**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.

**R**is the

*position vector*of the centre of mass.

*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.

**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})

*m*

_{1}x

_{1}+

*m*

_{2}

**x**

_{2}+

*m*

_{3}

**x**

_{3})/ (

*m*

_{1}+

*m*

_{2}+

*m*

_{3}),

*m*

_{1}y

_{1}+

*m*

_{2}

**y**

_{2}+

*m*

_{3}

**y**

_{3})/ (

*m*

_{1}+

*m*

_{2}+

*m*

_{3}) and

*m*

_{1}z

_{1}+

*m*

_{2}

**z**

_{2}+

*m*

_{3}

**z**

_{3})/ (

*m*

_{1}+

*m*

_{2}+

*m*

_{3})

**(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

**R**= (

*m*

_{1}

**r**

_{1}+

*m*

_{2}

**r**

_{2})/M

**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

*/*

^{L}_{3}

*/*

^{L}_{2}

^{3L}/

_{5}

^{2L}/

_{3}

^{ }

^{3L}/

_{4 }

*dx*of the rod at distance

*x*from the end A. The mass of this element is

*dm*

*= λdx =*(

*Kx/L*)

*dx*

**R**=

**Σ**

*m*

_{i}

**r**

_{i }/

**Σ**

*m*

_{i}

*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

*v*

*v*

*v*

*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)].

**(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

**R**= (

*m*

_{1}

**r**

_{1}+

*m*

_{2}

**r**

_{2}+

*m*

_{3}

**r**

_{3})/ (

*m*

_{1}+

*m*

_{2}+

*m*

_{3})

**R**= [1(2

**i**+

**j**+

**k**) + 2(3

**i**– 2

**j**+ 2

**k**) + 3(4

**i**–

**j**– 2

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

**i**– 6

**j**–

**k**)/ 6

**(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

^{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.

*m*

_{1}x

_{1}+

*m*

_{2}

**x**

_{2})/ (

*m*

_{1}+

*m*

_{2}) = (3×0.5 + 2×0.5)/(3+2)

*m*

_{1}y

_{1}+

*m*

_{2}

**y**

_{2})/ (

*m*

_{1}+

*m*

_{2 }) = [3×0.5 + 2×(–1)]/(3+2)