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What I want: I have a rubber rope which is $5m$ in length when not stressed and is able to stretch about $100\%$ (to $10m$ long). I want to accelerate a constant mass horizontally, which has negligible friction. I'd like to have a function that tells me the velocity of the mass dependent on time, so for instance velocity $1 s$ after releasing it.

What I did:​ I've done some measurements of forces of the rope when pulling it to different lengths. Of course, when pulling $0cm$ (total length $5m$) I got a force of $0N$. Here is a graph of my results.

http://i.imgur.com/vtEnACQ.png

$x-axis$: displacement of one end of the rope

$y-axis$: measured force

I was also able to do a regression and found a function which describes how much force I get after I pull a given length. I name this function $F(s)$ for Force dependent on displacement. From this, it's easy to get the acceleration function, which is $a(s) = F(s)/m$ with $m = mass$ of the object I want to accelerate. But now I'm stuck. I somehow need to get $a(t)$ instead of $a(s)$, thus the acceleration by time, not by length, so I can then integrate that to get $v(t)$.

How do I convert the dependency of the function?

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This question seems to be harder to answer than I expected. Please leave comments on how to make it clearer if you don't get it. –  user23224 Apr 17 '13 at 11:02
    
Can't you just measure the time by which it return to normal length? –  WInterfell Apr 22 '13 at 16:24
    
Unfortunaly not without a major effort. Also, I'd like to have the formula independent of the mass, so measuring with a specific mass is no solution. –  user23224 Apr 25 '13 at 11:21

2 Answers 2

You certainly want to numerically integrate your motion equation givent your expression of the force.

I will assume that the mass $m$ is attached at one side of the rope, while the other side is attached to a wall or something that won't move during the integration. Something like this, where the spring is in fact your rope, with $x$ the extension of the rope

spring-mass system

Then you can simply integrate (for example using Euler scheme) the equation $$\frac{d^2x}{dt^2} = \frac{F(x)}{m}$$ Here is some Ruby code I use whenever I want to get numerical solution of an ODE in mechanics:

##################################################
## Integration using Euler (mid-point) method knowing position x and speed xp=\dot{x}
##################################################

def integration(x,xp)
  dxp = force(x)*@dt
  dx  = (xp+dxp/2)*@dt
  return x+dx,xp+dxp
end


##################################################
## Force expression knowing position x
##################################################

def force(x)
  return -10*x
end


##################################################
## Effective integration loop
##################################################

@dt = 0.001  # time increment
t = 0  # initial time
x = 5  # initial position
xp= 0  # initial speed
10000.times do |i|
  t += @dt
  x,xp = integration(x,xp)
  puts "#{t}\t#{x}\t#{xp}"
end

Just plug in your initial conditions and force expression from your measures and you will have both position and speed along time (NB: I took $m=1$ in my code but you can add the correct value wherever you like in the integration function or in the force function [that shall be renamed "acceleration" then]).

You can also refine it to integrate over constant total time and output fewer results than you have integration steps (precision increase with smaller integration step @dt but it can become difficult to draw when you have a million points just to be more precise in the integration)

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With the acceleration he has of the form $a=A x^b$ when you start numerically have to take care of the initial $x=0$ condition, where you are left with zero acceleration and nothing moves. You need an implicit integration scheme for this to work right. –  ja72 Apr 25 '13 at 13:02
    
@ja72 Note that if you start at rest with no force, you should stay at the equilibrium if it is stable (which it is in this rubber rope case). –  JJ Fleck Apr 25 '13 at 13:33

When the acceleration is a function of position use the following

$$ a(x) = \frac{{\rm d}v}{{\rm d}t} = \frac{{\rm d}v}{{\rm d}x} \frac{{\rm d}x}{{\rm d}t} = \frac{{\rm d}v}{{\rm d}x} u $$

$$ \int a(x)\,{\rm d} x = \int u\,{\rm d} u = \frac{1}{2} u^2 + K_1 $$

which is solved for $u(x)$.

The the position is found from

$$ t = \int \frac{1}{u(x)}\,{\rm d} x + K_2 $$

which is solved for $x(t)$.

Example

$$ a(x) = A\,x^{b} $$ $$ \int A\,x^{b}\,{\rm d} x = \frac{A}{b+1} \left( x^{b+1}-1\right) = \frac{1}{2} u^2 + K_1 $$

when $u=0$ at $x=0$ then $K_1 = \mbox{-} \frac{A}{b+1}$ or

$$ u(x) = \sqrt{\frac{2 A}{b+1} x^{b+1} } $$

Then

$$ t = \int \frac{1}{\sqrt{\frac{2 A}{b+1} x^{b+1} }}\,{\rm d} x + K_2 = \sqrt{ \frac{2(b+1)}{A (b-1)^2}} \left( x^{\mbox{-}\frac{b-1}{2}}-1\right) + K_2 $$

and when $x=0$, $t=0$ then $K_2 = \sqrt{ \frac{2(b+1)}{A (b-1)^2}}$ or

$$ t = \sqrt{ \frac{2(b+1)}{A (b-1)^2}} x^{\mbox{-}\frac{b-1}{2}} $$

$$ x(t) = \left( \frac{t}{\sqrt{ \frac{2(b+1)}{A (b-1)^2}}}\right) ^{\mbox{-} \frac{2}{b-1}} = \left( \frac{A (b-1)^2}{2 (b+1)}\right)^{\mbox{-}\frac{1}{b-1}} \;t^\left({\mbox{-}\frac{2}{b-1}}\right)$$

If your mass was $m=1$ then $a=f(x)$ in graph and

$$ x=1.397882\, t^{3.058906} $$

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well, well, you just demonstrated that when you attach a mass to an elastic rope, it's flying away :) Do not forget that you made some (false) assumption on the sign of $A$ in your mathematical computations. –  JJ Fleck Apr 25 '13 at 13:36
    
Hmm, it depends on what the x-axis of the graph represents. Is it position or extension? Let me check again. –  ja72 Apr 25 '13 at 13:45
    
@JJFleck - If $x$ is extension the above is correct. If $x$ is position then acceleration is $a=A\,(-x)^b$ but with $x<0$. If you shoot to the right (positive direction), the you move the payload to the left (negative direction) to stretch the rope. Either way the math is the same, and only the initial conditions change. –  ja72 Apr 25 '13 at 13:54
    
Well, either I misunderstood severely the system (see my post), either your math will be wrong because $A<0$: the rope pull on the left. In your computation, the intermediate $u(x)$ (which is the speed, right ?) become imaginary, which is a problem, no ? –  JJ Fleck Apr 25 '13 at 14:15
    
This approach seems to be the one I was looking for, but I don't quite get it to be honest. It'd be nice if you could elaborate on ot a little further. –  user23224 Apr 25 '13 at 18:18

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