A circular hole of radius 3 text{ cm} is cut | vedclass

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(B) Let the mass per unit area of the disc be $\sigma$.Total mass of the original disc,$M = \pi R^2 \sigma$,where $R = 6 	ext{ cm}$.Mass of the scoop

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$1$

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(B) Let the mass per unit area of the disc be $\sigma$.Total mass of the original disc,$M = \pi R^2 \sigma$,where $R = 6 	ext{ cm}$.Mass of the scooped-out portion,$M' = \pi r^2 \sigma$,where $r = 3 	ext{ cm}$.We take the centre $O$ of the original disc as the origin $(0,0)$.The centre of mass of the original disc is at $(0,0)$.The centre of the hole is at $(3,0)$.The centre of mass of the remaining portion is given by:$x_{CM} = \frac{M x_1 - M' x_2}{M - M'}$Substituting the values:$x_{CM} = \frac{(\pi R^2 \sigma)(0) - (\pi r^2 \sigma)(3)}{\pi R^2 \sigma - \pi r^2 \sigma}$$x_{CM} = \frac{-3 \pi r^2 \sigma}{\pi \sigma (R^2 - r^2)} = \frac{-3 r^2}{R^2 - r^2}$$x_{CM} = \frac{-3 	imes 3^2}{6^2 - 3^2} = \frac{-3 	imes 9}{36 - 9} = \frac{-27}{27} = -1 	ext{ cm}$.The negative sign indicates that the centre of mass shifts $1 	ext{ cm}$ to the left of the original centre. The distance is $1 	ext{ cm}$.

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