thanks for theanswer of my post(on 3/9/2010).The question was:MI of a ring is mr^2.If we treat rings as elementsof a disc as well as of a hemisphere,then MI of a disc and of hemisphere should be same(becoz mass as well as perp. distance of every correspondingring in both from the axiswill be same,but actually not. What is wrong in the concept?(here axis is the axis passing through centre and perp. to plane in both cases)but if we donot use the concept : MI= ∫ dm.r^2 but simply see the relative positions of rings in disc and hemisphere(as someone has opened the disc to form a hemisphere),then what is wrong in that?MI(of both)=∑m.r^2 m=mass of every ring = same in both casesr=perp. distance of every ring from axis=same in both cases.

Dear neeru

In case of hemisphere you can use this concept because mass distribution will not be same throght out its radius.

in case of disk you will get uniform mass distribution  through out its area.

but if you will deal hemisphere as a disk then as you move away from the rotation axis then for same dr distance you will get more mass because hemisphere surface become more and more verticle . (mass distribution is not same in this case).

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