0
$\begingroup$

I've been going thru the book by Hodges and Roithmayr regarding dynamics theory and I can't seem to prove that: enter image description here

I have tried using multiple substitutions, including the fact that every component of the direction cosine matrix is equal to its cofactor, in multiple, different ways. But I feel like I've been walking in circles. The last attempt I made was, when using the definition of angular velocity:

enter image description here

And then writing every derivative of the coordinate system by substituting its direction cosines and substituting the cofactors inside. This led to another interesting thing:

enter image description here

Which may or may not be related to the proof I'm looking for, but I cannot work out how. My guess is that this would be a pretty standard equation to use, since angular velocity is such a big thing in dynamics, but I could not find it anywhere.

Improve this question
$\endgroup$
7
  • $\begingroup$ I think the "Maxwell–Betti reciprocal work theorem" may help. en.wikipedia.org/wiki/Betti%27s_theorem $\endgroup$
    – r13
    Apr 9, 2021 at 23:19
  • $\begingroup$ Could you please explain the $^A\vec\omega^B$ notation? I've never came across it. $\endgroup$
    – NMech ♦
    Apr 10, 2021 at 5:39
  • $\begingroup$ @NMech is the angular velocity of a body B in a reference frame A. $\endgroup$
    – Engels Leonhardt
    Apr 10, 2021 at 13:42
  • $\begingroup$ @r13 I have found the answer, should I post it myself? $\endgroup$
    – Engels Leonhardt
    Apr 10, 2021 at 13:43
  • $\begingroup$ I don't know if the rule permits that (answer the question of yourself) or not. But since nobody has provided a response with a pointed answer, I think it is considered a good service to the readers who have the same question. Also, it might draw further responses with great insight. So why not, I support posting your findings. $\endgroup$
    – r13
    Apr 10, 2021 at 13:52

1 Answer 1

Reset to default
0
$\begingroup$

Knowing that the derivative of a vector in frame A can be written as:

enter image description here

And similarly, the derivative of the same vector written in the B frame can be written as:

enter image description here

Substituting the second equation in the first one we have:

enter image description here

enter image description here

For the above to be true, either the vector being differentiated is a null vector (which is a trivial solution) or:

enter image description here

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.