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According to wikipedia, the vector equation describing velocity during circular motion is $ \mathbf{v} = \boldsymbol{\Omega} \times \mathbf{r} $, where $\Omega$ is the axis of rotation, with a magnitude representing the rotational velocity.

However, I'm working in 2 dimensions, and I'd prefer not to throw in a 3-dimensional vector just to make the notation look right. How can I write this formula so that its' valid for 2-vectors? I know I can write it like this, where $\omega$ is a scalar:

$\mathbf{v} = \begin{bmatrix}0 & -1 \\ 1 & 0 \end{bmatrix} \times \mathbf{r} \times \omega = \begin{bmatrix}0 & -\omega \\ \omega & 0 \end{bmatrix} \times \mathbf{r} $

But that feels like even more of a kludge. Which piece of notation am I missing?

2 Answers 2

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Looks like I can use the perpendicular vector notation ($\perp$):

$\mathbf{v} = \mathbf{r}^\perp \times \omega$

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In dimension $N$, the angular velocity has $ \frac{N (N-1)}{2}$ components. Only in three dimensions, it can be expressed as a pseudo-vector, and one can use ordinary vector calculus. In a general dimension, one has to represent the angular velocity as an antisymmetric 2-tensor. In the special case of two dimensions, the angular velocity is a pseudo-scalar and we can use diadic notation, to express the angular velocity as the bi-vector:

$\mathbf{\Omega} = \omega(\mathbf{\hat{x}}\mathbf{\hat{y}}-\mathbf{\hat{y}}\mathbf{\hat{x}})$

In this notation, the linear velocity becomes the dot product between the angular velocity bi-vector and the position vector:

$ \mathbf{v} = \mathbf{\Omega} . \mathbf{r}$

One can verify this notation in components:

$ v_x\mathbf{\hat{x}}+ v_y\mathbf{\hat{y}} = \omega(\mathbf{\hat{x}}\mathbf{\hat{y}}-\mathbf{\hat{y}}\mathbf{\hat{x}}). (x\mathbf{\hat{x}}+ y\mathbf{\hat{y}})$

$ = \omega x (\mathbf{\hat{x}}\mathbf{\hat{y}}.\mathbf{\hat{x}}-\mathbf{\hat{y}}\mathbf{\hat{x}}.\mathbf{\hat{x}}) + \omega y (\mathbf{\hat{x}}\mathbf{\hat{y}}.\mathbf{\hat{y}}-\mathbf{\hat{y}}\mathbf{\hat{x}}.\mathbf{\hat{y}})$

$ = -\omega x \mathbf{\hat{y}} + \omega y \mathbf{\hat{x}}$