Question

A square of side $4cm$ and of uniform thickness is divided into four equal squares. If one of them is cut off $(OECF)$, then the position of the centre of mass of the remaining portion from $O$ is

• A. $\frac{1}{\sqrt{3}}cm$
• B. $\frac{2}{\sqrt{3}}cm$
• C. $\frac{5}{\sqrt{3}}cm$
• D. $\frac{\sqrt{2}}{3}cm$

Answer 1

Answer: (D)

$W$ = Whole square $ABCD$
$C=$ Cut square $OECF$
$R=$ Remaining part of square i.e. shaded region.
Now center of mass of the whole square is at $O(0,0)$.
The $CM$ of cut square is at its centre namely $(1cm,-1cm)$ Let mass per unit area $=\sigma$
$x_W=\frac{m_C{x}_C+m_R{m}_R}{m_C+m_R}$
$0=\frac{\sigma \left(12c{m}^2\right)\cdot x_R+\sigma \left(4c{m}^2\right)\cdot (1cm)}{\sigma \left(16c{m}^2\right)}$

$\Rightarrow x_R=-\frac{1}{3}cm$
Similarly, $y_W=\frac{m_C{y}_C+m_R{y}_R}{m_C+m_R}$
$0=\frac{\sigma \left(12c{m}^2\right)\cdot \left(y_R\right)+\sigma \left(4c{m}^2\right)\cdot (-1cm)}{\sigma \left(16c{m}^2\right)}$
$\Rightarrow y_R=\frac{1}{3}cm$
Distance from origin of $\left(x_R,y_R\right)$
$=\sqrt{x_R^2+y_R^2}=\sqrt{{\left(\frac{1}{3}\right)}^2+{\left(\frac{1}{3}\right)}^2}$
$=\frac{\sqrt{2}}{3}cm$