dynamics of system of particles

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1. DYNAMICS OF A SYSTEM OF PARTICLES

SHREEJI ACADEMY OF PHYSICS TRILOK SIR

(9033619800)

1 Obtain the expression of Centre of Mass of

a system of particles in one dimension for two

particles only.

Centre of mass is the point where whole mass

of the system can be supposed to be

concentrated and motion of the system can be

defined in terms of the centre of mass.

As shown in Fig. consider two particles having

mass m1 and m2 lying on X-axis at distancesof x1and x2respectively from the origin (O).

The centre of mass of this system is that point

whose distance from origin O is given by

is also called as mass-weighted averageposition of . If both particles are ofthe same mass then

Thus the centre of mass of the two particles of

equal mass lies at the centre between the two

particles on the line joining the two particles.

2. Obtain the expression of Centre of Mass of

a system of n particles in one dimension for

two particles only. If n particles of mass m1, m2, m3,., mn are

lying on X-axis at distance x1,x2,x3,.,xn

respectively from the origin O, then the

centre of mass of the system of n particles is

Where total mass of the systemof n particles.

If the mass of each particle is same i.e. m then

3. Obtain the expression of centre of mass of

System of n-particles in three dimensions.

Figure shows a system of n particles in three

dimensions. Let the position vectors of the

particles of mass m1, m2, ,mn with respect

to the origin O of the coordinate system are

respectively. The position vectorof centre of mass of the system is given by

following equation.

where M= total mass of system of n particles

4. Explain the Motion of Centre of Mass. OR

Consider

and Obtain Consider the expression of centre of mass of

system of n particles .

If the mass of each particle of the system of n

particles does not change with time, then

differentiating this equation with respect to

time.


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is the velocity of centre of

mass, and are the velocities ofrespective particles.

where are the momenta of

respective particles and is the total momentum of the system of n

particles.

Equation 1 shows that the total momentum of

the system of particles is equal to the product

of total mass of the system and velocity of the

centre of mass of the system.

5. By considering the motion of centre of mass

explain Newtons second law of motion.

Considering that centre of mass is moving so

its equation of motion

Differentiating this equation with time

In the above equation are the

forces acting on the respective particles of the

system and is the resultant force on thesystem.

are the acceleration of therespective particles produced due to these

foreces.

6. Explain the interdependence of Newtons

laws of motion for a system of particles.

The forces acting on the particles of a system

are of two kinds. 1) Internal forces among the

particles of the system 2) External forces.

For a system of two particles as shown in the

fig, let external forces acting on particles 1 and

2 are respectively and , and the mutualforces of interaction acting between them are

and While discussing overall motion of the

system, all these forces are considered to be

acting on the centre of mass C

According to Newtons third law of motion

so the resultant internal forcesbecomes zero.

Thus in equation of resultant force on system

we include onlyexternal forces not internal forces.

The resultant external force acting on a system

is equal to the rate of change of total linear

momentum of the system. This is the


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molecules. When bomb explodes, the internal

energy associated with them is converted into

heat energy and remaining part in the form of

kinetic energy of the fragments.

Thus in this case the work is done at the cost

of internal energy which leads to the more

general form of the work energy theorem.

9. Explain the centre of mass of a rigid body.

A system of particles in which the relative

position of particles remain invariant is called

a rigid body.

The location of centre of mass of a rigid body

depends on the distribution of mass in the

body and the shape of the body.

The centre of mass of a rigid body can be

anywhere inside or outside the body. For e.g.

the centre of mass of a disc of uniform mass

distribution is at its geometric centre within

the matter, whereas the centre of mass of a

ring of uniform mass distribution lies at its

geometrical centre which is outside of its

matter.

The position of centre of mass of symmetric

bodies with uniform mass distribution can be

easily obtained theoretically. The centre of

mass C of certain symmetric bodies are shownin fig.

10. Explain theoretical method to estimate the

centre of mass of a solid body.

Consider a small volume element dV,

containing mass dm. It is called as mass

element. whose position vector is

In this way consider the whole body is divided

in to n small mass elements dm1, dm2,

dm3,,dmn having position vectors

respectively. Position vector of centre of mass of a solid

body is

This equation can be also represented as


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Representing this equation in terms of its

components,

=

11. Explain the experimental method of

estimation of the CM of a solid body having

uniform density and geometrical shape.

To estimate the CM of the solid body having

uniform density and geometrical shape, the

symmetry of the body useful.

Suppose we want to find the CM of the

triangular solid object as shown in the fig.

For this let imagine that the body is divided

into thin stripes parallel to one particular side

of triangle.

By the law of symmetry, the CM of the

uniformly shaped object lies at its geometrical

centre. Hence the CM of the triangle will be

lying along the median.

Similarly two such medians can be obtained

by assuming the object divided in to thin

stripes parallel to the other two sides of the

triangle.

Thus the point of concurrence of all three

medians will be the CM of the provided solid

object.

12. Obtian the CM of a thin rod of uniform

density with respect to its end.

As shown in the diagram, a rod of mass M,

length L and uniform linear mass density has

uniform cross section area.

Assume one end of the rod at the origin

keeping the rod along the positive X axis.

Consider a length element dx at the distance x

from the origin on the rod.

The mass per unit length of the rod (i.e. linear

mass density) is The mass for the length dx would be say dm =

By definition the CM of the rod should be

[

]

[

]

is the required equation. So the CM of the rod of uniform density lies in

its middle.

dynamics of system of particles

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