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SENIOR SECONDARY COURSE
PHYSICS
1
(CORE MODULES)
Coordinators
Dr
...
R
...
Dass
NATIONAL INSTITUTE OF OPEN SCHOOLING
A-25, INSTITUTIONAL AREA, SECTOR-62, NOIDA-201301 (UP)
COURSE DESIGN COMMITTEE
CHAIRMAN
Prof
...
C
...
A
...
Verma
Former Director, National
Physical Laboratory, Delhi,
160, Deepali Enclave
Pitampura, Delhi-34
Dr
...
)
Deptt
...
U
...
Oum Prakash Sharma
Asstt
...
L
...
Kothari
Prof
...
)
Delhi University
71, Vaishali, Delhi-11008
Dr
...
of Physics
IGNOU, Maidan Garhi
Delhi
Sh
...
S
...
)
BRMVB, Sr
...
School
Lajpat Nagar, New Delhi-110024
Dr
...
S
...
of Physics
IIT Roorkee
Sh
...
S
...
P
...
V
...
Bhatia
Prof
...
)
Delhi University
215, Sector-21, Faridabad
COURSE DEVELOPMENT TEAM
CHAIRMAN
Prof
...
C
...
V
...
Bhatia
215, Sector-21,
Faridabad
Prof
...
B
...
of Physics (Retd
...
K
...
Upadhyaya
Principal
Navodaya Vidyalaya
Rohilla Mohammadabad, (U
...
)
Dr
...
P
...
E
...
M
...
S
...
Garg
Former Pro-Vice Chancellor
IGNOU, Delhi
MEMBERS
Prof
...
B
...
of Physics (Retd
...
Anil Kumar
Principal
Pratibha Vikas Vidyalaya
Shalimar Bagh, Delhi
Sh
...
C
...
Director (Rtd
...
Sher Singh
Lecturer (Physics)
Navyug Public School
Lodhi Road, Delhi
Sh
...
S
...
)
BRMVB, Delhi-110024
GRAPHIC ILLUSTRATORS
Vijay Computer
1E, Pocket-1, Mayur Vihar
Phase-1, Delhi
Sh
...
As you go to higher classes, you
will appreciate that the method of sciences is characterised by objectivity, openness to change, innovation,
self-correction and dynamism
...
To encourage this, we have included a number of exercises and activities
...
This will also
provide you an opportunity to reflect on how a scientist works
...
But fundamental discoveries in rapid succession in the
early half of the 20th century brought in profound changes in our concepts of space, time, matter and
energy
...
Therefore, future development in knowledge society will heavily depend on
the availability of well trained scientific human capital endowed with entrepreneurship capabilities
...
The organisation of the course is generic
...
Out of two optional modules, which intend to develop professional competencies, you will be required
to opt for any one
...
This will also give you an opportunity to understand basic
physical principles
...
Your confidence in yourself and genuine interest in learning science should help you develop being an
independent learner with drive and initiative
...
So to ensure your active participation in teaching-learning as also to facilitate selfregulation and pacing, we have given questions in the body of each lesson
...
In curriculum design an effort has been made to put thematically coherent topics together for braviety
and completeness
...
You are therefore advised to make a
note of your difficulties and discuss them in the counselling sessions as well as amongst peers
...
It is sincerely hoped that their lives will inspire you as role
models to contribute your best!
Our best wishes are with you
...
After
making a comprehensive study, we found that our curriculum is more functional,
related to life situations and simple
...
We invited leading educationists of the country and under
their guidance, we have been able to revise and update the curriculum in the subject
of Physics
...
I hope you will find the new material interesting and exciting with lots of activities
to do
...
Let me wish you all a happy and successful future
...
R
...
A consistent format has been developed for self-study
...
Title is an advance organisor and conveys an idea about the contents of the lesson
...
Introduction highlights the contents of the lesson and correlates it with your prior
knowledge as well as the natural phenomena in operation in our immediate environment
...
Objectives relate the contents to your desired achievements after you have learnt the
lesson
...
Content of the lesson has been divided into sections and sub-sections depending on
thematic unity of concepts
...
After completing each section, answer intext questions and solve numerical
problems yourself
...
You
should continue reading a section till such time that you gain mastery over it
...
This indicates that it is important
...
Solved Examples will help you to understand the concepts and fix your ideas
...
Do them yourself and note
the main concept being taught through a particular example
...
These will help you to understand physics
by doing
...
Intext questions are based on the concepts discussed in every section
...
This will help you
progress
...
Answer these
your answers
to judge your
answers, turn
What you have learnt is essentially summary of the learning points for quick recapitulation
...
Terminal exercises in the form of short, long and numerical questions will help you to
develop a perspective of the subject, if you answer these meticulously
...
Answers to intext questions : These will help you to know how correctly you have
answered the intext questions
...
You may listen to these on FM Gyanvani or may buy the CDs
from Priced Publication Unit, NIOS
Video: Video programmes on certain elements related to your subject have been made to
clarify certain concepts
...
www
These are few selected websites that you can access for extended learning
...
Therefore you must develop regular study habit
...
You should earmark a well-ventilated and well-lighted space in
your home for your study
...
Overview of the Learning Material
Module - I
Motion, Force and Energy
1
...
3
...
5
...
7
...
Module - III
Thermal Physics
Units, Dimensions and Vectors
Motion in a straight line
Laws of Motion
Motion in a Plane
Gravitation
Work Energy and Power
Motion of Rigid Body
10
...
Thermodynamics
12
...
13
...
Wave Phenomena
Elastic Properties of Solids
Module - V
Electricity and Magnetism
2
23
...
16
...
18
...
Electromagnetic induction and
Alternating Current
24
...
26
...
28
...
Applications of
Semiconductor Devices
20
...
Dispersion and Scattering of Light
22
...
31
...
33
...
31
...
33
...
Module - I : Motion, Force and Energy
1
...
Motion in a straight line
26
3
...
Motion in a Plane
72
5
...
Work Energy and Power
120
7
...
Elastic Properties of Solids
182
9
...
Kinetic Theory of Gases
237
11
...
Heat Transfer and Solar Energy
273
Module - IV : Oscillations and Waves
13
...
Wave Phenomena
308
MODULE - I
MOTION, FORCE AND ENERGY
1
...
Motion in a straight line
3
...
Motion in a Plane
5
...
Work Energy and Power
7
...
Several times in the history of science, precise measurements have led to new discoveries
or important developments
...
For example, if you measure the length of your room, it is expressed in suitable units
...
The unit of a physical quantity is derived, by expressing it in base units fixed by
international agreement
...
You will learn that physical quantities can generally be divided in two groups: scalars and
vectors
...
The mathematical operations with vectors are somewhat different from those which you
have learnt so far and which apply to scalars
...
You will experience it in this course
...
3
...
1 Unit of Measurement
The laws of physics are expressed in terms of physical quantities such as distance, speed,
time, force, volume, electric current, etc
...
For example, time could be measured in minutes, hours or days
...
As another example, when we say that the distance
between Mumbai and Kolkata is nearly 2000 kilometres, we have for comparison a basic
unit in mind, called a kilometre
...
It is essential that all agree on the standard units,
so that when we say 100 kilometres, or 10 kilograms, or 10 hours, others understand what
we mean by them
...
Suppose you undertake an investigation on the solubility of a chemical in water
...
You communicate the
results of your investigation to a scientist friend in Japan
...
Do you now realize the need for agreed standards and units?
Remember that in science, the results of an investigation are considered
established only if they can be reproduced by investigations conducted elsewhere
under identical conditions
...
The following quote from
Manusmriti amply illustrates this point :
“The king should examine the weights and balance every six months to ensure true
measurements and to mark them with royal stamp
...
In Harappan Era, signs of systematic use of measurement are found in abundance :
the equally wide roads, bricks having dimensions in the ratio 4 : 2 : 1, Ivory scale in
Lothal with smallest division of 1
...
05, 0
...
2, 0
...
Akbar introduced gaz of 41 digits for measuring length
...
1 bigha was 60 gaz × 60 gaz
...
2
Units, Dimensions and Vectors
MODULE - 1
Motion, Force and Energy
1
...
1 The SI Units
With the need of agreed units in mind, the 14th General Conference on Weights and
Measures held in 1971, adopted seven base or fundamental units
...
The name SI is abbreviation for Système International d’Unités for the
International System of units
...
The SI
units along with their symbols are given in Table 1
...
Notes
Table 1
...
However, in scientific work we always use SI units
...
It is quite easy to handle because
the smaller and larger units of the base units are always submultiples or multiples of
ten
...
These are listed in
Table 1
...
Table 1
...
3 : Order of magnitude
of some masses
Mass
kg
Electron
10–30
Proton
10–27
Amino acid
10–25
Hemoglobin
10–22
Flu virus
10–19
Giant amoeba
10–8
Raindrop
10–6
Ant
10–2
Human being
102
Saturn 5 rocket
106
Pyramid
1010
Earth
1024
Sun
1030
Milky Way galaxy
1041
Universe
1052
3
...
3
and 1
...
Similarly, Table 1
...
1
...
2 Standards of Mass, Length and Time
Notes
Table 1
...
We define here standards of mass,
length and time
...
The standard
kilogram was established in 1887
...
The standard is kept in
the International Bureau of Weights and Measures in Paris,
France
...
1
...
For
India, the national prototype is the kilogram no
...
This is
maintained by the National Physical Laboratory, New Delhi (Fig
...
1)
...
It is defined in terms of a natural phenomenon:
One metre is defined as the distance travelled by light in vacuum in a time
interval
of
1/299792458 second
...
This definition of a second has helped in the development of a device called atomic
clock (Fig
...
2)
...
1
...
As of now, clock with an uncertainty of 5 parts in 1015 have been developed
...
You can
convert 1015s to years and get the astonishing result that this clock could run for 6 million
years and lose or gain less than a second
...
Researches are being conducted
today to improve upon this accuracy constantly
...
To give you an idea
of this technological achievement, if this clock were started at the time of the birth of the
universe, an event called the Big Bang, it would have lost or gained only two seconds till
now
...
In one experiment, he passed the air bubbles through liquid ammonia over red hot
copper contained in a tube and measured the density of pure nitrogen so obtained
...
The density of nitrogen obtained in second experiment was
found to be 0
...
The experiment suggested
that air has some other gas heavier than nitrogen present in it
...
Another example is the failed experiment of Michelson and Morley
...
4 fringe width in the interference
pattern obtained by the superposition of light waves travelling in the direction of
motion of the earth and those travelling in a transverse direction
...
Thus they
were expecting to measure the speed of earth with respect to ether and conclusively
prove that ether existed
...
This led to the concepts
of length contraction, time dilation etc and ultimately to the theory of relativity
...
Table 1
...
1
...
Derived Units
We have so far defined three basic units for the measurement of mass, length and time
...
These units
are called derived units
...
Another example is the interaction of the unit
of length with itself
...
Now are would like you to list all the physical quantities that you are familiar with and the
units in which they are expressed
...
5
5
MODULE - 1
Motion, Force and Energy
Physics
Some derived units have been given special names
...
6
...
6 : Examples of derived units with special names
Quantity
Force
newton
N
kg m s–2
Pressure
pascal
Pa
N m–2
Energy/work
joule
J
Nm
Power
Notes
Name
Symbol
Unit Symbol
watt
W
J s–1
One of the advantages of the SI system of units is that they form a coherent set in the
sense that the product or division of the SI units gives a unit which is also the SI unit of
some other derived quantity
...
Some care should be exercised in the order in which the units are
written
...
If by mistake we write it as mN,
it becomes millinewton, which is something entirely different
...
Otherwise, it is meaningless and, therefore, of no significance
...
1 : Anand, Rina and Kaif were asked by their teacher to measure the
volume of water in a beaker
...
Therefore, we do not know what it means
...
The second one is the only correct
answer
...
Note that the mass of a book, for example, can be expressed in kg or g
...
Nomenclature and Symbols
(i) Symbols for units should not contain a full stop and should remain the same in the
plural
...
or 7cms
...
g
...
(iii) When a prefix is placed before the symbol of a unit, the combination of prefix
and symbol should be considered as one symbol, which can be raised to a positive
6
Units, Dimensions and Vectors
or a negative power without using brackets, e
...
, µs–1, cm2, mA2
...
Similarly 1 poise = 1 g s–1cm–1 and not 1 g/s/cm
...
g
...
Notes
(vi) For convenience in reading of large numbers, the digits should be written in groups
of three starting from the right but no comma should be used: 1 532; 1 568 320
...
However, the failed experiment stirred the
scientific world to reconsider all old theories and led to a new world of physics
...
Through his stellar interferometer along with 100′′ Hookes telescope,
he made some precise measurements about stars
...
Solve the following questions
...
Intext Questions 1
...
The mass of the sun is 2 × 1030 kg
...
If the sun was
made only of protons, calculate the number of protons in the sun?
...
Earlier the wavelength of light was expressed in angstroms
...
Now the wavelength is expressed in nanometers
...
3
...
Express this frequency in GHz
...
4
...
3
...
2 Dimensions of Physical Quantities
Most physical quantities you would come across in this course can be expressed in terms
of five basic dimensions : mass (M), length (L), time (T), electrical current (I) and
temperature (θ)
...
Following examples show how dimensions of the physical quantities are
combinations of the powers of M, L and T :
(i)
Volume requires 3 measurements in length
...
(ii)
Density is mass divided by volume
...
(iii) Speed is distance travelled in unit time or length divided by time
...
(iv) Acceleration is change in velocity per unit time, i
...
, length per unit time per unit
time
...
(v)
Force is mass multiplied by acceleration
...
Similar considerations enable us to write dimensions of other physical quantities
...
Thus if dimension of x is L, then dimension of 3x will also be L
...
Remember that dimensions are not the same as the units
...
Dimensional analysis is the process of checking the dimensions of a quantity, or a
combination of quantities
...
Thus if x = p + q, then p and q will have the same dimensions as x
...
The following examples illustrate the use of dimensional analysis
...
2 : You know that the kinetic energy of a particle of mass m is
1
mv2 while
2
its potential energy is mgh, where v is the velocity of the particle, h is its height from the
ground and g is the acceleration due to gravity
...
e, energy, their dimensions must be the same
...
Solution : The dimensions of
8
1
mv2 are M
...
(Remember that the
2
Units, Dimensions and Vectors
numerical factors have no dimensions
...
LT–2
...
Clearly, the two expressions are the same and hence represent the same physical
quantity
...
Example 1
...
Let us
use dimensional analysis to find expression for the distance covered
...
Then we may
write
x ∝ tm
...
Comparing the powers of L and T on both sides, you will easily get n = 1, and m = 2
...
This is as far as we can go with dimensional analysis
...
To get the numerical factors, we have
to get input from experiment or theory
...
Besides numerical factors, other quantities which do not have dimensions are
angles and arguments of trigonometric functions (sine, cosine, etc), exponential
and logarithmic functions
...
In ex, x
is said to be the argument of the exponential function
...
Intext Questions 1
...
Experiments with a simple pendulum show that its time period depends on its length (l)
and the acceleration due to gravity (g)
...
...
Consider a particle moving in a circular orbit of radius r with velocity v and acceleration
a towards the centre of the orbit
...
...
You are given an equation: mv = Ft, where m is mass, v is speed, F is force and t is
time
...
...
9
9
MODULE - 1
Motion, Force and Energy
Physics
1
...
3
...
In one case, we need only to
state their magnitude with proper units and that gives their complete description
...
If we say that the mass of a ball is 50 g, we do not have to add anything to
the description of mass
...
Such quantities are called scalars
...
On the other hand, there are quantities which require both magnitude and direction for
their complete description
...
The statement that the velocity
of a train is 100 km h–1 does not make much sense unless we also tell the direction in
which the train is moving
...
We must specify not only the
magnitude of the force but also the direction in which the force is applied
...
A vector quantity has both magnitude and direction
...
1
...
If not, it is a scalar
...
1
...
3
...
Take vector AB in Fig
...
4
...
The arrow indicates its direction
...
Vector EF is a vector whose
Fig
...
4 : Directions and
magnitude is the same as that of vector CD, but its
magnitudes of vectors
direction is different
...
r
A vector is written with an arrow over the letter representing the vector, for example, A
...
In print, a vector is indicated by a bold
letter as A
...
This means that all vectors which are parallel to each other have the same
magnitude and point in the same direction are equal
...
1
...
We say A = B = C
...
10
Units, Dimensions and Vectors
MODULE - 1
Motion, Force and Energy
A vector (here D) which has the same
magnitude as A but has opposite
direction, is called negative of A, or
–A
...
A
B
C
D
For respresenting a physical vector
quantitatively, we have to invariably
choose a proportionality scale
...
1
...
between Delhi and Agra, which is about
300 km, is represented by choosing a scale 100 km = 1 cm, say
...
Notes
From the above we can say that if we translate a vector parallel to itself, it remains
unchanged
...
Let us sec how
...
3
...
For example,
two forces or two velocities can be added
...
Suppose we wish to add vectors A and B
...
1
...
For this
draw a line (say pq) parallel to vector A
...
e
...
Next draw vector B such that its tail coincides with the tip of
vector A
...
e
...
The sum of two vectors then is the vector from the tail of A to the
tip of B, i
...
the resultant will be represented in magnitude and direction by line pr
...
1
...
That is, A + B = B + A, as
shown in Fig
...
6 (b)
...
1
...
This gives us a rule for finding the resultant of two vectors :
If two vectors are represented in magnitude and direction by the two sides
of a triangle taken in order, the resultant is represented by the third side of
the triangle taken in the opposite order
...
3
...
In Fig
...
6(b), pr is the
resultant of A and B
...
Let us now learn to calculate resultant of more than
two vectors
...
A
+C
+
B
B
(A + B
)+C
A + (B
+ C)
B
The resultant of more than two vectors, say A, B
and C, can be found in the same manner as the sum
of two vectors
...
Alternatively, you could add B and C, and
then add A to (B + C) (Fig
...
7)
...
Thus, vector addition is
associative
...
Fig
...
7 : Addition of three vectors
in two different orders
Notes
A
Many a time, the point of application of vectors is the same
...
Let us now learn about it
...
3
...
1
...
To calculate the vector sum, we complete the parallelogram
...
Can you recognize that the diagonal PR is the sum vector A + B? It is called the resultant
of vectors A and B
...
Remember that vectors PQ and SR are equal to A, and vectors PS and QR are equal, to
B
...
Then in terms of magnitudes
S
A
R
B
A+
θ
P
R
B
B
θ
α
A
Q
T
Fig
...
8 : Parallelogram law of addition of vectors
(PR) 2 = (PT)2 + (RT)2
= (PQ + QT)2 + (RT)2
12
Units, Dimensions and Vectors
Motion, Force and Energy
= (PQ)2 + (QT)2 + 2PQ
...
QT
MODULE - 1
(1
...
QT
= (PQ)2 + (QR)2 + 2PQ
...
cosθ
Therefore, the magnitude of R is
R =
A 2 + B2 + 2AB
...
2)
For the direction of the vector R, we observe that
tanα =
RT
RT
Bsin θ
= PQ + QT =
PT
A + Bcos θ
(1
...
Special Cases
Now, let us consider as to what would be the resultant of two vectors when they are
parallel?
To find answer to this question, note that the angle between the two parallel vectors is zero
and the resultant is equal to the sum of their magnitudes and in the direction of these
vectors
...
What would be the magnitude
of the resultant? In this case, θ = 90º and cos θ = 0
...
What would be the direction of the
resultant?
Notice that tan α = B/A = 1
...
In this case α = 0
...
Example 1
...
Hamid is pulling the same cart in the south-west direction with a force of magnitude
50 N
...
Solution :
Here, magnitude of first force, say, A = 70 N
...
Angle θ between the two forces = 135 degrees
...
13
13
MODULE - 1
Motion, Force and Energy
Physics
So, the magnitude of the resultant is given by Eqn
...
2) :
(70)2 + (50)2 + 2 × 70 × 50 × cos(135)
R=
=
R
4900 + 2500 - 7000 × sin45
= 49
...
5 N
...
(1
...
1
...
00
Therefore, α = 45
...
Thus R makes an angle of 45º with 70 N force
...
1
...
1
...
5 Subtraction of Vectors
How do we subtract one vector from another?
If you recall that the difference of two vectors,
A – B, is actually equal to A + (–B), then you
can adopt the same method as for addition of
two vectors
...
1
...
Draw
vector –B from the tip of A
...
The resulting vector is the
difference (A – B)
...
Fig
...
10 : Subtraction of vector B
from vector A
Intext Questions 1
...
Make diagrams to show how to find the following vectors:
(a) B – A, (b) A + 2B, (c) A – 2B and (d) B – 2A
...
2
...
Determine
A + B and A – B
...
3
...
Find the resultant vector
...
14
Units, Dimensions and Vectors
1
...
4
...
This means that the magnitude of the
B
resultant vector is k |A|
...
If k is negative, the direction of
A
the new vector is opposite to its original
Fig
...
11: Projection of B on A
direction
...
But vector –3A is in a direction opposite to vector A, although its
magnitude is thrice that of vector A
...
4
...
B and is equal to AB cosθ,
where θ is the angle between the vectors
...
1
...
Therefore, the scalar
product of A and B is the product of magnitude of A with the length of the projection of B
along A
...
Since a dot
between the two vectors indicates the scalar product, it is also called the dot product
...
A familiar example of the scalar product is the work done when a force F acts on a body
moving at an angle to the direction of the force
...
Since dot product is a scalar, it is commutative: A
...
A = ABcosθ
...
(B + C) = A
...
C
...
4
...
We can draw a plane which
contains these two vectors
...
1
...
Then the vector product of these vectors, written as A × B, is
a vector, say C, whose magnitude is AB sinθ and whose direction is perpendicular to the
plane Ω
...
1
...
Imagine the fingers of your right hand curling from A to B along the smaller angle between
them
...
If you
follow this rule, you can easily see that direction of vector B × A is opposite to that of the
vector A × B
...
Since a cross
is inserted between the two vectors to indicate their vector product, the vector product is
also called the cross product
...
15
15
MODULE - 1
Physics
Motion, Force and Energy
C =A× B
B
π
θ
Notes
A
–C=B×A
(a)
(b)
Fig
...
12 (a) : Vector product of Vectors; (b) Direction of the product vector C =A × B is given
by the right hand rule
...
A familiar example of vector product is the angular momentum possessed by a rotating
body
...
Intext Questions 1
...
Suppose vector A is parallel to vector B
...
2
...
How is the direction of vector A
2
× B related to the direction of vector A × C
...
3
...
What happens
to the direction of vector C = A × B
...
4
...
Can you make vector C = A × B to point in exactly
opposite direction?
...
If vector A is along the x-axis and vector B is along the y-axis, what is the direction of
vector C = A × B? What happens to C if A is along the y-axis and B is along the
x-axis?
...
A and B are two mutually perpendicular vectors
...
B and (b) A × B
...
16
Units, Dimensions and Vectors
MODULE - 1
Motion, Force and Energy
1
...
