Question

If a square shape of side $5\text{cm}$ is cut from a square sheet of side $10\text{cm}$. Find the sum of $x$ and $y$ coordinates of the centre of mass of the remaining part as shown in the figure below. Consider the origin at the corner as shown in figure. Material is considered to be of uniform density.

• A. $11.66\text{cm}$
• B. $17\text{cm}$
• C. $6.6\text{cm}$
• D. $2.5\text{cm}$

Answer 1

The correct option is A $11.66\text{cm}$

Consider the removed square$(\text{side}=5\text{cm})$ as the negative mass superimposed on the square $(\text{side}=10\text{cm})$

Area of larger square
$A_1={10}^2{\text{cm}}^2$
& Area of smaller square
$A_2=5^2{\text{cm}}^2$

Their respective coordinates of COM are given by:
$(x_1,y_1)=(5\text{cm},5\text{cm})$
$(x_2,y_2)=(2.5\text{cm},2.5\text{cm})$

Let $\sigma$ be the areal density of the sheet.
For the COM of the remaining part :

$x_{\text{CM}}=\frac{\sigma A_1{x}_1-\sigma A_2{x}_2}{\sigma A_1-\sigma A_2}=\frac{\left(\right({10}^2\times 5)-(5^2\times 2.5\left)\right)}{({10}^2-5^2)}$
$=\frac{500-62.5}{100-25}=5.83\text{cm}$

Simillarly;
$y_{\text{CM}}=\frac{\sigma A_1{y}_1-\sigma A_2{y}_2}{\sigma A_1-\sigma A_2}=5.83\text{cm}$

Hence, sum of the coordinates of COM$=5.83+5.83=11.66\text{cm}$