What exactly is center of mass?

From Wikipedia,

In physics, the center of mass of a distribution of mass in space (sometimes referred to as the balance point) is the unique point where the weighted relative position of the distributed mass sums to zero. This is the point to which a force may be applied to cause a linear acceleration without an angular acceleration.

You can also think of the COM in mathematical terms, defined as follows $$ overrightarrow{R} = dfrac{1}{M}iiint_{U} rho (overrightarrow{r})overrightarrow{r}{rm d}V$$ where $overrightarrow{R}$ is the COM, $M$ is the mass, $rho$ is the density function, and $U$ is the region the solid occupies.

For the rest of my answer, I will focus on rigid bodies.

Also why is COM considered as that special point whre putting force on it produces translation motion instead of torque? Has it got to do with the distribution of force through the rest of body of an object?

When a rigid body rotates, it will rotate about its center of mass (it can also rotate about another axis, e.g. like the Earth, which orbits the Sun and rotates about its own axis, but for our purposes, we will just analyze rotation about COM). So if you apply a force at the center of mass, the torque about the center of mass is 0, thus, no angular accleration. As to why a free rigid body will always rotate about COM, see this post.

Is COM that point where if you apply force, that is equally distributed to rest of the atoms of the body of the object, no matter how far those atoms are from the COM?

Each particle will have the same acceleration (because pure translational motion), so yes, the force must be distributed to EVERY particle in the rigid body. Not sure what you mean by equally -- do note that more dense parts of the object (where mass is greater) will need a greater force to be accelerated. However, if the object has constant density, then yes, the force will be distributed equally.

Conversely, does applying force on any other point on the object beside COM mean that that force is not equally distributed to the rest of the body, which then produces torque instead of transnational motion?

When a body rotates, not all its particles will have the same velocities, hence the rotation. Therefore, the force will not be distributed evenly -- some particles will feel greater forces than others, such that each point experiences a different acceleration, which is why the body will rotate.

I will also add that it's easier to start a problem by thinking about the forces as acting on the body, rather than individual particles themselves. If you apply the force $overrightarrow{F}$ on a rigid body, it will accelerate with acceleration $overrightarrow{a}=overrightarrow{F}/m$. Then, depending on the vector $overrightarrow{r}$ (whose tail is at the COM and head is at the point where the force is applied), the rotation about the center of mass can be found using the torque, $overrightarrow{tau} = overrightarrow{r}timesoverrightarrow{F}$.

As for your question about cutting a ball through its COM: It can be cut in parts of equal masses, but that depends on the axis through which you cut. See this post for more.

If your object is a rigid body, then you can imagine the centre of mass to be a point "painted" on the object. The thing about the centre of mass is that it has a number of nice properties, because of how it is defined.

One of those nice properties that you have pointed out is that it behaves as if all the mass of the object was concentrated there. What does that mean? Consider an extend rigid body like a bat. If I was to toss the bat up into the air, each point on the bat is going to translate and rotate at the same time, and in general the motion of each point will be rather complicated. However, the centre of mass (which you can imagine to be a special point painted on the bat) will trace out a nice parabola. If we were to instead throw a point particle with the same mass as the bat with the same initial velocity, it will trace out the exact same path as the centre of mass of the bat. In other words, the motion of the bat can be broken down into a translation of the centre of mass, and a rotation about the centre of mass.

Now the centre of mass need not coincide with "a special atom" or anything. Just consider the centre of mass of a donut shaped thing, there's nothing there!

However, for concreteness, you can imagine that for the donut shaped object, the centre of the donut is filled by a imaginary substance that has no mass but is perfectly rigid. Then, if you were to try and accelerate the object by applying a force there, the object will translate without rotating.

