Timeline for Which higher-order terms require 4th-order integration of quadratically-constrained dynamics?
Current License: CC BY-SA 4.0
12 events
when toggle format | what | by | license | comment | |
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Jun 25 at 18:50 | vote | accept | Not a chance | ||
Jun 23 at 15:13 | history | edited | Not a chance | CC BY-SA 4.0 |
added quadratically to the title
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Jun 23 at 10:47 | answer | added | LPZ | timeline score: 2 | |
Jun 23 at 7:48 | history | edited | Not a chance | CC BY-SA 4.0 |
added image, made question more clear
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Jun 22 at 14:21 | history | edited | Kyle Kanos | CC BY-SA 4.0 |
boldsymbol is for greeks letters
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Jun 22 at 14:14 | answer | added | Kyle Kanos | timeline score: 0 | |
Jun 22 at 13:15 | comment | added | Kyle Kanos | FWIW, I suspect there's something screwy in your derivation as your EOM (your last equation) does not appear to line up with expected equations (cf. this Physics.SE post). | |
Jun 22 at 13:04 | comment | added | Kyle Kanos | Wouldn't that require $\ddot{g}=0$? Is that condition true? Also, why solve it in Cartesian instead of the more straight-forward spherical coordinates? | |
Jun 22 at 10:40 | comment | added | Not a chance | well after calling myself completely into question, yes. The first derivative being $2\dot xx + 2\dot yy + 2\dot zz$ | |
Jun 22 at 10:15 | comment | added | basics | Are you sure about this "Using the second time derivative of g we have $\dot{x}^2+\dot{y}^2+\dot{z}^2=−(\ddot{x}x+\ddot{y}y+\ddot{z}z)$? | |
Jun 22 at 9:34 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
added 6 characters in body; edited tags; edited title
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Jun 22 at 9:21 | history | asked | Not a chance | CC BY-SA 4.0 |