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Unformatted text preview: Version 133 exam3 Chiu (60180) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A uniform 9 kg rod with length 12 m has a frictionless pivot at one end. The rod is released from rest at an angle of 27 beneath the horizontal. 6 m 1 2 m 9 k g 27 What is the angular acceleration of the rod immediately after it is released? The moment of inertia of a rod about the center of mass is 1 12 mL 2 , where m is the mass of the rod and L is the length of the rod. The moment of inertia of a rod about either end is 1 3 mL 2 , and the acceleration of gravity is 9 . 8 m / s 2 . 1. 1.29793 2. 0.324483 3. 0.803557 4. 0.457454 5. 1.09148 6. 2.23819 7. 0.471867 8. 1.62242 9. 0.486772 10. 0.353627 Correct answer: 1 . 09148 rad / s 2 . Explanation: Let : m = 9 kg , L = 12 m , and = 27 . The rods moment of inertia about its end point is I = 1 3 mL 2 , so the angular accelera tion of the rod is = I = 1 2 mg L cos 1 3 mL 2 = 3 2 g L cos = 3 2 9 . 8 m / s 2 12 m cos27 = 1 . 09148 rad / s 2 . 002 10.0 points A(n) 65 kg boat that is 7 m in length is initially 8 . 9 m from the pier. A 47 kg child stands at the end of the boat closest to the pier. The child then notices a turtle on a rock at the far end of the boat and proceeds to walk to the far end of the boat to observe the turtle. Assume: There is no friction between boat and water. 8 . 9 m 7 m How far is the child from the pier when she reaches the far end of the boat? 1. 12.9625 2. 15.4528 3. 12.1874 4. 11.6375 5. 13.3273 6. 15.9 7. 14.8316 8. 12.6674 9. 13.98 10. 12.0465 Correct answer: 12 . 9625 m. Explanation: Version 133 exam3 Chiu (60180) 2 Let D = distance of the boat from the pier , = 8 . 9 m , L = length of the boat , = 7 m , M = mass of the boat , = 65 kg , and m = mass of the child , and = 47 kg , X = change in position of the boat . = final distance of child from pier . d = final distance of boat from pier . D L X d X We will use the pier as the origin of the xcoordinate. Initially, the center of mass of the boatchild system is x cm = parenleftbigg D + L 2 parenrightbigg M + D m M + m Finally, the center of mass of the boatchild system is x cm = parenleftbigg D + L 2 X parenrightbigg M + ( D + L X ) m M + m , where X is the change in position of the center of mass of the boat. Since the center of mass of the system does not move, we can equate the above two expressions for x cm parenleftbigg D + L 2 parenrightbigg M + D m M + m = parenleftbigg D + L 2 X parenrightbigg M + ( D + L X ) m M + m and, solving for X , we have parenleftbigg D + L 2 parenrightbigg M + D m = parenleftbigg D + L 2 X parenrightbigg M + ( D + L X ) m 0 =X M + ( L X ) m X ( M + m ) = Lm X = m m + M L = (47 kg) (47 kg) + (65 kg) (7 m) = 2 . 9375 m ....
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This note was uploaded on 03/11/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Friction

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