$(n - 1)$ equal point masses each of mass m are placed at the vertices of a regular n-polygon. The vacant vertex has a position vector a with respect to the centre of the polygon. Find the position vector of centre of mass.
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Key concept: poaition vector of center of mass for n particle system. If a system consists of n particles of masses $m_1, m_2, m_3, ......., mn,$ whose positions vectors are $\vec{\text{r}}_1,\vec{\text{r}}_2,\ ...., \vec{\text{r}}_\text{n}$ respectively. then position vector of centre of mass,
$\vec{\text{r}}=\frac{{\text{m}}_1\vec{\text{r}}_1+\text{m}_2\vec{\text{r}}_2+\text{m}_3\vec{\text{r}}_3+\ ....\ +\text{m}_\text{n}\vec{\text{r}}_\text{n}}{\text{m}_1+\text{m}_2+\text{m}_3+\ ...\ +\text{m}_\text{n}}$ Hence the center of mass of n particles is a weighted average of the position vectors of n particles making up the system. The centre of mass of a regular n-polygon lies at its geometrical centre. Let position vector of each centre of mass or regular n polygon is r. (n - 1) equal point masses each of mass m are placed at (n - 1) vertices of the regular n-polygon, therefore, for its centre of mass, $\vec{\text{r}}_\text{cm}=\frac{(\text{n}-1)\text{mb}+\text{ma}}{(\text{n}-1)\text{m}+\text{m}}=0$
$(\because$ Centre of mass lies at centre$)$ Here, b = distance of (n - 1) masses form centre of the polygon, a = the distance of point mass, placed at the vacant vertex. $\Rightarrow(\text{n}-1)\text{mb}+\text{ma}=0$
$\Rightarrow\text{b}=-\frac{\text{a}}{(\text{n}-1)}$ The negative sign indicates that $\vec{\text{b}}$ is in a direction opposite to $\vec{\text{a}}$
$\vec{\text{r}}=\frac{{\text{m}}_1\vec{\text{r}}_1+\text{m}_2\vec{\text{r}}_2+\text{m}_3\vec{\text{r}}_3+\ ....\ +\text{m}_\text{n}\vec{\text{r}}_\text{n}}{\text{m}_1+\text{m}_2+\text{m}_3+\ ...\ +\text{m}_\text{n}}$ Hence the center of mass of n particles is a weighted average of the position vectors of n particles making up the system. The centre of mass of a regular n-polygon lies at its geometrical centre. Let position vector of each centre of mass or regular n polygon is r. (n - 1) equal point masses each of mass m are placed at (n - 1) vertices of the regular n-polygon, therefore, for its centre of mass, $\vec{\text{r}}_\text{cm}=\frac{(\text{n}-1)\text{mb}+\text{ma}}{(\text{n}-1)\text{m}+\text{m}}=0$
$(\because$ Centre of mass lies at centre$)$ Here, b = distance of (n - 1) masses form centre of the polygon, a = the distance of point mass, placed at the vacant vertex. $\Rightarrow(\text{n}-1)\text{mb}+\text{ma}=0$
$\Rightarrow\text{b}=-\frac{\text{a}}{(\text{n}-1)}$ The negative sign indicates that $\vec{\text{b}}$ is in a direction opposite to $\vec{\text{a}}$
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