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 initial moment of inertia of a skater for arms and legs in is 0.90 kgm2 while with the final moment of inertia with arms out and one leg extended is 3.0 kgm2 . If she starts out spinning at 4.5 rev/s , what is her angular speed (in rev/s) when her arms and one leg open outward?

Explanation / Answer

apply the law of conservation of angular momentum, we get

I2w2 = I1w1

w2 = [i1/i2]w1

= [0.9 / 3][4.5]

= 1.35 rev/s

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