To find the centre of mass of rigid body why it is not possible to know$\sum {{m_i}\overrightarrow {{r_i}} } $ for all the particles ?
Mention the centre of mass of three particles which are not in line but have equal masses.
Consider a two particle system with particles having masses $m_1$ and $m_2$. If the first particle is pushed towards the center of mass through a distance $d$, by what distance should the second particle is moved, so as to keep the centre of mass at the same position?
Three particles of masses $50\, g$, $100\, g$ and $150\, g$ are placed at the vertices of an equilateral triangle of side $1\, m$ (as shown in the figure). The $(x, y)$ coordinates of the centre of mass will be
Center of mass of a system of three particles of masses $1, 2, 3\, kg$ is at the point $(1\, m, 2\, m, 3\, m)$ and center of mass of another group of two particles of masses $2 \,kg$ and $3\, kg$ is at point $(-1 \,m, 3\, m, -2\, m)$ . Where a $5\, kg$ particle should be placed, so that center of mass of the system of all these six particles shifts to center of mass of the first system?