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A gymnast is performing a floor routine. In a tumbling run she spins through the air, increasing her angular velocity from 3.00 to 5.20 rev/s while rotating through one-half of a revolution. How much time does this maneuver take?

2006-11-23 13:18:17 · 5 answers · asked by Any help? 1 in Science & Mathematics Physics

5 answers

If the initial angular velocity is 3.0 and the final is 5.2, then the average would be 4.1 rev/sec. The inverse of this is .244 seconds per revolution. 1-1/2 revolutions would take 1.5 * .244 and this would be .366 seconds.

2006-11-23 13:53:46 · answer #1 · answered by MrWiz 4 · 1 0

If the initial angular velocity is 3.0 and the final is 5.2, then the average would be 4.1 rev/sec. The inverse of this is .244 seconds per revolution. 1-1/2 revolutions would take 1.5 * .244 and this would be .366 seconds.

2006-11-23 16:25:43 · answer #2 · answered by praveenplp 2 · 0 0

is she doing a back-half or a front-half?
does this include the meneuvers before it (such as a roundoff back hand back half? or a front hand front tuck front half? or roundoff whip to back half? etc.)

ide say back-halves take longer. like 1 second for a front, and like 1.5 seconds for a back.

ps im a gymnast, so this is coming from how long it feels like

2006-11-23 13:30:49 · answer #3 · answered by x333 1 · 0 0

Not as long as it took to write the question!

2006-11-23 13:25:38 · answer #4 · answered by raredawn 4 · 0 0

t² = 2θ/α..........α = Δω²/2θ.........t² = 4θ²/Δω²---->

t = 2θ/Δω = 2(π)/(2.2*(2π)) = 1/2.2 = .4545 sec

2006-11-23 15:41:53 · answer #5 · answered by Steve 7 · 0 0

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