phyhw17 - valdez(vv689 – Homeowork 17 – florin...

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Unformatted text preview: valdez (vv689) – Homeowork 17 – florin – (58140) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A horizontal platform vibrates with simple harmonic motion in the horizontal direction with a period of 2 . 33 s. A body on the plat- form starts to slide when the amplitude of vibration reaches 0 . 334 m. Find the coefficient of static friction be- tween body and platform. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 0 . 247838. Explanation: Let : T = 2 . 33 s , A max = 0 . 334 m , and g = 9 . 8 m / s 2 . At each instant, there are two forces acting on the platform: the force responsible for the oscillation F = − kx and the force of friction F s = μN between the body and the platform. Applying Newton’s second law horizontally, summationdisplay F x = − kx + F s = ma platform The only force acting on the block in the horizontal direction is the frictional force, so F s = ma block If the block does not slide, its acceleration is the same as the platform: a block = a platform set = a The force of friction is F s = μN = μmg and from simple harmonic motion x = A cos ωt. a = d 2 x dt 2 = − Aω 2 cos ωt, so the maximum acceleration is a max = A max ω 2 and the coefficient of static friction is μ = A max ω 2 g . Since T = 2 π ω , μ = A max 4 π 2 g T 2 = (0 . 334 m)(4 π 2 ) (9 . 8 m / s 2 )(2 . 33 s) 2 = . 247838 . 002 (part 1 of 3) 10.0 points Consider a light rod of negligible mass and length L pivoted on a frictionless horizontal bearing at a point O . Attached to the end of the rod is a mass M 1 . Also, a second mass M 2 of equal size ( i.e. , M 1 = M 2 = M ) is attached to the rod parenleftbigg 2 5 L from the lower end parenrightbigg , as shown in the figure. 2 5 L L M 2 M 1 O θ What is the moment of inertia I about O ? 1. I = 17 16 M L 2 2. I = 13 9 M L 2 3. I = 73 64 M L 2 valdez (vv689) – Homeowork 17 – florin – (58140) 2 4. I = 89 64 M L 2 5. I = 34 25 M L 2 correct 6. I = 74 49 M L 2 7. I = 58 49 M L 2 8. I = 25 16 M L 2 9. I = 10 9 M L 2 10. I = 29 25 M L 2 Explanation: The momentum of inertia is I ≡ M d 2 . There are two masses with I M 1 = M L 2 and I M 2 = M parenleftbigg L − 2 5 L parenrightbigg 2 = M parenleftbigg 3 5 L parenrightbigg 2 = 9 25 M L 2 , so I = I M 1 + I M 2 = parenleftbigg 1 + 9 25 parenrightbigg M L 2 = 34 25 M L 2 . (1) 003 (part 2 of 3) 10.0 points If θ is in radians, what is Newton’s second law of rotational motion for this pendulum? Use the small angle approximation. 1. I d 2 θ dt 2 = − 9 7 M g Lθ 2. I d 2 θ dt 2 = − 5 3 M g Lθ 3. I d 2 θ dt 2 = − 6 5 M g Lθ 4. I d 2 θ dt 2 = − 12 9 M g Lθ 5. I d 2 θ dt 2 = − 3 2 M g Lθ 6. I d 2 θ dt 2 = − 12 7 M g Lθ 7. I d 2 θ dt 2 = − 16 9 M g Lθ 8. I d 2 θ dt 2 = − 9 5 M g Lθ 9. I d 2 θ dt 2 = − 8 5 M g Lθ correct 10. I d 2 θ dt 2 = − 4 3 M g Lθ Explanation: Torque: τ ≡ r F...
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This note was uploaded on 09/26/2010 for the course PHY 58230 taught by Professor Svhets during the Spring '10 term at University of Texas.

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phyhw17 - valdez(vv689 – Homeowork 17 – florin...

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