Skip to main content
14 events
when toggle format what by license comment
Apr 12, 2021 at 20:41 answer added Eli timeline score: 3
Apr 11, 2021 at 20:13 answer added Qmechanic timeline score: 4
Apr 11, 2021 at 17:53 comment added Pablo T. Indeed, @Eli, in that example $V=-mgL\cos(\Omega t - \theta)$ there is no dependence on the generalised velocities, hence it is a conservative potential and $V_0=V$, $V_1 = 0$.
Apr 11, 2021 at 17:37 comment added Eli the potential energy id not depending on the velocity i gave you example pendulum
Apr 11, 2021 at 17:31 history edited Pablo T. CC BY-SA 4.0
added 20 characters in body
Apr 11, 2021 at 17:30 comment added Pablo T. @Eli $V_0$ and $V_1$ are homogeneous polynomials of degree $0$ and $1$ on the generalised velocities. As a conservative $V$ only depends on $q^k,t$ and not in $\dot{q}^k$, it equals its own $V_0$, and $V_1 = 0$.
Apr 11, 2021 at 17:17 comment added Eli lets take pendulum that rotate ,the potential energy is $~V=-m\,g\,L\cos(\Omega\,t-\theta)~$ hence the time is not explicit you can't obtain $~V=V_0+V_1$ or what is your $V_0$ and $V_1$ in this case ?
Apr 11, 2021 at 17:07 history edited Pablo T. CC BY-SA 4.0
small edits made, inspired by a comment, to make the post slightly easier to understand.
Apr 11, 2021 at 17:05 comment added Pablo T. By $V$, @Eli, I meant a generalised potential, which can depend on the $\dot{q}$s. But, considering that's not relevant for my question, maybe I should've just assumed $V$ is a conservative potential $V(q,t)$ all along (it can still depend on $t$, tho).
Apr 11, 2021 at 14:35 comment added Eli "We would do the same for V" ?. V is not depending on the velocity and not explicit depending on the time. hence you can split it to $V_0+V_1+V_2$
Apr 10, 2021 at 17:05 history edited Pablo T. CC BY-SA 4.0
edited body
Apr 10, 2021 at 16:48 history edited Pablo T. CC BY-SA 4.0
added 7 characters in body
Apr 10, 2021 at 16:29 history edited Qmechanic CC BY-SA 4.0
edited tags
Apr 10, 2021 at 16:23 history asked Pablo T. CC BY-SA 4.0