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Unformatted text preview: 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 fnd 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 = Iω 2 2 mv 2 2 = µ 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 ±ind 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 . Thevalingam, Donald Homework 30 Due: Apr 30 2007, midnight Inst: Eslami 2 keywords: 003 (part 1 of 3) 10 points For any given rotational axis, the radius of gyration, K , of a rigid body is de±ned by the expression K 2 = I M , where M is the total mass of the body and I is the moment of inertia about the given axis. In other words, the radius of gyration is the distance between an imaginary point mass M , and the axis of rotation with I for the point mass about that axis is the same as for the rigid body. Find the radius of gyration of a solid disk of radius 3 . 93 m rotating about a central axis. Correct answer: 2 . 77893 m. Explanation: In all part the axis is the central axis, as stated in the problem. The moment of inertia of a solid disk with radius R and mass M is I d = M R 2 2 . Hence, K d = r M R 2 2 M = R 2 = (3 . 93 m) 2 = 2 . 77893 m . 004 (part 2 of 3) 10 points Find the radius of gyration of a uniform rod of length 7 . 73 m rotating about a central axis. Correct answer: 2 . 23146 m. Explanation: The moment of inertia of a uniform rod of length L and mass M is I r = M L 2 12 Hence, K r = r M L 2 12 M = L 6 3 = (7 . 73 m) 6 3 = 2 . 23146 m . ...
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