The movement of a chain

Each link is a rigid rod having three degrees of freedom (2 translational and 1 rotational). Let [itex](x_{k}, y_{k})[/itex] be the Cartesian coordinates of the CM of the kth rod and [itex]\theta_{k}[/itex] 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:

[tex]
x_{l k} = x_{k} - l \, \cos{\theta_{k}}
[/tex]

[tex]
y_{l k} = y_{k} - l \, \sin{\theta_{k}}
[/tex]

[tex]
x_{r k} = x_{k} + l \, \cos{\theta_{k}}
[/tex]

[tex]
y_{r k} = y_{k} + l \, \sin{\theta_{k}}
[/tex]

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:

[tex]
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)
[/tex]

[tex]
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)
[/tex]

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

[tex]
x_{l 0} = x_{0} - l \, \cos{\theta_{0}} \Rightarrow x_{0} = x_{l 0} + l \, \cos{\theta_{0}}
[/tex]

[tex]
y_{l 0} = y_{0} - l \, \ain{\theta_{0}} \Rightarrow y_{0} = y_{l 0} + l \, \sin{\theta_{0}}
[/tex]

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

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

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

The velocities of the CM of the kth rod are:

[tex]
\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
[/tex]

[tex]
\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
[/tex]

Thus, the kinetic energy of the translational motion is:

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

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:

[tex]
I_{0} = \frac{M \, L^{2}}{12}
[/tex]

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

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:

[tex]
L = T_{\mathrm{trans}} + T_{\mathrm{rot}}
[/tex]

and the equations of motion are:

[tex]
\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
[/tex]

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

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

 

Kinteric energy is:

[tex]
T = \frac{1}{2} \, M \, v^{2}
[/tex]

where

[tex]
v^{2} = v^{2}_{x} + v^{2}_{y}
[/tex]

is the speed of the CM squared.

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

[tex]
I_{0} = \frac{M L^{2}}{3}
[/tex]