# SOL30 - Thevalingam Donald Homework 30 Due midnight Inst...

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Thevalingam, Donald – Homework 30 – Due: Apr 30 2007, midnight – Inst: Eslami 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points The center of mass of a pitched baseball or radius 5 . 36 cm moves at 18 . 3 m / s. The ball spins about an axis through its center of mass with an angular speed of 50 . 2 rad / s. Calculate the ratio of the rotational energy to the translational kinetic energy. Treat the ball as a uniform sphere. Correct answer: 0 . 00864758 . Explanation: The wanted ratio is ratio = 2 2 mv 2 2 = 2 mr 2 5 ω 2 mv 2 = 2 r 2 ω 2 5 v 2 = 2(0 . 0536 m / cm) 2 (50 . 2 rad / s) 2 5(18 . 3 m / s) 2 = 0 . 00864758 keywords: 002 (part 1 of 1) 10 points A circular disk with a mass m and radius R is mounted at its center, about which it can rotate freely. The disk has moment of inertia I = 1 2 m R 2 . 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 = 1 2 m v 2 2. K = 3 4 m v 2 correct 3. K = m v 2 4. K = 4 5 m v 2 5. K = 1 3 m v 2 6. K = 5 2 m v 2 7. K = 3 2 m v 2 8. K = 2 3 m v 2 9. K = 5 4 m v 2 Explanation: The total kinetic energy is the sum of the kinetic energy of the hanging weight, K m = 1 2 m v 2 , plus the kinetic energy of rotation of the disk, K d = 1 2 I ω 2 = 1 4 m v 2 . Therefore, K = K g + K d = 1 2 m v 2 + 1 4 m v 2 = 3 4 m v 2 .

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