Here we calculate components of
a given vector along any set of coordinate axes
...
1
...
Let these components
be called Ax and Ay respectively
...
4)
Notes
(1
...
What about the components of vector
A along X and Y-axes (Fig
...
13)? If the angle between the X-axis and A is φ, then
AX = A cos φ
y
Y
AY
Ay
A
O
θ φ
x
Ax
AX
X
Fig
...
13 : Resolution of vector A along two sets of coordinates (x, y) and (X, Y)
AY = A sin φ
...
Note also that the
magnitude of vector A and its direction in terms of its components are given by
A = A x2 + A y2 =
AX + AY
2
2
(1
...
(1
...
1
...
As the name suggests, a unit
vector has unitary magnitude and has a specified direction
...
ˆ
ˆ
As an example, we can write vector A as A n where a cap on n (i
...
n ) denotes a unit
vector in the direction of A
...
17
17
MODULE - 1
Motion, Force and Energy
Physics
the direction of the vector; the magnitude has been taken care of by A
...
Unit vector along x-axis is denoted
ˆ
by ˆ , along y-axis by ˆ and along z-axis by k
...
j
i
Notes
(1
...
j
i
(1
...
10)
By the rules of scalar product you can show that
ˆ ˆ =1, ˆ ˆ =1, k
...
ˆ = 0, ˆ
...
i
j
...
k
(1
...
B = (Ax ˆ + Ay ˆ )
...
ˆ ) + AxBy ( ˆ
...
ˆ ) + AyBy ( ˆ
...
12)
Here, we have used the results contained in Eqn
...
11)
...
4 On a coordinate system (showing all
the four quadrants) show the following vectors:
B
C
j
j
j
i
i
A = 4ˆ + 0ˆ , B = 0ˆ + 5ˆ , C = 4ˆ + 5ˆ ,
i
j
D = 6ˆ – 4ˆ
...
Solution : The vectors are given in component form
...
1
...
All the vectors
are shown on the coordinate grid (Fig
...
14)
...
So, the magnitude of A = 4
...
(1
...
This quantity is zero, since Ay = 0
...
Vector B has x-component = 0, so it lies along the y-axis
and its magnitude is 5
...
Here, Cx = 4 and Cy = 5
...
The angle that it makes with the x-axis is tan (Cy /Cx) = 51
...
Similarly, the magnitude of D is D =
(0
...
7º (in the fourth quadrant)
...
Its direction is tan (Dy/Dx) = tan
Example 1
...
D for the vectors given in Example 1
...
Notes
Solution : The dot product of C with D can be found using Eqn
...
12):
C
...
The cross product of two vectors can also be written in terms of the unit vectors
...
For this remember that the angle between
i j
the unit vectors is a right angle
...
Sine of the angle between
them is one
...
Its direction is perpendicular to
j
i
the xy - plane containing ˆ and ˆ , which is the z-axis
...
And what is the unit vector in the positive
ˆ
z - direction
...
Therefore,
ˆ
ˆ × ˆ = k
...
i i j j
(1
...
14)
(1
...
6 Calculate the cross product of vectors C and D given in Example (1
...
Solution : We have
j
j
i
i
C × D = (4 ˆ + 5 ˆ ) × (6 ˆ – 4 ˆ )
j i
j
j
= 24 ( ˆ × ˆ ) –16 ( ˆ × ˆ ) + 30 ( ˆ × ˆ ) –20 ( ˆ × ˆ )
i i
i j
Using the results contained in Eqns
...
13 – 1
...
Since C and D are in the xy-plane, it is obvious that the cross product must be
perpendicular to this plane, that is, it must be in the z-direction
...
5
1
...
If its magnitude is 50 units, find its components in x, y directions
...
Find its components now
...
3
...
Two vectors A and B are given respectively as 3 ˆ – 4 ˆ and –2 ˆ + 6 ˆ
...
Find their magnitudes and the angles that they make with
the x-axis (see Fig
...
14)
...
3
...
Notes
...
Having
learnt vectors, we must now add this: For an equation to be correct, each term in it
must have the same character: either all of them be vectors or all of them be
scalars
...
The SI system has been accepted and followed universally for scientific
reporting
...
In addition to
base units, there are derived units
...
Dimensional analysis is a useful tool for
checking correctness of equations
...
A
scalar has only magnitude
...
Vectors are added according to the parallelogram rule
...
The vector product of two vectors is a vector perpendicular to the plane containing
the two vectors
...
Terminal Exercise
1
...
Meteors are small pieces of rock which enter the earth’s atmosphere occasionally
at very high speeds
...
The streak of light that is seen as a result is called a ‘shooting star’
...
3
...
It is the distance
covered by light in one year
...
Take speed of light as
3 × 108 m s–1
...
Check the
Units, Dimensions and Vectors
correctness of the expression using dimensional analysis
...
6
...
8
...
MODULE - 1
m1m 2
r2
where G is the universal constant of gravitation
...
Hamida is pushing a table in a certain direction with a force of magnitude 10 N
...
Calculate the magnitude of the resultant force on the table and its direction
...
Is it a scalar
or a vector? What is the nature of a physical quantity obtained as cross product of
two vectors?
John wants to pull a cart applying a force parallel to the ground
...
Who is correct and why?
Notes
j
j
i
i
Two vectors are given by 5 ˆ – 3 ˆ and 3 ˆ – 5 ˆ
...
Answers to Intext Questions
1
...
Mass of the sun = 2 × 1030 kg
Mass of a proton = 2 × 10–27 kg
( No of protons in the sun =
2
...
2 × 10 −27 kg
1 angstrom = 10–8 cm = 10–10 m
1 nanometer (nm) = 10–9 m
∴
1 nm/1 angstrom = 10–9 m /10–10 m = 10 so 1 nm = 10 Å
3
...
370 × 10–3 GHz
4
...
21
21
MODULE - 1
Motion, Force and Energy
Physics
1
...
Notes
Dimension of length = L
Dimension of time = T
Dimensions of g = LT–2
Let time period t be proportional to lα and gβ
Then, writing dimensions on both sides T = Lα (LT–2)β = Lα+β T–2β
Equating powers of L and T,
α + β = 0, 2β = –1 ⇒ β = –1/2 and α = 1/2
l
So, t α g
...
Dimensions of mv = MLT–1
Dimensions of Ft = MLT–2 T1 = MLT–1
Dimensions of both the sides are the same, therefore, the equation is dimensionally
correct
...
1
...
Suppose
A+
(a)
(b)
A
22
B
B
Units, Dimensions and Vectors
MODULE - 1
Motion, Force and Energy
A
–2A
A – 2B
(c)
–2B
(d)
B–
2A
B
Notes
2
...
| A + B | = 77 units
1
...
If A and B are parallel, the angle θ between them is zero
...
If they are antiparallel then the angle between them is 180o
...
2
...
Its magnitude may
change
...
23
23
MODULE - 1
Motion, Force and Energy
Physics
Since vectors A and B rotate without change in the plane containing them, the direction
of C = A × B will not change
...
Suppose initially the angle between A and B is between zero and 180o
...
After rotation through arbitrary
amounts, if the angle between them becomes > 180o, then C will drop underneath
but perpendicular to the plane
...
If A is along x-axis and B is along y-axis, then they are both in the xy plane
...
If A is along y-axis and B is along
x-axis, then C is along the negative z-axis
...
Notes
3
...
B = |A| |B| cos θ = 0 when θ = 90º
(b) A × B = |A| |B| sin θ = |A| |B| as sin θ = 1 at θ = 90º
1
...
When A makes an angle of 60o with the x-axis:
Ax = A cos 60 = 50
...
√3/2 = 50
...
866
= 43
...
0
...
3 units
Ay = 50 sin 30 = 50
...
2
...
1
...
Suppose A makes an angle θ with the x-axis, then
tan θ = – 4/3 ⇒ θ = tan–1(– 4/3)
= –53o 6′ or 306o 54′
after taking account of the quadrant in which the angle lies
...
The dot product of A and B:
j
j
i
i
A
...
(–2 ˆ +6 ˆ )
j j
i i
= –6( ˆ
...
ˆ ) = –30
j i
j j
i j
i i
because ˆ
...
ˆ = 1
The cross product of A and B:
j
j
i
i
A × B = (3 ˆ – 4 ˆ ) × (–2 ˆ + 6 ˆ )
ˆ
ˆ
ˆ
j i
i j
= 18 ( ˆ × ˆ ) + 8 ( ˆ × ˆ ) = 18 k – 8 k = 10 k
24
Units, Dimensions and Vectors
on using Eqs
...
14) and (1
...
So, the cross product is in the direction of z-axis,
since A and B lie in the xy plane
...
1 ly = 9
...
2
...
⎛1⎞
15
...
A
...
3
...
Humans, animals, vehicles can be seen
moving on land
...
Birds and aeroplanes
move in air
...
It is, therefore, quite apparent that we live in a world that is
very much in constant motion
...
Motion can be in a straight line(1D), in a plane(2D) or in
space(3D)
...
For example, motion of a car on a straight road, motion of a train on straight
rails, motion of a freely falling body, motion of a lift, and motion of an athlete running on a
straight track, etc
...
In the following lessons, you will
study the laws of motion, motion in plane and other types of motion
...
2
...
But the difference between the initial and final position vectors of a body is called its
displacement
...
Thus, the displacement is a vector quantity but
26
Motion in a Straight Line
distance is a scalar
...
Unlike
speed, velocity is a vector quantity
...
2
...
1 Average Velocity
MODULE - 1
Motion, Force and Energy
Notes
When an object travels a certain distance with different velocities, its motion is specified
by its average velocity
...
Let x1 and x2 be its positions at instants t1 and t2, respectively
...
1)
where x2 – x1 signifies change in position (denoted by ∆x) and t2 – t1 is the corresponding
change in time (denoted by ∆t)
...
Average velocity can be represented as vav
also
...
2)
If the motion is in the same direction along a straight line, the average speed is the same as
the magnitude of the average velocity
...
2)
...
Example 2
...
Calculate
the average velocity of the object over the time interval from 3s to 4s
...
It means that the constant of
proportionality (20) has dimensions ms–2
...
27
27
MODULE - 1
Physics
Motion, Force and Energy
x 1 = 20 × (3)2
= 20 × 9 = 180 m
Similarly, for t 2 = 4s
x 2 = 20 × (4)2
Notes
= 20 × 16 = 320 m
x2 − x1
140 m
(320 − 180) m
v =
=
= 1 s = 140 ms–1
(4 − 3) s
t2 − t1
∴
Hence, average velocity = 140 ms–1
...
2 : A person runs on a 300m circular track and comes back to the starting
point in 200s
...
Solution : Given,
Total length of the track = 300m
...
5 ms–1
200
As the person comes back to the same point, the displacement is zero
...
Note that in the above example, the average speed is not equal to the magnitude of the
average velocity
...
1
...
Thus it is
implied that the referred velocity is with respect to some reference point
...
Since all bodies are
in motion, we can say that every velocity is relative in nature
...
For example, if vA and
vB are the velocities of the two objects along a straight line, the relative velocity of B with
respect to A will be vB– vA
...
MODULE - 1
Motion, Force and Energy
Importance of Relative Velocity
The position and hence velocity of a body is specified in relation with some other
body
...
You will learn the equations of kinematics in this lesson
...
However, it can be simplified by invoking the concept of O
xA(0) xB(0)
relative motion
...
If body A moves
along positive x-direction with velocity vA and body B with velocity vB, then the
positions of points A and B after t seconds will be given by
x A(t) = xA(0) + vAt
xB(t) = xB(0) + vBt
Therefore, the relative separation of B from A will be
xBA(t) = xB(t) – xA(t) = xB(0) – xA(0) + (vB – vA) t
= xBA(0) + vBAt
where vBA = (vB – vA) is called the relative velocity of B with respect to A
...
Example 2
...
Another train B is moving from South to North with a
speed of 70km h–1
...
In the above example, you have seen that the relative velocity of one train with respect to
the other is equal to the sum of their respective velocities
...
But, if the other train were moving in the same direction as your train, it would
appear to be very slow
...
29
29
MODULE - 1
Motion, Force and Energy
Notes
Physics
2
...
3 Acceleration
While travelling in a bus or a car, you might have noticed that sometimes it speeds up and
sometimes it slows down
...
Just as the velocity is
defined as the time rate of change of displacement, the acceleration is defined as time
rate of change of velocity
...
In one
dimension, there is no need to use vector notation for acceleration as explained in the case
of velocity
...
3)
a = t − t = ∆t
2
1
In one dimensional motion, when the acceleration is in the same direction as the motion or
velocity (normally taken to be in the positive direction), the acceleration is positive
...
Then the acceleration is
taken as negative and is often called deceleration or retardation
...
Example 2
...
0 s
...
Solution : Given,
v 1 = 0 m s–1
v 2 = 12 m s–1
t = 3
...
0m s –1 )
3
...
0 m s–2
a =
Intext Questions 2
...
Is it possible for a moving body to have non-zero average speed but zero average
velocity during any given interval of time? If so, explain
...
2
...
Finding market closed, she came
back home at a speed of 10 km h–1
...
...
Can a moving body have zero relative velocity with respect to another body? Give an
example
...
4
...
0 m s–1 in the direction of motion of
the train
...
0 m s–1, calculate his
Motion in a Straight Line
(a) velocity as seen by passengers in the compartment, and (b) velocity with respect
to a person sitting on the platform
...
2
...
The different positions and corresponding
times can be plotted on a graph giving us a certain
curve
...
Generally, the time is represented along x-axis
whereas the position of the body is represented
along y-axis
...
What will
be its position after 1s, 2s, 3s, 4s and 5s? You will
find that the graph is a straight line parallel to the
time axis, as shown in Fig
...
1
30
20
10
1
2
3
time(s)
4
5
Fig
...
1 : Position-time graph for
a body at rest
2
...
1 Position-Time Graph for Uniform Motion
Now, let us consider a case where an object covers equal distances in equal intervals of
time
...
Time (t) in s
1
2
3
4
5
Position (x) in m
10
20
30
40
50
The graph is a straight line inclined with
the x-axis
...
Its position-time graph is a straight line
inclined to the time axis
...
50
40
position(m)
In order to plot this data, take time along
x-axis assuming 1cm as 1s, and position
along y-axis with a scale of 1cm to be
equal to 10m
...
2
...
2
...
31
31
MODULE - 1
Motion, Force and Energy
2
...
2 Position-Time Graph for Non-Uniform Motion
Let us now take an example of a train which starts from a station, speeds up and moves
with uniform velocity for certain duration and then slows down before steaming in the next
station
...
Such a motion is said to
be non-uniform motion
...
The position-time
B
graph for such an object is as shown in
Fig
...
3
...
2
...
Hence, the velocity of the body changes
continuously
...
Let us
learn to do so now
...
2
...
If it is a straight line parallel to the time axis, you can say that the body is at rest
(Fig
...
1)
...
2
...
A continuous curve implies continuously changing velocity
...
For determining the slope, we
choose two widely separated points (say A and B) on the straight line (Fig
...
2) and form a
triangle by drawing lines parallel to y-axis and x-axis
...
4)
Hence, average velocity of object equals the slope of the straight line AB
...
5 : The position - time graphs of two bodies A
and B are shown in Fig
...
4
...
32
Position (m) →
It shows that greater the value of the slope (∆x/∆t) of the straight line position - time
graph, more will be the average velocity
...
e
...
Any two
corresponding ∆x and ∆t intervals can be used to determine
A
the slope and thus the average velocity during that time internal
...
2
...
But in the case of non-uniform motion,
the position - time graph is a curved line, as shown in Fig
...
5
...
The velocity of the particle at
4
any instant of time or at some point of its path is called
∆x 0
its instantaneous velocity
...
As ∆t is made smaller and
∆t
smaller the average velocity approaches
instantaneous velocity
...
However, for uniform motion, the average
and instantaneous velocities are the same
...
2
...
6 : The position - time graph for the motion of an object for 20 seconds is
shown in Fig
...
6
...
5 s? Calculate the average
speed for this total journey
...
5 5
10
15 17
...
2
...
8 m s
Time
ii) During 5 s to 10 s, distance travelled = 12 – 4 = 8 m
∴ speed =
(12 – 4) m 8 m
–1
∴ speed = (10 – 5) s = 5 s = 1
...
5 s, distance travelled = 12 m
3
...
5 s = 4
...
Notes
Intext Questions 2
...
Draw the position-time graph for a motion with zero acceleration
...
2
...
Look at the graphs carefully and answer
the following questions
...
B
(ii) Who stays farther from the school?
A
700
...
500
400
300
(iv) Who moves faster?
...
Under what conditions is average velocity of a body equal to its instantaneous velocity?
...
distance →
4
...
2
C
A
B
time (t) →
(b)
Motion in a Straight Line
MODULE - 1
Motion, Force and Energy
2
...
While plotting a
velocity-time graph, generally the time is taken along the x-axis and the velocity along the
y-axis
...
3
...
e
...
The velocity-time graph for such a uniform motion is a
straight line parallel to the time axis, as shown in the Fig
...
6
...
2
...
2
...
3
...
The
velocity-time graph for such a motion is a straight
line inclined to the time axis
...
2
...
It is clear from
C
the graph that the velocity increases by equal
amounts in equal intervals of time
...
time (t)
However, it is also possible that the rate of Fig
...
8 : Velocity-time graph for a motion
with varying acceleration
variation in the velocity is not constant
...
In such a situation, the slope of the velocity-time graph will vary at every instant,
as shown in Fig
...
8
...
2
...
3 Interpretation of Velocity-Time Graph
Using v – t graph of the motion of a body, we can determine the distance travelled by it
and the acceleration of the body at different instants
...
3
...
2
...
2
...
The portion AB shows the
motion with constant acceleration, whereas the
portion CD shows the constantly retarded motion
...
e
...
For uniform motion, the distance travelled by the body
from time t1 to t2 is given by s = v (t2 – t1) = the area
under the curve between t1 and t2 Generalising this
result for Fig
...
7, we find that the distance travelled
by the body between time t1 and t2
s = area of trapezium KLMN
= (½) × (KL + MN) × KN
= (½) × (v1 + v2) × (t2 – t1)
(b) Determination of the acceleration of the body : We know that acceleration of a
body is the rate of change of its velocity with time
...
2
...
∆t
t2 − t1
If the time interval ∆t is made smaller and smaller, the average acceleration becomes
instantaneous acceleration
...
(ii) Calculate the distances travelled by these bodies
in first 3s
...
7 : The velocity-time graphs for three
different bodies A,B and C are shown in Fig
...
9(a)
...
2
...
3−0 3
∆t
(ii) The distance travelled by a body is equal to the area of the v-t graph
...
the distance travelled by B = Area OB′L
= (½) × 3 × 3 = 4
...
the distance travelled by C = (½) × 1 × 3 = 1
...
(iii) At the end of the journey, the maximum distance is travelled by B
...
(iv) Since v-t graph for each body is a straight line, instantaneous acceleration is equal to
average acceleration
...
80 m s–1 (approx
...
3
1
...
(i) Describe the
motion in terms of velocity, acceleration and distance
travelled (ii) Find the average speed
...
2
...
...
0
5
-10
15
22
t (s)
Using the adjoining v -t graph, calculate the (i)
average velocity, and (ii) average speed of the
particle for the time interval 0 – 22 seconds
...
...
37
37
MODULE - 1
Physics
Motion, Force and Energy
2
...
In the case of constant acceleration, the velocity acquired
and the distance travelled in a given time can be calculated by using one or more of three
equations
...
2
...
1 Equation of Uniform Motion
In order to derive these equations, let us take initial time to be zero i
...
t1 = 0
...
The initial position (x1) and initial velocity (v1) of an
object will now be represented by x0 and v0 and at time t they will be called x and v
(rather than x2 and v2)
...
(2
...
t
(2
...
4
...
As you know, by definition
Acceleration (a) =
v 2 − v1
Change in velocity
=
t2 − t1
Time taken
If at t1 = 0, v1 = v0 and at t2 = t, v2 = v
...
5)
(2
...
8 : A car starting from rest has an acceleration of 10ms–2
...
4
...
B
The distance travelled = area under v – t graph
v(m –1) →
s
v
Suppose that at t = 0, x1 = x0; v1 = v0 and at t = t,
x2 = x; v2 = v
...
2
...
7)
Example 2
...
Car B is following it with uniform velocity of 70 km h–1
...
5 km, the car B is given a decceleration of 20 km h–1
...
For car A, the distance travelled in t time, x = 60 × t
...
5
∴
or
(70 t – 10 t2) – (60 t) = 2
...
5 = 0
It gives t = ½ hour
∴
x = 70t – 10t2
= 70 × ½ – 10 × (½)2
= 35 – 2
...
5 km
...
39
39
MODULE - 1
Motion, Force and Energy
Physics
2
...
4 Third Equation of Uniformly Accelerated Motion
The third equation is used in a situation when the acceleration, position and initial velocity
are known, and the final velocity is desired but the time t is not known
...
(2
...
),
we can write
x – x0 = ½ (v + v0) t
...
(2
...
8)
Thus, the three equations for constant acceleration are
v = v0 + at
x = x0 + v0t + ½ at2
2
v2 = v 0 + 2a (x – x0)
and
Example 2
...
If initially she was at a position of 5m and had a velocity of 3m s–1, calculate
(i) the position and velocity at time t = 2s, and
(ii) the position of the motorcyclist when its velocity is 5ms–1
...
(i) Using Eqn
...
7)
x = x0 + v0t + ½ at2
= 5 + 3 × 2 + ½ × 4 × (2)2 = 19m
From Eqn
...
6)
v = v0 + at
= 3 + 4 × 2 = 11ms–1
Velocity, v = 11ms–1
...
Notes
2
...
Do you know why they come to the
earth and what type of path they follow? It happens because of the gravitational force of
the earth on them
...
Therefore, motion
under gravity is along a straight line
...
The free fall of a
body towards the earth is one of the most common examples of motion with constant
acceleration
...
Though the acceleration due to gravity
varies with altitude, for small distances compared to the earth’s radius, it may be taken
constant throughout the fall
...
The acceleration of a freely falling body due to gravity is denoted by g
...
8 ms–2
...
Galileo Galilei (1564 – 1642)
He was born at Pisa in Italy in 1564
...
He devised a telescope and used it for astronomical observations
...
He supported
the idea that the earth revolves around the sun
...
11 : A stone is dropped from a height of 50m and it falls freely
...
e
...
Solution : Given
Height h = 50 m and Initial velocity v0 = 0
Consider, initial position (y0) to be zero and the origin at the starting point
...
Since acceleration is downward in the negative
y-direction, the value of a = – g = –9
...
(i) From Eqn
...
7), we recall that
y = y0 + v0t + ½ at2
3
...
8 × (2)2
= –19
...
The negative sign shows that the distance is below the starting point in downward direction
...
8),
2
v 2 = v 0 + 2a (y – y0)
= 0 + 2 (–9
...
