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Apr 25, 2018 at 22:24 answer added AngusTheMan timeline score: 2
Apr 25, 2018 at 21:44 comment added Daniel Teixeira that's still a quadratic term, in some sense, which is the point
Apr 25, 2018 at 21:43 comment added ZeroTheHero There are so many assumptions here it's hard to know what to answer. In a rotating frame for instance, where energy is still conserved, the kinetic energy does not have the form $\frac{1}{2}m\dot{q}^2$ but will contain a Coriolis-type term proportional to powers of $\dot{q}+\vec\omega\times \vec r$ and thus the $\dot{q}$ term does not "vanish"
Apr 25, 2018 at 21:10 history edited Daniel Teixeira CC BY-SA 3.0
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Apr 25, 2018 at 20:43 history edited Qmechanic
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Apr 25, 2018 at 20:32 history edited Daniel Teixeira CC BY-SA 3.0
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Apr 25, 2018 at 20:30 comment added Daniel Teixeira @Qmechanic this is not a duplicate, I'm not relating Hamiltonian and Lagrangian mechanics, I'm asking if their relation (the question you mentioned) is the same as that between conservation of energy and the equations of motion.
Apr 25, 2018 at 20:27 history reopened Qmechanic classical-mechanics
Apr 25, 2018 at 20:26 comment added Qmechanic Possible duplicates: physics.stackexchange.com/q/105912/2451 and links therein.
Apr 25, 2018 at 20:26 history closed Qmechanic classical-mechanics Duplicate of Equivalence between Hamiltonian and Lagrangian Mechanics
Apr 25, 2018 at 20:25 history edited Qmechanic CC BY-SA 3.0
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Apr 25, 2018 at 19:38 history asked Daniel Teixeira CC BY-SA 3.0