Suppose that the moment of inertia of a skater with arms out and one leg extended is 2.9 kg⋅m2 and for arms and legs in is 0.90 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?

Question

Josie Stephenson

Mathematics

11 August, 07:46

Suppose that the moment of inertia of a skater with arms out and one leg extended is 2.9 kg⋅m2 and for arms and legs in is 0.90 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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Belen Arnold · Answer 1

1.5 rev/s Since angular momentum is conserved, the product of the moment of inertia and the rotational rate will remain a constant. So, assuming X as the desired rate of rotation we have: X * 2.9 = 4.8 * 0.90 Now solve for X X * 2.9 = 4.8 * 0.90 X * 2.9 = 4.32 X = 1.489655172 So when her arms and one leg go outward, her rotation slows to 1.489 rev/s. Rounding to 2 significant figures, gives 1.5 rev/s.