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14 April, 12:02

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 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?

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  1. 14 April, 12:58
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    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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