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TensorExerciseSolutions

# TensorExerciseSolutions - Tensor exercise Solutions Problem...

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Tensor exercise - Solutions Problem 1: Although d ˇ I dt 6 = 0 in general, show that ω · d ˇ I dt · ω = 0 . Solution: Since ˇ I = 3 i =1 I i e i e i , and ˙ I i = 0, we have d ˇ I dt = X i ( ˙ I i e i e i + I i ˙ e i e i + I i e i ˙ e i ) = X i ( I i ˙ e i e i + I i e i ˙ e i ) = X i I i ( ω × e i ) e i + I i e i ( ω × e i ) . Now contract the left leg of each term with ω : ω · d ˇ I dt = X i I i [ ω · ( ω × e i )] e i + I i ( ω · e i )( ω × e i ) . Each term in the equation is now just a vector. The first term on the right hand side vanishes [because ω · ( ω × e i ) = 0], but the second term does NOT vanish in general. (In the second term, we are taking the dot product of ω with e i , not with ω × e i !) So ω · d ˇ I dt = X i I i ( ω · e i )( ω × e i ) . Since ( ω · e i ) is just a scalar, and ( ω × e i ) · ω = 0, each term on the right hand side of the above equation is a vector perpendicular to ω , so the whole expression ω · d ˇ I dt is a vector perpendicular to ω . Therefore ω · d ˇ I dt · ω = 0.

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TensorExerciseSolutions - Tensor exercise Solutions Problem...

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