Question

Four squares $P,Q,R$ and $S$ each of side $4\text{cm}$ and uniform thickness are kept together as shown in the figure. Then the position of the centre of mass of the combination of squares with respect to the centre of mass of $P$ will be:

• A. ($2.4\text{cm},2.8\text{cm})$
• B. ($4\text{cm},4\text{cm})$
• C. ($4\text{cm},2\text{cm})$
• D. ($2\text{cm},2\text{cm})$

Answer 1

The correct option is D ($2\text{cm},2\text{cm})$



Replacing the squares with a point mass at their respective ${\text{COM}}^{\prime }s$.
$\text{COM}$ will lie at the geometric centre of uniform shapes. Thus,
$\text{COM}$ of square $P(x_1,y_1)=(2,2)$
$\text{COM}$ of square $Q(x_2,y_2)=(6,2)$
$\text{COM}$ of square $R(x_3,y_3)=(6,6)$
$\text{COM}$ of square $S(x_4,y_4)=(2,6)$
Area of each square $=16{\text{cm}}^2$

Let $(x_{\text{CM}},y_{\text{CM}})$ be the position of $\text{COM}$ of the system :
Then,
$x_{\text{CM}}=\frac{A_1{x}_1+A_2{x}_2+A_3{x}_3+A_4{x}_4}{A_1+A_2+A_3+A_4}$
$=\frac{16\times 2+16\times 6+16\times 6+16\times 2}{16+16+16+16}$
$\therefore x_{\text{CM}}=\frac{16(2+6+2+6)}{16\times 4}=4\text{cm}$

Similarly:
$y_{\text{CM}}=\frac{A_1{y}_1+A_1{y}_2+A_3{y}_3+A_4{y}_4}{A_1+A_2+A_3+A_4}$
$=\frac{16\times 2+16\times 2+16\times 6+16\times 6}{16\times 4}=4\text{cm}$

Therefore, position vector of $\text{COM}$ of system is:
$\overrightarrow{r_S}=4\hat{i}+4\hat{j}\text{cm}$

Position vector of $\text{COM}$ of system w.r.t to $\text{COM}$ of square $P$:
$\overrightarrow{r_{SP}}=\overrightarrow{r_S}-\overrightarrow{r_P}$
$=(4\hat{i}+4\hat{j})-(2\hat{i}+2\hat{j})$
$\therefore \overrightarrow{r_{SP}}=(2\hat{i}+2\hat{j})\text{cm}$

or, in cartesian coordinates, position of $\text{COM}$ w.r.t that of P can be represented by $(x,y)=(2\text{cm},2\text{cm})$