Question

I'm trying to derive the equation for the moment of inertia of a hollow cone, and i'm having some problems. Help?

Answer 1

fire away

Answer 2

Well, I'm trying to set the integral up, and I can't figure out how to take into account the fact that the cone slopes to a point in the integral. Thus far, I have the the triple integral of r^3 d(theta) dr dx, with limits 0 to height, b to a (where b is the inner radius and a is the outer radius), and 0 to 2pi. This, however, will give me the moment of a cylinder, because there is no information about, well, about the cone being a cone.

Answer 3

\[\rho \int\limits_{0}^{h}\int\limits_{b}^{a}\int\limits_{0}^{2\pi}r^3d \theta dr dx\]

Answer 4

thanks.

r is a function of z (you call it x, i know, but in cylindrical i am so used to calling it z ...)

let me try drawing something

Answer 5

Yeah, choice of axis is arbitrary, no worries.

Answer 6

actually this is a hollow cone, right, so the "r dr dtheta dz" formulation -- for volume -- will not work.

Answer 7

It should, if you define the limits of integration for r as the inner surface to the outer surface. I did this already where r is a function of x for a barrel with a curved surface, and it worked out, but I need to do it for a cone without plugging in specific numbers, and I'm having problems. But I'm definitely open to alternate strategies.

Answer 8

I think I just figured it out. My limits of integration for dr should be (R-(R/h)x (or R-(R/h)z in your coordinate system) for the outer surface and (r-(r/h)x for the inner surface.

Answer 9

the obvious and really simple approach is to revolve a line, say y = (R/h) x, around the x axis. that is super simple but it is still and arc length ds integration. and have mass spread over the area using an area density.

i will have a look at your idea though.

Answer 10

i think if you do it as a volume, you need to do separate integrations for 2 cones, one the outer and one for the inner.

for any typical cone, R has to sweep out from zero to that cone's radius over the course of the integral.

that is doable, albeit messy.

you then have to repeat it passing in the mr^2 dm bit to get the inertia, again 2 integrals.

Answer 11

please post solution if you do it.

Answer 12

When I integrate that for r, I get \[[R(1-\frac{ x }{ h }]^4 - [r(1-\frac{ x }{ h })]^4\], all over 4.

Answer 13

It looks like I can just distribute that exponent to the R and the r, pull them out of the integral, and the equation reduces to the (1-x/h)^4 terms, which means it reduces to zero. But that would make the integral equal to zero, so that's no good, and I'm wondering where the flaw in my logic is.

Answer 14

where delta is thickness of cone, this is for volume :

\[\int\limits_{\theta = 0}^{2 \pi}\int\limits_{z = 0}^{h}\int\limits_{r = 0}^{\frac{zR}{h}} r dr d \theta dz - \int\limits_{\theta = 0}^{2 \pi}\int\limits_{z = \frac{\delta H}{R}}^{h}\int\limits_{r = 0}^{\frac{zR}{h} - \delta } r dr d \theta dz\]

for Inertia:

\[\int\limits_{\theta = 0}^{2 \pi}\int\limits_{z = 0}^{h}\int\limits_{r = 0}^{\frac{zR}{h}} \sigma r^3 dr d \theta dz - \int\limits_{\theta = 0}^{2 \pi}\int\limits_{z = \frac{\delta H}{R}}^{h}\int\limits_{r = 0}^{\frac{zR}{h} - \delta } \sigma r^3 dr d \theta dz\]

Answer 15

\[2 \pi \rho \int\limits_{0}^{h}\int\limits_{r(1-h/x)}^{R(1-h/x}\int\limits_{0}^{2\pi}r^3d \theta dr dx\]

Answer 16

I don't understand zR/h.

Answer 17

r = 0 at origin and increases linearly with z, ie r = zR/h, so at height h, radius of cone is R.

Answer 18

i have done, and attach, the volume integrations I posted above --- and they work!!

[to be clear, delta is the wall thickness, h is height (typo in formula above also uses H but they are the same thing) and radius R.]

have compared them results from the standard formula applied to simple inner and outer cones. exactly same as per attached.

this means that the triple integrals MUST ALSO work, as the only difference is the extra sigma r^2 term in the integral.

however, it also means that the integration is even more turgid. the work involved doing it this way really is cracking a nut with a sledgehammer. but it was good fun proving that.

i will not be computing the inertia integrals because i know they work. they have to.

Answer 19

Simply start with a solid cone.

Then subtract the moment of inertia of the smaller from the bigger.

Answer 20

@Vincent-Lyon.Fr
this was supposed to be interesting rather than practical!

and it is most certainly not practical.