Can you tell me the formula for the moment of inertia of a frustrum of a cone please?

Answer 1

You need to provide some more information. Is it the moment of inertia about the central axis? Is it a cone of uniform density? Assuming yes, the moment of inertia is given by the intetgral of r^2 dm throughout the volume of the cone.

Consider a disc of thickness dy as a section of the cone. The moment of intertia of that disc depends on the vertical coordinate y and is

I(y)*dy ={ integral[r=R0 to r= R0-(R1-R0)(y/h)] of m*(r^2)*(2*pi*r)*dr }*dy

After integrating that, the overall moment is

I = integral[ y=0 to y=h] of i(y)dy

h = height of the frustrum
R0 = radius of the base
R1 = radius of the top
m is density of the cone material

Edit: I have worked this out, but it is very complicated with lots of chances for numerical error. My tentative solution (until I check for numerical errors) is:

I = (pi/10)*m*h*[26*R0^4 - 49*R0^3*R1 + 21*R0^2*R1^2 - 9*R0*R1^3 + (R1^4)]

Edit: found some coefficient errors and corrected the formula. I hope it is ok now.

Further edit (8/5/06):
No it wasn't ok; further checking revealed more mistakes; it's hard to keep track of all the terms. I will review further, please keep checking back. When I'm satisfied with the result i will tell you how to get a copy of the complete calculation so you can verify it for yourself.

Final edit (I hope) just made.

You can download the Word document that has the calculation from http://rapidshare.de/files/28314245/Moment_of_Inertia_Calculation.html

Select "free" download, wait for the countdown, enter the code and you will get the file.

Answer 2

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