Question

From a square of uniform density, a portion removed as shown in fig. Find the center of mass of the remaining portion if the side of the square is a.

Answer 1

$O_{x_1}=\frac{\frac{a}{2}+a}{3}=\frac{a}{2}$

$O_{y_1}=\frac{\frac{a}{2}+0+0}{3}=\frac{a}{6}$

$O_{x_2}=\frac{\frac{a}{2}+a}{3}=\frac{a}{2}$

$O_{y_2}=\frac{\frac{a}{2}+a+a}{3}=\frac{5a}{6}$

$O_{x_3}=\frac{\frac{a}{2}+0+0}{3}=\frac{a}{6}$

$O_{y_3}=\frac{\frac{a}{2}+a}{3}=\frac{a}{2}$

$X_{cm}=\frac{\frac{a}{2}\times m+\frac{a}{2}\times m+\frac{a}{6}\times m}{3m}=\frac{\frac{am}{2}+\frac{am}{2}+\frac{am}{6}}{3m}=\frac{3a+3a+a}{6\times 3}=\frac{7a}{18}$

$Y_{cm}=\frac{\frac{am}{6}+\frac{5am}{6}+\frac{am}{2}}{3m}=\frac{a+5a+3a}{6\times 3}=\frac{9a}{18}=\frac{a}{2}$

$(X_{cm},Y_{cm})=(\frac{7a}{18},\frac{a}{2})$