From a uniform circular disc of radius R and | vedclass

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(D) Let $\sigma$ be the mass per unit area of the disc.The total mass of the original disc is $M_{total} = 9M$.The radius of the original disc is $R$.

Vedclass · Accepted Answer

$4MR^2$

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(D) Let $\sigma$ be the mass per unit area of the disc.The total mass of the original disc is $M_{total} = 9M$.The radius of the original disc is $R$.The mass of the removed small disc of radius $r = \frac{R}{3}$ is:$m = \sigma 	imes \pi r^2 = \sigma 	imes \pi \left(\frac{R}{3}ight)^2 = \frac{\sigma \pi R^2}{9} = \frac{M_{total}}{9} = \frac{9M}{9} = M$.The moment of inertia of the complete disc of mass $9M$ about an axis passing through its centre $O$ and perpendicular to its plane is:$I_1 = \frac{1}{2}(9M)R^2 = \frac{9}{2}MR^2$.The moment of inertia of the removed disc of mass $M$ about its own centre $O'$ is:$I_{O'} = \frac{1}{2}M\left(\frac{R}{3}ight)^2 = \frac{1}{18}MR^2$.Using the parallel axis theorem,the moment of inertia of the removed disc about the axis passing through $O$ is:$I_2 = I_{O'} + M d^2$,where $d = \frac{2R}{3}$ is the distance between $O$ and $O'$.$I_2 = \frac{1}{18}MR^2 + M\left(\frac{2R}{3}ight)^2 = \frac{1}{18}MR^2 + \frac{4}{9}MR^2 = \left(\frac{1+8}{18}ight)MR^2 = \frac{9}{18}MR^2 = \frac{1}{2}MR^2$.The moment of inertia of the remaining disc is:$I = I_1 - I_2 = \frac{9}{2}MR^2 - \frac{1}{2}MR^2 = \frac{8}{2}MR^2 = 4MR^2$.

From a uniform circular disc of radius $R$ and mass $9M$,a small disc of radius $\frac{R}{3}$ is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through the centre of the original disc is

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