## anonymous one year ago mechanics

1. 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?

2. anonymous

|dw:1433245759143:dw|

3. ganeshie8

@Kainui @UnkleRhaukus

4. anonymous

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

5. anonymous

I'm not sure

6. anonymous

That's kind of what I'm asking

7. anonymous

then both of them are same..

8. IrishBoy123

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

9. anonymous

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

10. 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?

11. IrishBoy123

it's all very pompous, i agree

12. 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

13. anonymous

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

14. anonymous

nvm I figured out what they did in the later problem

15. 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