9 ms–1
(iii) Using v = v0 + at, at t = 3s, we get
∴
v = 0 + (–9
...
4 ms–1
This shows that the velocity of the stone at t = 3 s is 29
...
Note : It is important to mention here that in kinematic equations, we use certain
sign convention according to which quantities directed upwards and rightwards are
taken as positive and those downwards and leftward are taken as negative
...
Intext Questions 2
...
A body starting from rest covers a distance of 40 m in 4s with constant acceleration
along a straight line
...
...
A car moves along a straight road with constant aceleration of 5 ms–2
...
...
With what velocity should a body be thrown vertically upward so that it reaches a
height of 25 m? For how long will it be in the air?
...
A ball is thrown upward in the air
...
42
Motion in a Straight Line
MODULE - 1
Motion, Force and Energy
What You Have Learnt
The ratio of the displacement of an object to the time interval is known as average
velocity
...
Notes
The rate of change of the relative position of an object with respect to another object is
known as the relative velocity of that object with respect to the other
...
The position-time graph for a body at rest is a straight line parallel to the time axis
...
A body covering equal distance in equal intervals of time, however small, is said to be in
uniform motion
...
The slope of the position-time graph gives the average velocity
...
The area under the velocity-time graph gives the displacement of the body
...
The motion of a body can be described by following three equations :
(i)
v = v0 + at
(ii)
x = x0 + v0 t + ½ at2
(iii)
2
v 2 = v0 + 2a
...
Distinguish between average speed and average velocity
...
A car C moving with a speed of 65 km h–1 on a straight road is ahead of motorcycle
M moving with the speed of 80 km h–1 in the same direction
...
How long does a car take to travel 30m, if it accelerates from rest at a rate of
2
...
A motorcyclist covers half of the distance between two places at a speed of
30 km h–1 and the second half at the speed of 60 kmh–1
...
3
...
A duck, flying directly south for the winter, flies with a constant velocity of 20 km h–1
to a distance of 25 km
...
Bangalore is 1200km from New Delhi by air (straight line distance) and 1500 km by
train
...
Notes
7
...
0 s
...
A body with an initial velocity of 2
...
0 ms–2 for 3 seconds
...
A ball is released from rest from the top of a cliff
...
10
...
Taking beach as the reference (zero) level, upward as the
positive direction, draw the motion graphs
...
e
...
11
...
What will be the value of
the velocity and acceleration of the body at the highest point?
12
...
Will they reach the ground at the same time? Explain your answer
...
What happens to the uniform motion of a body when it is given an acceleration at right
angle to its motion?
14
...
1
1
...
When body returns to its initial postion its velocity is zero but speed is non-zero
...
89 km h , average velocity = 0
+
8 10
2
...
Yes, two cars moving with same velocity in the same direction, will have zero relative
velocity with respect to each other
...
(a) 1 m s–1
(b) 2 m s–1
2
...
See Fig
...
2
...
(i) A, (ii) B covers more distance, (iii) B, (iv) A, (v) When they are 3km from the
44
Motion in a Straight Line
MODULE - 1
Motion, Force and Energy
starting point of B
...
In the uniform motion
...
(a) is wrong, because the distance covered cannot decrease with time or become
zero
...
3
1
...
(b) Motion of the body between start and 5th seconds is uniformly accelerated
...
a=
15 − 0
= 3 m s–2
5−0
(c) Motion of the body between 5th and 10th second is a uniform motion (represented
by AB)
...
15 − 5 10
(d) Motion between 15th and 25th second is uniformly retarded
...
a =
0 − 15
= – 1
...
25 − 15
(ii) (a) Average speed =
Distance covered Area of OA BC
=
time taken
(25 − 0)
⎛1
⎞
⎛1
⎞
⎜ ×15 × 5) ⎟ + (15 × 10) + ⎜ × 15 × 10 ⎟ 525
⎠
⎝2
⎠=
= 10
...
=⎝2
25
50
(b) Deccelerated Velocity decreases with time
...
⎝ 2 ⎠
⎝ 2 ⎠
⎛ 185 ⎞ −1
∴ average speed = ⎜
⎟ ms = 8
...
⎝ 22 ⎠
⎛ 20 × 15 ⎞
⎛ 10 × 7 ⎞
Total displacement = ⎜
⎟m − ⎜
⎟ m = 115 m
...
22 m s–1
...
45
45
MODULE - 1
Motion, Force and Energy
Physics
2
...
Using x = x0 + v0t + ½ at2
40 =
Notes
1
× a × 16
2
⇒ a = 5 ms–2
2
Next using v2 = v 0 + 2a (x – x0)
v = 20 m s–1,
20 = 0 +
1
× 5 × t2 ⇒ t = 2 2 s
2
2
...
(2
...
(2
...
3
...
(2
...
6 m s–1
...
4
...
Answers to Terminal Exercises
2
...
5
...
40 ms–1
5
...
25 h
6
...
2
...
(i) 42 m (ii) 36 m
11
...
8 m s–2
...
But an important question is : what makes an
object to move? Or what causes a ball rolling along the ground to come to a stop? From
our everyday experience we know that we need to push or pull an object if we wish to
change its position in a room
...
A cricket ball has to be hit hard by a batter to send it across the boundary
for a six
...
There are, however, many situations where the cause behind an action is not visible
...
The concept of force developed in this lesson will be useful in different branches of
physics
...
The laws of motion
are fundamental and enable us to understand everyday phenomena
...
3
...
These objects
3
...
Similarly, an object moving with constant velocity has to be forced to change
its state of motion
...
Mass of a body is a
measure of its inertia
...
If it were not present, your books or classnotes
could mingle with those of your younger brother or sister
...
You must however recall that the state of rest or
of uniform motion of an object are not absolute
...
Observations show that the change in velocity of an object can
only be brought, if a net force acts on it
...
We use it in so many situations in our everyday
life
...
Though a
force is not visible, its effect can be seen or experienced
...
A balloon changes shape
depending on the magnitude of force acting on it
...
A force can set an object into
motion or it can bring a moving object to rest
...
(c) Forces can rotate a body about an axis
...
3
...
1 Force and Motion
Force is a vector quantity
...
Motion of a body is characterised by its displacement, velocity etc
...
For example, in the case of a body falling freely, the velocity of the body increases
continuously, till it hits the ground
...
From experience we know that a net non-zero force is required to change the state of a
body
...
If a net force acts on a body in motion, its velocity will increase in
magnitude, if the direction of the force and velocity are same
...
However, if a net force acts on a body in a direction perpendicular to its velocity,
the magnitude of velocity of the body remains constant (see Sec 4
...
Such a force changes
only the direction of velocity of the body
...
3
...
2 First Law of Motion
When we roll a marble on a smooth floor, it stops after some time
...
However, if we want it to move
continuously with the same velocity, a force will have to be constantly applied on it
...
Is there any net force acting on the marble or trolley in the situations
mentioned here?
Motion and Inertia
Notes
Galileo carried out experiments to prove that in the absence of any external force, a
body would continue to be in its state of rest or of uniform motion in a straight line
...
3
...
3
...
He argued
that if the plane is neither inclined upwards nor downwards (i
...
if it is a horizontal
plane surface), the motion of the body will neither be accelerated not retarded
...
Fig
...
1 : Motion of a body on inclined and horizontal planes
In another thought experiment, he considered two inclined planes facing each other,
as shown in Fig
...
2
...
3
...
The plane PQRS is very smooth and the ball is of marble
...
As the
inclination of the plane RS decreases, the balls moves a longer distance to rise to the
same height on the inclined plane (Fig
...
2b)
...
e
...
Fina P
l osition
Initia P ition
l os
P A
Fina P ition
l os
B S
P
Q
R
Q
R
(a
)
S
h
h
h
h
B
A
(b)
P
Where is thefina position?
l
h
Q
(c)
R
S
Fig
...
2 : Motion of a ball along planes inclined to each other
3
...
He studied at
Trinity College, Cambridge and became the most profound scientist
...
He enunciated the laws of
motion and the law of gravitation
...
His
contributions are of a classical nature and form the basis of the modern science
...
You may logically ask : Why is it necessary to apply a force continuously to the trolley to
keep it moving uniformly? We know that a forward force on the cart is needed for balancing
out the force of friction on the cart
...
Isaac Newton generalised Galileo’s conclusions in the form of a law known as Newton’s
first law of motion, which states that a body continues to be in a state of rest or of
uniform motion in a straight line unless it is acted upon by a net external force
...
A person in a running car is at rest with respect to another person
in the same car
...
For this reason, it is necessary to record measurements of changes in position,
velocity, acceleration and force with respect to a chosen frame of reference
...
This
nomenclature follows from the property of inertia of bodies due to which they tend to
preserve their state (of rest or of uniform linear motion)
...
Now you may like to take a break and answer the following questions
...
1
1
...
2
...
3
...
50
Laws of Motion
4
...
MODULE - 1
Motion, Force and Energy
...
2 Concept of Momentum
Notes
You must have seen that a fielder finds it difficult to stop a cricket ball moving with a large
velocity although its mass is small
...
These examples suggest that both, mass and
velocity of a body, are important, when we study the effect of force on the motion of the
body
...
Mathmatically, we write
p = mv
In SI units, momentum is measured in kg ms–1
...
The direction
of momentum vector is the same as the direction of velocity vector
...
The following examples illustrate this point
...
1 Aman weights 60 kg and travels with velocity 1
...
5 m s–1 towards Aman
...
Solution : For Aman
momentum = mass × velocity
= (60 kg) × (1
...
5 ms–1)
= – 60 kg ms–1
Note that the momenta of Aman and Manoj have the same magnitude but they are in
opposite directions
...
2 A 2 kg object is allowed to fall freely at t = 0 s
...
Solution : (a) As velocity of the object at t = 0 s is zero, the initial momentum of the object
will also be zero
...
8 ms–1 [use v = v0 + at] pointing downward
...
8 ms–1) = 19
...
3
...
6 m s–1 pointing downward
...
6 ms–1) = 39
...
Notes
Thus, we see that the momentum of a freely-falling body increases continuously in magnitude
and points in the same direction
...
3 A rubber ball of mass 0
...
Calculate the change in
momentum of the ball
...
In each case the magnitude of momentum is
(0
...
e
...
If we choose initial momentum vector to be along + x axis, the final momentum vector will
be along –x axis
...
Therefore, the change in momentum
of the ball, pf – pi = (–2 kg ms–1) – (2 kg ms–1) = – 4 kg ms–1
...
What causes this change in momentum of the ball?
In actual practice, a rubber ball rebounds from a rigid wall with a speed less than its speed
before the impact
...
3
...
Newton’s first law of motion suggests that no net external force acts on such a body
...
2 we have seen that the momentum of a ball falling freely under gravity
increases with time
...
Newton’s second law of motion gives
a quantitative relation between these three physical quantities
...
Change in momentum of the body takes place in the direction of net external
force acting on the body
...
By expressing momentum as a product of mass and velocity, we can rewrite this result as
⎛ ∆v ⎞
F = k m⎜
⎟
⎝ ∆t ⎠
F =kma
(as
∆v
= a)
∆t
Notes
(3
...
If these units are chosen
such that when the magnitude of m = 1 unit and a = 1 unit, the magnitude of F is also be 1
unit
...
1
...
e
...
(3
...
2)
In SI units, m = 1 kg, a = 1 m s–2
...
3)
This unit of force (i
...
, 1 kg m s–2) is called one newton
...
The SI unit of
force i
...
, a newton may thus, be defined as the force which will produce an acceleration
of 1 ms–2 in a mass of 1 kg
...
3 A ball of mass 0
...
Calculate the force which stops the ball, assuming it to be constant in
magnitude throughout
...
4 kg, initial velocity u = 20 ms–1, final velocity v = 0 m s–1 and
t = 10s
...
4 kg ( − 20 ms –1 )
m(v – u )
=
|F| = m|a| =
10 s
t
=
– 0
...
8 N
Here negative sign shows that force on the ball is in a direction opposite to that of its
motion
...
4 A constant force of magnitude 50 N is applied to a body of 10 kg moving
initially with a speed of 10 m s–1
...
3
...
We have to calculate t
...
–50 N
50 kg m s –2
It is important to note here that Newton’s second law of motion, as stated here is applicable
to bodies having constant mass
...
2
1
...
Which of them is moving
faster?
...
A boy throws up a ball with a velocity v0
...
(b) magnitude of the momentum of the ball?
...
When a ball falls from a height, its momentum increases
...
4
...
1 s on a 2 kg object initially at rest
...
2 s on a 2 kg
...
...
An object is moving at a constant speed in a circular path
...
...
4 Forces in Pairs
It is the gravitational pull of the earth, which allows an object to accelerate towards the
earth
...
Here, by
action and reaction we mean ‘forces of interaction’
...
One of them is called ‘action’ and the other is called
‘reaction’
...
3
...
1 Third Law of Motion
On the basis of his study of interactions between
bodies, Newton formulated third law of motion: To
every action, there is an equal and opposite
reaction
...
Thus,
when a book placed on a table exerts some force on
the table, the latter, also exerts a force of equal Fig 3
...
3
...
Do the forces F1 and F2 shown
the table exerts a force F2 on
here cancel out? It is important to note that F1 and
the book
...
The action and reaction in a given situation appear as a pair of forces
...
If one goes by the literal meaning of words, reaction always follows an action, whereas
action and reaction introduced in Newton’s third law exist simultaneously
...
Vectorially, if F12 is the force which object 1 experiences due to object 2 and F21 is the
force which object 2 experiences due to object 1, then according to Newton’s third law of
motion, we can write
F 12 = –F21
(3
...
4
...
Impulse is defined as the
product of force (F) and the time duration (∆t) for which the force is applied
...
e
...
∆t
If the initial and final velocities of body acted upon by a force F are u and v respectively
then we can write
3
...
∆t
∆t
= mv – m u
= pf – p i
Notes
= ∆p
That is, impulse is equal to change in linear momentum
...
Intext Questions 3
...
When a high jumper leaves the ground, where does the force which throws the jumper
upwards come from?
...
Identify the action - reaction forces in each of the following situations:
(a) A man kicks a football
...
(c) A ball hits a wall
...
“A person exerts a large force on an almirah to push it forward but he is not pushed
backward because the almirah exerts a small force on him”
...
...
5 Conservation of Momentum
It has been experimentally shown that if two bodies interact, the vector sum of their
momenta remains unchanged, provided the force of mutual interaction is the only force
acting on them
...
Generally, a number of bodies interacting with each other are said to be
forming a system
...
In an isolated system, the
vector sum of the momenta of bodies remains constant
...
Here, it follows that it is the total momentum of the bodies in an isolated system remains
unchanged but the momentum of individual bodies may change, in magnitude alone or
direction alone or both
...
56
Laws of Motion
Conservation of linear momentum is applicable in a wide range of phenomena such as
collisions, explosions, nuclear reactions, radioactive decay etc
...
5
...
(3
...
That is, the momentum of the body will remain unchanged
...
Newton’s third law can also be used to arrive at the same result
...
If FAB and FBA
are the forces which they exert on each other, then in accordance with Newton’s third law
FAB = – FBA
or
or
or
or
∆p A
∆t
=–
∆p B
∆t
∆pA + ∆pB = 0 or
∆p total = 0
p total = constant
That is, there is no change in the momentum of the system
...
3
...
2 A Few Illustrations of Conservation of Momentum
a) Recoil of a gun : When a bullet is fired from a gun, the gun recoils
...
Let
m be the mass of the bullet being fired from a gun of mass M
...
5)
Here, negative sign shows that v2 is in a direction opposite to v1
...
b) Collision : In a collision, we may regard the colliding bodies as forming a system
...
57
57
MODULE - 1
Motion, Force and Energy
Physics
system can be considered to be an isolated system
...
Collision of the striker with a coin of carrom or collision between the billiared balls may be
quite instructive for the study of collision between elastic bodies
...
5 : Two trolleys, each of mass m, coupled together are moving with initial
velocity v
...
What will be the velocity of the trolleys after the impact?
Solution : Let v′ be the velocity of the trolleys after the impact
...
Consider a bomb at rest initially which explodes into two fragments A and B
...
For this reason, the two fragments
will fly off in opposite directions
...
d) Rocket propulsion : Flight of a rocket is an important practical application of
conservation of momentum
...
The shell is provided with a nozzle through which high pressure
gases are made to escape
...
Due to their high pressure, these gases escape
from the nozzle at a high velocity and provide thrust to the rocket to go upward due to the
conservation of momentum of the system
...
If the increase in velocity of the rocket in t second is V, the increase in its momentum =
MV
...
e
...
6 Friction
You may have noticed that when a batter hits a ball to make it roll along the ground, the ball
does not continue to move forever
...
Thus,
the momentum of the ball, which was imparted to it during initial push, tends to be zero
...
Such a force, called the frictional force, exists whenever bodies in contact tend to move
with respect to each other
...
Notes
Force of friction is a contact force and always acts along the surfaces in a direction
opposite to that of the motion of the body
...
For this reason deliberate attempts are made to
make the surfaces rough or smooth depending upon the requirement
...
But then, it is only due to friction that we are able to walk, drive
vehicles and stop moving vehicles
...
It is therefore
said that friction is a necessary evil
...
6
...
To
illustrate this point, let us consider a block resting on some horizontal surface, as shown in
Fig
...
4
...
Initially the block does not
move
...
The force is called
the force of static friction and is represented by symbol fs
...
When Fext is increased further, the block starts to slide and is then subject to kinetic
friction
...
For this reason, the maximum
value of static friction fs between a pair of surfaces in contact will be larger than the
force of kinetic friction fk between them
...
3
...
For a given pair of surfaces in contact, you may like to know the factors on which fs(max)
and fk depend? It is an experimental fact that fs(max) is directly proportional to the normal
force FN
...
e
...
6)
where µs is called the coefficient of static friction
...
Refer to Fig
...
4
...
Since fs = Fext for fs < fs max, we can write
fs < µs FN
...
59
59
MODULE - 1
Motion, Force and Energy
Physics
It has also been experimentally found that maximum force of static friction between a
pair of surfaces is independent of the area of contact
...
3
...
3
...
3
...
In general, µs > µk
...
Values of
µs and µk for a given pair of materials
depend on the roughness of surfaces, there
cleanliness, temperature, humidity etc
...
6 A 2 kg block is resting on a horizontal surface
...
25
...
Solution :
Here m = 2 kg and µs= 0
...
From Eqn
...
6), we recall that
fs(max) = µsFN
= µs mg
= (0
...
8 ms–2)
= 4
...
Example 3
...
1
...
1) (5 kg) (9
...
9 kg ms–2 = 4
...
9 N
= 5
...
Fnet
= 5kg = 1
...
02 ms–2 in the direction of externally applied force
...
6
...
The motion of a wheel is different from sliding motion
...
The
friction in the case of rolling motion is known as rolling friction
...
For example, when steel wheels
roll over steel rails, rolling friction is about 1/100th of the sliding friction between steel and
steel
...
006 for steel on steel and
0
...
04 for rubber on concrete
...
1
Place a heavy book or a pile of books on a table and try to push them with your fingers
...
In which case
do you need less force? What do you conclude from your experience?
3
...
3 Methods of Reducing Friction
Wheel is considered to be greatest invention of
mankind for the simple reason that rolling is much
easier than sliding
...
In a ballbearing, steel balls are placed between two co-axial
cylinders, as shown in Fig
...
6
...
Here the rotation of the balls is almost
frictionless
...
Use of lubricants such as grease or oil between the
Fig
...
6 : Balls in the ball-bearing
3
...
In heavy machines, oil is made to flow
over moving parts
...
In fact, the presence of lubricants changes the nature of
friction from dry friction to fluid friction, which is considerably smaller than the former
...
It also prevents dust and dirt from getting collected on the moving parts
...
Shooting stars (meteors)
shine because of the heat generated by air-friction
...
This is why fishes have a special
shape and fast moving aeroplanes and vehicles are also given a fish-like shape,
called a stream-line shape
...
If
a car is run at a high speed, more fuel will have to be burnt to overcome the increased
fluid (air) friction
...
3
...
A diagram which shows all the forces acting on a
body in a given situation is called a free body diagram (FBD)
...
Draw a simple, neat diagram of the system as per the given description
...
Isolate the object of interest
...
3
...
4
...
(ii)
For obtaining a complete solution, you must have as many independent equations as the
number of unknowns
...
8 : Two blocks of masses m1 and m2 are connected by a string and placed on a
smooth horizontal surface
...
What will be the acceleration of the blocks and the tension in the string
connecting the two blocks (assuming it to be horizontal)?
Solution : Refer to Fig
...
7
...
On applying ΣF = ma in the component form to the
free body diagram of system of two bodies of masses m1 and m2, we get
62
Laws of Motion
Motion, Force and Energy
N
N – (m1 + m2) g = 0
and
MODULE - 1
m
1
F = (m1 + m2)a
T
T
m
2
F
m1 + m2
F
F
⇒ a = m +m
1
2
(m1 + m2) g
Notes
Fig 3
...
F
1
2
Fig 3
...
Example 3
...
Find the acceleration of the masses and the
tension in the string connecting them when the masses are released
...
The
acceleration of mass m2 will also be a only but upward
...
Let
T be the tension in the string connecting the two masses
...
g
Hm +m K
1
2
1
2
T
m2
T
a
m1g
T
T – m2g = m2a
a=
T
m1
FG 2m m IJ
H
K
a
m2g
Fig
...
9
1 2
T = m +m a
1
2
At this stage you can check the prediction of the results thus obtained for the extreme
values of the variables (i
...
m1 and m2)
...
Example 3
...
3
...
The coefficient of kinetic friction between the trolley and the
3
...
02
...
Notes
M
Solution : Fig (b) and (c) shows the free body diagrams of
the trolley and the block respectively
...
For the trolley,
F N = Mg and
FN
a
T
a
T
FK
T – fk = Ma where fk = µk FN
= µk Mg
So
m
(a)
T – µk Mg = Ma
For the block mg – T= ma
mg
Mg
(b)
(c)
Fig
...
10
...
(2)
On adding equations (1) and (2) we get mg – µk Mg = (M + m) a
or
a =
mg − µ k Mg
(2 kg) (9
...
02) (10 kg) (9
...
6 kg ms −2 − 1
...
47 ms–2
12 kg
So
a = 1
...
8 ms–2 – 1
...
33 ms–2)
So
T = 16
...
4
1
...
Show in a
diagram, various forces acting on the block
...
2
...
F
A B
What is the magnitude of force which block A exerts on block
Fig
...
11
B?
...
What will be the tension in the string when a 5kg object suspended from it is pulled up
with
MODULE - 1
Motion, Force and Energy
(a) a velocity of 2ms–1?
(b) an acceleration of 2ms–2?
...
8 Elementary Ideas of Inertial and Non Inertial Frames
To study motion in one dimension (i
...
in a straight line) a reference point (origin) is enough
...
This set of lines is called frame
of reference
...
The description of motion will change with the
change in the state of motion of the observer
...
A person standing on the platform will say that the box is at rest
...
