{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

PHY303Kquiz3soln (2005)

# PHY303Kquiz3soln (2005) - Johnson Matthew Quiz 3 Due Apr 6...

This preview shows pages 1–3. Sign up to view the full content.

Johnson, Matthew – Quiz 3 – Due: Apr 6 2005, 10:00 pm – Inst: Kleinman 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points 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 rests on the very top of the incline. The sphere has mass M and radius R . The acceleration of gravity is 9 . 8 m / s 2 . Hint: The moment of inertia of a sphere with respect to an axis through its center is 2 5 M R 2 . Choose the instantaneous axis through the contact point P as the axis of origin for the torque equation. R M μ θ h The acceleration of the center of mass is 1. a = 3 7 g cos θ . 2. a = 3 5 g sin θ . 3. a = 2 7 g sin θ . 4. a = 3 5 g cos θ . 5. a = 2 7 g cos θ . 6. a = 3 7 g sin θ . 7. a = 5 7 g sin θ . correct 8. a = 5 7 g cos θ . Explanation: Basic Concepts: X ~ F = m ~a X ~ τ = I ~α . m g cos θ N f m g sin θ With the origin at the point of contact, X τ : M g R sin θ = I α , (1) where I is obtained using the parallel-axis theorem I = 2 5 M R 2 + M R 2 = 7 5 M R 2 . (2) If the sphere rolls without slipping, α = a R . (3) Now substituting I from Eq. 2 and α from Eq. 3 into Eq. 1, we have M g R sin θ = 7 5 M R 2 a R , so a = 5 7 g sin θ . 002 (part 2 of 2) 10 points The minimum coefficient of friction such that the sphere rolls without slipping is 1. μ = 3 7 tan θ . 2. μ = 2 7 sin θ . 3. μ = 2 7 tan θ . correct 4. μ = 5 7 cos θ . 5. μ = 3 7 sin θ .

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

View Full Document
Johnson, Matthew – Quiz 3 – Due: Apr 6 2005, 10:00 pm – Inst: Kleinman 2 6. μ = 5 7 tan θ . 7. μ = 3 5 cos θ . 8. μ = 2 7 cos θ . Explanation: The net force along the direction of the incline is X F = M g sin θ - f = M 5 7 g sin θ , where f = μ N = μ M g cos θ . Then M g sin θ - μ M g cos θ = M 5 7 g sin θ , so μ = 2 7 tan θ . 003 (part 1 of 1) 10 points A 27 kg mass and a 14 kg mass are suspended by a pulley that has a radius of 12 cm and a mass of 7 kg. The cord has a negligible mass and causes the pulley to rotate without slipping. The pulley rotates without friction. The masses start from rest 2 . 8 m apart. Treat the pulley as a uniform disk. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 8 m 12 cm 7 kg ω 27 kg 14 kg Determine the speeds of the two masses as they pass each other. Correct answer: 2 . 83129 m / s. Explanation: Let : M = 7 kg , R = 12 cm , m 1 = 27 kg , m 2 = 14 kg , h = 2 . 8 m , v = ω R , I = 1 2 M R 2 , and K disk = 1 2 I ω 2 = 1 4 M v 2 . From conservation of energy K 1 + K 2 + K disk = - Δ U or m 1 v 2 2 + m 2 v 2 2 + M v 2 4 = ( m 1 - m 2 ) g h 2 m 1 + m 2 + M 2 v 2 = ( m 1 - m 2 ) g h , where h 2 is the height. Taking no slipping into account, we can solve for v v = s 2 ( m 1 - m 2 ) g h 2 ( m 1 + m 2 ) + M (1) = s 356 . 72 J 44 . 5 kg = 2 . 83129 m / s . Alternative Solution: The forces in the vertical directions for m 1 and m 2 , and the torque on the pulley, give us m 1 g - T 1 = + m 1 a m 2 g - T 2 = - m 2 a , so T 1 = m 1 ( g - a ) T 2 = m 2 ( g + a ) , and I α = [ T 1 - T 2 ] R 1 2 M R 2 a R · = h m 1 ( g - a )] R - [ m 2 ( g + a ) i R 1 2 M a = m 1 g - m 1 a - m 2 a - m 2 g a = 2 ( m 1 - m 2 ) M + 2 ( m 1 + m 2 ) .
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