Derive the equations of motion, including gear ratios

classical-mechanicscontrol theoryeuler-lagrange-equation

Continuing the topic Equation of motion including gear ratio

I have found some answers to questions, but not all. The following drawing gave me a hint.

enter image description here

Here are the equations of motion that include gear ratios.

Each of the equations corresponds to the motion of the corresponding generalized coordinate, and in this equation it remains unchanged, but the rest are "fitted" to it with the help of gear ratios.

But my question is: explain to me how the Lagrangian was composed to obtain such equations of motion and is this result correct at all?

That is, help reproduce the result.

I would be very glad and grateful for your help!

The code in Mathematica reproduces an attempt to obtain the same results, but to no avail.

Clear["Derivative"]

ClearAll["Global`*"]

K = Sum[1/2 Subscript[J, i] D[Subscript[\[Phi], i][t], t]^2, {i, 2}];


U = Sum[1/2 Subscript[c, 
     10 i + i + 
      1] (Subscript[\[Phi], i][t] - 
       Subscript[\[CapitalPhi], i + 1][t])^2, {i, 1}];

L = ReplaceAll[K - U, 
   Subscript[\[CapitalPhi], 2][
     t] -> (Subscript[r, 1]/Subscript[r, 2]) Subscript[\[Phi], 2][t]];

eq1 = D[D[L, Derivative[1][Subscript[\[Phi], 1]][t]], t] - 
    D[L, Subscript[\[Phi], 1][t]] == Subscript[T, 1][t];

eq1 = D[D[L, Derivative[1][Subscript[\[Phi], 2]][t]], t] - 
     D[L, Subscript[\[Phi], 2][t]] == Subscript[T, 2][t] // Simplify;

Best Answer

Assuming $d_1=0$ (It is not clear to me how the dissipation actuates here) we have

$$ \cases{ \phi_1^* r_1 = \phi_2^* r_2\\ \phi_2^* = \phi_2\\ K = \frac 12 J_1\dot\phi_1^2+\frac 12 J_2\dot \phi_2^2\\ U = \frac 12 c_1\left(\phi_1-\phi_2^*\right)^2\\ L = K - U\\ \left(\nabla_{\phi_1',\phi_2'}L\right)'-\nabla_{\phi_1,\phi_2}L=(T_1(t),T_2(t)) } $$

so we have finally

$$ \cases{ J_1\ddot\phi_1+c_1\phi_1= c_1\frac{r_2}{r_1}\phi_2 + T_1(t)\\ J_2\left(\frac{r_1}{r_2}\right)^2\ddot\phi_2 + c_1\phi_2=c_1\frac{r_1}{r_2}\phi_1+\left(\frac{r_1}{r_2}\right)^2T_2(t) } $$

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