Guest

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
148 Points
14 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.

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 Experts
Badiuddin


Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free