Question

from a uniform square plate the shaded portions areremoved as shown in the figure .Find the coordinates of center of mass of the remaiming plate X -and Y-axes and origin are shown in the figure​

Answer 1

As we know, the position of center of mass of the whole square plate is at $\sf{\left(\dfrac{a}{2},\ \dfrac{a}{2}\right)}$ and so its position vector is $\sf{\left<\dfrac{a}{2},\ \dfrac{a}{2}\right>.}$

Length of side of each square portion is $\sf{\dfrac{a}{4}.}$

Clearly we get that the position vector of the center of mass of the square portion in,

• 1st row and 1st column is $\sf{\left<\dfrac{a}{8},\ \dfrac{7a}{8}\right>.}$

• 2nd row and 2nd column is $\sf{\left<\dfrac{3a}{8},\ \dfrac{5a}{8}\right>.}$

• 3rd row and 3rd column is $\sf{\left<\dfrac{5a}{8},\ \dfrac{3a}{8}\right>.}$

Let mass of the square plate be M and each square portion be M'.

Area of square plate,

$\sf{\longrightarrow A=a^2}$

Area of a square portion,

$\sf{\longrightarrow A'=\left(\dfrac{a}{4}\right)^2=\dfrac{a^2}{16}}$

Since the square plate is uniform, the areal density of both the square plate and each square portion should be same, i.e.,

$\sf{\longrightarrow\dfrac{M}{A}=\dfrac{M'}{A'}}$

$\sf{\longrightarrow\dfrac{M}{a^2}=\dfrac{M'}{\left(\dfrac{a^2}{16}\right)}}$

$\sf{\longrightarrow M'=\dfrac{M}{16}}$

Then, position of center of mass of the remaining plate is,

$\longrightarrow \sf{\bar x}=\dfrac{\sf{M\left<\dfrac{a}{2},\ \dfrac{a}{2}\right>-M'\left<\dfrac{a}{8},\ \dfrac{7a}{8}\right>-M'\left<\dfrac{3a}{8},\ \dfrac{5a}{8}\right>-M'\left<\dfrac{5a}{8},\ \dfrac{3a}{8}\right>}}{\sf{M-3M'}}$

$\longrightarrow \sf{\bar x}=\dfrac{\sf{M\left<\dfrac{a}{2},\ \dfrac{a}{2}\right>-\dfrac{M}{16}\left<\dfrac{a}{8}+\dfrac{3a}{8}+\dfrac{5a}{8},\ \dfrac{7a}{8}+\dfrac{5a}{8}+\dfrac{3a}{8}\right>}}{\sf{M-\dfrac{3M}{16}}}$

$\longrightarrow \sf{\bar x}=\dfrac{\sf{\left<\dfrac{a}{2},\ \dfrac{a}{2}\right>-\dfrac{1}{16}\left<\dfrac{9a}{8},\ \dfrac{15a}{8}\right>}}{\sf{\dfrac{13}{16}}}$

$\longrightarrow \sf{\bar x}=\dfrac{\sf{\left<\dfrac{64a}{128},\ \dfrac{64a}{128}\right>-\left<\dfrac{9a}{128},\ \dfrac{15a}{128}\right>}}{\sf{\dfrac{13}{16}}}$

$\longrightarrow \sf{\bar x}=\dfrac{\sf{16}}{\sf{13}}\sf{\left<\dfrac{55a}{128},\ \dfrac{49a}{128}\right>}}$

$\longrightarrow\underline{\underline{\sf{\bar x}=\sf{\left<\dfrac{55a}{104},\ \dfrac{49a}{104}\right>}}}}$