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(A) Let the mass of the original full disc be $M$. Its center of mass is at the origin $(0, 0)$.The area of the original disc is $A = \pi R^2$. The ma

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$(0, -R/6)$

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(A) Let the mass of the original full disc be $M$. Its center of mass is at the origin $(0, 0)$.The area of the original disc is $A = \pi R^2$. The mass $M = \sigma \pi R^2$,where $\sigma$ is the surface mass density.The hole has radius $r = R/2$. Its area is $a = \pi (R/2)^2 = \pi R^2 / 4$. Its mass is $m = \sigma a = M/4$.The center of the hole is at $(0, R/2)$ because it touches the rim of the original disc.Treating the hole as a negative mass,the center of mass $(Y_{CM})$ of the remaining part is given by:$Y_{CM} = \frac{M(0) - m(R/2)}{M - m}$$Y_{CM} = \frac{0 - (M/4)(R/2)}{M - M/4} = \frac{-MR/8}{3M/4} = -\frac{MR}{8} 	imes \frac{4}{3M} = -R/6$.Since the system is symmetric about the $y$-axis,$X_{CM} = 0$.Thus,the center of mass of the remaining part is $(0, -R/6)$.

$A$ circular disc of radius $R$ is cut from a thin metal plate. $A$ hole of radius $R/2$ is cut from this disc such that it touches the rim of the original disc. Find the distance of the center of mass of the remaining part from the center of the original disc.

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