1_new - Ch9-HW2 florin (56950) This print-out should have...

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Ch9-HW2 – florin – (56950) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A man whose mass is 70 kg and a woman whose mass is 55 kg sit at opposite ends of a canoe 5 m long, whose mass is 45 kg. Assume the man is seated at x = 0 and the boat extends along the positive x axis with the woman at the other end. Where is the center of mass of the system consisting of man, woman, and canoe? Correct answer: 2.27941 m. Explanation: With the man (m 1 ) at the origin, we have x CM = m 1 x 1 + m 2 x 2 + m 3 x 3 m 1 + m 2 + m 3 = 0 + (55 kg)(5 m) + (45 kg)(2.5 m) 70 kg + 55 kg + 45 kg = 2.27941 m . 2 (part 2 of 2) 10.0 points Suppose that the man moves quickly to the center of the canoe and sits down there. How far does the canoe move in the water? Assume force of friction between water and the canoe is negligible. Correct answer: 1.02941 m. Explanation: Since friction is negligible, the net hori-zontal force on the system (Man+Woman+Canoe) is zero. Therefore, its center of mass momen-tum (and velocity) remain constant (which is zero in this case). So the center of mass of the system remains stationary. For this problem, let us take the origin to be at the left end of the canoe (note that it is not a stationary point). The new center of mass is at x CM = m 1 x 1 + m 2 x 2 + m 3 x 3 m 1 + m 2 + m 3 = (70 kg)(2.5 m) + (55 kg)(5 m) 70 kg + 55 kg + 45 kg (45 kg)(2.5 m) + 70 kg + 55 kg + 45 kg = 3.30882 m . At first, the center of mass was 2.27941 m from the left end. Now the center of mass is 3.30882 m from the left end. But the center of mass never moved! This means the boat moved 3.30882 m − 2.27941 m = 1.02941 m to the left as the man walked 1.02941 m to the right. 003 (part 1 of 4) 10.0 points Consider a rod of length L and mass m which is pivoted at one end. An object with mass m is attached to the free end of the rod. m C 27 L m Determine the moment of inertia of the system with respect to the pivot point. The acceleration of gravity g = 9.8 m/s 2 . Consider the mass at the end of the rod to be a point particle. 1. I = 3 2 m L 2 2. None of these. 3. I = 4 3 m L 2 correct 4. I = 5 4 m L 2
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Ch9-HW2 – florin – (56950) 2 5. I = 13 12 m L 2 6. I = L 2 7. I = 5 3 m L 2 Explanation: The moment of inertia of the rod with re-spect to the pivot point is I ROD = 1 3 m L 2 , and the moment of inertia of the mass m with respect to the pivot point is I MASS = m L 2 . Thus the moment of inertia of the system is I = I ROD + I M = 1 3 m L 2 + m L 2 = 4 3 m L 2 . 004 (part 2 of 4) 10.0 points The length C in the figure represents the lo-cation of the center-of-mass of the rod plus mass system. Determine the position of the center of mass from the pivot point. 005 (part 3 of 4) 10.0 points The unit is released from rest in the horizontal position. What is the kinetic energy of the unit when the rod momentarily has a vertical orienta-tion?
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