Answer to Question #249956 in Physics for wafa

Question

an ice skater starts spinning at a rate of 1.5 rev/s with arms extended. He then pulls his arms in close to his body, resulting in a decrease of his moment of inertia to three quarters of the initial value. what is the skater's final angular free

Answer 1

Apply conservation of angular momentum:

$$I_1\omega_1=I_2\omega_2,\space\omega=2\pi n,\\ I_1n_1=I_2n_2,\\\space\\ n_2=n_1\frac{I_1}{I_2}=n_1\frac{I_1}{3/4I_1}=2\text{ rev/s}.$$