Timeline for What makes us twist in a somersault?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jul 31, 2015 at 7:06 | vote | accept | monoid | ||
| Jul 29, 2015 at 17:34 | comment | added | CR Drost | I've added an extra section showing all of the effects that you neglect in order to fixate on the Coriolis force, which then describes a slight increase of your on-axis rotation coupled with a constant twist counterclockwise in the co-rotating coordinates. | |
| Jul 29, 2015 at 17:31 | history | edited | CR Drost | CC BY-SA 3.0 |
extra section.
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| Jul 29, 2015 at 15:03 | comment | added | CR Drost | Sure. Again, think of what it looks like from the back-flipper's perspective: your shoulders have a net motion "back-wards" (i.e. in the direction that your back points) relative to your center of mass. You adduct your arm and, ignoring center of mass changes (a "cartwheel" rotation), your shoulders go backwards at different speeds, creating the "twist". This is always going to be a twist which sends your right shoulder "front-ward" and your left shoulder "back-ward", so if your back is pointed "up" from a perspective on the ground, the left shoulder goes "more up" than your right. | |
| Jul 29, 2015 at 9:02 | comment | added | monoid | Could you explain, why lowering the left arm accelerates the left side? | |
| Jul 29, 2015 at 6:09 | comment | added | CR Drost | It's $(s + \Delta s, s + \frac{\Delta s}{2}, s)$, where all quantities are positive. Because you are face-down and your left side is going upward, it is also a counterclockwise twist in your own frame. But if you don't believe me, there should also be a straightforward derivation in the corotating frame (which just has a centrifugal and Coriolis force; the twist should come from the Coriolis force on the infalling mass. | |
| Jul 29, 2015 at 0:44 | comment | added | monoid | Oh, I think it should rather be $(s',s',s') \mapsto (s'-\Delta s',s - \frac{\Delta s'}{2},s')$. And then due to momentum of inertia, the left shoulder would be slower as the right one and therefore the gymnast twists counter-clockwise?! | |
| Jul 29, 2015 at 0:20 | comment | added | monoid | Adduction of the left limb lets you rotate around the outstretched arm. I do have problems understanding why twisting late should result in a counter-clockwise direction. Because using your explanation (I guess your axis of somersault rotation is through the hips) in a late position the gymnast is facing down, the velocities should be $(s',s',s')$. Lowering the left arm would result in $(s'+\Delta s', s' + \frac{\Delta s'}{2}, s')$. So the left should would rotate clockwise around the right arm? | |
| Jul 28, 2015 at 23:59 | comment | added | monoid | Yes, I am judging the direction of the twist from the ground, like seeing the mannequin on the picture I uploaded. | |
| Jul 28, 2015 at 23:52 | history | edited | CR Drost | CC BY-SA 3.0 |
added 765 characters in body
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| Jul 28, 2015 at 23:43 | history | answered | CR Drost | CC BY-SA 3.0 |