The centre of mass (

**R**) of a system of particles of masses*m*_{1},*m*_{2},*m*_{3},*m*_{4}, …etc. is defined by**R**= (

*m*

_{1}

**r**

_{1}+

*m*

_{2}

**r**

_{2}+

*m*

_{3}

**r**

_{3}+

*m*

_{4}

**r**

_{4}+ …etc.)/ (

*m*

_{1}+

*m*

_{2}+

*m*

_{3}+

*m*

_{4}+ …etc.) where

**r**

_{1},

**r**

_{2},

**r**

_{3},

**r**

_{4}etc. are the

*position vectors*of the particles of masses

*m*

_{1},

*m*

_{2},

*m*

_{3},

*m*

_{4}, …etc. respectively.

Note that

**R**is the*position vector*of the centre of mass.
The above expression for the

*position vector*of the centre of mass can be written in a compact form as**R**=

**Σ**

*m*

_{i}

**r**

_{i }/M where M =

**Σ**

*m*

_{i }which is the total mass of the system of particles. The value of ‘i’

_{.}should run from 1 to ‘n’ if there are ‘n’ particles in the system.

In the adjoining figure a system containing three point masses with position vectors

**r**_{1},**r**_{2}and**r**_{3 }is shown. The position vector of the centre of mass also is shown in the figure. If the masses of the particles are respectively*m*_{1},*m*_{2}and*m*_{3}the centre of mass has position vector**R**given by**R**= (

*m*

_{1}

**r**

_{1}+

*m*

_{2}

**r**

_{2}+

*m*

_{3}

**r**

_{3})/ (

*m*

_{1}+

*m*

_{2}+

*m*

_{3})

The x, y and z coordinates of the position of the centre of mass are evidently given respectively by

x = (

*m*_{1}x_{1}+*m*_{2}**x**_{2}+*m*_{3}**x**_{3})/ (*m*_{1}+*m*_{2}+*m*_{3}),
y = (

*m*_{1}y_{1}+*m*_{2}**y**_{2}+*m*_{3}**y**_{3})/ (*m*_{1}+*m*_{2}+*m*_{3}) and
z = (

*m*_{1}z_{1}+*m*_{2}**z**_{2}+*m*_{3}**z**_{3})/ (*m*_{1}+*m*_{2}+*m*_{3})
Even though we have considered a general case involving all the three components (x, y and z) for the position vectors of the particles and their centre of mass you will often find simple cases in which the particles are collinear or coplanar or having spherical symmetry in three dimensional arrangement. The following multiple choice questions will make you more confident:

**(1**) Two particles of masses 2 mg and 6 mg are separated by a distance of 6 cm. the distance of their centre of mass from the heavier particle is

(a) 1.5 cm

(b) 2 cm

(c) 3 cm

(d) 4 cm

(e) 4.5 cm

You may imagine the particles to be positioned on the X-axis, with the heavier one (of mass 6 mg) at the origin as shown.

6 mg ●----x-------------● 2 mg

If the centre of mass is at x we have

**R**= (*m*_{1}**r**_{1}+*m*_{2}**r**_{2})/M
Or, x = (6 mg ×0 + 2 mg×6 cm)/ 8 mg =

**1.5 cm**
You can easily arrive at this answer if you remember that in the case of a two particle system the moments of the masses about the centre of mass are equal. Therefore

6 × x = 2 ×(6 –x) from which x = 1.5 cm.

**(2)**A straight rod AB of length

*L*has non-uniform linear density,

*λ*=

*Kx/L*where

*K*is a constant and

*x*is the distance from end A. The distance of the centre of mass of the rod from the end A is

(a)

*/*^{L}_{3}
(b)

*/*^{L}_{2}
(c)

^{3L}/_{5}
(d)

^{2L}/_{3}
(e)

^{ }^{3L}/_{4 }
Consider an element of very small length

*dx*of the rod at distance*x*from the end A. The mass of this element is*dm**= λdx =*(*Kx/L*)*dx*
We have

**R**=**Σ***m*_{i}**r**_{i }/**Σ***m*_{i}
If

*X*is the distance of the centre of mass from the end A we can rewrite the above equation as*X =*[

