Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

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}}$

share|improve this answer
up vote 0 down vote accepted

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.