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Unformatted text preview: reddy (ar38357) – Rotational Inertia – clancy – (SCI4112) 1 This printout should have 8 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A nonuniform disk of mass M and radius R has its mass distributed in such a way that the mass per unit area is a function of the radial distance r from the center of the disk: σ ( r ) = b r , where b is a constant to be determined. What is the rotational inertia of this disk about an axis through the center of mass and perpendicular to the plane of the disk? The area differential can be written as a ring of radius r and thickness dr : dA = 2 π r dr . 1. 2 5 M R 2 2. 3 5 M R 2 correct 3. 1 5 M R 2 4. M R 2 5. 2 3 M R 2 6. 1 2 M R 2 7. 4 5 M R 2 8. 1 4 M R 2 Explanation: The mass per unit area is dM dA = σ ( r ) = b r , so M = integraldisplay dm = integraldisplay dM dA dA . dA = 2 π r dr , so dM dA dA = ( b r ) (2 π r dr ) = 2 π b r 2 dr and M = 2 π b integraldisplay R r 2 dr = 2 π b R 3 3 b = 3 M 2 π R 3 . Thus the moment of inertia is I CM = integraldisplay R r 2 dm = integraldisplay R r 2 dM dA dA = integraldisplay R r 2 (2 π b r 2 ) dr = 2 π b integraldisplay R r 4 dr = 2 π parenleftbigg 3 M 2 π R 3 parenrightbigg r 5 5 vextendsingle vextendsingle vextendsingle vextendsingle R = 3 M 5 R 3 ( R 5 0) = 3 5 M R 2 . 002 10.0 points A pendulum is made of a rod of mass 4 . 9 kg and length 2 . 8 m whose moment of inertia about its center of mass is 1 12 M L 2 and a thin cylindrical disk of mass 3 kg and radius 1 . 7 m whose moment of inertia about its center of mass is 1 2 M R 2 ....
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This note was uploaded on 11/21/2010 for the course PHYS 12231 taught by Professor Sontar during the Fall '10 term at Ill. Chicago.
 Fall '10
 Sontar
 Physics, Inertia, Mass

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