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From a circular disc of radius R and mas...

From a circular disc of radius R and mass 9 M , 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

A

`4 MR^(2)`

B

`(40)/(9)MR^(2)`

C

`10 MR^(2)`

D

`(37)/(9)MR^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Mass per unit area of disc `= (9M)/(piR^(2))`
Mass of removed portion of disc
`= (9M)/(piR^(2)) xx pi ((R )/(3))^(2) = M`
Moment of inertia of removed portion about an axis passing through centre of disc and perpendicular to the plane of disc, using theorem of parallel axis is
`I_(1) = (M)/(2) ((R )/(3))^(2) + M ((2R)/(3))^(2) = (1)/(2)MR^(2)`
when portion of disc would not have been removed, then the moment of inertia of complete
disc about the given axis is `I_(2) = (1)/(2)MR^(2)`
So moment of inertia of the disc with removed portion, about the given axis is
`I = I_(2) - I_(1) = (9)/(2)MR^(2) - (1)/(2)MR^(2) = 4MR^(2)`
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