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A hole of radius R/2 is cut from a thin ...

A hole of radius R/2 is cut from a thin circular plate of raduis R as shown in the figure. If the mass of the remaining plate is M, then find moment of inertia of the plate about an axis through O perpendicular to plane

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The correct Answer is:
`57//140 MR^(2)`
density of solid sphere `=(M)/((4)/(3) piR^(3) -(4)/(3) pi((R)/(2))^(3))`
`rho =(M)/((4)/(3) piR^(3) (1-(1)/(8))) =(8M)/(7 xx(4)/(3) pi R^(3))`
Let us fill the cavity with mass `m`'
`m' = rho xx (4)/(3) pi((R)/(2))^(3)`
`m' =(8M)/(7 xx(4)/(3)pi R^(3)) xx(4)/(3) pi(R^(3))/(8) =(M)/(7)`
when cavity is filled with mass of density `rho` than the sphere is complete and its `I` about `O`.
`I_(o) = I_( r) + I_("cavity") [I_(r) = I of remaining mass]`
`(2)/(5)((8M)/(7)) R^(2) =I_(r) +[(2)/(5)(M)/(7) ((R)/(2))^(2) +(M)/(7) xx((R)/(2))^(2)]`
(Parallel axis theorem)
or `(2)/(5)((8M)/(7))R^(2) -(7)/(5) xx(M)/(7) .(R^(2))/(4) =I_(r)`
or `(16)/(35) MR^(2) -(MR^(2))/(20) =I_(r)`
`((64 -7)/(140))MR^(2) =I_(r) (57)/(140) MR^(2) =I_(r)`
Alternatively
Let mass of complete sphere of radiys `R` is `M'` and of taken out sphere of radius `R//2` is `M''`. So,
`M + M'' = M'` ....(1)
& `M'' = M'//8` ...(2)
from (1) & (2)
`M'=(8M)/(7) & M'' =(M)/(7)`
`I = I_(r) + I_("cavity")` [`I_(r) = I` of remaining mass]
`(2)/(5) ((8M)/(7)) R^(2) =I_(r) + [(2)/(5) (M)/(7) ((R)/(2))^(2) + (M)/(7) xx ((R)/(2))^(2)]`
`I_(r) = (16)/(35) MR^(2) -(MR^(2))/(20) rArr I_(r) =(57)/(140) MR^(2)`.
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