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. $\frac{R}{5}(\hat{i}+\hat{j})$
• B. $\frac{R}{5}(\hat{i}-\hat{j})$
• C. $-\frac{3R}{14}(\hat{i}+\hat{j})$
• D. $\frac{3R}{14}(\hat{i}+\hat{j})$

Answer 1

The correct option is C $-\frac{3R}{14}(\hat{i}+\hat{j})$

Centre of mass of circular disc of radius $4R$ is at $(0,0)$

Centre of mass of the upper disc of radius $R$ is at $(0,3R)$

Centre of mass of the lower disc of radius $R$ is at $(3R,0)$

Let, mass of complete disc, $m_{4R}=M$

Since, the density of the cut-out discs are same as the discs of radius $4R$, so the mass of cut out disc will be

$m_R=\frac{M}{\pi (4R{)}^{2}t}\times \pi R^{2}t=\frac{M}{16}(t=\text{thickness of the disc}$

Since the small disc is cut off from the big disc, so we will take negative mass for that.

$m_R=-\frac{M}{16}$

Hence, the centre of mass of the new disc at $x-$axis is,

$\overline{x}=\frac{m_{4R}{x}_{4R}+m_R{x}_R+m_R{x}_R}{m_{4R}+m_R+m_R}$

$=\frac{M\left(0\right)-\frac{M}{16}\left(0\right)-\frac{M}{16}\left(3R\right)}{M-\frac{M}{16}-\frac{M}{16}}=-\frac{3R}{14}$

Similarly, centre of mass at $y-$ axis,

$\overline{y}=-\frac{3R}{14}$

Thus, position vector of centre of mass,

${\overrightarrow{r}}_{com}=-\frac{3R}{14}(\hat{i}+\hat{j})$

Hence, option $\left(C\right)$ is the correct answer.