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Unformatted text preview: d L dt = N . (9) Next, take the dot product of Eq.(9) with Ï‰ to get Ï‰ Â· d L dt = Ï‰ Â· N . (10) Now use the rate of change of a vector formula to rewrite the left dot product of Eq.(10) as Ï‰ Â· d L dt = Ï‰ Â· Î¼ d L dt Â¶ b + Ï‰ Â· ( Ï‰ Ã— L ) , (11) and recognize that Ï‰ Â· ( Ï‰ Ã— L ) = . Then, express Ï‰ and L as vectors in the principal axis body frame (with constant values of the principal moments I 1 , I 2 , and I 3 ) to expand the dot product as Ï‰ Â· Î¼ d L dt Â¶ b = ( Ï‰ 1 I 1 dÏ‰ 1 dt + Ï‰ 2 I 2 dÏ‰ 2 dt + Ï‰ 3 I 3 dÏ‰ 3 dt ) = 1 2 d dt ( I 1 Ï‰ 2 1 + I 2 Ï‰ 2 2 + I 3 Ï‰ 2 3 ) = dT dt , (12) and obtain the desired result, dT dt = Ï‰ Â· N . (13) The distinction between a space frame and body frame derivative in the f nal equation is super F uous because T is a scalar. 2...
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 Fall '10
 Wilemski
 mechanics, Derivative, Angular Momentum, Force, Inertia, Mass, Special Relativity, DT DT DT

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