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Moment of Inertia of a Point mass 
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#1
Mar2405, 01:33 PM

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Hello all,
I am trying to work with the moment of inertia formula to integrate around some different geometrical shapes I'm working with. I looked up the general formula which I found to be [tex]I = mr^2[/tex]. I know that I can figure out the moments of inertia for any shape I need by taking that point mass and integrating it in various ways to get the moment of inertia for whatever i want but I'm stuck with this formula because I'm not sure what to input the the mass variable. All of the shapes I'm working with have a uniform density (except for some having holes cut in them), but I'm not sure how to represent the particular material in the point mass equation. I can find the total mass of the object I'm working with because I know the density of the material, but how do I apply that information to the point mass formula? 


#2
Mar2405, 01:42 PM

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You can't...
[tex]I_{axis}=\int dm\ r^{2} [/tex] is the moment of inertia for a RIGID SOLID around a rotation axis...For a point mass,that "r" (distance between the mass element "dm" and the rotation axis) is ZERO,if the axis of rotation contains the point particle. Daniel. 


#3
Mar2405, 01:52 PM

P: 1,295

It sounds like you are missing this:
[tex] I = \int r^2 dm = \rho \int_{vol} r^2 dV [/tex] In the last step, I have used the density rho to convert the mass integral in to a volume integral. Then, integrate over the volume of the shape (with the r^2 inside the integral). 


#4
Mar2405, 01:53 PM

P: 547

Moment of Inertia of a Point mass



#5
Mar2405, 01:58 PM

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Yes,but only for bodies with finite size.IIRC,you asked about a moment of inertia or a point mass...Incidentally,to be fair,the mass volumic density for a point mass in the point [itex] \vec{R} [/itex] is:
[tex] \rho (\vec{r})=m\delta^{3}(\vec{r}\vec{R}) [/tex],so the volume integral would make sense (!!).It's the "r" which is IDENTICALLY ZERO,in the case of pointlike particles which would rotate around an axis which would contain them... Daniel. 


#6
Mar2405, 02:03 PM

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#7
Mar2405, 02:11 PM

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The moment of inertia for pointlike masses can be computed using the SUM.The integral formulas which u saw apply only to FINITE SIZE/CONTINUOUS bodies and to pointlike particles it yields the SUM...It's like computing the mass...You integrate density for a finite body and for a collection of pointlike particles,simply add the individual masses...
Again i stress that the integral formula does not apply to pointlike objects for which the rotation axis passes through the body.Incidentally,for pointlike masses,the integral formula >[itex] I=mr^{2} [/itex],because of the deltatype mass volumic density and the fact that the "r" is not zero,since the axis of rotation doesn't pass through the point particle... Daniel. Daniel. 


#8
Mar2405, 02:13 PM

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[tex]I_{axis}=\int_0^m dm\ r^{2} [/tex] Your differential mass is your density ([tex]\rho[/tex]) times your differential volume. In other words, if you add a certain amount of volume of a given mass, your overall mass changes by a given amount. The differential volume of your solid depends on its shape. For example, if you have a cylinder, the volume is: [tex]V=\pi r^2 h[/tex] If you're rotating your cylinder about the central axis, you don't care that much about your height  it's just a constant. You differentiate with respect to r to get your differential volume, or: [tex]dV=\pi h 2 r dr [/tex] That gets your variable of integration to the term you want: your radius. By substituting your density and differential volume for the differential mass, you get: [tex]I_{axis}= 2 \rho \pi h \int_0^r r^{3} dr\ [/tex] with density equal to mass/volume A cylinder is about the easiest to find. Other shapes use the same principle, though, even though you sometimes wind up having to integrate over each axes (for a triple integral). The key is to change the differential mass into a differential volume for the shape you're working with. 


#9
Mar2405, 02:47 PM

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#10
Mar2405, 02:59 PM

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Yes,to everything.That "r" is a distance,it can be in any (crvilinear) system of coordinates.For a revolution ellipsoid (the football),u can choose elliptic (Jacobi) coordinates.So the integral contains 3 variables,among which the ones which enter in the expression of "r".
Daniel. P.S.Think about the ball.It will be very simple to figure out which is "r" and what are the 3 variables. 


#11
Mar2405, 03:30 PM

P: 547

[tex]V=\pi\int_{\sqrt{5}}^{\sqrt{5}} \left(\frac{\sqrt{5x^2}}{2}\right)^2 dx [/tex] This will make a football shape that is centered on the origen. Now how do I use this integral to calculate the moment of inertia for the object? 


#12
Mar2405, 03:53 PM

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What is the mass volumic density and which are the variables it depends upon...?
Daniel. 


#13
Mar2405, 06:33 PM

P: 1,295

If you only know how to calculate the volume of solids of revolution, you are limited to those shapes to calculate the moment of inertia. Generally you would do a triple integration in either cartesian or spherical coordinates:
[tex] I = \int_{x_o}^{x_f }\int_{y_o}^{y_f} \int_{z_o}^{z_f} (x^2 + y^2 +z^2) \rho (x,y,z) dz dy dx[/tex] Of course, if you want the boundaries to be curves then the limits of integration should be functions. 


#14
Mar2405, 11:16 PM

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#15
Mar2405, 11:29 PM

P: 547

The only types of volume integration problems I've done in my calculus class are solids of revolution where you take a cross section from two or more functions that intersect each other and revolve the cross section around some axis of revolution (either x or y + offsets). I know how to do the solids of revolution using the disk, washer, and shell methods. One other thing I know how to do is find the volume of solids with known cross sections. For instance, I can take a region formed by one or more functions, and use shapes like triangles or squares to fill the region 3 dimensionally to generate a solid. Now, there is one specific thing I want to do where I may need to do this, but I'll ask you about my specific problem in a bit. But I want to understand how the math connects. I do have a general understanding of what you are describing to me and I thank you both for your patience in helping me to understand this. But I'm still stuck at understanding how to apply my knowledge of volume integrals to get moments of inertia of the solids. 


#16
Mar2605, 04:21 AM

P: 161

I've been there and done that and found that one stage of integration that could eliminate the "integral  dt" simplified the integration; It might be that th velocity part of momentum may be why you are confused. My integration used tubular increments that were concentric relative to the axis of spin. The element of time disappeared by considering only one complete 360 degree cycle of rotation. Incidentally I discovered that the angular momentum per cycle of a solid sphere was a function of the outer radius of the sphere raised to the fourth power. That figures: R^3 for the volume and R^1 in the mvR of angular momentum. The trouble was that I knew neither the radius of the point mass nor its density. Go figure! Cheers, Jim 


#17
Mar2605, 08:56 AM

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That's what you're trying to do for other shapes, find all of the point masses that happen to be the same distance away from the central axis. So, the shell method is the proper way to set the problem up. You also have to multiply by the density (which occurs when you're changing your limits from differential mass to differential volume). Edit: I should add one note about the density. The first time I did this, I was able to see that the cross sectional area was the key to finding the moment of inertia. In other words, for any shape, I could find the equivalent radius for a cylinder. I also knew that if you stacked two identical cylinders together, the new cylinder would still have the exact same moment of inertia as the two individual cylinders. In other words, if I found the moment of inertia for a single disk in the cylinder, I'd have the moment of inertia for the entire cylinder. That led me to believe I didn't care much about the mass  just plug it in as a constant and I'd be good to go. Since I'd compressed all of the cylinder's mass into a single disk, every single formula I came up with was wrong . A 10kg disk with a radius of 5 m has a substantially different moment of inertia than a 1000kg disk with a radius of 5 m . Easy fix once I figured out what I'd done wrong. 


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