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The kinetic energy & potential energy of a dynamical system are given by $$T = \sum_{i=1}^4 f_i(q_i)\left(\frac{dq_i}{dt}\right)^2$$ $$V =\sum_{i=1}^4 V_i(q_i)$$ where $q_i$'s are generalized coordinates; $i=1,\dots, 4$. To show that the Lagrange's equations then separate and the problem can be reduced to a quadrature.

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thanks for editting – user54843 Jan 10 '13 at 12:43
I would suggest you give this question a title which is a bit more informative. – Fabian Jan 10 '13 at 13:58


Note that the total Lagrangian separates, $L = T- V= \sum_{i=1}^{4} L_i$ with $$ L_i = f_i(q_i) \left(\frac{dq_i}{dt} \right)^2 - V_i(q_i).$$

So what can you say about the Euler-Lagrange equations $$ \frac{d}{dt} \frac{\partial}{\partial_{\dot q_i}} L = \frac{\partial}{\partial_{q_i}}L ?$$

Spoiler below:

As only $L_i$ depends on $q_i$ and $\dot q_i$, this is equivalent to $$ \tag{1}\frac{d}{dt} \frac{\partial}{\partial_{\dot q_i}} L_i = \frac{\partial}{\partial_{q_i}}L_i$$ and the equations separate. In the end, you have to solve a single equation per coordinate of the type (1). Due to the fact that $L_i$ does not depend explicitly on time the energy $$ E_i = T+ V =f_i(q_i) \left(\frac{dq_i}{dt} \right)^2 + V_i(q_i) $$ is conserved and this last equation can be integrated via separation of variables.

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