# System of Particles and Moment of Mass

I recently came across the definition of the Center of Mass of a system as the point about which the first moment of mass is zero.

Further, it defined Moment of Inertia as the second moment of mass.

My question is, What is this 'moment of mass'?

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Given some distribution or density $\rho(x),$ a moment is the 'expectation value' of some power of $x \in \mathbb{R}$. To be precise, the $n$-th moment $M_n$ is given by $$M_n = \int_{\mathbb{R}} x^n \rho(x) \mathrm{d}x.$$ In the mechanics case, $\rho(x)$ is simply the mass density.

You can extend this to vectors in $\mathbb{R}^d$ in a straightforward way; for example, for the moment of inertia you replace $x^2$ by $\mathbf{x}^2 = x_1^2 + \ldots x_d^2$ to obtain $$I = M_2 = \int_{\mathbb{R}^d} \mathbf{x}^2 \rho(\mathbf{x}) \mathrm{d}^dx$$ which should match the definition given in your mechanics textbook.

For the first moment of mass, you need to distinguish different directions. As you indicate, you can choose your coordinates such that

$$\int_{\mathbb{R}^d} x_i \,\rho(\mathbf{x}) \mathrm{d}^d x = 0$$ where $i$ runs over the coordinates. In three dimensions, you have $x_1 = x, x_2=y$ and $x_3=z.$

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Sorry for the extra answer. Your's went up as I was finishing up mine - which I'm trying to delete, since you said essentially the same thing. I would just like to add the link en.wikipedia.org/wiki/Moment_%28mathematics%29 –  Michael Brown Jan 1 at 14:31
Can you please elaborate on what you mean by 'expectation value'? –  AchiralSarkar Jan 1 at 14:36
Expectation value = average –  Michael Brown Jan 1 at 14:50
@AchiralSarkar: have you seen any probability theory? In that domain, people think of functions $\rho(x)$ as probability 'densities' (pdf's, wiki), for example Gaussian curves. Then the expectation value of some observable $f(x)$ is given by $\int f(x) \rho(x) \mathrm{d}x,$ i.e. the value of your observable $f$ at $x$ multiplied by the probability of getting $x$. In particular, moments are just the expectation values for $f(x) = x^n$. I hope this is clear! –  Vibert Jan 1 at 14:53
A probability measure is a slightly more formal term than probability density, but it expresses the same thing. People (textbooks, wiki) indeed use the term 'expected value' for the weighted average of the random variable itself, i.e. $\int x \rho(x) \mathrm{d}x.$ The expectation value I use above is the weighted average of any function $f$ of that random variable, see for example this reference. –  Vibert Jan 1 at 16:26
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I don't know whether this is right or wrong, coz its like bringing back the high school...

When physicists define a "moment of something".., then it necessarily means Distance $\times$ the "something". Moment of mass simply implies Distance $\times$ Mass.

For a system of $n$ particles, in order to obtain the center of mass - we consider a reference point. The effective mass times the distance to center of mass (which is a moment) will be equal to the sum of moments of individual masses. If $x_{c}$ is the distance from center of mass to the reference point, then... $$\sum_{i=1}^nm_i\ x_{c}=\sum_{i=1}^nm_i x_i$$

Hence, At the center of mass - plugging both equations to the left side, $$\sum_{i=1}^nm_i(x_{c}-x_i)=0$$ Well, I think this is the first moment of mass (which has also equated to zero).

As far as I can see in your definition,
I guess that the second moment roughly says, it is the moment of (moment of mass) which means $$I=\sum_{i=1}^nm_ix_i\times x_i\implies I=MX^2$$ For my luck, it also satisfies with the units $\text{kg m}^2$.

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Err... I actually defined second moment to be moment of [moment] because I had some tough maths. –  Crazy Buddy Jan 1 at 14:52
Well, your explanation is pretty straight forward for the first moment of mass but while talking about the second moment of mass, you say that it is the moment of (moment of mass). Since it is dimensionally correct , I guess, it is right but is it the correct of interpreting 'moment of inertia' which is a tensor when talking about a rigid body rotating in 3 D space? We ca get the components by your definition but what about the $M(X^2+Y^2)$ component of the matrix –  AchiralSarkar Jan 1 at 15:03
@AchiralSarkar: Well, Yeah. It need not be the actual definition though. Just an assumption of mine. BTW, I forgot to mention that I was using 2D (teasing the whole concept thereby). And, I don't like to see things getting more complex-y :-) –  Crazy Buddy Jan 1 at 15:06