Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given his

Question

Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.410 kg·m2. 6.18 kg · m2/s (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 drops to 2.40 rev/s. kg·m2 (c) Suppose instead he keeps his arms in and allows friction with the ice to slow him to 3.00 rev/s. What average torque was exerted if this takes 18.0 seconds? N·m

Explanation / Answer

(a) = 2*6 = 37.7 rad/s
angular momentum = I* = 0.41kg.m^2 * 37.7rad/s = 15.457 kg.m^2/s

(b) = 2*2.4 = 15.1 rad/s
moment of inertia I = 15.457 / 15.1 kg.m^2 = 1.02 kg.m^2

(c) f = 2*3 = 18.8 rad/s
angular accel = (f - i) / t = (18.8 - 37.7) / 18 rad/s^2 = -1.05 rad/s^2

Av torque T = I. = 0.37kg.m^2 * -1.05 rad/s^2 = -0.3885 N.m

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