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An ice skater has a moment of inertia of 5. 0 kg∙m2 when her arms are outstretched. At this time she is spinning at 3. 0 revolutions per second (rps). If she pulls in her arms and decreases her moment of inertia to 2. 0 kg∙m2, how fast will she be spinning?.

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Answer:

2440.24 J

Explanation:

Moment of inertia, I1 = 5 kg m^2

frequency, f1 = 3 rps

ω1 = 2 x π x f1 = 2 x π x 3 = 6 π rad/s

Moment of inertia, I2 = 2 kg m^2

Let the new frequency is f2.

ω2 = 2 x π x f2

here no external torque is applied, so the angular momentum remains constant.

I1 x ω1 = I2 x ω2

5 x 6 π = 2 x 2 x π x f2

f2 = 7.5 rps

ω2 = 2 x π x 7.5 = 15 π

Initial kinetic energy, K1 = 1/2 x I1 x ω1^2 = 0.5 x 5 x (6 π)² = 887.36 J

Final kinetic energy, K2 = 1/2 x I2 x ω2^2 = 0.5 x 3 x (15 π)² = 3327.6 J

Work done, W = Change in kinetic energy = 3327.6 - 887.36 = 2440.24 J

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