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(C) Let the surface mass density be $\sigma$. The area of the original disc is $A_1 = \pi (28)^2$ and the area of the removed disc is $A_2 = \pi (21)^

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$9$

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(C) Let the surface mass density be $\sigma$. The area of the original disc is $A_1 = \pi (28)^2$ and the area of the removed disc is $A_2 = \pi (21)^2$. The mass of the original disc is $M = \sigma A_1$ and the mass of the removed part is $m = \sigma A_2$.The centre of mass of the original disc is at the origin $(0,0)$. The centre of the removed disc is at $x = 28 - 21 = 7 \ cm$.Using the formula for the centre of mass of a system with a cavity: $X_{cm} = \frac{M X_1 - m x_2}{M - m}$.Here,$X_1 = 0$ and $x_2 = 7 \ cm$.$X_{cm} = \frac{(\sigma \pi 28^2)(0) - (\sigma \pi 21^2)(7)}{\sigma \pi 28^2 - \sigma \pi 21^2} = \frac{-21^2 	imes 7}{28^2 - 21^2}$.$X_{cm} = \frac{-441 	imes 7}{(28-21)(28+21)} = \frac{-441 	imes 7}{7 	imes 49} = \frac{-441}{49} = -9 \ cm$.The negative sign indicates the centre of mass shifts $9 \ cm$ away from the centre of the original disc,towards the side opposite the cavity.

$A$ circular plate of uniform thickness has a diameter of $56 \ cm$. $A$ circular part of diameter $42 \ cm$ is removed from one edge. What is the position of the centre of mass of the remaining part in $cm$?

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