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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: MOI small doubt..urgent..
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ananth_patri (595)

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From a circular disc of radius R and mass 9M a small disc of radius R/3 is removed..the moment of inertia of the remaining disc abt an axis perpendicular to the plane passing through the centre is ??? i have a small doubt.... after cal the MOI of the removed portion,,should we shift it to the centre of big disc b4 sutractin?? or just directly subtract???plz reply fast...its urgent...post u r answer.....


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ananth_patri (595)

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plz reply guys>????

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karthik2007 (3399)

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Understand that Moment of Inertia is the rotational analogue of mass. Just like how you can add up masses of individual bodies to get the total mass of the system, you can add up Moment of Inertia of various bodies about the same axis to get the net moment of Inertia of the system

As for the problem:

Find the MI of the big disc about the center. Then find the MI of the small disc about the center of the big disc using parallel axis theorem. Subtract the two to get the MI of the remaining part of the disc.

Will nip in at times to solve problems :)
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ananth_patri (595)

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well thanx for clearin my doubt... .....

The only place where you will find success before hard work is in the dictionary
A winner never quits.....a quitter never wins......
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sboosy (3063)

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\mbox{As karthik says} \\ \\ \mbox{Big disc about centre is} \ \frac{9MR^2}{2} \\ \\ \pi R^2 = 9M \ \Rightarrow \frac{\pi R^2}{9} = M \\ \\ \mbox{Removed disc about its centre is} \ \frac{MR^2}{18} \\ \\ \mbox{Thus about big disc
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