p0604 - 280 CHAPTER 6. SYSTEMS OF PARTICLES 6.4 Chapter 6,...

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Unformatted text preview: 280 CHAPTER 6. SYSTEMS OF PARTICLES 6.4 Chapter 6, Problem 4 Problem: A man of mass 4m is initially standing at one end of a canoe of mass m. Then, he moves to the opposite end of the canoe. The length of the canoe is L and it is perfectly symmetric about its center. (a) What distance does the canoe move if the frictional resistance of the water is negligible? (b) What is the final velocity of the canoe and the man? (c) How does your Part (b) answer change if the frictional resistance of the water is appreciable? ....................................................................................................................................................................................................................................................................................... . . . . . . . . . . . . . . . . ....... .... . . . . . . . . . . .... . .. ... . ... ................ . ... .. . .. . . . .... .............................................................................. ....................................................... .. ..... .. ............... . ... . . . . . . ............................................................................................................................................................................... .......................................................................................................................................................................................................................... .. .................. ..... ..... .. .................... ................. ................ ............................................................. . . .. . . . .... . . . ....................................................................................................................................................................................................................................................................... . ............................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................................... . . ...................................................................................................... .................... ................ . .. ................................................................................................................ .. . Solution: The key to this problem is in recognizing it as a two-particle system, the man being one particle and the canoe being the other. Initially, the “man-particle” is located at x ≈ L (see figure below). Because it is symmetric about its center, we can say the “canoe-particle” is located at x = 1 L. 2 L .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... • x H .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... • xf (a) Before he man moves he cen er of mass s oca ed a x= m 1 2L + 4mL 9 = L 5m 10 Af er he man moves o he o her end of he canoe he cen er of mass s oca ed a xf = m H + 1 L + 4mH 1 2 =H+ L 5m 10 Because here are no ex erna forces ac ng on he sys em he cen er of mass does no move Th s means x = xf so ha 1 9 L=H+ L 10 10 6.4. CHAPTER 6, PROBLEM 4 Therefore, the distance the canoe moves is 281 4 L 5 (b) Again noting that there are no external forces acting, the velocity of the center of mass is zero. This is the velocity of the man and the canoe after the man changes ends of the canoe. Thus, we have H= v= dx =0 dt (c) If the water has appreciable frictional resistance, Newton’s Second Law tells us that 5m dv =F dt where F is the friction force. Since the canoe moves to the right as the man changes ends, the friction force — which opposes the motion — is directed to the left. Thus, we can say 5m Therefore, the velocity of the system will be t dv = −|F | dt v (t) = − 0 |F | dt < 0 5m Therefore, the canoe and the man will be moving to the left after the man changes ends of the canoe. ...
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This note was uploaded on 12/16/2010 for the course AME 301 at USC.

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