I want to know what does the quantity - $\displaystyle \frac{1}{\text{M}}\int |\vec{\text{r}_{\text{CP}}}| \ \text{dm} $$\displaystyle \frac{1}{\text{M}}\int |\vec{\text{r}_{\text{PC}}}| \ \text{dm} $ signifies(where $\vec{\text{r}_{\text{CP}}}$$\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 = \frac{\omega}{\mu g}\left(\frac{k^2}{\frac{1}{M}\int |\vec{\text{r}_{\text{CP}}}| \ \text{dm} }\right) $$\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) $