How can one find the center of a right circular cone with height $h$ and radius $r$?
I've found these formulas:
$$M_{xy} = \iiint\limits_V z \rho (x,y,z) \, dx \, dy \, dz$$
$$M_{yz} = \iiint\limits_V x \rho (x,y,z) \, dx \, dy \, dz$$
$$M_{zx} = \iiint\limits_V y \rho (x,y,z) \, dx \, dy \, dz$$
$$M = \iiint\limits_V \rho (x,y,z) \, dx \, dy \, dz$$
We can assume that $\rho (x,y,z)=1$
A right circular cone with vertex pointing up, with height $h$ and base radius $r$ with base located at $z=0$ can be parameterized by:
$$x = \frac{h-u}{h} r \cos \varphi$$
$$y = \frac{h-u}{h} r \sin \varphi$$
$$z=u$$
$$u \in [0,h], \ \ \varphi \in [0, 2 \pi)$$
My problem is that I've calculated these integrals:
$$M_{xy} = \int_0 ^{2 \pi} \int_0 ^h u\,du\,d \varphi = \pi h^2$$
$$M_{yz} = \int_0 ^{2 \pi} \int_0^h \frac{h-u}{h}r \cos \varphi\, du \,d \varphi = 0$$ and similarly $M_{zx} = 0$.
$$M = \int \int \, du \, d \varphi = \int_0^{2 \pi}h \, d \varphi = 2h \pi.$$
There must be a mistake somewhere, because the center of mass of a right circular cone is at $\frac{3}{4}$ of its height.
Could you help me?
Thanks!
Best Answer
Because of the circular symmetry, it is obvious that the center of mass is on the $z$ axis. The $z$ component of the center of mass is $$ \frac{\int_V z \rho \;\mathrm{d}v}{\int_V \rho \;\mathrm{d}v} = \frac{M_{xy}}{M} $$ But you mis-calculated both $M$ and $M_{xy}$.
By far the easiest way to do these integrals is to work in cylindrical coordinates $(s, \phi, z)$ where the usual notation is to call the radial coordinate $r$ or $\rho$ but since the statement of the problem uses those two letters I use $s$ instead of $r$. The integrals are easy but one must remember that the volume element is $s\;\mathrm{d}s\;\mathrm{d}\phi\;\mathrm{d}z$ rather than just $\mathrm{d}s\;\mathrm{d}\phi\;\mathrm{d}z$. Take $\rho = 1$ as before:
\begin{align} M &= \int_{z=0}^{h} \int_{\phi=0}^{2\pi} \int_{s=0}^{r \frac{h-z}{h} } s\;\mathrm{d}s\;\mathrm{d}\phi\;\mathrm{d}z =\int_{z=0}^{h} \int_{\phi=0}^{2\pi} \frac{r^2}{2} \left(\frac{h-z}{h}\right)^2\;\mathrm{d}\phi\;\mathrm{d}z \\[8pt] &= \int_{z=0}^{h} \pi r^2\left(\frac{h-z}{h}\right)^2\;\mathrm{d}z = \frac{\pi r^2}{h^2} \int_{0}^{h} (z-h)^2\;\mathrm{d}z = \frac{\pi r^2}{3h^2} \left[ (z-h)^3 \right]_{0}^{h} \\ M &= \frac{\pi r^2 h}{3} \end{align}
\begin{align} M_{xy} &= \int_{z=0}^{h} \int_{\phi=0}^{2\pi} \int_{s=0}^{r \frac{h-z}{h} } z s\;\mathrm{d}s\;\mathrm{d}\phi\;\mathrm{d}z =\int_{z=0}^{h} \int_{\phi=0}^{2\pi} z \frac{r^2}{2} \left(\frac{h-z}{h}\right)^2 \;\mathrm{d}\phi\;\mathrm{d}z \\[8pt] &= \int_{z=0}^{h} \pi z r^2\left(\frac{h-z}{h}\right)^2 \;\mathrm{d}z = \frac{\pi r^2}{h^2} \int_{0}^{h} z(z-h)^2 \;\mathrm{d}z = \frac{\pi r^2}{h^2} \int_{0}^{h} \left( z^3 - 2 z^2 + z \right) \;\mathrm{d}z \\[8pt] &= \frac{\pi r^2}{h^2} \left[ \frac{z^4}{4} -\frac{2 z^3}{3} + \frac{z}2 \right]_{0}^{h} \\[8pt] &= \frac{\pi r^2 h}{3} = \frac{\pi r^2}{h^2} h^4 \left( \frac{1}{4} - \frac{2}{3} +\frac{1}{2} \right) \\[8pt] M_{xy} &= \frac{\pi r^2 h^2}{12} \end{align}
$$ z_{CM} = \frac{M_{xy}}{M} = \frac{h}{4} $$