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 3.2 kgm2 and for arms and legs in is 0.90 kgm2 . If she starts out spinning at 5.1 rev/s , what is her angular speed (in rev/s) when her arms and one leg open outward?

Explanation / Answer

This is based on the law of conservation of angular momentum.
Let I in and w in be the initial moment of inertia and initial angular momentum. Let I f and w f be the respective final data.
By the law of conservation of angular momentum, in the absence of external torque,
I f w f = I in w in
So w f = (I in/ I f )* w in = (0.9/3.2)*5.1 = 1.43 rev/s

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