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Moment of inertia of a Conic Frustum

  1. Aug 17, 2006 #1
    Does anyone know how to calculate the moment of inertia of a frustum? I need to workout the MOI of a taper, as in tapered cylinder bearings.

    Just as the cone in the picture below, the taper (frustum) is uniform and revolving about its longitudinal axis:

    [​IMG]

    I have found a couple of equations but I have nomenclature issues with them and I do not fully understand how to use them.

    This is the main ‘solution’ I found:

    http://answers.yahoo.com/question/index?qid=20060804212218AA5Ma9R

    There is a link to a RapidShare word file through that link that should explain it better, but my ISP (NTL) uses proxies and RapidShare therefore refuses to allow me to download anything from their site!

    Any help appreciated.
     
  2. jcsd
  3. Aug 17, 2006 #2
    I have been given a formula for the MOI of a frustum, as follows:


    (3/10)*M*[(R1)^5 - (R0)^5]/[(R1)^3 - (R0)^3]


    I have taken this equation and made an example calculation. I am not sure I am using it correctly, or using the correct SI units and in the correct way.

    Are the workings below correct, or at least tackled in the correct way?


    Mass = 0.01kg (10g)
    R1 = 0.008m (8mm)
    R0 = 0.005m (5mm)


    (3/10)*0.01*[0.005^5 – 0.008^5] / [0.005^3 – 0.008^3]

    = 0.3 * 0.01 * [32.768 – 3.125] / [0.512 – 0.125]

    = 0.3 * 0.01 * 29.643 / 0.387

    = 0.229790697 kg m² (MOI)

    Where:

    R1 = radius of frustum base
    R0 = radius of frustum top
     
  4. Aug 17, 2006 #3
    I am not so sure that is correct.

    I am not very sure at all about how to use numbers with a zero before the decimal place? For example: The formula for the MOI of a cone is 3/10MR². Using SI units, a mass of 0.01kg and a cone base radius of 0.008m is the calculation as follows:

    0.3 * 0.01 * 0.008 * 0.008

    Or is it:

    0.3 * 0.01 * 8 *8

    There is a very large difference between both answers.

    When dealing with mass figure to the power of, for example, five, is the calculation for 10 grams (0.01kg) as follows:

    0.01^5kg = 0.01 * 0.01 * 0.01 *0.01 * 0.01 = 1^-10kg

    Or is it:

    0.01^5kg = (10 * 10 * 10 * 10 *10) / 100 = 1000kg

    This is a basic, yet highly important question. I guess this sort of confusion and mix-up is the reason why aircraft run out of fuel over the mid Atlantic due to an error with the re-fuelers!
     
  5. Aug 18, 2006 #4
    Anyone, please?
     
  6. Aug 18, 2006 #5

    FredGarvin

    User Avatar
    Science Advisor

    As a first pass, you could simply break the bearing element down into a finite number of circles stacked on top of each other and sum them. This would be easy in something like Excel. In terms of the integral to find it, the first thing I have to ask is, since this is a rotating element, do you want mass moment of inertia or do you want polar moment of inertia?

    In regards to the integral you stated, on a quick look, I beleive you need to take out the 2*pi*r factor and add a dθ.

    Also, from the page where you got that graphic (Eric Weisstein's), you can see the calculation of the MOI for the cone. Simply calculate the MOI of the overall cone and then, if you take a small cone from the top to form the frustum, you can caluclate the MOI of the small cone. Simply subtract the MOI of the small cone from the large cone to get the MOI of the frustum.
     
    Last edited: Aug 18, 2006
  7. Aug 22, 2006 #6
    Many thanks for your reply.

    It is mass moment of inertia that I am interested in, not polar. I am not sure what you mean by breaking down the bearing element into a finite number of circles? I also do not follow what you are saying regarding the ‘2*pi*r’ factor?

    And I do not know why I missed the very simple ‘big cone – little top cone = frustum MOI’?! I have missed a lot of simple things like that of late. I seem to be experiencing a period of intellectual downtime! Someone must be feeding me stupid pills. Maybe I can put it down to stress; maybe I am a subject of HAARP, who knows? I hope my ‘lack of logic’ phase ends very soon.
     
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