A circular hole of radius R/2 is cut from a c | vedclass

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(A) $1$. The mass of the original disc is $M$ and its radius is $R$. The moment of inertia of the original disc about an axis through its center and p

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$\left( \frac{13}{32} \right) MR^2$

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(A) $1$. The mass of the original disc is $M$ and its radius is $R$. The moment of inertia of the original disc about an axis through its center and perpendicular to its plane is $I_{disc} = \frac{1}{2} MR^2$.$2$. The hole has a radius $r = R/2$. The area of the original disc is $A = \pi R^2$. The area of the hole is $a = \pi (R/2)^2 = \frac{\pi R^2}{4} = A/4$.$3$. The mass of the removed part is $m = M 	imes (a/A) = M/4$.$4$. The center of the hole is at a distance $d = R/2$ from the center of the disc. The moment of inertia of the hole about its own central axis is $I_{hole, cm} = \frac{1}{2} m r^2 = \frac{1}{2} (M/4) (R/2)^2 = \frac{1}{2} (M/4) (R^2/4) = \frac{1}{32} MR^2$.$5$. Using the parallel axis theorem,the moment of inertia of the hole about the center of the disc is $I_{hole} = I_{hole, cm} + md^2 = \frac{1}{32} MR^2 + (M/4)(R/2)^2 = \frac{1}{32} MR^2 + \frac{1}{16} MR^2 = \frac{3}{32} MR^2$.$6$. The moment of inertia of the remaining part is $I_{rem} = I_{disc} - I_{hole} = \frac{1}{2} MR^2 - \frac{3}{32} MR^2 = \frac{16}{32} MR^2 - \frac{3}{32} MR^2 = \frac{13}{32} MR^2$.

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