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# ex4 - Version 095 EX4 ditmire(58216 This print-out should...

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Version 095 – EX4 – ditmire – (58216) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A siphon consists of a flexible tube with the same cross section throughout the tube. It is used to drain a special liquid from a tank. Let the density of the liquid be 502 kg / m 3 and the diameter of the tube be 2 . 6 cm . Let the siphon discharge a distance 5 . 8 m below the surface of the liquid. h 5 . 8 m 2 . 6 cm What is the maximum height h ; i.e. , the vertical distance between the surface of the liquid to the point at the top of the siphon, for this siphon to work? The acceleration of grav- ity is 9 . 8 m / s 2 and P atm = 1 . 013 × 10 5 N / m 2 . 1. 11.95 2. 20.5911 3. 17.6094 4. 19.0014 5. 12.9533 6. 11.6667 7. 13.4418 8. 19.285 9. 12.3203 10. 14.7879 Correct answer: 20 . 5911 m. Explanation: Let : P atm = 1 . 013 × 10 5 N / m 2 , g = 9 . 8 m / s 2 , = 5 . 8 m , d = 2 . 6 cm , and ρ = 502 kg / m 3 . Applying Bernoulli’s equation at the sur- face of the liquid pool and at the top of the siphon gives P atm + 1 2 ρ v 2 A = P B + 1 2 ρ v 2 B + ρ g h . The limiting case occurs when velocity in the tube and the pressure at the top of the tube are essentially zero. Hence h = P atm ρ g = 1 . 013 × 10 5 N / m 2 (502 kg / m 3 ) (9 . 8 m / s 2 ) = 20 . 5911 m . 002 10.0 points A uniform plank of mass 1 . 48 kg and length 36 . 3 cm is pivoted at one end, and the op- posite end is attached to a spring of force constant 474 N / m. The height of the pivot has been adjusted so that the plank will be in equilibrium when it is horizontally oriented as in the figure. 36 . 3 cm 1 . 48 kg θ 474 N / m Find the period of small oscillation about the equilibrium point. 1. 0.202703 2. 0.290968 3. 0.182044 4. 0.325383 5. 0.3196 6. 0.279113 7. 0.258874 8. 0.379447 9. 0.228038 10. 0.285553

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Version 095 – EX4 – ditmire – (58216) 2 Correct answer: 0 . 202703 s. Explanation: Let : L = 36 . 3 cm , m = 1 . 48 kg , and k = 474 N / m . τ = I α Using the parallel axis theorem, the rotational inertia of the plank about the pivot point is I = 1 12 m L 2 + m L 2 2 = 1 3 m L 2 . (1) From the free-body diagram, after displace- ment by a small angle θ , we have τ = - m g L 2 + k y L cos θ = - m g L 2 + k L sin θ L cos θ , (2) where y = L sin θ is the displacement from equilibrium. Using I from Eq. (1), and for small θ , cos θ = 1 and sin θ = θ , we have m g L 2 - k L 2 θ = 1 3 m L 2 d 2 θ dt 2 . (3) Substitute θ = θ - m g 2 k L and d 2 θ dt 2 = d 2 θ dt 2 , we have d 2 θ dt 2 + 3 k m θ = 0 , (4) where the coe ffi cient of θ is ω 2 = 3 k m . There- fore T 2 π ω = 2 π m 3 k (5) = 2 π (1 . 48 kg) 3 (474 N / m) = 0 . 202703 s . Alternate Solution: Use conservation of energy. E = m g y 2 + 1 2 I ω 2 + 1 2 k y 2 = m g y 2 + 1 2 1 3 m L 2 ω 2 + 1 2 k y 2 = 1 2 m g y + 1 6 m v 2 + 1 2 k y 2 Di ff erentiate with respect to t . Note: Total energy is a constant, thus dE dt = 0 and use the chain rule for [ v ( t )] 2 and [ y ( t )] 2 . d E dt = d dt 1 2 m g y + 1 6 m v 2 + 1 2 k y 2 0 = 1 2 m g v + 1 6 m a v + 1 2 k y v 0 = m g + 1 3 m a + k y . By substitution we can write, d 2 y dt 2 + 3 k m y = 0 .
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ex4 - Version 095 EX4 ditmire(58216 This print-out should...

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