Analytical Mech Homework Solutions 122

Analytical Mech Homework Solutions 122 - 2 2 2 2 2 ma + ma...

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22 2 ˆˆ 32 2 55 5 5 ma ma ma ma Li j ωω ω   =⋅ +− + −+     G 2 4 3 () 2 2 65 ma =+ G j (d) From equation 9.1.32: 2 11 21 25 65 ma TL m a 5 = ⋅ += G G 9.2 ( a ) ( ) ˆ 3 ijk = ++ G 2 2 2 2 12 3 xx rod yy zz ma I Im a I I = == = = a -a a -a -a a 0 xy Ix y d m = −= since, for each rod, either x or or both are 0. The same is true for the other products of inertia. y From equation 9.1.29: ( ) 2 2 ˆ 3 3 Lm a i j k = ⋅+ + G From equation 9.1.32, 222 12 111 23 333 ma ma T 2 + = (b) From equation 9.1.10, with the moments of inertia equal to 2 2 3 ma and the products of inertia equal to 0: 2 cos cos cos 33 ma 2 I ma αβ γ + = (c) For the x-axis being any axis through the center of the
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This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.

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