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Newton’s Third Law of Motion
SOMNATH DATTA
656 “Snehalata, 13th Main, 4th Stage, T.K. Layout, Mysore 5700009, India
E.mail: [email protected]
http://sites.google.com/site/physicsforpleasure
1 A Slide Show on Newton’s Third Law
of Motion
1. Newton’s first and second laws tell us that a
body accelerates due to external forces. The net
external force equals the mass of the body
times its acceleration. These laws do not tell us
where this force comes from. Does the force
that drives this car come from the engine?
2. If so, why is this car stuck on wet clay road
when the engine is running in full throttle? The
force that pushes the car forward does not
come from the engine. This force comes from
the road due to friction between the tyre and
the road, and is called traction.
3. Does the force come from his muscles?
Physics Education • July − September 2009
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4. If so, why is he slipping? No runner is his
able to run on a slippery track in spite of his
muscle power. The force that drives the runner
forward comes from the ground when there is
sufficient friction between his feet and the
ground.
5. This car requires 40,000 N to speed up at the
rate of 10m/s2. This sprinter requires 300 N to
speed up at the rate of 5m/s2. How is the road
able to judge the individual force requirements
for the car and the sprinter?
6. Answer to this comes from Newton’s Third
Law of Motion:
•
•
Nothing in the universe can act without
being acted upon.
To every action there is an equal and
opposite reaction.
The first statement is somewhat
philosophical. The second one is vague. In
exact terms what the first statement means is
that whenever any object A offers a force F, it
itself experiences the same force back on itself
in the reverse direction.
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7. The second statement can be interpreted as
follows. Assume two objects A and B. A offers
a force on B directed from A to B. We denote
this force by FAB. The 3rd Law says B will
simultaneously offer a force FBA on A which is
directed from B to A. These two forces are
equal and opposite, even if the objects A and B
are grossly unequal in size and mass. Written
as vectors,
FBA = −FAB.
(1)
We can refer to any one of the pair of forces,
say the force FAB, as the action force and the
second one, i.e. FBA, as the reaction force.
Then Eq. (1) is a mathematical expression of
the second statement.
In the particular example shown these two
forces are repulsive forces.
8. Here they are attractive forces. A pulls B
with a force FAB. As a result A is itself being
pulled towards B with a force FBA, and FBA =
−FAB.
9. A runner can run only if the ground kicks
him forward. To get this kick the runner must
kick the ground backward. It is possible to kick
a solid and firm object like football or grass
with sufficient force, but it is difficult to kick a
loose object like air or slippery road with
sufficient force. Therefore the runner is unable
to run on a slippery road.
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10. It is possible to get the necessary force of
40,000 N from a dry ground because the tyre,
powered by the engine, can press against the
road with this force due to friction between the
two surfaces (of the tyre and the ground.) A
frictionless surface on the other hand will not
allow the tyre to rub with a large force and
because of this the car will not be able to
accelerate.
11. There is a common misconception about
the Third Law especially among the beginners.
They think that this law holds only when the
interacting bodies are in equilibrium since the
forces of action and reaction will then cancel
each other. This is entirely wrong. Here we
show two bodies A and B.
12. But the force FAB is acting on B only.
Therefore B is not in equilibrium.
13. The reaction force FBA is acting on A only.
Therefore A is not in equilibrium.
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14. Now see the complete picture. The forces
are equal and opposite. But they do not cancel
each other. Because they are acting on different
bodies.
15. We see that the third law holds between
any two bodies, whether at rest or in motion.
16. Let us illustrate the Third Law with a few
more examples. This stone of mass 1 kg is
falling to the ground in a parabolic trajectory
with acceleration g equal to 9.81 m/s2, like any
other stone. The gravitational pull on the stone
is m times g, that is, 9.81 N. Since the earth is
pulling the stone with force 9.81 N downward,
the stone is at the same time pulling the earth
with a force of the same magnitude of 9.81 N,
but directed upward.
17. Therefore the earth is also accelerating
toward the stone! However, since the mass of
the earth is very LARGE, its acceleration
(toward the stone) is very small.
