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Moment of inertia (integration)

  1. Oct 17, 2012 #1
    1. The problem statement, all variables and given/known data

    See attachment

    2. Relevant equations

    Integrating

    3. The attempt at a solution
    The answer is 16pi slug-ft^2.

    Now I know your suppose to integrate it. Its in 2D, so a double integral at most, maybe a single integral? I'm not very good with integrals and their limits in finding moments of inertia, so any guidance would be appreciated.
     

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  2. jcsd
  3. Oct 17, 2012 #2

    tiny-tim

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    hi xzibition8612! :smile:

    slice it into cylindrical shells of thickness dy :wink:
     
  4. Oct 17, 2012 #3

    SteamKing

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    The ellipse is being rotated about the x-axis, according to the problem statement, which also asks you to calculate the MOI about the x-axis.

    I think TT wanted to suggest trying to use cylindrical slices with thickness dx instead of dy.
     
  5. Oct 17, 2012 #4

    tiny-tim

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    you can use cylinders with thickness dx,

    or cylindrical shells with thickness dy :smile:
     
  6. Oct 18, 2012 #5
    let's keep it simple. So I slice that shape into thickness dx and then integrate it. That would only be 1 integral right? So its going to be integral of ..... dx.
    Now the problem is what do I integrate? What do you have to integrate to get moment of inertia? Area?
     
  7. Oct 18, 2012 #6

    tiny-tim

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    hi xzibition8612! :smile:
    no, you use the ready-made formula for moment of inertia of a cylinder … 1/2 mr2 :wink:

    (if you used dy and cylindrical shells, you'd use the obvious and easy-to-remember formula for moment of inertia of a cylindrical shell … mr2)
     
  8. Oct 20, 2012 #7
    so m=15 slug, but what is r?
    And I still don't understand how you can use 1/2mr^2 on the ellipsoid and not integrate. thanks.
     
  9. Oct 21, 2012 #8

    tiny-tim

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    no, m is the mass of the cylindrical shell (or the cylinder), r is its radius, and yes do you have to integrate!
     
  10. Oct 21, 2012 #9
    Ixx = ∫(1/2)mr2dm

    I think this equation is correct. Now how do you find the radius? And you can take the r2 and (1/2) out of the integral correct?
     
  11. Oct 21, 2012 #10

    tiny-tim

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    a cylinder is 1/2 mr2

    a cylindrical shell is, obviously, mr2

    "m" here is the mass of the cylinder or cylindrical shell, not the whole mass

    r is the distance from the x-axis, ie y, and no you can't take it outside the ∫, it isn't constant
     
  12. Oct 21, 2012 #11
    Ixx=∫∫mr2dmdr

    The limits of dm is 0 to 15 slugs. The limit of r is from -1 to 1 ft (for the range of dx). Is this correct?
     
    Last edited: Oct 21, 2012
  13. Oct 21, 2012 #12

    SteamKing

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    You are not setting up the integral for the moment of inertia properly.

    The density of the body is 15 slugs/cu.ft., not the total mass. The total mass of the body is its volume multiplied by the mass density.

    One way to set up the integral for the elliptical body is to slice it up into a series of circular disks, each of thickness dx running along the x-axis. Each disk will have a mass of dm, which in turn, is a function of dx and the radius at the location of the disk.
     
  14. Oct 21, 2012 #13
    Ixx=ρ∫mr2dm

    Ok, now dm is a function of dx and r, but I don't see/understand that and have no idea how to expand dm into dx and r......is it a double integral?
     
  15. Oct 21, 2012 #14

    SteamKing

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    The mass of a disk dm = ρ dV. dV = area of the disk * thickness = A dx
    The attachment shows the moment of inertia of the disk, but the mass of the disk is dm rather than M from the figure.

    What you want to do is add up the inertia of all of the disks along the x axis.
     

    Attached Files:

    • disk.pdf
      disk.pdf
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  16. Oct 22, 2012 #15
    Ixx=∫(1/2)ρR2Adx

    Limits of integration from -2 to 2 ft.

    Now how do you find area A and radius R? Are they of each element of the circular disk? No idea how to do this thanks.
     
  17. Oct 22, 2012 #16

    tiny-tim

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    you're slicing it into cylinders with thickness dx, so they have radius y :smile:

    (if you sliced it into cylindrical shells with thickness dy, they would have radius y and height x(y) )
     
  18. Oct 22, 2012 #17
    ok so y would have limits from -1 to 1 ft. I understand that. Problem is my equation is in terms of dx......so how would I incorporate this? A double integral?
     
  19. Oct 22, 2012 #18

    tiny-tim

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    no, a single

    your only "d" is dx …

    you've sliced it with slices perpendicular to the x axis, going along the x axis

    (and your integrand is already an area thingy, so that only leave one dimension, doesn't it? :wink:)
     
  20. Oct 22, 2012 #19
    hisefsdfsdf
     
    Last edited: Oct 22, 2012
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