Question

An ice skater starts spinning at a rate of 1.5 rev s-1 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 velocity?

Answer 1

Apply angular momentum conservation:

$$L_1=L_2,\\ I_1\omega_1=I_2\omega_2,\\\space\\ I_1n_1=I_2n_2,\\\space\\ I_1n_1=\bigg(I_1·\frac34\bigg)n_2,\\\space\\ n_2=n_1·\frac43=2\text{ rev/s}.$$