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# ch08f - CHAPTER 8 Systems of Particles and Conservation of...

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CHAPTER 8 Systems of Particles and Conservation of Momentum 1* · Give an example of a three-dimensional object that has no mass at its center of mass. A hollow sphere. 2 · Three point masses of 2 kg each are located on the x axis at the origin, x = 0.20 m, and x = 0.50 m. Find the center of mass of the system. Use Equ. 8-4; note that y cm = 0 x cm = [(2 × 0 + 2 × 0.2 + 2 × 0.5)/6] m = 0.233 m 3 · A 24-kg child is 20 m from an 86-kg adult. Where is the center of mass of this system? Take the origin at the position of the child. Use Equ. 8-4 x cm = (86 × 20/110) m = 15.6 m 4 · Three objects of 2 kg each are located in the xy plane at points (10 cm, 0), (0, 10 cm), and (10 cm, 10 cm). Find the location of the center of mass. Use Equ. 8-4 x cm = [(10 × 2 + 10 × 2)/6] cm = 6.67 cm; y cm = [(10 × 2 + 10 × 2)/6] cm = 6.67 cm 5* · Find the center of mass x cm of the three masses in Figure 8-46. Use Equ. 8-4 x cm = [(1 × 1 + 2 × 2 + 8 × 4)/11] m = 3.36 m 6 · Alley Oop’s club-ax consists of a symmetrical 8-kg stone attached to the end of a uniform 2.5-kg stick that is 98 cm long. The dimensions of the club-ax are shown in Figure 8-47. How far is the center of mass from the handle end of the club-ax? 1. Locate CM of stick and of stone 2. Use Equ. 8-4 By symmetry, x cm (stick) = 0.49 m; x cm (stone) = 0.89 m x cm = [(2.5 × 0.49 + 8 × 0.89)/10.5] m = 0.795 m 7 · Three balls A, B, and C , with masses of 3 kg, 1 kg, and 1 kg, respectively, are connected by massless rods. The balls are located as in Figure 8-48. What are the coordinates of the center of mass? Use Equ. 8-4 x cm = [(3 × 2 + 1 × 1 + 1 × 3)/5] m = 2 m y cm = [(3 × 2 + 1 × 1 + 1 × 0)/5] m = 1.4 m

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Chapter 8 Systems of Particles and Conservation of Momentum 8 · By symmetry, locate the center of mass of an equilateral triangle of side length a with one vertex on the y axis and the others at (- a /2, 0) and (+ a /2, 0). 1. Draw the triangle; assume vertex at y > 0 2. Locate the intersection of the bisectors; see sketch 3. Give the coordinates of CM x cm = 0; y cm = 1/2 a tan 30 o = 0.289 a 9* ·· The uniform sheet of plywood in Figure 8-49 has a mass of 20 kg. Find its center of mass. We shall consider this as two sheets, a square sheet of 3 m side length and mass m 1 and a rectangular sheet 1m × 2m with a mass of - m 2 . Let coordinate origin be at lower left hand corner of the sheet. Let σ be the surface density of the sheet. 1. Find x cm ( m 1 ), y cm ( m 1 ) and x cm ( m 2 ), y cm ( m 2 ) 2. Determine m 1 and m 2 3. Use Equ. 8-4 By symmetry, x cm ( m 1 ) = 1.5 m, y cm ( m 1 ) = 1.5 m and x cm ( m 2 ) = 1.5 m, y cm ( m 2 ) = 2.0 m m 1 = 9 σ kg, m 2 = 2 σ kg x cm = (9 σ × 1.5 - 2 σ × 1.5)/7 σ = 1.5 m y cm = (9 σ × 1.5 - 2 σ × 2.0)/7 σ = 1.36 m 10 ·· Show that the center of mass of a uniform semicircular disk of radius R is at a point (4/3 π ) R from the center of the circle. 1. The semicircular disk is shown; we also show here the surface element dA 2. Use Equ. 8-5 to find y cm ; x cm = 0 by symmetry. y cm = (1/ M ) y σ dA 3. y = r sin θ , dA = r d θ dr , and M = π R 2 σ /2; make the appropriate substitutions 4. y cm = R M 0 σ π 0 r 2 sin θ d θ dr = R M 0 2 σ r 2 dr = M 3 2 σ R 3 = π 3 4 R 0 11 ·· A baseball bat of length L has a peculiar linear density (mass per unit length) given by λ = λ 0 (1 + x 2 / L 2 ). Find the x
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ch08f - CHAPTER 8 Systems of Particles and Conservation of...

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