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Moment of inertia - cylinder

  1. Apr 19, 2008 #1
    Please, help me to solve this problem.

    I need to calculate moment of inertia, I, for this body on picture:
    - mass o the body is m
    - radius of the big cylinder is a
    - radius of the small cylinders is a/3


    Thanks for your help!
     

    Attached Files:

  2. jcsd
  3. Apr 19, 2008 #2
    Well you have that I=Integral(r^2*dm)
    Then u have the density=Mass/Volume (I'm supposing it's a cylinder)
    then dm=density*dV
    Then you solve I=Integral(r^2*density*dx*dy*dz) where r^2=x^2+y^2.
    now you have the full cylinder without the holes.
    Do the same for the holes and sum (with a minus of course)

    Good luck
     
  4. Apr 19, 2008 #3
    Ohh if you're having trouble with the holes just integrate but using r as a vector. So you do r=x+(cos(tethta), sen(theta)) where x is the vector from the oringin to the center of the wholes. Using that r just repeat it for the 4 circles (it's simetrical). Just remember that when integrating area in polar coordiantes dA=dx*dy=r*dr*dTheta

    Cheers
     
  5. Apr 19, 2008 #4
    Ohh and one more thing!! remember that in vectors r^2=inner product (r,r)
    :)
    If u get stressed check tubepolis.com for some funny videos jeje. Look for triger happy those r really fun.
     
  6. Apr 19, 2008 #5
    Thanks, will try it later when i will have some time! It seams logical! :)
     
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