But, what will be the description of the box by a person in a train having
acceleration (a)
...
Obviously, the first law of motion is failing for this observer
...
If the frame is
stationary or moving with a constant velocity with respect to the object under study (another
frame of reference), then in this frame law of inertia holds good
...
On the other hand, if the observer’s frame is accelerating, then
we call it non-inertial frame
...
In a rotating body, this
force is called centrifugal force
...
5
1
...
Will the free
surface of water remain horizontal as the train starts?
...
When a car is driven too fast around a curve it skids outwards
...
3
...
The particle is at a distance of
4 cm from the axis of rotation
...
65
65
MODULE - 1
Physics
Motion, Force and Energy
particle
...
4
...
...
In the reference frame attached to a freely falling body of mass 2 kg, what is the
magnitude and direction of inertial force on the body?
...
Newton’s first law states that a body remains in a state of rest or of uniform motion
in a straight line as long as net external force acting on it is zero
...
Newton’s second law states that the time rate of change of momentum of a body is
proportional to the resultant force acting on the body
...
Newton’s third law states that if two bodies A and B interact with each other, then
the force which body A exerts on body B will be equal and opposite to the force
which body B exerts on body A
...
Frictional force is the force which acts on a body when it attempts to slide, or roll
along a surface
...
The maximum force of static friction fs(max) between a body and a surface is proportional
to the normal force FN acting on the body
...
For a body sliding on some surface, the magnitude of the force of kinetic friction fk is
given by fk = µk FN where µk is the coefficient of kinetic friction for the surfaces in
contact
...
66
Laws of Motion
Newton’s laws of motion are applicable only in an inertial frame of reference
...
MODULE - 1
Motion, Force and Energy
For an object to be in static equilibrium, the vector sum of all the forces acting on it
must be zero
...
Notes
Terminal Exercise
1
...
Which of the following will always be in the direction of net external force acting on
the body?
(a) displacement
(b) velocity
(c) acceleration
(d) Change is momentum
...
(d) acceleration
3
...
5 kg ball is dropped from such a height that it takes 4s to reach the ground
...
4
...
5
...
2 kg falls through air with an acceleration of 6 ms–2
...
6
...
The
load covers a height of 5 m in 2 seconds
...
In a rocket
m changes with time
...
7
...
1 kg moving at 10 m s–1 is deflected by a wall at the same speed in
the direction shown
...
12
3
...
Mass of each bullet is 12 g
...
Notes
8
...
10
...
What will be the velocity of the body after
(a) 1 second from start? (b) 3 seconds from start?
11
...
13
12
...
2 kg block is resting on a horizontal surface
...
5
...
9 N ?
(c) 9
...
For a block on a surface the maximum force of static friction is 10N
...
What minimum force F is required to keep a 5 kg block at rest on an inclined plane of
inclination 300
...
25
...
Two blocks P and Q of masses m1 = 2 kg and m2 = 3 kg respectively are placed in
contact with each other on horizontal frictionless surface
...
Find the following
(a) acceleration of the blocks
(b) force which the block P exerts on block Q
...
Two blocks P and Q of masses m1 = 2 kg and m2 = 4 kg
are connected to a third block R of mass M as shown
68
m1
m2
P
Q
Fig 3
...
3
...
17
...
What will happen if a few
drops of oil are put on the rim?
18
...
How much distance will the block travel before coming to rest? The coefficient
of kinetic friction between the block and the incline plane is µk = 0
...
Notes
Take g = 10 m s–2 and use sin 370 = 0
...
8
...
1
1
...
The statement is true only for a body which was at rest before the application of
force
...
Inertial mass
3
...
4
...
It can also deform bodies
...
2
1
...
(a) Yes (b) No
...
Momentum of the falling ball increases because gravitational force acts on it in the
direction of its motion and hence velocity increases
...
In case (b) the change in momentum will be larger
...
⎜ as F ∝
⎝
5
...
Though the speed is constant, the velocity of the object changes due to change in
direction
...
3
...
The jumper is thrown upwards by the force which the ground exerts on the jumper
...
2
...
(b) The force with which earth pulls the moon is action and the force which the moon
exerts on the earth will be its reaction
...
3
...
No
...
The almirah moves when the push by the person
exceeds the frictional force between the almirah and the floor
...
On a slippery surface, he will not be able to push the almirah foward
...
4
Notes
FN
s
mg
in θ
θ
fs
θ
mg
mg cos θ
Fig
...
15
2
...
(a) (5 × 9
...
8) N = 59 N
3
...
Thus the total force acting on
water in the frame of reference attached to the train is
F=mg–ma
where m is the mass of the water and the glass
...
3
...
The surface of the water
takes up a position normal to F as shown
...
3
...
The
greater v is the larger r would be
...
Once again, the greater
is v, the larger will be r
...
04 m) = 9
...
(4) For an object to fly off centrifugal force (= centripetal force) should be just more
than the weight of a body
...
MODULE - 1
g / r will make objects fly off
...
Notes
Answers to Terminal Problems
1
...
(a) if internal forces developed within the material counter bank the external force
...
(b) It force is applied at right angles to the direction of motion of the body, the force
changes the direction of motion of body and not to speed
...
v = 0 + (–g) × 4
|v| = 40 m s–1
∴ ∆P = m (v – u) = (0
...
When 10 N force acts for 1s
...
0
...
250 N
...
27 N
10
...
(a) 0 N (b) 4
...
5 N
13
...
14
...
(a) 2 m s–2 (b) 6 N
16
...
20 m
3
...
Can you describe the motion of objects moving in a plane, i
...
To do so, we have to introduce certain new concepts
...
This motions is called a projectile motion
...
Such situations arise in projectile motion and circular motion
...
Objectives
After studying this lesson, you should be able to :
explain projectile motion and circular motion and give their examples;
derive expressions for the time of flight, range and maximum height of a projectile;
derive the equation of the trajectory of a projectile;
derive expressions for velocity and acceleration of a particle in circular motion;
and
define radial and tangential acceleration
...
1 Projectile Motion
The first breakthrough in the description of projectile motion was made by Galileo
...
This can be understood by doing the following activity
...
Project one of them horizontally from the top of building
...
What will you notice?
You will find that both the balls hit the ground at the same time
...
Moreover,
this takes place independent of its horizontal motion
...
MODULE - 1
Motion, Force and Energy
Notes
In other words, the two important properties of a projectile motion are :
(i) a constant horizontal velocity component
(i) a constant vertically downward acceleration component
...
Refer to Fig
...
1
...
According to Newton’s
vv
vH
second law there will be no
acceleration
in
the
vv
horizontal direction unless a
vH
C
horizontally directed force
D
acts on the ball
...
1: Curved path of a projectile
vv
friction of air, the only force
acting on the ball once it is free from the hand of the boy is the force of gravity
...
But as the ball moves with this
speed to the right, it also falls under the action of gravity as shown by the vector’s vv
2
2
representing the vertical component of the velocity
...
Having defined projectile motion, we would like to determine how high and how far
does a projectile go and for how long does it remain in air
...
4
...
1 Maximum Height, Time of Flight and Range of a Projectile
Let us analyse projectile motion to determine its maximum height, time of flight and range
...
We can characterise the
initial velocity of an object in projectile motion by its vertical and horizontal components
...
4
...
Let us assume that the initial position of the projectile is at the origin O at t = 0
...
73
73
MODULE - 1
Motion, Force and Energy
Physics
know, the coordinates of the origin are x = 0, y = 0
...
Its
components in the x and y directions are,
v ox = vo cos θ0
v oy = vo sin θ0
and
Notes
(4
...
1 b)
y
V0
O
hmax
θ0
x
R
Fig 4
...
Then
a x = 0; ay = –g = –9
...
2)
The negative sign for ay appears as the acceleration due to gravity is always in the negative
y direction in the chosen coordinate system
...
Therefore, we can use Eqns
...
6) and (2
...
These are given by
Horizontal motion
vx = vox,
since ax = 0
x = voxt = v0 cos θ0t
Vertical motion
vy = voy – g t = v0 sin θ0 – gt
y = voyt – ½g t2 = v0 sin θ0t – ½g t2
(4
...
3b)
(4
...
3d)
The vertical position and velocity components are also related through Eqn
...
10) as
2
2
– g y = ½ (v y – v oy )
(4
...
(4
...
And the vertical motion, given by Eqns
...
3c and d), is motion with constant
(downward) acceleration
...
Now, let us make use of these equations to know the maximum height, time of flight and
range of a projectile
...
At the instant when the projectile is at the
maximum height, the vertical component of its velocity is zero
...
Thus, putting vy = 0 in Eqns
...
3c and e), we get
MODULE - 1
Motion, Force and Energy
0 = voy – g t,
Notes
Thus the time taken to rise taken to the maximum height is given by
t =
v oy
g
=
v 0 sin θ0
g
(4
...
Therefore,
applying v2 – u2 = 2 a s = 2 g h, we get the expression for maximum height :
h=
2
v 0 sin 2 θ0
(as v = 0 and u = v0 sin θ)
2g
(4
...
This is a good
approximation for a projectile with a fairly low velocity
...
(4
...
This is termed as the time of flight
...
The time t given by Eq
...
4)
is the time for half the flight of the ball
...
6)
Finally we calculate the distance travelled horizontally by the projectile
...
(c) Range : The range R of a projectile is calculated simply by multiplying its time of flight
and horizontal velocity
...
(4
...
4), we get
R = (vox) (2 t)
= (v0 cos θ0)
2
= v0
(2 v0 sin θ0 )
g
(2sin θ0 cos θ0 )
g
Since 2 sin θ cos θ = sin 2θ, the range R is given by
R=
2
v0 sin 2θ0
g
(4
...
75
75
MODULE - 1
Motion, Force and Energy
Physics
From Eqn
...
7) you can see that the range of a projectile depends on
its initial speed v0, and
its direction given by θ0
...
Thus, for R to be maximum at a given speed v0, θ0 should be equal to 450
...
Example 4
...
6m
...
What was the maximum height of the hammer? How long did it
remain in the air? Ignore the height of the thrower’s hand above the ground
...
We take the
origin of the coordinate axes at the launch point
...
Thus we have from Eqn
...
7):
2
v0
R =
g
or
v0 =
Rg
It is given that R = 19
...
Putting g = 9
...
6m) × (9
...
8 2 ms–1 = 14
...
(4
...
6), respectively
...
(4
...
6), we get
(
)
2
⎛1⎞
9
...
9 m
Maximum height, h =
–2
2 × 9
...
8 2 m s –1
9
...
Solve the following problems
...
1
1
...
3 : Trajectories of a projectile
bomber plane
...
]
(e) A boat sails in a river
...
2
...
4
...
3
...
90 m
...
5 ms–1
...
78 ms–2
...
4
...
Can you recognise the shapes of
the trajectories of projectiles shown in Fig
...
1, 4
...
3
...
It is easy to determine the equation for the path or trajectory of a projectile
...
(4
...
3d) for x and y
...
(4
...
(4
...
77
77
MODULE - 1
Physics
Motion, Force and Energy
2
1 gx
y = voy x –
2
2 v ox
v ox
x ⎞
⎛
⎜ as t = v ⎟
ox ⎠
⎝
(4
...
(4
...
8a) becomes
y = (tan θ0) x –
Notes
g
x2
2(v 0 cos θ0 )2
(4
...
Eqn
...
8) is of the form y = a x + b x2, which is the equation of a parabola
...
In Fig 4
...
Eqns
...
5) to (4
...
For example,
these equations are used to calculate the launch speed and the angle of elevation required
to hit a target at a known range
...
To get a complete description, we
must include the rotation of the earth also
...
Now, let us summarise the important equations describing projectile motion launched from
a point (x0, y0) with a velocity v0 at an angle of elevation, θ0
...
9 a)
vx = v0 cos θ0
vy = v0 sin θ – g t
(4
...
9 c)
Equation of trajectory:
y = y0 + (tan θ) (x – x0) –
g
(x – x0)2
2(v0 cos θ0 )2
(4
...
The initial
coordinates are left unspecified as (x0, y0) rather than being placed at (0,0)
...
In projectile motion, the acceleration is constant both in magnitude
and direction
...
This is uniform circular motion, and you will
learn about in the following section
...
Disproved that
nature abhors vacuum, presented torricellis theorem
...
3 Circular Motion
Look at Fig
...
4a
...
The word ‘uniform’ refers
to constant speed
...
What about its
velocity? To find out velocity, recall the definition of average velocity and apply it to points
P1 and P2:
v av =
∆r
r2 – r1
=
∆t
t2 – t1
(4
...
The movement of gears, pulleys and wheels also involve circular motion
...
The most familiar example of uniform
circular motion are a point on a rotating fan blade or a grinding wheel moving at constant
speed
...
We have been benefitted immensely by the INSAT series of satellites
and other artificial satellites
...
4
...
1 Uniform Circular Motion
By definition, uniform circular motion is motion with constant speed in a circle
...
4
...
4
...
Now suppose you make the time interval ∆t smaller
3
...
What happens to ∆r? In particular, what is the
direction of ∆r? It approaches the tangent to the circle at point P1 as ∆t tends to zero
...
Can you say
why? This is because the direction of velocity is not constant
...
4
...
Because of this change
in velocity, uniform circular motion is accelerated motion
...
Let us learn about it in
some detail
...
Suppose at any instant its position is at A and its motion is directed along AX
...
Let r and r′ be the position vectors and v and v′ ; the velocities of the particle at A and B
′
respectively as shown in Fig
...
5 (a)
...
As the path of the particle is circular and velocity is along its
tangent, v is perpendicular to r and v′ is perpendicular to ∆r
...
e
...
∆t ⎠
⎝
Let the angle between the position vectors r and r′ be ∆θ
...
To determine the change in velocity ∆v due to the change in direction, consider a point O
outside the circle
...
As |v| = |v′ |, OP = OQ
...
You get a triangle OPQ (Fig
...
5b)
v
X
B
A
v′
Y
P
r′
r ∆θ
v
C
∆θ
O
(a)
Fig
...
5
80
∆v
v′
(b)
Q
Motion in a Plane
Now in triangle OPQ, sides OP and OQ represent velocity vectors v and v′ at A and B
′
respectively
...
In other words the change in the velocity equal to PQ in magnitude and direction
takes place as the particle moves from A to B in time ∆t
...
Then ∆ ACB and
∆POQ are isosceles triangles having their included angles equal
...
∆t
r
or
[as magnitudes of velocity vectors v1 and v2 = v (say)]
∆v
v2
=
∆t
r
or
But
∆v
is the acceleration of the particle
...
r
As ∆t is very small, ∆θ is also very small and ∠OPQ = ∠OQP = 1 right angle
...
Now AC is also
perpendicular to AX
...
It shows that the contripetal force at
any point acts towards the centre along the radius
...
In the absence of such a force, the body will move in a straight line path
...
Activity 4
...
Hold the other end with your
fingers and then try to whirl the stone in a horizontal or vertical circle
...
What happens when the speed of rotation is
3
...
What happens to the
stone when you leave the end of the string you were holding? How do you explain this?
Activity 4
...
Take help of some technical preson if required
...
4
...
How do you explain it?
Some Applications of Centripetal Force
(i) Centrifuges : These are spinning devices used for separating materials having
different densities
...
Therefore, it will move to outermost position in the vessel and hence can be
separated
...
In a chemistry
laboratory these are used for chemical analysis
...
4
...
Centripetal force, like gravitational force, is greater for the more dense substance
...
4
...
82
Motion in a Plane
MODULE - 1
Motion, Force and Energy
Notes
Fig
...
7: Mud or water on a fast-turning wheel flies off tangentially
(iii) Planetary motion : The Earth and the other planets revolving round the sun get
necessary centripetal force from the gravitational force between them and the sun
...
2 : Astronauts experience high acceleration in their flights in space
...
The capsule is whirled around in a circular path,
just like the way we whirl a stone tied to a string in a horizontal circle
...
Solution : The circumference of the circular path is 2π × (radius) = 2π × 15 m
...
Therefore,
24
speed of the capsule, v =
2πr 2π × 15 m
= (60/24) s = 38 ms–1
T
The magnitude of the centripetal acceleration
(38 ms –1 ) 2
v2
=
= 96 ms–2
a =
15 m
r
Note that centripetal acceleration is about 10 times the acceleration due to gravity
...
2
1
...
...
An athlete runs around a circular track with a speed of 9
...
What is the radius of the track?
...
The Fermi lab accelerator is one of the largest particle accelerators
...
0 km
3
...
99995% of the speed of light
...
...
4 Applications of Uniform Circular Motion
Notes
So far you have studied that an object moving in a circle is accelerating
...
From Newton’s second law we can say that
as the object in circular motion is accelerating, a net force must be acting on it
...
Then we will apply Newton’s laws of motion to uniform circular motion
...
Let us first determine the force acting on a particle that keeps it in uniform circular motion
...
From Newton’s
second law, the net external force acting on a particle is related to its acceleration by
mv 2
mv 2
(4
...
(4
...
An important thing to understand and remember
is that the term ‘centripetal force’ does not refer to a type of force of interaction like
the force of gravitation or electrical force
...
It does not tell us how this force is provided
...
For
example, in the motion of a planet around the sun, the centripetal force is provided by the
gravitational force between the two
...
You will understand
these ideas better when we apply them in certain concrete situations
...
4
...
Have you ever thought as to why does it happen?
You tend to be thrown out because enough centripetal force has not been provided to keep
you in the circular path
...
When you slow down, the needed centripetal force
decreases and you manage to complete this turn
...
4
...
To keep the car moving uniformly on the circular path, a force must act on it
directed towards the centre of the circle and its magnitude must be equal to
mv2/r
...
84
Motion in a Plane
MODULE - 1
Motion, Force and Energy
FN cos θ
A
FN
FN cos θ
FN sin θ
Fr
θ
X
FN sin θ
O
mg
(a)
FN
θ
Notes
mg
(c)
(b)
Fig
...
6 : A car taking a turn (a) on a level road; (b) on a banked road; and (c) Forces on the car with FN
resolved into its rectangular components
...
Now if the road is levelled, the force of friction between the road and the tyres provides
the necessary centripetal force to keep the car in circular path
...
The roads at curves are,
therefore, banked, where banking means the raising of the outer edge of the road above
the level of the inner edge (Fig
...
6)
...
For example, when car tyres are smooth or there is water or snow on
roads, the coefficient of friction becomes negligible
...
Let us now analyse the free body diagram for the car to obtain an expression for the angle
of banking, θ, which is adjusted for the sharpness of the curve and the maximum allowed
speed
...
The forces acting on the car are the car’s weight mg and FN, the force of normal
reaction
...
Thus, resolving
the force FN into its horizontal and vertical components, we can write
m v2
(4
...
20b)
FN cos θ = m g
We have two equations with two unknowns, i
...
, FN and θ
...
Dividing Eqn
...
20 a) by Eqn
...
20 b), we get
FN sin θ =
m v2 / r
v2
tan θ =
=
mg
rg
or
θ = tan
–1
v2
rg
(4
...
(4
...
(4
...
So even large trucks
and other heavy vehicles can ply on banked roads
...
85
85
MODULE - 1
Motion, Force and Energy
Notes
Physics
Secondly, θ should be greater for high speeds and for sharp curves (i
...
, for lower values
of r)
...
So a vehicle driver must drive within prescribed speed limits on
curves
...
Hence, there may be accidents
...
Vehicles can
maintain a stable circular path around curves, if their speed remains within this range
...
Let the typical speed
of a vehicle be 50 ms–1
...
(4
...
θ = tan
–1
(50 ms –1 )2
= tan–1 (0
...
8 ms –2 )
You may like to consider another application
...
4
...
4
...
In such situations, at the bottom of the loop, the pilots
feel as if they are being pressed to their seats by a force several times the force of gravity
...
Fig
...
8b shows the ‘free body’ diagram for the
pilot of mass m at the bottom of the loop
...
4
...
The forces acting on him are mg and the normal force N exerted by the seat
...
8 m s –2 × 1500 m) ⎥ = m g × 3
...
7
...
MODULE - 1
Motion, Force and Energy
Intext Questions 4
...
Aircrafts usually bank while taking a turn when flying
at a constant speed (Fig
...
8)
...
(Fa is the force exerted by the air on the aircraft)
...
What is the radius
of curvature of the turn? Take g = 10 ms–2
...
Calculate the maximum speed of a car which makes a turn of radius 100 m on a
horizontal road
...
90
...
...
An interesting act performed at variety shows is to swing a bucket of water in a
vertical circle such that water does not spill out while the bucket is inverted at the top
of the circle
...
Derive an expression for the minimum speed
of the bucket at the top of the circle in terms of its radius R
...
0 m
...
What You Have Learnt
Projectile motion is defined as the motion which has constant velocity in a certain
direction and constant acceleration in a direction perpendicular to that of velocity:
ax = 0
ay = – g
vx = v0 cos θ
vy = v0 sin θ –g t
x = x0 + (v0 cos θ) t
y = y0 + (v0 sin θ) –½ –g t2
2
v 0 sin 2θ
Height h =
g
Time of flight T =
2v 0 sin θ
g
2
v 0 sin 2θ
Range of the projectile R =
g
3
...
A particle
undergoing uniform circular motion in a circle of radius r at constant speed v has a
centripetal acceleration given by
Notes
v2
ar = –
ˆ
r r
where r is the unit vector directed from the centre of the circle to the particle
...
The centripetal force acting on the particle is given by
F = m ar =
m v2
ˆ = m r ω2
r r
Terminal Exercise
1
...
Explain why the outer rail is raised with respect to the inner rail on the curved portion
of a railway track?
3
...
A stone is thrown from the window of a bus moving on horizontal road
...
A string can sustain a maximum force of 100 N without breaking
...
Compute the maximum speed with which the body can be rotated without breaking
the string?
6
...
What will be
the centripetal acceleration when turning the curve?
7
...
At what distance from the gun will the bullet strike the ground?
8
...
What is the speed of the tip of
this hand?
9
...
In this problem consider a daredevil motor cycle rider trying to cross a
gap at a velocity of 100 km h–1
...
4
...
Let the angle of incline on either side be
450
...
10
...
Calculate
the vertical and horizontal components of the velocity, the maximum height that the
shell reaches, and its range
...
An aeroplane drops a food packet from a height of 2000 m above the ground while in
88
Motion in a Plane
horizontal flight at a constant speed of 200 kmh–1
...
4
...
4
...
A mass m moving in a circle at speed v on a frictionless table is attached to a hanging
mass M by a string through a hole in the table (Fig
...
10)
...
13
...
What is the
centripetal force on a passenger of mass m = 90 kg?
Answers to Intext Questions
4
...
(3) Maximum Range
2
v0
(9
...
23 m
9
...
23 m – 8
...
33 m
...
2
(1) (a) Yes (b) No (c) Yes
(d) No
The velocity and acceleration are not constant because their directions are changing
continuously
...
0 ms –1 ) 2
v2
v2
,r=
=
= 27 m
3 ms –2
r
α
c2
(3×108 ms –1 ) 2
=
r
10 ×103 m
= 9 × 1013 ms–2
a=
3
...
