HW17.sol - homework 17 RAHMAN, TARIQUE Due: Apr 2 2008,...

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homework 17 – RAHMAN, TARIQUE – Due: Apr 2 2008, 3:00 am 1 Question 1, chap 12, sect 5. part 1 of 1 10 points A circular disk, a ring, and a square, have the same mass M and width 2 r . disk 2 r ring 2 r square 2 r For the moment of inertia about their cen- ter of mass about an axis perpendicular to the plane of the paper, which of the follow- ing statements concerning their moments of inertia is true? 1. I square > I disk > I ring 2. I disk > I square > I ring 3. I ring > I disk > I square 4. I square > I ring > I disk correct 5. I ring > I square > I disk 6. I disk > I ring > I square Explanation: Basic Concept: Moment of inertia, rota- tional kinetic energy. Solution: In the ring, the same mass of the disk is concentrated at the maximum distance from the axis. There- fore I ring > I disk In the square, the same mass of the ring lies at distances which are at least the radius of the ring. Therefore I square > I ring thus I square > I ring > I disk Alternative Part 1: Using moments of iner- tia as found in the book and the parallel axis theorem, I disk = 1 2 mr 2 = 1 2 M r 2 I ring = mr 2 = M r 2 I rod cm = 1 12 mℓ 2 I square = 4 p 1 12 mℓ 2 + md 2 P m = M 4 = 2 r d = r I square = 4 b 1 12 p M 4 P (2 r ) 2 + p M 4 P r 2 B = 4 3 M r 2 , since 4 3 M r 2 > M r 2 > 1 2 M r 2 then I square > I ring > I disk . Question 2, chap 12, sect 5. part 1 of 1 10 points A solid sphere rolls along a horizontal, smooth surface at a constant linear speed without slipping. What is the ratio between the rotational kinetic energy about the center of the sphere and the sphere’s total kinetic energy? 1. None of these 2. 3 7 3. 5 3 4. 7 2 5. 2 7 correct 6. 3 5 7. 2 5
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homework 17 – RAHMAN, TARIQUE – Due: Apr 2 2008, 3:00 am 2 Explanation: Basic Concepts: KE trans = 1 2 mv 2 KE rot = 1 2 2 = 1 2 p 2 5 mr 2 P ω 2 v = Solution: KE rot = 1 2 p 2 5 mr 2 P ± v r ² 2 = 1 5 mv 2 and KE tot = KE trans + KE rot = 1 2 mv 2 + 1 5 mv 2 = 7 10 mv 2 Thus KE rot KE tot = 1 5 mv 2 7 10 mv 2 = p 1 5 Pp 10 7 P = 2 7 Question 3, chap 12, sect 5. part 1 of 1 10 points Two balls of equal mass at the bottom of an incline are rolled upward without slipping at the same initial velocity. One ball is solid and the other is a thin- walled hollow ball.
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This note was uploaded on 02/03/2010 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

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HW17.sol - homework 17 RAHMAN, TARIQUE Due: Apr 2 2008,...

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