# How can all particles of a rigid body have identical angular velocity when their position vector is not perpendicular to the axis?

Consider this spinning disk, in a 3D coordinate system. It is translated above the origin of the coordinate system and spinning around the axis given by this translation vector.

Every source I can find on the internet claims that angular velocity is the same for very particle of this rigid body, but at the same time it is given as the cross product of its positional vector (in relation to the origin) and its linear velocity vector [divided by the squared radius] (edited):

But in this case, the particles position vector is not perpendicular to the axis of rotation and therefore the angular velocities of all the particles should surely describe a cone around the axis. What am I getting wrong here? Is angular velocity calculated not with respect to the origin, but with respect to the projection of the particle on the axis?

• I'm fairly sure that that's not a correct formula for angular velocity. We do have that $\vec v = \vec \omega \times \vec r$ $-$ which makes sense, since velocity should indeed be proportional to the rotational speed and the distance from the axis of rotation. Now check whether that proportionality makes sense in your formula.
– Mew
Jun 3 '20 at 12:11
• I've added my source for the formula for omega.
– JMC
Jun 3 '20 at 12:15
• Please annotate where you edited your question, and please give sources for "Every source I can find on the internet claims that angular velocity is the same for very particle of this rigid body". For two particles on a rigid body, the angular velocity given to them by the rigid body is the same, but their angular velocities chosen about an arbitrary point not on the body will of course differ.
– Mew
Jun 3 '20 at 12:52
• I edited in a link to the wikipedia article in which the formula is found. The same formula is found in other source, as well as on SE. The same article includes the claim that angular velocity is equal for all points on the body.
– JMC
Jun 3 '20 at 13:02
• The wikipedia article is sloppy. In $\vec v=\vec\omega\times\vec r$, $\vec r$ is always the perpendicular displacement of the particle from the rotational axis. If, $\vec r$ is the position vector of the particle, then, as the article states, $\vec v$ is only the perpendicular component of the particle's velocity. Similarly, in $\vec\omega=\vec r\times\vec v/r^2$, $\vec r$ is the perpendicular displacement from the rotational axis. In your diagram, if $\vec a$ is the position of the centre of the disc, you would have $\vec\omega=(\vec r-\vec a)\times\vec v/(\vec r-\vec a)^2$. Jun 3 '20 at 13:39

To answer your question, the angular velocity of any constituent particle in a rotating rigid body is calculated from a perpendicular displacement from the rotation axis, and not from an arbitrary choice of your coordinate system's origin.

If it were the latter, then any angular velocity could be ascribed to that particle by a suitable choice of origin. And since the dynamics of a rotating rigid body are independent of coordinate system, the angular velocity vector cannot be so constructed.

More intuitively, angular mechanics is at its essence an evaluation of motion in terms of pure circular trajectories, as opposed to linear mechanics in which motion is evaluated in terms of straight-line trajectories. As such, the angular velocity vector encodes the idea that the particle moves in a circle in its very construction. Were you to take a single particle at position $$\vec r$$ and imparted a velocity $$\vec v$$, then without constraint the particle would move linearly (along a trajectory $$\vec{s}(t) = \vec r + t\,\vec v$$). If the particle were fixed to stay at locus $$r$$ from the origin $$O$$, then it would rotate about $$O$$. The rate of rotation is given by the angular velocity $$\vec\omega = \vec r \times \vec v/ r^2$$, which simply calculates the component of the velocity vector perpendicular to $$\vec r$$ as the velocity that is pertinent to circular motion: $$\vec \omega = \vec r \times \vec v/ r^2 = \frac{v \sin\theta}{r} \hat\Sigma$$ (where $$\hat \Sigma$$ is a basis vector perpendicular to the plane of motion; in fact, you can interpret this $$\hat\Sigma$$ to define your plane of motion). Thus, whenever angular velocities are calculated it is always the displacement from that centre of the circle that the particle will move along that is used (or, equivalently, the perpendicular displacement from the rotation axis), since the whole point of $$\vec \omega$$ is to describe motion about that point.

The equation $$\vec v =\vec\omega \times \vec r$$ applies to particles moving in circular paths about a common axis. $$\vec \omega$$ is the angular velocity of each particle about its own circle centre (a point on the axis). It is therefore a vector of magnitude $$\omega$$ parallel to the axis.

The equation $$\vec v =\vec\omega \times \vec r$$ then gives you the (tangential) velocity of one of these particles whose displacement from any point on the axis is $$\vec r$$. [One way to see how it works is to write $$\vec\omega \times \vec r$$ as $$\omega(\hat \omega \times \vec r)$$. Here, $$\hat \omega \times \vec r$$, in which $$\hat \omega$$ is the unit vector along the axis, is a vector of magnitude equal to the radius of the circle in which the particle is moving, and direction that of the particle velocity at that instant.]

I first came across the equation, used in this way, in Kibble: Classical Mechanics. I was confused at first because $$\vec \omega$$ is not (in general) taken about the same point for different particles, nor (in general) about the point on the axis from which the displacement is measured! But as long as you realise this, it all makes sense!