A uniform square plate of side a has a square | vedclass

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(A) Let the mass of the full square plate of side $a$ be $M$. Its center of mass is at $(0, 0)$.The area of the full plate is $A_1 = a^2$.The area of

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$(a/12, a/12)$

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(A) Let the mass of the full square plate of side $a$ be $M$. Its center of mass is at $(0, 0)$.The area of the full plate is $A_1 = a^2$.The area of the hole of side $a/2$ is $A_2 = (a/2)^2 = a^2/4$.The mass of the hole $m$ is proportional to its area: $m = M 	imes (A_2 / A_1) = M/4$.The center of mass of the hole is at $(a/4, a/4)$.The center of mass of the remaining part $(X_{cm}, Y_{cm})$ is given by:$X_{cm} = \frac{M_1 X_1 - m X_2}{M - m} = \frac{M(0) - (M/4)(a/4)}{M - M/4} = \frac{-Ma/16}{3M/4} = -a/12$.Since the hole is in the first quadrant,the center of mass of the remaining part shifts in the opposite direction,i.e.,to $(-a/12, -a/12)$.However,looking at the standard coordinate system where the hole is at $(a/4, a/4)$,the magnitude of the shift is $a/12$ in the negative direction. Thus,the center of mass is at $(-a/12, -a/12)$.

$A$ uniform square plate of side $a$ has a square hole of side $a/2$ cut out of it as shown in the figure. The hole is centered at $(a/4, a/4)$ from the center of the plate. The center of mass of the remaining part is:

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