Today Topic is **Motion of centre of mass**.

So i’m going to show you in this article what’s is meaning of

*motion of centre of mass*EXACTLY.Because most of students are confused to understand motion of centre of mass due to lack of true concept.

So after finishing this post you’ll learn all about exact concept and meaning of motion of centre of mass.

And this is very very important topic of Physics.

So today your all doubt and confusion will be clear here.

You will get best concept for motion of centre of mass in this article.

Let’s continue with me and dive in right now.

## Motion Of Centre Of Mass Class 11 Concept

Well, in previous post you have already studies about basic *concept of center of mass.*

For example what is centre of mass of a body.

And how to calculate centre of mass.

As well as importance and application of centre of mass.

If you have not read previous post then i’ll suggest you to check that link first So that you can better understand.

And i’m providing here the link you can refer the previous post

**Centre of mass Physics**. Now in this topic we will learn what Physics says about motion of centre of mass like velocity, acceleration.

So this’s right time to explain

**motion of centre of mass**Here is the deals:

When two bodies moving separately then what effect on motion of centre of mass so lets start.

Motion of Centre of mass |

A system has n particles and mass m1,m2,m3,m4……………………………..mn

As from the figure we see that if there are many

**point masses**in the system then its centre of mass is defined as rcom.

And position r1,r2,r3,r4………………………………rn

Then sum of all moments is define as

rcom = (m₁r₁+m₂r₂+m₃r₃+m₄r₄………………………..+mnrn)/(m₁+m₂+m₃+m₄+………..+mn)

now

*rcom is the position of centre of mass with respect to origin*.If you want velocity of centre of mass then differentiate this equation with respect to time on both side we will get velocity of centre of mass.

Hence Vcom = (m₁v₁+m₂v₂+m₃v₃+m₄v₄…………………….)/(m₁+m₂+m₃+m₄………..)

Now

**for acceleration of centre of mass**again differentiate the above vcom equation with respect to time on both side we will get acceleration of centre of mass.acom = (m₁a₁+m₂a₂+m₃a₃+m₄a₄+………………………..)/(m₁+m₂+m₃+m₄)

## Application of motion of centre of mass

From the above figure two ball of mass 1kg each having velocity 5m/s and 10m/s respectively.Considering as a system find the velocity of centre of mass.

Here velocity of both ball at an angle of 37⁰ so we have to take separate component for velocity along x axis and along y axis.

Now velocity of first ball along x axis = 5cos37⁰ =5*4/5 = 4m/s

similarly velocity of second ball along x axis = 10cos37 = 10*4/5 = 8m/s

Now apply formula Vcom =( m1v1+m2v2)/(m1+m2)

Hence velocity of centre of mass along x axis = Vcomx = (1*4+1*8)/(1+1) = 12/2 = 6m/s

Similarly velocity along y axis for first ball = 5sin37⁰ = 5*3/5 = 3m/s for second ball -10sin37⁰ = -6m/s

Vcomy = (1*3-1*6)/(1+1) = -3/2 m/s

We have already learn vector in details previous post if you have problem refer vector post.

Hence velocity of centre of mass Vcom = Vcomxi^ +Vcomyj^ = 6i^ -3/2j^

So net velocity of centre of mass will be Vcom = √6²+(3/2)² m/s

Now for angle tanϴ = Vcomy/Vcomx = -1.5/6

As from the question initially both mass are kept then its initial centre of mass will be as shown.

If mass of 10kg will move toward right them centre of mass will adjust itself and will try to move left.

But from question centre of mass should remain stationary 20kg mass have to move left that need to find.

You know that centre of mass rcom = (m₁r₁ +m₂r₂ )/(m₁+m₂)

now partial differentiate this equation you will get

**drcom = (m₁dr₁ +m₂dr₂)/(m₁+m₂) this equation is very very important remember forever.**

Here drcom is change in centre of mass but in this case no any change centre of mass remains stationary.

so drcom = 0 so put the value in above equation

drcom = (m₁dr₁ +m₂dr₂)/(m₁+m₂) here dr₁ = 2cm

0 = 10*2 +20dr₂

0 = 20 +20dr₂ or dr₂ = -20/20 = -1cm hence 20kg mass need to move 1cm towards left .

Which is opposite of the movement of 10kg mass to remain centre of mass remains stationary.

### Important point for motion of centre of mass

*This is very important concept for motion of centre of mass*.

You know that acceleration of centre of mass for large number of point masses is defined as.

acom = (m₁a₁+m₂a₂+m₃a₃+……………..+mₙaₙ)/(m₁+m₂+m₃+……….+mₙ)

Now after arranging the denumerator we can write as.

(m₁+m₂+m₃+……….+mₙ)acom = (m₁a₁+m₂a₂+m₃a₃+……………..+mₙaₙ)

Here it is clear that in Newton’s second law Fnet = M*acom

where acom is acceleration of centre of mass or Fnet =M*acom = m₁a₁+m₂a₂+m₃a₃+…………..+mₙaₙ

→ → → → →

Fnet = F₁+F₂+F₃+……………+Fₙ

### ***** Important conclusion for motion of centre of mass

Is this possible the net Force on system is zero, But individual particle of the system is doing motion ?

What is your opinion ?

Well, the correct answer is yes.

lets understand this important point.

If Fnet = 0 it does not mean all forces F1, F2 ,F3 …… are zero.

it may be F1,F2, F3 are having some values but after adding this value in vector method Fnet becomes zero.

acom = Fnet/M hence it clear that if Fnet is zero then centre of mass acom will must be zero.

This is 100% true if the Fnet acting on system is zero then acceleration of centre of mass must be zero.

If acceleration of center of mass is zero then it does not mean that each mass of the acceleration a₁, a₂,a₃…….. are zero.

Because individually each acceleration will have some value but after adding it becomes zero.

Now if acceleration of Center of mass acom is zero.

Its mean that there is no change in velocity of centre of mass.

So velocity of centre of mass is constant.

But it does not mean that every individual particle velocity is constant .

Because their acceleration is not zero hence it velocity may change.

Are you getting ?

So these all above points are very very important for key concept about motion of centre of mass.

But it will take sometime to digest you.

So understand carefully and practice key concept.

**#Conclusion:**

Centre of mass position is conserved during a motion. “If no external force is applied”. Internal force within the system. And external force which comes from outside of the system. For example

**Earth Moon system**. There will be a centre of mass closer to Earth because mass of Earth is heavy than Moon. So both Earth and Moon is consider as a system. Because**centre of mass is always closer to heavier mass**. But Remember “centre of mass never rotate”. Earth and Moon rotate about centre of mass freely. We will continue in next post. I hope you have enjoyed to learn**motion of centre of mass**if you like comment and share thanks for learning and sharing keep reading more and more.Dated 4th Oct 2018