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Moment of Inertia of a Point mass

by Jdo300
Tags: inertia, mass, moment, point
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Jdo300
#1
Mar24-05, 01:33 PM
P: 547
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?
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dextercioby
#2
Mar24-05, 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.
Crosson
#3
Mar24-05, 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).

Jdo300
#4
Mar24-05, 01:53 PM
P: 547
Moment of Inertia of a Point mass

Quote Quote by dextercioby
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...

Daniel.
Hmmm.... could you explain to me what you mean? I'm still trying to understand the whole concept. I saw another equation with a [tex]\rho[/tex] in it. would that be the right formula to use? All I'm trying to do is use a basic formula that I can put into integrals and rotate different ways to calculate the moment of inertia for various, custom 3D solids. and the only information I have about the material itself is its density. I could calculate the total mass of the solid using volllume integration if necessary.
dextercioby
#5
Mar24-05, 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.
Jdo300
#6
Mar24-05, 02:03 PM
P: 547
Quote Quote by Crosson
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).
Ahhh thatís the integral I saw. Now as far as calculating the volume, I know how to do integrals for 3D solids by rotating cross sections around an axis using the disk, washer, or shell method, but how can I apply this to the integral you gave me? I'm trying to understand how that particular formula you gave me works. Can I use the [tex]\rho[/tex] in place of the m in the first equation as a point mass equation like [tex]I = \rho r^2[/tex]?? I understand the whole idea of moment of inertia as taking an infinitely small dot of the material and calculating the moments of all the infinitely small dots inside a solid and adding them together to get the total moment of inertia. But I'm trying to understand what the basic equation for the dot is so I can do the integrations. thatís where Iím getting stuck at right now.
dextercioby
#7
Mar24-05, 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 delta-type 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.
BobG
#8
Mar24-05, 02:13 PM
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Quote Quote by Jdo300
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?
You have to sum up all of the point masses that exist in the body, or, in other words, the equation dextercioby gave.

[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.
Jdo300
#9
Mar24-05, 02:47 PM
P: 547
Quote Quote by BobG
You have to sum up all of the point masses that exist in the body, or, in other words, the equation dextercioby gave.

[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.
OK, I think I understand conceptually what you are saying. So the volume integral with the dV in it should be replaced with the volume integral for whatever the solid I'm working with is? Does this method work for objects that are not symmetrical about the axis of rotation? Like for example: what if I wanted to revolve a football around a rotation axis but rotate the football widthwise not lengthwise. I could make a volume integral using a oval function and revolve the cross-section to make my 3D football solid. Once I have this, would I just take the volume integral I made and replace dV with it? How would this work since my volume integral wouldn't have an r in it? (I would use dx in stead)
dextercioby
#10
Mar24-05, 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.
Jdo300
#11
Mar24-05, 03:30 PM
P: 547
Quote Quote by dextercioby
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.
Ok, I've never used the jacobi coordinate system before. I'm going to be defining most of my volumes in either polar or rectangular form (I can do parametric too if necessary). OK, I'm going to show you an example integral for the football I was talking about. Could you please show me how to relate it to the volume integral to calculate the moment of inertia? I'm doing it in rectangular coordinates so I'm not sure how I would get an r out of it:

[tex]V=\pi\int_{-\sqrt{5}}^{\sqrt{5}} \left(\frac{\sqrt{5-x^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?
dextercioby
#12
Mar24-05, 03:53 PM
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What is the mass volumic density and which are the variables it depends upon...?

Daniel.
Crosson
#13
Mar24-05, 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.
Jdo300
#14
Mar24-05, 11:16 PM
P: 547
Quote Quote by dextercioby
What is the mass volumic density and which are the variables it depends upon...?

Daniel.
Ummm... I don't have a specific mass volume density for this. I just was using it as an example to see how to do this. As far as the variables it depends on, I'm not sure I completely understand what you are asking. I can say that I'm using the disk method to get the volume of the football in that perticular integral and that dx is the thickness of the diskss being that the disks are infinately thin. I have the feeling that is not what you are asking though.
Jdo300
#15
Mar24-05, 11:29 PM
P: 547
Quote Quote by Crosson
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.
Hi, thanks for that information. But Iím still quite confused as to how to manipulate that particular integral to get the moment of inertia. But before I continue, I should let you know what I do know how to do because I think that some of what you have been showing me, I'm not understanding the way you think.

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.
NEOclassic
#16
Mar26-05, 04:21 AM
P: 161
Quote Quote by Jdo300
Hi, thanks for that information. But Iím still quite confused as to how to manipulate that particular integral to get the moment of inertia. But before I continue, I should let you know what I do know how to do because I think that some of what you have been showing me, I'm not understanding the way you think..
Hi Jdo,

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
BobG
#17
Mar26-05, 08:56 AM
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Quote Quote by Jdo300
I know how to do the solids of revolution using the disk, washer, and shell methods.
That's good. Remember what you're really doing - finding the moment of inertia for each point mass and then summing them together. All the point masses that are the same distance away have the same moment of inertia. If you google the formulas for moments of inertia, you'll notice the formula for finding moment of inertia of a ring about it's central axis is your base equation: [tex]I=mr^2[/tex]. All of the points in the ring are the exact same distance away from the rotational axis.

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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