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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? |
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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. |
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