A thin uniform circular disc has mass 9M and | vedclass

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(B) Let the mass of the original disc be $M_{total} = 9M$ and its radius be $R$. The surface mass density is $\sigma = \frac{9M}{\pi R^2}$.The mass of

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$I_2 - I_1$

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(B) Let the mass of the original disc be $M_{total} = 9M$ and its radius be $R$. The surface mass density is $\sigma = \frac{9M}{\pi R^2}$.The mass of the cut-out portion of radius $r = R/3$ is $m = \sigma 	imes \pi r^2 = \frac{9M}{\pi R^2} 	imes \pi (R/3)^2 = M$.The moment of inertia of the full disc about the axis through $O$ is $I_{total} = \frac{1}{2} (9M) R^2 = 4.5 MR^2$.The moment of inertia of the cut-out portion about its own center is $I_{cm} = \frac{1}{2} m r^2 = \frac{1}{2} M (R/3)^2 = \frac{MR^2}{18}$.Using the parallel axis theorem,the moment of inertia of the cut-out portion about the axis through $O$ is $I_{cut} = I_{cm} + m d^2$,where $d = 2R/3$ is the distance from $O$ to the center of the cut-out part.$I_{cut} = \frac{MR^2}{18} + M(2R/3)^2 = \frac{MR^2}{18} + \frac{4MR^2}{9} = \frac{MR^2 + 8MR^2}{18} = \frac{9MR^2}{18} = 0.5 MR^2$.The moment of inertia of the remaining part is $I_{rem} = I_{total} - I_{cut} = 4.5 MR^2 - 0.5 MR^2 = 4 MR^2$.Given the options,if $I_1$ is the moment of inertia of the cut-out part about $O$ and $I_2$ is the moment of inertia of the full disc about $O$,the remaining part is $I_2 - I_1$.

$A$ thin uniform circular disc has mass $9M$ and radius $R$. $A$ small circular part of radius $R/3$ is cut from the disc as shown in the figure. Calculate the moment of inertia of the remaining part about an axis passing through the center $O$ and perpendicular to the plane of the disc.

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