Question

From the circular disc of radius $4R$ two small discs of radius $R$ are cut off. The centre of mass of the new structure will be at

• A. $\hat{i}\frac{R}{5}+\hat{j}\frac{R}{5}$
• B. $-\hat{i}\frac{R}{5}+\hat{j}\frac{R}{5}$
• C. $-\hat{i}\frac{R}{5}-\hat{j}\frac{R}{5}$
• D. $none$ $of$ $these$

Answer 1

The correct option is D $none$ $of$ $these$

Two small discs which are taken out can be considered as having negative area for calculation of $CM$.
$A_1=\pi (4R{)}^2$;
$A_2=-\pi (R{)}^2$;
$A_3=-\pi (R{)}^2$;
Position vector for the COM of disc of radius $4R=\overrightarrow{r_1}$;
Position vector for the COM of disc-1 of radius $R=\overrightarrow{r_2}$;
Position vector for the COM of disc-2 of radius $R=\overrightarrow{r_3}$;

$\overrightarrow{r_{cm}}=\frac{A_1\overrightarrow{r_1}+A_2\overrightarrow{r_2}+A_3\overrightarrow{r_3}}{A_1+A_2+A_3}$;

$\overrightarrow{r_1}=0\hat{i}+0\hat{j}$;
$\overrightarrow{r_2}=3R\hat{i}+0\hat{j}$;
$\overrightarrow{r_3}=0\hat{i}+3R\hat{j}$;
Substituting $\overrightarrow{r_1},\overrightarrow{r_2},\overrightarrow{r_3}$in$\overrightarrow{r_{cm}}$ we get,

$\overrightarrow{r_{cm}}=\frac{\pi \times \left(4R{)}^2\right)\times 0-\pi \times (R{)}^2\times 3R\hat{i}-\pi \times (R{)}^2\times 3R\hat{j}}{\pi \times (4R{)}^2-\pi \times (R{)}^2-\pi \times (R{)}^2}=-\frac{3R}{14}\hat{i}-\frac{3R}{14}\hat{j}$