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chap 9 - SYSTEMS OF PARTICLES 9 EXERCISES Section 9.1...

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9.1 SYSTEMS OF PARTICLES E XERCISES Section 9.1 Center of Mass 12. Take the x axis along the seesaw in the direction of the father, with origin at the center. The center of mass of the child and her father is at the origin, so x m x m x cm c c f f = = + 0 , where the masses are given, and x c = − ( . )/ 3 5 2 m (half the length of the seesaw in the negative x direction). Thus, x m x m f c c f = − = = / ( / )( . ) . 28 65 1 75 75 4 m cm from the center. 13. I NTERPRET This problem is about center of mass. Our system consists of three masses located at the vertices of an equilateral triangle. If two masses are known and the location of the center of mass is also known, then the third mass can be calculated. D EVELOP The center of mass of a system of particles is given by Equation 9.2: r m r m m r M i i i i i i i i cm = = We shall choose x-y coordinates with origin (0,0) at the midpoint of the base. With this arrangement, the center of the mass is located at x cm = 0 and y y cm = 3 2 / , where y 3 is the position of the third mass (and of course, y y 1 2 0 = = for the equal masses m m m 1 2 = = on the base). E VALUATE Using Equation 9.2, the y coordinate of the center of mass is y m y m y m y m m m m m m y m cm = + + + + = + + + 1 1 2 2 3 3 1 2 3 3 3 0 0 ( ) ( ) m m m y m m + = + 3 3 3 2 2 ( ) cm Solving for m 3 , we have 2 2 3 3 m m m + = , or m m 3 2 = . A SSESS From symmetry consideration, it is apparent that x cm = 0. On the other hand, we have m m m + = 2 at the bottom two vertices of the triangle. Since y y cm = 3 2 / , i.e., y cm is halfway to the top vertex, we expect the mass there to be 2 m (See Example 9.2). 14. With origin at the barbell’s center, x 1 75 = − cm, and x 2 75 = cm. For part (a): x cm = + + ( )( ) ( )( ) 60 75 60 75 60 60 kg cm kg cm kg kg = 0 For part (b) : x cm = + + ( )( ) ( )( ) 50 75 80 75 50 80 kg cm kg cm kg kg cm = 17 3 . (toward the heavier mass). The addition of 75 cm to the above coordinates transforms these answers to the frame of Example 9.1. 15. I NTERPRET This problem is about locating the center of mass. Our system consists of three equal masses located at the vertices of an equilateral triangle of side L . 9
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9.2 Chapter 9 D EVELOP We take x-y coordinates with origin at the center of one side as shown. The center of mass of a system of particles is given by Equation 9.2: r m r m m r M i i i i i i i i cm = = E VALUATE From the symmetry (for every mass at x , there is an equal mass at x ), we have x cm = 0. As for y cm , since y = 0 for the two masses on the x -axis, and y L L 3 3 2 60 = ° = sin for the third mass, Equation 9.2 gives y m y m y m y m m m m m mL m cm / = + + + + = + + 1 1 2 2 3 3 1 2 3 0 0 3 2 ( ) ( ) + + = = m m L L 3 6 0 289 . A SSESS From symmetry consideration, it is apparent that x cm = 0. On the other hand, we have m m m + = 2 at the bottom two vertices of the triangle, and m at the top of the vertex. Therefore, we should expect y cm to be one third of y 3 . This indeed is the case, as y cm can be rewritten as y y cm = 3 3 / .
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