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How do I find the moment of inertia for a curve? 
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#1
Mar2709, 09:53 PM

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1. The problem statement, all variables and given/known data
Say I am given some curve f(x,y) (revolved around some axis), how do I find the moment of inertia about an axis? I know how to find the moment of inertia of things like a uniform rod, ring and sphere using [tex]I=\int r^2 dm[/tex] I believe I am supposed to to pick an elemental piece such that the revolved element is through the axis I want. But if I use I=[itex]\int[/itex]r^{2} dm, I don't get anywhere. I've various places that I am to use a double integral or even a triple integral. But I don't know how to set these up to compute the moment of inertia. 


#2
Mar2709, 09:58 PM

P: 619

Have you met up with the Theorems of Pappus yet?



#3
Mar2709, 09:59 PM

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#4
Mar2709, 10:44 PM

P: 619

How do I find the moment of inertia for a curve?
All right, then you will have to do something similar. Think in cylindrical coordinates, with the axis of revolution as the polar axis of the cylindrical coordinates. Let's suppose that your given curve is y = f(x). We want to use y as the maximum radius, and x as the z value in the cylindrical coordinate system, so the volume element is
dv = r dr dth dz where r is the radius to a point inside the volume, 0<=r<=f(z) th is the angle theta that measures angle around the z axis, 0<=th<=2*pi z is the original x value range Then make your triple integration with all the proper limits and this should give you the volume. 


#5
Mar2709, 10:48 PM

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what if I need to find the moment of inertia of a plane figure such as a triangle or rectangle? 


#6
Mar2709, 10:50 PM

P: 619

That is a different problem, solved in a different way. Get through this one for now.



#7
Mar2709, 11:02 PM

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#8
Mar2809, 08:26 AM

P: 619

In this problem, x is the radius, so solve the curve for x = f(y). Then use that as the upper limit of integration on r.



#9
Mar2809, 01:43 PM

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[tex] I_y= \pho \int_0 ^{0.03} \int_0 ^{2 \pi} \int_0 ^{\frac{y^3}{9}} r^3 dr d\theta dz [/tex] But I do not understand how x is the radius here if the curve shows that the x distance is not constant.Also why then would they give the distances 3cm and 3cm (vertically)? 


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