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Work Power Energy Web: http://www.locuseducation.org

PHYSICS 1LOCUS

INTRODUCTION:

Any body (or an assembly of bodies) represents, in fact, a system of mass points, or particles. If a systemchanges with time, it is said that its statevaries. The state of a system is defined by specifying the positions and

velocities of all constituent particles.

Experience shows that if the laws of forces acting on a system and the state of the system at a certain initial

moment are known, the motion equations can help predict the subsequent behaviour of the system.

However, an analysis of a systems behaviour by the use of motion equations requires so much effort (due

to the complexity of the system itself), that a comprehensive solution seems to be practically impossible. Moreover,

such an approach is absolutely out of the question if the laws of acting forces are not known. Besides, there are

some problems in which the accurate consideration of motion of individual particles is meaningless (example: gas).

Under these circumstances the following question naturally comes up: are there any general principles followingfrom Newtons laws that would help avoid these difficulties by opening up some new approaches to the solution of

the problem.

It appears that such principles exist. They are called conservation laws.

As we have already discussed, the state of a system varies in the course of time as that system moves.

However, there are some quantities,state functions, which possess the very important and remarkable property of

retaining their values constant with time. Among these constant quantities, energy, linear momentumand angular

momentumplay the most significant role.

The laws of conservation of energy, momentum and angular momentum fall into the category of the most

fundamental principles of physics. These laws have become even more significant since it was discovered that theygo beyond the scope of mechanics and represent universal laws of nature. In any case, no phenomena have been

observed so far which do not obey these laws. They work reliably in all quarters: in the field of elementary particles,

in outer space, in atomic physics and in solid state physics.

Having made possible a new approach to treating various mechanical phenomena, the conservation laws

turned into a powerful and efficient tool of research used by physicists. The importance of the conservation principles

is due to several reasons:

1. The conservation laws do into depend on either the paths of particles or the nature of acting forces.

Consequently, they allow us to draw some general and essential conclusions about the properties of

various mechanical processes without restoring to their detailed analysis by means of motion equations.

2. Since the conservation laws do not depend on the nature of the acting forces, they may be applied

even when the forces are not known. In these cases the conservation laws are the only and

indispensable tool of research. This is the present trend in physics of elementary particles.

3. Even when the forces are known precisely, the conservation laws can help substantially to solve

many problems of motion of particles. Although all these problems can be solved with the use of

motion equations (and the conservation laws provide no additional information in this case), the

utilization of the conservation laws very often allows the solution to be obtained in the most straight-

forward and elegant fashion, obviating cumbersome and tedious calculations.

We shall begin examining the conservation laws with the energy conservation law, having introduced theconcept of energyand work.


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PHYSICS 2LOCUS

ENERGY:

It is possible to give a numerical rating, called energy, to the state of a physical system. The total energy is

found by adding up contributions from characteristics of the system such as motion of objects in it, heating of the

objects, and the relative positions of objects that interact via forces. The total energy of a closed system always

remains constant. Energy can not be created or destroyed, but only transferred from one system to another.Energy comes in a variety of forms, and physicists didnt discover all of them right away. They had to start

somewhere, so they picked one form of energy to use as a standard for creating a numerical energy scale. One

practical approach is to defined an energy unit based on heating of water. The SI unit of energy is the joule, J, named

after the British physicist James Joule. One joule is the amount of energy required in order to heat 0.24 g of water

by 1C.

Note that heat, which is a form of energy, is completely different from temperature. In standard, formal

terminology, there is another, finer distinction. The word heatis used only to indicate an amount of energy that is

transferred, whereas thermal energyindicates an amount of energy contained in an object.

Once a numerical scale of energy bas been established for some form of energy such as heat, it can easily beextended to other types of energy. For instance, the energy stored in one gallon of gasoline can be determined by

putting some gasoline and some water in an insulated chamber, igniting the gas, and measuring the rise in the waters

temperature. (The fact that the apparatus is known as abomb calorimeterwill give you some idea of how dangerous

these experiments are if you dont take the right safety precautions). Here are some examples of other types of

energy that can be measured using the same units of joules.

