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Thread: Flywheel inertia analysis

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    mklotz mklotz's Avatar
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    Flywheel inertia analysis

    Everyone knows that the mass at the periphery of a flywheel makes more of a contribution to the moment of inertia than mass nearer the rotation axis.

    Today I decided to grab my stubby little pencil and work out quantitatively just how much more that contribution is.

    On the offhand chance that some of you engine builders may find the results useful, I offer them up here for your entertainment.


    Suppose you have a circular disk of material of density rho. Its radius is "R" and its thickness is "T". Then its mass, "M", is given by:

    M = rho * pi * R^2 * T

    and its moment of inertia, "I" is given by:

    I = (1/2) * M * R^2

    Now suppose we cut this disk into two pieces by removing a circular disk of radius "r" from the middle, leaving an annulus of inner radius "r" and outer radius "R".

    We have then:

    md = mass of disk = rho * pi * r^2 * T

    Id = moment of disk = (1/2) * md * r^2

    and for the annulus

    ma = rho * pi * [(R^2 - r^2)] * T

    Ia = ma * (r^2 + R^2)/2

    As a sanity check, when r = 0 (i.e., no inner disk) , ma should equal M and Ia should equal I. A quick check will show that to be true. When r = R, (no annulus), md should equal M and Id should equal I. Again, they do. A further check, left as an exercise for the student, shows that:

    Ia + Id = I

    for any value of r, as it should.

    We're interested in the ratio of Ia to I since it's this ratio that quantifies how much more important the outer portion of the flywheel is to the inertia.

    Ia/I = {[R^2 - r^2] * [r^2 + R^2]} / R^4

    Ia/Id = [R^4 - r^4] / R^4


    In the code box below the number in the first row is r/R, the size of the disk removed in terms of the size of the original solid disk. The number in the second row is the moment of the resulting annulus expressed as a percentage of the moment of the original disk.

    We can see that if we cut away a disk with a radius half the size of the original radius, the moment of the resulting annulus is still 93.75% of the original moment. Cutting away half the center of the disk loses us only 6.25% of the moment. That radius squared effect is very powerful.

    Code:
     r/R    Ia/I (%)
    
      0.000  100.000
      0.100  99.990
      0.200  99.840
      0.300  99.190
      0.400  97.440
      0.500  93.750
      0.600  87.040
      0.700  75.990
      0.800  59.040
      0.900  34.390
      1.000   0.000


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  2. The Following 3 Users Say Thank You to mklotz For This Useful Post:

    Moby Duck (08-27-2016), Paul Jones (09-09-2016), PJs (07-19-2016)

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