Particle dynamics applied to geometrically parametric line

  • #1
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Main Question or Discussion Point

I have an odd, yet intriguing question. I want to describe a particle that is constrained to move along a straight line in the x direction. The location of the particle can be described with theoretically infinite parametric parameters, d, along the line. For example:
R = [d 0 0]T
R = [d2 0 0]T
R = [d3 0 0]T
all describe the location of a particle in the x direction. These parameters are dependent on time and are a parametric generalized coordinate of units distance, distance,1/2 and distance1/3. I am interested in describing the motion of the particle to a simple input force along the x direction. The first case is very simple:
[tex]\dot{\bf{R}}= [\dot{d},0,0]^T[/tex]
[tex]\ddot{\bf{R}}= [\ddot{d},0,0]^T[/tex]
and newton's law gives that:
[tex]\dot{\bf{L}} = m*\ddot{\bf{R}}= m\ddot{d} = {F}[/tex]
Simple, easy peasy right? But the other parametric descriptions are more complicated, and don't appear to yield the correct results.
[tex]{\bf{R}}= [{d^2},0,0]^T[/tex]
[tex]\dot{\bf{R}}= [2\dot{d}d,0,0]^T[/tex]
[tex]\ddot{\bf{R}}= [2\ddot{d}d+2\dot{d}^2,0,0]^T[/tex]
[tex]\dot{\bf{L}} = m*\ddot{\bf{R}}= m(2\ddot{d}d+2\dot{d}^2) = {F}[/tex]
which gives the final second order differential equation:
[tex]\ddot{d}= (F-2\dot{d}^2)/(2md)[/tex]
Of interesting note is that this equation can not be solved for zero initial conditions. Also of interest is that the behavior of the system is dependent upon d, which is not the case with the first model of the system. To get the coordinates of the particle, one would of course have to square the numerical solution to the differential equation for d because
[tex]{\bf{R}}= [{d^2},0,0]^T[/tex]
I have also done this derivation utilizing Lagrange's equations and I obtained the exact same differential equation. What am I doing wrong?
 

Answers and Replies

  • #2
Simon Bridge
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I'm not sure of your reasoning.
You are trying to explore the case of a particle whose position x depends on a function of some, presumably known, parameter d ... x=f(d) [all others nill]
For something like a spring, the position depends on the square root of the potential energy since the potential energy depends on the square of the position ... vis: ##\vec r = \frac{2}{k}(\sqrt{U},0,0)## ... so you could sort-of use U as a parameter for x.
If ##x=d^2## then you are considering a hypothetical physical property that depends on the square root of position.
The lagrangian depends on both the kinetic and potential energies - so your use of L above feels off to me.

Your first example gives F=ma because ##a=\ddot x = \ddot d## since, in this case, ##x=d##.
In your second example ##x=d^2## so ##a=\ddot (d^2)## so ##F= m(2d\dot d) = m(2\dot d^2 + 2d\ddot d) = ma## which is correct Newtonian mechanics.
But you say this is incorrect: please explain.
 
  • #3
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The motivation behind this is an exercise in defining motion for a particle on any parametric curve where the parameter of the curve is geometric rather than time. Once applied in 1 space P(d) = [d2,0,0], I could look toward application in 2 and 3 space P(d) = [d2,d3,0] for a particle following any parametric trajectory. With regards to the lagrangian, I did it as a check to make sure my Newtonian mechanics were correct and I got the same result. I seemed to have figured it out. When simulating the two equations, I appeared to get different results but I realized that I was not properly setting up my initial conditions because I left them the same. I am off by around 0.1 but it is likely this is a numerical computation artifact.
 
  • #4
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It's also really agitating me that the problem can not be solved for zero initial conditions in the d2 case.
 
  • #5
Simon Bridge
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Have you solved the problem in post #1 or not?
What is the problem that "cannot be solved for zero initial conditions"?
If you will not be clear, I cannot help you.
 
  • #6
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Have you solved the problem in post #1 or not?
What is the problem that "cannot be solved for zero initial conditions"?
If you will not be clear, I cannot help you.
I have solved my initial problem. The equations of motion I presented and that you confirmed were correct. I was lamenting that because the "d" term accompanies the second derivative term in
[tex] 2d\ddot{d} [/tex]
"d" will move into the denominator and therefore, the equation of motion will be unsolvable with an initial condition of d=0 as the second derivative will be undefined.
[tex] \ddot{d} = (F-2m\dot{d}^2)/(2md) [/tex]
This discontinuity in the second derivative appears to be an unfortunate side effect of the power term in the parametric curve definition. I am trying to determine if there is anyway around it.
 
  • #7
Simon Bridge
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It is a nonlinear DE of form ##\ddot y + y(\dot y)^2 = f(t): \ddot y(0) = \dot y(0) = 0## ... you'd normally look for a way to change the variable to get rid of the nonlinearity. In your case the obvious approach is to substitute ##x=d^2## to get ##m\ddot x = F(t): x=\dot x = 0## ... then substitute back when you have the solution for x(t).
 

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