Type of energy:

chemical energy released by burning

energy required to break an object

energy required to melt a solid substance chemical energy released by digesting food

raising a mass against the force of gravity

nuclear energy released in fission, etc.

Textbooks often give the impression that a sophisticated physics concept was created by one person who

had an inspiration one day, but in reality it is more in the nature of science to rough out an idea and then gradually

refine it over many years. The idea of energy was tinkered with from the early 1800s on, and new types of energy

kept getting added to the list.

To establish a new form of energy, a physicist has to

(1) show that it could be converted to and from other forms of energy, and(2) show that it is related to some definite measurable property of the object, for example its temperature,

motion, position relative to another object, or being in a solid or liquid state.

For example, energy is released when a piece of iron is stoked in water, so apparently there is some form of

energy already stored in the iron. The release of this energy can also be related to a definite measurable property of

the chunk of metal: it turns reddish-orange. There has been a chemical change in its physical state, which we call

rusting.

Although the list of types of energy kept getting longer and longer, it was clear that many of the types were

just variations on a theme. There is an obvious similarity between the energy needed to melt ice and to melt butter,

or between rusting of iron and many other chemical reactions. All the types of energy can be reduced to a very smallnumber by simplifications.


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PHYSICS 3LOCUS

WORK:

The concept of work :The mass contained in closed sytsem is a conserved quantity, but if the system is not closed,

we also have ways of measuring the amount of mass that goes in or out.

We often have a system that is not closed, and would like to know how much energy comes in or out.

Energy, however, is not a physical substance like water, so energy transfer can not be measured by the same kind of

meter which we used to measure flow of water. How can we tell, for instance, how much useful energy a tractor can

put out on one tank of gas?

The law of conservation of energy guarantees that all the chemical energy in the gasoline will reappear in

some form, but not necessarily in a form that is useful for doing farm work. Tractors, like cars, are extremely

inefficient, and typically 90% of the energy they consume is converted directly into heat, which is carried away by

the exhaust and the air flowing over the radiator. We wish to distinguish energy that comes out directly as heat from

the energy that serves to accelerate a trailer or to plow a filed, so we defined a technical meaning of the ordinary

workto express the distinction.

Definition of work : Work is the amount of energy transferred into or out of a system, not taking into

account energy transferred by heat conduction.

[Based on this definition, is work a vector, or a scalar? What are its units?]

The conduction of heat is to be distinguished from heating by friction. When a hot potato heats up your hands by

conduction, the energy transfer occurs without any force, but when friction heats your cars brake shoes, there is a

force involved. The transfer of energy with and without a force are measured by completely different methods, so

we wish to include heat transfer by frictional heating under the definition of work, but not heat transfer by conduction.

The definition of work could thus be restated as the amount of energy transferred by forces.

Work done by a constant force:

Work done by a constant force is defined as product of the force and the component of the displacement

along the direction of the force.

FS mm

fig 5.1.

F

Consider the situation shown in figure 5.1. A constant forceFis applied on a block of mass malong thehorizontal direction. If the block moves by a distance Son the horizontal surface on which it is placed, as shown in

figure, then the work done by the forceFis defined as

=w F S ...(1).

If the force and the displacement are not along the same direction, as shown in figure 5.2, then work done

by forceFis calculated by multiplying the force and the component of the displacement along the force, as shown

in figure 5.3, therefore, for the given case, work done by forceFis

F

S mm

fig. 5.2

F

!

Scos

!

mm S!

F(const.)

fig. 5.3


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PHYSICS 4LOCUS

= cosw F S

...(2)

Here you should note that work done by the forceFcan also be written as

( cos )w F S=

and we know that Fcosis the component ofFalong the displacement, as shown in figure 5.4.

