Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.1 kg⋅m2 and for arms and legs in is 0.90 kg⋅m2 . If she starts out spinning at 4.0 rev/s , what is her angular speed (in rev/s) when her arms and one leg open outward?

The angular speed (in rev/s) when her arms and one leg open outward is 1.161 rev/s

Explanation:

Given;

moment of inertia of a skater with arms out, = 3.1 kg.m²

moment of inertia of a skater with arms in, = 0.9 kg.m²

inward angular speed, = 4 rev/s

The angular momentum of the skater when her arms are out and one leg extended is equal to her angular momentum when her arms and legs are in.

Therefore, the angular speed (in rev/s) when her arms and one leg open outward is 1.161 rev/s

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