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Unformatted text preview: lee (cyl327) – HW09 – gentle – (56245) 1 This printout should have 33 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A ladybug sits at the outer edge of a merry goround, and a gentleman bug sits halfway between her and the axis of rotation. The merrygoround makes a complete revolution once each second. What is the gentleman bug’s angular speed? 1. the same as the ladybug’s correct 2. twice the ladybug’s 3. impossible to determine 4. half the ladybug’s Explanation: Angular speed is the same for every point on the merrygoround. 002 (part 1 of 5) 10.0 points A wheel starts from rest with constant angu lar acceleration of 4 . 8 rad / s 2 . After 6 . 8 s, what is its angular velocity? Correct answer: 32 . 64 rad / s. Explanation: Let : ω = 0 , α = 4 . 8 rad / s 2 , and t = 6 . 8 s . The angular velocity is ω = ω + αt = αt = (4 . 8 rad / s 2 ) (6 . 8 s) = 32 . 64 rad / s . 003 (part 2 of 5) 10.0 points Through what angle did the wheel turn? Correct answer: 110 . 976 rad. Explanation: Δ θ = ω Δ t + 1 2 α (Δ t ) 2 = 1 2 α (Δ t ) 2 = 1 2 (4 . 8 rad / s 2 ) (6 . 8 s) 2 = 110 . 976 rad . 004 (part 3 of 5) 10.0 points Through how many revolutions did the wheel turn? Correct answer: 17 . 6624 rev. Explanation: Δ θ = (110 . 976 rad) parenleftbigg 1 rev 2 π rad parenrightbigg = 17 . 6624 rev . 005 (part 4 of 5) 10.0 points What is the speed of a point 0 . 6 m from the axis of rotation? Correct answer: 19 . 584 m / s. Explanation: Let : r = 0 . 6 m . The tangential speed is v = r ω = (0 . 6 m) (32 . 64 rad / s) = 17 . 6624 rev . 006 (part 5 of 5) 10.0 points What is the acceleration of a point 0 . 6 m from the axis of rotation? Correct answer: 639 . 228 m / s 2 . Explanation: Let : r = 0 . 6 m . lee (cyl327) – HW09 – gentle – (56245) 2 The acceleration has two components, the tangential and the centripetal acceleration, which act perpendicular to each other: a = radicalBig a 2 t + a 2 c = radicalBig ( r α ) 2 + ( r ω 2 ) 2 = r radicalbig α 2 + ω 4 = (0 . 6 m) × radicalBig (4 . 8 rad / s 2 ) 2 + (32 . 64 rad / s) 4 = 19 . 584 m / s . keywords: 007 (part 1 of 3) 10.0 points A car accelerates uniformly from rest and covers a distance of 61 m in 8 . 3 s. If the diameter of a tire is 49 cm, find the angular acceleration of the wheel. Correct answer: 7 . 22832 rad / s 2 . Explanation: Let : Δ x = 61 m , Δ t = 8 . 3 s , and R = 24 . 5 cm = 0 . 245 m . From kinematics, Δ x = 1 2 a Δ t 2 a = 2 Δ x Δ t 2 , so a = Rα α = a R = 2 Δ x R (Δ t ) 2 = 2 (61 m) (0 . 245 m) (8 . 3 s) 2 = 7 . 22832 rad / s 2 . 008 (part 2 of 3) 10.0 points Find the final angular velocity of one of the car’s wheels....
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This note was uploaded on 02/28/2012 for the course PHY 316 taught by Professor Markert during the Fall '10 term at University of Texas.
 Fall '10
 MARKERT
 mechanics

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