# ch09 - 369 Chapter 9 1 We use Eq 9-5 to solve for 3 3 x y(a...

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Unformatted text preview: 369 Chapter 9 1. We use Eq. 9-5 to solve for 3 3 ( , ). x y (a) The x coordinate of the system’s center of mass is: 3 1 1 2 2 3 3 com 1 2 3 (2.00 kg)( 1.20 m) 4.00 kg 0.600 m 3.00 kg 2.00 kg 4.00 kg 3.00 kg 0.500 m. x m x m x m x x m m m Solving the equation yields x 3 = –1.50 m. (b) The y coordinate of the system’s center of mass is: 3 1 1 2 2 3 3 com 1 2 3 (2.00 kg)(0.500 m) 4.00 kg 0.750 m 3.00 kg 2.00 kg 4.00 kg 3.00 kg 0.700 m. y m y m y m y y m m m Solving the equation yields y 3 = –1.43 m. 2. Our notation is as follows: x 1 = 0 and y 1 = 0 are the coordinates of the m 1 = 2.0 kg particle; x 2 = 2.0 m and y 2 = 1.0 m are the coordinates of the m 2 = 4.0 kg particle; and x 3 = 1.0 m and y 3 = 2.0 m are the coordinates of the m 3 = 8.0 kg particle. (a) The x coordinate of the center of mass is 1 1 2 2 3 3 com 1 2 3 4.0 kg 2.0 m 8.0 kg 1.0 m 1.1 m. 2.0 kg 4.0 kg 8.0 kg m x m x m x x m m m (b) The y coordinate of the center of mass is 1 1 2 2 3 3 com 1 2 3 4.0 kg 1.0 m 8.0 kg 2.0 m 1.4 m. 2.0 kg 4.0 kg 8.0 kg m x m x m x y m m m (c) As the mass of m 3 , the topmost particle, is increased, the center of mass shifts toward that particle. As we approach the limit where m 3 is infinitely more massive than the others, the center of mass becomes infinitesimally close to the position of m 3 . 3. We use Eq. 9-5 to locate the coordinates. CHAPTER 9 370 (a) By symmetry x com = – d 1 /2 = –(13 cm)/2 = – 6.5 cm. The negative value is due to our choice of the origin. (b) We find y com as com, com, com, cm, com 3 3 3 3 11 cm / 2 7.85 g/cm 3 11 cm / 2 2.7 g/cm 8.3 cm. 7.85 g/cm 2.7 g/cm i i a a i i i a a a i a i i a a m y m y V y V y y m m V V (c) Again by symmetry, we have z com = (2.8 cm)/2 = 1.4 cm. 4. We will refer to the arrangement as a “table.” We locate the coordinate origin at the left end of the tabletop (as shown in Fig. 9-37). With + x rightward and + y upward, then the center of mass of the right leg is at ( x,y ) = (+ L , – L /2), the center of mass of the left leg is at ( x,y ) = (0, – L /2), and the center of mass of the tabletop is at ( x,y ) = ( L /2, 0). (a) The x coordinate of the (whole table) center of mass is com 3 / 2 3 2 M L M M L L x M M M . With L = 24 cm, we have x com = (24 cm)/2 = 12 cm....
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ch09 - 369 Chapter 9 1 We use Eq 9-5 to solve for 3 3 x y(a...

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