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Unformatted text preview: CHAPTER 10 SYSTEMS OF PARTICLES ActivPhysics can help with these problems: Activities 6.6, 6.7 Section 101: Center of Mass Problem 1. A 28kg child sits at one end of a 3.5mlong seesaw. Where should her 65kg father sit so the center of mass will be at the center of the seesaw? Solution Take the xaxis 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 cm = 0 = m c x c + m f x f , where the masses are given, and x c = (3 . 5 m) / 2 (half the length of the seesaw in the negative x direction). Thus, x f = m c x c /m f = (28 / 65)(1 . 75 m) = 75 . 4 cm from the center. Problem 2. Two particles of equal mass m are at the vertices of the base of an equilateral triangle. The center of mass of the triangle is midway between the base and the third vertex. What is the mass at the third vertex? Solution Choose x y coordinates with origin at the midpoint of the base, as shown in the solution to Problem 5. In this case, y cm = 1 2 y 3 , where y 3 is the position of the third mass (and of course, y 1 = y 2 = 0 for the equal masses m 1 = m 2 = m on the base). The y component of Equation 102 gives y cm = m 3 y 3 / ( m 1 + m 2 + m 3 ) = m 3 (2 y cm ) / (2 m + m 3 ) , so m 3 = 2 m. Problem 3. Four trucks, with masses indicated in Fig. 1023, are on a rectangular barge of mass 35 Mg whose center of mass is at its center. The trucks individual centers of mass are located 25 m apart on the barges long dimension and 10 m apart on the short dimension, as shown. Where is the center of mass of the entire system? Express in relation to the truck at lower left. y x 35 Mg 12.5 m figure 1023 Top view of four trucks on a barge, with truck masses given. Dots mark individual trucks centers of mass (Problems 3, 8). Solution As explained in the text (see Figs. 108 and 9), in order to find the center of mass of this system, the trucks and the barge can be treated as point masses located at their centers of mass. With x y axes as shown superimposed on Fig. 1023, x cm = (18 Mg)(0) + (23 Mg)(25 m) + (19 Mg)(25 m) + (11 Mg)(0) + (35 Mg)(12 . 5 m) (18 + 23 + 19 + 11 + 35) Mg = 14 . 0 m , y cm = (19 Mg + 11 Mg)(10 m) + (35 Mg)(5 m) 106 Mg = 4 . 48 m. Problem 4. Rework Example 101 with the origin at the center of the barbell, showing that the physical location of the center of mass is independent of the choice of coordinate system. Solution With origin at the barbells center, x 1 = 75 cm , and x 2 = 75 cm. For part (a): x cm = (60 kg)( 75 cm) + (60 kg)(75 cm) 60 kg + 60 kg = 0 (at the center). For part (b): x cm = (50 kg)( 75 cm) + (80 kg)(75 cm) 50 kg + 80 kg = 17 . 3 cm 136 CHAPTER 10 (toward the heavier mass). The addition of 75 cm to the above coordinates transforms these answers to the frame of Example 101....
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This note was uploaded on 10/31/2010 for the course PHYSICS 69054 taught by Professor Huang during the Fall '09 term at Berkeley.
 Fall '09
 Huang

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