# Where is the angular velocity when linear velocity takes place and vice versa?

We can see the body is moving in a straight line with a linear velocity 10 m/s. After 4 second it is 50 m away from the observer. And assume it to continue to move straight.

Also by the relation v=rω when there is linear velocity there has to be angular velocity.

Now where exactly the angular velocity is taking place . The radius is changing every time so how does exactly the angular velocity comes in picture. I am not getting the picture that how does the angular velocity will look like or takes place. Thank you .

• Also by the relation $v=rω$ when there is linear velocity there has to be angular velocity. No, of course not, that formula works for circular motion (or motion with curvature).
– Gert
Feb 23, 2021 at 19:00
• It's also clear that there's no centripetal acceleration, so there's no $r$ and no $\omega$.
– Gert
Feb 23, 2021 at 20:43
• @Gert angular velocity can be defined using the same formula in more general situations, such as this one. The reason we only talk about it in the context of circular motion is because that's generally the only time it's useful. Feb 23, 2021 at 22:17
• @Sandejo Just because something can be defined doesn't mean it should, and I think it's misleading to say you can always define an angular acceleration regardless of whether the motion is circular. If the motion is not circular, the only true statement you can make generally is that we are always free to express velocities in the polar coordinate basis. Feb 23, 2021 at 23:31
• Feb 23, 2021 at 23:31

Angular velocity depends on the origin you select. Let $$\vec v$$ be the linear velocity of a particle at point P. Let O be the point selected as the origin. The angular velocity $$\vec \omega$$ of the particle at P with respect to O is $$\vec \omega = {{\vec r \times \vec v} \over {r^2}}$$ where $$\vec r$$ is the position vector from point O to point P, and $$\times$$ is the vector cross product. If O is selected such that $$\vec r$$ is parallel to $$\vec v$$, than $$\vec \omega$$ is zero.
• If r x v changes then $\omega$ can change even with v constant. It depends on the origin and the r and v vectors. Feb 24, 2021 at 15:03