3
(1) This is similar to the case of banking of roads
...
Fig
...
11 shows the free body diagram
...
3 m
–2
g tan θ0
⎝ 10 ms × tan 30 ⎠
L
L cos θ
θ
30º
θ
L sin θ
mg
Fig
...
11
(2) The force of friction provides the necessary centripetal acceleration :
mv 2
r
Since the road is horizontal N – mg
F s = µsN =
mv 2
Thus µs mg =
r
or
v 2 = µsg r
or v = (0
...
(3) Refer to Fig
...
12 showing the free body diagram for the bucket at the top of the
circle
...
At the top of the circle
...
4
...
4
...
For
R = 1
...
2 ms–1
Answers to Terminal Problems
5
...
2 ms–2
7
...
1
...
77
...
vx = 250 3 ms–1
vy = 250 ms–1
Vertical height = 500 m
Horizontal range = 3125 m
11
...
9 m
12
...
125 N
3
...
Since times immemorial, human beings
have wondered about this phenomenon
...
He proposed that the gravitational force is responsible for bodies being
attracted to the earth
...
It is a universal force, that is, it is
present everywhere in the universe
...
In this lesson you will learn Newton’s law of gravitation
...
This acceleration, called acceleration
due to gravity, is not constant on the earth
...
You will also study Kepler’s laws of planetary motion and orbits of artificial satellites of various kinds in this lesson
...
Objectives
After studying this lesson, you should be able to:
state the law of gravitation;
calculate the value of acceleration due to gravity of a heavenly body;
analyse the variation in the value of the acceleration due to gravity with height,
depth and latitude;
identify the force responsible for planetary motion and state Kepler’s laws of
planetary motion;
calculate the orbital velocity and the escape velocity;
explain how an artificial satellite is launched;
distinguish between polar and equatorial satellites;
state conditions for a satellite to be a geostationary satellite;
calculate the height of a geostationary satellite and list their applications; and
state the achievements of India in the field of satellite technology
...
1 Law of Gravitation
It is said that Newton was sitting under a tree
when an apple fell on the ground
...
He asked
himself: Could it be the same force which
keeps the moon in orbit around the earth?
Newton argued that at every point in its orbit,
Earth
the moon would have flown along a tangent,
but is held back to the orbit by some force
Fig
...
1 : At each point on its orbit, the moon (Fig
...
1)
...
Kepler’s laws that the force between the Sun
and planets varies as 1/r2
...
Then he generalised the idea to formulate the universal law of gravitation as
...
Thus, if m1 and m2 are the masses of the two particles, and r is the
distance between them, the magnitude of the force F is given by
...
1)
The constant of proportionality, G , is called the universal constant of gravitation
...
This means that if the force between two particles is F on the earth, the force between
these particles kept at the same distance anywhere in the universe would be the same
...
It is also one of the fundamental forces of nature
...
Also, the force is along the line joining the
two particles
...
(5
...
As
stated, the gravitational force acts along the line
F12
F21
m1
m2
r12
Fig
...
2 : The masses m1 and m2 are
placed at a distance r12 from
eact other
...
3
...
That is, m2 attracts m1 with a force which is along the line
joining the two particles (Fig
...
2)
...
2)
ˆ
Here r12 is a unit vector from m1 to m2
In a similar way, we may write the force exerted by m2 on m1 as
F 21 = – G
m1m2
ˆ
r
2
r21 21
(5
...
(5
...
3) we find that
F12 = – F21
(5
...
Remember that r12 and r21
have unit magnitude
...
Unless specified, in this lesson we would use only the magnitude of the gravitational force
...
It was determined by Cavendish for the first time about
100 years later
...
67 × 10–11 Nm2 kg–2
...
Example 5
...
Show that the force acting on a planet is inversely proportional to the
square of the distance
...
(In reality, the orbits
are nearly circular
...
Since v = rω =
2πr
, where T is the period, we can rewrite
T
above expression as
⎛ 2πr ⎞
F =m ⎜
⎝ T ⎟
⎠
or
94
F =
4π2 mr
T2
2
r
Gravitation
But T2∝ r3 or
MODULE - 1
Motion, Force and Energy
T2 = Kr3 (Kelpler’s 3rd law)
where K is a constant of proportionality
...
2
3
r
r
K
K
Kr
4π2 m
is constant for a planet)
r
K
Before proceedins further, it is better that you check your progress
...
1
1
...
3 days
...
5 days; it is this period that is used to fix the duration
of a month in some calendars)
...
84 × 108 m (60 times
the earth’s radius)
...
8 ms–2 divided by 3600, to take account of the
variation of the gravity as 1/r2
...
2
...
(5
...
...
Using Eqn
...
1), show that G may be defined as the magnitude of force between
two masses of 1 kg each separated by a distance of 1 m
...
4
...
What
happens to F if (i) the distance is doubled without any change in masses, (ii) the
distance remains the same but each mass is doubled, (iii) the distance is doubled and
each mass is also doubled?
...
Two bodies having masses 50 kg and 60 kg are seperated by a distance of 1m
...
...
2 Acceleration Due to Gravity
From Newton’s second law of motion you know that a force F exerted on an object
produces an acceleration a in the object according to the relation
F = ma
(5
...
e
...
The acceleration produced by the
3
...
It is denoted by the symbol
g
...
(5
...
6)
where M is the mass of the earth and R is its radius
...
(5
...
6), we get
mg = G
or
g =G
mM
R2
M
R2
(5
...
It is this direction that we call vertical
...
5
...
The
B
direction perpendicular to the vertical is called the
horizontal direction
...
(5
...
On the surface of the earth, the value of
g is taken as 9
...
Given the mass and the radius of a satellite or a
planet, we can use Eqn
...
7) to find the acceleration
due to the gravitational attraction of that satellite or
planet
...
5
...
(5
...
The acceleration due to gravity produced in a body is independent of its mass
...
If we drop these
balls from a certain height at the same time, both would reach the ground
simultaneously
...
1
Take a piece of paper and a small pebble
...
Observe the path followed by the two bodies and note the times at which they
touch the ground
...
Release them
simultaneously from a height and observe the time at which they touch
the ground
...
Till sixteenth century it was a common belief that a heavy body falls
faster than a light body
...
It is said that he went up to the top
of the Tower of Pisa and released simultaneously two iron balls of considerably
different masses
...
But when feather
and a stone were made to fall simultaneously, they reached the ground at different
times
...
He said that if there were no air, the two bodies would
fall together
...
Remember that the
moon has no atmosphere and so no air
...
For
small heights above the surface of the earth, the acceleration due to gravity does not
change much
...
s = ut + (
and
(5
...
A body falling with an acceleration equal
to g is said to be in free-fall
...
(5
...
So, a simple experiment like dropping a heavy coin from a height and
measuring its time of fall with the help of an accurate stop watch could give us the value
of g
...
45 s
...
However, in the laboratory you would determine g by an
indirect method, using a simple pendulum
...
When we consider two discreet particles or mass points, the separation between them is
just the distance between them
...
This is a point such that, as far as the
gravitational effect is concerned, we may replace the whole body by just this point and the
effect would be the same
...
That is
why we choose the center of the earth to measure distances to other bodies
...
3
...
But this point is outside the mass of the body
...
Where is your own centre of gravity located? Assuming that we have a
regular shape, it would be at the centre of our body, somewhere beneath the navel
...
This is
a point at which the whole mass of the body can be assumed to be concentrated
...
The use of centre of gravity, or the center of mass, makes our calculations extremely
simple
...
You should remember that G and g represent different physical quantities
...
You may like to answer a few questions to check your progress
...
2
1
...
97 × 1024 kg and its mean radius is 6
...
Calculate
the value of g at the surface of the earth
...
2
...
Compare values of g at the poles and at the equator
...
3
...
What is the direction of g when (i) the particle is going up, (ii)
when it is at the top of its journey, (iii) when it is coming down, and (iv) when it has
come back to the ground?
...
The mass of the moon is 7
...
74 × 106 m
...
...
3 Variation in the Value of g
5
...
1 Variation with Height
The quantity R2 in the denominator on the right hand side of Eqn
...
7) suggests that the
magnitude of g decreases as square of the distance from the centre of the earth
increases
...
However, if
98
Gravitation
the distance h above the surface of the earth, called altitude, is small compared with the
radius of the earth, the value of g, denoted by gh, is given by
gh =
MODULE - 1
Motion, Force and Energy
GM
( R + h) 2
Notes
GM
= R 2 ⎛1 + h ⎞
⎜
⎟
⎝ R⎠
2
g
= ⎛1 + h ⎞
⎜
⎟
⎝
R⎠
2
(5
...
Therefore,
2
2
g
h⎞
2h
⎛
⎛ h⎞
1+ ⎟ = 1 +
+ ⎜ ⎟
gh = ⎜ R ⎠
⎝
⎝ R⎠
R
Since (h/R) is a small quantity, (h/R)2 will be a still smaller quantity
...
Thus
gh =
g
⎛ 2h ⎞
⎜1 + ⎟
⎝
R⎠
(5
...
Example 5
...
Let us calculate the
value of g at an altitude of 10 km
...
8 ms–2
...
(5
...
8 ms –2
=
= 9
...
⎛ 2
...
003
⎜1 + 6400 km ⎟
⎝
⎠
5
...
2
...
5
...
Let us assume that the earth is
a sphere of uniform density ρ
...
Draw a sphere of radius (r – d)
...
99
99
MODULE - 1
Motion, Force and Energy
Notes
Physics
r
...
That is, the net force on
the particle at P due to the matter in the shell is zero
...
The mass M′ of the sphere of radius
(r – d) is
4π
ρ (R – d)3
3
M′ =
r=
–
d d
P
R
(5
...
5
...
11)
Note that as d increases, (R – d) decreases
...
At d = R, that is, at the centre of the earth, the acceleration
due to gravity will vanish
...
Therefore, acceleration due to gravity is linearly proportional to r
...
5
...
–2
g→
9
...
45 ms–2
O
R
2R
r→
Fig
...
5 : Variation of g with distance from the centre of the earth
We can express gd in terms of the value at the surface by realizing that at d = 0, we get the
surface value: g =
4πG
ρR
...
12)
On the basis of Eqns
...
9) and (5
...
100
Gravitation
MODULE - 1
Motion, Force and Energy
Internal Structure of the Earth
45
63
km
3490
Core km
Notes
1
...
01 × 10 kg
25km
CRUST
3
...
5
...
Three prominent layers of the earth are
shown along with their estimated masses
...
5
...
The top surface layer is very light
...
So, the value of g increases up to a certain depth and then
starts decreasing
...
5
...
3 Variation of g with Latitude
You know that the earth rotates about its axis
...
In the absence of gravity, all these particles would be
flying off the earth along the tangents to their circular orbits
...
You also know that to keep a particle in
circular motion, it must be supplied centripetal force
...
As a result, the force of attraction of the earth on
objects on its surface is slightly reduced
...
At poles, the effect vanishes completely
...
If gλ denotes the value of g at latitude λ and
g is the value at the poles, then
gλ = g – Rω2 cosλ,
(5
...
You can easily see that at
the poles, λ = 90 degrees, and hence gλ = g
...
3 : Let us calculate the value of g at the poles
...
357 × 106 m
The mass of the earth = 5
...
(5
...
67 × 10–11 × 5
...
357 × 106)2] ms–2
= 9
...
101
101
MODULE - 1
Motion, Force and Energy
Physics
Example 5
...
Solution :
The period of rotation of the earth, T = 24 hours = (24 × 60 × 60) s
∴ frequency of the earth’s rotation = 1/T
angular frequency of the earth ω = 2π/T = 2π/(24 × 60 × 60)
Notes
= 7
...
371 × 106 × (7
...
5 = 0
...
853 – 0
...
836 ms–2
Intext Questions 5
...
At what height must we go so that the value of g becomes half of what it is at the
surface of the earth?
...
At what depth would the value of g be 80% of what it is on the surface of the earth?
...
The latitude of Delhi is approximately 30 degrees north
...
...
A satellite orbits the earth at an altitude of 1000 km
...
(5
...
Which
method do you consider better for this case and why?
...
4 Weight and Mass
The force with which a body is pulled towards the earth is called its weight
...
14)
Since weight is a force, its unit is newton
...
8 ms–2 = 490 N
...
The weight is maximum at the poles and minimum at the equator
...
The weight
decreases when we go to higher altitudes or inside the earth
...
Mass is an intrinsic property of a body
...
MODULE - 1
Motion, Force and Energy
Note: In everyday life we often use mass and weight interchangeably
...
Activity 5
...
Plot a graph showing the weight against distance
...
Try the following questions to consolidate your ideas on mass and weight
...
4
1
...
In what way would your weight and mass be affected?
...
Compare your weight at Mars with that on the earth? What happens to your mass?
Take the mass of Mars = 6 × 1023 kg and its radius as 4
...
...
You must have seen two types of balances for weighing objects
...
In one pan, we place the object to be weighed and in the other we
place weights
...
Here the object to be weighed is
suspended from the hook at the end of a spring and reading is taken on a scale
...
Now you take them to the moon
...
5
...
Greek
astronomers lent great support to this notion
...
However, Polish Astronomer Copernicus in the 15th century proposed that all the planets
revolved around the Sun
...
Another European astronomer, Tycho Brahe, collected a lot of
observations on the motion of planets
...
3
...
Tycho religiously collected the
data of the positions of various planets on the daily basis for more
than 20 years
...
On the basis of his analysis,
Kepler arrived at the three laws of planetary motion
...
For his assertion that the earth revolved around the Sun, Galileo came into conflict
with the church because the Christian authorities believed that the earth was at the
centre of the universe
...
Interestingly, Galileo was
freed from that blame recently by the present Pope
...
These are:
1
...
5
...
(An ellipse
has two foci
...
5
...
If the time taken by the
planet to move from point A to B is the same as from point C to D, then according to the
second law of Kepler, the areas AOB and COD are equal
...
The area swept by the line joining the planet to the sun in unit time is constant through
out the orbit (Fig 5
...
The square of the period of revolution of a planet around the sun is proportional to the
cube of its average distance from the Sun
...
Let us look at the third law a little more carefully
...
1)
...
15)
The constant of proportionality cancels out when we divide the relation for one planet by
the relation for the second planet
...
For example, it can be
used to get T2, if we know T1, r1 and r2
...
5 : Calculate the orbital period of planet mercury, if its distance from the
Sun is 57
...
You are given that the distance of the earth from the Sun is
1
...
Solution : We know that the orbital period of the earth is 365
...
So, T1 = 365
...
5 × 1011 m
...
9×109 m for mercury
...
25) 2 × (57
...
5 × 1011 ) 3 m 3
r1
= 87
...
In the same manner you can find the orbital periods of other planets
...
You can also check your results with numbers in Table 5
...
Table 5
...
387
0
...
0
1
...
2
9
...
2
30
...
4
Radius
(x103 km)
Mass
(Earth Masses)
2
...
05
6
...
39
71
...
00
25
...
3
1
...
53
0
...
00
0
...
8
95
...
50
17
...
002
Kepler’s laws apply to any system where the force binding the system is gravitational in
nature
...
They also apply to the earth
and its satellites like the moon and artificial satellites
...
6 : A satellite has an orbital period equal to one day
...
) Calculate its height from the earth’s surface, given that
3
...
3 days
...
With respect to the earth, which itself is in orbit round the Sun, the orbital period of
the moon is about 29
...
]
Solution : A geostationary satellite has a period T2 equal to 1 day
...
3
days and r1 = 60 RE, T2 = 1 day
...
(5
...
3 day
⎦
= 6
...
Remember that the distance of the satellite is taken from the centre of the earth
...
6 RE
...
6 RE
...
6 by the radius of the earth in km
...
5
...
If the orbital period of a planet is T and
its distance from the Sun is r, then it covers a distance 2πr in time T
...
16)
There is another way also to calculate the orbital velocity
...
This force must be supplied by the
force of gravitation between the Sun and the planet
...
Equating the two forces, we get
r2
gravitational force on the planet is
mυorb
r
2
=
G M
r2
s
,
so that,
G Ms
(5
...
The orbital velocity
depends only on the distance from the Sun
...
(5
...
(5
...
v orb =
Intext Questions 5
...
Many planetary systems have been discovered in our Galaxy
...
106
Gravitation
2
...
Which one of them has the longer period? If the time
period of the former is 90 min, find the time period of the latter
...
3
...
It is orbiting the Sun at a distance of 86 AU
...
It is equal to 1
...
) Calculate its orbital period in years
...
4
...
...
Using Eqns
...
16) and (5
...
...
6 Escape Velocity
You now know that a ball thrown upwards always comes back due to the force of gravity
...
If you have
a friend with great physical power, ask him to throw the ball upwards
...
You may then ask: Is it possible
for an object to escape the pull of the earth? The answer is ‘yes’
...
It is defined as the minimum velocity
required by an object to escape the gravitational pull of the earth
...
It will also depend on
the radius of the body, because smaller the radius, stronger is the gravitational force
...
18)
where M is the mass of the earth and R is its radius
...
It is not that the force of gravity ceases to act when an object is launched with escape
velocity
...
Both the velocity of the object as well as the force of gravity
acting on it decrease as the object goes up
...
Hence the object escapes the pull of gravity
...
3
...
6
1
...
97 × 1024 kg and its radius is 6371 km
...
...
Suppose the earth shrunk suddenly to one-fourth its radius without any change in its
mass
...
3
...
What would be the escape velocity from this planet in terms of the escape
velocity from the earth?
...
7 Artificial Satellites
A cricket match is played in Sydney in Australia but we can watch it live in India
...
Have you ever wondered what makes it
possible? All this is made possible by artificial satellites orbiting the earth
...
If you project a body at an angle to
the horizontal, it follows a parabolic path
...
What happens is shown in Fig
...
8
...
Eventually, the projectile goes into an orbit around the
earth
...
Remember that such satellites are man-made
and launched with a particular purpose in mind
...
Earth
Fig
...
8 : A projectile to orbit the earth
In order to put a satellite in orbit, it is first lifted to a height of about 200 km to minimize loss
of energy due to friction in the atmosphere of the earth
...
The orbit of an artificial satellite also obeys Kepler’s laws because the controlling force is
108
Gravitation
gravitational force between the satellite and the earth
...
Remember that the orbital velocity of an artificial satellite has to be less than the escape
velocity; otherwise it will break free of the gravitational field of the earth and will not orbit
around the earth
...
19)
2
Artificial satellites have generally two types of orbits (Fig
...
9) depending on the purpose
for which the satellite is launched
...
The altitude of these orbits is about 800 km
...
As a result, it moves to a lower height where the density is high
...
The time period of polar satellites is around 100 minutes
...
During repeated crossing, the satellite can scan the whole earth as it spins
about its axis (Fig
...
10)
...
Polar plane
GN
equator
equatorial plane
GS
Fig
...
9: Equitorial and polar orbits
Satellites used for communications are put in equatorial orbits at high altitudes
...
Their height, as you saw in
Example 5
...
Since their orbital period matches that of the
earth, they appear to be hovering above the same spot on the earth
...
Since a geo-synchronous satellite observes the same spot on the earth
all the time, it can also be used for monitoring any peculiar happening that takes a long time
to develop, such as severe storms and hurricanes
...
109
109
MODULE - 1
Physics
Motion, Force and Energy
Descending orbit
West looking
Notes
Ascending orbit
East looking
Fig
...
10: A sun synchromous satellite scanning the earth
Applications of Satellites
Artificial satellites have been very useful to mankind
...
Weather Forecasting : The satellites collect all kinds of data which is useful
in forecasting long term and short term weather
...
For a country like India, where so much depends on timely
rains, the satellite data is used to watch the onset and progress of monsoon
...
2
...
This is of great help in locating our own position
if we have forgotten our way and are lost
...
3
...
Apart from television signals, telephone and radio signals
are also transmitted
...
4
...
You must have heard of Hubble Space Telescope and Chandra X-Ray Telescope
...
So there is hardly any reduction in
its intensity
...
Recently, a group of Europeon scientists have observed an earth like planet outside our solar system at a distance of 20 light years
...
110
Monitoring Military Activities : Artificial satellites are used to keep an eye
on the enemy troop movement
...
Gravitation
MODULE - 1
Motion, Force and Energy
Vikram Ambalal Sarabhai
Born in a family of industrialists at Ahmedabad, Gujarat, India
...
His
initial work on time variation of cosmic rays brought him laurels in
scientific fraternity
...
Notes
5
...
1 Indian Space Research Organization
India is a very large and populous country
...
So, weather forecast is an important
task that the government has to perform
...
Then much of our area remains unexplored for minerals, oil and gas
...
With this in
view, the Government of India set up in 1969 the Indian Space Research Organization
(ISRO) under the dynamic leadership of Dr
...
Dr
...
ISRO has pursued a very vigorous programme
to develop space systems for communication, television broadcasting, meteorological
services, remote sensing and scientific research
...
5
...
5
...
In fact, it has launched satellites for other countries like Germany, Belgium
and Korea
...
Its scientific programme
includes studies of
(i) climate, environment and global change,
(ii) upper atmosphere,
(iii) astronomy and astrophysics, and
(iv) Indian Ocean
...
It is being used to educate both young and adult students living in remote places
...
Fig
...
11: PSLV
Fig
...
12: GSLV
3
...
7
1
...
Suppose people living this desire to put in orbit a Mars synchronous satellite
...
6 hours
...
4 × 1023 kg and 3400 km, respectively
...
2
...
...
It varies as
the product of their masses and inversely as the square of distance between them
...
The force of gravitation of the earth attracts all bodies towards it
...
8 ms–2
...
The acceleration due to gravity varies with height, depth and latitude
...
Kepler’s first law states that the orbit of a planet is elliptic with sun at one of its foci
...
Kepler’s third law states that the square of the orbital period of a planet is proportional
to the cube of its mean distance from the Sun
...
The orbital velocity of a satellite depends on its distance from the earth
...
You have learnt that the gravitational attraction is mutual
...
We set up an experiment on earth to measure the force of gravitation between two
particles placed at a certain distance apart
...
We
take the same set up to the moon and perform the experiment again
...
Suppose the earth expands to twice its size without any change in its mass
...
Suppose the earth loses its gravity suddenly
...
Refer to Fig
...
6 which shows the structure of the earth
...
Notes
6
...
7
...
Calculate your weight on the moon
...
A polar satellite is placed at a height of 800 km from earth’s surface
...
Answers to Intext Questions
5
...
Moon’s time period T = 27
...
3 × 24 × 3600 s
Radius of moon’s orbit R= 3
...
=
R
T2
T2
4π 2 × 3
...
3 × 24 × 3600) 2 s 2
4π2 × 3
...
3 × 2
...
6) 2
=
...
8
ms–2
3600
3
...
00272 ms–2
2
...
