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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

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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).

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  • $\begingroup$ This explanation gives great insight. Thanks $\endgroup$ – Nikhil Kumar May 1 at 12:15
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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.

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  • $\begingroup$ Now I remember about free vectors and localized vectors. Thanks for this explanation. $\endgroup$ – Nikhil Kumar May 1 at 12:16

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