Question

From the circular disk of radius $4R$, two small discs of radius $R$ each are cut off. If $R=14\text{cm}$, then find the location of the center of mass of the new structure (in $\text{cm})$. (Take $\text{O}$ to be the origin)

• A. $(-14,-14)$
• B. $(-3,-3)$
• C. $(-56,-3)$
• D. $(0,0)$

Answer 1

The correct option is B $(-3,-3)$


Area of each removed part $A_2=A_3=\pi R^2$
Area of full circle $A_1=16\pi R^2$
Coordinates of centres of mass of removed parts are shown in the figure.
$(x_1,y_1)=(0,0),(x_2,y_2)=(3R,0),(x_3,y_3)=(0,3R)$

Then, x - coordinate of COM of remaining structure:
$x_{COM}=\frac{A_1{x}_1-A_2{x}_2-A_3{x}_3}{A_1-A_2-A_3}$
$=\frac{16\pi R^2\left(0\right)-\pi R^2\times 3R-\pi R^2\times 0}{14\pi R^2}$
$=\frac{-3\pi R^3}{14\pi R^2}$
$\therefore x_{COM}=\frac{-3R}{14}=-3\text{cm}$

Similarly
y-coordinate of the new structure:
$y_{COM}=\frac{A_1{y}_1+A_2{y}_2+A_3{y}_3}{A_1+A_2+A_3}$
$=\frac{16\pi R^2\left(0\right)-\pi R^2\times 0-\pi R^2\times 3R}{14\pi R^2}$
$\therefore y_{\text{COM}}=\frac{-3R}{14}=-3\text{cm}$

i.e $(x_{COM},y_{COM})=(-3,-3).$