F = G
Force × r 2
Nm 2
=
2
(mass)
kg 2
m1 m2
r2
If m1 = 1kg, m2 = 1kg, r = 1m, then F = G
or G is equal to the force between two masses of 1kg each placed at a distance of 1m
from each other
4
...
(ii) F α m1m2, if m1 and m2 are both doubled then F becomes 4 times
...
50 kg × 60 kg
Nm 2
–11
5
...
68 × 10
1 m2
kg 2
= 6
...
kg 2
1 m2
= 6
...
2
1
...
67 × 10–11
Nm 2
5
...
kg (6
...
97 × 59
...
371 × 6
...
81 m s–2
114
Gravitation
MODULE - 1
Motion, Force and Energy
2
...
67 × 10
–11
5
...
6 2
2
kg 2 (6
...
97 × 59
...
371 × 6
...
81 ms–2
Similarly,
6
...
7
N
geguator = 6
...
378 kg = 9
...
The value of g is always vertically downwards
...
g moon= 6
...
3 ×10 22 kg
Nm2
× (1
...
67 × 7
...
74 × 1
...
61 m s–2
5
...
Let g at distance r from the centre of the earth be called g1
...
412 R
∴ Height from earth’s surface= 1
...
4142 R
2
...
Suppose at depth d,
g is called gd
...
115
115
MODULE - 1
Physics
Motion, Force and Energy
0
...
2 R
3
...
3, we calculated ω = 7
...
37 × 106 × (7
...
3 2 = 0
...
853 m s2
g at poles
(Calculated in example 5
...
853 ms–2 – 0
...
824 ms–2
4
...
9),
g h=
=
g
9
...
81 m s –2
= 7
...
67 × 10–11
Nm 2
5
...
kg 2 (7
...
33 ms–2
This gives more accurate results because formula (5
...
In this
case h is not << R
...
4
1
...
So, your weight on moon will
become 1/6th of your weight on the earth
...
2
...
3 × 106 m
116
Gravitation
MODULE - 1
Motion, Force and Energy
Nm 2
∴
g Mars = G
M
= 6
...
R2
kg 2
6 ×1023 kg
= 2
...
3×106 )2 m 2
Weight on Mars
m
...
16
Weight on Earth = m
...
81 = 0
...
Mass remains constant
...
Balances with two pans actually compare masses because g acts on both the pans and
gets cancelled
...
The balance with two pans gives the same reading on the moon as on the earth
...
5
...
Yes
...
2
...
Let T1 = 90 min,
r 1 = 1000 km + 6371 km
r 2 = 2000 km + 6371 km
[ From the centre of the earth]
∴
T12
...
9 min
3
...
, rearth = 1 AU
2
sedna
T
(1yr) 2 (86 AU)3
(86)3 yr 2
=
(1AU)3
∴ Tsedna=797
...
If v is the orbital velocity of the satellite of mass m at a distance r from the centre of the
earth, then equating centripltal force with the gravitational force, we have
3
...
5
...
(5
...
17),
Notes
GM
4π 2 r 2
4π2
...
or T2 α r3
...
6
1
...
67 × 10 –11
Nm 2 5
...
kg 2 6
...
67 × 5
...
371
= 11
...
3 kms–1
2
...
3
...
5
...
(R + h) 2 =
T
( R + h) 2
⇒ (R + h)3 =
118
GM 2
T
4π2
Gravitation
=
6
...
4 × 10 × (14
...
14) 2
23
2
= 8370 × 1018 m
R + h = 20300 km
h = 26900 km
Notes
2
...
also work
...
125 N
5
...
5 ms–2
7
...
T
500
N , mass 50 kg on moon as well as on earth
6
1
1 h , v = 7
...
119
119
MODULE - 1
Physics
Motion, Force and Energy
6
Notes
WORK ENERGY AND POWER
You know that motion of objects arises due to application of force and is described by
Newton’s laws of motion
...
In this lesson, you will learn the concepts of work
and energy
...
Primitive man used muscular energy to do work
...
With the invention of various kinds of machines, the
ability to do work increased greatly
...
Energy and work are, therefore, closely linked
...
e
...
The rate of doing work is known as power
...
120
Work Energy and Power
MODULE - 1
Motion, Force and Energy
6
...
When you study, you do
mental work
...
But in science, work has a definite
meaning
...
The work is defined in the following way :
Notes
Let us suppose that a constant force F acting on an object results in displacement d i
...
moves it by a distance d along a straight line on a horizontal surface, as shown in Fig
...
1
...
F
F
θ
d
Fig 6
...
The direction of force
makes an angle θ with the horizontal direction
...
Then work done by force F is given by
W = F cosθ
...
1)
In vector form, the work done is given by:
W = F
...
2)
Note that if d = 0, W = 0
...
Also note that though both force and displacement
are vectors, work is a scalar
...
1
You and your friends may try to push the wall of a room
...
Thus we say that no work is done
...
(6
...
If the applied force is in newton and displacement
is in metre, then the unit of work is joule
...
metre = Nm
(6
...
One joule is defined, as the work done by a force of one newton when it produces a
displacement of one metre
...
3
...
1 : Find the dimensional formula of work
...
It is related to
joule as
1kWh = 3
...
Example 6
...
Calculate the work done in moving the object by 2m in the horizontal direction
...
(6
...
3 : A person lifts 5 kg potatoes from the ground floor to a height of 4m to
bring it to first floor
...
Solution : Since the potatoes are lifted, work is being done against gravity
...
8 ms–²
= 49 N
Work done = 49 × 4 (Nm)
= 196 J
6
...
1 Positive and Negative Work
As you have seen, work done is defined by Eqn
...
2), where the angle θ between the
force and the displacement is also important
...
Consider the examples given below:
Fig
...
2 (a) shows a car moving in + x direction and a force F is applied in the same
direction
...
The force and the displacement both are
in the same direction, i
...
θ = 0º
...
4)
The work is this case is positive
...
6
...
a) A force F is applied in the direction of the
moving car
...
b) A force F is applied in opposite direction so that the
car comes to rest after some distance
...
2 (b) shows the same car moving in the +x direction, but the force F is applied in
the opposite direction to stop the car
...
Therefore,
W = Fd cos 180º
...
5)
Hence, the work done by the force is negative
...
From the above examples, we can conclude that
a) When we press the accelerator of the car, the force is in the direction of motion of the
car
...
The work done is positive
...
The car loses speed and may finally come to rest
...
c) In case the applied force and displacement are as right angles, i
...
θ = 90º, no work is
said to be done
...
1
...
6
...
6
...
The weight of the object is mg in both cases
...
In Fig
...
3 (a), the work is done against the force mg (downwards) and the displacement
is upward (θ = 180º)
...
123
123
MODULE - 1
Physics
Motion, Force and Energy
mg
Notes
h
h
F
mg
(b)
(a)
Fig 6
...
In the Fig
...
3(b), the mass is being lowered
...
Therefore, the work done
W = Fd cos 0º
= + mgh
(6
...
When the object is
lifted up, the work done by the gravitational force is negative but the work done by the
person lifting the object is positive
...
In both of these cases, it is assumed that the object is being moved without
acceleration
...
1
1
...
Calculate the work
done by this force on the particle
...
2
...
Work done by a force is
a) zero
b) negative
c) positive
...
124
A bag of grains of mass 2 kg
...
Work Energy and Power
MODULE - 1
Motion, Force and Energy
a) How much work is done by the lift force?
b) How much work is done by the force of gravity?
...
j
j
A force F = (2 ˆ + 3 ˆ ) N produces a displacements d = (– ˆ + 2 ˆ ) m
...
Notes
...
j
j
A force F = (5 ˆ +3 ˆ ) acts on a particle to give a displacement d = (3 ˆ + 4 ˆ ) m
i
i
a) Calculate the magnitude of displacement
b) Calculate the magnitude of force
...
6
...
This
may not always be true
...
Let us now consider a case in which the magnitude of force F(x)
changes with the position x of the object
...
Let us assume that the displacement is from xi to xf , where xi and xf are the initial
and final positions
...
In fact, ∆x is taken so small that the force F(x) can be
assumed to be constant over each such interval
...
7)
F
∆x
x→
xi
xf
x→
Fig 6
...
The
variation of force with distance is shown by the solid curve (arbitrary) and work done is
numerically equal to the shaded area
...
6
...
The total
work done by the force between xi and xf is the sum of all such areas (area of all strips
added together):
3
...
8)
Notes
The width of the strips can be made as small as possible so that the areas of all strips
added together are equal to the total area enclosed between xi and xf
...
9)
lim ∆x → 0
6
...
1 Work done by a Spring
A very simple example of a variable force is the force exerted by a spring
...
Fig
...
5(a) shows the equilibrium position of a light spring whose one end is attached to a
rigid wall and the other end is attached to a block of mass m
...
We take x-axis along the horizontal direction
...
The spring is now compressed (or elongated) by an external force F
...
This force Fs
keeps increasing with increasing x and becomes equal to F when the compression (or
elongation) is maximum at x = xm
...
Since the direction of Fs is always opposite to compression (or extension), it is
written as :
F = Fs= – kx
(6
...
6
...
a) The relaxed position of the spring’s, free end at x = 0; b) The
spring is compressed by applying external force F and c) Pulled or elongated by an
external force F
...
126
Work Energy and Power
Let us now calculate the work done and also examine, if it is positive or negative
...
Hence, the work done by the external force is
positive
...
e
...
The work done by the
spring force is negative
...
At x = 0, the
force Fs = 0
...
Since the variation of the force is linear with displacement, the average force during
Fs
⎛ 0 + Fs ⎞
...
displacement
=
Fs
...
Hence
W=
=
1
k x × xm
2 m
1
k x 2m
2
(6
...
It is shown in Fig
...
6
...
6
...
3
...
12)
This is the same as that obtained analytically in Eqn
...
11)
Activity 6
...
6
...
Now attach a block of mass m
to the lower end of the spring
...
Measure the extension
...
7 (b)
...
This is because
the spring force (restoring force) acting
upwards balances the weight mg of the block
in equilibrium state
...
6
...
F s = k
...
s
k =
mg
s
(6
...
4: A mass of 2 kg is attached to a light spring of force constant
k =100 Nm–1
...
Solution:
m = 2 kg
W=
=
1
kx²
2
1
× 100 × (0
...
6
...
= 50 × 0
...
5 J
As explained earlier, the work doen by the restioning force in the spring = – 0
...
128
Work Energy and Power
MODULE - 1
Motion, Force and Energy
Intext Questions 6
...
Define spring constant
...
...
A force of 10 N extends a spring by 1cm
...
6
...
In such calculations, we did
not consider whether the work is done in one second or in one hour
...
For example, a man
may take several hours to load a truck with cement bags, whereas a machine may do this
work in much less time
...
The rate at which work is done is called power
...
14)
If the rate of doing work is not constant, this rate may vary
...
15)
∆W
∆t
= joule/ second = watt
Thus, the SI unit of power is watt
...
The power of an agent doing work is 1W, if one joule of work is done by it in one second
...
1 kW = 103 W,
and
1 MW= 106W
3
...
This paved the way
for industrial revolution
...
SI unit of power
watt is named in his honour
...
Example 6
...
Solution : Since
P=
work
time
= Force ×
Distance
Time
∴ Dimension of P = [Mass] × [Acceleration] ×
⎡ L⎤
[Distance]
[Time]
⎡L⎤
= [M] × ⎢ 2 ⎥ × ⎢ ⎥
⎣T⎦
⎣T ⎦
–3
= [ML²T ]
You may have heard electricians discussing the power of a machine in terms of the horse
power, abbreviated as hp
...
It is a larger unit:
1hp = 746 W
(6
...
One such unit of work is
kilowatt hour
...
kilowatt
...
hour
= 10³ W
...
6× 106 J
=
Or
1 kWh = 3
...
17)
The electrical energy that is consumed in homes is measured in kilowatt-hour
...
Intext Questions 6
...
A body of mass 100 kg is lifted through a distance of 8 m in 10s
...
...
MODULE - 1
Motion, Force and Energy
Convert 10 horse power into kilo watt
...
6
...
If a system (object) has energy, it
has ability to do work
...
It can push an object which comes on its way to some distance
...
All moving objects possess energy because they can do work before
they come to rest
...
Kinetic energy is the
energy of an object because of its motion
...
This force produces a uniform
acceleration a such that F = ma
...
This speed
becomes v2 at another instant of time t2
...
Using Equations of Motion, we can write
2
v 2 = v12 + 2as
or
a =
2
v 2 – v12
2s
(6
...
19)
1 2
1
mv 2 and K1 = mv12 respectively denote the final and initial kinetic energies
...
Kinetic Energy is a scalar quantity
...
131
131
MODULE - 1
Motion, Force and Energy
Physics
the speed
...
It is the total value
1
m v 2 that determines the kinetic energy
...
6 : A body of mass 10 kg is initially moving with a speed of 4
...
A
force of 30 N is now applied on the body for 2 seconds
...
Work-Energy Theorem
The work-energy theorem states that the work done by the resultant of all forces
acting on a body is equal to the change in kinetic energy of the body
...
4
1
...
2-
What happens to the kinetic energy of a particle if
a) The speed v of the particle is made 2v
...
3-
A particle moving with a kinetic energy 3
...
Calculate the maximum compression of the spring
...
4-
A car of mass 1000 kg is moving at a speed of 90 kmh–1
...
What is the average
force applied by brakes? If the car stops in 25s after braking , calculate the
average power of the brakes?
...
If an external force does 375 J of work in compressing a spring, how much work is
done by the spring itself?
...
6 Potential Energy
In the previous section we have discussed that a moving object has kinetic energy associated
3
...
Objects possess another kind of energy due to their position in space
...
Familiar example is the Gravitational Potential Energy
possessed by a body in Gravitational Field
...
6
...
1 Potential Energy in Gravitational Field
Suppose that a person lifts a mass m from a given height
h1 to a height h2 above the earth’s surface
...
The mass has been displaced by a
distance h = (h2 – h1 ) against the force of gravity
...
Therefore, the work done by the person is
Notes
W = force × distance
= mgh
Fig
...
9: Object of mass m
originally at height h1
above the earth’s surface is moved to a
height h2
...
20)
The work is positive and is stored in mass m as energy
...
It has capacity to do
work
...
For example, it can lift
another mass if properly connected by a string, which is
passing over a pulley
...
The
important concept is the change in height, i
...
(h2 – h1)
...
Any point in space
can be chosen as a point of zero potential energy
...
Example 6
...
The total mass of the load and the
truck together is 100,000 kg
...
What average power must the engine produce to lift the material?
Solution :
W = mgh
= (100,000 kg) × (9
...
8× 7× 107 J
= 68
...
6 × 107 J
=
3600 s
134
Work Energy and Power
= 1
...
91× 105
= 2
...
746
Example 6
...
In a power station,
1000 × 10³ kg water falls through a height of 51 m in one second
...
E
...
8 ms–2) × (51 m)
...
8 × 51 × 106 J
= 500 × 106 J
Water loses all its potential energy
...
Therefore
W = Force × distance
= mg × h
= 1000 × 103 × (9
...
In practice,
there is the always some loss in machines
...
6
...
2 Potential energy of springs
You now know that an external force is required to compress or stretch a given spring
...
6
...
Let there be a spring of force constant k
...
From Eqn
...
11) we recall that work done by the
external force to compress the spring is given by
W=
1
kx²
2
3
...
When the spring is left free, it
bounces back and the elastic potential energy of the spring is converted into kinetic energy
of the mass m
...
6
...
Examples are Electrical Energy, Thermal Energy, Gravitational Energy, Chemical
Energy and Nuclear Energy etc
...
There is a very fundamental law about energy
...
It states, “ The total energy of an
isolated system always remains constant
...
It can
be converted from one form to other
...
In an isolated system, if there is any loss of energy of one form, there is a gain of an equal
amount of another form of energy
...
The
universe is also an isolated system as there is nothing beyond this
...
It is a law of great importance
...
h1
In a Thermal Power Station, the chemical energy of
coal is changed into electrical energy
...
In these machines, the
electrical energy changes into mechanical energy, light
energy or thermal energy
...
It applies to systems ranging
from big planets and stars to the smallest nuclear
particles
...
6
...
It
is then lowered to a height
h2 at point P
...
We now wish to test the validity of the law of
conservation of energy in case of mechanical energy,
which is of immediate interest
...
The work done is mgh,
which is stored in the object as potential energy
...
Let us calculate the energy of this object when it has
fallen through a distance h1
...
10)
...
When the object falls freely, it gets accelerated and gains in speed
...
21)
Work Energy and Power
MODULE - 1
Motion, Force and Energy
where u is the initial speed at the height h1, i
...
u = 0 and s = h1
...
E =
1
mv²
2
Notes
m
× 2gh1
2
= mgh1
=
(6
...
23)
This is same as the potential energy at the highest point
...
(b) Conservation of Mechanical Energy for a Mass Oscillating on a Spring
Fig
...
11 shows a spring whose one end is fixed to a rigid wall and the other end is
connected to a wooden block lying on a smooth horizontal table
...
A block of mass m moving with speed v along the line of
the spring collides with the spring at the free end, and compresses it by xm
...
It is
2
the kinetic energy of the mass
...
At the point of
maximum compression
...
The total energy now is
1 2
k xm and the kinetic energy of
2
1 2
k xm
...
24)
v
x
0
x0
0
xm
Fig
...
11 : A block of mass m moving with velocity v on a horizontal surface collides with the
spring
...
K
...
E (Before collision) = K
...
+ P
...
(After collision)
3
...
25)
i
...
, the total energy is conserved
...
On the contrary, it is
obtained by transformation of mass into energy
...
Example 6
...
5 kg slides down a smooth curved surface and falls
through a vertical height of 2
...
12)
...
Solution :
i) Potential energy at A = mgh
= (0
...
8) × 2
...
9 × 2
...
25 J
A
The kinetic energy at A = 0 and
Total Energy = 12
...
2
...
E
...
E
...
25 J
1
mv²
2
Fig
...
12 : A block slides on a curved
surface
...
E
...
E
...
1
The total energy (P
...
+ K
...
) at B =
∴
2
mv² = 12
...
25 × 2
0
...
25 × 4
v² = 49
...
0 ms–1
Note: This can also be obtained from the equations of motion:
138
Work Energy and Power
MODULE - 1
Motion, Force and Energy
2
v² = v 0 + 2gx
= 0 + 2 × 9
...
5
v² = 49
v = 7 ms–1
Notes
6
...
4 Conservative and dissipative (Non conservative) Forces
(a) Conservative forces
We have seen that the work done by the gravitational force acting on an object depends on
the product of the weight of the object and its vertical
B
1
displacement
...
13), the work done by
gravity depends on the vertical separation between
the two points
...
When a force obeys this
A
2
rule, it is called a conservative force
...
A conservative force has a property that the work
done by a conservative force is independent of
path
...
13 (a)
B
1
WAB (along 1) = WAB (along 2)
Fig
...
13 (b) shows the same two positions of the
object
...
By definition,
the work done by a conservative force along path 1 is
equal and opposite to the work done along the path 2
...
27)
A
2
(b)
Fig
...
13 : a) The object is moved
from A to B along two
different paths
...
This result brings out an important property of the conservative force in that the work
done by a conservative force on an object is zero when the object moves around a
closed path and returns back to its starting point
...
Fig
...
14 shows a
rough horizontal surface
...
After moving a certain distance along a straight line, the block stops at the point B
...
It has neither kinetic energy nor
2
3
...
It has lost all its energy
...
Work has been done against the frictional force or we can say
that force of friction has done negative work on the block
...
The block with the same kinetic energy E is now taken
from A to B through a longer path 2
...
It may stop much
before reaching B
...
Thus, it canbe said that the work done depends on the path
...
6
...
It starts with the same speed υ at A but now
moves along a different path 2
...
5
1
...
The length of the sides AB = 3m, BC = 4m
and AC=5m
...
What is the
change in potential energy of the block when
a) it is taken from A to B
b) from B to C
Fig
...
15
c) from C to A
d) How much work is done by gravitational force in moving the mass form B to C
(positive or Negative work)?
...
A ball of mass 0
...
Solve
the following questions by applying work-energy principle
...
140
C
Fig
...
16
Work Energy and Power
3
...
A block at the top of an inclined plane slides down
...
The mass of the block
is 2 kg
...
6 J
...
How much
is the magnitude of the frictional force?
MODULE - 1
Motion, Force and Energy
2m
B
The Figure shows two curves A and B between
energy E and displacement x of the bob of a simple
pendulum
...
E
...
6
...
5
...
6
...
The system is a closed system which implies that
no external force acts on it
...
When two bodies interact, it is termed as collision
...
Let us start with a collision of two balls and to make the analysis simpler, let there be a
“head-on” or “central collision”
...
The collisions are of two kinds :
(i) Perfectly Elastic Collision: If the forces of interaction between the two bodies are
conservative, the total kinetic energy is conserved i
...
the total kinetic energy before
collision is same as that after the collision
...
(ii) Perfectly Inelastic collision: When two colliding bodies stick together after the
collision and move as one single unit, it is termed as perfectly inelastic collision
...
You should remember that the momentum is conserved in all types of collisions
...
6
...
1 Elastic Collision (Head-on)
Let two balls A and B having masses mA and mB respectively collide “head-on”, as shown
in Fig
...
19)
...
3
...
6
...
28)
For conservation of kinetic energy
1
1
1
1
2
2
2
mA v Ai + mBi = mA v Af + mB v Bf
2
2
2
2
(6
...
(6
...
29)]
...
Therefore, we quote the results only
(vBf – vAf ) = – (vBi – vAi)
v Af =
(6
...
31)
(mB – mA ) v Bi
2mAv Ai
+
(mA + mB )
mA + mB
(6
...
CASE I : Suppose that the two balls colliding with each other are identical i
...
mA= mB=
m
...
(6
...
32) will drop out resulting in
v Af = vBi
and
(6
...
34)
That is, if two identical balls collide “head-on”, their velocities after collision get interchanged
...
ii) the velocity of B is same as that of A before collision
...
Then vAf = 0 and vBf = vAi
After collision, A comes to rest and B moves with the velocity of A before collision
...
Complete
loss of kinetic energy or partial loss of kinetic energy (mA # mB) by A is same as the gain
in the kinetic energy of B
...
Notes
CASE II : The second interesting case is that of collision of two particles of unequal
masses
...
From Eqns
...
31) and (6
...
The light particle returns back on
its path with a velocity equal to its the initial velocity
...
These results find applications in Physics of atoms, as for example in the case where an
α – particle hits a heavy nucleus such as uranium
...
6
1
Two hard balls collide when one of them is at rest
...
2
...
B and C are in contact and at
rest
...
After collision, what will be the velocities of A, B and
C separately? Explain
...
6
...
3
...
A is moving in + x direction
with speed 50ms–1 and B is moving in –x
direction with speed 40ms–1
...
–1
uA = 50 ms
–x
A
–1
uB = 40 ms
B
+x →
Fig
...
21
...
143
143
MODULE - 1
Motion, Force and Energy
Physics
4
...
