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moment di inerzia

  • Thread starter Thread starter b243970
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b243970

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question:

why the moment of inertia of a homogeneous cylinder internal radius r and outer radius r vale (m(r^2+r^2))/2? ? ?

According to me, integrating an infinitesimal cylindrical shell from r to r should be (m(r^2-r^2))/2;

and also considering the moment of inertia of a cylinder full of radius r, (mr^2)/2, less that of a cylinder full of radius r and negative mass would anyway i=(m(r^2-r^2))/2,

Then why isn't it? I searched around but on the books and on the internet I found only the formula put there so without development. Give me your sleep.
:frown:
 
I knew that between me and the truth there was an immon ca...ata that I was doing: )

thanks.:finger:
 

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