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Rings Vs Disks In Moment of Moment of Inertia Integral

by teroenza
Tags: disks, inertia, integral, moment, rings
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teroenza
#1
Mar5-12, 09:10 AM
P: 128
1. The problem statement, all variables and given/known data
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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vela
#2
Mar5-12, 12:00 PM
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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.
teroenza
#3
Mar5-12, 08:23 PM
P: 128
Ok thank you.


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