COSMOS Chapter 9 Solution 185

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 185. m1 = ρ ta 2 From Problem 9.170 Now I x = ( I x )1 + ( I x )2 = m2 = 1 ρ ta 2 2 1 1 1 5  ρ ta 2 a 2 +  ρ ta 2  a 2 = ρ ta 4 3 6 2 12  ( ) ( )1 + ( I y )2 = 13 ( ρta2 ) a2 + 16  12 ρta2  ( a2 + a2 ) = 12 ρta4 Iy = Iy I z = ( I z )1 + ( I z )2 = Now note that symmetry implies 1 11 3  ρ ta 2 a 2 + a 2 +  ρ ta 2  a 2 = ρ ta 4 3 6 2 4  ( )( ) ( I x′y′ )1 = ( I y′z′ )1 = ( I z′x′ )1 = 0 ( I x′y′ )2 = ( I y′z′ )2 = 0 Have I uv = I u′v′ + mu v continued COSMOS: Complete Online Solutions Manual Organization System  a  a  1 I xy = m1x1 y1 + m2 x2 y2 = ρ ta 2    = ρ ta 4  2  2  4 Then I yx = m1 y1z1 +0 m2 y2 z02 0 ( )2 + m2 z2 x2  I zx = m1z1x1 +  I z ′x′  ( I z′x′ )2 = − 721 ρ ta 4 From Problem 9.170 I zx = − Then 1 1 1  1  1  ρ ta 4 +  ρ ta 2  a  a  = ρ ta 4 72 2 3 3 24     (a) Equation 9.56 ( ) ( ) ) K 3 − I x + I y + I z K 2 + I x I y + I y I z + I z I x − I xy − I yz − I zx K ( 2 2 2 − I x I y I z − I x I yz − I y I zx − I z I xy − 2I xy I yz I zx = 0 Substituting 2   5  1   1  3   3  5   1 2  5  1 3  1   K 3 −  + +  ρ ta 4  K 2 +     +    +    −   − 0 −    ρ ta 4  24    12 2 4     12  2   2  4   4  12   4  ( 2 2   5  1  3    1  1   3  1  −      − 0 −    −    − 0  ρ ta 4 12 2 4  2  24   4  4        ( Solving numerically...  3 )  = 0  K = ρ ta 4ζ Simplifying and letting yields 2 )  K 5 3 ζ3 − ζ2 + ζ 1 = 0.203032 479 125 ζ − =0 576 1152 or K1 = 0.203ρ ta 4
ζ 2 = 0.698281 or K 2 = 0.698ρ ta 4
ζ 3 = 0.765354 or K3 = 0.765ρ ta 4
(b) Equations 9.54a and 9.54b ( I x − K )( λx ) − I xy ( λ y ) − I zx ( λz ) = 0 ( )( ) − I xy ( λx ) + I y − K λ y − I yz ( λz ) = 0 continued COSMOS: Complete Online Solutions Manual Organization System Substituting K1  5 1   12 − 0.203032  ( λx )1 − 4 λ y   ( )1 − 241 ( λz )1  ρta 4 = 0   1 1   − ( λx )1 +  − 0.203032  λ y 4 2    ( )1 − 0 ρ ta 4 = 0  or ( λ y )1 = 0.841842 ( λx )1 and ( λz )1 = 0.0761800 ( λx )1 Equation 9.57 Substituting ( λx )12 + ( λ y )1 + ( λz )12 2 ( λx )12 + 0.841842 ( λx )1  or 2 =1 2 + 0.0761800 ( λx )  = 1 ( λx )1 = 0.763715 then (θ x )1 = 40.2°
( λ y )1 = 0.642927 then (θ y )1 = 50.0°
( λz )1 = 0.0581798 then (θ z )1 = 86.7°
Substituting K 2  5 1   12 − 0.698281 ( λx )2 − 4 λ y    ( )2 − 241 ( λz )2  ρta4 = 0   1 1   − ( λx )2 +  − 0.698281 λ y 2   4 ( )2 − 0 ρ ta 4 = 0 or and Then  ( λ y )2 = −1.260837 ( λx )2 ( λz )2 = 0.806278 ( λx )2 ( λx )22 +  −1.260837 ( λx )2  or 2 2 + 0.806278 ( λx )2  = 1 ( λx )2 = 0.555573 ( λ y )2 = − 0.700487 ( λz )2 = 0.447946 then then then (θ x )2 = 56.2°
(θ y )2 = 134.5°
(θ z )2 = 63.4°
continued COSMOS: Complete Online Solutions Manual Organization System Substituting K 3  5 1   12 − 0.765354  ( λx )3 − 4 λ y   ( )3 − 241 ( λz )3  ρta 4 = 0   1 1   − ( λx )3 +  − 0.765354  λ y 2   4 ( )3 − 0 ρta 4 = 0 or ( λ y )3 = − 0.942138 ( λx )3 And Then ( λz )3 = − 2.71567 ( λx )3 ( λx )32 +  − 0.942138 ( λx )3  or (c)  2 2 +  − 2.71567 ( λx )3  = 1 ( λx )3 = 0.328576 then (θ x )3 = 70.8°
( λ y )3 = − 0.309564 then (θ y )3 = 108.0°
( λz )3 then (θ z )3 = − 0.892304 = 153.2°