From a circular disc of radius R,a square is | vedclass

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(A) Let the mass of the original circular disc be $M_1 = \sigma \pi R^2$ and its center of mass be at the origin $(0,0)$.Let the mass of the square cu

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$\frac{R}{4\pi - 2}$

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(A) Let the mass of the original circular disc be $M_1 = \sigma \pi R^2$ and its center of mass be at the origin $(0,0)$.Let the mass of the square cut out be $M_2$. The diagonal of the square is $R$,so its side length $a$ satisfies $a\sqrt{2} = R$,which means $a = R/\sqrt{2}$.The area of the square is $A_2 = a^2 = R^2/2$. Thus,$M_2 = \sigma (R^2/2)$.The center of mass of the square is at the center of the square,which is at a distance $d = a/2 = R/(2\sqrt{2})$ from the center of the disc along the diagonal.Let the diagonal lie along the $x$-axis. The center of mass of the square is at $(x_2, y_2) = (R/(2\sqrt{2}), R/(2\sqrt{2}))$.The center of mass of the remaining part $(X_{CM}, Y_{CM})$ is given by:$X_{CM} = \frac{M_1 X_1 - M_2 x_2}{M_1 - M_2} = \frac{0 - (\sigma R^2/2)(R/(2\sqrt{2}))}{\sigma \pi R^2 - \sigma R^2/2} = \frac{-R/(4\sqrt{2})}{\pi - 1/2} = \frac{-R}{4\sqrt{2}(\pi - 0.5)} = \frac{-R}{4\sqrt{2}\pi - 2\sqrt{2}}$.The distance from the center is $\sqrt{X_{CM}^2 + Y_{CM}^2} = \sqrt{2} |X_{CM}| = \sqrt{2} \frac{R}{4\sqrt{2}\pi - 2\sqrt{2}} = \frac{R}{4\pi - 2}$.

From a circular disc of radius $R$,a square is cut out with a radius as its diagonal. The center of mass of the remaining part is at a distance (from the centre) of:

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