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(B) Let the centre of the regular $n$-polygon be at the origin $(0, 0)$.For a regular $n$-polygon with $n$ equal masses $m$ placed at all its vertices

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$-a/(n-1)$

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(B) Let the centre of the regular $n$-polygon be at the origin $(0, 0)$.For a regular $n$-polygon with $n$ equal masses $m$ placed at all its vertices,the centre of mass is at the centre of the polygon,i.e.,$R_{CM} = 0$.Let $r_i$ be the position vector of the $i$-th vertex. Then,$\sum_{i=1}^{n} m r_i = 0$.We are given that $(n-1)$ masses are placed at the vertices,and one vertex with position vector $a$ is vacant.Let $R$ be the position vector of the centre of mass of these $(n-1)$ masses.The total mass of the system is $(n-1)m$.The equation for the centre of mass is: $R = \frac{\sum_{i=1}^{n-1} m r_i}{(n-1)m}$.From the property of the regular polygon,the sum of all position vectors is $\sum_{i=1}^{n} r_i = 0$.Therefore,$\sum_{i=1}^{n-1} r_i + a = 0$,which implies $\sum_{i=1}^{n-1} r_i = -a$.Substituting this into the centre of mass equation:$R = \frac{m(-a)}{(n-1)m} = -\frac{a}{n-1}$.

$(n - 1)$ 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 the centre of mass.

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