The discussion revolves around finding the center of mass of a homogeneous solid cube with side length L, focusing on the analytical approach in three dimensions.
The discussion is ongoing, with participants exploring different approaches and confirming the consistency of the calculations across dimensions. Some guidance has been provided on how to extend the two-dimensional method to three dimensions, but no consensus has been reached on the final presentation of the solution.
Participants are working within the constraints of a homework assignment, focusing on analytical methods without providing complete solutions. There is an emphasis on understanding the reasoning behind the calculations rather than simply arriving at an answer.
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 said: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.
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 said: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.
Yes, you'd end up with that. Nothing would change.Davidllerenav said: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?
Ok. Then the aswer would be just to write thos three calculations and that's it?andrewkirk said:Yes, you'd end up with that. Nothing would change.