Here's an intuitive explanation for your other query. Consider the setup in the picture (a rod), which can be approximated crudely as 3 balls joined by two springs. If I were to give the centre ball a flick, then it will start moving forward and drag the other two balls along, but because of the symmetry there is no rotation. On the other hand, if I weas to flick the ball on the right, then it will start dragging the balls on the left along with it, but it drags the one closer to it more than the one further from it, so the whole thing starts to rotate. Another thing to note is that to create pure translational motion, the force need not be applied at the com, instead the condition is that the line of action of the force coincides with the com. In that case the force produces no torque wr to the com and the result is pure translational motion.

I would start by giving an example that can be found in elementary books (Introduction to Mechanics Klepner & Kolenkow ).

Consider a Bola which consists of three balls of stone or iron connected by thongs. The gaucho whirls the bola in the air and hurls it at the animal. What can we say about its motion?

Since the constraining forces depend on the instantaneous positions of all three balls, it is a real problem even to write the equation of motion of one ball. However, the total momentum obeys the simple equation :

$$frac{dmathbf{P}}{dt}=mathbf{F}_{ext}=mathbf{f}_1^{ext}+mathbf{f}_2^{ext}+mathbf{f}_3^{ext}$$ $$=m_1mathbf{g}+m_2mathbf{g}+m_3mathbf{g}$$

or $$frac{dmathbf{P}}{dt}=Mmathbf{g}$$

This equation represents an important first step in finding the detailed motion. The equation is identical to that of a single particle of mass $M$ with momentum $mathbf{P}$. This is instinctively clear to the gaucho when he hurls the bola; although it is a complicated system, he need only aim it like a single mass.

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Now If we are concern with the exact position and rotational motion of each ball, treating the problem as a single mass may not be the best idea. But if we are just concern about the overall motion of the bola then the concept of single might be a good idea. This single example can not tell you the importance of the center of mass but It gives the overall idea.

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my question then becomes that what exactly is COM?

The exact definition of the center of mass can found here.

How come it is deemed as if all the mass of the object is concentrated in that single point?

It's not we are going to concentrate the whole object at a single point. I would rather say, The replacement of a complecated system like bola to a single mass which tells you the overall motion of the system. Now We never confuse ourselves about How we replace or how we can assume a single point with mass like of planet Because that doesn't matter here We are not concerned with the internal or atomic structure or stability of these things because this doesn't affect the motion.

why is COM considered as that special point where putting force on it produces translation motion instead of torque?

As I said at the moment we concern with the rotation, We need to take care of other things. Now It turn out that even when we concern with the rotational motion of the object, the Center of mass turns out to be imporatant concept due to following reasons.

Chasles’ theorem asserts that it is always possible to represent an arbi-trary displacement of a rigid body by a translation of its center of mass plus a rotation around its center of mass.

We of course not going to detail of it. But this theorem alone give idea of imporatance of center of mass.

All other question are more like a misconception that should be clear from above. Again center of mass is replacement of complicated mass distribution to simple point mass when we are not concern with rotations.

The answer is correct, the center of mass does not divide the object into to equal masses. The center of mass is rather the average position of the distribution of mass. These two are different (if each mass at position $x_i $ were equal, the center of mass is the average $x_i$ and the point of equal mass is the median of the $x_i$'s). Imagine two balls of different masses along a line, joined by a rigid massless wire. The center of mass will be somewhere along the line, and the masses on the two sides are still different, regardless of the COM position. Actually, any point not on the masses themselves will divide the object into different masses.

The position of the center of mass is:

$x_{COM}=sum x_im_i/sum m_i$

and from here it is straightforward to see that there will be no torque in a uniform force field. The total torque will be:

$tau=sum x_iF_i=sum x_im_ig=Mg x_{COM}$

which is zero if the center of mass is at the origin.

It is an imaginary point, if the object were hollow it would be outside the mass of the object, but if you calculate the torque about that point (the origin of coordinates does not need to be on the object itself), it still will result in a value of zero.

The individual forces are applied on the different parts of the body, not in the COM, it only looks as if it were applied in the COM for some purposes, like in the result above: the total torque is equivalent to all forces being applied to the COM.