Question

From a square sheet of uniform density, a portion is removed shown shade in Fig Find the center of mass of the remaining portion if the side of the square is a

• A. $\left[(\frac{7}{18},\frac{a}{2})\right]$
• B. $\left[(\frac{8}{18},\frac{a}{2})\right]$
• C. $\left[(\frac{9}{18},\frac{a}{2})\right]$
• D. $\left[(\frac{71}{18},\frac{a}{2})\right]$

Answer 1

The correct option is A $\left[(\frac{7}{18},\frac{a}{2})\right]$

According to figure

$o{x}_1=\frac{\frac{a}{2}+a}{3}=\frac{a}{2}$

$o{y}_1=\frac{\frac{a}{2}+0+0}{3}=\frac{a}{6}$

$o{x}_2=\frac{\frac{a}{2}+a}{3}=\frac{a}{2}$

$o{y}_2=\frac{\frac{a}{2}+a+a}{3}=\frac{5a}{6}$

$o{x}_3=\frac{\frac{a}{2}+0+0}{3}=\frac{a}{2}$

$o{y}_3=\frac{a+\frac{a}{2}}{3}=\frac{a}{2}$

Now, center of mass

$X_{cm}=\frac{\frac{a}{2}m+\frac{a}{2}m+\frac{a}{6}m}{3m}$

$X_{cm}=\frac{3a+3a+a}{18}$

$X_{cm}=\frac{7a}{18}$

$Y_{cm}=\frac{\frac{ma}{6}+\frac{5ma}{6}+\frac{ma}{2}}{3m}$

$Y_{cm}=\frac{a+5a+3a}{18}$

$Y_{cm}=\frac{9a}{18}=\frac{a}{2}$

Hence, this is the required solution