My goal is to show that $\frac{^Bd}{dt}\vec{\omega}^{B/N}$ = $\frac{^Nd}{dt}\vec{\omega}^{B/N}$, where $\vec{\omega}^{B/N}$ is an angulr-velocity vector representing the rotation of frame $B$ with respect to an inertial reference frame $N$, and $\frac{^Bd}{dt}$ is the time derivative in the $B$ frame (same with $N$ and $\frac{^Nd}{dt}$).

I need to prove this using the Transport Theorem in the form $\frac{^Nd}{dt}\vec{v} = \frac{^Bd}{dt}\vec{v}\ +\ \vec{\omega}^{B/N}\times\vec{v}$, or $\frac{^Bd}{dt}\vec{v} = \frac{^Nd}{dt}\vec{v}\ +\ \vec{\omega}^{N/B}\times\vec{v}$.

It's useful to note that since the frame $B$ is the frame of the object moving with velocity $\vec{v}$, $\frac{^Bd}{dt}\vec{v} = \frac{^Bd}{dt}(\frac{^Bd}{dt}\vec{v}) = 0$, and that $\vec{\omega}^{B/N} = -\vec{\omega}^{N/B}$.

To do this I tried doing all four combinations of two time derivatives. Here are my results:

  • $\frac{^Nd}{dt}(\frac{^Nd}{dt}\vec{v}) = \frac{^Bd}{dt}\vec{\omega}^{B/N}\times \vec{v} + \vec{\omega}^{B/N}\times(\vec{\omega}^{B/N}\times\vec{v})$

  • $\frac{^Bd}{dt}(\frac{^Bd}{dt}\vec{v}) = 0 = \frac{^Nd}{dt}(\frac{^Nd}{dt}\vec{v}) + \frac{^Nd}{dt}\vec{\omega}^{N/B}\times\vec{v} + 2\,\vec{\omega}^{N/B}\times\frac{^Nd}{dt}\vec{v} + \vec{\omega}^{N/B}\times(\vec{\omega}^{N/B}\times\vec{v})$

  • $\frac{^Nd}{dt}(\frac{^Bd}{dt}\vec{v}) = 0 = \frac{^Bd}{dt}(\frac{^Nd}{dt}\vec{v}) + \frac{^Bd}{dt}\vec{\omega}^{N/B}\times\vec{v} + \vec{\omega}^{B/N}\times\frac{^Nd}{dt}\vec{v} + \vec{\omega}^{B/N}\times(\vec{\omega}^{N/B}\times\vec{v})$

  • $\frac{^Bd}{dt}(\frac{^Nd}{dt}\vec{v}) = \frac{^Nd}{dt}\vec{\omega}^{B/N}\times\vec{v} + \vec{\omega}^{B/N}\times\frac{^Nd}{dt}\vec{v} + \vec{\omega}^{N/B}\times(\vec{\omega}^{B/N}\times\vec{v})$

I haven't been able to use these to prove $\frac{^Bd}{dt}\vec{\omega}^{B/N}$ = $\frac{^Nd}{dt}\vec{\omega}^{B/N}$. I must have done some of the differentiation wrong or messed up with the double Transport Theorem.

Can anyone see any mistake of mine or point me in the right direction?



${}^a \dot{\boldsymbol{\omega}}_{b/a} = {}^b\dot{\boldsymbol{\omega}}_{b/a} + \boldsymbol{\omega}_{b/a} \times \boldsymbol{\omega}_{b/a} = {}^b\dot{\boldsymbol{\omega}}_{b/a}$ where $\bf{v} \times \bf{v} = 0 $. Simple.


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