explain moment of inertia of hollow  cone

4 years ago

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Dear Amrit pal



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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Badiuddin



4 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 

3 years ago
                                        

 how to find monent of inertia of solide cone?

2 years ago

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