# Moment of Inertia of (Right) Cone

1. Jul 1, 2007

### Demon-Machine

1. The problem statement, all variables and given/known data
Find the moment of inertia of the right circular cone or radius r and height h with respect to its axis, and in terms of its mass.

*As of this point, I am supposed to use solids of revolution, and so I need to rotate a line about an axis, and find the moment of inertia with respect to that axis. I only know basic integration at this point.

2. Relevant equations
The textbook given equations for moment of inertia of the solid of revolution about the x-axis and y-axis are:

$$I_{y}=2\pi\rho\int_a^b yx^3\ dx$$
$$I_{x}=2\pi\rho\int_c^d xy^3\ dy$$

3. The attempt at a solution
I set up a line with the equation $$y=(\frac{h}{r} )x$$ to be rotated about the y-axis. Applying the formula i get:
$$I_{y}=2\pi\rho\int_a^b\ yx^3 dx$$
$$I_{y}=2\pi\rho\int_0^r\ ((\frac{h}{r} )x)x^3 dx$$
$$I_{y}=(\frac{2\pi\rho\ h}{r})\int_0^r\ x^4 dx$$ which simplifies to:
$$I_{y}=(\frac{2\pi\rho\ h}{r})\cdot\frac{r^5}{5} = \frac{2\pi\rho hr^4}{5}$$

Since the volume of this cone is $$\frac{\pi\rho hr^2}{3}$$
$$I_{y}= \frac{2\pi\rho hr^4}{5}= \frac{6r^2}{5}\cdot \frac{\pi\rho hr^2}{3}$$
$$I_{y}= \frac{6mr^2}{5}$$ Which is wrong because it is too large, and because the correct answer is $$I_{y}= \frac{3mr^2}{10}$$.

I've done this a few times and feel stupid because I still can't find my mistake. I suspect it is an obvious one, or has to do with simply "plugging in" info into the formula.

I am learning some calculus on my own over the summer, so I don't have anyone to ask other than the members of this forum. As this is my first post on this forum, I hope some of the members will make any suggestions as to my posting/question. Thanks.

Last edited: Jul 1, 2007
2. Jul 2, 2007

### Dick

Look at this: $$y=(\frac{h}{r} )x$$. It says that at x=0 the height of your 'cone' is 0 and at x=r it is h. And you are rotating around the origin. This doesn't sound quite like a cone to me.

3. Jul 3, 2007

### Demon-Machine

Hmm, I was probably unclear in explaining how I set up the problem. Since the book I am using only covers plain old two-variable equations (no multivariable stuff like $$z^2=x^2+y^2$$), the only way I know how to calculate volumes of solids is to rotate equations (or the areas bounded by those equations) around the axis. A better way to calculate the moment of inertia is probably with a triple integral, but I don't know how to do that yet. This means that rotating the line x=3 about the y-axis from y=0 to y=3 (the equivalent of rotating the rectangular area bounded by x=0, x=3, y=0, and y=3) generates a cylinder of radius 3 and height 3. (I'm not trying to teach you this stuff, just trying to give the context of how I have learned it).

Then, for a specific example in the text, the question states: "To generate a right circular cone of radius a and altitude h, we may rotate a right triangle about one of its legs. Placing the leg of length h along the x-axis, we rotate the right triangle whose hypotenuse is given by $$y=(\frac{a}{h} )x$$ about the x-axis." The only difference between this question setup and my setup is that I'm rotating about the y-axis, and so the line $$y=(\frac{h}{r} )x$$ should be just fine for generating a cone that is rotated around the y-axis unless I am missing some other glaringly obvious mistake. Besides, I tried the question by setting up a cone as in the previous example, and I still get the same (incorrect) result.

I'm not sure what you mean, because I am not rotating around the origin, but about the y-axis. Also, the cone does have a height of 0 when x=0 and a height of h at x=r. The x seems to me to only act as a sort of parameter for this stuff, and is kind of arbitrary because I only need to integrate from 0 to r to obtain the cone of the desired dimensions. Anyway, its late and I'm not thinking straight, so please point out any errors in my above reasoning. Oh, and thanks for taking interest Dick.

4. Jul 3, 2007

### Dick

You aren't seeing the geometry you are integrating correctly - possibly because the formulas you are working with don't make that explicit. In your formula, x is the distance from the axis of the cone. Then you are integrating dy from 0 to x*(h/r). Your solid doesn't look like a cone - it looks like a cylinder with a cone subtracted. Change x*(h/r) to h*(1-x/r) and you will get the book answer.

5. Jul 3, 2007

### Demon-Machine

Ahah! Now I see exactly where I went wrong; I can now see two ways to solve the problem. I can set it up as you suggested, with a line with the equation $$y=h(1-\frac{x}{r} )$$ and then use my "formula" to integrate from 0 to r.
$$I_{y}=2\pi\rho\int_a^b yx^3\ dx$$
$$I_{y}=2\pi\rho\int_0^r (h(1-\frac{x}{r} ))x^3\ dx$$
$$I_{y}=2\pi\rho h\int_0^r x^3-\frac{x^4}{r} dx$$
$$I_{y}=\frac{\pi\rho hr^4}{10} =\frac{3r^2}{10} \cdot \frac{\pi\rho hr^2}{3}$$
$$I_{y}=\frac{3}{10} mr^2$$

Alternately, I could realize that the y in the formula actually symbolizes the height of the representative element, and that it is therefore equal to h-y for the line $$y=(\frac{h}{r} )x$$. This means
$$I_{y}=2\pi\rho\int_a^b yx^3\ dx$$
$$I_{y}=2\pi\rho\int_0^r (h-y)x^3\ dx$$
$$I_{y}=2\pi\rho\int_0^r (h-((\frac{h}{r} )x))x^3\ dx$$
$$I_{y}=2\pi\rho h\int_0^r x^3-\frac{x^4}{r} dx$$ and this is the same as above. This is messy though, and I think your way is the more correct option, Dick.

Glad that issue was cleared up; that's the second time the textbook has introduced a misleading formula (or perhaps I am just misleading myself). Thanks for the help and the great first impression of the forum!