The velocity if the bullet before collision is 90m/s
...
b) Calculate the kinetic energies before and after the collision?
c) Is it an elastic collision or inelastic collision?
Notes
d) How much energy is lost in collision?
...
In an elastic collision between two balls, does the kinetic energy of each ball change
after collision?
...
d = Fd cosθ
Where θ is the angle between F and d
...
Work is a scalar
quantity
...
Work done by elastic force obeying Hooke’s law is W =
1
kx² where k is force
2
constant of the elastic material (spring or wire)
...
x is compression or elongation of the spring
...
)
Power is the time rate of doing work
...
e
...
Kinetic energy of mass m moving with speed v is E =
1
mv²
...
2
The Work-Energy Theorem states that the work done by all forces is equal to the
change in the kinetic energy of the object
...
In other words the mechanical energy is conserved under conservative forces
...
MODULE - 1
Motion, Force and Energy
Work done by a conservative force does not depend on the path of the moving object
...
Work done is path dependent for a non-conservative force
...
Notes
The potential energy of a particle is the energy because of its position in space in a
conservative field
...
It has a value
1
kx², where k is spring constant and x is diplacement
...
It is called the
gravitational potential energy
...
The reference level of zero potential energy is arbitrary
...
The total energy always remains constant
...
The kinetic energy is also conserved in elastic collision while it is not conserved in
inelastic collision
...
If two particles have the same kinetic energy, are their momenta also same? Explain
...
A particle in motion collides with another one at rest
...
Does the total mechanical energy of a system remain constant when dissipative forces
work on the system?
4
...
(a) At what point is the kinetic energy maximum?
(b) At what point is the potenital energy maximum?
5
...
Calculate the maximum compression of the spring
...
What will be the compression of the spring in question 5 at the moment when kinetic
energy of the block is equal to twice the elastic potential energy of the spring?
7
...
Calculate the electrical energy consumed in 30
days if the bulb is lighted for 12 hours per day
...
1000kg of water falls every second from a height of 120m
...
Calculate the power of the generator assuming
no losses
...
145
145
MODULE - 1
Motion, Force and Energy
Physics
9
...
The driver applies brakes to
stop the car
...
Calculate the average power of the
brakes
...
A 400g ball moving with speed 5 m s–1 has elastic head-on collision with another ball
of mass 600g initially at rest
...
Notes
11
...
It hits a 20kg wooden
block at rest and gets embedded into the block
...
An object of mass 6kg
...
A horizontal force of 15N
is constantly applied on the object
...
(a) How much work does the applied force do?
(b) What is the kinetic energy of the block after 10 seconds?
(c) What is the magnitude and direction of the frictional force (if there is any)?
(d) How much energy is lost during motion?
13
...
Diameter BC = 50cm
...
Calculate it’s
(a) Potential energy at A relative to B
...
D
R E
2
(b) Speed at the point B (Lowest point)
...
The force constant of a spring is 400N/m
...
0cm (b) from x = 4
...
0 cm, where x = 0 is the
relaxed position of the spring
...
The mass of a car is 1000kg
...
0seconds
...
(b) The work done on the car by the engine
...
1
1
...
Hence no work is
done by the force
...
(a) Work done is zero (i) when there is no displacement of the object
...
When a mass moves on a horizontal plane the work done by gravitation force is
zero
...
MODULE - 1
Motion, Force and Energy
(c) When a particle moves in the direction of force, the work done by force is positive
...
(a) W = mgh = 2 × 9
...
F = (2 ˆ + 3 ˆ )
i
Notes
j
d = (– ˆ + 2 ˆ )
i
j
j
= (2 ˆ + 3 ˆ )
...
d
–2 + 6 = 4
j
5
...
83
(c) W = F
...
(3 ˆ + 4 ˆ )
i
i
= 15 + 12 = 27 J
6
...
Spring constant is defined as the restoring force per unit displacement
...
2
...
F = ⎜100 ⎟ (0
...
W =
1 100N ⎛ 5
5 ⎞ 2
1 2
×
×⎜
×
m
kx =
⎝ 100 100 ⎟
⎠
2
m
2
= 1
...
25 J
...
3
1
...
8 × 8)
=
J = 784W
...
10 H
...
46 kW
6
...
k
...
=
1
mv2
...
147
147
MODULE - 1
Physics
Motion, Force and Energy
(i) m is never negative
(ii) v2 is always positive
...
(a) K
...
E becomes 4 times and E becomes 4E
Notes
(b) When m becomes
E
m
, E becomes
2
2
P
...
of spring =
1 2
kx = 3
...
6 2 × 3
...
04m
180
k
3
...
2m = 20cm
...
v2 = u2 – 2as Final velocity is zero and initial velocity is
∴
25 × 25
u2
= a = 2 × 15 = 20
...
83= 20830N
...
Work done by external force = 375 J
Work done by spring = – 375 J
6
...
(a) O, no change in P
...
(b) Change in P
...
= mgh = 2 × 9
...
4 J
(c) Change in P
...
= 78
...
(d) – 78
...
2
...
E
...
5 × 9
...
6 J
K
...
at B =
v2 =
1
mv2 = 19
...
6 × 2
0
...
4 ⇒ v = 8
...
5 × 9
...
0 J (+ positive)
W = +49 J
3
...
E
...
8 × 1
= 19
...
E
...
6 J
Energy lost = 19
...
6 = 4J
This loss is due to frictional force
4J = F × d = F × 2
F = 2N
4
...
E
...
The P
...
is min at x = 0 and max at x = xm
...
E
...
5
...
6
...
(a) No, because, it will go against the low of conservation of linear momentum
...
2
...
3
...
149
149
MODULE - 1
Physics
= – 35ms–1
...
Thus ball A returns back with a velocity of 35 ms–1 and ball B moves on with a velocity
of 20 ms–1
...
(a) 1
...
(b) 81 J and 1
...
42 J
5
...
Answers to Terminal Problem
5
...
6
...
707 m
7
...
6 kW
8
...
2 mega watt
9
...
1
19
m s–1,
ms–1
4
6
11
...
25 m s–1
(b) 1249
...
(a) 1500 J
(b) 1200 J
(c) 3 N opposite to the direction of motion
(d) 300 J
5 m s–1
13
...
625 J
14
...
72 J
(b) 0
...
(a) 37
...
125 × 105 J
(c) 0
...
This simplification is quite useful for learning the laws of mechanics
...
A tiny pebble contains millions of particles
...
You will also study an important concept of physics, the angular momentum
...
This has very important
implications in physics
...
Objectives
After studying this lesson, you should be able to :
define the centre of mass of a rigid body;
explain why motion of a rigid body is a combination of translational and rotational
motions;
define moment of inertia and state theorems of parallel and perpendicular axes;
define torque and find the direction of rotation produced by it;
write the equation of motion of a rigid body;
state the principle of conservation of angular momentum; and
calculate the velocity acquired by a rigid body at the end of its motion on an
inclined plane
...
151
151
MODULE - 1
Motion, Force and Energy
Notes
Physics
7
...
In practice, when extended bodies interact with each other and the distances
between them are very large compared to their sizes, their sizes can be ignored and they
may be treated as point masses
...
So, stars can be considered as point masses
...
But when we have to consider the rotation of a body
about an axis, the size of the body becomes important
...
Such a system of particles is called a rigid body
...
This definition implies that the shape of a rigid body is preserved during its motion
...
Therefore, in nature there
is nothing like a perfectly rigid body
...
A cricket ball, a wooden block, a steel disc, even the earth
and the moon could be considered as rigid bodies in this lesson
...
You may now like to check what you have understood about a rigid body
...
1
1
...
The rods are firmly attached to each other
...
2
...
...
2 Centre of Mass (C
...
) of a Rigid Body
Before we deal with rigid bodies consisting of several particles, let us consider a simpler
case
...
Can we consider this system as a rigid body?
In this system, the distance between the two particles is fixed
...
Suppose that the two particles are at heights z1 and z2 from a horizontal surface (Fig
...
1)
...
The force on each particle will be mg
...
We have now to find a point C somewhere in the system so that
152
Motion of Rigid Body
if a force 2mg acts at that point located at a
height z from the horizontal surface, the motion
of the system would be the same as with two
forces
...
The potential
energy of the particle at C is 2mgz
...
7
...
1)
z1 + z2
2
(7
...
If the two masses were
unequal, this point would not have been in the middle
...
(7
...
3)
so that
m1 z1+ m2 z2
z = (m + m )
1
2
(7
...
As such, it is a mathematical
tool and there is no physical point as CM
...
Example 7
...
Solution : m1= m and m2= 2 m, Then Eqn
...
4) gives
z =
m z1 + 2 m z2
z1 + 2 z2
( m + 2 m) =
3
When a body consists of several particles, we generalise Eqn (7
...
7
...
x =
m1 + m2 +
...
5)
i
i =1
3
...
6)
M
x
N
z
Fig
...
2 : C
...
of a body
consisting of several
particles
and
z=
∑m
i
i =1
zi
(7
...
i=1
Why should we define CM so precisely?
Recall that the rate of change of displacement is velocity, and the rate of change of
velocity is acceleration
...
(7
...
(7
...
Similar equations can be
written for accelerations along y- and z-axes
...
(7
...
m1 a1 is therefore the sum of all forces
acting on particle 1
...
The right hand
side is, thus, the total force acting on the body
...
Some forces can be due to sources
outside the body
...
A familiar example is the
force of gravity
...
These are called internal forces
...
In the case of a rigid body, the sum of the internal forces is zero because they cancel each
other in pairs
...
In the light of this, we may write Eqn
...
9) as
(7
...
Note the simplification introduced in the derivation by defining the
centre of mass
...
The fact
154
Motion of Rigid Body
MODULE - 1
that the motion of the CM is determined by the external forces and that the internal forces
have no role in this at all leads to very interesting consequences
...
Can you recall what path is traced by a
projectile?
Y
The motion is along a parabolic path
...
The
explosion is caused by the internal forces
...
The centre of mass of
Path of
the projectile, therefore, continues to be the
CM
same parabola on which the bomb would have
moved if it had not exploded (Fig
...
3)
...
7
...
Notes
You might have now understood the importance of the concept of centre of mass of a
rigid body
...
Let
us therefore see how the centre of mass of a system is obtained by taking a simple
example
...
2 : Suppose four masses, 1
...
0 kg, 3
...
0 kg are located at the
corners of a square of side 1
...
Locate its
centre of mass?
Solution : We can always make the square
lie in a plane
...
Further, let us assume that one of the corners
coincides with the origin of the coordinate
system and the sides are along the x and y
axes
...
0,0), m3 (1
...
0) and m4 (0, 1
...
7
...
From Eqns
...
5) and (7
...
0)
(0, 0
...
7
(0
...
5, 0)
(1
...
0)
(1
...
7
...
0 × 10 + 3
...
0 × 0
...
...
0 + 3
...
0
...
5 m
and
y =
10 × 0 + 2
...
0 × 10 + 4
...
...
m
10 + 2
...
0 + 4
...
= 0
...
5 m, 0
...
7
...
Note that the CM
is not at the centre of the square although the square is a symmetrical figure
...
3
...
2
...
The fact that
all the particles of a rigid body have same mass and are uniformly distributed makes things
somewhat simpler
...
But even such calculations are
beyond the scope of this course
...
1 the position of CM of some regular, symmetrical bodies
...
(ii) When two bodies revolve around each other, they actually revolve
around their common centre of mass
...
1 Centres of Mass of some regular symmetrical bodies
Figure
Position of Centre of Mass
Triangular plate
Point of intersection of the three medians
Regular polygon and circular plate
At the geometrical centre of the figure
Cylinder and sphere
At the geometrical centre of the figure
Pyramid and cone
On line joining vertex with centre of base and at
h/4 of the height measured from the base
...
The Earth-Sun system also revolves around its common
centre of mass
...
MODULE - 1
Motion, Force and Energy
Now it is time to check your progress
...
2
1
...
0 kg,
2
...
0 kg, 4
...
0 kg
...
7
...
...
If three particles of masses m1 =1 kg, m2 =
2 kg, and m3 = 3 kg are situated at the
corners of an equilateral triangle of side
1
...
...
Notes
Show that the ratio of the distances of the
Fig
...
5
two particles from their common centre of
mass is inversely proportional to the ratio of their masses
...
7
...
7
...
Since all the particles execute identical
motion, its centre of mass must also be tracing out an identical path
...
We have seen that this motion is given by Eqn
...
10) :
M a = Fext
Do you now see the advantage of defining the centre of mass of a body? With its help, the
translational motion of body can be described by an equation for a single particle having
mass equal to the mass of the whole body
...
To understand
the concept clearly, perform the following activities
...
157
157
MODULE - 1
Physics
Motion, Force and Energy
Activity 7
...
Make two or three marks
on any of its surfaces
...
Note the paths traced by the
marks
...
7
...
You can easily see that the lengths
of the paths are also equal
...
7
...
Activity 7
...
Take a cylindrical piece of wood
...
Now roll the cylinder slowly
on the floor, keeping the plane face towards you
...
7
...
So, the body
has performed both translational and rotational
motion
...
7
...
The most convenient point to fix
for this purpose is the CM of the body
...
The
handle of the stone moves in a circular path
...
7
...
The motion of a rigid body in which all its constituent
particles describe concentric circular paths is known
as rotational motion
...
7
...
7
...
You are familiar with such
equations
...
The rotational motion can
be obtained by keeping a point of the
body fixed so that it cannot have any
translational motion
...
The rotation is then about an axis
passing through the CM
...
7
...
You have studied in earlier
lessons that the mass of the body plays a very
important role
...
Can we
define a similar quantity for rotational motion
also? Let us find out
...
3
...
7
...
Suppose it rotates about an axis through this point
(Fig
...
10)
...
are located at distances r1, r2, r3
...
Then particle 1 has kinetic
2
2
energy (½) m1 v1
...
By
adding the kinetic energies of all the particles, we get the total energy of the body
...
i =1
=
⎛1⎞
∑⎜ 2 ⎟ m v
⎝ ⎠
i
2
i
(7
...
You have studied in lesson 4 that angular speed (ω) is related to linear speed (v) through
the equation v = r ω
...
(7
...
12)
i =1
Note that we have not put the subscript i with w because all the particles of a rigid body
have the same angular speed
...
(7
...
13)
The quantity
3
...
14)
i
is called the moment of inertia of the body
...
3 : Four particles of mass m each are located at the corners of a square of
side L
...
Solution : Simple geometry tells us that the distance of each particle from the axis of
rotation is r =L 2
...
7
...
2
= 2 m L2
It is important to remember that moment of inertia is defined with reference to an axis of
rotation
...
In the present case, I is the moment of inertia about an axis passing
through the centre of the square and normal to the plane containing four perfect masses
...
7
...
The moment of inertia of a rigid body is often written as
I = M K2
(7
...
The radius of gyration is that distance from the axis of rotation where the whole
mass of the body can be assumed to be placed to get the same moment of inertia
which the body actually has
...
If the distribution
of mass changes, the moment of inertia will also change
...
3
...
Then the moment of inertia of the system about the axis through C and
perpendicular to the plane of square would be
2
2
2
2
I = m r + 2m r + m r + 2m r
= 6m r2
Note that moment of inertia has changed from 2mL2 to 3 mL2
...
2 Moments of inertia of a few regular and uniform bodies
...
(7
...
Can you draw any analogy? You will note that in rotational motion, the role
of mass has been taken over by the moment of inertia and the angular speed has replaced
the linear speed
...
Physical significance of moment of inertia
The physical significance of moment of inertia is that it performs the same role in
rotational motion that the mass does in linear motion
...
161
161
MODULE - 1
Motion, Force and Energy
Notes
Physics
Just as the mass of a body resists change in its state of linear motion, the moment of
inertia resists a change in its rotational motion
...
Most machines, which produce rotational motion
have as one of their components a disc which has a very large moment of inertia
...
The disc with a large
moment of inertia is called a flywheel
...
Because of its large
moment of inertia, the flywheel resists this attempt
...
Similarly, it works against the attempts to suddenly reduce the speed, and allows
only a gradual decrease in the speed
...
We have seen that in rotational motion, angular velocity is analogous to linear velocity in
linear motion
...
B
...
If it is rotating with a constant angular velocity ω,
as shown, then it will turn through an angle θ in t seconds such that
O
θ =ωt
θ
7
...
The following equations describe its rotational motion:
Fig
...
12 :Rotation of
7
...
about a
Similarly, we can write
P′
fixed nail
1
α t2
2
ω f 2 = ωi 2 + 2 α θ
θ = ωi t +
7
...
16(d)
where θ is angular dispalcement in t seconds
...
Example 7
...
7
...
It is initially at rest
...
By what angle would the line OP move
in 2 s if it had a uniform angular acceleration of 2
...
P
O
Fig
...
13 : Rotation of bicycle wheel
162
Motion of Rigid Body
MODULE - 1
Motion, Force and Energy
Solution : Angular displacement of line OP is given by
θ = ω0 t + (½) α t2
= 0 + (½) × (2
...
× 4 s2
= 5 rad
We have mentioned above that for rotational motion of a rigid body, its CM is kept fixed
...
But many a time, we
consider points other than the CM
...
But then the axis of rotation will pass through this fixed
point
...
The relation between the two moments of inertia
can be obtained using the theorems of moment of inertia
...
3
...
These are :
P
C
d
(i) the theorem of parallel axes, and
(ii) the theorem of perpendicular axes
...
(i) Theorem of parallel axes
Fig
...
14 : Parallel axes
through CM and
another point P
Suppose the given rigid body rotates about an axis passing through any point P other than
the centre of mass
...
Theorem of
parallel axis states that the moment of inertia about an axis parallel to the axis passing
through its centre of mass is equal to the moment of inertia about its centre of mass
plus the product of mass and square of the perpendicular distance between the
parallel axes
...
17)
where M is the mass of the body and d is the distance between the two axes (Fig
...
12)
...
(ii) Theorem of perpendicular axes
Let us choose three mutually perpendicular axes, two of which, say x and y are in the
plane of the body, and the third, the z–axis, is perpendicular to the plane (Fig
...
13)
...
3
...
7
...
7
...
(7
...
7
...
From Table 7
...
The theorem of perpendicular axes tells us
that this must be equal to the sum of the moments of inertia about two diameters which are
perpendicular to each other as well as to the central axis
...
This means that all the diameters are equivalent and any two perpendicular diameters
may be chosen
...
(7
...
Let us now take a point P on the rim
...
The distance between the two axes is obviously equal to R
...
It is given by
I tan = M R2 + M R2 = 2 M R2
...
2 have been computed using the
theorems of parallel and perpendicular axes
...
3
...
3
Have you ever noticed that it is easy to open the door by
applying force at a point far away from the hinges? What
O
happens if you try to open a door by applying force near the
hinges? Carry out this activity a few times
...
F
θ
Why is it so? Similarly, for turning a screw we use a spanner
B
A
with a long handle
...
7
...
164
Motion of Rigid Body
Suppose O is a fixed point in the body and it can rotate about an axis passing through this
point (Fig
...
17)
...
If
AB passes through the point O, the force F will not be able to rotate the body
...
The turning effect of a force is called torque
...
19)
MODULE - 1
Motion, Force and Energy
Notes
where s is the distance between the axis of rotation and the line along which the force is
applied
...
The torque
is actually a vector quantity
...
(7
...
20)
S
which gives both magnitude and direction of the torque
...
7
...
7
...
If we extend the thumb of the right hand at
right angles to the fingers and curl the fingers so as to point from r to F through the smaller
angle, the direction in which thumb points is the direction of τ
...
7
...
This corresponds to clockwise rotation of
the body
...
5 : Fig
...
19 shows a bicycle pedal
...
(i) What torque do you produce? (ii) Where should
your foot be for generating maximum torque?
F
B
F
(b)
(a
)
Fig
...
19 : A bicycle pedal (a) at the top when τ = 0; (b) when τ is maximum
Solution : (i) When your foot is at the top, the line of action of the force passes through
the centre of the pedal
...
(ii) To get maximum torque, sinθ must have its maximum value, that is θ must be 90º
...
3
...
7
...
Do you see any
correspondence between the role of
torque in the rotational motion and the role
of force in the linear motion? Consider two
forces of equal magnitude acting on a
body in opposite directions (Fig
...
20)
...
The two
torques on the body have magnitudes
τ 1 = (a + b) F
τ 2 = a F
...
Therefore,
the magnitude of the net turning effect on the body
is in the direction of the larger torque, which in
this case is τ1 :
τ = τ1 – τ2 = bF
(7
...
7
...
There is another useful expression for torque which clarifies its correspondence with
force in linear motion
...
7
...
Obviously, a particle like P is rotating about the axis in a circle of radius r
...
The radial force is the centripetal force m ω2 r, which
keeps the particle in the circular path
...
Its magnitude
is m a, where a is the tangential acceleration
...
Do you know why? The tangential force produces a torque of magnitude m
a r
...
If we consider all the particles of the body, we can write
i=n
⎛
2⎞
τ = ∑ mi ri2 α = ⎜ ∑ mi r1 ⎟ α
i=0
⎝ i
⎠
= I α
...
22)
Motion of Rigid Body
MODULE - 1
Motion, Force and Energy
because α is same for all the particles at a given instant
...
A list of corresponding quantities in rotational
motion and linear motion is given in Table 7
...
With the help of this table, you can write any
equation for rotational motion if you know its corresponding equation in linear motion
...
3 : Corresponding quantities in rotational and translational motions
Translational Motion
Notes
Rotation about a Fixed Axis
Angular displacement
θ
Angular velocity
ω=
dθ
dt
Angular acceleration
α=
dω
dt
M
F=ma
Moment of inertia
Torque
I
τ=Iα
Work
W=
Work
W=
Kinetic energy
½M v2
Kinetic energy
(½) I ω2
Power
P = Fv
Power
P=τω
Linear momentum
Mv
Angular momentum
Iω
Displacement
x
Velocity
v =
Acceleration
a=
Mass
Force
dx
dt
dv
dt
z
F dx
z
τ dθ
With the help of Eqn
...
22) we can calculate the angular
acceleration produced in a body by a given torque
...
6 : A uniform disc of mass 1
...
1m can rotate about an axis passing through its centre and
normal to its plane without friction
...
1 kg hangs at its end
(Fig
...
22)
...
Take
g = 10 ms–2
Solution : (i) If R and M denote the radius and mass of the
disc, from Table 7
...
If F denotes the magnitude of force (=
m g) due to the mass at the end of the string then τ = F R
...
(7
...
1kg
Fig
...
22
α = τ /I = FR/I = 2F/MR
2
2 × (0
...
(1
...
1 m)
3
...
(7
...
Since the initial angular
velocity is zero, we have
θ = (½) × 20 × 1
...
0 = 20 rad s–1
Notes
Now, you may like to check your progress
...
Intext Questions 7
...
1
...
Calculate the moment of inertia about an axis
Axis
along passing through one of the corners and
the
perpendicular to the plane of the square
...
