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23 Feb 2010 13:04:37 IST
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moment of inertia of hollow cone
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moment of inertia of hollow cone


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Sagar Saxena's Avatar

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27 Feb 2010 11:04:52 IST
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hii

7837-833_7662_120px-Moment_of_inertia_cone.svg.png


consider this hollow cone ,now cut a ring of thickness dx at a distance x from the origin ,

   so radius of the ring will be  r'=xtanΘ         where Θ is the half angel of cone


so moment of inertia of this ring is  dI =dm r'2

   dm =[ M/πrL ]2πr'dL  where L is slant hight

        =[2M/rL]r'dx/cosΘ

so dI =[2M/rL]r'3dx/cosΘ

        =tan3Θ/cosΘ [2M/rL]x3dx

        =tan3Θ [2M/rh]x3dx

I =oh tan3Θ [2M/rh]x3dx

I=mr2/2                     !!!!!

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27 Feb 2010 12:57:02 IST
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 HAVE A SALUTE FROM ME                                          PLEASE SEE A PROBLEM  IN PHYSICAL CHEMISTRY AS UNDER AND REPLY

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27 Feb 2010 12:58:04 IST
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The figures given below show the location of atoms in three crystallographic planes in a FCC lattice. Draw the unit cell for the corresponding structures and identify these planes in your diagram.                                                                                                                                                                                                        

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27 Feb 2010 15:52:58 IST
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 SOME ONE REPLY

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27 Feb 2010 15:54:30 IST
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 PLEASE   REPLY MY QUESTION




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