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# HW17.sol - homework 17 RAHMAN TARIQUE Due Apr 2 2008 3:00...

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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 AlternativePart 1: Using moments of iner- tia as found in the book and the parallel axis theorem, I disk = 1 2 m r 2 = 1 2 M r 2 I ring = m r 2 = M r 2 I rod cm = 1 12 m ℓ 2 I square = 4 parenleftbigg 1 12 m ℓ 2 + m d 2 parenrightbigg m = M 4 = 2 r d = r I square = 4 bracketleftbigg 1 12 parenleftbigg M 4 parenrightbigg (2 r ) 2 + parenleftbigg M 4 parenrightbigg r 2 bracketrightbigg = 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 parenleftbigg 2 5 mr 2 parenrightbigg ω 2 v = Solution: KE rot = 1 2 parenleftbigg 2 5 mr 2 parenrightbigg parenleftBig v r parenrightBig 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 = parenleftbigg 1 5 parenrightbigg parenleftbigg 10 7 parenrightbigg = 2 7 Question 3, chap 12, sect 5.
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HW17.sol - homework 17 RAHMAN TARIQUE Due Apr 2 2008 3:00...

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