# Relative linear velocity of a particle to a rotating object

I am trying to calculate the "relative linear velocity" of a particle moving over a rotating object.

According to this paper (section 2.2) I am reading the relative linear velocity is calculated by:

$$v'= v - \omega r$$

Where $v$ is the velocity of the particle, $\omega$ the angular velocity of the rotating object and $r$ the radius of the object.

What I don't understand is why subtract? For high angular velocities with low particle velocities that will produce a negative velocity. The particle's surely not traveling backwards relatively, if the particle is traveling in the same direction as the rotation?

• can you use vector notations please . it seems the situation you are talking of is that $v$ is velocity of axis of rotation and $r\omega$ is in opposite direction as that of axis of rotation. – user31782 Dec 18 '13 at 15:38
• Could you reference the paper? – Kyle Kanos Dec 18 '13 at 15:41
• @KyleKanos This is the paper sciencedirect.com/science/article/pii/S0304389411006637 Section 2.2.2 is where the above equation comes from. – Lerp Dec 18 '13 at 15:45 $$\vec v = \vec{\omega} \times\vec{r}$$ when object has no translational motion but if the object has both translational and rotational motion then $\vec v$ will be measured from Center-of-momentum frame. in the frame where the C.M is translating at velocity $\vec v_{C.M}$ the velocity of any point $P$ revolving at angular velocity $\vec \omega$ can be found by using Galilean transformation i.e $\vec v^{'}= \vec v + \vec {v_{C.M}}$. (assuming $v,v_{C.M}<<c$)
in your case point $P$ is moving in the direction that is opposite to the direction of motion of C.M that's why subtraction is being done , if $P$ was moving in the same direction as that of $v_{C.M}$ then addition would have been done.
Suppose we have a particle moving with a velocity $\vec{v}_1$ and a object which has no linear velocity, but is rotating with angular velocity $\vec{\omega}$. Now suppose we look at a point at a displacement $\vec{r}$ from the axis of rotation of this rotating velocity. The velocity of this point is $\vec{v}_2 = \vec{\omega} \times \vec{r}$.
Now the velocity of the first object relative to the point on the disc is the difference of these two velocity. This makes sense because if they are moving with the same velocity then the relative velocity is zero. Thus we find $\vec{v}_{rel} = \vec{v}_1 - \vec{v}_2 = \vec{v}_1 -\vec{\omega} \times \vec{r}$.