A skater spins with a roatation of 3 cycles per second. If she draws in her arms so that rotational inertia is reduced to 80% of her former amount, what will be her spin rate?
Ans: 1) dL / dt = 0
2) L=L'
3)Iw=I'w'
w = 2pi * 3revs/ 1 sec= 6pi rad/sec
assume I = 100% = 1
I' = reduction by 80% = 20% = 2/10
4) solve for w' ...... = Iw/I' = (1*6pi)/(0.2)
w' = 6pi/0.2 = 94.2477 rad/second
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