Suppose you take 2 similar tennis balls and put them a bit apart, you see that the centre of mass of the two balls would be their centre. If one of these balls was heavier, the centre of mass will shift towards the heavier ball. But if you take a cricket bat, the centre of mass would be below the centre of the bat, in the lower half.Â Let us study more about the motion of centre of mass.

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## Motion of Center of Mass

Let us talk about centre of mass of a system of particles.

Whenever we talk about motion of an object, we usuallyÂ talk about velocity with which the object is moving or the acceleration with which the object is moving. As we know the centre of mass is denoted by **x** andÂ **y.**

X= m_{iÂ }x_{i}Â /M

Y = m_{iÂ }y_{i} /M

X =Â Î£Â \( \frac{m_i}{x_i} \)

Y =Â Î£Â \( \frac{m_i}{y_i} \)

The position vector of the centre of mass can be written as

Î£Â \( \frac{m_i r_i}{M} \)

â‡’ m_{1Â }r_{1Â }+ m_{2Â }r_{2Â }+………. +m_{n}r_{n}

Differntiating on both the sides,

MÂ \( \frac{dr}{dt} \) = M_{1Â }\( \frac{dr_1}{dt} \) + M_{2Â }\( \frac{dr_2}{dt} \) +Â M_{nÂ }\( \frac{dr_n}{dt} \)

Change of displacement of time is the velocity.

mv =Â m_{1Â }v_{1Â }+ m_{2}v_{2Â }+………..Â m_{n}v_{n}

where v_{1Â }is the velocity of the particle and v = \( \frac{dr}{dt} \) is the velocity of centre of mass.

V =Â Î£Â \( \frac{m_i v_i}{M} \)

This is an expression for velocity of centre of mass.

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

- Introduction to Rotational Dynamics
- Vector Product of Two Vectors
- 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

### Acceleration of System of Particles

Now we know that, mv =Â m_{1Â }v_{1Â }+ m_{2}v_{2Â }+ ………..+m_{n}v_{n}, differentiating on both the sides we get,

MÂ \( \frac{dv}{dt} \) = M_{1Â }\( \frac{dv_1}{dt} \) = M_{2Â }\( \frac{dv_2}{dt} \) +Â M_{nÂ }\( \frac{dv_n}{dt} \)

Now the rate of change of velocity is,

MA = m_{1}a_{1Â }+ m_{2}a_{2Â }+…….+m_{n}a_{n}

A =Â \( \frac{dV}{dt} \) is the acceleration of centre of mass of system of particles. The force of particle is given by F So, MA = F_{1Â }+ F_{2Â }+….+Â F_{n}

MA = F_{ext}

which is the sum of all external forces acting on the particles of the system. This is how we determine the velocity and acceleration of the centre of mass of the system of particles. Hence we conclude that the centre of the mass of the system of particles moves as if all the mass system was concentrated at the centre of mass and all external forces were applied to that point.

## Questions For You

Q1.Â Two objects P andÂ Q initially at rest move towards each other under a mutual force of attraction. At the instant when the velocity ofÂ P is V and that of Q is 2v the velocity of the centre of mass of the system is

- v
- 2v
- 3v
- 1.5v
- Zero

Solution: E. Since they move due to the mutual interaction between two objects so, the centre of mass remains the same and its velocity is zero.

Q2.Â A ball kept in a closed container moves in it making collision with the walls. The container is on a smooth surface. The velocity of the centre of mass of

- the ball remains fixed.
- ball relative to container remains the same.
- container remains fixed.
- ball container and ball remains fixed.

Solution: D. Since the container is closed, the collisions made by the ball with the walls of a containerÂ will not affect the mass of container and in turn, there is no change in velocity of the centre of mass.

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