COSMOS Chapter 9 Solution 25 | Answer Key - Edubirdie

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

2019
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415

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1
Author:

gaarmaakbn9h

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 25. J O ≡ I 0 = ∫ r 2dA (a) Have dA = Where Then R  3π J O = ∫R 2 r 2  1  2 3π rdr 2  r  dr  R = 3π 4 2 3π 4 r R2 − R14 = 8 8 R1 ( ) JO = I x = 3 ( I x )1 so that ( )1 + ( I y )2 + ( I y )3 = 3 ( I y )1 Iy = Iy ( I x )1 = ( I y )1 Symmetry implies Ix = I y Then Then ) ( I x )1 = ( I x )2 = ( I x )3 By inspection Now ( I x = ( I x )1 + ( I x )2 + ( I x )3 (b) Now Similarly, 3π 4 R2 − R14 8 JO = I x + I y Ix = I y = JO 3π 4 = R2 − R14 2 16 ( ) or I x = I y = 3π 4 R2 − R14 16 ( )

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