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MI of a ring is mr^2. If we treat rings as elements of 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 corresponding ring in both from the axis will 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)

MI  of a  ring is mr^2.


If  we  treat rings as  elements of 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 corresponding ring in both from  the axis will 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)

Grade:11

1 Answers

Badiuddin askIITians.ismu Expert
147 Points
11 years ago

Dear neeru

it is because of that for same mass and radius surface area of disk and hemisphear is not same

for disk we right =dm = (M/πR2 ) 2πrdr


and for the hemisphear =dm = (M/2πR2) (2πRsinΘ)(RdΘ)


here we should consider that in case of dish ,element area is a area of ring which is  a plane surface

but in case of hemisphear element area is not plane urface it is a curved surface.

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