Analytical Mech Homework Solutions 125

Analytical Mech Homework Solutions 125 - 0 = I 3 33 + 1 2 3...

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() 333 123 2 1 0 I II ωω ωωω =+ ± Adding, 0 111 2 2 2 II I ω =+ + ±± ± 222 11 2 2 33 1 0 2 d III dt  +   From equation 9.2.5, [] rot d T dt = 0 rot T constant = Multiplying equations 9.3.5 by I , 22 I , and I , respectively: 2 1123 3 2 0 I ± I 2 2123 1 3 0 I ± I 2 3123 2 1 0 I ± I Adding, 0 22 2 1 0 2 d dt + From equation 9.2.4, 2 d L dt = 0 2 L constant = 9.8 From equations 9.3.5 for zero torque … 2 3 3 2 0 I ± 13 1 3 0 I ± From the perpendicular axis theorem, 312 I = + 231 0 I I ± 132 0 I I =− ± Multiplying by 1 1 I and 2 2 I , respectively: 123 0 ± 0 ± Adding, 1 2 1 2 d dt 0 =+ = + ±± 12 constant += If I I = , from the third of Euler’s equations … ( ) 0 I = ± 3 constant = 9.9 (a) From symmetry … 3 s I I = and I = = From the perpendicular axis theorem,
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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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