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Unformatted text preview: Version PREVIEW – Practice Exam 4 – chelikowsky – (59005) 1 This printout should have 22 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. Box on an Incline 001 10.0 points A box with its center of mass o ff center as indicated by the dot is placed on an inclined plane. In which orientation does the box tip over? 1. 2. correct 3. 4. 5. None of the orientations will cause the box to tip over. Explanation: In order to tip over, the box must pivot about its bottom left corner. The only orien tation where gravity creates a torque sufficient to tip the block is Bowling the Ball 002 10.0 points A bowling ball has a mass of 2 . 6 kg, a moment of inertia of 0 . 120224 kg · m 2 , and a radius of . 34 m. It rolls along the lane without slipping at a linear speed of 3 . 5 m / s. What is the kinetic energy of the rolling ball? 1. @@@ 2. @@@ 3. @@@ 4. @@@ 5. @@@ 6. @@@ 7. @@@ 8. @@@ 9. @@@ 10. @@@ Correct answer: 22 . 295 J. Explanation: Let : v = 3 . 5 m / s , r = 0 . 34 m , m = 2 . 6 kg , and I = 0 . 120224 kg · m 2 . E = E rot + E kin = 1 2 I ω 2 + 1 2 mv 2 = 1 2 I v 2 r 2 + 1 2 mv 2 = 1 2 (0 . 120224 kg · m 2 ) (3 . 5 m / s) 2 (0 . 34 m) 2 + 1 2 (2 . 6 kg)(3 . 5 m / s) 2 = 22 . 295 J . Dynamics of a Top Version PREVIEW – Practice Exam 4 – chelikowsky – (59005) 2 003 10.0 points The top in the figure has moment of inertia . 000401 kg · m 2 and is initially at rest. It is free to rotate about the stationary axis AA . A string wrapped around a peg along the axis of the top is pulled in such a manner as to maintain a constant tension of 6 . 17 N. F A A' If the string does not slip while being un wound, what is the angular speed of the top after 64 . 3 cm of string has been pulled o ff of the peg? 1. @@@ 2. @@@ 3. @@@ 4. @@@ 5. @@@ 6. @@@ 7. @@@ 8. @@@ 9. @@@ 10. @@@ Correct answer: 140 . 667 rad / s. Explanation: Let : I = 0 . 000401 kg · m 2 , F = 6 . 17 N , and L = 64 . 3 cm . The top starts from rest, so by conservation of energy, W = Δ K = I ω 2 f 2 I ω 2 2 F L = I ω 2 f 2 ω f = 2 W I = 2 F L I = 2(6 . 17 N)(64 . 3 cm) . 000401 kg · m 2 = 140 . 667 rad / s . Airplane Momentum 004 (part 1 of 2) 10.0 points An airplane of mass 23575 kg flies level to the ground at an altitude of 15 km with a constant speed of 185 m / s relative to the Earth. What is the magnitude of the airplane’s angular momentum relative to a ground ob server directly below the airplane in kg · m 2 / s? 1. @@@ 2. @@@ 3. @@@ 4. @@@ 5. @@@ 6. @@@ 7. @@@ 8. @@@ 9. @@@ 10. @@@ Correct answer: 6 . 54206 × 10 10 kg · m 2 / s. Explanation: Since the observer is directly below the air plane, L = hmv 005 (part 2 of 2) 10.0 points Does this value change as the airplane contin ues its motion along a straight line?...
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 Fall '09
 Chelikowsky
 Simple Harmonic Motion, Correct Answer, Amax ω

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