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oldquiz3_07 - quiz 03 BROWNING ADAM Due 10:00 pm Question 1...

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quiz 03 – BROWNING, ADAM – Due: Nov 14 2007, 10:00 pm 1 Question 1, chap 9, sect 1. part 1 of 1 10 points Given: G = 6 . 673 × 10 11 N m 2 / kg 2 Two balls, each with a mass of 0.854 kg, exert a gravitational force of 8 . 49 × 10 11 N on each other. How far apart are the balls? 1. 0 . 669566 m 2. 0 . 690561 m 3. 0 . 712316 m 4. 0 . 734363 m 5. 0 . 75712 m correct 6. 0 . 780597 m 7. 0 . 806763 m Explanation: Basic Concept: F g = G m 1 m 2 r 2 Given: m 1 = m 2 = 0 . 854 kg F g = 8 . 49 × 10 11 N G = 6 . 673 × 10 11 N · m 2 kg 2 Solution: r 2 = G m 1 m 2 F g = parenleftBig 6 . 673 × 10 11 N · m 2 kg 2 parenrightBig (0 . 854 kg) 2 8 . 49 × 10 11 N = 0 . 57323 m 2 Thus r = 0 . 57323 m 2 = 0 . 75712 m Question 2, chap 12, sect 5. part 1 of 2 10 points A wheel is formed from a hoop and seven equally spaced spokes. The hoop’s radius is the length of each spoke. 0 . 13 kg 4 . 9 kg 0 . 21 m There are seven equally spaced spokes. Each spoke has mass 0 . 13 kg. The hoop’s mass is 4 . 9 kg. The hoop’s radius is 0 . 21 m. Determine the moment of inertia of the wheel about an axis through its center and perpendicular to the plane of the wheel. 1. 0 . 183629 kg m 2 2. 0 . 189382 kg m 2 3. 0 . 196735 kg m 2 4. 0 . 205845 kg m 2 5. 0 . 2124 kg m 2 6. 0 . 22016 kg m 2 7. 0 . 229467 kg m 2 correct 8. 0 . 237695 kg m 2 9. 0 . 245318 kg m 2 10. 0 . 253063 kg m 2 Explanation: Let : n = 7 spokes , m = 0 . 13 kg , M = 4 . 9 kg , and R = 0 . 21 m . Each spoke contributes to the total moment of inertia as a thin rod pivoted at one end, hence I = M R 2 + n m R 2 3 = bracketleftBig M + n m 3 bracketrightBig R 2 = bracketleftbigg (4 . 9 kg) + 7 (0 . 13 kg) 3 bracketrightbigg (0 . 21 m) 2 = 0 . 229467 kg m 2 . Question 3, chap 12, sect 5. part 2 of 2 10 points
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quiz 03 – BROWNING, ADAM – Due: Nov 14 2007, 10:00 pm 2 Determine the moment of inertia of the wheel about an axis through its rim and per- pendicular to the plane of the wheel. 1. 0 . 454873 kg m 2 2. 0 . 4704 kg m 2 3. 0 . 485688 kg m 2 correct 4. 0 . 502059 kg m 2 5. 0 . 517915 kg m 2 6. 0 . 534876 kg m 2 7. 0 . 558918 kg m 2 8. 0 . 583093 kg m 2 9. 0 . 6027 kg m 2 10. 0 . 623779 kg m 2 Explanation: From the parallel axis theorem I rim = I + ( M + n m ) R 2 = (0 . 229467 kg m 2 ) + [(4 . 9 kg) + 7 (0 . 13 kg)] (0 . 21 m) 2 = 0 . 485688 kg m 2 . Question 4, chap 10, sect 2. part 1 of 1 10 points Pat builds a track for his model car out of wood. The track is 5 cm wide (along the z coordinate), 1 m high (along the y coordi- nate) and 3 . 2 m long (along the x coordinate starting from x = 0). The runway is cut such that it forms part of the left-hand side of a parabola, see figure below. x 1 m y 3 . 2 m y = 1 (10 . 24 m) [ x (3 . 2 m)] 2 5 cm Locate the horizontal position of the center of gravity of this track. 1. 0 . 6 m 2. 0 . 625 m 3. 0 . 65 m 4. 0 . 675 m 5. 0 . 7 m 6. 0 . 725 m 7. 0 . 75 m 8. 0 . 775 m 9. 0 . 8 m correct 10. 0 . 825 m Explanation: Let : a = 3 . 2 m , b = a 2 = 10 . 24 m , z = 5 cm , and σ = surface density . Basic Concepts: vectorr CM = integraltext vectorr ρ dV integraltext ρ dV = integraltext vectorr ρ dV M x 1 m y 3 . 2 m y = 1 (10 . 24 m) [ x (3 . 2 m)] 2 5 cm x dx Solution: Let σ represent the mass-per- face area. A vertical strip at position x , with width dx and height ( x a ) 2 b , has mass dm = σ ( x a ) 2 b dx . The total mass is M = integraldisplay dm = integraldisplay 3 . 2 m x =0 σ ( x a ) 2 b dx
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quiz 03 – BROWNING, ADAM – Due: Nov 14 2007, 10:00 pm 3 = σ b integraldisplay 3 . 2 m 0 m ( x 2 2 a x + b ) dx = σ b bracketleftbigg x 3 3 a x 2 + b x bracketrightbigg vextendsingle vextendsingle vextendsingle vextendsingle 3 . 2 m 0 m .
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