{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Test #3-solutions

# Test #3-solutions - Version 110/ABCDC Test#3...

This preview shows pages 1–4. Sign up to view the full content.

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(part1of2)5.0points 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 a R = 1 2 M a R 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 a R 002(part2of2)5.0points 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 = μ M g , so τ = f R = I α μ M 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 = 0 . 558795 . 003(part1of2)5.0points

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 , 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 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: m g cos θ N f m g 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 . Alternate Axis: With the origin at the center of the sphere, I = 2 5 M R 2 . summationdisplay τ : f R = I α f = I α R = parenleftbigg 2 5 M R parenrightbigg a R summationdisplay F bardbl : M g sin θ - f = M a M g sin θ - 2 5 M a = M a g sin θ = 7 5 a a = 5 7 g sin θ . 004(part2of2)5.0points What is the minimum coefficient of friction such that the sphere rolls without slipping? 1. μ = 3 7 sin θ 2. μ = 5 7 cos θ 3. μ = 2 7 tan θ correct 4. μ = 5 7 tan θ 5. μ = 2 7 sin θ
Version 110/ABCDC – Test #3 – Antoniewicz – (57420) 3 6. μ = 3 5 cos θ 7. μ = 3 7 tan θ 8. μ = 2 7 cos θ Explanation: The net force along the direction of the incline is summationdisplay F = M g sin θ - f = M parenleftbigg 5 7 g sin θ

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern