Chapter_9_09 - Chapter 9 9.2 Figure 9-37 shows a three...

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Chapter 9 9.2 Figure 9-37 shows a three particle system with masses m 1 = 3.0 kg , m 2 = 4.0 kg , and m 3 = 8.0 kg . The scales are set by x s = 2.0 m and y s = 2.0 m . What are (a) the x coordinate and (b) the y coordinate of the system’s center of mass? (c) If m 3 is gradually increased, does the center of mass shift toward or away from that particle or does it remain stationary. We begin by writing the coordinates of each mass and then we use those coordinates to find the cm position. m 1 : (0,0) m 2 : (2,1) m 3 : (1,2) x cm = m 1 x 1 + m 2 x 2 + m 3 x 3 m 1 + m 2 + m 3 = 1.067 m y cm = m 1 y 1 + m 2 y 2 + m 3 y 3 m 1 + m 2 + m 3 = 1.333 m As m 3 is increased, the cm position will move toward the position of m 3 . 9.3 What are (a) the x coordinate and (b) the y coordinate of the center of mass for the uniform plate shown in Fig 9-38 For continuous objects, we can often use symmetry to find the cm. In the drawing below, we have labeled the location of the center of each piece of the plate. We can now consider each piece of the plate as a point mass at the center. The mass of each plate is proportional to the area.
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2L 4L 3L 4L 2L 2L L (2L, 2.5L) (-L, -0.5L) (L, -3L) x y x cm = 1 Total Area Area i x i = (2 L 7 L ) ( L ) + (4 L L ) 2 L + (2 L 2 L ) L (2 L 7 L ) + (4 L L ) + (2 L 2 L ) = 2 L 3 22 L 2 = 1 11 L = 0.455 cm
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y cm = 1 Total Area Area i y i = (2 L 7 L ) ( 0.5 L ) + (4 L L ) 2.5 L + (2 L 2 L ) ( 3 L ) (2 L 7 L ) + (4 L L ) + (2 L 2 L ) = 9 L 3 22 L 2 = 9 22 L = 2.045 cm 9.7 The figure shows a slab with dimensions d 1 = 11.0 cm , d 2 = 2.80 cm , and d 3 = 11.0 cm . Half the slab consists of aluminum (density = 2.70 g / cm 3 ) and half consists of iron density = 7.85 g / cm 3 ) . What are the coordinates of the center of mass. We begin finding the coordinates of the center of mass of each slab (Aluminum and Iron). x Al = d 3 2 = 6.50 cm y Al = d 1 + d 1 2 = 16.5 cm z Al = d 2 2 = 1.40 cm x Fe = d 3 2 = 6.50 cm y Fe = d 1 2 = 5.50 cm z Fe = d 2 2 = 1.40 cm The volume of each slab is the same. We use the volume to find the mass of each slab. V = d 1 d 2 d 3 m Al = ρ Al V m Fe = ρ Fe V
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x cm = m Al x Al + m Fe x Fe m Al + m Fe = ρ Al V x Al + ρ Fe V x Fe ρ Al V + ρ Fe V = ρ Al x Al + ρ Fe x Fe ρ Al + ρ Fe = 2.70 ⋅− 6.50 + 7.85 ⋅− 6.50 2.70 + 7.85 = 6.50 y cm = m Al y Al + m Fe y Fe m Al + m Fe = ρ Al y Al + ρ Fe y Fe ρ Al + ρ Fe = 2.70 16.5 + 7.85 5.50 2.70 + 7.85 = 8.32 cm z cm = m Al z Al + m Fe a Fe m Al + m Fe = ρ Al z Al + ρ Fe z Fe ρ Al + ρ Fe = 2.70 1.40 + 7.85 1.40 2.70 + 7.85 = 1.40
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