Analytical Mech Homework Solutions 136

Analytical Mech Homework Solutions 136 - a ( b2 + c 2 ) 1 m...

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() [] ( ) 22 2 222 1 2 3 ab c m Ta b c b abc ca b ω a c + =+ ++ + ( 2 22 2 2 6 m T abc bac cab  + + +  ) + 2 3 m b a c b c + With the origin at one corner, from the parallel axis theorem: 4 33 xx mb c m I c b c + + = + 4 3 yy m I ac , 4 3 zz m I xy I xydm xy dV ρ =− ∫∫ 000 8 xaybzc xy xyz I xydxdydz a b c ρρ === ∫∫∫ , so 222 8 m a b c abc == xy I mab = − xy I mac , yz I mbc = − 4 3 4 3 4 3 b c ab ac I m ab a c bc ac bc a b +−  + −− + I 9.23 (See Figure 9.7.1) ()()() zx y z zz z LL L L ′′ =++ ( ) ( ) sin cos sin sin cos zy z s L I I S θ θφ =+= + ± 9.24 (See Figure 9.7.1) If the top precesses without nutation, it must do so at = D where V D is a minimum of ( ) V
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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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