Question

From a uniform disc of radius R, a disc of radius $\frac{R}{2}$ is scooped out such that they have the common tangent. Find the centre of mass of the length remaining part

Answer 1

If the initial disc had a mass $M$, The cutout part would have a man $m_2=M/4$ (assuming man is uniforming distributed)

The center of mass of initial disc,

$\overrightarrow{r_1}=O\hat{i}+O\hat{j}$ and that of cutout disc

$\overrightarrow{r_2}=(\frac{R}{2}\hat{i}+O\hat{j})$ [from fig.]

Thus, $COM$ of remaining part,

$\overrightarrow{r}=\frac{m_1\overrightarrow{r_1}-m_2\overrightarrow{r_2}}{m_1-m_2}=\frac{M(O\hat{i}+O\hat{j})-\frac{M}{4}(\frac{R}{2}\hat{i}+O\hat{j})}{M-\frac{M}{4}}$

$=\frac{-\frac{MR}{8}\hat{i}+O\hat{j}}{\frac{3m}{4}}=(-\frac{R}{6}\hat{i}+O\hat{j})$