1
$\begingroup$

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) $

$\endgroup$
  • $\begingroup$ looks like the center of mass. $\endgroup$ – philip_0008 Aug 28 '16 at 17:49
  • $\begingroup$ "like" center of mass. COM is after vector addition. This is scalar addition. $\endgroup$ – Kartik Sharma Aug 28 '16 at 17:55
  • 1
    $\begingroup$ Mind sharing your equations because I think there is a mistake somewhere. $\endgroup$ – John Alexiou Aug 29 '16 at 3:56
  • $\begingroup$ @ja72 I've added. You can check. $\endgroup$ – Kartik Sharma Aug 29 '16 at 16:58
1
$\begingroup$


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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So, basically it isn't an important engineering/mechanical quantity? $\endgroup$ – Kartik Sharma Aug 30 '16 at 16:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.