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217
18. Another example: A football player kicks a
football. Therefore according to the 3rd Law,
the football kicks the player back.
19. The kick is an impulsive force, lasting for a
very short time, say, 101 of a second. Taking
the mass of the ball to be ¼ kg, we have
calculated the kick force to be 250 N.
20. The 3rd Law tells us that the ball also kicks
the player back in the opposite direction.
However, since the mass of the player is 100
kg, i.e., 400 times the mass of the ball, his
acceleration is 400 times less and lasts exactly
for 101 second.
21. Here we see the complete picture. The
mutual kick forces are equal and opposite. But
they do not cancel each other, because they act
on different bodies.
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Physics Education • July − September 2009
22. Let us try to understand how this horse is
able to pull the cart.
23. The cart requires a force of 2500 N to
accelerate. But there is a backward frictional
drag on the wheels. Therefore the horse must
pull the cart with a force of 3000 N.
24. The horse needs 1500 N to accelerate.
According to the 3rd Law, the cart is pulling
horse back with force 3000 N. Therefore the
ground must provide the necessary traction
force of 4500 N in the forward direction.
25. How does the horse get this 4500 N
traction force? By pressing against the ground
with this force, according to the 3rd Law. Also
since the wheels are experiencing frictional
drag of 500 N backward, the ground is at the
same time dragged forward with 500 N, again
due to the 3rd Law. The resulting acceleration
of the ground is very small, since the mass of
the earth is very LARGE.
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219
26. Now see the complete picture. For every
force there is an equal and opposite reaction
force. But the forces on any one of the three
bodies are not in equilibrium. Therefore, the
cart, the horse and the ground are all
accelerating (though in different degrees).
27. As final example we consider the motion of
the earth and the moon due to the gravitational
attraction between the two. Note the distance
between the centres of the earth and the moon,
and the masses of the earth and the moon. The
gravitational force with which the earth is
pulling the moon is 20.6 × 1019 N.
28. As a result the moon is accelerating toward
the earth and is moving in a circle.
29. The earth is experiencing a force of exactly
the same magnitude of 20.6 × 1019 N but
directed toward the moon. However, the mass
of the earth is about 83 times the mass of the
moon. Therefore the earth is undergoing a
much smaller acceleration of 3.44 × 10−5m/s2.
As a result the centre of the earth is also
moving in a circle, a much smaller circle.
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Physics Education • July − September 2009
30. Both the moon and the earth are
accelerating toward each other according to the
Third Law of Motion. The centre of each is
moving in a circle, one circle being much
larger than the other because one mass is much
smaller than the other.
2 Free Body Diagrams
Newton’s first law of motion is difficult to
believe. Why should anyone believe that an
object keeps moving forever along a straight
line in the absence of external interferences,
when our everyday encounters seem to
contradict the statement?
Similarly Newton’s third law of motion is
rarely believed. It is easy to “imagine” equal
and opposite action-reaction forces when, for
example, a stone is resting on a table. But how
to trust the same statement when the table
breaks and the stone crashes down?
The confusion is understandable, and
persists. The reason − our apathy to explain the
theory properly with effective drawings. For
understanding concepts and principles of
mechanics drawing freebody diagrams is an
essential first step. Without this Newton’s laws
of motion can never be appreciated.
• A car and the ground in Slide 10;
• A falling stone and the earth in Slides 1617;
• A football player, football and the ground
in Slides 19- 21;
• A cart, a horse pulling it, and the ground in
Slides 23-26;
• The earth and the moon in Slides 27-29.
2.2 The Golden Rules for Understanding
Newton’s Laws of Motion
For understanding Newton’s laws of motion
properly, especially their applications to
extended objects, it is essential that the
following golden rules be kept in mind.
1.
2.1 What is a Freebody Diagram?
2.
A freebody diagram (FBD) is a pictorial
formulation of the mechanics problem through
figures in which each object in a system of
interacting bodies is shown separately and the
forces acting on this body (and on this body
alone) are clearly delineated and also shown
separately. The slides presented in the previous
section have shown many examples of FBDs.
