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(A) The moment of inertia of a complete uniform circular disc of mass $M_{total}$ and radius $R$ about an axis passing through its center and perpendi

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$ \frac{1}{2}MR^2 $

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(A) The moment of inertia of a complete uniform circular disc of mass $M_{total}$ and radius $R$ about an axis passing through its center and perpendicular to its plane is given by $I = \frac{1}{2}M_{total}R^2$.Since the disc is uniform,the mass is distributed proportionally to the area. $A$ quarter sector has $\frac{1}{4}$ of the total area,so its mass $M$ is $\frac{1}{4}M_{total}$,which means $M_{total} = 4M$.The moment of inertia of the quarter sector about the same axis is $\frac{1}{4}$ of the moment of inertia of the full disc.Therefore,$I_{sector} = \frac{1}{4} 	imes (\frac{1}{2} M_{total} R^2) = \frac{1}{4} 	imes (\frac{1}{2} 	imes 4M 	imes R^2) = \frac{1}{2}MR^2$.

$A$ quarter circular sector is cut from a uniform circular disc. The mass of this sector is $M$. It is rotated about an axis passing through the center of the original disc and perpendicular to its plane. The moment of inertia of this sector about the axis of rotation is:

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