Test#3-solutions - Version 110/ABCDC – Test#3 – Antoniewicz –(57420 1 This print-out should have 14 questions Multiple-choice questions may

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Unformatted text preview: Version 110/ABCDC – Test #3 – Antoniewicz – (57420) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 5.0 points A constant horizontal force of 230 N is applied to a lawn roller in the form of a uniform solid cylinder of radius 0 . 55 m and mass 14 kg . M R F If the roller rolls without slipping, find the acceleration of the center of mass. The accel- eration of gravity is 9 . 8 m / s 2 . 1. 4.44444 2. 5.2381 3. 4.7619 4. 11.2821 5. 6.66667 6. 7.22222 7. 12.3077 8. 10.9524 9. 6.19048 10. 8.88889 Correct answer: 10 . 9524 m / s 2 . Explanation: Let : F = 230 N , R = 0 . 55 m , m = 14 kg , and g = 9 . 8 m / s 2 . If we choose the center of mass as the axis, summationdisplay F = F- f = M a summationdisplay τ = f R = I cm α ( F- M a ) R = parenleftbigg 1 2 M R 2 parenrightbigg parenleftBig a R parenrightBig F R- M aR = 1 2 M aR F = 3 2 M a a = 2 F 3 M = 230 N 3 (14 kg) = 10 . 9524 m / s 2 . Alternate Solution: Choosing the con- tact point as the axis and using the parallel axis theorem I = 1 2 M R 2 + M R 2 = 3 2 M R 2 , τ = F R = I α F R = 3 2 M R 2 parenleftBig a R parenrightBig = 3 2 M aR 002 (part 2 of 2) 5.0 points Find the minimum coefficient of friction nec- essary to prevent slipping. 1. 0.226757 2. 0.510204 3. 0.283447 4. 0.544218 5. 0.430839 6. 0.418629 7. 0.470958 8. 0.558795 9. 0.816327 10. 0.623583 Correct answer: 0 . 558795. Explanation: When there is no slipping, f = μM g , so τ = f R = I α μM g R = I α = 1 2 M R 2 α = 1 2 M R 2 parenleftBig a R parenrightBig μ = 1 2 a g = 1 2 10 . 9524 m / s 2 9 . 8 m / s 2 μ = . 558795 . 003 (part 1 of 2) 5.0 points Version 110/ABCDC – Test #3 – Antoniewicz – (57420) 2 Consider the problem of the solid sphere rolling down an incline without slipping. The incline has an angle θ , the sphere’s length up the incline is ℓ , and its height is h . At the beginning, the sphere of mass M and radius R rests on the very top of the incline. M μ ℓ θ h What is the acceleration of the center of mass? The moment of inertia of a sphere with respect to an axis through its center is 2 5 M R 2 . 1. a = 2 7 g cos θ 2. a = 3 5 g cos θ 3. a = 2 7 g sin θ 4. a = 5 7 g sin θ correct 5. a = 3 7 g sin θ 6. a = 5 7 g cos θ 7. a = 3 7 g cos θ 8. a = 3 5 g sin θ Explanation: Consider the forces acting on the sphere: mg cos θ N f mg sin θ Using the parallel-axis theorem I = 2 5 M R 2 + M R 2 = 7 5 M R 2 , and because the sphere rolls without slipping, α = a R . With the origin at the point of contact between the sphere and the incline surface, summationdisplay τ : M g R sin θ = I α M g R sin θ = 7 5 M R 2 a R a = 5 7 g sin θ ....
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This note was uploaded on 05/02/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Test#3-solutions - Version 110/ABCDC – Test#3 – Antoniewicz –(57420 1 This print-out should have 14 questions Multiple-choice questions may

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