Consider a particle performing circular motion in an anticlockwise sense, along a circle of radius r. In a very small time interval δt, the particle moves from point A to point B through a distance δs and its angular position changes by δθ.
δθ = \(\frac{arc\,AB}{radius}\) = \(\frac{δs}r\)
As δt \(\longrightarrow\) 0, B will be very close to A and displacement \(\vec{AB}\) = \(\vec{δs}\) will be a straight line perpendicular to radius vector \(\vec{OA}\) = \(\vec{r}\).
By the right hand rule of cross product,
\(\vec{δs}\) = \(\vec{δθ}\) x \(\vec{r}\)
The Line vector \(\vec{v}\) of the particle is the time rate of displacement and its angular velocity \(\vec{\omega}\) is the time rate of angular displacement.
Therefore, from Eq. (1).
\(\vec{v}\) = \(\vec{\omega}\) x \(\vec{r}\)
Since \(\vec{ds}\) is tangential, the instantaneous linear velocity \(\vec{v}\) of a particle performing circular motion is along the tangent to the path, in the sense of motion of the particle.
\(\vec{v}\), \(\vec{\omega}\) and \(\vec{r}\) are mutually perpendicular, so that in magnitude, v = ωr.
