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anonymous
 one year ago
mechanics
anonymous
 one year ago
mechanics

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433245222553:dw XYZ is the fixed system of coordinates and xyz is the timevarying system of coordinates Given \[\vec A=A_{1}i+A_{2}j+A_{3}k\] Also given \[\frac{d \vec A}{dt}_{f}=\frac{d \vec A}{dt}_{m}+\vec \omega \times \vec A\] where \[\frac{d \vec A}{dt}_{f}\] and \[\frac{d \vec A}{dt}_{m}\] are time derivatives of vector A with respect to the fixed and variying system of coordinates respectively Now the part I don't understand is Let \[D_{f}\] and \[D_{m}\] be symbolic time derivative operators in the fixed and moving systems respectively. Demonstrate the operator equivalence \[D_{f} \equiv D_{m} + \vec \omega \times\] What is the triple equal sign and where is there an unecessary cross next to omega?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433245759143:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0@Kainui @UnkleRhaukus

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0isnt the last eqn the same equation written in other notations?? and that after that cross must be A(vector)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's kind of what I'm asking

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then both of them are same..

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0"\(\equiv\)" : If A is *defined* to be B, then some people write A\(\equiv\)B. \(\vec A \) is missing at end last line

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I've written it exactly as it was written in the pdf, there was no A at the end

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Here's what they've done : \[D_{f} \vec A=D_{m} \vec A+\vec \omega \times \vec A=(D_{m}+\vec \omega \times) \vec A\]\[\implies D_{f} \equiv D_{m}+\vec \omega \times\] I don't understand the need of this definition?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0it's all very pompous, i agree

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Another thing they've done later in another problem is \[\omega \times D_{m} \vec A\] becomes \[(D_{m} \omega) \times \vec A\] Isn't that wrong?You're suppose to do the derivative of A but they switch it with omega

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry there's an arrow with the omega in both those 2 steps

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0nvm I figured out what they did in the later problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0last 3rd post was a box product and we can replace dot by cross.. that wont change the value of box as the determinant's 3rd row will be replaced by 1st row
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