JEE ADVANCED CENTER OF MASS

ARJUNA JEE AIR D1 (2026)
Centre of Mass & System of Particles
CENTRE OF MASS
For a system of particles centre of mass is that point at which its total mass is supposed to be
concentrated. The centre of mass of an object is a point that represents the entire body and moves in the
same way as a point mass having mass equal to that of the object, when subjected to the same external
forces that act on the object.
Centre of mass of system of discrete particles y
Total mass of the body: M = m1 + m2 + ......+ mn
m1(x1y1z1)
m r + m2 r2 + m3r3 + ... 1
Then RCM = 11 =  mi ri m2(x2y2z2)
m1 + m2 + m3 + ... M r1 m3(x3y3z3)
r2 r3
co-ordinates of centre of mass: mn(xnynzn)
1 1 1 rn
xcm = mixi, ycm = miyi and zcm = mizi (0,0,0)
x
M M M
z
Centre of mass of continuous distribution of particles:
If the system has continuous distribution of mass, treating the mass element dm at position as a point
1
mass and replacing summation by integration RCM =
M 
rdm. .

1 1 1 y
So that xcm =
M 
xdm, ycm =
M 
y dm and zcm =
M
z dm 
If co-ordinates of particles of mass m1, m2, .......are dm
(x1, y1, z1), (x2, y2, z2) ..... r

then position vector of their centre of mass is x
(0,0,0)
R = x iˆ + y ˆj + z kˆ
CM cm cm cm z

m1 (x1ˆi + y1 ˆj+ z1 k)
ˆ + m (x ˆi + y ˆj+ z k)
2 2 2 2
ˆ + m (x iˆ + y ˆj + z kˆ) + ...
3 3 3 3
=
m1 + m2 + m3 + ...

(m1x1 + m2 x 2 + ......)iˆ+ (m1y1 + m2 y2 ...) ˆj+ (m1 z1 + m2 z2 + ...)kˆ
=
m1 + m2 + m3 + ...
 m x + m2 x 2 + .....   m1 y1 + m2 y2 + .....   m1 z1 + m2 z2 + ..... 
So, xcm =  1 1  , ycm =   , zcm =  
 m1 + m2 + m3 + .....   m1 + m2 + .....   m1 + m2 + ..... 

The centre of mass after removal of a part of a body:
If a portion of a body is taken out, the remaining portion may be considered as,
Original mass (M)–mass of the removed part (m)={original mass (M)} + {–mass of the removed part (m)}
Mx − mx ' My − my ' Mz − mz '
The formula changes to : xCM = , yCM = , zCM =
M −m M −m M −m
Where x ', y ' and z ' represent the coordinates of the centre of mass of the removed part.

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, CENTRE OF MASS & SYSTEM OF PARTICLES

Do yourself –1:
(i) Two particles of mass 1 kg and 2 kg are located at x = 0 and x = 3 m. Find the position of their centre
of mass.

(ii) The position vector of three particles of mass m1 = 1 kg, m2 = 2 kg and m3 = 3 kg are
r1 = (iˆ + 4 ˆj + kˆ) m , r2 = (iˆ + ˆj + kˆ) m and r3 = (2iˆ − ˆj − 2kˆ) m respectively. Find the position vector of
their centre of mass.

(iii) Half of the rectangular plate shown in figure is made of a material of density ρ1 and the other half of
density ρ2. The length of the plate is L. Locate the centre of mass of the plate.




(iv) The centre of mass of rigid body always lie inside the body. Is this statement true or false?

(v) The centre of mass always lie on the axis of symmetry if it exists. Is this statement true of false?

(vi) If all the particles of a system lie in y-z plane, the x-coordinate of the centre of mass will be zero. Is
this statement true or not?



