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

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.

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

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

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

• looks like the center of mass. – philip_0008 Aug 28 '16 at 17:49
• "like" center of mass. COM is after vector addition. This is scalar addition. – Kartik Sharma Aug 28 '16 at 17:55
• Mind sharing your equations because I think there is a mistake somewhere. – ja72 Aug 29 '16 at 3:56
• @ja72 I've added. You can check. – Kartik Sharma Aug 29 '16 at 16:58

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.