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Unformatted text preview: Version 110/ABCDC – Test #3 – Antoniewicz – (57420) 1 This printout should have 14 questions. Multiplechoice 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 parallelaxis 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.
 Spring '08
 Turner
 Physics, Force

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