# Use Transport Theorem to prove equality of angular acceleration vectors.

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?

Thanks.

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