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If we have two reference systems, $N$ and $B$, with common origins $O_N=O_B$ and $B$ being rotating around $N$ with angular velocity $\omega_{B|N}$, the time derivates of any vector $\vec{u}$ in both systems are related by the next theorem:

$$\frac{d \vec{u} }{dt}_N=\frac{d \vec{u} }{dt} _{B}+ \vec{\omega}_{B|N} \wedge \vec{u} $$

In wich reference frame are expressed the vectors $\vec{\omega}$ and $\vec{u} $ from the second term of the RHS: in the basis of $N$ or in the basis of $B$?

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  • $\begingroup$ surely your textbook will make this clear. $\endgroup$ – ZeroTheHero Apr 12 '18 at 15:39
  • $\begingroup$ No, it doesn't, I ask here because of that $\endgroup$ – Quaerendo May 4 '18 at 7:25
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The angular velocity $\vec{\omega}$ is that of the $B$ rotating about $N$. The vector $\vec{u}$ on the right-hand side is expressed in terms of the basis of $B$.

One way to convince yourself about it could be by considering $\vec{u} = u_x\hat{i} + u_y\hat{j} + u_z\hat{k}$ with $u_x, u_y, u_z$ as constants. Then, in the basis of $B$, the vector $\vec{u}$ remains a constant. However, it appears time-varying in the basis of $N$. For, $$\frac{d\tilde{u}_x}{dt} = \omega_y u_z - \omega_z u_y,$$ where I have chosen to write the components of $\vec{u}$ in the basis of $N$ with a tilde on top.

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We could rewrite the equation in the following way:

$$\frac{d {}^{B}\vec{u} }{dt}_N=\frac{d {}^{B}\vec{u} }{dt} _{B}+ {}^{B}\vec{\omega}_{B|N} \wedge {}^{B}\vec{u} $$

The angular velocity vector $\vec{\omega}_{B|N}$ is typically written in the B frame.

However, it is not necessary for the vector $\vec{u}$ to be written in the B coordinate frame, because $\vec{u}$ is simply one of the infinity of possible components of the unique vector $\vec{u}$. Rather, components can be written in any arbitrary coordinate frame.

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