17 - bhappu (nkb284) Homeowork 17 orin (58140) This...

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bhappu (nkb284) – Homeowork 17 – forin – (58140) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices be±ore answering. 001 10.0 points A horizontal plat±orm vibrates with simple harmonic motion in the horizontal direction with a period o± 2 . 38 s. A body on the plat- ±orm starts to slide when the amplitude o± vibration reaches 0 . 337 m. ²ind the coe³cient o± static ±riction be- tween body and plat±orm. The acceleration o± gravity is 9 . 8m / s 2 . Correct answer: 0 . 239668. Explanation: Let : T =2 . 38 s , A max =0 . 337 m , and g =9 . / s 2 . At each instant, there are two ±orces acting on the plat±orm: the ±orce responsible ±or the oscillation F = - kx and the ±orce o± ±riction F s = μN between the body and the plat±orm. Applying Newton’s second law horizontally, ± F x = - + F s = ma platform The only ±orce acting on the block in the horizontal direction is the ±rictional ±orce, so F s = block I± the block does not slide, its acceleration is the same as the plat±orm: a block = a platform set = a The ±orce o± ±riction is F s = = μmg and ±rom simple harmonic motion x = A cos ωt . a = d 2 x dt 2 = - 2 cos ωt , so the maximum acceleration is a max = A max ω 2 and the coe³cient o± static ±riction is μ = A max ω 2 g . Since T = 2 π ω , μ = A max 4 π 2 gT 2 = (0 . 337 m)(4 π 2 ) (9 . / s 2 )(2 . 38 s) 2 = 0 . 239668 . 002 (part 1 o± 3) 10.0 points Consider a light rod o± negligible mass and length L pivoted on a ±rictionless horizontal bearing at a point O. Attached to the end o± the rod is a mass M 1 . Also, a second mass M 2 o± equal size ( i.e. , M 1 = M 2 = M ) is attached to the rod ² 7 9 L ±rom the lower end ³ , as shown in the Fgure. 7 9 L M 2 M 1 O θ What is the moment o± inertia I about O ? 1. I = 130 81 ML 2 2. I = 106 81 2 3. I = 26 25 2
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bhappu (nkb284) – Homeowork 17 – forin – (58140) 2 4. I = 58 49 ML 2 5. I = 85 81 2 correct 6. I = 17 16 2 7. I = 13 9 2 8. I = 74 49 2 9. I = 34 25 2 10. I = 65 49 2 Explanation: The momentum oF inertia is I Md 2 . There are two masses with I M 1 = 2 and I M 2 = M ± L - 7 9 L ² 2 = M ± 2 9 L ² 2 = 4 81 2 , so I = I M 1 + I M 2 = ± 1+ 4 81 ² 2 = 85 81 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 Lθ 2. I d 2 θ dt 2 = - 14 9 3. I d 2 θ dt 2 = - 10 7 4. I d 2 θ dt 2 = - 13 8 5. I d 2 θ dt 2 = - 5 4 6. I d 2 θ dt 2 = - 6 5 7. I d 2 θ dt 2 = - 11 8 8. I d 2 θ dt 2 = - 16 9 9. I d 2 θ dt 2 = - 11 9 correct 10. I d 2 θ dt 2 = - 7 5 Explanation: Torque: τ rF sin φ The relationship between torque and angu- lar acceleration is τ = I α.
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17 - bhappu (nkb284) Homeowork 17 orin (58140) This...

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