Timeline for Pendulum forces in orbit
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Feb 11, 2015 at 1:33 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
added 1 character in body; edited tags
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| Feb 11, 2015 at 1:19 | vote | accept | ThePlanMan | ||
| Feb 11, 2015 at 1:19 | answer | added | docscience | timeline score: 1 | |
| Feb 11, 2015 at 1:11 | comment | added | ThePlanMan | Ah ok I've got it now. Could you compile your comments into an answer so I can mark this question as answered please! | |
| Feb 11, 2015 at 1:09 | comment | added | docscience | Last comment: Yes, but only if acted on through the block's center of mass. | |
| Feb 11, 2015 at 1:07 | comment | added | ThePlanMan | Ok thanks. So assuming the force on the square was perpendicular to the surface at the point of attachment of the string would there be any rotation of the square? | |
| Feb 11, 2015 at 1:05 | comment | added | docscience | I should have added, the reason there would be tension in the spring is that the ball has inertia, and resists the leftward acceleration when the string pulls on it. | |
| Feb 11, 2015 at 0:44 | comment | added | docscience | There would be tension. If you accelerate the block to the left, the ball would follow under tension of the string. Then you have essentially 'created' gravity. The dynamics would be a harmonic oscillator (pendulum) with natural frequency of SQRT(L/a) where L is the string length and a is the acceleration. This all assumes the string is inelastic. | |
| Feb 11, 2015 at 0:40 | comment | added | ThePlanMan | So (from the image in the question) if the square is forced leftward, but no force is applied to the sphere there would be no tension? | |
| Feb 11, 2015 at 0:36 | comment | added | docscience | It's a '2-body rigid dynamics' problem, but highly non-linear since the two masses will only create tension in the string when they are somehow forced in opposite directions from one another. When that's not happening they are essentially separate bodies acting on their own with slack in the string. Predicting motion really depend on where forces are applied. And whether the string is inelastic or elastic. If elastic the string can provide restoring force if tension occurs. | |
| Feb 11, 2015 at 0:30 | comment | added | ThePlanMan | So what dynamics govern this problem then? | |
| Feb 11, 2015 at 0:19 | comment | added | docscience | If you are in orbit with this device at zero-g (free fall) then technically its not a 'pendulum' anymore. For a pendulum its assumed you have gravity as a restoring force. No gravity, no restoring force, no pendulum. | |
| Feb 10, 2015 at 22:38 | history | asked | ThePlanMan | CC BY-SA 3.0 |