physicscatalyst.com logo






Center of mass : System of particles and Collision




(1) Introduction

  • Until now we have focused on describing motion of a single particle in one, two or three dimensions. By particle we mean to say that it has a size negligible in comparison to the path raveled by it.
  • When we studied Law's of motion we have applied them even to the bodies having finite size imagining that motion of such bodies can be described in terms of motion of particles.
  • While doing so we have ignored the the internal structure of such bodies. Any real body we encounter in our daily life has a finite size and idealized model of particle is inadequate when we deal with motion of real bodies of finite size .
  • Real bodies of finite size can also be regarded as the system of particles. While studying system of particles we will not concentrate on each and every particle of the system instead we will consider the motion of system as a whole.
  • Large number of problems involving extended bodies or real bodies of finite size can be solved by considering them as Rigid Bodies. We define rigid body as a body having definite and unchanging shape.
  • A rigid body is a rigid assembly of particles with fixed inter-particle distances. In actual bodies deformations do occur but we neglect them for the sake of simplicity.
  • In this chapter we will study about center of mass of system of particles, motion of center of mass and about collisions.

Introductory concepts

Let us learn about few important definitions before moving further

  1. Particle :- A particle is defined as an object whose mass is finite and it has a size that is negligible in comparison to the path raveled by it.
  2. System:- A system is a collection of a very large number of particles which mutually interact with one another. So, a body of finite size can be regarded as a system because it is made up of large number of particles interacting with one another.
  3. Internal forces:- The mutual forces exerted by the particles of system on one another are called internal forces.
  4. External forces:- The outside forces exerted on an object by any external agency is called external forces. Such a force changes the velocity of an object.

(2) Center of mass

  • Consider a body consisting of large number of particles whose mass is equal to the total mass of all the particles. When such a body undergoes a translational motion the displacement is produced in each and every particle of the body with respect to their original position.
  • If this body is executing motion under the effect of some external forces acting on it then it has been found that there is a point in the system , where whole mass of the system is supposed to be concentrated. The nature the motion executed by the system remains unaltered when force acting on the system is directly applied to this point. Such a point of the system is called center of mass of the system.
  • Hence for any system Center of mass is the point where whole mass of the system can be supposed to be concentrated and motion of the system can be defined in terms of the center of mass.

Difference between center of mass and center of gravity

The center of mass of a body i a point where whole mass of a body may be assumed to be concentrated for describing its translational motion. On the other hand the center of gravity is the point at which the resultant of the gravitational forces on all the particles of the body acts.
Please note that for many objects, these two points are in exactly the same place. But they're only the same when the gravitational field is uniform across an object. For example in uniform gravitational field such as that of the earth on a small body , the center of gravity coincides with the center of mass.




(3) Center of mass Formula

(i) Position of center of mass for two particle system

consider a system of two point masses (or particles) $m_1$ and $m_2$ , whose position vectors at time $t$ with reference to the origin $O$ of the inertial frame are $\vec{r_1}$ and $\vec{r_2}$ respectively as shown below in the figure.

centre of mass
The total force $(\vec{F_1})_{tot}$ that is acting on point mass $m_1$ consists of two parts
(i) A force $(\vec{F_1})_{ext}$ which appears because of some agency acting on the system
(ii) A force $\vec{F_{12}}$ which appears because of point charge $m_2$. This force is internal force of the two particle system.
So, Total force is
\begin{equation} (\vec{F_1})_{tot}=\vec{F_{12}}+(\vec{F_1})_{ext} \tag{1} \end{equation} Similarly for point mass $m_2$ \begin{equation} (\vec{F_2})_{tot}=\vec{F_{21}}+(\vec{F_2})_{ext} \tag{2} \end{equation} Now we will write the equation of motion for point mass $m_1$ using Newton's second law of motion
\begin{equation} \frac{d}{dt}(m_1\vec{v_1})=\vec{F_1})_{tot} \tag{3} \end{equation} Similarly for point mass $m_2$ \begin{equation} \frac{d}{dt}(m_2\vec{v_2})=\vec{F_2})_{tot} \tag{4} \end{equation} Adding equations (3) and (4) we get
\begin{equation} \frac{d}{dt}(m_1\vec{v_1})+\frac{d}{dt}(m_2\vec{v_2}) = \vec{F_1})_{tot} +\vec{F_2})_{tot} \\ \frac{d}{dt}(m_1\vec{v_1} + m_2\vec{v_2}) = \vec{F_1})_{tot} +\vec{F_2})_{tot} \end{equation} From equation (1) and (2) \begin{equation} \frac{d}{dt}(m_1\vec{v_1} + m_2\vec{v_2}) = \vec{F_{12}}+(\vec{F_1})_{ext} + \vec{F_{21}}+(\vec{F_2})_{ext} \end{equation} But from Newton's third law of motion we know that $\vec{F_{21}}=-\vec{F_{12}}$ so now we have
\begin{equation} \frac{d}{dt}(m_1\vec{v_1} + m_2\vec{v_2}) = (\vec{F_1})_{ext} + (\vec{F_2})_{ext} \\ \end{equation} Or,
\begin{equation} \frac{d}{dt}(m_1\vec{v_1} + m_2\vec{v_2}) = \vec{F} \tag{5} \end{equation} where, $\vec{F}=(\vec{F_1})_{ext} + (\vec{F_2})_{ext}$ is the total external force.
You must remember that $(\vec{F_1})_{ext}$ and $(\vec{F_2})_{ext}$ are acting on different points of the system and we are adding them as free vectors.
Now the velocity vectors are given by \begin{equation} \vec{v_1}=\frac{d\vec{r_1}}{dt} \quad \text{and} \quad \vec{v_2}=\frac{d\vec{r_2}}{dt} \end{equation} So, we have
\begin{align} m_1\vec{v_1} + m_2\vec{v_2} &= m_1\frac{d\vec{r_1}}{dt} + m_2\frac{d\vec{r_2}}{dt} \\ &=\frac{d}{dt}(m_1\vec{r_1})+\frac{d}{dt}(m_2\vec{r_2}) \\ &= \frac{d}{dt}(m_1\vec{r_1} + m_2\vec{r_2}) \end{align} From equation (5) we have \begin{equation} \vec{F}=\frac{d^2}{dt^2}(m_1\vec{r_1} + m_2\vec{r_2}) \end{equation} or,
\begin{equation} \vec{F}=M\frac{d^2}{dt^2}\frac{(m_1\vec{r_1} + m_2\vec{r_2})}{M} \end{equation} This equation is clearly the equation of motion of a hypothetical object of mass $M=m_1+m_2$. The position of this point at any time is given by position vector $\vec{R_{CM}}$ such that
\begin{equation} \vec{R_{CM}}=\frac{(m_1\vec{r_1} + m_2\vec{r_2})}{M} \tag{6} \end{equation} the point whose position is defined by $\vec{R_{CM}}$ is called the center of mass of two particle system. So, equation 6 gives the center of mass formula for a two particle system at any time $t$. It is a point at which the total external force is supposed to be acting.

