The coordinates of the center of mass of a un | vedclass

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(D) Divide the lamina into two rectangular plates: Plate-$1$ and Plate-$2$.Plate-$1$ has dimensions $1 \; m 	imes 3 \; m$,so its area is $A_{1} = 3 \

Vedclass · Accepted Answer

$(0.75 \; m, 1.75 \; m)$

Vedclass · Answer

(D) Divide the lamina into two rectangular plates: Plate-$1$ and Plate-$2$.Plate-$1$ has dimensions $1 \; m 	imes 3 \; m$,so its area is $A_{1} = 3 \; m^{2}$.Plate-$2$ has dimensions $1 \; m 	imes 1 \; m$,so its area is $A_{2} = 1 \; m^{2}$.Since the lamina is uniform,the mass is proportional to the area. Total area $A = A_{1} + A_{2} = 4 \; m^{2}$.Given total mass $M = 4 \; kg$,the mass of each part is $m_{1} = 3 \; kg$ and $m_{2} = 1 \; kg$.The center of mass of Plate-$1$ is at $(x_{1}, y_{1}) = (0.5 \; m, 1.5 \; m)$.The center of mass of Plate-$2$ is at $(x_{2}, y_{2}) = (1.5 \; m, 2.5 \; m)$.The $x$-coordinate of the center of mass is $x_{cm} = \frac{m_{1}x_{1} + m_{2}x_{2}}{m_{1} + m_{2}} = \frac{3 	imes 0.5 + 1 	imes 1.5}{4} = \frac{1.5 + 1.5}{4} = 0.75 \; m$.The $y$-coordinate of the center of mass is $y_{cm} = \frac{m_{1}y_{1} + m_{2}y_{2}}{m_{1} + m_{2}} = \frac{3 	imes 1.5 + 1 	imes 2.5}{4} = \frac{4.5 + 2.5}{4} = \frac{7}{4} = 1.75 \; m$.Thus,the coordinates are $(0.75 \; m, 1.75 \; m)$.

The coordinates of the center of mass of a uniform flag-shaped lamina (thin flat plate) of mass $4 \; kg$ (the coordinates of the same are shown in the figure) are:

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