# Proof 25: Centre-of-mass

An object’s centre-of-mass is a point $\vec r_\text{com}=(x_\text{com},y_\text{com})$ that “represents” the object, taking into account all the “particles” it consists of as well as how heavy all those “particles” are. The $\vec r_\text{com}$ is a mass-weighed average position of all “particles”.

Were there no masses to take into account (if we were looking for the centre-of-geometry), then the usual well-known average of all positions is simply ‘all positions added together and divided with the total number‘. Say, we have 4 “particles” in an object. What point is “in the middle” of them all?

$$\vec r_\text{centre-of-geometry}=\frac{\vec r_1+ \vec r_2 + \vec r_3 + \vec r_4}{4}= \frac 14\vec r_1+ \frac 14 \vec r_2 + \frac 14 \vec r_3 + \frac 14 \vec r_4$$

Each “particle” has one fourth of the influence; they are all equally influential. Now, if you of some reason wanted, say, “particle” 3 to have double the influence, you can take this into account by simply including it twice – as if we imagine two identical “particles” at position $\vec r_3$:

$$\vec r_\text{weighed centre}=\frac{\vec r_1+ \vec r_2 + \vec r_3+ \vec r_3 + \vec r_4}{5}= \frac{\vec r_1+ \vec r_2 + 2\vec r_3 + \vec r_4}{5}$$

With “particle” 3 acting as two “particles” – weighing as two “particles” – the total weighed number of “particles” is now 5. “Particle” 1, 2 and 4 have weight 1, while “particle” 3 has weight 2. In total 5. Clearly, the total weighed number of particles is the sum of all individual weights $\sum \text{weights}$:

$$\vec r_\text{weighed centre}= \frac{\vec r_1+ \vec r_2 + 2\vec r_3 + \vec r_4}{ \sum \text{weights} }= \frac{\vec r_1+ \vec r_2 + 2\vec r_3 + \vec r_4}{ 1+1+2+1} = \frac 15\vec r_1+ \frac 15 \vec r_2 + \frac 25 \vec r_3 + \frac 15 \vec r_4$$

“Particle” 3 here accounts for $\frac 25$ (two out of five) of the total weighed number of “particles”. Had the weight been not 2 but, say, 2.5, then the pattern still fits:

$$\vec r_\text{weighed centre}= \frac{\vec r_1+ \vec r_2 + 2.5\vec r_3 + \vec r_4}{ 1+1+2.5+1} = \frac{\vec r_1+ \vec r_2 + 2.5\vec r_3 + \vec r_4}{ 5.5} = \frac 1{5.5}\vec r_1+ \frac 1{5.5} \vec r_2 + \frac {2.5} {5.5} \vec r_3 + \frac1{5.5} \vec r_4$$

The pattern of the numerator is clearly that regardless of the weight, the positions and weights of each “particle” are multiplied, and then it is all added up:

$$\vec r_\text{weighed centre}= \frac{\text{weight}_1\cdot \vec r_1+ \text{weight}_2 \cdot \vec r_2 + \text{weight}_3 \cdot \vec r_3 + \text{weight}_4 \cdot \vec r_4}{ \sum \text{weights} } = \frac{\sum \text{weight} \cdot \vec r}{ \sum \text{weights} }$$

By added a weighing, we are skewing the result towards “particles” that weigh more. That weighing could be anything – energy density, temperature, mass and the like. With mass being the weighing factor, we get the centre-of-mass:

$$\vec r_\text{com}= \frac{\sum m\vec r}{ \sum m} \qquad\blacksquare$$

## Proof 26: Parallel-axis theorem

When rotating about the centre-of-mass, the moment-of-inertia $I_\text{com}$ is often easier to find. When the axis-of-rotation is somewhere else, we’ll below derive a formula that…

## Proof 24: Newton’s 2nd law in rotation

Get some friends to help you push the stuck carousel with your daughter sitting in it. Each applies a force on her. Friction and other…

## Proof 17: Power rules of algebra

Definitions: The pattern looks as if reducing the exponent means removing an $a$ term: \vdots\\a^4=a\cdot a \cdot a \cdot a\\a^3=a\cdot a \cdot a\\ a^2=a\cdot a…

## Proof 28: Plate’s moment-of-inerta

Plate rotating perpendicularly about its centre-of-mass With a single point having the moment-of-inertia $I_\text{point}=r^2m$ ($m$ and $r$ are mass and distance from the axis-of-rotation), infinitely…

## Proof 27: Rod’s moment-of-inertia

A rod rotating perpendicularly about its centre-of-mass With a single point having the moment-of-inertia $I_\text{point}=r^2m$ ($m$ and $r$ are mass and distance from the axis-of-rotation),…