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(C) Let the radius of the original circular plate be $R = 28\, cm$ and the radius of the removed circular part be $r = 21\, cm$.The center of the orig

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$9$

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(C) Let the radius of the original circular plate be $R = 28\, cm$ and the radius of the removed circular part be $r = 21\, cm$.The center of the original plate is at $(0, 0)$. The center of the removed part is at $(R - r, 0) = (28 - 21, 0) = (7, 0)$.The mass of the original plate is $M = \sigma \pi R^2$ and the mass of the removed part is $m = \sigma \pi r^2$,where $\sigma$ is the surface mass density.The center of mass $X_{cm}$ of the remaining part is given by $X_{cm} = \frac{M X_1 - m X_2}{M - m}$.Substituting the values: $X_{cm} = \frac{(\sigma \pi R^2)(0) - (\sigma \pi r^2)(7)}{\sigma \pi R^2 - \sigma \pi r^2}$.$X_{cm} = \frac{-r^2 	imes 7}{R^2 - r^2} = \frac{-(21)^2 	imes 7}{(28)^2 - (21)^2}$.$X_{cm} = \frac{-441 	imes 7}{784 - 441} = \frac{-3087}{343} = -9\, cm$.The distance from the origin is $|-9| = 9\, cm$.

$A$ circular plate of uniform thickness with a diameter of $56\, cm$ has its center at the origin. $A$ circular part with a diameter of $42\, cm$ is removed from one edge. What is the distance of the center of mass of the remaining part in $cm$?

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