Moment of inertia tensor of hollow cone

[Lawrencel2]

Homework Statement

Find the inertia tensor for a uniform, thin hollow cone, such as an ice cream cone, of mass M, height h, and base radius R, spinning about its pointed end.

Homework Equations

$I_{zz} = \sum x^{2}+y^{2}$

$\rho = \sqrt{x^{2}+y^{2}}$

The Attempt at a Solution

I first tried to think of this as a bunch of little rings
$area = 2\pi\rho dz$ as rho is changing as height increases
i defined rho as $\rho=\frac{z R}{h}$
but how would i set up this integral? My book shows a few examples of doing it with a solid but i don't know how to do it with an area. so far i have
$\int \frac{2\pi R^{3}}{h^{3}} z^{3}dz$
I know that that the $I_{zz}= \frac{MR^{2}}{2}$ just need help getting started

 

[tiny-tim]

Hi Lawrencel2!

I first tried to think of this as a bunch of little rings
$area = 2\pi\rho dz$ as rho is changing as height increases

they aren't little rings, they're little volumes, all at a distance r from the axis, and with thickness t and height dr*R/h

 

[brainpushups]

Another way to think about the moment of inertia of a conical shell is this: If the thickness of the shell is infinitesimally small then the rings do not overlap. This is just like splitting a solid disc up unto little rings. As Lawrencel2 mentioned, the moment of inertia of the hollow cone is MR^2/2, which is also that of a solid disc.Now here's my problem. How does one do this when the walls of the cone are not thin (say inner radius b and outer radius a)?

First, I'd like to explain how I did this for a hollow sphere because I think the same approach should work here.

My approach for the sphere was to change the limits for the radius to be from b to a instead of from 0 to R. When the integration is said and done you end up with

$I_{zz}$=$\frac{2M(a^{5}-b^{5})}{5(a^{3}-b^{3})}$

This can be factored to give

$I_{zz}$=$\frac{2M(a^{4}+a^{3}b+a^{2}b^{2}+a b^{3}+b^{4})}{(a^{2}+ab+b^{2})}$

If you then let the inner radius b equal the outer radius a you will get the familiar result that a thin hollow sphere has MoI equal to 2/3MR^{2}
Now, the same approach for the hollow cone. Its the same set up for a solid cone

Integrate[$r^{3} dr d$$\phi$ dz] using cylindrical polar coordinates, but now the limits on r will be from bz/h to az/h.

I end up with $\frac{1}{10}$ $\rho$ $π h(a^{4}-b^{4})$

Where $\rho$ is the density and h is the height. Plugging in M/V for the density I get (the volume of a conical shell is V= $\frac{π h(a^{3}-b^{3})}{(3a}$
$\frac{3 M a}{(a^{3}-b^{3} π h}$ $\frac{1}{10}$ $\rho$ π h$(a^{4}-b^{4})$

or

$\frac{2Ma(a^{3}+a^{2}b+a b^{2}+a b^{3})}{(a^{2}+ab+b^{2})}$

By the way, how the heck does this volume formula V= $\frac{π h(a^{3}-b^{3})}{(3a}$ for a cone come about?! I looked it up and I'm puzzled by it.This answer is not correct and does not reduce to the thin conical shell The correct expression for the MoI (assuming Wolfram Alpha is correct) is

$\frac{3(a^{5}-b^{5})}{(10(a^{3}-b^{3})}$

Which does reduce to the correct thin walled answer.

Can anyone point out where my mistake(s) are?

 

[brainpushups]

Another question: how can I correctly set up the other moment of inertia integrals $I_{xx}$ and $I_{yy}$ (they will be the same because of the symmetry) for a hollow cone? I have no problem doing this if the cone is solid, but I'm having trouble figuring it out for a hollow one.

Thanks in advance!

 

[paranormal]

To find the moment of inertia tensor for a hollow cone, we can use the parallel axis theorem, which states that the moment of inertia of a body about an axis is equal to the moment of inertia of the body about a parallel axis through its center of mass, plus the product of its mass and the square of the distance between the two axes. In this case, the cone is spinning about its pointed end, so the parallel axis theorem can be written as:

I_{zz} = I_{cm} + Md^{2}

Where I_{zz} is the moment of inertia about the pointed end, I_{cm} is the moment of inertia about the center of mass, and d is the distance between the two axes.

To find I_{cm}, we can use the formula for the moment of inertia of a thin hollow cone about its central axis, which is given by:

I_{cm} = \frac{3MR^{2}}{10}

To find d, we can use the Pythagorean theorem to calculate the distance between the pointed end and the center of mass. Let's call this distance h_{cm}. Using the given height and base radius, we can calculate h_{cm} as:

h_{cm} = \sqrt{R^{2} + \frac{4h^{2}}{9}}

Now, we can substitute our values into the equation for the parallel axis theorem and solve for I_{zz}:

I_{zz} = \frac{3MR^{2}}{10} + M\left(\sqrt{R^{2} + \frac{4h^{2}}{9}}\right)^{2} = \frac{3MR^{2}}{10} + M\left(R^{2} + \frac{4h^{2}}{9}\right) = \frac{3MR^{2}}{10} + MR^{2} + \frac{4Mh^{2}}{9} = \frac{13MR^{2}}{10} + \frac{4Mh^{2}}{9}

Therefore, the moment of inertia tensor for a hollow cone spinning about its pointed end is:

I = \begin{pmatrix} \frac{13MR^{2}}{10} + \frac{4Mh^{2}}{9} & 0 & 0 \\ 0 & \frac{13MR^{2}}{