# Center Of Mass IIT JEE Study Material

Centre of mass is the point where all the mass of an object is concentrated. If you support a body from its center of mass, the net torque on the body will be zero and the body will be in static equilibrium. The center of gravity is based on weight, while the center of mass is based on mass. When gravitational field across the mass is uniform, the center of mass coincides with the center of gravity.

## Centre of mass (COM) of an object having ‘n’ Discrete Particles:

Consider an object having ‘n’ point masses m1 , m2, . . . . . . . . mn and r1, r2, . . . . . . rn be its position vector from origin then,

$\mathbf{\overrightarrow{r}_{cm}\;=\;\frac{m_{1}\;\overrightarrow{r}_{1}\;+\;m_{2}\overrightarrow{r}_{2}\;+\;.\;.\;.\;.\;.\;+\;m_{n}\;\overrightarrow{r}_{n}}{m_{1}\;+\;m_{2}\;+\;.\;.\;.\;.\;.\;.\;+m_{n}}}$

## Centre of Mass of some common systems:

A. System with two point masses m1 and m2, the centre of mass (COM) lies closer to the heavier mass.

B. Rectangular plate: [L = length, b = breadth]

$\mathbf{x_{c}\;=\;\frac{b}{2}\;\;and\;\;y_{c}\;=\;\frac{L}{2}}$

C. Triangular Plate: [h = height]

$\mathbf{y_{c}\;=\;\frac{h}{3}}$ [At Centroid]

D. Semi-circular Ring: [R = Radius of Ring]

$\mathbf{y_{c}\;=\;\frac{2\;R}{\pi }\;\;and\;\;x_{c}\;=\;At\;\;Center\;(O)}$

E. Semi-circular Disc: [R = Radius of Disc]

$\mathbf{y_{c}\;=\;\frac{4\;R}{3\;\pi }\;\;and\;\;x_{c}\;=\;At\;\;Center\;(O)}$

F. Hemispherical shell: [R = Radius of Hemisphere]

$\mathbf{y_{c}\;=\;\frac{R}{2}\;\;and\;\;x_{c}\;=\;At\;\;Center\;(O)}$

G. Solid hemisphere: [R = Radius of Hemisphere]

$\mathbf{y_{c}\;=\;\frac{3\;R}{8}\;\;and\;\;x_{c}\;=\;At\;\;Center\;(O)}$

H. Solid circular cone: [h = height of cone]

$\mathbf{y_{c}\;=\;\frac{h}{4}}$

I. Hollow circular cone: [h = height of cone]

$\mathbf{y_{c}\;=\;\frac{h}{3}}$

Having a good knowledge of center of mass helps you to understand mechanics better which lays the foundation for higher physics. Some of the FAQ’s related to this topic is discussed below.

1. How to calculate the center of mass of a two-particle system?

Sol: Let there be two masses m and M, separated by a distance d.

Total mass of the system is $(m ~+ ~M)$. The distance of the center of mass of the system from the mass $m$ will be:

$\frac{Md}{M~+~m}$

And the distance of the center of mass of the system from the mass $M$ will be:

$\frac{md}{M~+~m}$

This explains that the center of mass of a two-particle system will always be on the side of heavier mass. We can also say that the center of mass divides internally the line joining the two particles in the inverse ratio of their masses.

2.How to calculate the center of mass of a group of particles?

Sol:

Suppose there are n particles having masses $m_1, m_2, m_3, …………m_n$ with $x_1, x_2, x_3 ……………. x_n$ as $x$ coordinates, and $y_1, y_2, y_3 …………… y_n$ as $y$ coordinates respectively. The x coordinate of center of mass is given as:

$X$ = $\frac{m_1 x_1~+~ m_2 x_2~+~ m_3 x_3~+~ …..m_n x_n}{m_1~+~ m_2~+~m_3~+~ …..~m_n}$

And, the $y$ coordinate of center of mass is:

$Y$ = $\frac{m_1 y_1~+~ m_2 y_2~+~ m_3 y_3~+~ …..m_n y_n}{m_1~+~ m_2~+~m_3~+~ …..m_n}$

3. Explain the motion of the center of mass.

Sol:

Imagine a system consisting of several masses, and let the collective mass of the system be $M$. Let $F$ be the net external force acting on the system. Acceleration of the center of mass:

$a_{cm}$ = $\frac{F}{M}$

So we can say that internal forces don’t affect the motion of the center of mass, and if there is no external force on the system, the center of mass has no acceleration. If there is no external force acting on the system, and its center of mass is at rest initially. Then the center of mass remains fixed even when the particles individually move and accelerate. Thus the motion of the center of mass of a system is identical to the motion of a single particle of mass equal to the mass of the given system, acted upon by the same external forces that act on the system. To learn more about the center of mass, get connected to our mentors here at Byju’s.

#### Practise This Question

What are the products of the reaction, CH3COOH+Na2CO3 ?