Note:- It should be noted that it is not necessary that there may be a material particle at the center of mass of the system. Center of mass of a system is a hypothetical point and it may lie inside the system or outside it.

Center of mass formula observation

  1. The position vector $\vec{R_{CM}}$ of the center of mass $C$ of two particles is given by
    $$\vec{R_{CM}}=\frac{m_1\vec{r_1}+m_2\vec{r_2}}{m_1+m_2}$$ From above equation we can see that the position vector of a system of particles is the weighted average of the position vectors of the particles of which the system is made up of. Each particle in the system makes contribution proportional to its mass.
  2. From above equation we have
    $$(m_1+m_2)\vec{R_{CM}}=m_1\vec{r_1}+m_2\vec{r_2}$$ Here we can see that the product of the total mass of the system and the position vector of its center of mass is equal to the sum of the products of individual masses and their respective position vectors.
  3. Again if $m_1=m_2=m$ , then
    $$\vec{R_{CM}}=\frac{\vec{r_1}+\vec{r_2}}{2}$$ Thus , center of mass of two equal masses lie exactly at the center of the line joining the two masses.
  4. If $(x_1,y_1)$ and $(x_2,y_2)$ are the coordinates of the locations of two particles , then coordinates of their center of mass is given by
    $$x_{CM}=\frac{m_1x_1 + m_2x_2}{m_1 +x_1}$$ $$y_{CM}=\frac{m_1y_1 + m_2y_2}{m_1 +x_1}$$

(ii)Position of center of mass for many particle system

  • Consider a many particle system made up of number of particles as shown below in the figure. Let m1 , m2 , m3 , . . . . . . . . . . . . . , mn be the masses of the particles of system and their respective position vectors w.r.t. origin are r1 , r2 , r3 , . . . . . . . . . . . . . . . . . , rn.


    centre of mass formula
  • Because of the definition of center of mass

    Center of mass formula

Some conceptual questions on center of mass


Question 1 Is center of mass a hypothetical point or it really exists?
Answer The concept of center of mass of a system enable us to discuss overall motion of the system. It is defined as a point at which the entire mass of the bodies is supposed to be concentrated. So it is a hypothetical point only. So it is not necessary that the total mass of the system is to be exactly present at the center of mass. Hence we can say that it is a hypothetical point.
Question 2 How is the center of mass depends on the relative distance between the particles?
Answer Consider we have to find out the center of mass of the two masses having mass m Kg and 5m Kg. Now center of mass will lie closer to the heavier body (5m Kg) because center of mass is a point about which the whole mass of the system is balanced.
Question 3 Write two factors on which center of mass of a body does not depend.
Answer density, velocity
Question 4 Give the location of a center of mass of a ring? does it it lie inside the body?
Answer The center of mass of the ring lies in the center of the circle outlined by the ring. It is at the middle of the ring and does not lie inside the body. This is because, center of mass is a mathematical point where the mass of a body seems to be concentrated at.
Question 5 Define motion of center of mass of a fire cracker that explode in air before and after exposition"?
Answer A fire cracker that explodes in air, in the absence of air drag , the center of mass of the fragments would continue to follow the original parabolic path, until the fragmented parts began to touch the ground.

Some other resources you can look for reference are
Hope you like this Center of mass notes and these help you in your exams. If you like this article please share it among your friends.

Go Back to Class 11 Maths Home page Go Back to Class 11 Physics Home page


link to us