Timeline for Choice of Generalized Coordinates
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jul 22 at 14:15 | comment | added | Georgy Zhukov | The textbook is does not imply any such simplifications, which is the source of my confusion. It is the textbook for most classical mechanics courses in American Ph.D. programs, but it is not (by far) the best mechanics book I've read. Your choice of generalized coordinates certainly clarifies certain parts of the problem, though, so thank you very much! | |
| Jul 22 at 14:13 | comment | added | jalex | I am not familiar in detail with this specific textbook. But it might be that $\phi$ is fixed, or that when it comes to velocities $\dot s$ and $\dot t$ are not independent as they are projections of the speed of the disk center which is constrained to be tangent the the path the contact point makes. These kind of problems and complex in general, and the author might expect some simplifications that are discussed in the book that I am not familiar with at this point. | |
| Jul 22 at 14:09 | comment | added | Georgy Zhukov | I wonder if the problem as phrased in the textbook has a typo, then. | |
| Jul 22 at 14:08 | comment | added | jalex | The problem has 3DOF as there are two position variables describing the contact point location on the ramp, and one orientation variable $\phi$ describing the bearing direction (azimuth angle). | |
| Jul 22 at 14:05 | history | edited | jalex | CC BY-SA 4.0 |
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| Jul 22 at 14:01 | comment | added | Georgy Zhukov | So, having written down the angular and translational velocities in these terms, how would you formulate Lagrange equations of motion to solve the (apparently, according to the wording, 2D) problem? | |
| Jul 22 at 13:57 | comment | added | jalex | See my update above. You had defined the velocity in terms of the angular speeds trying to bake-in the no-slip constraint. But if you consider the contact point position as part of the gen. coordinates then the velocity is only the time derivative of positions and the angles play no role here. A separate no-slip condition needs to be considered for the solution, and it will be implemented in the kinetic energy and not in the potential energy where you had the difficulty. | |
| Jul 22 at 13:53 | history | edited | jalex | CC BY-SA 4.0 |
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| Jul 22 at 13:36 | comment | added | Georgy Zhukov | I agree, but I'm having trouble seeing how that clarifies things. | |
| Jul 22 at 13:12 | comment | added | jalex | It is a feature of generalized coordinates to treat positional and velocity coordinates as independent. | |
| Jul 22 at 12:47 | comment | added | Georgy Zhukov | That makes sense! But using $s$ and $t$, it seems like writing down the kinetic energy of the system (in particular the rotational kinetic energies of the wheel about the two axes the wheel can rotate) might be difficult. | |
| Jul 22 at 12:37 | history | answered | jalex | CC BY-SA 4.0 |