F cos !mm

S!

F(const.)

work done by= force along displacement displacement

F

fig. 5.4

Hence,work done by a constant force can also be defined as the product of the displacement and

the component of the force along the displacement.

In vector form equations (1) and (2) can be generalized as

=!!

w F S ...(3)

Hence, work done by a constant force is the scalar (dot) product of the force and the displacement.HereI would like to emphasize that

!Sis the displacement of the point of application of the force

!F.

Note: For a constant force!

F, in equation (3) you should notice the following:

* When i.e., 90 , 0.fF S, w = =!!

* When (0,90 ), 0.fw >* When (90 ,180 ), 0.fw 0

2 (3cos 2) 0m

u gll

+ [Using equation (ii)]

2 (3cos 2) 0u gl +


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PHYSICS 40LOCUS

2 (2 3cos )u gl

2 maximum value of (2 3cos )u gl

2 5u glat = (topmost position)

'2-3cos ' has its maximum value

"

5u gl

You should not that if 5u gl= , then at the topmost position T

becomes zero but vis nonzero. For 5u gl= at = ,equation (i) gives,

v gl=

and equation (i) gives,

2/v !

T= 0mg

!%&

= 5u gl

fig. 5.36

T= 0

Therefore, for 5 ,u gl= at the topmost position gravity aloneis providing the centripetal acceleration. This fact also gives

2vmg m

l=

v gl= velocity of the ball when it reaches the lowermost position is

5u gl= 2 2

using loss in P.E.

1 1(2 ) 5

2 2

i f f iE E k k

mu mv mg l u gl

= = + = + =

If the speed of the ball at the lowermost position, u, is greater than 5 ,gl then its speed at the topmost position is

also greater than gland hence more centripetal force is required. In this case both tension and gravity contribute

to the centripetal force and hence T> 0.CASE B:

BALL OSCILLATES WITH ANGULAR APTITUDE, 0, SMALLER THAN /2:

Rearrange equations (i) and (ii) to get

2 2( 2 ) 2 cosv u gl gl = + ...(ii)

and2( 2 ) 3 cos

lT u gl gl

m

= +

...(iv)

For this case we have [0, /2). Therefore from equation (iii) and (iv) it is clear that if2

2u gl> then neither vnor Tbecomes zero in this interval of . That is neither the ball stops nor the string becomes slack in this region. We


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PHYSICS 41LOCUS

can also say that if 2 2u gl> then string would definitely cross the horizontal position (i.e., = /2 position).Therefore if 2 < 2u glthen the ball can not deflect by /2 and it would oscillate about its lower most position with

angular amplitude smaller than /2. In such a case, it is clear from equations (iii) and (iv) that speed of the ballvanishes before the tension in the string. That is, in such a case tension in the string is never zero and speed of the ball

is zero at the extreme position (the position from where it starts returning back towards lowermost position). These

facts could also be explained in the following way:

The position of the ball is shown at some arbitrary angular position

in figure 5.37(a). If vbe the speed of the ball at this position then

it is obvious from the figure that at this moment tension force is

balancing the radially outward component of gravity as well as

providing the required centripetal acceleration to the ball and hence

it is greater than cosmg (radially outward component of gravity).

Here tangential component of gravity, sin ,mg is retarding the

upward (along the circle) motion of the ball as shown in figure

5.37(a). When speed of the ball becomes zero at some angle 0,then also tension has to balance the radially outward component of

gravity, 0cosmg , and hence it can not be zero, as shown in figure

5.37(b). In the same figure you should notice that, at

0, = sinmg would accelerate the ball towards its lowermost position.

fig. 5.37 (a)

( > cos )T mg !

!

mgsi

n!

v

T2v

!

mg

cos!

!

0

mg

cos!

0

mgs

in!0

v=0

T

fig. 5.37 (b)

( = cos )T mg !