Some of the important ones are the following
3.
• A runner and the ground in Slide 9 (The
gravity force has been suppressed);
Physics Education • July − September 2009
Clearly identify the “system” whose
motion you are going to analyze.
Sometimes it may be useful to draw a
boundary around this system in order to
identify it and isolate it from all
neighbouring systems.
All the forces that are coming from outside
this system you have just identified − by
this I mean the forces that are coming from
outside the boundary you have drawn - are
to be called external forces. Forces of
“interaction” among objects inside the
system that is, forces coming from objects
that are confined within the boundary you
have drawn − are all internal forces.
Therefore draw a freebody diagram (FBD)
of all the external forces. You must
remember that FBDs are a must for
understanding mechanics. It is impossible
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to appreciate the principles of mechanics,
the forces and how they guide the motion
of objects, without drawing FBDs.
Add up vectorially all the external forces
acting on the system. In some cases this
vector sum may be zero. In that case a
central point of the object, called the
Centre of Mass, to be abbreviated as CM,
will be either stationary or moving with
constant velocity (i.e., constant speed
along a straight line.)
4.
We shall refer to the motion of the CM as
the bulk translatory motion of the system.
If the vector sum of the external forces is
non-zero, then that vector sum Ftotal will be
called the Resultant Force on the system.
The bulk translatory motion of the system
is determined by Newton’s second law of
motion, which can now be written as
M r = Ftotal
(2)
where M is the total mass of all the component
parts of the system, and r is the radius vector of
the CM of the system, so that r is the
acceleration of the CM.
We have illustrated Rules 1,2,3 in Figures
1. In the Top Box we have shown three objects
A, B and C interacting with one another, and
forming a system. We assume that this system
of three objects is not subjected to any
“external force”. Individually the objects A, B
and C are moving and accelerating only due to
the forces of interaction among themselves.
In frame (a) we have depicted each one of
them as a single system, by drawing
boundaries (with dotted lines) around each.
The forces of interaction are (F1,F′1), (F2,F′2),
(F3,F′3). The pair of forces within parentheses
are the “equal and opposite” action-reaction
pairs. For example if F′1 is the action force,
then its counterpart F1 is the corresponding
(equal and opposite) reaction force. Similarly if
F3 is the action force, then its counterpart F′3 is
the corresponding (equal and opposite) reaction
force. Note also that we have drawn “shooting”
lines on each one of the three objects indicating
that the objects are all moving objects having
velocities va, vb, vc respectively.
Each one of the three systems A,B,C is
subjected to the following external forces:
(1) external forces F1,F2 on A;
(2) external forces F′1,F′3 on B;
(3) external forces F′2,F3 on C.
In frame (b) we have clubbed A and B
forming a single system AB. Now it is an
interaction between two systems namely, AB
and C. The only external force on AB is F′, and
the only external force on C is F = −F′.
In frame (c) we have taken A, B and C
together as a single system ABC. There are no
external forces on this system ABC.
We shall now illustrate Rule 4 in the form
of the following equations of motion.
System
A
Net external force on the system
Ftotal = F1 + F2
Eqn of Motion
ma ra = F1 + F2
B
Ftotal = F′1 + F′3
mb rb = F′1 + F′3
C
Ftotal = F′2 + F3
mc rc = F′2 + F3
AB
Ftotal = −F = −(F′2+F3) = F2+ F′3
(ma+mb) rab = F′=−F
ABC
Ftotal = 0
(ma+mb+mc) rabc =0
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(3)
Physics Education • July − September 2009
Figure 1. Forces of interaction between systems of material objects.
In the above table ma,mb,mc represent the
masses of the systems A, B, C respectively,
and ra, rb, rc, rab, rabc represent the locations of
CMs of the systems A, B, C, AB, ABC
respectively.
In the Bottom Box of Figure 1 we have
shown a stone falling under the force of
gravity. The bulk motion of the stone is
determined by the gravity force mg. We had
shown an FBD of this object in slide 16 of the
previous section. However, we have repeated
the FBD in frame (a).
