Question

Two discs of radii $4\text{cm}$ and $2\text{cm}$ respectively are attached as shown in the figure. The distance of the new centre of mass of the system from ${\text{C}}_1$ is

• A. $1.2\text{cm}$
• B. $2.4\text{cm}$
• C. $5.2\text{cm}$
• D. $3.2\text{cm}$

Answer 1

The correct option is A $1.2\text{cm}$

Let the position of centre of mass of disc 1, ${\text{C}}_1=(0,0)$
Then, position of centre of mass of disc 2, ${\text{C}}_2=(6,0)$
Area of disc 1, $A_1=\pi r_1^2$ $=(4\times {10}^{-2}{)}^2\pi =16\pi \times {10}^{-4}{\text{m}}^2$
Area of disc 2, $A_2=\pi r_2^2$ $=(2\times {10}^{-2}{)}^2\pi =4\pi \times {10}^{-4}{\text{m}}^2$
Let the distance of new centre of mass from ${\text{C}}_1$ be $r_{cm}$
Then,
$r_{cm}=\frac{A{r}_1+A_2{r}_2}{A_1+A_2}=\frac{16\pi \times {10}^{-4}\left(0\right)+(4\pi \times {10}^{-4})\left(6\right)}{16\pi \times {10}^{-4}+4\pi \times {10}^{-4}}$
$=\frac{24\pi \times {10}^{-4}}{20\pi \times {10}^{-4}}=1.2\text{cm}$