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### Moment of Inertia of a Solid Cone

```
Date: 02/03/99 at 04:25:35
From: Anonymous
Subject: MI of Solid Cone

How do you find the Moment of Inertia of a solid cone, in terms of
height h and base radius r? I know how to find the MI of a solid
sphere; do I use the same method? If so, what do I take as the
arbitrary disc radius? Should I model the cone as sitting on the axis,
with its apex at (0,0) and its height as the maximum x along the
x-axis?
```

```
Date: 02/03/99 at 14:37:05
From: Doctor Anthony
Subject: Re: MI of Solid Cone

You must, of course, specify about which axis you want the moment of
inertia. For the MI about the major axis you would use the MI of a disk
about an axis perpendicular to the disc and integrate from 0 to h. Have
the cone with its axis along the x-axis and its vertex at the origin.

If base radius is r and height h, then the radius at distance x from
the origin is given by (r/h)x.

The MI of a disc of mass m and radius a perpendicular to its plane is
ma^2/2

For our elementary disk at point x the volume is pi.radius^2.dx

= pi.(r/h)^2.x^2.dx

and the MI of this elementary disc is

(1/2)pi.(r/h)^2.x^2.(r/h)^2.x^2.dx

= (pi/2)(r/h)^4.x^4.dx and integrating between 0 and h we get

= (pi/2)(r/h)^4.h^5/5

= (pi/2)r^4.(h/5)

The volume of the whole cone is (1/3)pi.r^2.h

So we have MI  =  (1/3)pi.r^2.h[(3/10).r^2]

=  (3/10)M.r^2,   where M = mass of cone

---------------------------------------------------------------------
Moment of Inertia about axis through Vertex perpendicular to axis of
cone
---------------------------------------------------------------------
Now we use the MI of a flat disk about a diameter and then use the
parallel axis theorem to get MI about the y-axis.

The MI of a disk about a diameter is ma^2/4

So the MI of the elementary disk about the diameter is

(1/4)pi.(r/h)^2.x^2.(r/h)^2.x^2.dx

and the MI about the y-axis is

(pi/4)(r/h)^4.x^4.dx + pi.(r/h)^2.x^2. x^2.dx

= (pi/4)(r/h)^4.x^4.dx + i.(r/h)^2.x^4.dx

and integrating,

= (pi/4)(r/h)^4.(h^5/5)  + pi.(r/h)^2.h^5/5

= (pi/20)r^2.h[r^2 + 4h^2]

= (3/20)M[r^2 + 4h^2]

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Physics/Chemistry

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