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 m^2. 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 1.75 rev/s. 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 19.0 seconds? (Indicate the direction with the sign of your answer. Assume that the skater's rotation is in the positive direction.) Consider the angular momentum principle in its general form. Check the values you use for the known variables.

Explanation / Answer

(c)

The change in angular velocity is:

3.0 rev/s-6.0 rev/s=-3.0 rev/s=-3.0*2*pi= -18.8 rad/s

angular accel ? = (?f - ?i) / t = (-18.8 ) / 19 rad/s^2 = -0.992 rad/s^2

Av torque T = I.? = 0.41kg.m^2 * -0.992 rad/s^2 = -0.41 N.m

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