive. How did you know where to go from here, [itex]\frac {dx}{dt} = u + v \frac {x}{L}[/itex]? What do we gain from the transformation that happens after

    [tex]\frac {dx}{dt} = \frac {dL}{dt} \frac {dx}{dL}[/tex]

    ? Frankly I'm almost totally lost, but I'll venture a guess at saying that this equation says, "The infinitismal distance that the ant walks up the band over an infinitismal time period equals the velocity of the band times the velocity of the point x on the band."

    [tex]v = \frac {dL}{dt}[/tex]

    This is another intuitive step, and I'm not sure where we are after we substitute it in to get

    [tex]v \frac {dx}{dL} = u + v \frac {x}{L}[/tex]

    Are we trying to find the point when the ant's velocity ([itex]u + v \frac {x}{L}[/itex]) equals the tip of the rubber band's velocity ([itex]v \frac {dx}{dL}[/itex])?

    If you're still reading this I'm surprised but very thankful. I appreciate it, Tide.
  12. Aug 16, 2005 #11


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    Great start!


    You seem to have the right idea! The basic statement is that, relative to the wall, the ant's speed is equal to the speed of the rubber upon which it is standing (assuming it is not slipping!) plus the speed at which it is walking relative to the band. That's what

    [tex]\frac {dx}{dt} = u + w(x)[/tex]

    means. The equation

    [tex]\frac {dx}{dt} = u + v \frac {x}{L}[/tex]

    is a bit of a challenge because the right side depends on both x and t so it's not easy to integrate directly. Rather than getting bogged down in a lot of algebra trying to do that I noticed that L has a simple time dependence so I decided to make it the independent variable. This is where

    [tex]v \frac {dx}{dL} = u + v \frac {x}{L}[/tex]

    came from. Notice that if you set u = 0 then it just represents the relationship between x and L of points on the rubber band even if it does look complicated.

    I showed you one way of solving that equation in my previous post. Here's another that may be more revealing. First, rewrite the equation as

    [tex]\frac {dx}{dL} - \frac {x}{L} = \frac {u}{v}[/tex]

    You may notice the left side somewhat resembles the derivative of a product or quotient (notice the signs of the terms on the left are different!) In fact, dividing both sides by [itex]1/L[/itex] gives

    [tex]\frac {1}{L} \frac {dx}{dL} - \frac {x}{L^2} = \frac {1}{L} \frac {u}{v}[/tex]

    and now the left side is the derivative of [itex]x/L[/itex] so

    [tex]\frac {d}{dL} \frac {x}{L} = \frac {1}{L} \frac {u}{v}[/tex]

    so that

    [tex]d \frac {x}{L} = \frac {u}{v} \frac {dL}{L}[/tex]

    which can be integrated.
  13. Aug 16, 2005 #12
    Hey, tide, thanks for the help, this is in fact part of my highschool calculus based physics class, so I posted it here... Personally I came out with an equation that looked like this... (D for delta)
    x(t+Dt) = x(t) + va*Dt + vc*x(t)/(1+t*vc)*Dt
    then differentiated, messed up a few equations and solved some things... and ultimately got Lo*e^(u/v)-1 but did not divide by v...what you're showing is I'm now sure absolutely right.. so thanks again for the help!
  14. Aug 16, 2005 #13


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    Great job! Incidentally, you should check your units for consistency. If you're looking for time then your units should reflect that and getting units of length is a clue you've messed up on the algebra somewhere along the line.
  15. Aug 17, 2005 #14
    Tide thank you very much for your willingness to explain these things to me. I know for sure now that I lack the ability to problem solve using advanced math, and I'm sure it's because I don't feel confident with some fundamentals. I hope I can learn the fundamentals that it takes to be as good as you are with equations some day...it doesn't help that I'm really out of practice (last college calc class was 5+ years ago). I should've committed myself to the concepts better back then.

    For the moment, could you answer this? How does this equation represent the relationship between the velocity of the tip of the band and the velocity of any point on the band?

    (1) [tex]v \frac {dx}{dL} = v \frac {x}{L}[/tex]

    Assuming you divide by v you're left with

    (2) [tex]\frac {dx}{dL} = \frac {x}{L}[/tex]

    Maybe it's the notation I'm becomming confused with? They didn't stress Leibniz notation when I was in school...frankly I think they should've instead of simply saying, "The derivative is f(x) with a prime, f'(x), see?"

    If I'm not mistaken does equation (2) mean the same thing as equation (3) below if you assign it some function lable?

    (3) [tex]f(x) = \frac {x}{L}[/tex]

    Or would f(x) actually be the antiderivative of [itex]\frac {x}{L}[/itex] with respect to L?

    (4) [tex]f(x) = x\ ln(L) + C[/tex]
    Last edited: Aug 17, 2005
  16. Aug 17, 2005 #15
    I think I've hijacked this thread enough and I feel guilty, so I'm going to trundle off to the Calc forum to ask some more relevant questions there...
  17. Aug 17, 2005 #16


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    [tex]\frac {dx}{dL} = \frac {x}{L}[/tex]


    [tex]\frac {x}{L} = \frac {x_0}{L_0}[/tex]

    which you can see by rewriting it as

    [tex]\frac {dx}{x} = \frac {dL}{L}[/tex]

    Integrate both sides to find

    [tex]ln \frac {x}{x_0} = ln \frac {L}{L_0}[/tex]

  18. Aug 17, 2005 #17
    Hey Tide, the symmetry of this equation (1) here unnerves me at this point:

    (1) [tex]\frac {dx}{dL} = \frac {x}{L}[/tex]

    Is it true to say

    [tex]\frac {df}{dL} = \frac {x}{L}[/tex]

    where [itex]f[/itex] is the antiderivative of [itex]\frac {x}{L}[/itex] with respect to L? I'm getting the variable x and the function x messed up in the symmetrical equation (1) and it's making it hard for me to really understand what it means...
  19. Aug 17, 2005 #18


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    Not quite. The equation says that the derivative of x with respect to L is the same as x/L. In the notation you're accustomed to it would be written as

    [tex]f{ '} = \frac {f}{L}[/tex]

    where the prime means differentiation with respect to L. In other words,
    the antiderivative of f/L is equal to f.
  20. Aug 17, 2005 #19
    Interesting...spoken in plain english this equation says, "The derivative of f with respect to L is f itself divided by L." Because I'm used to seeing nice cushy, friendly equations like this,

    [tex]f(x) = x^2+3[/tex]

    I don't think I've ever seen an equation like the one you posted there. It has f on both sides, for god's sake!! In fact...now that I look back I realize that this is the keystone that's been bothering me all along. The equation you initially wrote for the ant on the rubber band was:

    [tex]\frac {dx}{dt} = u + v \frac {x}{L}[/tex]

    and essentially you're defining the derivative of x in terms that use x itself...I don't get this. Or rather, I kept mixing up what 'x' meant. Did it mean a point on the rubber band that's distance x from the wall? Or a change in the position of the ant who's walking on the rubber band? Or both?
  21. Aug 17, 2005 #20
    You know what, after changing the way the equation looks from (1) to (2)

    (1) [tex]\frac {dx}{dt} = u + v \frac {x}{L}[/tex]

    (2) [tex]dx = (u + v \frac {x}{L})dt[/tex]

    It's rather easy to see that you have an equation to represent the ant's progress. Because the progress of the ant down the rubber band equals rate times time. The rate of the ant is its speed relative to the rubber band (u) plus the speed of the point where it stands on the rubber band at the moment ([itex]v\frac{x}{L}[/itex]), and the time is the infinitismal quantity dt.
    Last edited: Aug 17, 2005
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