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Unformatted text preview: d L dt = N . (9) Next, take the dot product of Eq.(9) with ω to get ω · d L dt = ω · N . (10) Now use the rate of change of a vector formula to rewrite the left dot product of Eq.(10) as ω · d L dt = ω · μ d L dt ¶ b + ω · ( ω × L ) , (11) and recognize that ω · ( ω × L ) = . Then, express ω 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 ω · μ d L dt ¶ b = ( ω 1 I 1 dω 1 dt + ω 2 I 2 dω 2 dt + ω 3 I 3 dω 3 dt ) = 1 2 d dt ( I 1 ω 2 1 + I 2 ω 2 2 + I 3 ω 2 3 ) = dT dt , (12) and obtain the desired result, dT dt = ω · 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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This note was uploaded on 12/19/2010 for the course PHYSICS ph 409 taught by Professor Wilemski during the Fall '10 term at Missouri S&T.
 Fall '10
 Wilemski
 mechanics, Inertia, Mass

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