If the coordinates of center of mass are:
$$ x_{\text{CM}} = \frac{m_1x_1 + m_2 x_2 + \cdots}{m_1 + m_2 + \cdots} $$
I am confused with mass multiplying with distances that gives another distance. Why does that happen? I mean, how does this work? Please explain.
I just want to know how they derive this equation.It just seems too arbitrary to me.Whether they did any experiment and found this or it was just a miracle that they found an equation for the position of center of mass randomly.
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The easiest way to see this is to start with all masses the same; choose $m_i=1$ for all $i$. Then the centre of mass is $$x_{CM} = \frac{x_1+x_2+\cdots+x_n}{1+1+\cdots+1} = \frac1n\sum_1^n x_i,$$ which is just the average of the $x_i$.
Now suppose the mass at $x_1$ is $3$ instead of $1$. Chop that object into three smaller ones, each having mass $1$; they're all still located at $x_1$. In that case, the centre of mass will be $$x_{CM} = \frac{(x_1+x_1+x_1)+x_2+\cdots+x_n}{(1+1+1)+1+\cdots+1}= \frac{3x_1+x_2+\cdots+x_n}{3+1+\cdots+1}.$$
In other words, you can think of the mass as a "repetition" of an object. Then it's just a matter of generalizing this concept to masses that aren't whole numbers.
On
According to Wikipedia center of mass of a distribution of mass in space is defined as the unique point where the weighted relative position of the distributed mass sums to zero. You can read more about mechanical properties of the center of the mass in the article. Here we derive the required formula for the center of the mass of one-dimensional system consisting of finitely many masses $m_1,\dots, m_n$ placed at points $x_1,\dots, x_n$, respectively. Given a point $x$, the sum $S(x)$ of the weighted relative positions $x-x_i$ of the distributed mass equals $\sum_{i=1}^n m_i(x-x_i)$. For the center $x_{CM}$ of the mass of the system $S(x)=0$, which easily yields the required formula.
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The point that's being missed here is that "center of mass" is not a real thing - it is a mathematical construction that makes computations easier. The center of mass formula isn't derived, it's defined. So it is more prudent to ask the question: why do we use center of mass, instead of what is center off mass. So what do we find so useful about the center of mass? For one, in an isolated system, its equation of motion is always simple. Let $\mathbf{r}_1,...,\mathbf{r}_n$ be the positions of a system of particles in space with masses $m_1,...,m_n$. Then, $$\frac{\mathrm{d}}{\mathrm{d}t}\mathbf{r}_{\text{C.O.M}}=\frac{\sum_{i=1}^{n}m_i \dot{\mathbf{r}}_i}{\sum_{i=1}^{n}m_i}$$ In the denominator of the R.H.S is the total mass, which is constant. In the numerator is simply the total momentum of the system, which, since the system is isolated, is conserved. Thus, $$\frac{\mathrm{d}}{\mathrm{d}t}\mathbf{r}_{\text{C.O.M}}=\text{const.}$$ Thus the equation of motion for the center of mass is always constant or linear. It can also be shown that for a continuous rigid body in a constant force field, the resultant torque around the center of mass is $0$. An admittedly tedious calculation, but I can outline it if you wish. It can also be shown with a similarly tedious calculation that the gravitational force felt from all of the mass elements of a continuous object in space is the same as if we were to replace the continuous object with a point-particle of identical mass at the center of mass of the continuous object.
The point is, sometimes the mathematical equations we see are not a realization of a real thing written in mathematics, rather an abstract but useful construction that allows us to make accurate predictions of real observed phenomena.
You can see how it works in $1D$ most easily. Imagine you have a weight of $5$ at $2$ and a weight of $10$ at $3$. You would think the CM should be $2/3$ of the way over because the second weight is twice as large. In fact $$x_{\text{CM}} = \frac{2\cdot 5 + 3 \cdot 10}{5+10}=\frac {40}{15}=\frac 83$$ It is just the weighted average of all the masses.