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A ballet dancer spins about a vertical a...

A ballet dancer spins about a vertical axis at `60 rpm` with arms outstretched. When her arms are folded the angular frequency increases to `90 rpm`. Find the change in her moment of inertia.

Text Solution

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By the principle of conservation of angular momentum `Ixx60=I_(2)xx90`
Final moment of inertia, `I_(2)=(2I)/3`
Change in moment of inertia `=I-(2I)/3=I/3`
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