COSMOS Chapter 9 Solution 119 | Answer Key - Edubirdie

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

2024
Views:

72

Pages:

1
Author:

Tyshawn Shaffer

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 119. mass = m = ρV = ρ tA First note 1   = ρ t ( 2a )( a ) + ( 2a )( a )  2   = 3ρ ta 2 I mass = ρ tI area Also = m I area 3a 2 I x, area = ( I x )1, area + ( I x )2, area (a) Now = 1 1 ( 2a )( a )3 + ( 2a )( a )3 3 12 = 5 4 a 6 I x, mass = Then m 5 × a4 2 6 3a or I x, mass = 5 ma 2 18 (b) Have I z, area = ( I z )1, area + ( I z )2, area 2 1 1 1 3 3 1    =  ( a )( 2a )  +  ( a )( 2a ) + ( 2a )( a )  2a + × 2a   2 3 3   36    = 10a 4 Then I z, mass = = Finally, m × 10a 4 3a 2 10 2 ma 3 I y, mass = I x, mass + I z, mass = 5 10 2 ma 2 + ma 18 3 = 65 2 ma 18 or I y, mass = 3.61ma 2

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