# Center of mass of a pyramid

For an electromagnetic exercise, I need to find the center of mass of a pyramid. The pyramid is made of a square base of lenth a and four equilateral triangles with sides that mesure a long to. This is what I came up with for the moment but something seems to be wrong, when I try with some values, the results aren't right.

The summit of the pyramid is A, the four corners of the base are BCDE, the center of mass is o and the center of the base H. I used the fact that the center of mass is the center of the sphere going through all summits of the pyramid. p is the distance between H and O. $$CH=\frac{a*\sqrt{2}}{2}, AC^2=AH^2+HC^2 \Rightarrow AH^2=a^2-\frac{a^2}{2}=\frac{a^2}{2}$$ O is the center of mass so OA=OC $$OC^2=p^2*\frac{a.\sqrt{2}}{2}, OA=\frac{a^2}{2}-p$$ $$OC=OA \Rightarrow OC^2=OA^2$$ $$p^2*\frac{a.\sqrt{2}}{2}=\frac{a^4}{4}-a^2p+p^2$$ $$a^2p=\frac{a^4}{4}-\frac{a.\sqrt{2}}{2}$$ $$p=\frac{a^4-a*2\sqrt{2}}{4*a^2}$$ Is there a mistake because when I replace a with a value, the result doesn't seem right. Thanks in advance. Image of the pyramid with all the points

The center of mass of a pyramid is obtained as follows:

1.Join the vertex with the center of mass of the base.

2.Find the point on the above segment at height $\frac{1}{4} \times$ the height of the pyramid. ( so it divides this segment in ratio $1 \colon 3$.

(So it's not at the center of the sphere in general)

• What is the vertex? I'm a french student and I'm not used to do maths in english. Does that mean that O is placed at 1/4 of HA ? – Pierre Trott Sep 23 '17 at 15:27
• @Pierre Trott: summit in your words. A summit, H center of square, then center of mass $O$ is a point on $AH$ so that $OH= \frac{1}{4}AH$ – orangeskid Sep 23 '17 at 15:37
• @Pierre Trott: This works for any pyramid, as long as $H$ is the center of mass of the base. – orangeskid Sep 23 '17 at 15:43
• Why don't we get the same result as the answer above? Or is it ? – Pierre Trott Sep 23 '17 at 15:46

The center of mass of a regular tetrahedron is the only center of a regular tetrahedron, i.e. the point splitting any median (line joining a vertex with the centroid of the opposite face) into two segments with lengths proportional to $\frac{1}{4},\frac{3}{4}$. By Cavalieri's principle it follows that the center of mass of half a octahedron lies at one fourth of the height relative to the square base. No integrals are really needed.

First let's calculate the height $$h$$ of the pyramid. The height of the equilateral triangle of side $$a$$ is $$\frac{a\sqrt{3}}{2}$$ so by the Pythagorean theorem we get:

$$h = \sqrt{\left(\frac{a\sqrt{3}}{2}\right)^2 - \left(\frac{a}{2}\right)^2} = \frac{a}{\sqrt2}$$

Let's place the pyramid in the Cartesian coordinate system such that the center of the square base is at $$(0,0,0)$$ and the vertex is at $$\left(0,0,\frac{a}{\sqrt2}\right)$$. Since the pyramid is symmetrical around the $$z$$-axis, it is clear that the center of mass will be on the $$z$$-axis. We just have to calculate its $$z$$ coordinate. Let $$\overline{z}$$ be the coordinate, $$M$$ the mass of the pyramid and $$\rho$$ its density.

We have $$\overline{z} = \frac{1}{M}\iiint_V z\cdot\rho(x,y,z)\,dV = \frac{\rho}{M}\iiint_V z\,dV = \frac1V \iiint_V z\,dv$$

since the density $$\rho$$ is constant.

To integrate over the volume of the pyramid, notice that for $$z \in \left[0,\frac{a}{\sqrt2}\right]$$, we have to integrate over the square $$\left[-\left(\frac{a}2 - \frac{z}{\sqrt2}\right), \left(\frac{a}2 - \frac{z}{\sqrt2}\right)\right]^2$$. You can see this from the similarity of the triangles $$\Delta(0,0,0)\left(\frac{a}{2},0,0\right)\left(0,0,\frac{a}{\sqrt2}\right)$$ and $$\Delta(0,0,z)\left(x,0,z\right)\left(0,0,\frac{a}{\sqrt2}\right)$$ and solving for $$x$$ from here.

\begin{align} \iiint_V z\,dV &= \int_0^{\frac{a}{\sqrt2}} \int_{-\left(\frac{a}2 - \frac{z}{\sqrt2}\right)}^{\left(\frac{a}2 - \frac{z}{\sqrt2}\right)} \int_{-\left(\frac{a}2 - \frac{z}{\sqrt2}\right)}^{\left(\frac{a}2 - \frac{z}{\sqrt2}\right)} z \,dxdydz\\ &= \int_0^{\frac{a}{\sqrt2}} \int_{-\left(\frac{a}2 - \frac{z}{\sqrt2}\right)}^{\left(\frac{a}2 - \frac{z}{\sqrt2}\right)} z(a-z\sqrt2) \,dydz\\ &= \int_0^{\frac{a}{\sqrt2}} z(a-z\sqrt2)^2 \,dz\\ &= \frac{a^4}{24} \end{align}

The volume is:

$$V = \frac{1}{3}Bh = \frac13 a^2 \cdot \frac{a}{\sqrt2} = \frac{a^3}{3\sqrt2}$$

Therefore,

$$\overline{z} = \frac1{V}\iiint_V z\,dV = \frac{a}{4\sqrt2}$$

The center of mass is at $$\left(0,0,\frac{a}{4\sqrt2}\right) = \left(0,0,\frac{h}4\right)$$.