Timeline for Why does a particle initially at rest at origin with acceleration as square of its $x$ coordinate ever move?
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
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| Jan 16 at 10:49 | history | reopened |
Michael Seifert John Rennie gandalf61 |
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| Jan 16 at 2:06 | review | Reopen votes | |||
| Jan 16 at 10:49 | |||||
| Dec 13, 2023 at 21:42 | history | left closed in review | Qmechanic♦ | Original close reason(s) were not resolved | |
| Dec 13, 2023 at 20:10 | comment | added | John Doty | @brainfreeze By "practical" you mean "physical". It is only possible in the imaginary world of mathematical modeling. | |
| Dec 13, 2023 at 16:03 | comment | added | Ján Lalinský | In this concrete case you've got the solution function $x(t)$ wrong (it does not obey initial conditions). But in general, similar examples exist which really do manifest the problem you touch on: that there is no unique solution, the particle may stay at the origin, or it may not. This happens for certain different potentials, or functions $a(x)$, such as in the case of Norton's dome. | |
| Dec 13, 2023 at 12:31 | review | Reopen votes | |||
| Dec 13, 2023 at 21:42 | |||||
| Dec 13, 2023 at 11:10 | history | closed |
hft Jon Custer Miyase |
Needs details or clarity | |
| Dec 13, 2023 at 2:23 | vote | accept | brainfreeze | ||
| Dec 12, 2023 at 19:09 | answer | added | John Doty | timeline score: 0 | |
| Dec 12, 2023 at 19:08 | review | Close votes | |||
| Dec 13, 2023 at 11:10 | |||||
| Dec 12, 2023 at 18:54 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
added 1 character in body; edited tags
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| Dec 12, 2023 at 17:55 | comment | added | Michael Seifert | @brainfreeze: That wasn't my point (or ProfRob's point) at all; the point is that your solution doesn't obey the initial conditions See my answer below. | |
| Dec 12, 2023 at 17:54 | answer | added | Michael Seifert | timeline score: 5 | |
| Dec 12, 2023 at 16:50 | comment | added | brainfreeze | @MichaelSeifert If I ignore any practical concerns like technology, precision, etc., I think it should be theoretically possible to create this situation though. Like if an object is kept on some really long digital ruler which measures its position, and a machine is fitted on the object which provides force such that the device gets accelerated along the digital ruler with the magnitude equal to square of the reading of the scale. | |
| Dec 11, 2023 at 17:33 | comment | added | Qmechanic♦ | Possible duplicates: Infinite series of derivatives of position when starting from rest, How does anything move? and links therein. | |
| Dec 11, 2023 at 17:32 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
edited tags; edited tags
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| Dec 11, 2023 at 16:48 | answer | added | SuchAgoodDoge | timeline score: 1 | |
| Dec 11, 2023 at 16:40 | comment | added | Michael Seifert | Exactly. You found a solution to the ODE that arises from Newton's Second Law ($\ddot{x} = x^2$). It's just not the solution that's consistent with the initial conditions you asked for. | |
| Dec 11, 2023 at 15:59 | comment | added | ProfRob | But you specified that $x=0$ when $t=0$ - how does that tally with $x = 6/t^2$ ? | |
| Dec 11, 2023 at 15:52 | history | edited | brainfreeze | CC BY-SA 4.0 |
edited title
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| S Dec 11, 2023 at 15:49 | review | First questions | |||
| Dec 11, 2023 at 16:38 | |||||
| S Dec 11, 2023 at 15:49 | history | asked | brainfreeze | CC BY-SA 4.0 |