What is $\displaystyle \frac{1}{ \text{M}} \int |\vec{\text{r}_{\text{PC}}}| \ \text{dm} $?

Question

I want to know what does the quantity - $\displaystyle \frac{1}{\text{M}}\int |\vec{\text{r}_{\text{PC}}}| \ \text{dm} $ signifies(where $\vec{\text{r}_{\text{PC}}}$ is the vector joining center of mass to a point $\text{P}$ in a rigid body.

Motivation behind it is inertia about center of mass seems like the RMS distance, so what can be this "normal statistical mean" mean?

Actually, I was working on a problem to find out the time taken by various bodies to stop rotating given same $\omega$ on a rough table and I arrived at something like -

$\displaystyle T_s = \dfrac{\omega}{\mu g}\left(\dfrac{k^2}{\dfrac{1}{M}\displaystyle\int |\vec{\text{r}_{\text{PC}}}| \ \text{dm} }\right) $, where $k$ is radius of gyration.

Working:

If we put a body rotating with $\omega$ about COM on a rough table, friction will all over it. So, we'll have to take tangential friction torque over the body.

Let's just make $\text{dm}$ elements in the body at $\vec{\text{r}_{\text{PC}}}$. Then,

Normal Force : $\text{dN} = \text{dmg}$

$\text{df} = \mu\text{dmg}$

$\text{d}\tau = \mu\text{dmg} |\vec{\text{r}_{\text{PC}}}| $, since friction is acting tangential to the rotating axis.

$\tau_{net} = \int{\text{d}\tau} = \int{\mu \text{g}|\vec{\text{r}_{\text{PC}}}|\text{dm}}$

$0 = \omega - \dfrac{\tau_{net}}{\text{I}_{\text{CM}}}(T_s)$

$\displaystyle T_s = \dfrac{\omega}{\mu g}\left(\dfrac{k^2}{\dfrac{1}{M}\displaystyle \int |\vec{\text{r}_{\text{PC}}}| \ \text{dm} }\right) $

Answer 1

It's something radius-of-gyration-like, but perhaps lesser than radius of gyration. For the radius of gyration, the integral form is $$k = \sqrt{\frac{I_{CM}}{M}} = \sqrt{\frac{1}{M}\int |r_{PC}|^2dm}$$ which reads "root-mean-square distance of the object's parts from axis" (source)
So, $$\frac{1}{M}\int |{r_{PC}}|dm$$ IMO should read "mean distance" or "average distance" of the object's parts from axis.