_{0}∫

^{L}*dm x*]/ [

_{0}∫

*]*

^{L }dm
= [

_{0}∫^{L}*Kx*^{2}*dx*^{ }*/L*]/ [_{0}∫^{L}*Kxdx*^{ }*/L*] = (*L*^{3}/3)/ (*L*^{2}/2) =^{2L}/_{3}**(3)**A radioactive nucleus of mass

*M*moving along the positive x-direction with speed

*v*emits an α-particle of mass

*m*. If the α-particle proceeds along the positive y-direction, the centre of mass of the system (made of the daughter nucleus and the α-particle) will

(a) remain at rest

(b) move along the positive x-direction with speed less than

*v*
(c) move along the positive x-direction with speed greater than

*v*
(d) move in a direction inclined to the positive x-direction

(e) move along the positive x-direction with speed equal to

*v*
The state of rest or of uniform motion of the centre of mass of a system of particles can be changed by

*external*forces only. Since the α-emission is produced by*internal*forces, the centre of mass is unperturbed and it will continue to move along the positive x-direction with speed equal to*v*[Option (e)].
It will be interesting to note that the centre of mass of a shell which explodes in mid air will continue to move along the parabolic path originally followed by the unexploded shell.

**(4)**Three homogeneous solid spheres of masses 1 kg, 2 kg and 4 kg are arranged with their centres at (2

**i**+

**j**+

**k**), (3

**i**– 2

**j**+ 2

**k**) and (4

**i**–

**j**– 2

**k**) respectively where

**i**,

**j**,

**k**are unit vectors in the x, y and z directions. All distances are in metre. The y-coordinate of the centre of mass of the system of spheres is

(a) 3.3 m

(b) 2.6 m

(c) 0.33 m

(d) – 1 m

(e) – 2 m

By symmetry the centre of mass of a homogeneous sphere is its geometric centre. The three spheres can therefore be treated as three poin masses (for finding the centre of mass).

The position vector of the centre of mass is given by

**R**= (

*m*

_{1}

**r**

_{1}+

*m*

_{2}

**r**

_{2}+

*m*

_{3}

**r**

_{3})/ (

*m*

_{1}+

*m*

_{2}+

*m*

_{3})

Substituting proper values,

**R**= [1(2

**i**+

**j**+

**k**) + 2(3

**i**– 2

**j**+ 2

**k**) + 3(4

**i**–

**j**– 2

**k**)] /(1+2+3)

= (20

**i**– 6**j**–**k**)/ 6
The y-component of the position vector of the centre of mass is –1 metre. Therefore the correct option is (d).

**(5)**The figure shows a T-shaped portion cut from a plane sheet of uniform thickness (lamina) having areal density of 1 kgm

^{–2}. Imagine the T to be made of the horizontal rectangular portion of sides 3 m and 1 m and the vertical rectangular portion of sides 2 m and 1 m. Neglect the thickness of the sheet. The centre of mass of the entire T with respect to the coordinate system shown in the figure is at the point

(a) (0.5, – 0.5, 0)

(b) (0.5, – 0.1, 0)

(c) (0.5, – 0.4, 0)

(d) (0.4, – 0.2, 0)

(e) (0.5, 0, 0)

The horizontal rectangle has its centre of mass at its centre A and the vertical rectangle has its centre of mass at its centre B. The mass of the horizontal rectangular portion is 3 kg since its area is 3 m

^{2}. The mass of the vertical rectangular portion is 2 kg since its area is 2 m^{2}. The entire T can be imagined to be reduced to two particles of masses 3 kg and 2 kg located at A and B respectively. Evidently the coordinates of A and B are (0.5, 0.5, 0) and (0.5, –1, 0). The Z coordinate is zero since the T is placed in the XY plane.
The centre of mass has x coordinate given by

x = (

*m*_{1}x_{1}+*m*_{2}**x**_{2})/ (*m*_{1}+*m*_{2}) = (3×0.5 + 2×0.5)/(3+2)
Or, x = 0.5 m

Similarly the centre of mass has y coordinate given by

y = (

*m*_{1}y_{1}+*m*_{2}**y**_{2})/ (*m*_{1}+*m*_{2 }) = [3×0.5 + 2×(–1)]/(3+2)
Or, y = – 0.1 m. The centre of mass is therefore at the point (0.5, – 0.1, 0) given in option (b).

You will find additional questions (with solution) in this section at physicsplus

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