# angular velocity as a tensor rather than a vector

I was looking at a very simplistic case of rotational dynamics yesterday, which described the in-plane rotation of a rod of length $L$ about its center of mass. The position vector of a point at the edge of the rod was defined as $$\vec{P}=\left(\begin{matrix}\frac{L}{2}\cos\theta\\ \frac{L}{2}\sin\theta\\ 0\end{matrix}\right).$$ The angular velocity was defined as $$\vec{\omega}=\left(\begin{matrix}0\\ 0\\ {\theta}'\end{matrix}\right).$$ Obviously the velocity was defined as $\vec{v}=\vec{\omega}\times\vec{P}$ and the cross-product of the vectors was defined as $$\vec{v}=\det \left[A\right];$$ matrix $A$ was defined as the cross-product matrix $$[A]=\left(\begin{matrix}\hat{x} &\hat{y} &\hat{z}\\0 &0 &{\theta}^{'}\\ \frac{L}{2}\cos\theta &\frac{L}{2}\sin\theta &0\end{matrix}\right).$$ Of course it works out, but my question is if I defined $$\vec{P}=\left(\begin{matrix}\frac{L}{2}\cos\theta \\ \frac{L}{2}\sin\theta\end{matrix}\right)$$ and rotational velocity as a tensor $\langle\omega\rangle$ such that $$\langle \omega\rangle= \left(\begin{matrix}0 &{-{\theta}}^{'}\\ {\theta}^{'} &0\end{matrix}\right),$$ then I could easily avoid using the third $z$ co-ordinate into the picture and also satisfy the linear map $$\vec{v}=\langle\omega\rangle\vec{P}.$$ Why then is $\omega$ forced to be a vector and not a tensor? Does it solve some purpose in physics or does it make the maths simple?

• While one usually sees $\vec{\omega}=\dot{\theta} \hat{z}$ in introductory examples, this is not required and in more advanced examples it is not true. For example, the axis of rotation of a spinning top will generally not point straight up but will rather precess around the $z$-axis. So the cross product definition is more general than the tensor definition you give. – Semiclassical Nov 20 '16 at 23:53