ohw 22 optional-solutions

ohw 22 optional-solutions - Bodet (ngb299) ohw 22 optional...

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Bodet (ngb299) – ohw 22 optional – turner – (57340) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 of 3) 10.0 points As a result oF Friction, the angular speed oF a wheel c hanges with time according to dt = ω 0 e - σt , where ω 0 and σ are constants. The angular speed changes From an initial angular speed oF 3 . 87 rad / sto3 . 17 rad / sin33 . 93 s . Determine the magnitude oF the angular acceleration aFter 2 . 68 s. Correct answer: 0 . 0224014 rad / s 2 . Explanation: Let : ω 0 =3 . 87 rad / s , t 0 =0 , ω 2 =3 . 17 rad / s , t 2 =33 . 93 s , and t 3 =2 . 68 s . The equation oF motion is ω = ω 0 e - σt , so ω 2 ω 0 = e - σt 2 ln ± ω 2 ω 0 ² = - σt 2 σ = - ln ± ω 2 ω 0 ² t 2 = - ln ± 3 . 17 rad / s 3 . 87 rad / s ² 33 . 93 s =0 . 00588043 s - 1 . Thus the angular acceleration at t 3 is α ( t 3 )= dt = ω 0 ( - σ ) e - σt 3 = - (3 . 87 rad / s) (0 . 00588043 s - 1 ) × e - (0 . 00588043 s - 1 )(2 . 68 s) = - 0 . 0224014 rad / s 2 ± ( t 3 ) ± = 0 . 0224014 rad / s 2 . 002 (part 2 of 3) 10.0 points How many revolutions does the wheel make aFter 2 . 48 s ? Correct answer: 1 . 51642 rev. Explanation: Let : t f =2 . 48 s . θ = - ³ t f 0 ω 0 e - σt dt = - ω 0 σ ³ t f 0 e - σt ( - σdt ) = - ω 0 σ e - σt ´ ´ ´ t f 0 = ω 0 σ ( 1 - e - σt f ) = 3 . 87 rad / s 0 . 00588043 s - 1 × µ 1 - e - (0 . 00588043 s - 1 )(2 . 48 s) =9 . 52794 rad , so the number oF revolutions is n = θ 2 π = 9 . 52794 rad 2 π = 1 . 51642 rev . 003 (part 3 of 3) 10.0 points ±ind the number oF revolutions it makes be- Fore coming to rest. Correct answer: 104 . 742 rev. Explanation: θ = ω 0 σ ( 1 - e - σt f )
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Bodet (ngb299) – ohw 22 optional – turner – (57340) 2 For t = , this reduces to θ end = ω 0 σ = 3 . 87 rad / s 0 . 00588043 s - 1 1rev 2 π rad = 104 . 742 rev . 004 (part 1 of 2) 10.0 points Consider a 6 kg square which has its mass concentrated along its perimeter, with each side of length 4 m. d d What is the moment of inertia of the square about an axis perpendicular to the plane of the square at its center of mass? Use the parallel axis theorem and divide the square into parts. The moment of inertia of a rod rotated about its CM is I CM rod = 1 12 md 2 . Correct answer: 32 kg · m 2 . Explanation: Let : m rod = 6kg 4 and $ =4m . By the parallel axis theorem, the moment of inertia of each rod about the center of the square is I rod square = 1 12 m rod d 2 + m rod ± d 2 ² 2 = 1 3 m rod d 2 . Since there are four sides to the square, the moment of inertia of the square about its center of mass is I CM square =4 I rod square =4 ± 1 3 m rod d 2 ² = 4 3 m rod $ 2 = 4 3 ( 6kg 4 )(4m) 2 = 32 kg · m 2 . 005 (part 2 of 2) 10.0 points
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ohw 22 optional-solutions - Bodet (ngb299) ohw 22 optional...

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