I'm having some trouble with a simple classical mechanics problem, where I need to calculate the center of mass of a cone whose base radius is $a$ and height $h$..!
I know the required equation. But, I think that I may be making a mistake either with my integral bounds or that $r$ at the last..! $$z_{cm} = \frac{1}{M}\int_0^h \int_0^{2\pi} \int_0^{a(1-z/h)} r \cdot r \:dr d\phi dz$$
'Cause, once I work this out, I obtain $a \over 2$ instead of $h \over 4$...!
Could someone help me?
I am not sure about this formula. Lets start by taking the vertex of the solid cone to be $O(0, 0, 0)$ in cylindrical coordinates ($r$, $\theta$, $z$). Then take the height of the cone to be $h$ and the base of the cone to have radius $a$. In this case the we know that
$$r = \frac{a}{h} z.$$
The formula for the center of mass of this cone can be written as
$$Mz_{m} = \int^{h}_{0} z \mathrm{d}m,$$
where $M$ is the total mass of the (solid) cone and $z_{m}$ is the location of the center of mass. We can write $\mathrm{d}m$ as
$$\mathrm{d}m = \pi \rho \frac{a^{2}}{h^{2}}z^{2}\mathrm{d}z,$$
where we have considered $\mathrm{d}m$ to be the mass of a thin disk at height $z$ and of radius $r$, with thickenss $\mathrm{d}z$. Now we can write the full equation for the center of mass as
$$Mz_{m} = \pi\rho\int^{h}_{0}\frac{a^{2}}{h^{2}}z^{3}\mathrm{d}z,$$
this becomes
$$Mz_{m} = \rho Vz_{m} = \frac{1}{4}\pi\rho a^{2}h^{2}.$$
We know that the volume of a cone $V = \frac{1}{3}\pi a^{2} h$, so we find
$$z_{m} \rho \frac{1}{3}\pi a^{2} h = \frac{1}{4}\pi\rho a^{2}h^{2},$$
so
$$z_{m} = \frac{3}{4} h.$$
Which is the distance from the vertex of the cone.
I hope this helps.
I see the problem you have here, change the $r^2\, \mathrm{d}r\, \mathrm{d}\phi\, \mathrm{d}z$ there to $r \, \mathrm{d}r\, \mathrm{d}\phi\, \mathrm{d}z$, then you should get the correct answer
The volume element is $ (dr)*(rd \phi)*(dz) $. Hence, the extra r in your integrand should be eliminated.
