Nice 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 much work will she have to do to pull her arms in?

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