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Unformatted text preview: Morby, Grant – Quiz 3 – Due: Apr 5 2006, 10:00 pm – Inst: Drummond 1 This printout should have 20 questions. Multiplechoice 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 A cylinder with moment of inertia I 1 rotates with angular speed ω about a frictionless ver tical axle. A second cylinder, with moment of inertia I 2 = 1 2 I 1 , initially not rotating, drops onto the first cylinder. Since the surfaces are rough, the two eventually reach the same an gular speed ω . 1 2 I 1 I 2 I 1 ω Before ω After Calculate the final angular speed ω with respect to the initial angular speed ω . 1. ω = 4 7 ω 2. None of these 3. ω = 2 3 ω correct 4. ω = 3 4 ω 5. ω = 1 3 ω 6. ω = 4 5 ω 7. ω = 3 5 ω 8. ω = 2 5 ω 9. ω = 3 7 ω Explanation: X ~ L = const From conservation of angular momentum I 1 ω = ( I 1 + I 2 ) ω , so ω = I 1 I 1 + I 2 ω (1) = I 1 I 1 + 1 2 I 1 ω = 2 3 ω , since K is less than K , kinetic energy is lost. 002 (part 1 of 1) 10 points A uniform bar of length L and weight W is attached to a wall with a hinge that exerts on the bar a horizontal force H x and a vertical force H y . The bar is held by a cord that makes a 90 ◦ angle with respect to bar and angle θ with respect to wall. The acceleration of gravity g = 9 . 8 m / s 2 . L 90 ◦ W θ What is the magnitude of the horizontal force H x on the pivot? 1. H x = 1 2 W sin 2 θ Morby, Grant – Quiz 3 – Due: Apr 5 2006, 10:00 pm – Inst: Drummond 2 2. H x = 1 2 W cos θ 3. H x = 1 2 W sin θ 4. H x = 1 2 W tan θ 5. H x = 1 2 W cos 2 θ 6. H x = 1 2 W sin θ cos θ correct Explanation: Analyzing the torques on the bar, with the hinge at the axis of rotation, we have X τ = L T µ L 2 sin θ ¶ W = 0 , so, T = 1 2 W sin θ . Analyzing the force on the bar, we have X F x = H x T cos θ = 0 . Put T into this equation and get H x = µ 1 2 W sin θ ¶ cos θ . 003 (part 1 of 2) 10 points A student sits on a rotating stool holding two 4 . 1 kg masses. When his arms are extended horizontally, the masses are 0 . 88 m from the axis of rotation, and he rotates with an angu lar velocity of 2 . 3 rad / sec. The student then pulls the weights horizontally to a shorter dis tance 0 . 4 m from the rotation axis and his angular velocity increases to ω 2 . ω i ω f For simplicity, assume the student himself plus the stool he sits on have constant com bined moment of inertia I s = 3 . 1 kg m 2 . Find the new angular velocity ω 2 of the student after he has pulled in the weights. Correct answer: 4 . 92638 rad / s. Explanation: Let : M = 4 . 1 kg , R 1 = 0 . 88 m , ω 1 = 2 . 3 rad / sec , R 2 = 0 . 4 m , and ω 2 = 4 . 92638 rad / sec , As the student moves his arms, his moment of inertia changes from I 1 = I s + 2 m R 2 1 = (3 . 1 kg m 2 ) + 2(4 . 1 kg) (0 . 88 m) 2 = 9 . 45008 kg m 2 to I 2 = (3 . 1 kg m 2 ) + 2(4 . 1 kg) (0 . 4 m) 2 = 4 . 412 kg m 2 , but his angular momentum is conserved,...
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This note was uploaded on 03/02/2009 for the course PHY 58235 taught by Professor Kleinman during the Spring '09 term at University of Texas.
 Spring '09
 KLEINMAN
 Physics

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