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Saturday, 31 March 2018

motion of centre of mass

Motion of centre of mass

Today our topic is all about motion of centre of mass very very important topic most of the students confused in motion of centre of mass due to lack of key concept of motion of centre of mass, So your all doubt and confusion will be clear here. You will get best concept for motion of centre of mass in this topic.
In previous post we have already studies about basic concept of center of mass, calculation of centre of mass, importance and application of centre of mass.




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

                                                            

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.

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 we 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

                                                                     
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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 
                                                          
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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.

We know that centre of mass rcom = (m₁r₁ +m₂r₂ )/(m₁+m₂)
now partial differentiate this equation we 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 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, We 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)
                                                                  
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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 ? 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.

This above point is very very important for key concept, it will take sometime to digest you but ready carefully and understand key concept.

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          


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