# Why is velocity not constant for a rotating body but angular momentum is?

I'm having trouble expressing my confusion but I hope you'll bear with me.

Considering a wheel rotating in a uniform circular motion.

Splitting the wheel into a series of concentric rings of $dr$ in thickness, such that if we took the ring and bent them into a line it would be a very thin rectangle.

• The angular velocity of the constituent particles in a rotating object are uniform.

• The velocity of each of the constituent particles in each of the rings would be different. This agrees with the first statement $v = r \dot \theta$ with $\dot \theta =$ constant. With variable $r$, ring velocity must then be different per ring.

• However, if there is no net angular acceleration, then angular momentum is conserved. That means $r^2 \dot \theta = const$ yet $r \dot \theta$ is not. So in $L = (r \dot \theta)m r$, I have trouble intuitively understanding how this makes sense mathematically or intuitively.

Is my confusion clear? There seems to be an asymmetry here in my mind. We have a constant quality $\dot \theta$ and a quantity $r$ which, when multiplied by $\dot \theta$ gives a variable quantity (which works fine intuitively for me, as particles on the edge of the object have to sweep out equal angles in the same time despite being farther away necessitates a higher velocity for my first bullet point to hold true as sweeping out some angle farther away requires more distance than closer to the center), which makes me think having something vary proportionately to $r$ makes it not constant. Now, a variable quantity $r \dot \theta$ multiplied by $r$ is a constant. How can I address this with my previous intuitions?

Perhaps my intuitions themselves are mere coincidental observations, which means they can't be used to understand other things unrelated, but I'm wondering if there's anything I'm getting at that is touching on a key insight. I can happily accept velocity is not a constant quality, but then cannot say that angular momentum being conserved in the same vein is an obvious next step considering this uniformly rotating system. Why does $v \ne constant$ become constant when multiplied by $mr$?

The key point is that the meaning of conservation of some quantity is that it is a constant in time. For example, the conservation of angular momentum is formulated as

$$\dot{L}=0$$

In your case, it means that the angular momentum of each ring is a constant in time, but this constant may differ between different rings. As you said, the angular momentum of each ring of mass $dm=\sigma2\pi rdr$ is

$$dL=dm\omega r^2=2\pi\sigma\omega r^3dr$$

and thus the total angular momentum of the object is

$$L=\int dL=2\pi\sigma\omega\int_{0}^{R}r^3dr=\frac{1}{2}\pi\sigma\omega R^4$$

Observe that $m=\sigma\pi R^2$ is the mass of the disc, so

$$L=\frac{1}{2}mR^2\omega\equiv I\omega$$

Here $I=\frac{1}{2}mR^2$ is the moment of inertia of a disc of mass $m$ and radius $R$.

• This constant may differ between different rings? Forgive me if this sounds blatantly oblivious, but how is angular momentum conserved for the object if different sections of it have different angular momenta? – sangstar Nov 14 '17 at 10:13
• It is conserved for each ring separately, and as a consequence also for the object as a whole. If $I$ is the moment of inertia of the object and $\omega$ is its angular velocity, then $L=I\omega$ is the angular momentum of the whole object and it is constant in time. – eranreches Nov 14 '17 at 10:17
• So angular momentum on the disk is uniform then? Or is it a function of $R$? That would leave me to believe the constant may differ between different rings, at least. – sangstar Nov 14 '17 at 10:46
• Given a rotating disc with some parameters, its angular momentum is just a constant. $R$ is the radius of the disc - it is not a variable parameter. On the other hand, $r$ is a coordinate variable that you relate to the rings - different rings have different $r$. I'll try to clarify what I've written in the answer: You can divide the disc into rings, each has constant value of angular momentum $dL$ (which depends of $r$, the radius of the specific ring). When you add up these values using integration, you get the angular momentum of the whole disc. – eranreches Nov 14 '17 at 10:52
• I see. So it's almost as if I'm considering a rotating particle in each ring - each of them will be moving at different velocities and at different coordinate variable $r$, so thus different but constant angular momenta. Adding them up in small contributions of $dL$ will be the angular momentum of the entire disk. – sangstar Nov 14 '17 at 10:56