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# hw11 - al-ameri(aoa434 Hwk10 Stokes(19102 This print-out...

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al-ameri (aoa434) – Hwk10 – Stokes – (19102) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This is Online Homework No 10. It is due before 6.00 am on Tuesday 21 April, Abu Dhabi Time. (Monday 9.00 pm in Texas) 001 10.0 points A homogeneous cylinder of radius 20 cm and mass 50 kg is rolling without slipping along a horizontal floor at 2 m / s. How much work was required to give it this motion? Correct answer: 0 . 15 kJ. Explanation: Let : r = 20 cm , m = 50 kg , and v = 2 m / s . Because the cylinder rolls without slipping, the translational speed is v = r ω and the total kinetic energy of the cylinder is K total = 1 2 m v 2 + 1 2 I ω 2 = 1 2 m v 2 + 1 2 parenleftbigg 1 2 m r 2 parenrightbigg parenleftBig v r parenrightBig 2 = 3 4 m v 2 = 3 4 (50 kg) (2 m / s) 2 · 1 kJ 1000 J = 0 . 15 kJ , which is also the work required to give the cylinder this motion. keywords: 002 (part 1 of 2) 10.0 points A skater spins with an angular speed of 13 . 8 rad / s with his arms outstretched. He lowers his arms, decreasing his moment of in- ertia from 47 kg · m 2 to 43 kg · m 2 . a) Calculate his initial rotational kinetic energy. Correct answer: 4475 . 34 J. Explanation: Basic Concepts: L = KE = 1 2 2 Given: I i = 47 kg · m 2 I f = 43 kg · m 2 ω i = 13 . 8 rad / s Solution: KE i = 1 2 I i ω 2 i = 1 2 (47 kg · m 2 )(13 . 8 rad / s) 2 = 4475 . 34 J 003 (part 2 of 2) 10.0 points b) Calculate his final rotational kinetic en- ergy. Correct answer: 4891 . 65 J. Explanation: Solution: Angular momentum of the system is con- served, so I f ω f = I i ω i ω f = I i ω i I f KE f = 1 2 I f parenleftbigg I i ω i I f parenrightbigg 2 = 1 2 I 2 i ω 2 i I f = 1 2 (47 kg · m 2 ) 2 (13 . 8 rad / s) 2 43 kg · m 2 = 4891 . 65 J 004 (part 1 of 2) 10.0 points A cylinder with moment of inertia 26 . 3 kg m 2 rotates with angular velocity 5 . 93 rad / s on a

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al-ameri (aoa434) – Hwk10 – Stokes – (19102) 2 frictionless vertical axle. A second cylinder, with moment of inertia 25 . 2 kg m 2 , initially not rotating, drops onto the first cylinder and remains in contact. Since the surfaces are rough, the two eventually reach the same an- gular velocity. I 2 I 1 ω 0 Before ω After Calculate the final angular velocity. Correct answer: 3 . 02833 rad / s. Explanation: From conservation of angular momentum ( I 1 + I 2 ) ω = I 1 ω 0 , or ω = I 1 I 1 + I 2 ω 0 = (26 . 3 kg m 2 ) (26 . 3 kg m 2 ) + (25 . 2 kg m 2 ) (5 . 93 rad / s) = 3 . 02833 rad / s . 005 (part 2 of 2) 10.0 points Show that energy is lost in this situation by calculating the ratio of the final to the initial kinetic energy. Correct answer: 0 . 51068. Explanation: K i = 1 2 I 1 ω 0 2 and K f = 1 2 ( I 1 + I 2 ) ω 2 , Therefore K f = 1 2 ( I 1 + I 2 ) ω 2 = 1 2 ( I 1 + I 2 ) parenleftbigg I 1 I 1 + I 2 parenrightbigg 2 ω 2 0 = parenleftbigg I 1 I 1 + I 2 parenrightbigg 1 2 I 1 ω 2 0 = parenleftbigg I 1 I 1 + I 2 parenrightbigg K i .
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