2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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)

$\endgroup$
  • $\begingroup$ 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 ? $\endgroup$ – Pierre Trott Sep 23 '17 at 15:27
  • $\begingroup$ @Pierre Trott: summit in your words. A summit, H center of square, then center of mass $O$ is a point on $AH$ so that $OH= \frac{1}{4}AH$ $\endgroup$ – orangeskid Sep 23 '17 at 15:37
  • $\begingroup$ @Pierre Trott: This works for any pyramid, as long as $H$ is the center of mass of the base. $\endgroup$ – orangeskid Sep 23 '17 at 15:43
  • $\begingroup$ Why don't we get the same result as the answer above? Or is it ? $\endgroup$ – Pierre Trott Sep 23 '17 at 15:46
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $\left(0,0,\frac{a}{4\sqrt2}\right) = \left(0,0,\frac{h}4\right)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.