I was just reviewing classical mechanics problems on Khan Academy and got the following question:
I knew that the angular speeds at points II, III, and IV must be equal, but I'm not sure how you'd prove that the angular speed at point I is equal as well, since the usual formula $ω=\frac{v}{r}$ just gives $ω=\frac{0}{0}$.
The point of the question is to demonstrate that the angular velocity $\omega$ of the spinning disk, measured from it's centre, is the same for all points on the disk. You can imagine placing an infinitely long rod parallel to the plane of the disc, with one end of the rod positioned at the centre of the disk and the other end infinitely far away. When the disk is spinning, the rod sweeps out an angle $\phi$ at all points of the rod (including the point at the centre of the disk). Now the angular velocity is the rate of change of $\phi$ with respect to time:$$\omega=\frac{d\phi(t)}{dt}.$$Notice how $\phi$ has only $t$ dependence. To show that this is the same for all points on the disk, let $r$ be the distance from the centre of the disk to some point on the disk. Now taking the derivative of $\omega$ with respect to $r$ gives us$$\frac{d\omega}{dr}=\frac{d}{dr}\left(\frac{d\phi(t)}{dt}\right)=0.$$This means that $\omega$ is the same for all points on the disk (analogous to the infinite rod sweeping out the same angle along every point).
Mathematically, you are right that the angular velocity of the zero-dimensional centre point is not defined. This is expressed in the formula you provide, where you approach a zero denominator.
Physically though, to all practical purposes, the centre point can be considered as rotating - you can imagine zooming in down to the centre atom and still define such rotation. From a physical perspective you can argue that every point rotates at the same angular velocity, not just based on the given formula, but simply from the perspective of the object being idealised as perfectly rigid. For a rigid physical object, the angular velocity of any point is equal in order to maintain the material bonds from the definition of rigid.
The relationship between tangential velocity and angular velocity can also be expressed like this:
$$v = \omega R$$
Start with an initial angular velocity $\omega_0$. As the radius approaches zero, the tamgential velocity approaches zero but $\omega$ remains constant. As a result, the dimensionless point retains its angular velocity $\omega_0$ but has no tangential velocity.
Take the equation $v = \omega r$. If $r = 0$, then $v = 0$. Rearranging we get $$\omega = \frac{0}{0},$$
which is indeterminate. That does not mean that $\omega$ is zero at the centre when r =0 and neither does it mean that it is something impossible like infinite. Indeterminate just means the maths cannot give us a definite answer without additional information. All we can do is is make the most reasonable assumption that $\omega$ at the centre is the same as it is everywhere else on the disk.
However, let's step back a minute and consider that we are now talking about the angular velocity of a point with zero radius. Is that reasonable? It cannot possibly be an extended particle like an atom. We are essentially asking about the angular velocity of nothing, which is a pointless question. On the other hand, we talk about the properties of an electron which is commonly said to be a dimensionless point particle, but when we look deeper, there are good arguments that an electron is not really a point particle. See this article.
Now consider this. We agree any point on the disk that is not exactly at the centre of the disk has the same non-zero angular velocity $\omega$. So where exactly, is the exact centre of the disk? Well the Heisenberg uncertainty principle tells us we can not be exactly certain of its location. If we use our best estimate or calculation of where the exact centre of the disk is and assume there is some physical but dimensionless point particle at that location that has properties like mass, then the uncertainty principle tells us the point particle is not exactly at that location, but is in a superposition of locations about that point. Statistics tells us the probability of the 'particle' being exactly at the centre is practically zero. This hypothetical, but probably none existent point particle, cannot be exactly at the centre of the disk and so it is at a non zero distance from the centre and shares the same angular velocity as the rest of the disk.
Angular velocity is defined as the angle covered per unit time, or to be more precise the angular displacement per unit time, in a rotating disk all the points on the disc complete one rotation in the same time. This implies that they cover 360 degree or 2pi radians in the same time, hence according to the core definition of angular velocity, their angular velocities must be equal
Angular rotation is an emergent property of the collection of particles that move together and not a property of any individual particle.
Angular rotation is shared for the entire body describing the rotational motion of all the particles that are fixed to it.
This is used to describe the velocity of each particle as
$$ \vec{v}_i = \vec{\omega} \times \vec{r}_i \tag{1}$$
where $\vec{r}_i$ is the position of each particle with respect to the center of rotation.
See, you have to combine the rotational velocity $\vec{\omega}$ with the concept of axis of rotation, and hence the position $\vec{r}_i$ relative to this axis, to get the motion of each particle.
Full disclosure, the full motion of each particle can always be decomposed as a rotation $\vec{\omega}$ in combination with a parallel velocity $\vec{v}_{\parallel}$ such that $$\vec{v}_i = \vec{\omega} \times \vec{r}_i + \vec{v}_{\parallel} \tag{2} $$
