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                   explain moment of inertia of hollow  cone


5 years ago

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                                        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 =o∫h tan3Θ [2M/rh]x3dx
I=mr2/2
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed  solution very  quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin


5 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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