Why in some cases there is a term of only kinetic energy in the Lagrangian while is some cases there are both the terms showing both the kinetic energy and the potential energy in the Lagrangian? why this is so?
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$\begingroup$ Do you see this in a textbook? Which eqs.? $\endgroup$– Qmechanic ♦Commented Jan 9, 2018 at 7:24
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$\begingroup$ Yes. in the first chapter of the book'' introduction to classical mechanics by Goldstein. there are some examples one of them is about motion of a single particle in free space using cartesian coordinate/ a bead slliding on a uniformly rotating wire in a force free space , another is about atwood machines. in the former one there is no term of potential energy while in the later one there is. $\endgroup$– ADRCommented Jan 9, 2018 at 7:41
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2 Answers
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If the potential energy is the same everywhere - like for an object on a horizontal surface, then you don't include it in the lagrangian (or more formally, you set it to 0 everywhere)
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$\begingroup$ Thanks . I have another question related to lagrangian formulation. In some of examples the lagrange's equation is taken equal to some quantity fore example in the case of the motion of a single particle in space using cartesian co ordinates lagrange's equation is take equal to force while in the case of atwood's machine lagrange's equation is taken equal to zero. what is the reason ? $\endgroup$– ADRCommented Jan 9, 2018 at 7:32
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If a generalized force has a generalized potential, one can include it in Lagrange equations $$\frac{d}{dt}\frac{\partial (T-U)}{\partial \dot{q}^j}-\frac{\partial (T-U)}{\partial q^j}~=~Q_j,\qquad j~\in \{1,\ldots, n\}, \tag{L}$$ via $Q_j$ or via $U$, but of course not in both. That would be double counting. See also this related Phys.SE post.
answered Jan 9, 2018 at 10:10