HW #9-solutions

# HW #9-solutions - wolz(cmw2833 HW#9 Antoniewicz(57420 This...

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wolz (cmw2833) – HW #9 – Antoniewicz – (57420) 2 For t = , this reduces to θ end = ω 0 σ = 2 . 67 rad / s 0 . 189557 s 1 1 rev 2 π rad = 2 . 24178 rev . 004 10.0 points The net work done in accelerating a propeller from rest to an angular speed of 204 rad/s is 1286 . 8 J. What is the moment of inertia of the pro- peller? Correct answer: 0 . 0618416 kg · m 2 . Explanation: Let : W net = 1286 . 8 J and ω f = 204 rad / s . Since ω 0 = 0 , W net = Δ KE = 1 2 I ( ω 2 f ω 2 i ) = 1 2 I ω 2 f I = 2 W net ω 2 f = 2 (1286 . 8 J) (204 rad / s) 2 = 0 . 0618416 kg · m 2 . keywords: 005 10.0 points A uniform solid disk of radius 7 . 93 m and mass 63 . 1 kg is free to rotate on a frictionless pivot through a point on its rim. 7 . 93 m Pivot If the disk is released from rest in the position shown by the solid circle, what is the speed of its center of mass when the disk reaches the position indicated by the dashed circle? The acceleration of gravity is 9 . 8 m / s 2 .
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HW #9-solutions - wolz(cmw2833 HW#9 Antoniewicz(57420 This...

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