COSMOS Chapter 9 Solution 165 - Edubirdie

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

2019
Views:

420

Pages:

1
Author:

Lily Callahan

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 165. Iy = From Figure 9.28 1 2 ma 2 and using the parallel-axis theorem Ix = Iz = 2 1 1 h m 3a 2 + h 2 + m   = m 3a 2 + 4h 2 12 12 2 ( ) ( ) I xy = I yz = I zx = 0 Symmetry implies For convenience, let point A lie in the yz plane. Then 1 λOA = 2 h + a2 ( hj + ak ) With the mass products of inertia equal to zero, Equation (9.46) reduces to 0 I OA = I xλx2 + I yλ y2 + I z λz2  1 h = ma 2   2 2 2  h +a 2  1 m 3a 2 + 4h 2  + 12  ( )  a  2 2  h +a or IOA =    2 1 10h 2 + 3a 2 ma 2 2 12 h + a2 Note: For point A located at an arbitrary point on the perimeter of the top surface, λOA is given by λOA = 1 h2 + a 2 ( a cos φ i + hj + a sin φ k ) which results in the same expression for I OA

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