I've seen some problems solved in a weird way, I just want to be sure: the whole kinetic energy has to be in the lagrangian, right? For example, if we have a particle fixed in a plane with spherical coordinates $(r,\theta,\phi)=(r_0,\theta_0,\phi)$, and that plane is rotating with a constant angular velocity $\omega=(0,0,\omega_0)$, so that the $\phi$ coordinate of the particle is $\phi=\omega t$, then that term of the kinetic energy: in spherical coordinates: $\frac{1}{2}mr^2\omega^2\sin^2(\theta)$, that term has to go in the lagrangian, am I right? And does it have to go in the expression of the total mechanical energy?
$\begingroup$
$\endgroup$
2
asked Jan 27, 2013 at 19:15
-
$\begingroup$ Could you clarify what exactly is rotating - the particle or the plane? How is the angle $\theta$ defined? $\endgroup$– JoeCommented Jan 27, 2013 at 19:25
-
$\begingroup$ @Joe I edited it to try to explain it better. $\endgroup$– MyUserIsThisCommented Jan 27, 2013 at 19:29
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
0
If you can form the problem in such a way as the is a contribution to the kinetic energy that can not change during the time modeled then you can drop that term.
Why? Because the Euler-Lagrange equation is only concerned with differentials and a added or subtracted constant does not affect them.
The simplest example of this is writing the problem in a pair of reference frames where the system is at rest in one and in constant linear motion in another. The KE of the center of mass does not affect the outcome.
answered Jan 27, 2013 at 19:28
dmckee --- ex-moderator kittendmckee --- ex-moderator kitten
86.7k9 gold badges175 silver badges346 bronze badges