HW3 Solution

HW3 Solution - lew (dl9564) hk3 Opyrchal (41104) 1 This...

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Unformatted text preview: lew (dl9564) hk3 Opyrchal (41104) 1 This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A circular disk with moment of inertia I = 1 2 mR 2 mass m and radius R is mounted at its center, about which it can rotate freely. A light cord wrapped around it supports a weight mg . R I m T g Find the total kinetic energy of the system when the weight is moving at a speed v . 1. K = 5 4 mv 2 2. K = 2 3 mv 2 3. K = 1 2 mv 2 4. K = 1 3 mv 2 5. K = 3 2 mv 2 6. K = mv 2 7. K = 3 4 mv 2 correct 8. K = 5 2 mv 2 9. K = 4 5 mv 2 Explanation: v = r , so the total kinetic energy is K tot = K m + K rot = 1 2 mv 2 + 1 2 I 2 = 1 2 mv 2 + 1 2 parenleftbigg 1 2 mR 2 parenrightbigg parenleftBig v R parenrightBig 2 = 3 4 mv 2 . keywords: 002 10.0 points A 1.5 kg bicycle tire of radius 0.33 m starts from rest and rolls down from the top of a hill that is 15.0 m high. What is the translational speed of the tire when it reaches the bottom of the hill? As- sume that the tire is a hoop with I = mr 2 . The acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 12 . 1305 m / s. Explanation: Let : m = 1 . 5 kg , r = 0 . 33 m , h = 15 . 0 m , and g = 9 . 81 m / s 2 . The moment of inertia of the solid cylinder is I = 1 2 mr 2 v i = 0 m/s and h f = 0 m, so by conservation of energy, U i = K f U i = K trans,f + K rot,f...
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This note was uploaded on 01/22/2009 for the course PHYS Phys 106 taught by Professor Opyrachal during the Fall '08 term at NJIT.

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HW3 Solution - lew (dl9564) hk3 Opyrchal (41104) 1 This...

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