Seven identical discs each of mass M and radi | vedclass

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(D) The moment of inertia $(I)$ of a single circular disc about an axis passing through its center and perpendicular to its plane is $I_{cm} = \frac{M

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$\frac{55}{2} MR^2$

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(D) The moment of inertia $(I)$ of a single circular disc about an axis passing through its center and perpendicular to its plane is $I_{cm} = \frac{MR^2}{2}$.For the system of seven discs,the total moment of inertia about the axis passing through the center of the central disc (point $O$) and normal to the plane is the sum of the moment of inertia of the central disc and the six surrounding discs.$1$. For the central disc: The axis passes through its center,so its moment of inertia is $I_1 = \frac{MR^2}{2}$.$2$. For the six surrounding discs: The distance between the center of the central disc and the center of any surrounding disc is $d = 2R$. Using the parallel axis theorem,the moment of inertia of each surrounding disc about the central axis is $I_2 = I_{cm} + Md^2 = \frac{MR^2}{2} + M(2R)^2 = \frac{MR^2}{2} + 4MR^2 = \frac{9MR^2}{2}$.Since there are six such surrounding discs,their total moment of inertia is $6 	imes I_2 = 6 	imes \frac{9MR^2}{2} = 27MR^2$.Total moment of inertia $I = I_1 + 6I_2 = \frac{MR^2}{2} + 27MR^2 = \frac{MR^2 + 54MR^2}{2} = \frac{55MR^2}{2}$.

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