A skater makes one revolution per second, f = w/(2p) = 1, with her arms and leg stretched (left). By lowering her leg and bringi ng her arms close to her chest (right) she reduces her moment of inertia from I to I′ = 0.4I. How many revolutions per second, f′ = w¢/(2p), is she making now? Show all your work.
2 answers:
Answer:
f'=w'/2p=2.49rev/s
Explanation:
The angular momentum L must be equal for both cases.
Hence you have that
L before = L after
I = moment of inertia
w =angular momentum
Hence
f'=w'/2p=2.49rev/s
hope this helps!!
Answer:
Answer is given in the attachment.
Explanation:
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