two nonholonomic particles in mutual potential


by coolnessitself
Tags: mutual, nonholonomic, particles, potential
coolnessitself
coolnessitself is offline
#1
Nov18-10, 01:37 PM
P: 32
Hi all,
1. The problem statement, all variables and given/known data
Not actually homework, but hopefully it fits well in this forum.
I have two particles that can only move in the direction of their respective heading angles. I'm trying to relate a radially-symmetric mutual potential between the particles to the acceleration of this heading angle. For example,
[tex]
\begin{bmatrix}
\dot{x}_i \\
\dot{y}_i \\
\dot{z}_i \\
\dot{\theta}_i \\
\dot{\Theta}_i \\
\dot{\phi}_i \\
\dot{\Phi}_i
\end{bmatrix} = \begin{bmatrix} v\sin\theta_i\cos\phi_i \\ v\sin\theta_i\sin\phi_i \\ v\cos\theta_i \\ \Theta_i \\ f(r) \\ \Phi_i \\ g(r)\end{bmatrix}
[/tex]
for the distance r between 1 and 2. Normally we might say that [tex]\ddot{r} = -\nabla G(r)[/tex], but here in order for r to change, the heading angles must change. So my question is how to find f(r) and g(r) (the acceleration of the heading angles) so that the same behavior occurs as with [tex]\ddot{r} = -\nabla G(r)[/tex]. If we require that the direction of r is constant throughout this acceleration, ie [tex]\dot{r} = (\ldots) \hat{r} + 0\hat{\theta} + 0\hat{\phi}[/tex] then it seems to me like there should be a unique solution (if one exists).

2. Relevant equations



3. The attempt at a solution
If the direction of r doesn't change,
[tex] (\dot{z}_2 - \dot{z}_1)r = (z_2 - z_1)\dot{r}[/tex] and [tex] (\dot{y}_2 - \dot{y}_1)(x_2-x_1) = (y_2-y_1)(\dot{x}_2 - \dot{x}_1)[/tex]. My hope would be to then expand out [tex]\ddot{r}[/tex] until equations for [tex]\ddot{\theta}[/tex] and [tex]\ddot{\phi}[/tex] appear, then use that condition to find one solution. Something like
[tex] (\dot{r})^2 + r\ddot{r} = (\dot{x}_2 - \dot{x}_1)^2 + (x_2-x_1)(\ddot{x}_2 - \ddot{x}_1) + (\dot{y}_2 - \dot{y}_1)^2 + (y_2-y_1)(\ddot{y}_2 - \ddot{y}_1) + (\dot{z}_2 - \dot{z}_1)^2 + (z_2-z_1)(\ddot{z}_2 - \ddot{z}_1)[/tex]
But after plugging in [tex]\dot{x}[/tex], etc. with phis and thetas, It doesn't seem to lead anywhere.
So I get the feeling I'm going about this incorrectly. Physics gurus, if you have any suggestions for an alternative approaches to finding this heading-angle-potential, I'd appreciate your input!
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