Ice skaters often end their performances with spin turns, where they spin very f

Question

Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms folded in and legs together. Upon ending, their arms extend outward, proclaiming their finish. Not quite as noticeably, one leg goes out as well. Suppose that the moment of inertia of a skater with arms out and one leg extended is 2.9 kg middot m^2 and for arms and legs in is 0.90 kg middot m^2 If she starts out spinning at 6.0 rev/s, what is her angular speed (in rev/s) when her arms and one leg open outward? Express your answer using two significant figures.

Explanation / Answer

we have conservation of angular momentum which is

in the absence of external torque the product of MOment of inertia and angular speed remains constant

so I1w1=I2w2

now 0.9X6=2.9Xw2 now w2=1.8620rev/sec

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