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Problem 7-10.5.41 : An ice skater is spinning at 5.2 rev/'s and has a moment of inertia of .24 kg - m”. Part (o) Calculate the angular momentum, in kilogram meters squared per second, of the ice skater spinning at 5.7 rev/s. To calculate the angular momentum, we first need to convert the given angular velocity from revolutions per second to radians per second. 52rev 2xrad 1s 1 rev w w — (5.2-2x) rad/s Next, we can apply the relation between the angular velocity, moment of inertia, and angular momentum to find the answer. Ly Iw Ly~ 024kg -m?.(5.2-2x) rad/s Ly~ 7841kg -m?/s Part (b) He reduces his rate of rotation by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia (in kilogram meters squared) if his rate of rotation decreases to /. 75 rev/s. To find the answer, we can use the conservation of momentum. Let's begin by converting the given angular velocity from revolutions per second to radians per second. 0.75rev 2xrrad w2 1s 1 rev wy — (0.75 - 2x) rad/s Now we can use our results from part (a) and the conservation of momentum to get the correct answer. Ly Ly Iw = Lw, Lo w2 024kg -m?.(5.2-2x) rad/s (0.75 - 2x) rad/s I, I, — 1.664kg -m?