I've been studying centres of masses for different solids and laminas in class recently, and it seems from the exercises that I have done that the centre of mass of a right regular-polygon-based pyramid of uniform density is always situated at $\frac{h}{4}$ vertically away from the base, eg for when the the base is a circle and the pyramid is a cone and for a square based pyramid, as well as a regular tetrahedron. In all the cases I am talking about, the top vertice of the pyramid is along the axis of symmetry, ie directly above the centre of the base.
I have been wondering if there is any way to prove this for the general case of any regular-polygon-based pyramid?
Thanks for your help.
Yes, the centroid of an $n$-dimensional pyramid or cone is always $1/(n+1)$ of the distance along the line from the centroid of its $(n-1)$-dimensional base to its apex.
If the base is $B\subset\mathbb R^{n-1}$, the pyramid $P\subset\mathbb R^n$ with height $h$ can be parametrized as
$$P=\{(1-t)(x_1,x_2,\cdots,x_{n-1},0)+t(0,0,\cdots,0,h)\mid(x_1,x_2,\cdots,x_{n-1})\in B,\;t\in[0,1]\}.$$
That is, any $\vec y\in P$ can be written as a linear interpolation of some point $\vec x$ in the base and the apex point $h\vec e_n=(0,0,\cdots,0,h)$.
The centroid of $B$ is the average position of all points in $B$, calculated as the integral
$$\vec c_B=\frac{\int_{\vec x\in B}\vec x\,dx_1dx_2\cdots dx_{n-1}}{\int_{\vec x\in B}\,dx_1dx_2\cdots dx_{n-1}},$$
and similarly the centroid of $P$ which we want to find is
$$\vec c_P=\frac{\int_{\vec y\in P}\vec y\,dy_1dy_2\cdots dy_{n-1}dy_n}{\int_{\vec y\in P}\,dy_1dy_2\cdots dy_{n-1}dy_n}.$$
We need to relate the volume element of $P$ to the volume element of $B$, using the Jacobian of the coordinate system $(x_1,x_2,\cdots,x_{n-1},t)\mapsto(y_1,y_2,\cdots,y_{n-1},y_n)$:
$$\vec y=(1-t)\vec x+th\vec e_n$$
$$\frac{\partial\vec y}{\partial x_1}=(1-t)\vec e_1,\quad\frac{\partial\vec y}{\partial x_2}=(1-t)\vec e_2,\quad\cdots,\quad\frac{\partial\vec y}{\partial x_{n-1}}=(1-t)\vec e_ {n-1},\quad\frac{\partial\vec y}{\partial t}=h\vec e_n-\vec x$$
$$J=\begin{bmatrix}(1-t)&0&0&\cdots&0&-x_1\\0&(1-t)&0&\cdots&0&-x_2\\0&0&(1-t)&\cdots&0&-x_3\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&\cdots&(1-t)&-x_{n-1}\\0&0&0&\cdots&0&h\end{bmatrix}$$
$$\det J=(1-t)^{n-1}h$$
$$dy_1dy_2\cdots dy_{n-1}dy_n=(1-t)^{n-1}h\,dx_1dx_2\cdots dx_{n-1}dt.$$
Integrating this immediately gives the pyramid's volume $|P|=\frac1n|B|h$. Proceeding to calculate the centroid,
$$\vec c_P=\frac{1}{|P|}\int_{t\in[0,1]}\int_{\vec x\in B}\big((1-t)\vec x+th\vec e_n\big)(1-t)^{n-1}h\,dx_1dx_2\cdots dx_{n-1}dt$$
$$=\frac{1}{|P|}\int_{t\in[0,1]}\bigg((1-t)^n\int_{\vec x\in B}\vec x\,dx_1dx_2\cdots dx_{n-1}+t(1-t)^{n-1}h\vec e_n\int_{\vec x\in B}\,dx_1dx_2\cdots dx_{n-1}\bigg)h\,dt$$
$$=\frac{1}{|P|}\int_{t\in[0,1]}\bigg((1-t)^n|B|\vec c_B+t(1-t)^{n-1}h\vec e_n|B|\bigg)h\,dt$$
$$=\frac{|B|h}{|P|}\bigg(\frac{1}{n+1}\vec c_B+\frac{1}{n(n+1)}h\vec e_n\bigg)$$
$$=\frac{n}{n+1}\vec c_B+\frac{1}{n+1}h\vec e_n.$$
So we can write $\vec c_P=(1-t)\vec c_B+th\vec e_n$ where $t=1/(n+1)$.