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

Options

A. `4 MR^2`
B. `(40)/(9) MR^2`
C. `40 MR^2`
D. `(37)/(9) MR^2`

Answer

A

Explanation

(a) Mass per unit area of disc `(9 M)/(pi R^2)`
Mass of removed portion of disc
=`(9 M)/(pi R^2) xx pi ((R )/(3))^2 = M`
Moment of inertia of removed portion about an axis passing through centre of disc and perpendicular the plane of disc, using theorem of parallel axes is.
`I_1 = (M)/(2)((R )/(3))^2 + M((2 R)/(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 = (9)/(2) MR^2`
So moment of inertia of the disc with remove portion, about the given axis is.
`I = I_2 - I_1 = (9)/(2) MR^2 -(1)/(2) MR^2 = 4 MR^2`.
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