The center of mass of a homogeneous solid cube with side length L can be analytically determined to be at the coordinates (L/2, L/2, L/2) when the cube is centered at the origin. The calculations for the x, y, and z components utilize the formula for center of mass in three dimensions, which involves integrating the density over the volume. Specifically, for each dimension, the center of mass is calculated using the integrals Xcm = (∫σx dA)/(∫σ dA), Ycm = (∫σy dA)/(∫σ dA), and Zcm = (∫σz dA)/(∫σ dA), confirming that the center of mass remains at (L/2, L/2, L/2).
PREREQUISITESStudents studying physics, particularly those focusing on mechanics and calculus, as well as educators teaching concepts related to center of mass and integration techniques.
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.