0

If 2 2u gl= then according to equations (iii) and (iv), vand Tboth vanishe simultaneously at = /2, as shown in figure

5.38. In this case the ball oscillates with angular amplitude /2.

mg

fig. 5.38(ball just reaches thehorizontal position)

&/2

T = 0v = 0

2u gl=

CASE C:

BALL CROSSES THE HORIZONTAL POSITION BUT DOES NOT COMPLETE THE VERTICAL

CIRCLE.

From the previous two cases it must be clear to you that if uis greater than 2glbut smaller than 5glthen the ball would deflect more than /2 but it can not complete the vertical circle. Therefore, it would leave thecircular path for an angle greater than /2 but smaller than .

T=0

mg(

u

fig. 5.39

vFrom equations (iii) and (iv) it can be concluded that if2 2u gl> and /2 > then Tvanishes before v (only if2 5u gl< ), i.e., at some angle (as mentioned above) the string

becomes slack but the ball still has some nonzero speed, as shown

in figure 5.39. After this particular position gravity is the only force

acting on the moving ball and it has already left the circular path

(because the slack string means that the distance of the ball from

the point of suspension is smaller than the length of the string),

therefore, motion of the ball would be equivalent to that of a

projectile moving in a parabolic path.


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PHYSICS 42LOCUS

ALTERNATE APPROACH:

II

I

III

IV

Vertical line

Horizontalline

fig. 5.40

Let us divide the vertical circle under consideration in four quadrants,

as shown in figure 5.40. Now, we will analyze the each quadrant

separately.

Quadrant I:

!

mgsi

n!

v

T2v

!

mg

cos!

u fig. 5.41

Some arbitrary position of the ball in this quadrant is shown in

figure 5.41. From this figure it is clear that the radial component of

the gravity is in the opposite direction of the radial acceleration of

the ball and hence in this quadrant tension has two roles: first is to

balance the radial component of gravity and second is to provide

the necessary centripetal acceleration to the ball, as discussed before.

Hence, in this quadrant tension can never be zero.

If the balljust managesto reach the horizontal position then at this moment alongwith the speed of the ball tension

in the thread also becomes zero because at this position the radial acceleration and the radial component of gravity

both are zero, as shown in figure 5.38 and hence there is no requirement of tension in the thread. Using work energy

theorem or conservation of mechanical energy we can prove that for this to happen ushould be 2 .gl

Quadrant II:

Some arbitrary position of the ball when it is in quadrant II is shown in figure 5.42(a). It is obvious from the figure

that as the ball moves up, its speed decreases and component of gravity along the radially inward direction increases

and hence requirement of tension force becomes less and less as the ball moves up in this quadrant. Eventually at the

highest point of the circle the speed of the ball becomes minimum and the contribution of gravity in centripetal force

becomes maximum and hence at this position requirement of tension is minimum, as shown in figure 5.42(b). Therefore

this position can be defined as thecritical position, because if the ball crosses this position successfully, i.e., if the

string is taut in this position, it would always be taut or we can say that the ball would complete the vertical circle.

fig. 5.42 (b)

Tmg

2/v !

O

v

(requirement of tension is minimum at this position, i.e.,chance of slacking of string is maximum at this position)

fig. 5.42 (a)

T

mgsin

!

!

2/v !

O

mg

cos!

v


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PHYSICS 43LOCUS

If the ball is just completing the vertical circle we can assume zero tension at the critical position (topmost

point) because gravity is there to provide the required centripetal force to the ball. In such a case if vbe the speed

of the ball at the highest position, then

2mvmg

l

=

2v gl=

and if ube the speed of the ball when it reaches the lowermost position, then using

loss in . .f ik k P E = +

we get,2 21 1 (2 )

2 2mu mv mg l = +

2 4u gl gl = +

5u gl=

Hence, to complete the vertical circle, 5u gl.

