Particle dynamics applied to geometrically parametric line

In summary, the particle's position x depends on a function of some, presumably known, parameter d ... x=f(d) [all others nill] but the equations of motion are not solvable for zero initial conditions.
  • #1
malaxtom
4
0
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?
 
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  • #2
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
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
It's also really agitating me that the problem can not be solved for zero initial conditions in the d2 case.
 
  • #5
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
Simon Bridge said:
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
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).
 

1. What is particle dynamics?

Particle dynamics is the study of the behavior and motion of particles, such as atoms, molecules, and subatomic particles, within a given system. It involves analyzing the forces acting on these particles and how they affect their motion.

2. How is particle dynamics applied to geometrically parametric line?

In this context, particle dynamics is used to study the motion of particles along a geometrically parametric line, which is a line that can be described using a set of parameters. By analyzing the forces acting on the particles and their interactions with the line, we can understand how the particles move along the line.

3. What is the significance of studying particle dynamics applied to geometrically parametric line?

This type of study is important for understanding the behavior of particles in various systems, such as chemical reactions, fluid dynamics, and material science. It also allows us to predict the behavior of particles in these systems and make improvements or optimizations.

4. What are some examples of geometrically parametric lines?

Geometrically parametric lines can include curves, such as circles, ellipses, and parabolas, as well as more complex shapes like spirals and helices. These lines can also be described in terms of mathematical equations and parameters.

5. How is particle dynamics used in real-world applications?

Particle dynamics applied to geometrically parametric lines is used in a variety of fields, such as engineering, physics, chemistry, and biology. It helps us understand and predict the behavior of particles in these systems, which can then be applied to create more efficient and effective technologies and processes.

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