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Taylor Relies on these relations

  1. $v = \omega \times r$
  2. $\frac{d}{dt}Q = \omega \times Q$

To show that

$a = a' + 2 \omega \times v' + \omega \times \omega \times r' + \alpha \times r' $

So we take the product rule of (1) and get:

$a = \dot{\omega} \times r + \omega \times \dot{r}$

The first terms is where I run into problems, which is my question, because

$\dot{\omega} \times r = \alpha \times r$

but shouldn't we use (2) on $\omega$ since its position can be written as

$\omega = \omega \hat{u} = \omega_x \bar{x} + \omega_y \bar{y} + \omega_z \bar{z}$

Hence, we should instead get

$\dot{\omega} \times r = \omega \times \omega \times r$

What makes $\omega$ so special that we avoid using property (2)? Does it have to do with the fact that $\omega = \omega '$? I saw something about Euler angles using $\omega$ written as a position vector

Edward
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1 Answers1

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First of all, the correct relation to start from is
$$\frac{dQ}{dt}= \frac{dQ’}{dt}+ \omega \times Q’, $$ where $Q$ is any vector.
Without this you will not be able to derive your formula.

Now, to your question. You have to remember that order matters when taking the cross product. We have $\omega\times(\omega\times r)\not\equiv 0$, but $(\omega\times \omega) \times r \equiv0$. Keep this in mind in your last equation and use the correct relation for the derivative, and you should be able to derive you formula.

e4f5
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