Note:

* To just complete the vertical circle you can not assume zero speed at the highest point. Why so? Try to

answer it on your own.

* The only difference between the analysis of the ball in quadrant II and III is that when the ball is moving in

quadrant II, it is speeding down and while it is moving in quadrant III, it is speeding up. Similar argument canbe given for quadrants I and IV.

SUMMARY:

* When 2u gl : The ball oscillates with angular amplitude, 0 /2.

* When 2 5gl u gl< < : The ball leaves the vertical circular path at some position (which woulddepend upon u) in the IInd quadrant and thereafter moves in a parabolic

path.

* When 5u gl : The ball moves in a complete vertical circular path.


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PHYSICS 44LOCUS

!

F

l

m

1. An object of mass mis tied to a string of length land a

variable forceFis applied on it which brings the stringgradually at angle with the vertical. Find the work done

by the forceF. [Solving using potential energy method]

2. A body is dropped from a certain height. When it lost an

amount of P.E. U, it aquires a velocity v. The mass of the

body is :

(a) 2 / U v (b) 22 /v U

(c) 2v/U (d) 2/2 .U v

3. A particle of mass mis attached to a light string of length l, the other end is fixed. Initially the string is kept

horizontal and the particle is given an upward velocity v. The particle is just able to complete a circle.

(a) The string becomes slack when the particle reaches its highest point

(b) The velocity of the particle becomes zero at the highest point

(c) The kinetic energy of the ball in initial position was21 .

2mv mgl =

(d) The particle again passes through the initial position.

4. A small block of mass m slides along the frictionless loop-the-loop track

shown in fig. (a) If it starts from rest atP, what is the resultant force actingon it at Q? (b) At what height above the bottom of the loop should the

block be released so that the force it exerts against the track at the top of

the loop is equal to its weight?

RQ

5R

P

5. A simple pendulum of length l, the mass of whose bob is m, is

observed to have a speed 0v when the cord makes the angle 0

with the vertical 0(0 /2), < < as in fig. In terms of g and theforegoing given quantities, determine (a) the total mechanical energy

of the system; (b) the speed1v of the bob when it is at its lowest

position; (c) the least value 2v that 0v could have if the cord is toachieve a horizontal position during the motion; (d) the speed 3v

such that if 0 2v v> the pendulum will not oscillate but rather willcontinue to move around in a vertical circle.

! l

m

0

0v

6. A chain of length land mass mlies on the surface of a smooth hemisphere of radiusR> lwith one end tied to

the top of the hemisphere. Find the gravitational potential energy of the chain.

R

ZERO P. E. LEVEL


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PHYSICS 45LOCUS

7. A smooth sphere of radiusRis made to translate in a straight line with a constant acceleration a. A particle

kept on the top of the sphere is released from there at zero speed with respect to the sphere. Find the speed

of the particle with respect to the sphere as a function of the angle it slides. [solve using potential energymentod]

8. A block rests on an inclined plane as shown in figure. A spring to which it is attached via a pulley is beingpulled downward with gradually increasing force. The value of s is known. Find the potential energy Uof

the spring at the moment when the block begins to move.

m

k

F!

s

9. The particle min figure is moving in a vertical circle of radiusRinside a

track. There is no friction. When mis at its lowest position, its speed is 0v .

(a) What is the minimum value mv to 0v for which mwill go completely

around the circle without losing contact with the track? (b) Suppose 0v is

0.775 .mv The particle will move up the track to some point atPat which

it will lose contact with the track and travel along a path shown roughly by

the dashed line. Find the angular position of pointP.

R

P

v0

m

!

10. A chain of length land mass mlies on the surface of a smooth sphere of radiusR> lwith one end tied to thetop of the sphere.

(a) Find the gravitational potential energy of the chain with reference level at the centre of the sphere.

(b) Suppose the chain is released and slides down the sphere. Find the kinetic energy of the chain, when

it has slide through an angle .

