Solving cross product equation with first variable unknown

Question

I am trying to solve a practical problem for a mechanical engineering application: finding the angular acceleration given measurements from the linear acceleration and the angular rate in two different points A and B. In a solid rigid, with points A and B fixed (no Coriolis forces) this is related by:

\begin{equation} \vec{a_B} = \vec{a_A} + \vec{\alpha} \times \vec{r} + \vec{\omega}\times(\vec{\omega} \times \vec{r}) \end{equation}

Where $\vec{a_A}$, $\vec{a_B}$ represent the linear accelerations of $A$ and $B$ respectively, $\vec{\alpha}$ the angular acceleration, $\vec{\omega}$ the angular rate, $\vec{r}$ the vector from $A$ to $B$, and for all what matters here they are all vectors or pseudovectors on $\mathbb{R}^3$. Re-arranging:

\begin{equation} \vec{\alpha} \times \vec{r} = \vec{a_B}- \vec{a_A} - \vec{\omega}\times(\vec{\omega} \times \vec{r}) \end{equation}

and renaming the rhs for convenience to $\vec{c}$

\begin{equation} \vec{\alpha} \times \vec{r} = \vec{c} \end{equation}

When trying to solve the system of 3 equations resulting from the cross product through Gauss elimination, I find that the system is linearly dependent and I can't isolate an $\alpha_i$ component without cancelling out the whole row.

What is the geometrical intuition behind this result? If the unknown variable was any of the linear accelerations $\vec{a_A}$, or $\vec{a_B}$, solving it would be trivial. Why having the unknown variable involved in a cross product makes the problem unsolvable?

Answer 1

The intuition is that adding a multiple of $\vec{r}$ to $\vec{\alpha}$ preserves $\vec{\alpha}\times\vec{r}$, so $\vec{\alpha}$ isn't unique. You can't solve for $\vec{\alpha}$ without some more information.

Answer 2

Let me provide a set of solutions, rather than an unique solution; maybe this would be useful to solve your problem. Let me also drop the arrows over variables for clarity. Then, we need to solve $\alpha \times r = c$ for $\alpha$. From cross product anticommutativity we have \begin{equation} -r \times \alpha = c . \end{equation} Then, using the theorem proved in Solve the vector cross product equation, we have \begin{equation} \alpha = \dfrac{c\times -r}{\|-r\|^2} - kr = \dfrac{r\times c}{\|r\|^2} - kr , \end{equation} for aribtrary scalar $k$.