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hw25 - Muraj Hamza Homework 25 Due 4:00 am Inst Florin This...

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Muraj, Hamza – Homework 25 – Due: Mar 31 2006, 4:00 am – Inst: Florin 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points A solid sphere of radius 21 cm is positioned at the top of an incline that makes 28 angle with the horizontal. This initial position of the sphere is a vertical distance 2 . 5 m above its position when at the bottom of the incline. The sphere is released and moves down the incline. The acceleration of gravity is 9 . 8 m / s 2 . Hint: The moment of inertia of a sphere with respect to an axis through its center is 2 5 M R 2 . 21 cm M μ 28 2 . 5 m Calculate the speed of the sphere when it reaches the bottom of the incline in the case where it rolls without slipping. Correct answer: 5 . 91608 m / s. Explanation: Basic Concepts: X E = K trans + K rot + U gravity = const From conservation of energy we have M g h = 1 2 M v 2 1 + 1 2 I ω 2 = 1 2 M v 2 1 + 1 2 2 5 M R 2 v 2 1 R 2 = 7 10 M v 2 1 . Therefore, v 1 = r 10 7 g h = r 10 7 (9 . 8 m / s 2 ) (2 . 5 m) = 5 . 91608 m / s . 002 (part 2 of 2) 10 points Calculate the speed of the sphere when it reaches the bottom of the incline in the case where it slips frictionlessly without rolling. Correct answer: 7 m / s. Explanation: From conservation of energy we have M g h = 1 2 M v 2 2 . Therefore, v 2 = p 2 g h = q 2 (9 . 8 m / s 2 ) (2 . 5 m) = 7 m / s . 003 (part 1 of 3) 10 points Consider a thick cylindrical shell rolling down along an incline. It has a mass m , an outer ra- dius R and a moment of inertia I cm = 4 5 mR 2 about its center of mass. The incline is at an angle θ to the horizontal. Its length is s and its height is h . s h θ v A A The equation of motion τ = as the shell is rolling down the incline is given by(using the contact point as the pivot point): 1. mgR = I cm α 2. mgR sin θ = ( I cm + mR 2 ) α correct 3. mgR tan θ = I cm α 4. mgR = ( I cm + mR 2 ) α 5. mgR sin θ = I cm α
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