(c) Find the tangential accelerationdv

dtof the chain when the chain starts sliding down.

11. A spherical ball of mass mis kept at the highest point in the space

between two fixed, concentric spheresAandB (see figure). The

smaller sphereAhas a radiusRand the space between the twospheres has a width d. The ball has a diameter very slightly less

than d. All surfaces are frictionless. The ball is given a gentle push

(towards the right in the figure). The angle made by the radius vector

of the ball with the upward vertical is denoted by (shown in thefigure)

!O

R

SphereB

SphereA

d

(a) Express the total normal reaction forces exerted by the spheres on the ball as a function of angle .

(b) LetAN and BN denote the magnitudes of the normal reaction forces on the ball exerted by the

spheresAandB, respectively. Sketch the variations ofAN and BN as functions of cosin the range

0 by drawing two separate graphs, taking coson the horizontal axes. Also sketch thevariations ofN

AandN

Bas functions of .


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PHYSICS 46LOCUS

CONSERVATIVE FORCES AND POTENTIAL ENERGY AGAIN:

The interaction of a particle with surrounding bodies can be described in two ways : by means of forces or

through the use of notion of potential energy. In classical mechanics both ways are extensively used. The first

approach, however, is more general because of its applicability to forces in the case of which the potential energy is

impossible to introduce (i.e., nonconservative forces). As to the second method, it can be utilized only in the case ofconservative forces.

Our objective is to establish the relationship between potential energy and the force of the conservative

field, or putting it more precisely, to define the conservative field of forces ( )F r! !

from a given potential energy ( )U r!

as a function of a position of a particle in the field.

We have learnt by now that the work performed by conservative forces on a particle during the displacement

of the particle from one point in the field to another may be described as the decreaseof the potential energy of the

particle, that is,

.conw U= The same can be said about the elementary displacement dr

!as well:

condw dU =

or = ! !F dr dU ...(37)

Recalling that ,tdw F dr F ds= = ! !

where ds dr = !

is the elementary length covered along the path and tFis the

tangential component of ,F!

we shall rewrite equation (37) as

.tF ds dU =

Hence, = t

UF

s

...(38)

i.e., the projection of the conservative force at a given point in the direction of the displacement dr!

equals the

derivative of the potential energy Uwith respect to a given direction, taken with the opposite sign. The designation

of apartialderivative / s emphasizes the fact of differentiating with respect to a finite direction.

The displacement dr!

can be resolved along any direction and, specifically, along thex,y,zcoordinate axes. For

example, if displacement dr!

is parallel to thexaxis, it may be described as .dr dxi=!

The work performed by theconservative forceF

!over the displacement dr

!parallel to thexaxis is

( ) ,xF dr F dxi F dx = = ! !!

where xF is thex-component of the force .F!

Substituting the last expression into equation (38), we get

=xU

Fx

...(39)

where the partial derivative symbol implies that in the process of differentiating ( , , )U x y z should be considered as

a function of only one variable,x, while all other variables are assumed constant. It is obvious that the equations for

yF and zF are similar to that for .xF So, having reversed the sign of the partial derivatives of the function Uwith

respect tox,y,z, we obtain the components , andx y zF F F of the conservative force .F!


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PHYSICS 47LOCUS

Hence, we have

2

x yF F i F j F k= + +!

= + +

! U U UF i j k

x y z

...(40)

The quantity in parentheses is referred to as the scalar gradient of the function Uand is denoted by grad Uor

U. Generally the second, more convenient, designation where (nabla) signifies the operator

i j kx y z

= + +

is used. Consequently, we can write,

=!F U ..(41)

i.e., the conservative force F!

is equal to the potential energy gradient, taken with the minus sign. Put simply, the

conservative forceF!

is equal to the antigradient of potential energy.

The potential energy of a particle in certain conservative field has the following form:

(a) U(x,y) = xy, where is a constant;

(b) ( )U r!

