Chap 9 - SYSTEMS OF PARTICLES 9 EXERCISES Section 9.1 Center of Mass 12 Take the x axis along the seesaw in the direction of the father with origin

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9.1 SYSTEMS OF PARTICLES EXERCISES 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 xm x m x cm c c f f == + 0, where the masses are given, and x c =− (. ) / 35 2 m (half the length of the seesaw in the negative x direction). Thus, x m fc c f = = /( / ) ( .). 28 65 1 75 75 4 mc m from the center. 13. INTERPRET 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. DEVELOP The center of mass of a system of particles is given by Equation 9.2: r rr r mr m M ii 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 yy cm = 3 2 /, where y 3 is the position of the third mass (and of course, 12 0 for the equal masses mmm on the base). EVALUATE Using Equation 9.2, the y coordinate of the center of mass is y my y m cm = ++ = + 11 2 2 33 123 00 () mm + = + 3 3 3 2 2 cm Solving for m 3 , we have 22 m += , or 3 2 = . ASSESS From symmetry consideration, it is apparent that x cm = 0. On the other hand, we have mm m 2 at the bottom two vertices of the triangle. Since 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. INTERPRET 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 DEVELOP 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 rr r mr m M ii i i i i cm == EVALUATE 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 yL L 3 3 2 60 = sin for the third mass, Equation 9.2 gives y my mmm L m cm / = ++ = 11 2 2 33 123 00 3 2 () + + mm LL 3 6 0 289 . ASSESS From symmetry consideration, it is apparent that x cm = 0. On the other hand, we have mm 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 yy cm = 3 3 /. 16. With origin at the center of the Earth, x MM r cm Em = + + = × + (. ) (. ) ( 0 7 35 3 85 10 597 5 km 735 4680 .) = km, or about 1690 km below the Earth’s surface.
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This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.

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Chap 9 - SYSTEMS OF PARTICLES 9 EXERCISES Section 9.1 Center of Mass 12 Take the x axis along the seesaw in the direction of the father with origin

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