### Centre of mass of system of the discrete particles

The centre of mass of an object is a point that represents the entire body and moves in the same way as a point mass having mass equal to that of the object, when subjected to the same external forces that act on the object. That is, if the resultant force acting on an object (or system of objects) of mass m is F^{->} , the acceleration of the centre of mass of the object (or system) is given by a_{cm} = ( F/m).

If the system is composed of a system of particles having masses m_{1} , m_{2} , m_{3} and so on , at co-ordinates (x_{1}, y_{1}, z_{1}) , (x_{2}, y_{2}, z_{2}) and so on , then the co-ordinates of the centre of mass are given by

where Σm_{i} = M = total mass and the summations extend over all masses composing the object

**Illustration :** Find out the centre of mass of a system of three particles of 1 kg, 2 kg and 3 kg respectively; kept at the vertices of an equilateral triangle of side 1 m.

**Solution**: Assume the origin of the co-ordinate system at the position of the 1kg mass

= 2/3

= √3/6

**Exercise : **A thin uniform lamina is in the form of a circular disc of radius R. From it a circular hole is cut off having exactly half the radius of the lamina and touching the lamina’s circumference. Find the centre of mass of the remaining part

##### Centre of mass of continuous distribution of particles

Centre of mass of a body having continuous distribution of particles (mass) is given by

**Illustration :** Find the centre of mass of a uniform semi-circular ring of radius R and mass M.

**Solution:** Consider the centre of the ring as origin. Consider a differential element of length dl of the ring whose radius vector makes an angle θ with the x axis. Let λ be the mass per unit length. Then mass of this element is dm = λRdθ

x_{cm} = by symmetry,

y_{cm} = 2R/π

**Exercise :** Find the centre of mass of a uniform semi-circular plate of radius r and mass M.