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Moment of Inertia of Solid Sphere

  1. Jan 4, 2010 #1
    This is not homework, just my practise question for an exam.

    I keep getting an answer of 3/5MR^2 when the correct one is 2/5MR^2, and that was only because I fluked it by deciding to change my intergrand a little bit.

    I've some sources including one from hyperphysics which suggests considering the sphere as a stack of disks. I'm completely stumped. I've used double intergrals before, but in this one the integrands I would have never though of using them - how did they get them?

    If someone could please run through me how to solve this sphere I would be grateful. No weird short cuts please...
     
  2. jcsd
  3. Jan 4, 2010 #2

    diazona

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    Which integrands are these that you're confused about?
     
  4. Jan 4, 2010 #3
    y^2 = (R^2 - x^2)

    I used this to substitute and integrate with respect to x , and I got 21/5MR^2...

    WTF?

    -.-
     
  5. Jan 4, 2010 #4

    berkeman

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    Staff: Mentor

    Could you please show us the integral and limits that you've set up? Integrating the disks seems like the easiest way to do it. What are you using for the moment of inertia of a disk?
     
  6. Jan 4, 2010 #5
    The latex feature is annoying to use, so I'm unable to post what I did with symbols and stuff.

    Basically I considered a small disks which make up the pyramid. Disk has dimensions y, R and x where y^2 = R^2 - x^2.

    In the equation with I = inte(r^2) dm, r becomes y, so y^2, since we are moving through y, and every other value changes as we change why. This is where I use y^2 = R^2 - x^2 and start to solve.

    dm I know.
     
  7. Jan 4, 2010 #6

    rock.freak667

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    If you are considering a disc, then the moment of inertia of that disc should be

    dI=1/2 y2 dm


    Your disc should also have a thickness of 'dx' so that you can compute the volume 'dV'.
     
  8. Jan 4, 2010 #7
    I tried my above way again and got 6/5MR^2....
     
  9. Jan 4, 2010 #8

    berkeman

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    Without a sketch, it's hard to be sure, but I think you are mixing up some axes or directions, etc.

    I drew the sphere as seen from the side, and labeled the horizontal axis x, and the vertical axis y. I stacked horizontal disks up to make the sphere (like you called half of the sphere a pyramid). It's true that you can just integrate from y=0 to y=R and double the answer to get the full sphere's moment of inertia.

    So on this drawing, mark the disk that is about half-way up the y axis. You see that the radius of that disk is x, and it is given correctly by the equation you wrote. That x is the radius that you use in the area calculation of the disk, which is used to get the moment of inertia calculation for the disk (since dm is related to the radius of the disk, dy and the volume density of the sphere).

    So maybe try setting up the integration (or the half-integration, which you will end up doubling in the end) that way, and be careful not to mix up the sphere's radius R with the radii of the disks x(y). Does that get you a lot closer to the correct answer?
     
  10. Jan 5, 2010 #9
    Well I was following the one from Hyper physics and begun to understand it but I still somehow get a wrong numerator. It's not that much of a problem since if this was a 10 mark question, I would still get like 8/10 for my method and technical prowess.

    It's all a load of bollocks really... Why is it that every different question referring to center of mass or moment of inertia introduces its own little trick or catch that varies with the problem? WTF, this is just what I expect from a mathematics. A load of crap, designed to confuse you into insanity. You are supposed to identify a infinitesimally small piece of mass with certain dimensions that are increasing. As you increase it, you analyze how the volume/area changes with the mass and density is constant. Choosing variables to integrate should not be difficult nor should anyone have a problem with finding the main dimension with which to move through the continous object in order to build up the object.

    Why is it then, that this train of thought fails with every other question, and somehow you are 'expected' to get the joke and answer the question correctly? o_O
     
  11. Jan 5, 2010 #10

    diazona

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    I don't know about the rest of the people here, but I never had any problems computing the moment of inertia of a sphere, or any other regular shape, using the general formula
    [tex]I = \int r^2 \mathrm{d}m[/tex]
    No tricks or catches.

    Give us a link to this Hyperphysics page you're following, and post your work IN FULL. With a sketch. Without seeing everything you did, we're reduced to just guessing at where you might have made a mistake, but if you show your work, I have a feeling we'll see that there are just one or two little errors.
     
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