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refer to this figure

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 .

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  • $\begingroup$ 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). $\endgroup$
    – Gert
    Commented Feb 23, 2021 at 19:00
  • $\begingroup$ It's also clear that there's no centripetal acceleration, so there's no $r$ and no $\omega$. $\endgroup$
    – Gert
    Commented Feb 23, 2021 at 20:43
  • $\begingroup$ @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. $\endgroup$
    – Sandejo
    Commented Feb 23, 2021 at 22:17
  • $\begingroup$ @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. $\endgroup$
    – Richard Myers
    Commented Feb 23, 2021 at 23:31
  • $\begingroup$ related: physics.stackexchange.com/q/616572/226902 $\endgroup$
    – Quillo
    Commented Feb 23, 2021 at 23:31

2 Answers 2

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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.

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  • $\begingroup$ Is omega changing every second as the body is going in linear velocity $\endgroup$
    – Shiv
    Commented Feb 24, 2021 at 14:01
  • $\begingroup$ If r x v changes then $\omega$ can change even with v constant. It depends on the origin and the r and v vectors. $\endgroup$
    – John Darby
    Commented Feb 24, 2021 at 15:03
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Linear velocity can be considered as the velocity of center of mass and the and forget about v=rw here because the there is no angular velocity in the car or any vehicle here. The angular velocity is only with the wheels but thats the part of a the system. And we dont count the internal forces of the system(torque in wheels) so we only take the linear velocity of the car but if the angular velocity of the wheels was known then we could have calculated the v by v= rw. And that v would have been the linear velocity of center of mass of the car which is only doing linear motion. Similarly if u need angular velocity of wheels u can get it by v= rw where v is 10 as in question and r is the radius of wheel. .. And v= rw doesnt mean that when there is angular velocity there has to be a angular and whatever because the center of mass does have a linear velocity but not the angular one since it passes from the axis of rotation. And in your question angular velocity is of wheels but car does not have angular velocity. And radius is constant. The r is the radius of wheel. Use it to find angular velocity of wheel.. ... Plz feel free to clear your doubts regarding rotational or rolling motion.

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