mechanics

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- anonymous

mechanics

- chestercat

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- anonymous

|dw:1433245222553:dw|
XYZ is the fixed system of coordinates and
xyz is the time-varying 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

|dw:1433245759143:dw|

- ganeshie8

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## More answers

- anonymous

isnt the last eqn the same equation written in other notations??
and that after that cross must be A(vector)

- anonymous

I'm not sure

- anonymous

That's kind of what I'm asking

- anonymous

then both of them are same..

- IrishBoy123

"\(\equiv\)" : If A is *defined* to be B, then some people write A\(\equiv\)B.
\(\vec A \) is missing at end last line

- anonymous

I've written it exactly as it was written in the pdf, there was no A at the end

- anonymous

Here'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

it's all very pompous, i agree

- anonymous

Another 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

sorry there's an arrow with the omega in both those 2 steps

- anonymous

nvm I figured out what they did in the later problem

- anonymous

last 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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