Chapter09 - 1 We use Eq 9-5 to solve for x3 y3(a The x...

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1. We use Eq. 9-5 to solve for 33 (, ) . x y (a) The x coordinates of the system’s center of mass is: () ( ) 3 11 2 2 com 123 (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 mx x mmm −+ + ++ == + + =− Solving the equation yields x 3 = –1.50 m. (b) The y coordinates of the system’s center of mass is: ( ) 3 com (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 my my my y +− + + + Solving the equation yields y 3 = –1.43 m.
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2. Our notation is as follows: x 1 = 0 and y 1 = 0 are the coordinates of the m 1 = 3.0 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 () ( ) ( ) 11 2 2 33 com 123 0 4.0 kg 2.0 m 8.0 kg 1.0 m 1.1 m. 3.0 kg 4.0 kg 8.0 kg mx x mmm ++ == = + + (b) The y coordinate of the center of mass is ( ) ( ) com 0 4.0 kg 1.0 m 8.0 kg 2.0 m 1.3m. 3.0 kg 4.0 kg 8.0 kg my my my y = + + (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 .
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3. Since the plate is uniform, we can split it up into three rectangular pieces, with the mass of each piece being proportional to its area and its center of mass being at its geometric center. We’ll refer to the large 35 cm × 10 cm piece (shown to the left of the y axis in Fig. 9-38) as section 1; it has 63.6% of the total area and its center of mass is at ( x 1 ,y 1 ) = ( 5.0 cm, 2.5 cm). The top 20 cm × 5 cm piece (section 2, in the first quadrant) has 18.2% of the total area; its center of mass is at ( x 2 , y 2 ) = (10 cm, 12.5 cm). The bottom 10 cm x 10 cm piece (section 3) also has 18.2% of the total area; its center of mass is at ( x 3 , y 3 ) = (5 cm, 15 cm). (a) The x coordinate of the center of mass for the plate is x com = (0.636) x 1 + (0.182) x 2 + (0.182) x 3 = – 0.45 cm . (b) The y coordinate of the center of mass for the plate is y com = (0.636) y 1 + (0.182) y 2 + (0.182) y 3 = – 2.0 cm .
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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 03 / 2 0.5 3 MLM ML x L MM M ++ + + == . With L = 22 cm, we have x com = 11 cm. (b) The y coordinate of the (whole table) center of mass is ()( ) /2 3 0 35 M L y −+ , or y com = – 4.4 cm. From the coordinates, we see that the whole table center of mass is a small distance 4.4 cm directly below the middle of the tabletop.
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Thus, () 11 NN com NH 14.0067 3.803 10 m 3.13 10 m 3 14.0067 3 1.00797 my y mm × == = × ++ where Appendix F has been used to find the masses. 5. (a) By symmetry the center of mass is located on the axis of symmetry of the molecule – the y axis. Therefore x com = 0.
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This note was uploaded on 05/11/2010 for the course PHYS 2101,2102 taught by Professor Giammanco during the Spring '10 term at LSU.

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Chapter09 - 1 We use Eq 9-5 to solve for x3 y3(a The x...

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