From a circular disc of radius R,a triangular | vedclass

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(D) Let the area of the full disc be $A_1 = \pi R^2$ and its center of mass be at the origin $(0, 0)$.The triangular portion cut out is an isosceles r

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$\frac{R}{3(\pi - 1)}$

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(D) Let the area of the full disc be $A_1 = \pi R^2$ and its center of mass be at the origin $(0, 0)$.The triangular portion cut out is an isosceles right-angled triangle with base $2R$ and height $R$.The area of the triangle is $A_2 = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 2R 	imes R = R^2$.The center of mass of this triangle lies at a distance $h/3$ from the base,i.e.,$y_2 = R/3$ above the center $O$.The center of mass of the remaining portion is given by $Y_{cm} = \frac{A_1 Y_1 - A_2 Y_2}{A_1 - A_2}$.Since $Y_1 = 0$,we have $Y_{cm} = \frac{(\pi R^2)(0) - (R^2)(R/3)}{\pi R^2 - R^2} = \frac{-R^3/3}{R^2(\pi - 1)} = \frac{-R}{3(\pi - 1)}$.The magnitude of the distance from the center $O$ is $\frac{R}{3(\pi - 1)}$.

From a circular disc of radius $R$,a triangular portion is cut as shown in the figure. The distance of the center of mass $(COM)$ of the remaining disc from the center of the disc $O$ is:

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