ex4 - Version 095 EX4 ditmire (58216) 1 This print-out...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 . 8m 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 = 502 kg / m 3 . Applying Bernoullis equation at the sur- face of the liquid pool and at the top of the siphon gives P atm + 1 2 v 2 A = P B + 1 2 v 2 B + 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 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. 3 6 . 3 cm 1 . 4 8 k g 474N / 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 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 . = I Using the parallel axis theorem, the rotational inertia of the plank about the pivot point is I = 1 12 mL 2 + m L 2 2 = 1 3 mL 2 . (1) From the free-body diagram, after displace- ment by a small angle , we have =- mg L 2 + k y L cos =- mg L 2 + k L sin L cos , (2) where y = L sin is the displacement from equilibrium. Using I from Eq. (1), and for small , cos = 1 and sin = , we have mg L 2- k L 2 = 1 3 mL 2 d 2 dt 2 . (3) Substitute = - mg 2 k L and d 2 dt 2 = d 2 dt 2 , we have d 2 dt 2 + 3 k m = 0 , (4) where the coefficient of is 2 = 3 k m . There- fore T 2 = 2 m 3 k (5) = 2 (1 . 48 kg) 3 (474 N / m) = 0 . 202703 s ....
View Full Document

This note was uploaded on 11/03/2010 for the course PHYSICS 303 taught by Professor Shih during the Spring '10 term at University of Texas at Austin.

Page1 / 13

ex4 - Version 095 EX4 ditmire (58216) 1 This print-out...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online