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exam3-solutions

# exam3-solutions - Version 133 exam3 Chiu(60180 This...

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Version 133 – exam3 – Chiu – (60180) 1 This print-out should have 16 questions. Multiple-choice 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 12 m 9 kg 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 m L 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 m L 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 θ = 27 . The rod’s moment of inertia about its end- point is I = 1 3 m L 2 , so the angular accelera- tion of the rod is α = τ I = 1 2 m g L cos θ 1 3 m L 2 = 3 2 g L cos θ = 3 2 9 . 8 m / s 2 12 m cos 27 = 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:

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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 x -coordinate. Initially, the center of mass of the boat-child system is x cm = parenleftbigg D + L 2 parenrightbigg M + D m M + m Finally, the center of mass of the boat-child 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 ) = L m X = m m + M L = (47 kg) (47 kg) + (65 kg) × (7 m) = 2 . 9375 m . The final distance of the child from the pier is = D + L - X = (8 . 9 m) + (7 m) - (2 . 9375 m) = 12 . 9625 m . The final distance d of the near end of the boat to the pier is d = D - X = (8 . 9 m) - (2 . 9375 m) = 5 . 9625 m . 003 10.0 points A uniform rod of mass 2 . 3 kg is 20 m long. The rod is pivoted about a horizontal, frictionless pin at the end of a thin extension (of negligible mass) a distance 20 m from the center of mass of the rod. Initially the rod makes an angle of 45 with the horizontal. The rod is released from rest at an angle of 45 with the
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exam3-solutions - Version 133 exam3 Chiu(60180 This...

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