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I came across this question while having conversation with one person. We know that Center of Gravity of a solid cube is at the intersection of connecting the opposite vertex of the cube. Suppose, you have a hollow cube, filled with water/air. Now, how do you find the center of gravity of that object?

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closed as too localized by Waffle's Crazy Peanut, Nic, Manishearth Apr 27 '13 at 8:45

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Hi Learner. Welcome to Physics.SE. How do you find... - Do you ask for some kinda formula-suggestion..? – Waffle's Crazy Peanut Apr 24 '13 at 8:25
intuition is enuough – Learner Apr 24 '13 at 8:27

What you call "centre of gravity" is more commonly called center of mass (COM). The general formula is not too hard:

$$\mathbf{r}_{COM} = \frac{\sum \mathbf{r}_i m_i } {\sum m_i}$$

where $\mathbf{r}_i$ are the position vectors of all individual masses $m_i$.

In words: multiply a mass by its offset w.r.t. some coordinate system. Do this for all the masses you might have in a system, and add everything up. Divide the end result by the total mass of the system, e.g., the sum of all the masses alone.

Now, if you do not have individual masses but a certain density distribution throughout an arbitrary volume $V$, just turn the summation into an integral:

$$\mathbf{r}_{COM} = \frac{\int\int\int{\rho(\mathbf{r})\mathbf{r}dV}}{\int\int\int{\rho(\mathbf{r})dV}}$$

Lastly, the COM of a system of different objects of which you already know the COM and total mass $M$ (like your cube, and the cubical body of water it contains), can be computed by substituting their masses and relative positions into the first equation.

If the hollow cube of mass 1 with sides 1 is completely filled with water of mass 3, the calculation is simple. The hollow cube has its $\mathbf{r}_{COM}$ at the intersection of the diagonals, as does the body of water. Taking a coordinate system in the centre of the cube:

$$ \frac{1 \cdot \mathbf{0} + 3 \cdot \mathbf{0} } {1 + 3} = \mathbf{0} $$

so, the COM is simply in the centre of the cube.

Filling the cube only with a third of the water changes things. The centre of mass for each 1/3 slice of water lies in the center of that slice, so, the centre of mass of only the bottom 1/3 slice of water lies at $z=-1/3$:

$$ \frac{1 \cdot \mathbf{0} + \frac33 \cdot [0\,, 0\,, -\frac13] } {1 + \frac33} = [0\,, 0\,, -\frac{1}{6}] $$

Where $[x\,, y\,, z]$ indicates a vector at coordinates $x$, $y$ and $z$, respectively.

So, the COM now lies at 1/6 down from the centre of mass of the cube alone.

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I didn't get that. Center of Mass is commonly Center of Gravity? – idiosincrasia23 Apr 24 '13 at 13:34
Center of mass and center of gravity is not the same thing. Admitted, they often are at the same point though. The center of gravity is the integral of the force of gravity, while the center of mass is the integral over the mass density. If you have a non-uniform acceleration of gravity they are different (such as for very tall buildings). – S. Gammelmark Apr 24 '13 at 13:49
@S.Gammelmark: Hmm, didn't know that, thanks. I can see why the distinction is useful. Perhaps you should adjust the wiki article on CoM (because centre of gravity just links to CoM) However, I think the OP simply meant the centre of mass :) – Rody Oldenhuis Apr 25 '13 at 5:31
There is actually a link at wikipedia: Centers of gravity in non-uniform fields. Turns out that I was not entirely accurate, since my description only applies to fields with a constant direction through space, not for general non-uniform fields. In any case, I think you are correct, that the OP was referring to CoM. – S. Gammelmark Apr 25 '13 at 7:47

Actually, a hollow cube (assuming it's hollowed out symmetrically) completely filled with water/air just has its center of gravity at the intersection of the space diagonals while a solid cube doesn't generally have that. It only has that if the mass is distributed homogeneously. And in a hollow cube completely filled with air/water, the mass is distributed homogeneously, so the center of gravity is where you expect it to be.

Now, if you only fill the hollow cube with water halfway, the situation is different. The center of gravity must then obviously move down. How far down depends on the mass of the left-over faces of the hollow cube and the mass of the water. The lower the mass of the faces, the closer the cg of the whole will be to the cg of the water itself (which is at the intersection of the space diagonals of the water cuboid). If the mass of the faces, however, is far greater than that of the water, the cg will only move down ever so slightly.

A hollow cube filled with air will always have its center of gravity at the intersection of the space diagonals since air is a gas which always spreads out homogeneously. You might now also see why - as Rody Oldenhuis points out in his answer - we more often refer to the center of gravity as the center of mass; it is determined completely by the mass distribution of the object.

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