The stone is composed of a large number of
atoms, of the order of, say, 1025. These atoms
are pulling or pushing each other. Yet, they are
all internal forces and, therefore, have no effect
on the bulk motion of the stone.
Now, “imagine” a boundary line Σ dividing
the stone into parts A and B, which we shall
refer to as A-stone and B-stone respectively.
Then there are action-reaction forces (F,F′)
Physics Education • July − September 2009
between the two parts, which are always equal
and opposite. We have shown the FBD of both
the parts in frame (b). The CM of A-stone is
moving under the effect of (1) the gravity force
mag, and (2) the force F on it coming from B.
These two forces cause the falling of the Astone under gravity, along with rotational
motion of the CM.1
In the same manner the CM of the B-stone
is moving under the effect of (1) the gravity
force mbg, and (2) the reaction force F′ = −F on
it coming from A. These two forces together
cause the falling under gravity along with
rotation of the CM of B.
We shall now apply Rule 4 to the falling
stone and its two parts and obtain the following
equations of motion.
1. There can also be oscillatory motion of the CMs
of A and B, due to vibration induced by the forces F
and F′ respectively.
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System
Whole stone
A-stone
B-stone
Net external force on the system
Ftotal = Mg
Ftotal =mag+ F
Ftotal = mbg − F
In the above table (M = ma + mb, r), (ma,
ra), (mb, rb) represent (mass, location of the
CM) of the whole stone, A-stone, B-stone
respectively.
The reader should note here an important
point. The only external force on the whole
stone is the gravitational force. Hence the point
C (representing the CM of the stone) falls in a
parabolic path like an ordinary projectile. On
the other hand A-stone and B-stone are
subjected to extra forces F and −F
respectively, besides the force of gravity. Their
CMs Ca and Cb follow “nonparabolic” paths.
Eqn of Motion
M r = Mg
ma r a = mag+F
r
mb b= mbg+F′ = mbg−F
(4)
4 Every Real Force has a Parent
In Figures 2-5 we have cited a number of
instances in which the third law presents us
with a paradox. We have illustrated graphically
each situation by identifying the forces with
FBDs, and in two cases listing them in a
tabular form. This, we hope, will facilitate
clarification of the Second and the Third Laws
of Motion.
We have presented in Tables 1 and 2 the
“Forces at Work” corresponding to two of the
examples cited, namely, (a) Diving, and (b)
Tug of War. In the right column of each table
you will see the forces on the subsystems. The
action-reaction forces on these subsystems
taken together add up to zero, so that the net
force on the “Total” system is nil.
In writing the vectors in Table 2 we have
taken the forces T1,T2,F1,F2 to be in the
direction of the positive X axis (i.e.,
rightward), and the forces N1,N2,M1g,M2g to be
in the direction of the positive Z axis (i.e.,
upward).
Newton’s first law of motion gives a
qualitative meaning of force. A force is
something which makes a particle, a system,
deviate from its “most natural” motion,
namely, moving in a straight line with constant
velocity. The second law pinpoints the concept
further by painting force in the colour of a
vector, a directed straight line with a specified
length − equal to the rate of change of
momentum of the object. The third law refines
the concept of force further by making the
grand statement, “force is an interaction
between two objects in which each one is
equal.”
The earth pulls a mosquito with the same
force with which the mosquito pulls the earth.
A proton pulls an electron with the same force
with which the electron (about two thousand
times lighter) pulls the proton. These
statements are as profound as the shock and
disbelief they generate in our minds. Yet the
statements are true.
However, the reader must not miss a crucial
assumption in the statements of the laws of
motion. Velocity, momentum, acceleration are
frame dependent quantities. They have no
meaning unless you have specified a frame of
reference. What frame? The frame in which the
statements are valid.
That reads like the old joke: which came
first, the egg or the chicken?
However, the assumption of an inertial
frame is not an absurdity. The concept of a
family of inertial frames (the name of the
frames in which the laws of motion hold) lies
at the foundation of classical mechanics.
