5
$\begingroup$

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?

$\endgroup$
6
  • 2
    $\begingroup$ Hello there Coolcrab and please keep an eye on our Homework policy before asking about homework questions, 'cause they're discouraged here..! Sorry :) $\endgroup$ – Waffle's Crazy Peanut Sep 29 '12 at 10:16
  • $\begingroup$ They are not homework, and I did do the sum just making a mistake somewhere. And asking for help with that.. $\endgroup$ – Coolcrab Sep 29 '12 at 10:55
  • 1
    $\begingroup$ What!? $a/2 := h/4$!? I don't think so... Increasing $a$ has no effect on $h$, this assertation is plane false. $\endgroup$ – MoonKnight Sep 29 '12 at 11:14
  • 1
    $\begingroup$ Ofc its not, and thats my problem. It's either my bounds or the r should be a z. I'm not sure. $\endgroup$ – Coolcrab Sep 29 '12 at 11:16
  • $\begingroup$ @Killercam: Hi guys, I understood it. But, I just assumed that the center of mass would be somewhere along the axis of the cone...! Isn't that right? $\endgroup$ – Waffle's Crazy Peanut Sep 29 '12 at 11:36
8
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

$\endgroup$
1
2
$\begingroup$

The volume element is $ (dr)*(rd \phi)*(dz) $. Hence, the extra r in your integrand should be eliminated.

$\endgroup$

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