= ,a r! !

where a!

is a constant vector and r!

is the position vector of the particle in the field.

Find the conservative field force corresponding to each of these cases.

Solution:(a) We have,

U UF i jx y

= +

!

( ) ( ) xy xyi jx y

= +

x yy i x jx y

= +

( )yi xj= +

(b) We have, x y za a i a j a k = + +!

and

r xi yj zk = + +!

Therefore,

U a r= ! !


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PHYSICS 48LOCUS

; thenx y za x a y a z = + +

U U UF i j kx y y

= + +

!

( )x y za i a j a k = + +

a= !

A conservative forceF(x) acts on a 1.0 kg particle that moves along thex-axis. The potential energy U(x) is given

as 2( ) ( )U x a x b= + where a= 20 J, b= 2 m. andxis in meters. Atx= 5.0 m the particle has a kinetic energy

of 20J. It is known that there is no other force acting on the system. Based upon this information, answer thefollowing questions:

(a) What is the mechanical energy of the system?

(b) What is the range ofxin which the particle can move?

(c) What is the maximum kinetic energy of the particle and the position where it occurs?

(d) What is the equilibrium position of the particle?

Solution: We have,2( ) ( )U x a x b= +

220 ( 2)x= +

We know that atx= 5.0, K.E. = 20 J, therefore, mechanical energy of the particle,

E= Potential energy, U+ kinetic energy,k

(at 5) (at 5)U x k x= = + =As there is no nonconservative force

acting on the particle, its mechanical

energy is conserved, i.e.,E at 5 is equal to the at any .x E x

=

220 (5 2) 20= + +

49 . [Ans.(a)]J=

We know that,

E= U+ k

k E U=

249 20 ( 2)x = +

229 ( 2)x=


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PHYSICS 49LOCUS

As the particle moves on thex-axis, its kinetic energy can never be negative, therefore, we have,

0k

229 ( 2) 0x

2( 2) 29x

2 4 4 29 0x x +

2 4 25 0x x

[ ]3.38, 7.38x [Ans (b)]

When the particle has maximum kinetic energy, we have,

0dk

dx =

229 ( 2)

0d x

dx

=

0 2 ( 2) (1) 0x =

2x=

i.e., at 2x= , the particle has maximum kinetic energy which is equal to 29 J. [Ans. (c)]

As the conservative force is antigradient of potential energy, we have,

( ) dU

F xdx

=

220 ( 2)d x

dx

+ =

[ ]0 2 ( 2) (1)x= +

4 2x= .

When the particle is in equilibrium, net force on it must be zero. As the only force acting on the particle isF(x), in

equilibrium position

( ) 0F x =

4 2 0x =

2 . [Ans.(d)]x m=


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PHYSICS 50LOCUS

ALTERNATE METHOD:

Let us solve this problem graphically. For that first calculate mechanical energy,E. Following the same procedure as

we did in the last method, we get

E= 49J.

and we already have,2( ) 20 ( 2) .U x x= +

7.382O3.38

20

49

E J=49

U J( )

X( )m

fig. 5.43

The plots ofEand U(x) are shown in figure 5.43. From the graph

it is clear that atx= 7.38 andx= 3.38 Ubecomes equal toE,

hence, at these positions k= 0 (becauseE= k+ U). Forx> 7.38

andx< 3.38,Ubecome greater thanEand hence kwould acquire

ve value, which is never possible. Therefore, particle can move

only forxgreater than 3.38 and smaller than 7.38.

As the sum of kand Uis constant, kwould be maximum when U

is minimum. Therefore, atx=2, kis maximum and is equal to49 U= 49 20 = 29J.

Atx = 2, slope of Uis zero, therefore, atx= 2,F(x), which

negative of slope of U, is also zero. Therefore,x= 2 is equilibrium

position of the particle.

NATURE OF EQUILIBRIUM

Whenever a conservative force is the only force acting on a particle, equilibrium positions of the particle can be

determined from the graph of Uwhen plotted against the position of the particle.

