Rings Vs Disks In Moment of Moment of Inertia Integral

AI Thread Summary
The discussion centers on the differing volume elements used in center of mass and moment of inertia calculations for a cone. For center of mass, the volume element is expressed as dv = (pi*r^2)*dz, while for moment of inertia, it is dv = (2*pi*r)*dr*dz, utilizing rings instead of disks. The teacher explains that rings are preferred in moment of inertia problems because they concentrate mass at a specific radius. The distinction arises from the different quantities being calculated: center of mass involves averaging z, while moment of inertia involves averaging r^2. Understanding these differences clarifies why the volume elements vary in these contexts.
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Homework Statement


Not really homework, but putting it here to be safe. While doing a center of mass calculation of a cone centered about the origin (tip touching origin, z axis through it's center) my teacher put on the board that the volume element was

dv= (pi*r^2)*dz using the area of a disk. The variable r was then eliminated by substituting r as a function of z.

However, when doing a moment of inertia problem, he used

dv= (2*pi*r)*dr*dz using rings of width dr

I asked him why the differentials were different and he said, because when doing moment of inertia problems, we use rings because they have their mass concentrated at one radius.

Wouldn't the same argument require us to use rings for the center of mass calculation, which is also dependent upon r ?

He is changing the differentials to avoid making people do triple integrals. Which I prefer here because I can get the correct answer in both cases by setting up the proper triple integral and working it out. I just want to understand it his way in case a test is more easily worked that way.

Thank you
 
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In the center of mass calculation, you're finding ##\langle z \rangle## whereas in the moment of inertia calculation, you're finding ##\langle r^2 \rangle##. If you set up the triple integrals and then integrate so that z or r is the last variable of integration, you'll see you end up with the two different volume elements.
 
Ok thank you.
 
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