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Physics Education • July − September 2009
3 Further Examples of FBDs
Figure 2: Free body diagrams of diving. The action-reaction pair of forces are shown with the same
roman bold letter appearing twice, once with prime and once again without it. The forces at play are
the tension forces (T,T′) between the board and the left pillar; the compressive (pressure) forces
(P,P′) between the board and the left pillar; the force R′ with which the diver kicks the board and the
kickback reaction force R he receives, the gravitational forces (G1,G′1), (G2,G′2), (G3,G′3), (G4,G′4).
Drawings on Left show Normal Diving. Drawings on Right show accident in diving. The board gets
uprooted from one of the supporting pillars, and the diver tumbles into the water. In this case the
pair of forces T,T′ disappear, and they disappear simultaneously.
Physics Education • July − September 2009
225
Figure 3. We have now written each pair of action reaction forces with the same symbol (even
though they have opposite signs.) For example (T1,T1) are action-reaction forces of tension. Top :
Tug of War. Note that the tension forces T1,T2 are not equal in magnitude, because the rope has
considerable mass m and it is accelerating. The contestant on the right is pushing the ground with
friction force F1, and a downward force N1, and at the same time pulling the earth upward with a
gravitational force M1g. The earth is repaying him back with the same force in opposite direction.
Bottom : Boy jumping off a boat. In Figure (b) we have shown the forces acting on the “boat + boy”
system. Figures (c) and (d) show the FBDs of the boat and the boy separately. We have removed
the earth from our “system”.
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Physics Education • July − September 2009
Figure 4. Top: FBD of an unfortunate man who is sinking into a quicksand. He sinks because the
gravitational force downward is larger in magnitude than the upward reaction force provided by the
ground. We have not shown the action-reaction counterpart of this gravitational force (pulling the
earth upward.) Bottom : Action-reaction forces between a balloon and the escaping gas. We have
marked out a rectangular column of the gas which is pushed downward with a gas-pressure force
of magnitude F1. This causes an upward reaction force of the same magnitude acting on the
balloon. The balloon accelerates upward because this reaction force F1 is larger than the weight of
the balloon (same as the downward force of gravity on it. The lateral pressure forces F2, F3, F4, F5
are balanced causing no sidewise acceleration of the gas. Their reaction counterparts are similarly
balanced causing no sidewise acceleration of the balloon.
Physics Education • July − September 2009
227
Figure 5. A locomotive steam engine hauling a train consisting of only two coaches. We assume
that the train is moving with constant speed, so that the net force on the train is zero. (a) General
picture. The train is moving from right to left as indicated by the arrow at the front and by the
direction in which the Driving Wheels are turning. (b) FBD of the engine and the coaches. Note that
the driving wheels, connected to the piston (reciprocating inside the steam chamber), shown as a
box) by the connecting rod, are pressing the rails backward thereby getting the traction force Ft,
directed forward. It is this traction force that moves the train. The other wheels are subjected to
rolling frictions Fr1,Fr2,Fr3,Fr4, directed backward, and trying to slow down the train. T1,T2 are
tensions in the chains that connect the engine to coach 1, and coach 1 to coach 2 respectively. (c)
FBD of the whole train. (d) FBD of the rails. We have suppressed the gravity forces. In the boxes
we have indicated the force equalities. When the train accelerates, the equality symbol “=”
becomes “greater than” symbol “>”.
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Physics Education • July − September 2009
Table 1. Forces at work in diving
System
Diver
Diving board
Left pillar
Right pillar
Earth
Total
External force on the system
R+G1
T+G2+P+R′
T′+G4
P′+G3
G′1+G′2+G′3+G′4
0
(5)
Table 2. Forces at work in tug of war
System
Player-right
Player-left
Rope
Earth
Total
Net external force on the system
−T1+F1+N1−M1g
T2−F2+N2−M2g
T1 −T2
−F1+F2−N1−N2+M1g+M2g
0
Anyone who applies Newton’s laws of
motion must be standing in an inertial frame.
The space capsule in which the modern day
heroes − the astronauts − are floating is the best
example of an inertial frame. In contrast, the
earth is not.
