Center of Mass of a Variable System

First we must think of the situation in phases.

1. The can is filled with soda and it's center of mass is at the center. This is true because of the shape of the can, the soda is evenly distributed over the volume which means there's an equal amount of soda above the point of gravity as there is below it.

2. When part of the Soda is consumed it means that situation 1 above has changed and there isn't equal mass of soda, thus the center of gravity has to shift to the region of higher mass, which is down.

3. When the soda is consumed, all that remains is just the mass of the can now, which again due to the shape is evenly distributed and thus the Centre of Mass has to return to the original position, the Centre.

Note that the Center of Mass of full can must be same as that of empty can as in case of full can the mass is distributed evenly across the can. As the can is getting emptied the concentration of mass at bottom increases which causes the Center of Mass to go down and when the mass of soda left gets less than the mass of the can itself, the Center of Mass starts moving upwards until it reaches it original position when can is emptied.

When it's completely full, its center of mass is at ~h/2, because vertically it's sort of/approximately "homogeneous". As you start drinking from it, the center of mass goes down with the soda level. But at some point it reaches a minimum and starts going back up, because the soda weight gets more and more negligible compared to the can. Until finally when you empty the can, the center of mass returns to that approximately ~h/2.

As the formula that tells us how far the co-center of mass is from one of the centers of mass goes:

$d_{m_1}=\frac {m_2 \cdot l}{m_1+m_2}$ ------①

while l is the distance between the two centers of mass.

Let the mass of the can be M, the mass of soda be m, the height of the can be H,

then height of the center of the can will be

$h_M=\frac{H}{2}$ ------②

The mass of soda m can be represented as

$m=ρSh=a \cdot h$ ------③

while h is the height of the soda-level.

The height of the center of mass of soda can express as

$h_m=\frac{h}{2}$ ------④

apply②③④to①, we get

$d_{co}=\frac {m \cdot {\frac {(H-h)}{2}}}{M+m}={\frac {a}{2}} \cdot {\frac { h \cdot {(H-h)}}{M+a \cdot h}}$ ------⑤

which is the distance of the co-mass-center below the center of the can. h is defined in [0,H].

Compare the rate of change of the equation, we can see that $d_{co}$ will go up first, but then it comes down, and when h=0, $d_{co}$ =0. This means that the height of the co-center will first go down, then goes up, until eventually it gets back to the original point.

Centre of mass : a point representing the mean position of the matter in a body or system.

When we haven't started drinking, it is given that centre of mass (CoM) is at middle.

As we start consuming the soda, the matter of the system gets concentrated at the bottom as more amount of mass is there,

Therefore CoM also moves down to ensure that equal amount of matter is there on all it sides.

As we finish the soda , we are only left with the can , and obviously CoM of an empty can is in the middle of it ( if the can's mass isn't concentrated to one of its sides ).

The can surrounds a cylinder of soda. So long as there is plenty of soda in the can, the mass of the system is dominated by the mass of this cylinder, and the centre of mass is closely approximated by the centre of mass of the soda-cylinder. So (to begin with) as soda is drunk the height of this cylinder is reduced and so the height of the centre of mass (which is half way along the cylinder) is also reduced.

However as more and more soda is drunk, the mass of the can begins to dominate and we need to think about what happens now. Well, we know that when the can is empty, the height of the centre of mass of the can will be half the height of the can, close to the original centre of mass of the full can. Recognising that the height of the centre of mass changes continuously as we change the height of the soda column, we can conclude that it must first of all decrease, then level off, and then start climbing up to finish near its original position.

https://brilliant.org/problems/avoiding-an-accidental-bottoms-up/?ref_id=1488337

I have a problem where you have to actually find how much water to add to a glass to minimize the center of gravity.