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

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

Given ice skaters initial moment of inertia is I1 = 0.80 kg m2 (arms and legs in ) and I2 = 3.4 kg m2 (with arms out and one leg extended) initial angular speed is w1 =4.6 rev/s final angular speed w2 = ? from conservation of angular momentum we know that angular momentum is L = I*W L1 = L2 I1*W1 = I2*W2 W2 = I1*W1 /I2 W2 = 0.80*4.6 / 3.4 rev/s W2 = 1.0824 rev/s

ID: 1783120 • Letter: I

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.

Part A

Suppose that the initial moment of inertia of a skater for arms and legs in is 0.80 kgm2 while with the final moment of inertia with arms out and one leg extended is 3.4 kgm2 . If she starts out spinning at 4.6 rev/s , what is her angular speed (in rev/s) when her arms and one leg open outward?

Explanation / Answer

Given ice skaters

initial

moment of inertia is I1 = 0.80 kg m2 (arms and legs in )

and I2 = 3.4 kg m2 (with arms out and one leg extended)

initial angular speed is w1 =4.6 rev/s

final angular speed w2 = ?

from conservation of angular momentum

we know that angular momentum is L = I*W

L1 = L2

I1*W1 = I2*W2

W2 = I1*W1 /I2

W2 = 0.80*4.6 / 3.4 rev/s

W2 = 1.0824 rev/s

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

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

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