(a) Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.400 kg ? m2 . (b) He reduces his rate of spin (his angular velocity) by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia if his angular velocity decreases to 1.25 rev/s. (c) Suppose instead he keeps his arms in and allows friction of the ice to slow him to 3.00 rev/s. What average torque was exerted if this takes 15.0 s?

Step-by-step solution Step 1 of 5 The formula for angular momentum of ice skater is, Here, is the angular momentum of the skater, is the moment of inertia of the skater and is the angular velocity. Step 2 of 5 (a) Convert angular velocity in rad/s. The angular momentum of skater is, Substitute for and 37.7 rad/s for . Hence, the angular momentum of the skater is . Step 3 of 5 (b) Convert angular velocity in rad/s. The new moment of inertia is calculated by the angular momentum conservation law as, Here, is the initial moment of inertia of the system, is the final moment of inertia of the system, is the initial angular velocity and is the final angular velocity. Substitute for L and for . Hence, the new moment of inertia is .