Ball hits a rod So let's imagine we have a homogeneous rod of length L and mass M and a tiny ball of mass m, all lying on a smooth horizontal table. Rod is stationary and ball is projected toward the rod at the right angle to the rod. The impact is inelastic and objects stick together. I know that they will rotate around new CM of the system comprised of the ball and rod. However what I don't understand is more philosophical. How do they "know" that they need to rotate around the centre of mass? Is there any other explanation different than that they need to rotate around the point where the moment of inertia is smallest, so that energy required to rotate it is the smallest?
 A: I haven't done the maths on this, but the object will rotate on its centre of mass because that is the point at which all of the forces balance out.
The first thing to realise is that unless the non-hit end of the rod is fixed or resting on some surface with friction then I don't believe that that end will actually travel backwards (happy to be corrected on this). Yes, it will rotate, but that rotation is relative to the new forward motion.
As the ball hits the rod, the force of impact is lateral. This will displace some of the atoms stretching the bonds to the adjacent atoms and creating tension. This tension is in the direction of the bonds, so it will mostly be parallel to the length of the rod but with some small lateral component. The lateral tension will pull back on the atoms in motion restricting their acceleration and the lengthwise tension will add an acceleration towards the centre of the rod, so their motion will no longer be perpendicular. As the force travels down the rod, it will decrease (as energy has been expended in setting the atoms in motion), so the lateral component will also decrease and hence the impulse imparted to the rod will decrease and the atoms will move slower. At the far end of the rod, the lengthwise tension will be pulling that end to the centre due to the inertia of the rest of the rod. Obviously, we can't have two ends of the rod travelling at different speeds unless they rotate. The rotation creates an acceleration towards the centre which will be balanced out at the centre of mass. There are complexities that I haven't thought through such as the fact that force will take time to travel the length of the rod. This will probably result in a vibration in the rod (assuming it isn't perfectly rigid), but eventually it will settle down.
So in summary, the rod and ball communicate their centre-of-mass through the tension in the rod (if you want to think of it that way) and the centre-of-mass is where those tensions cancel out.
There are equations to go with all of this, but that was too long ago for me.
A: The bodies do not particularly "know" what the center of mass is. The motion in and around the center of mass is the result of linear and angular momentum being conserved for the known physical forces. So if you desire to use "anthropomorphic language", it is nature who knows that it is not allowed to let bodies interact in a way that requires creating linear or angular momentum out of nothing. More physically speaking, the nature of the forces dictates linear and angular momentum conservation, not vice-versa. The conservation laws are just a secondary consequence of forces being spatially homogeneous and isotropic. This is clarified in Noether's theorem, if you are interested in diving any deeper into the foundations.
Allow me another picture: suppose you invest in stocks, and your neighbour does too. Then the market laws are constructed in such a way that what you buy has to be sold by your neighbour. Otherwise no deal will take place. Likewise, the money that is exchanged in the deal sums to zero. Of course, we have to spare "non-conserving" actions like credit, central bank policy or financial derivatives, in order for this picture to work. But in any case, neither you nor your neighbour needs to "know" where the money or the stocks go, although for a market with only two participants it is impossible to ignore it. However, if there are more market participants, you might well lose track of where the money/stocks go, and yet you cannot help doing the right thing in the sense of "conservation of money" (unless you start printing counterfeit money).
A: 
I know that they will rotate around new CM.

The above statement is not correct. The center of rotation is not the center of mass. The ball imparts both translational and rotational momentum to the rod. The result is that the combined center of mass is going to translate and rotate at the same time.
Only under very specific circumstances with appropriate initial conditions where both the ball and the rod at in motion initially you can have a pure rotation after impact, or in a different case, no motion at all.
So considering the general case, your question becomes moot, as there is no special motion that results in general.
There is a specific relationship between the instant center of rotation and the point of impact.

*

*Consider the perpendicular distance $d$ between the line of motion of the ball (where its momentum lies) and the combined center of mass.

*Consider the distance $c$ between the new instant center of rotation.

*Consider the combined mass $m_{\rm tot}$ and combined moment of inertia $I_{\rm tot}$ after the impact about the new center of mass.

The instant center of rotation is going to be on the other side of the center of mass as the impact point at a distance
$$ c = \frac{I_{\rm tot}}{m_{\rm tot}\, d} $$
With this relationship you see that the closer the impact point is to the center of mass, the less $d$ is, the further away the center of rotation is. The limiting case is when $d=0$, the center of rotation is at infinity, which means the resulting motion is a pure translation.
The other limiting case where the resulting motion is a pure rotation about the center of mass, $c=0$, could only occur if the impacting point is at infinity which is non-sensical in this scenario. This cannot happen as stated in the op.
A: from the collision equation you obtain
$$v_{\text{cm}}={\frac {mu}{m+M}}\\
\omega_{\text{cm}}=-{\frac {M\,m\,u\,\rho}{ \left( m+M \right) I_{{R}}}}$$
from here :
$$m\,\dot v_{cm}=0\\
I_R\,\dot \omega_{cm}=0$$
where

*

*M rod mass

*m ball mass

*$I_R~$ rod inertia

*u ball start velocity

*$\rho~$ distance to the center of mass where the ball hit the rod

*$v_{\text{cm}}~$ CM velocity after the collision

*$\omega_{\text{cm}}~$ CM angular velocity   after the collision

for inelastic collision the velocity of the mass m $~v_m~$ is  equal to the velocity of the CM and the  conservation of the linear  momentum must fulfilled. (this is also valid for elastic collision)
$$M\,v_{\text{cm}} +m(\,v_{cm}-u)=0$$
