How do I calculate the individual interactions between two links of a chain?

[ComputerPsi]

I want to simulate the (probably chaotic) two dimensional movement of a chain, given that there is no gravity, and all of the links of the chain have some constant mass. Additionally, there is an assumption that the chain cannot collapse - all of the links of the chain will always be touching at their tips. Basically, one would move one side of the chain up and down and observe how this affects the rest of the chain.

I am trying to find some type of physics formulas that allow me to understand the individual interactions between two links of a chain. After one link moves at a certain degree into some direction with some energy, how much energy, in what direction will go into the second link? How will the first link be affected from the second link being connected to it? etc..
I'm not very interested in a formula that describes the chain as a whole. Instead I'm interested in the specific interaction between two links of the chain.

Does anybody know the generic names for these formulas, or some place that references them?

Thank you,
Ven

 

[Dickfore]

Each link is a rigid rod having three degrees of freedom (2 translational and 1 rotational). Let $(x_{k}, y_{k})$ be the Cartesian coordinates of the CM of the kth rod and $\theta_{k}$ is the angle the rod's axis builds with the positive x-axis. Therefore, the left and right ends of the kth rod have coordinates:

$$x_{l k} = x_{k} - l \, \cos{\theta_{k}}$$

$$y_{l k} = y_{k} - l \, \sin{\theta_{k}}$$

$$x_{r k} = x_{k} + l \, \cos{\theta_{k}}$$

$$y_{r k} = y_{k} + l \, \sin{\theta_{k}}$$

But, the left end of the (k+1)st rod is the same as the right end of the kth rod, so we have the relation:

$$x_{k + 1} - l \, \cos{\theta_{k + 1}} = x_{k} + l \, \cos{\theta_{k}} \Rightarrow x_{k + 1} - x_{k} = l \, \left( \cos{\theta_{k}} + \cos{\theta_{k +1}} \right)$$

$$y_{k + 1} - l \, \sin{\theta_{k + 1}} = y_{k} + l \, \sin{\theta_{k}} \Rightarrow y_{k + 1} - y_{k} = l \, \left( \sin{\theta_{k}} + \sin{\theta_{k +1}} \right)$$

Knowing the coordinates of the left hand end of the 0th rod:

$$x_{l 0} = x_{0} - l \, \cos{\theta_{0}} \Rightarrow x_{0} = x_{l 0} + l \, \cos{\theta_{0}}$$

$$y_{l 0} = y_{0} - l \, \ain{\theta_{0}} \Rightarrow y_{0} = y_{l 0} + l \, \sin{\theta_{0}}$$

We can get the CM coordinates of the kth rod as:

$$x_{k} = x_{l 0} + 2 \, l \, \sum_{j = 0}^{k - 1}{\cos{\theta_{j}}} + l \, \cos{\theta_{k}}, \; k \ge 1$$

$$y_{k} = y_{l 0} + 2 \, l \, \sum_{j = 0}^{k - 1}{\sin{\theta_{j}}} + l \, \sin{\theta_{k}}, \; k \ge 1$$

The velocities of the CM of the kth rod are:

$$\dot{x}_{k} = \dot{x}_{l 0} - 2 \, l \, \sum_{j = 0}^{k - 1}{\dot{\theta}_{j} \, \sin{\theta_{j}}} - l \, \dot{\theta}_{k} \, \sin{\theta_{k}}, \; k \ge 1$$

$$\dot{y}_{k} = \dot{y}_{l 0} + 2 \, l \, \sum_{j = 0}^{k - 1}{\dot{\theta}_{j} \, \cos{\theta_{j}}} + l \, \dot{\theta}_{k} \, \cos{\theta_{k}}, \; k \ge 1$$

Thus, the kinetic energy of the translational motion is:

$$T_{\mathrm{trans}} = \frac{M}{2} \, \sum_{i = 0}^{N - 1}{\left(\dot{x}^{2}_{i} + \dot{y}^{2}_{i}\right)}$$

Substitute the upper equations and do some regrouping of terms! I suppose you want to consider the non-trivial case when there is more than 1 rod.

Added to this kinetic energy, there is also the kinetic energy due to rotational motion of each rod. The moment of inertia of each rod is:

$$I_{0} = \frac{M \, L^{2}}{12}$$

$$T_{\mathrm{rot}} = \frac{I_{0}}{2} \, \sum_{i = 0}^{N - 1}{\dot{\theta}^{2}_{i}}$$

If there is no potential energy due to the change in the configuration of the chain and there is no potential energy due to external fields, then the total Lagrangian of the system is:

$$L = T_{\mathrm{trans}} + T_{\mathrm{rot}}$$

and the equations of motion are:

$$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}_{k}}\right) - \frac{\partial L}{\partial \theta_{k}} = 0, 0 \le k \le N - 1$$

$$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}_{l 0}}\right) - \frac{\partial L}{\partial x_{l 0}} = 0$$

$$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{y}_{l 0}}\right) - \frac{\partial L}{\partial y_{l 0}} = 0$$

 

[ComputerPsi]

Wow. Thank you Dickfore for that input. I've started to look at the top part of the equations.. it's a lot to take in, but I'll figure it out due some time.
For now I just have one small question to confirm what I have been thinking about for the top formulas:
The variable l. That's the length of each rod, correct?

Thanks,
Ven

 

[ComputerPsi]

...After thinking about it, it seems (X, Y) represent the "center" of each rod and the variable l is actually 1/2 of the length of each rod. Is this correct?

 

[ComputerPsi]

Thus, the kinetic energy of the translational motion is:

$$T_{\mathrm{trans}} = \frac{M}{2} \, \sum_{i = 0}^{N - 1}{\left(\dot{x}^{2}_{i} + \dot{y}^{2}_{i}\right)}$$

How did you get this? Especially the M / 2 part..

 

[Dickfore]

Kinteric energy is:

$$T = \frac{1}{2} \, M \, v^{2}$$

where

$$v^{2} = v^{2}_{x} + v^{2}_{y}$$

is the speed of the CM squared.

You are right about the previous post, the length of the rod is actually $2 l$, making the formula for the moment of inertia of each rod (with respect to an axis passing through the CM):

$$I_{0} = \frac{M L^{2}}{3}$$