2

Sorry for boring you, my friends. I am haunted by re-deducing an expression of translational acceleration in ANSYS theory reference about acceleration effect. It takes the form of: $$\lbrace a_t^r \rbrace = \lbrace \omega \rbrace \times \left( \lbrace \omega \rbrace \times \lbrace r \rbrace \right)+\lbrace \dot{\omega} \rbrace \times \lbrace r \rbrace + 2 \lbrace \Omega \rbrace \times \left( \lbrace \omega \rbrace \times \lbrace r \rbrace \right)+\Omega \times \left( \Omega \times \lbrace\lbrace R \rbrace + \lbrace r \rbrace\rbrace\right )+\dot{\Omega} \times \lbrace\lbrace R \rbrace + \lbrace r \rbrace\rbrace$$

I could not reformulated it following the given context.

The formulation is in the following link: https://www.sharcnet.ca/Software/Ansys/17.0/en-us/help/ans_thry/thy_tool1.html

fig

Can anyone throw some light on this and tell the physical meaning for the different terms?

Thank you in advance for taking a look.

John Alexiou
  • 38,341
Zihan Shen
  • 81
  • Do you know that to differentiate a changing vector $A(t)$ on a rotating body you do $$\frac{{\rm d}A}{{\rm d}t} = \frac{\partial A}{\partial t} + \Omega \times A$$? The above is the results of the differentiation of velocity on a rotating body. – John Alexiou Jun 14 '16 at 15:46
  • See my answer here : http://physics.stackexchange.com/questions/67053/velocity-in-a-turning-reference-frame?rq=1. May be a help for you. – Frobenius Jun 14 '16 at 16:09
  • Yes, actually, during the deduction the above derivative formula was applied. But unfortunately, the middle term was missing in my trail. I supposed that the total motion is the addition of the global rotation and the local rotation in body frame by my imagination. And probably , it is why I couldn't figure it out. Could you please help me illustrate the correct and logical procedure to deal this kind of problem? Thanks in advance. – Zihan Shen Jun 14 '16 at 16:11
  • Related: http://physics.stackexchange.com/a/80449/392 – John Alexiou Jun 14 '16 at 19:41

0 Answers0