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Moment of inertia due to remaining disc= Moment of inertia due to disc - moment of inertia due to disc having radius R/3
Due To disc I= 1/2 MR^2
Due to disc having Radius R/3
Mass of R/3 radius disc= M/pie r^2 × pie (R/3)^2
= M/9
I= 1/2 M ( R/3)^2 /9+ M ( R-R/3)^2 /9( parallel axis theorem)
=1/9( 1/2 M × R^2/9 + M ( 2R/3)^2)
= 1/9(MR^2/18 + M× 4R^2/9)
= 1/9(MR^2 + 8MR^2)/18)
= 1/9(MR^2/2)
= MR^2/18
So I= MR^2/2 - MR^2/18
= 9MR^2 - MR^2)/18
= 8MR^2/18
= 4MR^2/9
Thanks
Due To disc I= 1/2 MR^2
Due to disc having Radius R/3
Mass of R/3 radius disc= M/pie r^2 × pie (R/3)^2
= M/9
I= 1/2 M ( R/3)^2 /9+ M ( R-R/3)^2 /9( parallel axis theorem)
=1/9( 1/2 M × R^2/9 + M ( 2R/3)^2)
= 1/9(MR^2/18 + M× 4R^2/9)
= 1/9(MR^2 + 8MR^2)/18)
= 1/9(MR^2/2)
= MR^2/18
So I= MR^2/2 - MR^2/18
= 9MR^2 - MR^2)/18
= 8MR^2/18
= 4MR^2/9
Thanks
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