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

2019
Views:

71

Pages:

2
Author:

Dixie Summers

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 127. dy = dx First note  dy  1+    dx  Then 2 1 2 2  2 −x  a 3 − x 3      2 2 2 − = 1 + x 3 a3 − x3      − 1 3 2  a 3 =   x 2  dy  dm = m′dL = m′ 1 +   dx  dx  For the element shown 1  a 3 = m′   dx x 1 Then Now a 1  2 3 3 m = ∫ dm = m′ 1 dx = m′a 3  x 3  = m′a 2 2   0 x3 1 3  2   2 3   a a 2 3 3 I x = ∫ y dm = ∫ 0  a − x   m′ 1 dx       x3    ∫ 0a a3  2 1 4 1 2 5 a = m′a 3 ∫ 0a  1 − 3a 3 x 3 + 3a 3 x − x 3  dx    3  x  = 1 m′a 3 a 2 4 4 2 8 3 9 3 3 3 3 2 3 3 2 3  a x − a x + a x − x  4 2 8   2 0  3 9 3 3 3 = m′a3  − + −  = m′a3  2 4 2 8 8 1 2 ma 4 1 I y = ma 2 4 or I x = Symmetry implies continued COSMOS: Complete Online Solutions Manual Organization System Alternative Solution  2 I y = ∫ x dm = = = Also 1 m′a 3 ∫ 0a 1  1 5  a 3 3 ′ ′ x  m 1 dx  = m a ∫ 0 x dx  x3    2 a3 a 8 3  3 ×  x 3  = m′a3 8  8  0 1 2 ma 4 ( ) I z = ∫ x 2 + y 2 dm = I y + I x or I z = 1 2 ma 2