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Dirk the ice skater spins at 4.51 rev/s and has moment of inertia is 0.56 kg...

Dirk the ice skater spins at 4.51 rev/s and has moment of inertia is 0.56 kg ⋅ m2 .  If he decreases his rate of spin to 2.45 rev/s by spreading his arms, what is his new moment of inertia?

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Answer #1

By conservation of angular momentum

0.56×4.51 = × 2.45

= 1.03 kg.m2

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