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# exam 3 - Version 177 THIRD EXAM Kleinman(58225 This...

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Version 177 – THIRD EXAM – Kleinman – (58225) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A “synchronous” satellite, which always re- mains above the same point on a planet’s equator, is put in orbit about a planet similar to Jupiter. This planet rotates once every 6 . 2 h, has a mass of 1 . 8 × 10 27 kg and a radius of 6 . 99 × 10 7 m. Given that G = 6 . 67 × 10 11 N m 2 / kg 2 , calculate how far above Jupiter’s surface the satellite must be. 1. 44953000.0 2. 81645800.0 3. 76853300.0 4. 85979500.0 5. 39311100.0 6. 59331700.0 7. 40706800.0 8. 67358100.0 9. 89669100.0 10. 68586000.0 Correct answer: 4 . 4953 × 10 7 m. Explanation: Basic Concepts: Solution: According to Kepler’s third law: T 2 = 4 π 2 G M r 3 where r is the radius of the satellite’s orbit. Thus, solving for r r = braceleftbigg G M T 2 4 π 2 bracerightbigg 1 3 = { 6 . 67 × 10 11 N m 2 / kg 2 } 1 3 × braceleftbigg (1 . 8 × 10 27 kg) (22320 s) 2 4 (3 . 14159) 2 bracerightbigg 1 3 = 1 . 14853 × 10 8 m . Now, the altitude h of the satellite (measured from the surface of Jupiter) is h = r - R = (1 . 14853 × 10 8 m) - (6 . 99 × 10 7 m) = 4 . 4953 × 10 7 m . 002 10.0 points A cylindrical stone column of diameter 2 R = 0 . 775 m and height H = 2 . 36 m is transported in standing position by a dolly. a side view When the dolly accelerates or decelerates slowly enough, the column stands upright, but when the dolly’s acceleration magnitude exceed a critical value a c , the column top- ples over. (For a > + a c the column topples backward; for a < - a c the column toppes forward.) Calculate the magnitude of the critical ac- celeration a c of the dolly. The acceleration of gravity is 9 . 8 m / s 2 . 1. 2.81542 2. 1.50285 3. 2.13185 4. 4.80161 5. 3.52699 6. 5.03425 7. 3.16129 8. 2.31763 9. 3.21822 10. 4.55897 Correct answer: 3 . 21822 m / s 2 . Explanation: Let : g = 9 . 8 m / s 2 , R = 0 . 3875 m , and H = 2 . 36 m .

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Version 177 – THIRD EXAM – Kleinman – (58225) 2 In the non-inertial frame of the accelerating dolly, the column is subject to the horizontal inertial force vector F in = - mvectorg . Together, the gravity and the inertial force combine into the apparent weight force vector W app = m ( vectorg - vectora ) in the direction θ = arctan parenleftbigg a g parenrightbigg from the vertical. From the torque point of view, this appar- ent weight force applies at the center of mass of the column. The column is stable in the vertical position when the line of this force goes through the column’s base CM W app but when this line misses the base, the column topples over CM W app For the critical acceleration a c , the line goes through the edge of the base, hence the direc- tion of the apparent weight force must deviate from the vertical by the angle θ c where tan θ c = R h cm = 2 R H . Consequently, the critical acceleration of the dolly is a c = g tan θ c = g 2 R H = (9 . 8 m / s 2 ) 2 (0 . 3875 m) 2 . 36 m = 3 . 21822 m / s 2 . 003 10.0 points An Atwood machine is constructed using a hoop with spokes of negligible mass. The 1 . 9 kg mass of the pulley is concentrated on its rim, which is a distance 24 . 6 cm from the axle. The mass on the right is 1 . 39 kg and on the left is 1 . 95 kg.
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