Linear velocity is cross product of angular velocity and position

Question

Why is linear velocity is cross product of angular velocity and position?

Answer 1

If you take $\bf d\theta$ to be the (infinitesimal) angle swept about the rotation axis (about axial vector $\bf\omega$ in diagram) after a small displacement $d\bf r_j'$ then looking at the diagram, you can form the cross product $$\tag 1d\bf r_j'=d\theta\times r_j'$$

If this displacement occurs in a time interval given by $dt$ and you divide both sides of equation (1) by $dt$ you get $$\bf\frac{d\bf r_j'}{dt}=\frac{d\theta}{dt}\times r_j'$$ or $$\bf v=\omega\times r$$

Answer 2

2D case

the components of the position vector to the point P, given in inertial system are

\begin{align*} &\vec{R}=\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}= \underbrace{\begin{bmatrix} \cos(\phi(t)) & -\sin(\phi(t)) & 0 \\ \sin(\phi(t)) & \cos(\phi(t)) & 0 \\ 0 & 0 &1 \\ \end{bmatrix}}_{\mathbf S}\,\begin{bmatrix} u_x \\ u_y \\ 0 \\ \end{bmatrix} \end{align*}

thus the velocity

\begin{align*} &\vec{v}=\frac{d}{dt}\,\vec{R}=\frac{d}{dt}\,\phi(t) \begin{bmatrix} -\sin(\phi(t)) & -\cos(\phi(t)) & 0 \\ \cos(\phi(t)) & -\sin(\phi(t)) & 0 \\ 0 & 0 &0 \\ \end{bmatrix}\,\begin{bmatrix} u_x \\ u_y \\ 0 \\ \end{bmatrix}=\vec{\omega}\times\vec{R}\quad,\text{where}\\ & \vec{\omega}= \begin{bmatrix} 0 \\ 0 \\ \frac{d}{dt}\,\phi(t) \\ \end{bmatrix} \end{align*}

3D case \begin{align*} & \vec{R}=\mathbf{S}\left[~\phi_x~,\phi_y,~\phi_z~\right]\,\vec{u} \quad\Rightarrow\\ &\vec v=\vec{\dot{R}}=\underbrace{\left[\frac{\partial\,\mathbf{S} }{\partial \phi_x}\,\dot{\phi}_x+ \frac{\partial\,\mathbf{S} }{\partial \phi_y}\,\dot{\phi}_y+ \frac{\partial \,\mathbf{S}}{\partial \phi_z}\,\dot{\phi}_z\right]}_{\mathbf{\dot{S}}}\,\vec u \overset{!}{=}\vec{\omega}\,\times\vec{R}\\ &\text{where}\quad \left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}}\\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right]=\mathbf{\dot{S}}\mathbf{S}^T \end{align*}

• $\mathbf S~$ is the transformation matrix between body-system and inertial-system

Answer 3

It's true by definition, there's really nothing more to it than that.

Angular velocity does not have to be represented by a vector, though it can be in 3 dimensions. We choose to represent it by the vector

$$\vec{w} = \vec{r} \times \vec{v}\tag1$$

And therefore following the cyclic properties of any cross-product, it must be true that

$$\vec{v} = \vec{w} \times \vec{r}\tag2$$

Answer 4

Angular velocity is a vector quantity and has both a magnitude and a direction. The direction is the same as the the angular displacement direction from which we defined the angular velocity. Where r is position vector and linear velocity is also vector quantity. The resultent of vector product of any two vectors is a vector quantity.