A uniform square plate is shown in the figure | vedclass

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(C) Let the square plate be placed in the $XY$-plane with its center at the origin $O(0,0)$.Let the four corners be located at $(a, a), (-a, a), (-a,

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On $OX'$

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(C) Let the square plate be placed in the $XY$-plane with its center at the origin $O(0,0)$.Let the four corners be located at $(a, a), (-a, a), (-a, -a),$ and $(a, -a)$.Let the squares $1, 2, 3,$ and $4$ be at these corners respectively.When squares $1$ and $3$ (diagonally opposite) are removed,the remaining system is symmetric about the line passing through the centers of squares $2$ and $4$.This line corresponds to the diagonal $Y = -X$ (or $OX'$ axis depending on the coordinate orientation).Due to the symmetry of the remaining mass distribution about this diagonal,the center of mass must lie on this line of symmetry.Therefore,the center of mass will be located on the axis $OX'$.

$A$ uniform square plate is shown in the figure. Four identical squares are removed from its corners. If squares $1$ and $3$ are removed,where will the center of mass $(C.M.)$ be located?

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