A uniform square plate has a side of length 2 | vedclass

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Question

(A) Let the mass of the original square plate be $M$ and its side length be $2R$. The area is $A = (2R)^2 = 4R^2$. The centre of mass is at the origin

Vedclass · Accepted Answer

$\frac{\pi R}{\sqrt{2}(16-\pi)}$

Vedclass · Answer

(A) Let the mass of the original square plate be $M$ and its side length be $2R$. The area is $A = (2R)^2 = 4R^2$. The centre of mass is at the origin $(0,0)$.When a circular piece of radius $r = R/2$ is removed from a quadrant,its area is $A' = \pi r^2 = \pi (R/2)^2 = \pi R^2 / 4$.The mass of the removed circular piece is $m = M 	imes (A'/A) = M 	imes (\pi R^2 / 4) / (4R^2) = M \pi / 16$.The centre of mass of the removed circular piece is at $(R/2, R/2)$ relative to the centre of the square.The shift in the centre of mass $\Delta x$ is given by $\Delta x = \frac{m \cdot d}{M - m}$,where $d$ is the distance of the centre of the circle from the centre of the square. The distance $d = \sqrt{(R/2)^2 + (R/2)^2} = \frac{R}{\sqrt{2}}$.Substituting the values: $\Delta x = \frac{(M \pi / 16) \cdot (R / \sqrt{2})}{M - M \pi / 16} = \frac{M \pi R / (16 \sqrt{2})}{M(16 - \pi) / 16} = \frac{\pi R}{\sqrt{2}(16 - \pi)}$.Thus,the correct option is $A$.

$A$ uniform square plate has a side of length $2R$. $A$ circular piece of maximum possible area is cut and removed from one of the quadrants of the plate as shown in the figure. Calculate the shift in the centre of mass of the plate.

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