(17%) Problem 5: An ice skater is spinning at 62 revis and has a moment of inert
ID: 2040887 • Letter: #
Question
(17%) Problem 5: An ice skater is spinning at 62 revis and has a moment of inertia of 0.28 kg . ma ? 33% Part (a) Calculate the angular momentum, in killogram meters squared per second, of the ice skater spinning at 6 2 revis. in kilogram meters cotan asinacos0 atano acotansinh0 cosho tanho cotanho Degrees O Radians detailed view I give upt Ilintst 0% deluction per hint. Itists ,emaining: 1. Feedback: 1% deductin per feedback. 33% Part (b) He reduces his rate of rotation by extending his arms and increasing his moment of inertia. Find te value of his moned inertia (in kilogram meters squared) if his rate of rotation decreases to 0.75 revls. F4 a dallo s ction of the ice to slow him to 325 reis. what is the magnitude of the 33% Part (c) Suppose instead he keeps his arms in average torque that was exerted, in N m, if this takes 19 s?Explanation / Answer
a)w = 6.2 rev/s = 38.95 rad/s
L = I w = 0.28 * 38.95
angular momentum = 10.91 kg.m2/s
b) w' = 0.75 rev/s = 4.71 rad/s
Li = Lf
10.91 = If * 4.71
moment of inertia If = 2.315 kg.m2
c) 3.25 rev/s = 20.42 rad/s
alpha = (w - w0) / t = (38.95 - 20.42) / 19 = 0.975 rad/s2
T = I * alpha = 0.28 * 0.975
magnitude of average torque = 0.273 N.m
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