Complex objects have particles that show mechanism differently. When we work on a system of particles we need to know its centre of mass to calculate the mechanics of oddly shaped objects. Rigid bodies constitute a system of particles which govern its motion and equilibrium. With the Centre of Mass, we can effortlessly understand the mechanism of complicated objects. Let’s see how?

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## Centre of Mass

The centre of mass (CoM) is the point relative to the system of particles in an object. This is that point of the system of particles that embarks the average position of the system in relation to the mass of the object. At the centre of mass,Â theÂ weightedÂ massÂ gives a sumÂ equal to zero. It is the point where any uniform forceÂ appliedÂ on the object acts.In other words, a particle’sÂ centreÂ of massÂ isÂ the point where Newton’s law of motions applies perfectly. When force is applied to the centre of mass, the object as a system of particles moves in the direction of force without rotating. No matter what the shape of the object, the centre of mass helps understand the mechanism of force and motion of that object.

**Browse more Topics under System Of Particles And Rotational Dynamics**

- Introduction to Rotational Dynamics
- Vector Product of Two Vectors
- Motion of Centre of Mass
- Moment of Inertia
- Theorems of Parallel and Perpendicular Axis
- Rolling Motion
- Angular Velocity and Angular Acceleration
- Linear Momentum of System of Particles
- Torque and Angular Momentum
- Equilibrium of a Rigid Body
- Angular Momentum in Case of Rotation About a Fixed Axis
- Dynamics of Rotational Motion About a Fixed Axis
- Kinematics of Rotation MotionÂ about a Fixed Axis

## Centre of Mass for Two Particles

For a system of two particles with equal masses, CoM is the point that lies exactly in the middle of both.

In the figure above, we take two particles with masses m_{1} and m_{2} respectively lying on the x-axis. The distance of both the particles from the centre O is d_{1} and d_{2} respectively. The CoM of this two system of particles is at point C lying at a distance D from point O. In this system of two particles D can be written as:

D = m_{1}d_{1Â }+ m_{2}d_{2Â }/ m_{1Â }+ m_{2}

From the equation D, can be taken as the mass-weightedÂ mean of d1 and d2. Now, let us presume that the particles in the system have equal masses. Hence, m_{1}=m_{2Â }= m, in this case,

D = (md_{1Â }+ md_{2})/ 2m = m (d_{1Â }+ d_{2}) / 2m

D =Â (d_{1Â }+ d_{2}) / 2

From the equation above we get the centre of mass of two particles with equal masses. From the above equation, it is clear that the CoM of two particles lies in the midway of both.

## Centre of Mass for n Particles

For a system of n particles, the centre of mass, according to its definition is:

D = m_{1}d_{1Â }+ m_{2}d_{2Â }+ m_{3}d_{3Â }+…… m_{n}d_{nÂ }/ m_{1Â }+ m_{2Â }+ m_{3Â }+….. m_{nmÂ }=Â âˆ‘m_{i}d_{i} / m_{i}

m_{iÂ }here is the sum of the masses of the particles.

## Centre of Mass for Three Particles at Different Positions

It is not necessary that the system of particles spoken about for the centre of mass lie on the same axis of a straight line. There may be cases when particles lie on different lines away from each other. What shall be the CoM in such case?Â Since these particles are not in aÂ straight line hence there must be more than two particles.

Let’s take the case of three particles lying on different axes x and y. Here, we represent the positions of the particles as (d_{1},e_{1}), (d_{2},e_{2}) and (d_{3},e_{3}) along x and y-axes. Â The CoM C for the system of three particles is located at D and E and are given as:

D = m_{1}d_{1Â }+ m_{2}d_{2Â }+ m_{3}d_{3Â }/ m_{1Â }+ m_{2Â }+ m_{3}

E = m_{1}e_{1Â }+ m_{2}e_{2Â }+ m_{3}e_{3Â }/ m_{1Â }+ m_{2Â }+ m_{3}

Now, if the particles have equal masses m_{1Â }= m_{2Â }= m_{3Â }= m

then, D = m (d_{1Â }+ d_{2Â }+ d_{3}) / 3m = d_{1Â }+ d_{2Â }+ d_{3Â }/ 3

and E = m (e_{1Â }+ e_{2Â }+ e_{3}) / 3m = Â e_{1Â }+ e_{2Â }+ e_{3Â }/ 3. From the above equations, we come to a conclusion that for a system of three particles, the centre of mass lies at the centroid of the triangle formed by these three particles.

### Centre of Mass for a System of Particles Outside a Plane

As said earlier, Centre of Mass helps in calculating the mechanism of forces for complex objects, which encompass particles lying at different positions. Now, if the object has a complex shape and the system of particles rather than lying in a plane are distributed variably in such situation calculating the centre of mass depends on the position of the particles.

Here, since the particles lie in different planes we take the CoM of such particles to lie on D, E and F. So,

D =Â âˆ‘m_{i}d_{i }/ M

E = âˆ‘m_{i}e_{iÂ }/ M

F = âˆ‘m_{i}f_{i }/ M

âˆ‘m_{iÂ }here is the sum of the total mass of the system of particles, whileÂ m_{iÂ }is the mass of i^{thÂ }particle and the position of respective particles is given by d_{i},e_{i}Â and f_{i}. Using this information of position vectors we combine the above equations, to get,

R =Â âˆ‘m_{i}r_{iÂ }/ M

R is the position vector of the CoM and r_{i} is the position vector of the i^{thÂ }particle.

## Centre of Mass of Homogeneous Bodies

Homogeneous bodies are those objects which have a uniformly distributed mass around the body as a whole. A few examples for homogeneous bodies are spheres, rings etc. These rigid bodies have a regular shape and for calculating the CoM for these we keep in mind the symmetry between the particles. According to our symmetric considerations, we can assume that the CoM for these regular bodies lies at their geometric centres.

## Solved Examples for You

Question: Two spheres of mass MÂ and 7MÂ are connected by a rod whose mass is negligible, and the distance between the centres of each sphere is d.Â How far from the centre of the 7MÂ sphere is the Centre of Mass for this object?

- d / 8
- d / 2
- 7d / 8
- 6d / 7

Solution: Option (A) d/8. Point C represents the centre of mass of the system.

Given : Â m1Â =Â 7M ;Â m2Â =Â M, x2 = d

Using, xÂ =Â m_{1}x_{1}Â +Â m_{2}x_{2} /Â m_{1}+m_{2
}xÂ =Â (7M)Â Ã—Â 0Â +Â Md /Â 7MÂ +Â MÂ =Â d /Â 8

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