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Unformatted text preview: Version One – Homework 8 – savrasov – 39822 – Nov 13, 2007 1 This printout should have 7 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Solid Sphere on an Incline 05 11:02, trigonometry, numeric, > 1 min, nor mal. 001 (part 1 of 1) 1 points A solid sphere of radius 40 cm is positioned at the top of an incline that makes 25 ◦ angle with the horizontal. This initial position of the sphere is a vertical distance 4 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 . 40 cm M μ ‘ 25 ◦ 4m Calculate the speed of the sphere when it reaches the bottom of the incline in the case where it rolls without slipping. Correct answer: 7 . 48331 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 µ 2 5 M R 2 ¶ µ 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 ) (4 m) = 7 . 48331 m / s . keywords: String Around a Cylinder 11:02, calculus, numeric, > 1 min, normal. 002 (part 1 of 1) 2 points A string is wound around a solid cylindrical spool of mass M = 5 kg and radius R = 0 . 1 m and is tied to the ceiling as shown in the figure. The acceleration of gravity is 9 . 8 m / s 2 . R M h=from bottom of cylinder to ground If the spool is released from rest and rolls on the string, and the distance to the floor is h = 5 m, how long will it take the spool to hit the floor? Correct answer: 1 . 23718 s. Explanation: Basic Concepts: Energy Conservation, K tot = K rot + K trans , where K rot is the kinetic energy of rotational motion about the center of mass and K trans is the kinetic energy of translational motion. Solution: mg T Version One – Homework 8 – savrasov – 39822 – Nov 13, 2007 2 From the force diagram, the force and torque about the center of mass are X F y = T M g = M ( a CM ) (1) and X τ center = + T R = I α = 1 2 M R...
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This note was uploaded on 01/10/2010 for the course PHY na taught by Professor Richardscalettar during the Fall '09 term at UC Davis.
 Fall '09
 RichardScalettar
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