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

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wolz (cmw2833) – HW #9 – Antoniewicz – (57420) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 of 3) 4.0 points As a result oF Friction, the angular speed oF a wheel c hanges with time according to d θ d t = ω 0 e σ t , where ω 0 and σ are constants. The angular speed changes From an initial angular speed oF 2 . 67 rad / s to 1 . 27 rad / s in 3 . 92 s . Determine the magnitude oF the angular acceleration aFter 1 . 39 s. Correct answer: 0 . 388884 rad / s 2 . Explanation: Let : ω 0 = 2 . 67 rad / s , t 0 = 0 , ω 2 = 1 . 27 rad / s , t 2 = 3 . 92 s , and t 3 = 1 . 39 s . The equation oF motion is ω = ω 0 e σ t , so ω 2 ω 0 = e σ t 2 ln p ω 2 ω 0 P = σ t 2 σ = ln p ω 2 ω 0 P t 2 = ln p 1 . 27 rad / s 2 . 67 rad / s P 3 . 92 s = 0 . 189557 s 1 . Thus the angular acceleration at t 3 is α ( t 3 ) = d ω dt = ω 0 ( σ ) e σ t 3 = (2 . 67 rad / s) (0 . 189557 s 1 ) × e (0 . 189557 s - 1 ) (1 . 39 s) = 0 . 388884 rad / s 2 b ( t 3 ) b = 0 . 388884 rad / s 2 . 002 (part 2 of 3) 3.0 points How many revolutions does the wheel make aFter 2 . 91 s ? Correct answer: 0 . 950465 rev. Explanation: Let : t f = 2 . 91 s . θ = i t f 0 ω 0 e σ t dt = ω 0 σ i t f 0 e σ t ( σ dt ) = ω 0 σ e σ t v v v t f 0 = ω 0 σ ( 1 e σ t f ) = 2 . 67 rad / s 0 . 189557 s 1 × b 1 e (0 . 189557 s - 1 ) (2 . 91 s) B = 5 . 97195 rad , so the number oF revolutions is n = θ 2 π = 5 . 97195 rad 2 π = 0 . 950465 rev . 003 (part 3 of 3) 3.0 points ±ind the number oF revolutions it makes be- Fore coming to rest. Correct answer: 2 . 24178 rev. Explanation: θ = ω 0 σ ( 1 e σ t f )
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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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