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**Centre of Mass**

- Every physical system has associated with it a certain point whose motion characterizes the motion of the whole system.
- When the system moves under some external forces, then this point moves as if the entire mass of the system is concentrated at this point and also the external force is applied at this point for translational motion. This point is called the centre of mass of the system.

**Centre of Mass of a System of `N' Discrete Particles**

Consider a system of N point masses m_{1}, m_{2}, m_{3}, .................. m_{n} 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.

Further, and

So, the cartesian co-ordinates of the COM will be

x_{com} =

or x_{COM} =

Similarly, y_{COM} = and

Note:

- 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.

**Position of COM of two Particles**

Consider two particles of masses m_{1} and m_{2} separated by a distance *l* as shown in figure.

Let us assume that m_{1} is placed at origin and m_{2} is placed at position (*l*, 0) and the distance of centre of mass from m_{1} & m_{2} is r_{1} & r_{2} respectively.

So x_{COM} =

r_{1} = = ...(1)

r_{2} = = ...(2)

From the above discussion, we see that

r_{1} = r_{2} = if m_{1} = m_{2}, i.e., COM lies midway between the two particles of equal masses.

Similarly, r_{1} > r_{2} if m_{1} < m_{2} and r_{1} < r_{2} if m_{2} < m_{1} i.e., COM is nearer to the particle having larger mass.

From equation (1) & (2),

m_{1}r_{1} = m_{2}r_{2} ...(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. **

**Ans.**

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

r_{1} = distance of COM from the particle of mass 1 kg = x

and r_{2} = distance of COM from the particle of mass 2 kg = (3 -x)

Using

or

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

x_{2kg} = 5 cos 37° = 4 m

y_{2kg} = 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

x_{com} = = = =

y_{com} = = = 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 r_{1} & r_{2} respectively. Then from relation

m_{1}r_{1} = m_{2}r_{2}

2r_{1} = 4r_{2} ...(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 m_{1}r_{1} = m_{2}r_{2}'

⇒ m_{1}(r_{1} + x) = m_{2} (r_{2} + y)

2(r_{1} + x) = 4(r_{2} + 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

m_{1}x = m_{2}y

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 r_{1} & r_{2} distance apart from mass m_{1} & m_{2} respectively as shown in figure.

from relation m_{1} r_{1} = m_{2} r_{2}

⇒ (1) r_{1} = 2r_{2}

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 m_{1}r_{1}' = m_{2}r_{2}'

⇒ 1 (x + r_{1} + 2) = 2 (10 + r_{2} - 2)

⇒ x + r_{1} + 2 = 20 + 2r_{2} - 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

**Centre of Mass of a Continuous Mass Distribution**

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

...(i)

= 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.

=

Note :

- 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**

- Suppose a rod of mass M and length L is lying along the x-axis with its one end at x = 0 and the other at x = L. Mass per unit length of the rod l =
- Hence, dm, (the mass of the element dx situated at x = x is) = l dx
- The coordinates of the element dx are (x, 0, 0). Therefore, x-coordinate of COM of the rod will be

x_{COM}= = = - The y-coordinate of COM is

y_{COM}= = 0 - Similarly, z
_{COM}= 0

i.e., the coordinates of COM of the rod are , i.e, it lies at the centre of the rod.

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