A uniform circular disc has radius r. A squar | vedclass

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(A) Let the mass of the circular disc be $M$ and its radius be $r$. The area of the disc is $A = \pi r^2$.Let the origin $(0,0)$ be the centre of the

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

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(A) Let the mass of the circular disc be $M$ and its radius be $r$. The area of the disc is $A = \pi r^2$.Let the origin $(0,0)$ be the centre of the disc.$A$ square with diagonal $d = r$ is cut from it. The side of the square $a$ is given by $a = d / \sqrt{2} = r / \sqrt{2}$.The area of the square is $A_s = a^2 = r^2 / 2$.The mass of the square portion is $m = M 	imes (A_s / A) = M 	imes (r^2 / 2) / (\pi r^2) = M / (2 \pi)$.The centre of mass of the square is at a distance $d_s = r/2$ from the centre of the disc.Using the formula for the centre of mass of the remaining part: $X_{cm} = (M_1 X_1 - M_2 X_2) / (M_1 - M_2)$.Here,$M_1 = M$,$X_1 = 0$,$M_2 = m = M / (2 \pi)$,and $X_2 = r/2$.$X_{cm} = (M 	imes 0 - (M / 2 \pi) 	imes (r / 2)) / (M - M / 2 \pi)$.$X_{cm} = (-Mr / 4 \pi) / (M(1 - 1 / 2 \pi)) = (-r / 4 \pi) / ((2 \pi - 1) / 2 \pi) = -r / (2(2 \pi - 1)) = r / (2 - 4 \pi)$.

$A$ uniform circular disc has radius $r$. $A$ square portion of diagonal $r$ is cut from it. The centre of mass of the remaining disc from the centre of the disc is

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