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1 June, 09:51

Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.5 kg⋅m2 and for arms and legs in is 0.70 kg⋅m2. if she starts out spinning at 4.8 rev/s, what is her angular speed (in rev/s) when her arms and one leg open outward?

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  1. 1 June, 10:52
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    Given:

    I₁ = 0.70 kg-m², the moment of inertia with arms and legs in

    I₂ = 3.5 kg-m², the moment of inertia with arms and a leg out.

    ω₁ = 4.8 rev/s, the angular speed with arms and legs in.

    That is,

    ω₁ = (4.8 rev/s) * (2π rad/rev) = 30.159 rad/s

    Let ω₂ = the angular speed with arms and a leg out.

    Because momentum is conserved, therefore

    I₂ω₂ = I₁ω₁

    ω₂ = (I₁/I₂) ω₁

    = (0.7/3.5) * (30.159)

    = 6.032 rad/s

    ω₂ = (6.032 rad/s) * (1 / (2π) rev/rad) = 0.96 rev/s

    Answer: 0.96 rev/s
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