Consider a system of N point masses m1, m2, m3, .................. mn whose position vectors from origin O are given by ........... respectively. Then the position vector of the centre of mass C of the system is given by:
where, is called the moment of mass of particle with respect to origin.
is the total mass of the system.
So, the cartesian co-ordinates of the COM will be
or xCOM =
Similarly, yCOM = and
- If the origin is taken at the centre of mass then = 0. hence, the COM is the point about which the sum of "mass moments" of the system is zero.
- If we change the origin then changes. So also changes but exact location of center of mass does not change.
Consider two particles of masses m1 and m2 separated by a distance l as shown in figure.
Let us assume that m1 is placed at origin and m2 is placed at position (l, 0) and the distance of centre of mass from m1 & m2 is r1 & r2 respectively.
So xCOM =
r1 = = ...(1)
r2 = = ...(2)
From the above discussion, we see that
r1 = r2 = if m1 = m2, i.e., COM lies midway between the two particles of equal masses.
Similarly, r1 > r2 if m1 < m2 and r1 < r2 if m2 < m1 i.e., COM is nearer to the particle having larger mass.
From equation (1) & (2),
m1r1 = m2r2 ...(3)
Centre of mass of two particle system lie on the line joining the centre of mass of two particle system.
Q.1. Two particle of mass 1 kg and 2 kg are located at x = 0 and x = 3 m. Find the position of their centre of mass.
Since, both the particles lie on x-axis, the COM will also lie on x-axis. Let the COM is located at x = x, then
r1 = distance of COM from the particle of mass 1 kg = x
and r2 = distance of COM from the particle of mass 2 kg = (3 -x)
or x = 2 m
thus, the COM of the two particles is located at x = 2m.
Q.2. Two particle of mass 4 kg & 2 kg are located as shown in figure then find out the position of centre of mass.
Ans. First find out the position of 2 kg mass
x2kg = 5 cos 37° = 4 m
y2kg = 5 sin 37° = 3 m
So these system is like two particle system of mass 4 kg and 2kg are located (0, 0) and (4, 3) respectively, then
xcom = = = =
ycom = = = 1 m
So position of C.O.M is
Q.3. Two particles of mass 2 kg and 4 kg lie on the same line. If 4 kg is displaced rightwards by 5 m then by what distance 2 kg should be move for which centre of mass will remain at the same position.
Ans. Let us assume that C.O.M. lie at point C and the distance of C from 2 kg and 4 kg particles are r1 & r2 respectively. Then from relation
m1r1 = m2r2
2r1 = 4r2 ...(i)
Now 4 kg is displaced rightwards by 5 m then assume 2 kg is displaced leftwards by x distance to keep the C.O.M. at rest.
from relation m1r1 = m2r2'
⇒ m1(r1 + x) = m2 (r2 + y)
2(r1 + x) = 4(r2 + 5) ...(ii)
2x = 20
x = 10 m
To keep the C.O.M at rest 2 kg displaced 10 m left wards
Alter: If centre of mass is at rest then we can write
m1x = m2y
2 × x = 4 × 5
x = 10 m.
Q.4. Two particles of mass 1 kg and 2 kg lie on the same line. If 2 kg is displaced 10 m rightwards then by what distance 1 kg should displaced so that centre of mass will displaced 2m right wards.
Ans. Initially let us assume that C.O.M is at point C which is r1 & r2 distance apart from mass m1 & m2 respectively as shown in figure.
from relation m1 r1 = m2 r2
⇒ (1) r1 = 2r2
Now 2kg is displaced 10 m rightwards then we assume that 1 kg is displaced x m leftward to move the C.O.M 2m rightwards.
So from relation m1r1' = m2r2'
⇒ 1 (x + r1 + 2) = 2 (10 + r2 - 2)
⇒ x + r1 + 2 = 20 + 2r2 - 4 ...(ii)
from eq. (i) & (ii) x = 14 m (leftwards).
Q.5. Three particles of mass 1 kg, 2 kg, and 3 kg are placed at the corners A, B and C respectively of an equilateral triangle ABC of edge 1 m. Find the distance of their centre of mass from A.
Ans. Assume that 1 kg mass is placed at origin as shown in figure.
co-ordinate of A = (0, 0)
co-ordinate of B = (1 cos60°,1 sin60°) =
co-ordinate of C = (1, 0)
Position of centre of mass
distance of C.O.M from point
For continuous mass distribution the centre of mass can be located by replacing summation sign with an integral sign. Proper limits for the integral are chosen according to the situation
= M (mass of the body)
here x,y,z in the numerator of the eq. (i) is the coordinate of the centre of mass of the dm mass.
- If an object has symmetric mass distribution about x axis then y coordinate of COM is zero and vice-versa.
➢ Centre of Mass of a Uniform Rod