Question

From a uniform square plate, one fourth part is removed as shown in figure. The centre of mass of remaining part will lie on line joining

• A. $OA$
• B. $OB$
• C. $OC$
• D. $OD$

Answer 1

The correct option is A $OA$

For a laminar shape, when its some portion is removed then, new centre of mass is given by
${\overrightarrow{r}}_{\text{com}}=\frac{A_1{\overrightarrow{r}}_1-A_2{\overrightarrow{r}}_2}{A_1-A_2}...\left(1\right)$
$A_1\to$Larger area
$A_2\to$smaller area
Taking origin of at $B$, Y and X-axis along $AB$ and $BC$ respectively.


From figure,
$A_1=a^2$
$\overrightarrow{r_1}=\frac{a}{2}\hat{i}+\frac{a}{2}\hat{j}$
$A_2=\frac{a^2}{4}$
${\overrightarrow{r}}_2=\frac{3a}{4}\hat{i}+\frac{a}{4}\hat{j}$
Since $x-$coordinate of smaller plate's $\text{COM}$ is
$x_2=\frac{a}{2}+\frac{a}{4}=\frac{3a}{4}$
and $y_2=\frac{a}{4}$
Substituting the values in Eq.$\left(1\right)$,
${\overrightarrow{r}}_{\text{com}}=\frac{a^2[\frac{a}{2}\hat{i}+\frac{a}{2}\hat{j}]-\frac{a^2}{4}[\frac{3a}{4}+\frac{a}{4}\hat{j}]}{a^2-\frac{a^2}{4}}$
${\overrightarrow{r}}_{\text{com}}=\frac{a^2[\frac{a}{2}\hat{i}+\frac{a}{2}\hat{j}-\frac{3a}{16}\hat{i}-\frac{a}{16}\hat{j}]}{a^2\times [1-\frac{1}{4}]}$
${\overrightarrow{r}}_{\text{com}}=\frac{a^2[\frac{5a}{16}\hat{i}+\frac{7a}{16}\hat{j}]}{a^2\times \frac{3}{4}}$
${\overrightarrow{r}}_{\text{com}}=\frac{\frac{5a\hat{i}+7a\hat{j}}{16}}{\frac{3}{4}}$

${\overrightarrow{r}}_{\text{com}}=\frac{5a}{12}\hat{i}+\frac{7a}{12}\hat{j}$
$\therefore {\overrightarrow{r}}_{\text{com}}=0.42a\hat{i}+0.58a\hat{j}$


Hence centre of mass of remaining portion will lie on $OA$,as represented by its position vector.