A skater has 2.34 kg m^2 at 0.800 rev/s with arms out, and 0.363 kg m^2 at 5.16 rev/s with arms in. What are her rotational kinetic energies before and after?

1. Rotational kinetic energy is one half I omega squared, with omega in rad/s. 2. Before, omega = (0.800 rev/s)(2 pi rad/rev) = 5.03 rad/s. 3. So KE = (0.5)(2.34 kg m^2)(5.03 rad/s) squared = (0.5)(2.34)(25.3) = 29.6 J. 4. After, omega prime = (5.16 rev/s)(2 pi rad/rev) = 32.4 rad/s. 5. So KE prime = (0.5)(0.363 kg m^2)(32.4 rad/s) squared = (0.5)(0.363)(1051) = 191 J. 6. The energy rises even though angular momentum is unchanged, because omega enters squared. 7. Reversing the two values would have her lose energy while speeding up, which cannot happen. _Source: OpenStax College Physics (CC BY 4.0), Ch 10 "Rotational Motion and Angular Momentum", section 10.5 Angular Momentum and Its Conservation_