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NISA [10]
3 years ago
14

Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms f

olded 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 kgm2 and for arms and legs in is 0.90 kgm2 . If she starts out spinning at 4.5rev/s, what is her angular speed (in rev/s) when her arms and one leg open outward?
Physics
1 answer:
nekit [7.7K]3 years ago
4 0

Answer:

Her angular speed (in rev/s) when her arms and one leg open outward is 1.4 rev/s

Explanation:

given information:

moment inertia of arm and leg when in, I₁ = 0.9 kgm²

moment inertia of arm and leg when extended, I₂ = 2.9 kgm²

angular speed when in, ω₁ = 4.5 rev/s

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

L₁ = L₂

I₁ω₁ = I₂ω₂

ω₂ = I₁ω₁/I₂

     = 0.9 x 4.5/2,9

     = 1.4 rev/s

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