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

2022
Views:

171

Pages:

2
Author:

Arielle Armstrong

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 123. 3 2a = k ( 2a ) x = 2a At y= Then 1 3 x 4a 2 ( dm = ρ π r 2dx Now k = or 1 4a 2 ) 2 πρ 6  1  x dx = πρ  2 x3  dx = 16a 4  4a  πρ m= Then 16a 4 = or πρ = 2a 6 ∫a x dx πρ 1 4 16a 7 x 7 2a = a πρ  127 7 7 2a ) − ( a )  = πρ a3 4 ( 112a   112 112m 127a3 2 (a) Now 1 1 1    πρ 6  d I x =  r 2  dm =  2 x3   x dx  4 2 2 4 a      16a  = Then Ix = = 1 6 112m x6 7m × × x dx = x12dx 4 3 4 11 32a 127a 16a 4064a 2a 7m 7m 1 13 2a 12 ∫ x dx = 4064a11 13 x 4064a11 a a 7m  57337 2 ma = 1.0853ma 2 ( 2a )13 − ( a )13  = 52832 52832a11  or I x = 1.085ma 2 continued COSMOS: Complete Online Solutions Manual Organization System (b) Have 2 1  1  πρ 6 1   dI y =  r 2 + x 2  dm =   2 x3  + x 2  x dx 4 4    16a  4  4a = 1 112m  1 12  × x + x8  dx  4 16a 127a3  64a 4  2a 7m 2a  1 12 7m  1 1   x + x8  dx = x13 + x9  Then I y = 7 ∫a  4 7 4 9 a 127a 127a  832a  64a  = 7m  1 1 1 1 9 13 9 13 2a ) + ( 2a ) − a − (a)  7  4( 4( ) 9 9 127a  832a 832a  = 7m  8191 9 511 9  a + a  = 3.67211ma 2 7  832 9 127a   or I y = 3.67 ma 2