2. Centre of Mass of Two Particle and n Particle System

Centre of mass of two particles and n particles Centre of mass: COM of a body is a point where the entire mass of the body can be supposed to be concentrated. Two particle system Suppose there are two masses m1 and m2 placed to a distance x1 and x 2 from origin then COM always lies in between two masses as shown in the diagram. Position of COM from origin is x and it will be given by-

x

m1 x1  m2 x2 m1  m2

If m1 = m2 then COM will be

x

x1  x2 2

Thus, for two particles system of equal mass the centre of mass lies exactly midway between them.

N particle system Case 1: All particles lying in a straight line If we have n particles of masses m1, m2,...mn respectively, along a straight line taken as the x- axis, then by definition the position of the centre of the mass of the system of particles is given by:-

x

m1 x1  m2 x2 ....  mn xn m1  m2 ....  mn

Case 1: Particles not lying in a straight line: Suppose that we have three particles, not lying in a straight line. We may define x and y axes in the plane in which the particles lie and represent the positions of the three particles by coordinates (x1,y1), (x2,y2) and (x3,y3) respectively. Let the masses of the three particles be m1, m2 and m3 respectively. The centre of mass P of the system of the three particles is defined and located by the coordinates (x,y) as shown in figure is given by:-

x

m1 x1  m2 x2  m3 x3 m1  m2  m3

or x 

m x m

i i i

,y

m1 y1  m2 y2  m3 y3 m1  m2  m3

or

y

m y m i

i

i

We can generalize it for n particle system in 2D or space (3D).

m x  m2 x2 ....  mn xn x 1 1 m1  m2 ....  mn

mx  x m

i i

or

i

Similarly:

m y y m i

mz  z m

i i

i

and

i

i

Here we can write

m  M i

summations of all masses

If all masses are equal then

x

x

i

n

,

y

y

i

n

and z 

z

i

n

If the origin of the frame of reference (the coordinate system) is chosen to be the centre of mass then

 m x  0  m y  0 and  m z i i

,

i i

i i

0

Now suppose there is a rigid body. The number of particles (atoms or molecules) in such a body is so large that it is impossible to carry out the summations over individual particles in above equations. Since the spacing of the particles is small, we can treat the body as a continuous distribution of mass. We subdivide the body into n small elements of masses i.e.

dm1, dm2 , dm3 .....dmi......dmn

over distances

x1, x2 ,.......xi ..... xn

from the origin. Then COM will be:

 (dm ) x x  dm i

i

or

i

ydm  y M

xdm  x M

as dm is very small mass . Similarly-

HERE

 dm  M i

zdm  z M

If we take (x,y,z) as a position vector r then we can write COM shortly in vector form as-

rdm  r M

Example: 1. A uniform disc of radius R is put over another uniform disc of radius 2R of the same thickness and density. The peripheries of the two discs touch each other. Locate the centre of mass of the system. Answer:Let the centre of the bigger disc be the origin. 2R = Radius of bigger disc R = Radius of smaller disc

m1   R 2  T   m2   (2 R ) 2  T   where T = Thickness of the two discs

 =Density of two discs The position of the centre of mass

(

m1 x1  m1 x2 m1 y1  m2 y2 , ) m1  m2 m1  m2

Here

x1  R , x2  0 , y1  0 , y2  0 , y1 , y2 is position vector of bigger disc.  R2T  R  0 0  R 2T  R R ( 2 , )  ( ,0)  ( ,0)  R T    (2R)2 T  m1  m2 5 R 2T  5 At R/5 from the centre of bigger disc towards the centre of smaller disc.