# 17 - bhappu(nkb284 Homeowork 17 orin(58140 This print-out...

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bhappu (nkb284) – 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 . 38 s. A body on the plat- form starts to slide when the amplitude of vibration reaches 0 . 337 m. Find the coe ffi cient of static friction be- tween body and platform. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 0 . 239668. Explanation: Let : T = 2 . 38 s , A max = 0 . 337 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 = μ N between the body and the platform. Applying Newton’s second law horizontally, F x = - kx + F s = m a platform The only force acting on the block in the horizontal direction is the frictional force, so F s = m a 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 = μ N = μ m g and from simple harmonic motion x = A cos ω t . a = d 2 x dt 2 = - A ω 2 cos ω t , so the maximum acceleration is a max = A max ω 2 and the coe ffi cient 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 . 337 m)(4 π 2 ) (9 . 8 m / s 2 )(2 . 38 s) 2 = 0 . 239668 . 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 7 9 L from the lower end , as shown in the figure. 7 9 L L M 2 M 1 O θ What is the moment of inertia I about O ? 1. I = 130 81 M L 2 2. I = 106 81 M L 2 3. I = 26 25 M L 2

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bhappu (nkb284) – Homeowork 17 – florin – (58140) 2 4. I = 58 49 M L 2 5. I = 85 81 M L 2 correct 6. I = 17 16 M L 2 7. I = 13 9 M L 2 8. I = 74 49 M L 2 9. I = 34 25 M L 2 10. I = 65 49 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 L - 7 9 L 2 = M 2 9 L 2 = 4 81 M L 2 , so I = I M 1 + I M 2 = 1 + 4 81 M L 2 = 85 81 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 = - 12 9 M g L θ 2. I d 2 θ dt 2 = - 14 9 M g L θ 3. I d 2 θ dt 2 = - 10 7 M g L θ 4. I d 2 θ dt 2 = - 13 8 M g L θ 5. I d 2 θ dt 2 = - 5 4 M g L θ 6. I d 2 θ dt 2 = - 6 5 M g L θ 7. I d 2 θ dt 2 = - 11 8 M g L θ 8. I d 2 θ dt 2 = - 16 9 M g L θ 9. I d 2 θ dt 2 = - 11 9 M g L θ correct 10. I d 2 θ dt 2 = - 7 5 M g L θ Explanation: Torque: τ r F sin φ The relationship between torque and angu- lar acceleration is τ = I α .
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