(a) Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given

Question

(a) Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.380 kg · m2.

= ____ 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 2.15 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 22.0 seconds? (Indicate the direction with the sign of your answer. Assume that the skater's rotation is in the positive direction.)

= ____ N · m

Explanation / Answer

(a) = 2*6 = 37.7 rad/s
angular momentum = I*

= 0.38kg.m^2 * 37.7rad/s

= 14.236 kg.m^2/s

(b) = 2*2.15 = 13.502 rad/s
moment of inertia I = 14.236 / 13.502 kg.m^2

= 1.0543 kg.m^2

(c) f = 2*3 = 18.8 rad/s
angular accel = (f - i) / t = (18.8 - 37.7) / 22

= -0.8572 rad/s^2

Av torque T = I. = 0.38kg.m^2 * -0.8572 rad/s^2

= -0.3257 N.m

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