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Let be
L=(x^2 + y^2)x* +2xyy*
where x* = dx/dt and y* = dy/dt. Which physical system is referred it to? why?
Thanks
L=(x^2 + y^2)x* +2xyy*
where x* = dx/dt and y* = dy/dt. Which physical system is referred it to? why?
Thanks
See hereLet be
L=(x^2 + y^2)x* +2xyy*
where x* = dx/dt and y* = dy/dt. Which physical system is referred it to? why?
Thanks
You should also read the link in my previous post. We do not provide solutions to students' homework problems here. It teaches the student very little (if anything) and is unethical. Please stick to giving only hints to problems, and only after the student has shown some effort.$$
L= (x^2+y^2)\dot{x}+ 2xy\dot{y}
$$
Your Lagrangian is a total derivative, and so I don't see any dynamics at all ...
$$
L = \frac{d}{dt}\Big(\frac{1}{3}x^3+xy^2\Big)
$$
Yes, thanks for that,You should also read the link in my previous post. We do not provide solutions to students' homework problems here. It teaches the student very little (if anything) and is unethical. Please stick to giving only hints to problems, and only after the student has shown some effort.
No problem. In future, please include descriptions of what you've tried (like the above) in the "attempt at solution" part of your posts.Yes, sorry about that, I get it (and thanks for the reply). The thing is this is from an exam problem, and after thinking about it, I couldn't reach to any plausible answer. I showed that it fits with the Euler-Lagrange equations, and then I tried writing it in polar coordinates (r, theta), maybe I could see that way what kind of system was it. I know the Lagrangian for several systems, but none of them fit with this one, so I didn't know even how to make a hypothesis. Thanks.
I don't see how. As a counterexample, consider [itex]L=xy\dot{x}\dot{y}[/itex].I see... And what about the fact that L doesn't have a term with x and y only? Can it be used to end up with the same conclusion? I mean, can I say that L is pure kinetic energy, and V=0, and so that F=0? Thanks
A particle with only kinetic energy and no potential energy DOES have dynamics ...I see... And what about the fact that L doesn't have a term with x and y only? Can it be used to end up with the same conclusion? I mean, can I say that L is pure kinetic energy, and V=0, and so that F=0? Thanks