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Homework 26 - homework 26 BAUTISTA ALDO Due Apr 3 2006 4:00...

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homework 26 – BAUTISTA, ALDO – Due: Apr 3 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 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 5 g cos θ . 2. a = 5 7 g cos θ . 3. a = 3 7 g sin θ . 4. a = 3 5 g sin θ . 5. a = 2 7 g cos θ . 6. a = 3 7 g cos θ . 7. a = 5 7 g sin θ . correct 8. a = 2 7 g sin 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 . Question 2 Part 2 of 2. 10 points. The minimum coefficient of friction such that the sphere rolls without slipping is 1. μ = 5 7 cos θ . 2. μ = 3 5 cos θ . 3. μ = 2 7 sin θ . 4. μ = 2 7 tan θ . correct 5. μ = 3 7 tan θ .
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homework 26 – BAUTISTA, ALDO – Due: Apr 3 2006, 4:00 am 2 6. μ = 3 7 sin θ . 7. μ = 2 7 cos θ . 8. μ = 5 7 tan 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 is the minimum “no slipping” criterion. Then M g sin θ - μ M g cos θ = M 5 7 g sin θ μ cos θ = 1 - 5 7 sin θ , so μ = 2 7 tan θ . Question 3 Part 1 of 3. 10 points. Consider a thick cylindrical shell rolling down along an incline. It has a mass m , an outer radius R and a moment of iner- tia I cm = 4 5 mR 2 about its center of mass. The incline is at an angle θ to the hori- zontal. Its length is s and its height is h . s h θ v A A The equation of motion τ = as the shell is rolling down the incline is given by(using the contact point as the pivot point): 1. mgR = ( I cm + mR 2 ) α 2. mgR tan θ = ( I cm + mR 2 ) α 3. mgR tan θ = I cm α 4. mgR sin θ = ( I cm + mR 2 ) α correct 5. mgR cot θ = ( I cm + mR 2 ) α 6. mgR = I cm α 7. mgR cot θ = I cm α 8. mgR cos θ = ( I cm + mR 2 ) α 9. mgR cos θ = I cm α 10.
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