How to find the center of mass of a cube?

[Davidllerenav]

Homework Statement

Find the center of mass of a homogeneous solid cube with side $L$ analytically.

Homework Equations

None.

The Attempt at a Solution

I don't understand how to find the center of mass on three dimensions. I know that since it is homogeneous, if I center the cube on the origin, then the center of mass would be on $(0,0,0)$, but how do I find this analytically?

 

[andrewkirk]

How would you do it in two dimensions? If you think about the technique you use in that case, it should become apparent how you might extend that to three dimensions.

 

[Davidllerenav]

How would you do it in two dimensions? If you think about the technique you use in that case, it should become apparent how you might extend that to three dimensions.

In two dimensions I would no center it at the origin, I use $(0,0)$ as one of the corners of the square, then I would find the $x$ component by $X_{cm}=\frac{\int\sigma x dA}{\int\sigma dA}$ and the $y$ component by $Y_{cm}=\frac{\int\sigma y dA}{\int\sigma dA}$. So $X_{cm}=\frac{\int_{0}^L Lxdx }{\int_{0}^L Ldx}=\frac{L\int_{0}^L xdx}{L\int_{0}^L dx}=\frac{\frac{x^2}{2}}{L}=\frac{L}{2}$, the same would be for $y$, so $Y_{cm}=\frac{\int_{0}^L Lydy }{\int_{0}^L Ldy}=\frac{L\int_{0}^L ydy}{L\int_{0}^L dy}=\frac{\frac{y^2}{2}}{L}=\frac{L}{2}$, so the center of mass would be on the point $\left(\frac{L}{2},\frac{L}{2}\right)$. How should I do this with three dimensions?

 

[andrewkirk]

Just do the same calculation for the z dimension as you did for the x and y dimensions, using (0,0,0) as one of the corners of the cube, the cube lying entirely in the first octant of the space (all three coordinates positive) and with each edge of the cube parallel to one of the three axes.

 

[Davidllerenav]

Just do the same calculation for the z dimension as you did for the x and y dimensions, using (0,0,0) as one of the corners of the cube, the cube lying entirely in the first octant of the space (all three coordinates positive) and with each edge of the cube parallel to one of the three axes.

So I would end up with $L/2$ again? Would be something like $Z_{cm}=\frac{\int_{0}^L Lzdz }{\int_{0}^L Ldz}=\frac{L\int_{0}^L ydz}{L\int_{0}^L dz}=\frac{\frac{z^2}{2}}{L}=\frac{L}{2}$? Or would something change?

 

[andrewkirk]

So I would end up with $L/2$ again? Would be something like $Z_{cm}=\frac{\int_{0}^L Lzdz }{\int_{0}^L Ldz}=\frac{L\int_{0}^L ydz}{L\int_{0}^L dz}=\frac{\frac{z^2}{2}}{L}=\frac{L}{2}$? Or would something change?

Yes, you'd end up with that. Nothing would change.

 

[Davidllerenav]

Yes, you'd end up with that. Nothing would change.

Ok. Then the aswer would be just to write thos three calculations and that's it?