What are the forces acting on a missile shot
out from an earth station? Along with the main
force, the force of gravity, there are other
forces, the centrifugal force and the Coriolis
force. These “other forces” are not forces of
interaction. And Newton’s third law of motion
does not apply to them.
Imagine yourself sitting in a fast
accelerating train (better still imagine yourself
sitting inside a jet plane as it accelerates along
the runway before taking off.) You feel as if
someone is pushing you backward, against the
chair in which you are sitting. Let us refer to
such forces as force type l.
Think of another sensation. You are on a
rotating platform of a merry −go-round − your
first stop on entering a carnival park. If you are
sitting on a chair on the go-round, you feel that
someone is trying to throw you outward, away
Physics Education • July − September 2009
(6)
from the centre, but the chair is saving you
from such disaster. Even stranger will be your
feeling if you try to walk on the platform.
Besides the outward “throw out” force, called
the centrifugal force, you will also feel a
“transverse force” (i.e. perpendicular to the
direction you are walking in) making you feel
dizzy. This sideward force is called Coriolis
force. Let us refer to these two forces as force
types 2 and 3)
The force types 1,2 and 3, are not forces of
action-reaction. They completely disappear
when the platform you are on (train, aeroplane,
merry-go-round) comes to a stop. People club
such forces under fancy names : fictitious
forces, pseudo-forces.
The pseudo forces are clearly distinct from
the real forces.
All real forces have parents. They originate
from some material source, and reach out to
other material objects altering their motions in
accordance with Newton’s laws.
In contrast, the pseudo-forces do not have
parents, i.e., they do not originate from
material sources. We mistake them to be forces
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because we are always thinking our frames to
be inertial even when they are not.
Physicists do not feel shy of applying
Newton’s laws of motion even when sitting in
noninertial frames. They make up for their
errors by inventing the fictitious pseudo-forces.
References
5 Mach’s Principle
1.
Pseudo-forces are not as untouchable, bereft of
respectability, as their category may indicate.
They have a trail of glorious history. It was
Ernst Mach’s deep insight that brought
pseudoforces on par with real forces by
suggesting that the pseudo-forces originate
from the mighty galaxies of an accelerating
universe (when your frame is accelerating with
respect to an inertial frame, the universe is also
accelerating backward with respect to your
frame.) Albert Einstein was so influenced by
this profound message that he removed all
distinctions between inertial and non-inertial
frames and proposed a theory of gravitation in
which pseudo-forces are no different from
gravitational forces.
Acknowledgment
The author gratefully acknowledges the help
provided by Michael Murphy and Professor
A.V. Gopala Rao help in installing Linux for
preparation of the manuscript.
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There is one full chapter (18 pages long) on FBDs,
with a wide variety of examples and illustrated
drawings in the following book which the author
found in the “References” of the article “Friction” in
Wikipedia. http://en.wikipedia.org/wiki/Friction.
Andy Ruina, Rudra Pratap, Introduction to
Statics and Dynamics, Pre-print for Oxford
Univ Press, January 2002.
Books on engineering mechanics usually contain
many examples of FBDs. See for example the
following book.
2.
F.P. Beer, E.R. Johnston, Vector Mechanics for
Engineers, McGraw Hill, N.Y. (1962).
Rolling friction and traction forces are well
explained in
3.
A.N. Mateev, Mechanics and Theory of
Relativity, Mir Publishers, Moscow (1989).
The drawings in this article were all drawn by the
author. In particular the drawings appearing in
Section 1 (Slide Show) and the drawings appearing
in Fig. 2 were published in this journal in 1986.
4.
S. Datta, A Film Strip on Newton’s Third Law
of Motion, Physics Education, Vol 2, No. 4
(January-March 1986), pp 30-40.
The drawings appearing in Figures 3 and 4 were
part of a Teacher’s Guide handwritten by the author
in December 1989 at Kavaratti, capital of
Lakshadweep islands, as part of a teacher training
programme organized by the Regional College of
Education (NCERT), Mysore.
Physics Education • July − September 2009