# Point of application of tangential velocity in rotational motion

The tangential velocity of a particle in a rigid body is given by: $$\vec{v}=\vec{\omega}\times \vec{r}$$. Since the cross product is perpendicular to both $$\vec{\omega}$$ and $$\vec{r}$$, the velocity $$\vec{v}$$ will be tangential to the circular path. While finding a cross product, is there a specific point through which the resulting product should pass? This doubt occurred to me because, the resulting $$\vec{v}$$ is drawn through the particle

## 2 Answers

Cross product of two vectors does not "pass through a point". The result of cross product is a vector and just as $$\omega$$ or $$\mathbf r$$, it is part of vector space, not an object in physical space.

However, since $$\mathbf r$$ represents a point in physical space, both $$\mathbf r$$ and $$\omega \times \mathbf r$$ can be regarded as functions of position in physical space. It is customary to draw an arrow representing $$\mathbf v$$ in the $$\mathbf r$$ vector space in such a way that the tail of the arrow is at the point $$\mathbf r$$, but this is only a visualization tool. The vector $$\mathbf v$$ is not fixed to that point mathematically and in some cases it is downrigh misleading to draw it that way (like in derivation of centripetal acceleration on a curved trajectory).

• This explanation gives great insight. Thanks – Nikhil Kumar May 1 '19 at 12:15

v is a global vector unlike r which holds meaning even when translated. It doesn't need to pass through any point. The reason v is drawn through the particle is to illustrate that it represents the velocity of the particle at r in the rigid body and not somewhere else.

• Now I remember about free vectors and localized vectors. Thanks for this explanation. – Nikhil Kumar May 1 '19 at 12:16