Consider the case shown in figure 5.44. In this figure potential

energy, U, of a particle under the action of a conservative force

F(x) is plotted against the position of the particle,x.

It is obvious from the graph of Uthat at 1x x= it has a maxima andat 2x x= it has a minima. Therefore, at both 1x and 2x derivativeof Uis zero and henceF(x) is zero. Consequently

1x and 2x are

equilibrium positions of the particle. Therefore, we can say that

extrema of Uoccur at equilibrium positions of the particle.

fig. 5.44

x2Ox

U

x1

Now, let us analyze the force on the particle when it is in the vicinity

of one of its equilibrium positions or it is in the vicinity of an extrema

of its potential energy.

For values ofxvery close to 1x but smaller than 1x derivative of

F x( ) F x( )x1

xO

U

fig 5.45

Uis positive because its tangent makes an acute angle with the +ve

direction of thex-axis. Therefore, in this region the force on the

particle is along negativexdirection (or we can say that it is away

from1x ), as shown in figure 5.45. Similarly, it can be justified that

for values ofxvery close to1

x but greater than 1x force,F(x), is

positive, i.e., it acts away from1x , as shown in figure 5.45. Now,

suppose a particle in equilibrium atx=1x is slightly displaced from

this position the either side and released under the action of

conservative force,F(x), only. What would happen now? The force


51/70


PHYSICS 51LOCUS

on the particle would act away from the equilibrium position

x=1x and the particle would never come back to this equilibrium

position. Such an equilibrium position is defined as unstable

equilibrium position.

Now, let us analyze the behaviour ofF(x) in the vicinity of a minimaofU. In figure 5.46

2x x= is a minima of U. Arguments similar towhat I provided in the previous paragraph lead to the fact that in

the vicinity of2x (a minima of potential energy) conservative force,

F(x), acts always towards2x . Therefore, if a particle is displaced

from2x on either side and released from rest and thereafter only

conservative force,F(x), acts upon it, it will return back to its

equilibrium position2( )x x= . (A little more thought would give an

idea of oscillation about equilibrium position, which I dont want to

discuss here.) Such an equilibrium position is defined as a stable

equilibrium position.

F x( ) F x( )x2

xo

y

fig 5.46

Therefore, the position where potential energy, U, Posesses a maxima (first derivative of Uis zero and second

derivative is negative) is an unstable equilibrium position and the position where potential energy posesses a minima

(first derivative ofUis zero and second derivative is positive) is a stable equilibrium position. Therefore, in example

20,x= 2, was a stable equilibrium position.

The potential energy of a particle in a certain field has the form 2/ / ,U a r b r = where aand bare positive constants,ris the distance from the centre of the field. Find:

(a) the value of 0r corresponding to the equilibrium position of the particle, examine whether this

positionis stable;

(b) the range of the attraction force.

Solution:(a) At equilibrium position:

0dU

dr=

2( / / )

0d a r b r

dr

=

3 22

0a b

r r

+ =

2abr

=


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PHYSICS 52LOCUS

2a

rb

=

0 2 /r a b=

Now, ( )2 3 2

2

2a bd

d dU dr d U r r

dr dr dr

+ = =

4 3

6 2a b

dr r=

At 0,r r=2

2 4 3

0 0

6 2d U a b

dr r r =

4 3

4 3

6 2

16 8

a b b b

a a

=

4 4

3 3

3 1

8 4

b b

a a=

4

3

1

8

b

a=

Therefore, at2

0 2,d U

r rdr

= is positive and hence this is an unstable equilibrium position.

(b) Radial component of force, ( ),rF r is antigradient of Uwith respect tor. therefore,

( )rdU

F rdr

=

3 2

2a b

r r

=

When this force is attractive, it must be along radically inward direction and hence it should be negative. That is,

( ) 0rF r

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