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HW3 Solution

# HW3 Solution - lew(dl9564 hk3 Opyrchal(41104 This print-out...

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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 m R 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 m g . R I ω m T g Find the total kinetic energy of the system when the weight is moving at a speed v . 1. K = 5 4 m v 2 2. K = 2 3 m v 2 3. K = 1 2 m v 2 4. K = 1 3 m v 2 5. K = 3 2 m v 2 6. K = m v 2 7. K = 3 4 m v 2 correct 8. K = 5 2 m v 2 9. K = 4 5 m v 2 Explanation: v = r ω , so the total kinetic energy is K tot = K m + K rot = 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 . 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 m r 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 m g h = 1 2 m v 2 f + 1 2 I ω 2 f m g h = 1 2 m v 2 f + 1 2 m r 2

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HW3 Solution - lew(dl9564 hk3 Opyrchal(41104 This print-out...

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