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Unformatted text preview: garcia (jg46766) Problem Set 10 lacosse (S2107) 1 This printout should have 40 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. This problem set includes questions for both Chapters 11 and 12. Post any questions on my website, or see me. We are entering the home stretch and hang in there!! 001 (part 1 of 2) 10.0 points You are playing with a YoYo which you can describe as a uniform disk with mass m and radius r . The string of the YoYo has a length L . The YoYo rolls down vertically. h r m Determine the tension T of the string at a height h below the original position h as indicated in the sketch. 1. T = 1 3 mg correct 2. T = r L mg 3. T = 5 2 mg 4. T = 1 4 mg 5. T = 2 5 mg 6. T = 1 2 mg 7. T = 3 mg 8. T = L h r mg 9. T = mg 10. T = 7 4 mg Explanation: Apply F = ma to the center of mass of the disk, F net = summationdisplay F : mg T = ma . (1) Apply = I to the YoYo (where the torque is calculated about the center), T r = I = 1 2 mr 2 = 1 2 mr a , so (2) T = 1 2 ma . (3) In the 3 rd step we used a = r . Substituting Eq. 3 into 1 leads to, mg 1 2 ma = ma , so a = 2 3 g , and T = 1 2 ma = 1 3 mg . 002 (part 2 of 2) 10.0 points After the YoYo has been elastically reflected at length L , it moves upward. Determine the speed of its center when it is at a height h below the original position h . 1. v = radicalbigg r L g h 2. v = radicalbig 2 g h 3. v = radicalbigg 1 8 g h 4. v = radicalbig g h g L 5. v = radicalbigg 4 3 g h correct 6. v = radicalbigg 3 2 g h 7. v = radicalbig g h 8. v = radicalbigg 1 2 g h 9. v = radicalbigg 5 3 g h garcia (jg46766) Problem Set 10 lacosse (S2107) 2 10. v = radicalbig 3 g h Explanation: The application of workenergy equation gives mg h = 1 2 mv 2 + 1 2 I 2 . But 1 2 I 2 = 1 2 parenleftbigg 1 2 mr 2 parenrightbigg 2 = 1 4 mv 2 . So mg h = parenleftbigg 1 2 + 1 4 parenrightbigg mv 2 = v = radicalbigg 4 3 g h. 003 (part 1 of 2) 10.0 points A particle is located at the vector position vectorr = (1 . 6 m) + (4 . 2 m) and the force acting on it is vector F = (2 . 5 N) + (3 . 2 N) . What is the magnitude of the torque about the origin? Correct answer: 5 . 38 N m. Explanation: Basic Concept: vector = vectorr vector F Solution: Since neither position of the par ticle, nor the force acting on the particle have the zcomponents, the torque acting on the particle has only zcomponent: vector = [ xF y y F x ] k = [(1 . 6 m) (3 . 2 N) (4 . 2 m) (2 . 5 N)] k = [ 5 . 38 N m] k . 004 (part 2 of 2) 10.0 points What is the magnitude of the torque about the point having coordinates [ a, b ] = [(3 m) , (8 m)]?...
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This note was uploaded on 03/28/2011 for the course PHYSICS 101 taught by Professor Carter during the Spring '11 term at University of Texas at Austin.
 Spring '11
 carter
 Physics

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