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anonymous

  • one year ago

mechanics

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  1. anonymous
    • one year ago
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    |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?

  2. anonymous
    • one year ago
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    |dw:1433245759143:dw|

  3. ganeshie8
    • one year ago
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    @Kainui @UnkleRhaukus

  4. anonymous
    • one year ago
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    isnt the last eqn the same equation written in other notations?? and that after that cross must be A(vector)

  5. anonymous
    • one year ago
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    I'm not sure

  6. anonymous
    • one year ago
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    That's kind of what I'm asking

  7. anonymous
    • one year ago
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    then both of them are same..

  8. IrishBoy123
    • one year ago
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    "\(\equiv\)" : If A is *defined* to be B, then some people write A\(\equiv\)B. \(\vec A \) is missing at end last line

  9. anonymous
    • one year ago
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    I've written it exactly as it was written in the pdf, there was no A at the end

  10. anonymous
    • one year ago
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    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?

  11. IrishBoy123
    • one year ago
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    it's all very pompous, i agree

  12. anonymous
    • one year ago
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    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

  13. anonymous
    • one year ago
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    sorry there's an arrow with the omega in both those 2 steps

  14. anonymous
    • one year ago
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    nvm I figured out what they did in the later problem

  15. anonymous
    • one year ago
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    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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