### 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 $\vec{F}$ , the acceleration of the centre of mass of the object (or system) is given by $a_{cm} = \frac{F}{m}$

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

$ \displaystyle X_{cm} =\frac{\sum m_i x_i}{\sum m_i} $

$ \displaystyle Y_{cm} =\frac{\sum m_i y_i}{\sum m_i} $

$ \displaystyle Z_{cm} =\frac{\sum m_i z_i}{\sum m_i} $

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

Solved Example : 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

$ \displaystyle X_{cm} =\frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} $

$ \displaystyle X_{cm} = \frac{1\times 0 + 2(1/2)+3\times 1}{1+2+3} $

= 2/3

$ \displaystyle Y_{cm} =\frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} $

$ \displaystyle Y_{cm} = \frac{1\times 0 + 2(\sqrt3 /2)+3\times 0}{1+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

$ \displaystyle X_{cm} = \frac{\int x dm}{\int dm} $

$ \displaystyle Y_{cm} = \frac{\int y dm}{\int dm} $

$ \displaystyle Z_{cm} = \frac{\int z dm}{\int dm} $

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,

$ \displaystyle Y_{cm} = \frac{\int y dm}{m} $

$ \displaystyle Y_{cm} = \frac{1}{m}\int_{0}^{\pi} (Rsin\theta)\lambda R d\theta $

$ \displaystyle Y_{cm} = \frac{\lambda R^2}{m}\int_{0}^{\pi} (sin\theta) d\theta $

$ \displaystyle Y_{cm} = \frac{2 R}{\pi} $

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

Sol: Let M = Mass of the semicircular disc

The centre of mass of the semicircular disc is

$\large r_{cm} = \frac{\int y_{cm} dm}{\int dm}$

Where y_{cm} = C.M. of the semicircular ring = 2y/π

$\large dm = \frac{M}{\pi r^2/2}(\pi y dy) = \frac{2 M}{r^2}y dy$

$\Large r_{cm} = \frac{\int_{0}^{r} (\frac{2 y}{\pi}) (\frac{2 M}{r^2}y dy)}{M}$

$\Large = \frac{4}{\pi r^2} \int_{0}^{r}y^2 dy $

$\large = \frac{4}{\pi r^2} [\frac{y^3}{3}]_{0}^{r}$

$\large = \frac{4 r}{3 \pi}$

### Also Read :

Velocity & Acceleration of Centre of mass Conservation of linear momentum & its Applications |