M and R are the mass and radius of a disc. A | vedclass

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(A) The moment of inertia of the complete disc of mass $M$ and radius $R$ about an axis passing through its centre $O$ and perpendicular to its plane

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) The moment of inertia of the complete disc of mass $M$ and radius $R$ about an axis passing through its centre $O$ and perpendicular to its plane is $I_1 = \frac{1}{2} MR^2$.The mass of the removed disc of radius $r = R/3$ is $m = \frac{M}{\pi R^2} 	imes \pi (R/3)^2 = \frac{M}{9}$.The moment of inertia of the removed disc about the same axis passing through $O$ is calculated using the parallel axis theorem: $I_2 = I_{cm} + md^2$,where $I_{cm} = \frac{1}{2} m r^2$ and $d = R - r = R - R/3 = 2R/3$.$I_2 = \frac{1}{2} \left(\frac{M}{9}ight) \left(\frac{R}{3}ight)^2 + \left(\frac{M}{9}ight) \left(\frac{2R}{3}ight)^2 = \frac{MR^2}{162} + \frac{4MR^2}{81} = \frac{MR^2 + 8MR^2}{162} = \frac{9MR^2}{162} = \frac{MR^2}{18}$.The moment of inertia of the remaining part is $I = I_1 - I_2 = \frac{1}{2} MR^2 - \frac{1}{18} MR^2 = \frac{9-1}{18} MR^2 = \frac{8}{18} MR^2 = \frac{4}{9} MR^2$.Comparing this with $\frac{4}{x} MR^2$,we get $x = 9$.

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