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Moment of inertia & area density

  1. Jul 17, 2008 #1

    dgm

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    Hey there Physics Forums!!

    I kept getting links to you guys cropping up in Google searches, so when I got stuck with something I figured I'd register and ask here -- seems like a pretty knowledgeable. :)

    Anyhow, I'm trying to calculate the mass moment of inertia for a flat, triangle-shaped plate. The axis of rotation goes through the center of mass of the plate, and is parallel to the Z axis if the vertices of the triangle are defined on the X and Y axes. According to:

    http://www.efunda.com/math/areas/triangle.cfm

    the formula for this is
    http://www.efunda.com/math/areas/images/triangle12.gif .

    Now, this is assuming that the area density is equal to 1. If I want to find the actual mass moment of inertia of the triangle, and I know the area density, would I just multiply the final result by the area density? I'm skeptical that it's that simple. Would I need to isolate the formula for the are of a triangle out of that equation, multiply it by the area density coefficient, and then plug that back into the equation by working backwards (if that even makes sense :| )?

    Any help is really appreciated!

    Peace and love,
    dgm
     
  2. jcsd
  3. Jul 17, 2008 #2

    CompuChip

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    First of all, welcome to PF.
    Second of all, I hope you have done some serious calculus, as you will be integrating your brains out on this one :smile:

    In general, the moment of inertia is defined as
    [tex]I = \int r^2 \, \mathrm dm = \iiint_V r^2 \rho(\vec r) \, \mathrm dV[/tex] (3D)
    or
    [tex]I = \iint_S r^2 \rho(\vec r) \, \mathrm dA[/tex] (2D)
    so you'd explicitly have to do the integral. When the mass density is constant, e.g.
    [tex]\rho(\vec r) \equiv \frac{m}{V} [/tex]
    you can take it outside the integral. But if the mass density varies, you will have to take it into account while doing the integration. Now a triangle isn't immediately the most easy shape to do this on (it's much more easier to have something where you can integrate in polar coordinates since r is already a "good" coordinate and you don't have to express the [itex]r^2[/itex] in terms of the integration variables) but depending on the mass density and the shape of the triangle you can work it out.

    Probably a good first step would be to take the mass density constant (m / A with A the area of the triangle, or if you want to give it a height h, m / V with V = h * A) and see if you can recover the original formula you quoted.

    Hope that gets you started, let us know how it goes :smile:
     
  4. Jul 17, 2008 #3

    tiny-tim

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    Welcome to PF!

    Hi dgm ! Welcome to PF! :smile:

    Yes, if the density is uniform, just multiply by the density.

    Not actually following what you mean by "isolate the formula for the area". :confused:

    If the density changes, then you have to go right back to the basic definition of moment of inertia, and start from there. :smile:
     
  5. Jul 17, 2008 #4

    dgm

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    Wow! Thanks a lot for the help... luckily my bodies are all of uniform density (I'm not a math wizard by any means)...

    Just for future reference, what exactly are "integrals" and what is "integration"? I looked it up just now and the definitions I found were beyond my comprehension. :/
     
  6. Jul 18, 2008 #5

    CompuChip

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    That's always a problem here, to guess from someones initial post at which level to answer the question. Since I have usually seen the question you asked in the context of an undergraduate (university level) physics course, chances were you did know about integration, so sorry for confusing you with that if you haven't seen it yet.

    Let me just briefly say something about integration "for future reference".
    Usually integration is introduced as the "opposite" of differentation (and certain integrals also called "anti-derivatives"). But in physics it is very important, because integration has to do with surfaces and summing. In this case, suppose that you have a non-uniform mass density function, which gives you the mass density at any point on the surface. What you could do, for example, to calculate the mass, is this: divide the surface in very small patches, and assume that the patch has a mass equal to the mass density at the center point (for example) times the area of the patch (hopefully you see that's reasonable, by the definition of mass density). Then you can find the mass of the entire surface by summing the masses of all these patches. (Note: if the mass density is constant, you can also first sum all the areas of the patches and then multiply by the mass density, so you just get mass density * area as in the posts above). Of course, this is an approximation, but the smaller the patches are (but also, the more of them there are), the more accurate the approximation will become. Integration is, if you want, a mathematical limit procedure in which the approximation becomes exact (e.g. we get infinitesimally small patches).

    For now, you can forget about all this, but if you ever want to so something serious in physics you will re-encouter integrals, precisely because we always want to do such summations over non-smooth surfaces (e.g. areas under curved graphs, densities over curved surfaces or volumes, ...). Maybe you can check out this page to see a simpler case: in the image you see how the area enclosed by the curve and the horizontal axis is approximated by rectangles. You can check for yourself that if you increase the number of rectangles on the third line and hit "Replot", that the approximation will get closer and closer to the actual value (they're shown below the plot). I hope you see that in the limit in which we have infinitely many rectangles which are infinitesimally small ("of zero width") we will get the actual area under the curve. This is sort of a "continuous sum" called the "integral" of the curve.
     
    Last edited: Jul 18, 2008
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