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From a circular disc of radius R and mass 9M, a small disc of radius R/3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is |
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Answer» `4M^(2)` `THEREFORE` Mass of removed portion `=(9M)/(piR^(2))xxpi(R/3)^(2)=M` LET moment of INERTIA of removed portion = `I_(1)` `thereforeI_(1)=M/2(R/3)^(2)+M((2R)/3)^(2)`, by theorem of parallel or `I_(1)=(MR^(2))/2` Let `I_(2)` = Moment of inertia of the whole disc `I_(2)=(9MR^(2))/2` `therefore` Let I = Moment of inertia of remaining disc `thereforeI=I_(2)-I_(1)` or `I=(9MR^(2))/2-(MR^(2))/2=(8MR^(2))/2=4MR^(2)` or `I=4MR^2` |
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