Four particles of masses $1\,kg, 2 \,kg, 3 \,kg$ and $4\, kg$ are placed at the four vertices $A, B, C$ and $D$ of a square of side $1\, m$ as shown in the figure. The coordinates of the centre of mass of the particles are:

$A$ straight rod of length $L$ has one of its ends at the origin and the other at $x = L$. If the mass per unit length of the rod is given by $\lambda = Ax$ (where $A$ is a constant),then where is its center of mass from the origin?

What is the centre of mass?

The coordinates of the positions of particles of mass $7 \, g$,$4 \, g$,and $10 \, g$ are $(1, 5, -3) \, cm$,$(2, 5, 7) \, cm$,and $(3, 3, -1) \, cm$ respectively. The position of the centre of mass of the system would be:

The balls $A, B$ and $C$ of masses $50 \ g, 100 \ g$ and $150 \ g$,respectively,are placed at the vertices of an equilateral triangle. The length of each side is $1 \ m$. If $A$ is placed at $(0,0)$ and $B$ is placed at $(1,0) \ m$,find the coordinates $(x, y)$ for the centre of mass of this system of the balls.

Three identical spheres,each of mass $M$,are placed at the corners of a right-angled triangle with mutually perpendicular sides equal to $2 \; m$ (see figure). Taking the point of intersection of the two mutually perpendicular sides as the origin,find the position vector of the center of mass.