Timeline for What's the Lagrangian for a freely rotating rod?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Jun 6, 2018 at 19:19 | vote | accept | WillG | ||
| Jun 6, 2018 at 19:05 | comment | added | Javier | @WillG that should be right, though I haven't done the math to check it. You can probably relate it to some property of the inertia tensor. | |
| Jun 6, 2018 at 19:04 | comment | added | WillG | Right, the non-uniqueness of $\vec\omega$ is what really vexes me. I guess this is only true because of the symmetry of a rod, and maybe this same symmetry is what leads to the same answer regardless of the choice of $\vec\omega$. | |
| Jun 6, 2018 at 14:30 | comment | added | Javier | @WillG Well, in the general case this would be more or less equivalent to just calculating the angular velocity. The case of the rod is a bit tricky because I think the angular velocity is not uniquely defined, since the rod doesn't care about rotations about itself. Everything should still work out as long as you take some valid angular velocity, though. | |
| Jun 5, 2018 at 21:59 | comment | added | WillG | @ Javier Thanks, this is convincing. But if the object didn't have such nice symmetry, and you were given the inertia tensor instead, would the method transfer to such a case? Slightly different question I know, but I'm just curious about how this works in general. | |
| Jun 5, 2018 at 21:07 | history | answered | Javier | CC BY-SA 4.0 |