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

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Unformatted text preview: 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 . The spheres 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 h The acceleration of the center of mass is 1. a = 3 5 g cos . 2. a = 5 7 g cos . 3. a = 3 7 g sin . 4. a = 3 5 g sin . 5. a = 2 7 g cos . 6. a = 3 7 g cos . 7. a = 5 7 g sin . correct 8. a = 2 7 g sin . Explanation: Basic Concepts: X ~ F = m ~a X ~ = I ~ . m g cos N f m g sin With the origin at the point of contact, X : M g R sin = I , (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, = a R . (3) Now substituting I from Eq. 2 and from Eq. 3 into Eq. 1, we have M g R sin = 7 5 M R 2 a R , so a = 5 7 g sin . Question 2 Part 2 of 2. 10 points. The minimum coefficient of friction such that the sphere rolls without slipping is 1. = 5 7 cos . 2. = 3 5 cos . 3. = 2 7 sin . 4. = 2 7 tan . correct 5. = 3 7 tan . homework 26 BAUTISTA, ALDO Due: Apr 3 2006, 4:00 am 2 6. = 3 7 sin . 7. = 2 7 cos . 8. = 5 7 tan . Explanation: The net force along the direction of the incline is X F = M g sin - f = M 5 7 g sin , where f N = M g cos is the minimum no slipping criterion. Then M g sin - M g cos = M 5 7 g sin cos = 1- 5 7 sin , so = 2 7 tan . 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 to the hori- zontal. Its length is s and its height is h ....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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

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