Motion of centre of mass
m1r1 + m2 r2 + m3r3 + ...
As for a system of particles, position of centre of mass is RCM =
m1 + m2 + m3 + ...
dr1 dr dr
m1 + m2 2 + m3 3 + ... m v + m2v2 + m3v3 + .....
So,
d
dt
(
RCM = ) dt dt dt
m1 + m2 + m3 + ...
 VCM = 1 1
m1 + m2 + m3 + .....
We can write MvCM = m1v1 + m2v2 + ... = p1 + p2 + p3 + .....[ p = mv ]
MvCM = pCM [ p = pi CM ]

m a + m2 a2 + ...
Similarly, acceleration aCM =
d
dt
( )
VCM = 1 1
m1 + m2 + .....
important points



CM CM
CM
CM
A B C D
• There may or may not be any mass present physically at centre of mass (See Figure A, B, C)
• Centre of mass may be inside or outside of the body (See figure A, B, C)
• Position of centre of mass depends on the shape of the body. (See figure A, B, C)
• For a given shape it depends on the distribution of mass of within the body and is closer to massive
part. (See figure A,C)
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• For symmetrical bodies having homogeneous distribution of mass it coincides with centre of
symmetry of geometrical centre. (See figure B,D).
• If we know the centre of mass of parts of the system and their masses, we can get the combined
centre of mass by treating the parts as point particles placed at their respective centre of masses.
• It is independent of the co-ordinate system, e.g., the centre of mass of a ring is at its centre whatever
be the co-ordinate system.
• If the origin of co-ordinate system is at centre of mass, i.e., RCM = 0 , then by definition.
1
mi ri = 0  mi ri = 0
M
The sum of the moments of the masses of a system about its centre of mass is always zero.

Do yourself – 2:
(i) A projectile is fired at a speed of 100 m/s at an angle of 37º above the horizontal. At the highest point,
the projectile breaks into two parts of mass ratio 1 : 3, the smaller coming to rest. Find the distance
from the launching point to the point where the heavier piece lands.

(ii) In a boat of mass 4 M and length  on a frictionless water surface. Two men A (mass = M) and

B (mass 2M) are standing on the two opposite ends. Now A travels a distance /4 relative to boat

towards its centre and B moves a distance 3/4 relative to boat and meet A. Find the distance travelled
by the boat on water till A and B meet.

(iii) A block A (mass = 4M) is placed on the top of a wedge B of base length  (mass = 20 M) as shown in
figure. When the system is released from rest. Find the distance moved by the wedge B till the block A
reaches ground. Assume all surfaces are frictionless.




(iv) An isolated particle of mass m is moving in a horizontal xy plane, along x-axis, at a certain height
above ground. It suddenly explodes into two fragments of masses m/4 and 3m/4. An instant later, the
smaller fragment is at y = + 15 cm. Find the position of heaver fragment at this instant.
Example
a a 3
Three bodies of equal masses are placed at (0, 0), (a, 0) and at  ,  . Find out the co-ordinates of
2 2 
centre of mass.




Solution
a a 3
0 m + a  m +  m 0 m + 0 m + m
2 a 2 a 3
xCM = = , yCM = =
m+m+m 2 m+m+m 6
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Example
Calculate the position of the centre of mass of a system consisting of two particles of masses m 1 and m2
separated by a distance L apart, from m1.




Solution
Treating the line joining the two particles as x axis
m1  0 + m2  L m2 L
= , yCM = 0 zCM = 0
m1 + m2 m1 + m2


Example
If the linear density of a rod of length L varies as λ = A + Bx, compute position of its centre of mass.
Solution
Let the x–axis be along the length of the rod and origin at one of its end as shown in figure. As rod is
along x–axis, for all points on it y and z will be zero so, yCM = 0 and zCM = 0 i.e., centre of mass will be
on the rod.
Now consider an element of rod of length dx at a distance x from the origin, mass of this element
dm = λdx = (A + Bx)dx
L L

 xdm  x(A+ Bx)dx AL2 BL3
+
3 = L(3A+ 2BL)
so, xCM = 0L = 0L = 2
BL2 3(2A+ BL)
0
dm 
0
(A+ Bx)dx AL +
2
Note: (i) If the rod is of uniform density then λ = A = constant & B = 0 then XCM= L/2
(ii) If the density of rod varies linearly with x, then λ = Bx and A = 0 then XCM = 2L/3



Example
Two bodies of masses m1 and m2 (<m1) are connected to the ends of a massless cord and allowed to move
as shown in. The pulley is both massless and frictionless. Calculate the acceleration of the centre of mass.




Solution
If a is the acceleration of m1, then – a is the acceleration of m2 then
Net force  m1 − m2 
acceleration of each body a = = g
Net mass  m1 + m2 

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