 # explain moment of inertia of hollow  cone Badiuddin askIITians.ismu Expert
148 Points
12 years ago

Dear Amrit pal 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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Regards,

10 years ago

The problem can also be solved like this:- AS we are calculating the moment of inertia along the vertical axis, i guess we can actually bring down all the circular rings forming the cone to the base level so that we get a circular disk and the moment of inertia of a circular disk is (MR^2)/2. I want to know if i am right...

I also want to know the moment of inertia of the cone along one of its base diameters. How do i do it??(i got the answer as (3MR^3)/4L.... i am not too sure about it)

Thank You!

Revanth.C