A square plate of side 12 cm is shown in the | vedclass

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Question

(A) Let the mass of the original square plate be $M$. Let the mass of the small square piece cut out be $m$. The mass of the remaining part is $(M - m

Vedclass · Accepted Answer

$\left( - \frac{5m}{M - m}, - \frac{5m}{M - m} \right) \ cm$

Vedclass · Answer

(A) Let the mass of the original square plate be $M$. Let the mass of the small square piece cut out be $m$. The mass of the remaining part is $(M - m)$.The center of mass of the original plate of mass $M$ is at $(0, 0)$.The center of mass of the cut-out square piece of mass $m$ is at $(5, 5) \ cm$ (since the side is $12 \ cm$,the corner is at $(6, 6)$,and the center of a $2 \ cm$ square at that corner is at $(6-1, 6-1) = (5, 5)$).Let the center of mass of the remaining part of mass $(M - m)$ be $(x, y)$.Using the principle of center of mass: $M(0, 0) = m(5, 5) + (M - m)(x, y)$.This gives $m(5, 5) + (M - m)(x, y) = (0, 0)$.Equating the $x$ and $y$ components:$5m + (M - m)x = 0 \implies x = - \frac{5m}{M - m}$.$5m + (M - m)y = 0 \implies y = - \frac{5m}{M - m}$.Thus,the center of mass is $\left( - \frac{5m}{M - m}, - \frac{5m}{M - m} \right) \ cm$.

$A$ square plate of side $12 \ cm$ is shown in the figure. If a square of side $2 \ cm$ is cut from one of its corners,where will the center of mass of the remaining part be with respect to the center of the original square? The plate has uniform thickness and density.

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