ohw 22 optional-solutions

ohw 22 optional-solutions - Bodet(ngb299 ohw 22 optional...

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Bodet (ngb299) – ohw 22 optional – turner – (57340) 2 For t = , this reduces to θ end = ω 0 σ = 3 . 87 rad / s 0 . 00588043 s - 1 1rev 2 π rad = 104 . 742 rev . 004 (part 1 of 2) 10.0 points Consider a 6 kg square which has its mass concentrated along its perimeter, with each side of length 4 m. d d What is the moment of inertia of the square about an axis perpendicular to the plane of the square at its center of mass? Use the parallel axis theorem and divide the square into parts. The moment of inertia of a rod rotated about its CM is I CM rod = 1 12 md 2 . Correct answer: 32 kg · m 2 . Explanation: Let : m rod = 6kg 4 and \$ =4m . By the parallel axis theorem, the moment of inertia of each rod about the center of the square is I rod square = 1 12 m rod d 2 + m rod ± d 2 ² 2 = 1 3 m rod d 2 . Since there are four sides to the square, the moment of inertia of the square about its center of mass is I CM square =4 I rod square =4 ± 1 3 m rod d 2 ² = 4 3 m rod \$ 2 = 4 3 ( 6kg 4 )(4m) 2 = 32 kg · m 2 . 005 (part 2 of 2) 10.0 points
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This note was uploaded on 10/06/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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ohw 22 optional-solutions - Bodet(ngb299 ohw 22 optional...

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