p0601 - 274 CHAPTER 6. SYSTEMS OF PARTICLES 6.1 Chapter 6,...

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274 CHAPTER 6. SYSTEMS OF PARTICLES 6.1 Chapter 6, Problem 1 Problem: At a given moment in time, a system of five particles is in motion as shown. Compute the position and velocity of the center of mass. Compute the angular momentum of the system relative to the origin. 0 λ 2 λ 3 λ 4 λ x 0 λ 2 λ 3 λ 4 λ y im i x i y i v x i v y i 1 m 00 vv 22 m λ 3 λ 33 m 2 λ 0 v v 44 m 3 λ v 55 m 4 λ 2 λ 0 2 v •• v 1 v 2 v 3 v 4 v 5 ......................................................................... . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .... ............................................................................................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solution: First, note that the total mass of the system, M ,is M = 5 3 i =1 m i = m +2 m +3 m +4 m +5 m =15 m The x component of the center of mass is x = 1 M 5 3 i =1 m i x i = 1 15 m [ m · 0+2 m · λ m · 2 λ m · 3 λ m · 4 λ ]= 8 3 λ The y component of the center of mass is y = 1 M 5 3 i =1 m i y i = 1 15 m [ m · m · 3 λ m · 0+4 m · 0+5 m · 2 λ 16 15 λ The x velocity component of the center of mass is v x = 1 M 5 3 i =1 m i v x i = 1 15 m [ m · v m · v m · ( v )+4 m · m · 0] = 0 The y velocity component of the center of mass is v y = 1 M 5 3 i =1 m i v y i = 1 15 m [ m · v m · v m · v m · v m · ( 2 v )] = 0 Thus, the position and velocity of the center of mass are as follows. r = 8 3 λ i + 16 15 λ j and v = 0 Turning to angular momentum, it is the sum of the angular momentum of all five particles, viz., H o = 5 3 i =1 m i r i × v i = 5 3 i =1 H i
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6.1. CHAPTER 6, PROBLEM 1 275 Now, noting that the motion is two dimensional, we have H i = m i e e e e e e e ij k x i y i 0 v x i v y i 0 e e e e e e e = m i ( x i v y i y i v x i ) k Proceeding particle by particle, we find the following.
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p0601 - 274 CHAPTER 6. SYSTEMS OF PARTICLES 6.1 Chapter 6,...

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