Question

Three discs of the same mass and same radius $R$ are placed as shown in figure. Find the $COM$ of the system.

• A. $(\frac{R}{\sqrt{3}},0)$
• B. $(0,\frac{R}{\sqrt{3}})$
• C. $(\frac{R}{3},0)$
• D. $(0,\frac{R}{3})$

Answer 1

The correct option is A $(\frac{R}{\sqrt{3}},0)$

From symmetry, $y_{COM}$ will be zero.
As we know,
$x_{COM}=\frac{A_1{x}_1+A_2{x}_2+A_3{x}_3}{A_1+A_2+A_3}$
Here, $A_1=A_2=A_3=\pi R^2$
$x_1=x_{COM}$ for disc $1$ from origin $\text{O}=0$
$x_2=x_{COM}$ for disc $2$ from origin $\text{O}=0$

From the figure, in $\Delta OAB$ by Pythagoras theorem


$O{A}^2+O{B}^2=A{B}^2$
Here $OA=R,AB=2R$
$R^2+O{B}^2=(2R{)}^2$
$\Rightarrow O{B}^2=4R^2-R^2$
$\Rightarrow OB=\sqrt{3}R$

$\therefore x_3=x_{COM}$ for disc 3 $=\sqrt{3}R$

Therefore for the system of discs,
$x_{COM}=\frac{\pi R^2\left(0\right)+\pi R^2\left(0\right)+\pi R^2\left(\sqrt{3}R\right)}{\pi R^2+\pi R^2+\pi R^2}$
$x_{COM}=\frac{\sqrt{3}R}{3}=\frac{R}{\sqrt{3}}$
$COM$ of the system of discs will be $\frac{R}{\sqrt{3}}$ from the origin towards right.