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# HW10 - homework 10 GAUTHIER JOSEPH Due Nov 9 2007 1:00 am...

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homework 10 – GAUTHIER, JOSEPH – Due: Nov 9 2007, 1:00 am 2 1. I = 13 81 M L 2 2. I = 13 75 M L 2 3. I = 19 147 M L 2 4. I = 19 81 M L 2 5. I = 19 192 M L 2 correct 6. I = 31 147 M L 2 7. I = 43 192 M L 2 8. I = 7 81 M L 2 9. I = 13 147 M L 2 10. I = 1 9 M L 2 Explanation: Basic Concepts: Using the parallel axis theorem, the momentum of inertia is I M d 2 . In this case there are two masses with I O = I cm + M D 2 , where D = parenleftbigg 5 8 1 2 parenrightbigg L = 1 8 L , so (1) I O = M bracketleftBigg 1 12 + parenleftbigg 1 8 parenrightbigg 2 bracketrightBigg L 2 = 19 192 M L 2 . (2) Question 3, chap 11, sect 3. part 2 of 3 10 points Hint: Use the small angle approximation. If θ is in radians, then Newton’s second law for rotational motion for this pendulum is given by 1. d 2 θ dt 2 = 45 26 g L θ 2. d 2 θ dt 2 = 9 14 g L θ 3. d 2 θ dt 2 = 105 62 g L θ 4. d 2 θ dt 2 = 3 2 g L θ 5. d 2 θ dt 2 = 12 7 g L θ 6. d 2 θ dt 2 = 63 38 g L θ 7. d 2 θ dt 2 = 21 26 g L θ 8. d 2 θ dt 2 = 24 19 g L θ correct 9.
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