COSMOS Chapter 9 Solution 111 | Answer Key - Edubirdie

University:
Michigan State University
Course:
ERG 160 | Success in Science, Technology, Engineering, and Mathematics
Academic year:

2023
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198

Pages:

1
Author:

salcignofi13

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 111. First observe that for a given area A and origin O of a rectangular coordinate system, the values of I ave and R are the same for all orientations of the coordinate axes. Shown below is a Mohr’s circle, with the moments of inertia, I x′ and I y′ , and the product of inertia. I x′y′ , having been computed for an arbitrary orientation of the x′y′ axes. From the Mohr’s circle I x′ = I ave + R cos 2θ I y′ = I ave − R cos 2θ I x′y′ = R sin 2θ I x′ I y′ − I x2′y′ Then, forming the expression I x′ I y′ − I x2′y′ = ( I ave + R cos 2θ )( I ave − R cos 2θ ) − ( R sin 2θ ) ( ) ( 2 = I ave − R 2 cos 2 2θ − R 2 sin 2 2θ 2 ) 2 = I ave − R 2 which is a constant ∴ I x′ I y′ − I x2′y′ is independent of the orientation of the coordinate axes Q.E.D. Shown is a Mohr’s circle, with line OA, of length L, the required tangent. Noting that OAC is a right angle, it follows that 2 L2 = I ave − R2 or L2 = I x′ I y′ − I x2′y′ Q.E.D.

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