Verify your result by using
the theorem of perpendicular axes
...
m
Calculate the radius of gyration of a solid sphere if the axis is a tangent to the sphere
...
2)
...
4 Angular Momentum
From Table 7
...
To understand its physical significance, we would like you to do an activity
...
4
If you can get hold of a stool which can rotate without much friction, you can perform an
interesting experiment
...
Make the
stool rotate fast
...
Ask your friend to stretch her arms and
measure the speed again
...
Let us try to understand why we expect a change in the speed of rotation of the stool in
two cases : sitting with folded and stretched hands
...
All the points of
the body describe circular paths about the axis of rotation with the centres of the paths on
the axis and have angular velocity ω
...
7
...
Its linear velocity is vi = riω and its momentum
is therefore mi ri ω
...
If we sum this product for
all the particles of the body, we get
L = ∑ mi ω ri ri =
i
F mrI
GH Σ JK ω
vi
P
Notes
ri
O
i
i
=Iω
Motion, Force and Energy
z
2
i
MODULE - 1
(7
...
7
...
Like the linear momentum, the angular
momentum is also a vector quantity
...
(7
...
It is important to remember
that I must refer to the same axis
...
Therefore, the rate of change
of angular momentum is equal to torque
...
24)
7
...
1 Conservation of angular momentum
Eqn
...
24) shows that if there is no net torque acting on the body,
dL
= 0
...
e
...
This is the principle of conservation of angular momentum
...
The principle of conservation of angular momentum allows us to answer questions such
as : How the direction of toy umbrella floating in air remains fixed? The trick is to make it
rotate and thereby impart it some angular momentum
...
Its angular momentum is then constant
...
Thus, the direction of
the toy umbrella remains fixed while it is in air
...
When the arms
are stretched, she causes the moment of inertia of the system to increase
...
(7
...
Similarly, when she folds
her arms, the moment of inertia of the system decreases
...
Note that the change is basically caused by the change in the moment of
inertia due to change in distance of particles from the axis of rotation
...
169
169
MODULE - 1
Motion, Force and Energy
Physics
Let us look at a few more examples of conservation of angular momentum
...
The ball is set rotating by applying a torque
on it
...
When there is no external torque, whatever angular
momentum the ball has acquired must be conserved
...
2), its angular momentum is given by
Notes
L =
2
M R2 ω
5
(7
...
Imagine now that the radius of the ball somehow decreases
...
This
is what really happens to some stars, such as those which become pulsars (see Box on
page 176)
...
(7
...
If instead
of radius, the moment of inertia of the system changes some
how, ω will change again
...
7
...
Scientists have observed very small and irregular
variations in the period of rotation of the earth about its
axis, i
...
the length of the day
...
Due to changes in
weather, large scale movement in the air of the earth’s
atmosphere takes place
...
Since
the angular momentum of the earth L = I ω must be
conserved, a change in I means a change in rotational
speed of the earth, or in the length of the day
...
You must have seen divers
jumping off the diving boards during swimming events in national or international events
such as Asian Games, Olympics or National meets
...
When she is in
air, there is no torque acting on her and therefore her angular momentum must be conserved
...
7
...
If she unfolds her body, her moment of inertia increases and she must
rotate slowly
...
Example 7
...
She holds a 2
...
0 m from the axis of rotation of the
170
Motion of Rigid Body
system
...
Calculate a) the initial
angular velocity in rad s–1, b) the angular velocity after the objects are brought to a distance
of 0
...
(d)
If the kinetic energy has increased, what is the cause of this increase? (Assume that the
moment of inertia of Shiela and platform ISP stays constant at 1
...
)
Motion, Force and Energy
Notes
Solution : (a) 1 rotation = 2π radian
∴ initial angular velocity ω =
MODULE - 1
10 × 2π radian
= 1
...
The initial
moment of inertia
I i = I SP+ m ri 2 + m ri 2
= 1
...
0 kg) × (1 m2) + (2
...
0 kg m2
...
2 m, final moment of inertia
...
0 kg m2 + 2
...
2)2 m2 + 2
...
2)2 m2
= 1
...
Conservation of angular momentum requires that
I i ωi = I f ωf
or
I i ωi
ωf = I
f
(5
...
05 rad s –1
=
1
...
5 rad s–1
Suppose the change in kinetic energy of rotation is ∆E
...
16 kg m2 × (4
...
0 kg m2
2
2
× (1
...
05J
Since final kinetic energy is higher than the initial kinetic energy, there is an increase in the
kinetic energy of the system
...
This
work goes into the system and increases its kinetic energy
...
171
171
MODULE - 1
Physics
Motion, Force and Energy
Intext Questions 7
...
A hydrogen molecule consists of two identical atoms, each of mass m and separated
by a fixed distance d
...
Calculate the angular momentum of the molecule
...
2
...
0 kg and radius 20 cm is rotated about one of its
diameters at an angular speed of 10 rad s–1
...
...
A wheel is rotating at an angular speed ω about its axis which is kept vertical
...
These two wheels then rotate with a common speed
...
...
It is said that the earth was formed from a contracting gas cloud
...
What was then its
period of rotation on its own axis?
...
5 Simultaneous Rotational and Translational Motions
We have already noted that if a point in a rigid body is not fixed, it can possess rotational
motion as well as translational motion
...
Imagine
P
the motion of an automobile wheel on a plane horizontal
surface
...
7
...
Fix your attention at a point P and at the centre
C
C of the circular face
...
As it rolls, you would notice that point P
rotates round the point C
...
So the wheel has both the
rotational and translational motions
...
7
...
26)
Motion of Rigid Body
where M is the mass
...
27)
where I is the moment of inertia
...
An interesting case, where both translational
and rotational motion are involved, is the motion of a body on an inclined plane
...
8 : Suppose a rigid body has mass M, radius R and moment of inertia I
...
7
...
At the end of its journey, it has
acquired a linear speed v and an angular speed ω
...
Obtain the value of v in terms of h
...
7
...
Therefore,
(½) Mv
2
+ ½ I ω2 = M g h
(7
...
Inserting this
expression is Eqn
...
28), we get
v2
1
1
M v2 +
I 2 =Mgh
2
2
R
(7
...
Table 7
...
Eqn
...
29) then gives
1
1 M R2 v 2
M v2 +
=Mgh
2
2
R2
or
v = gh
(7
...
Its means that a hoop of any material and any
radius rolls down with the same speed on the inclined plane
...
173
173
MODULE - 1
Physics
Motion, Force and Energy
Intext Questions 7
...
A solid sphere rolls down a slope without slipping
...
Notes
2
...
What fraction of its
kinetic energy is translational? What is the magnitude of its velocity after falling
through a height h?
...
A uniform sphere of mass 2 kg and radius 10cm is released from rest on an inclined
plane which makes an angle of 300 with the horizontal
...
...
These are called pulsars
...
The pulses are periodic and the periodicity is extremely
precise
...
Such
short time periods show that the stars are rotating very fast
...
(The neutrons and protons are the building
blocks of the atomic nuclei
...
These stars
represent the last stage in their life
...
The radius of a typical neutron star is only 10 km
...
The Sun rotates on its axis with a period of about
25 days
...
In order to conserve its angular momentum, the Sun will
have to rotate with a period as short as the fraction of a millisecond
...
The equation of translational motion far a rigid body may be written in the same form
as for a single particle in terms of the motion of its centre of mass
...
The moment of inertia about an axis of rotation is defined as
Σm
i
i
ri2
...
174
Motion of Rigid Body
The turning effect of a force F on a rigid body is given by the torque τ = r × F
...
The magnitude of turning effect
of torque is equal to the product of one of the forces and the perpendicular distance
between the line of action of forces
...
When no net torque acts on a body, the angular momentum of the body remains
constant
...
Terminal Exercise
1
...
Does this mean that the earth does not attract other particles?
2
...
In a molecule of carbon monoxide (CO), the nuclei of the two atoms are
1
...
Locate of the centre of mass of the molecule
...
A grinding wheel of mass 5
...
20 m is rotating with an angular
speed of 100 rad s–1
...
Through what distance would it
have to be dropped in free fall to acquire this kinetic energy? (Take g = 10
...
5
...
0 m is rotating about a fixed axis with an initial angular speed
of 2rev s–1
...
(a) Compute the angular velocity after 2 seconds
...
A wheel rotating at an angular speed of 20 rads–1 is brought to rest by a constant
torque in 4
...
If the moment of inertia of the wheel about the axis of rotation
is 0
...
7
...
The moment of inertia of wheel A is
5 × 10–2 kg m2, and that of wheel B is 0
...
Wheel A is set spinning at
600 rev min–1
...
A clutch now acts to join A and B so that
they must spin together
...
175
175
MODULE - 1
Motion, Force and Energy
Physics
8
...
Suggest a method to detect the hollow one
...
The moment of inertia of a wheel is 1000 kg m2
...
At some instant of time, its angular speed is 10 rad s–1
...
Calculate the torque applied to the wheel and the change in its kinetic
energy
...
A disc of radius 10 cm and mass 1kg is rotating about its own axis
...
During the first second it rotates through 2
...
Find the
angle rotated during the next second
...
1
1
...
2
...
Any disturbance can change the distance between sand particles
...
7
...
The coordinates of given five masses are A (–1, –1), B (–5, –1), C (6, 3), D (2, 6)
and E (–3, 0) and their masses are 1 kg, 2kg, 3kg, 4kg and 5kg respectively
...
-1× 1 – 5 × 2 + 6 × 3 + 2 × 4 – 3 × 5
=0
1+ 2 + 3+ 4 + 5
–1× 1 − 1 × 2 + 3 × 3 + 4 × 6 + 0 × 5
30
=
= 2
...
Consider axes to be as shown with 2 kg mass at the origin
...
5 + 3 × 1 3
...
5 m
x =
1+ 2 + 3
6
3
+ 3× 0
3
2
m
=
1+ 2 + 3
12
1 kg
2 × 0 + 1×
y =
2 kg
⎛ 3
...
176
X
3 kg
Let the two particles be along the x-axis and let their x-coordinates be o and x
...
The distance of m2 from CM is
m2 x
m1 x
x – X = x – m +m = m +m
1
2
1
2
∴
Notes
m2
X
= m
x+X
1
Thus, the distances from the CM are inversely proportional to their masses
...
3
1
...
I
...
Now,
according to the theorem of perpendicular axes, MI about SP (2mr2) + MI about QP
2 m r2 should be equal to MI about the axis through P and perpendicular to the plane
of the square (4 m r2)
...
2
...
I
...
5
5
If radius fo gyration is K, then M K2 =
Radius of gyration K = R
7
M R2
...
4
1
...
m d 2ω
2
Angular momentum about an axis of rotation (diameter)
...
177
177
MODULE - 1
Physics
Motion, Force and Energy
m r2
as M
...
(0
...
2 kg m2 s–1
...
I
...
⎛
2 m 2⎞
m r2 ω = ⎜ m r + r ⎟ ω1
⎝
2 ⎠
ω=
4
...
According to the
conservation of angular momentum
...
25 T
Thus, period of revolution of earth in the past T0 = 6
...
7
...
Using (I =
2
M R2), Eqn
...
29) for a solid sphere
5
1
1
mv2 + I ω2 = m g h
2
2
or,
v2
1
1
2
mv2 +
× m r2
...
It gives v =
2
...
E
Q ω = v /r
178
10
...
2 = m v2
2
2
2
2
4
R
2
Motion of Rigid Body
MODULE - 1
Motion, Force and Energy
1
m v2
2
2
Hence, fraction of translational K
...
=
=
3
3
m v2
4
Proceeding as in Q
...
At a distance 0
...
4
...
5 m
5
...
30 J
7
...
T = 5 × 104 N m, KE = 5 × 106 J
(b) 20 π rad
(b) Ei = 5 Ef
(c) 25 m s–1
(d) 1280 m s–2
(c) 49 π N m
10
...
5 rad, τ = 5 × 10–2 J
3
...
Do not send your assignment to NIOS
1
...
(1)
2
...
Can the law of conservation of linear momentum be applied for a body falling under gravity ? Explain
...
Why is the handle on a door provided at the largest possible distances form the hinges ?
(1)
5
...
Draw velocity time graph of a body moving in a straight line under a constant force
...
What is the radius of gyration of a disc of radius 20 cm, rotating about an axis passing through its center
and normal to its plane?
(2)
8
...
A vector A of magnitude 10 units and another vector B of magnitude 6 units make an angle of 600 with
each other
...
(1)
(1)
(2)
10
...
5 kg ball with a maximum speed of 10m s-1
...
The displacement of a particle is given by y = at +bt2, where a and b are constants and t is time
...
12
...
Calculate the speed of its tip
...
If by some freak of nature the earth collapses to 1/8th of its present volume, what would be the duration
of a day ?
...
(4)
14
...
You may take the distance between the sun and the earth as 1
...
A block of mass 2 kg is placed on plane surface
...
The
block is just at the verge of sliding when the inclination of the plans is 300, calculate the acceleration with
(4)
which the bock will slide down when the inclination of the plane is 450
...
A constant force of 20 N acts for 2s on a body of mass 2 kg initially at rest
...
Draw a load-extension graph for a spring
...
Two masses of 3 kg and 5 kg one attached to a massless string and the string is passed
over a frictionless pulley as shown in fig
...
(4)
19
...
Determine (i) The center of mass of the system
...
(5)
3kg
5kg
20
...
Find
the magnitude and direction of motion of each body after collision
...
181
181
MODULE - II
MECHANICS OF SOLIDS
AND FLUIDS
8
...
Properties of Fluids
MODULE - 2
Physics
Mechanics of Solids
and Fluids
8
Notes
ELASTIC PROPERTIES OF
SOLIDS
I n the previous lessons you have studied the effect of force on a body to produce
displacement
...
For
example, when a suitable force is applied on a spring, you will find that its shape as well as
size changes
...
Now apply a
force on some objects like wet modelling clay or molten wax
...
Thus some objects regain their original shape and size whereas others do not
...
The elastic property of materials is of vital importance in our daily life
...
We use this property to find the strength of beams for construction of
buildings and bridges
...
Objectives
After studying this lesson, you should be able to :
distinguish between three states of matter on the basis of molecular theory;
distinguish between elastic and plastic bodies;
distinguish between stress and pressure;
study stress-strain curve for an elastic solid ; and
define Young’s modulus, bulk modulus, modulus of rigidity and Poisson’s ratio
...
1 Molecular Theory of Matter : Inter-Molecular Forces
We know that matter is made up of atoms and molecules
...
The interaction forces between molecules
are known as inter-molecular forces
...
8
...
When the separation is
large, the force between
distance R two molecules is attractive
O
and weak
...
8
...
value and beyond this, the
force becomes repulsive
...
This separation is called
equilibrium separation
...
When R < R0 , a repulsive force will act between
them
...
Due to high intermolecular forces, they are almost fixed at their positions
...
In liquids, the average separation between the molecules is somewhat larger
( 10–8 m)
...
You can understand now why a liquid does not
have fixed shape
...
In gases, the intermolecular separation is significantly larger and the molecular force is
very weak (almost negligible)
...
That is why gases do not have fixed shape and size
...
He lived around
6th century B
...
He resided at Prabhasa (near Allahabad)
...
They
are eternal and indestructible
...
If two
atoms combine to form a molecule, it is called duyanuka and a triatomic molecule is
called triyanuka
...
The size of atom was also estimated
...
8
...
e
...
The extent of deformation depends on
3
...
When the
deforming forces are withdrawn,
the body tries to regain its original
shape and size
...
Fig 8
...
2), you
will observe that its shape changes
...
The property of matter to regain its original shape and size after removal of the deforming
forces is called elasticity
...
2
...
On the other hand, if it completely retains its modified form even
on removing the deforming force, i
...
shows no tendency to recover the deformation, it is
said to be perfectly plastic
...
There exists no perfectly elastic or perfectly plastic body in nature
...
Here it can be added that the object which opposes the deformation more
is more elastic
...
You might have seen
the processes such as stamping, bending and hammering of metal pieces
...
The phenomenon of elasticity can be explained in terms of inter-molecular forces
...
2
...
Each atom is acted upon by forces due to neighbouring atoms
...
When the
body is deformed, the atoms are displaced from their original positions and the interatomic distances change
...
e
...
However, if inter–atomic
separation decreases (i
...
R < R0), strong repulsive forces develop
...
The behaviour of atoms in a solid
can be compared to a system in which balls are connected with springs
...
184
Elastic Properties of Solids
MODULE - 2
Mechanics of Solids
and Fluids
8
...
3 Stress
When an external force or system of forces is applied on a body, it undergoes a change in
the shape or size according to nature of the forces
...
The internal restoring force opposes the deforming force
...
Notes
In equilibrium, the restoring force is equal in magnitude and opposite in direction to the
external deforming force
...
If the magnitude of deforming force is F and it
acts on area A, we can write
Stress =
Stress =
or
deforming force ( F )
restoring force
=
area ( A)
area
F
A
(8
...
The stress may be longitudinal, normal or shearing
...
(i)
Longitudinal Stress : If the deforming forces are along the length of the body, we
call the stress produced as longitudinal stress, as shown in its two forms in Fig 8
...
3 (b)
...
3 (a) : Tensile stress; (b) Compressive stress
(ii) Normal Stress : If the deforming forces are applied uniformly and normally all over
the surface of the body so that the change in its volume occurs without change in
shape (Fig
...
4), we call the stress produced as normal stress
...
Deforming
force per unit area normal to the surface is called pressure while restoring force developed
inside the body per unit area normal to the surface is known as stress
...
8
...
185
185
MODULE - 2
Mechanics of Solids
and Fluids
Physics
(iii) Shearing Stress : If the deforming forces act tangentially or parallel to the surface
(Fig 8
...
An example of shearing stress is shown in Fig 8
...
Its opposite face is held fixed by the force of friction
...
8
...
2
...
In general, the strain is
defined as the change in dimension (e
...
length, shape or volume) per unit
dimension of the body
...
Depending on the kind of stress applied, strains are of three types : (i) linear strain,(ii)
volume (bulk) strain, and (iii) shearing strain
...
8
...
8
...
7) without change of shape of the body, then
∆p
Volume strain =
change in volume
∆V
=
original volume
V
→
∆p
∆V
V
∆p
Fig
...
7: Volume strain
186
Elastic Properties of Solids
(i)
MODULE - 2
Shearing strain: When the deforming forces are tangential (Fig 8
...
(The angle θ is usually very small
...
8
...
2
...
8
...
Let us study the regions and points
Elastic limit
C E
B
A
F
Breaking Point
Plastic behaviour
Elastic behaviour
Permanent Set
O
D
Strain
X
Fig
...
9: Stress-strain curve for a steel wire
on this curve that are of particular importance
...
(ii) Elastic Limit : If we increase the strain a little beyond A, the stress is not linearly
proportional to strain
...
e
...
The maximum value of strain for
which a body(wire) shows elastic property is called elastic limit
...
(iii) Point C : When the wire is stretched beyond the limit B, the strain increases more
rapidly and the body becomes plastic
...
The material follows dotted line
CD on the graph on gradual reduction of load
...
After point E on the curve, no extension is recoverable
...
187
187
MODULE - 2
Mechanics of Solids
and Fluids
Notes
Physics
(iv) Breaking point F : Beyond point E, strain increases very rapidly and near point F,
the length of the wire increases continuously even without increasing of load
...
This is called the breaking point or fracture point and the
corresponding stress is known as breaking stress
...
Within the elastic limit, the maximum stress which an object can be subjected to is called
working stress and the ratio between working stress and breaking stress is called factor
of safety
...
K, it is taken 10, in USA it is 5
...
If large
deformation takes place between the elastic limit and the breaking point, the material is
called ductile
...
g
...
8
...
6 Stress-Strain Curve for Rubber
When we stretch a rubber cord to a few times its natural length, it returns to its original
length after removal of the forces
...
Substances having large strain are called elastomers
...
The stress-strain curve for rubber is
distinctly different from that of a metallic wire
...
8
...
Firstly, you can observe that there is no region of proportionality
...
The work done by the material in returning to
its original shape is less than the work done by the deforming force
...
(You can feel it by touching the
rubber band with your lips
...
Elastic hysteresis has an important application in shock absorbers
...
A body is said to be more elastic if on
applying a large deforming force on
it, the strain produced in the body is
small
...
But to produce same strain in the two
wires, significantly higher stress is
required in the steel wire than in rubber
wire
...
2
...
Steel is more Elastic than Rubber
0
2
4
6
8
Fig
...
10: Stress-strain curve for rubber
Strain
Elastic Properties of Solids
magnitude of internal restoring force produced in steel is higher than that in rubber
...
MODULE - 2
Mechanics of Solids
and Fluids
Example 8
...
0 m and cross
sectional area 0
...
The wire is stretched by 0
...
Calculate the (i) tensile stress,
and (ii) strain in the wire
...
80 ms–2
...
80 ms – 2 )
=
0
...
8 x 107 Nm–2
0
...
0 m
l
= 0
...
2 : Calculate the maximum length of a steel wire that can be suspended
without breaking under its own weight, if its breaking stress = 4
...
9 × 103 kg m–3 and g = 9
...
Therefore, the breaking stress
developed in the wire due to its own weight
W
= ρlg
...
0 x 108 Nm–2
...
0 ×108 Nm– 2
=
(7
...
8 ms – 2 )
= 0
...
Now it is time to take a break and check your understanding
Intext Questions 8
...
What will be the nature of inter-atomic forces when deforming force applied on an
object (i) increases, (ii) decreases the inter-atomic separation?
...
189
189
MODULE - 2
Mechanics of Solids
and Fluids
Physics
2
...
3
...
For large deformations what will be the changes in this ratio?
Notes
...
Under what conditions, a stress is known as breaking stress ?
...
If mass of 4 kg is attached to the end of a vertical wire of length 4 m with a diameter
0
...
60 mm
...
8
...
According to
this law: Within elastic limit, stress is directly proportional to corresponding strain
...
e
...
2)
This constant of proportionality E is a measure of elasticity of the substance and is called
modulus of elasticity
...
Its value is independent of the stress and strain
but depends on the nature of the material
...
Activity 8
...
8
...
steel
spring
Add 100 g load at a time on the bottom of the hanger in steps
...
Measure the extension
...
Plot a graph between load and extension
...
Repeat this activity with rubber and other materials
...
8
...
8
...
You would have seen that when some
object is placed on the pan, the length of the spring increases
...
e
...
Robert Hooke
(1635 – 1703)
MODULE - 2
Mechanics of Solids
and Fluids
Notes
Robert Hooke, experimental genius of seventeenth century, was a
contemporary of Sir Isaac Newton
...
Among
other accomplishments he has to his credit the invention of a universal joint, an early proto type of the respirator, the iris diaphragm, anchor escapement and balancing spring for clocks
...
He formulated Hooke’s law of eleasticity and correct
theory of combustion
...
8
...
1 Moduli of Elasticity
In previous sections, you have learnt that there are three kinds of strain
...
These
are Young’s modulus, Bulk Modulus and Modulus of rigidity corresponding to linear
strain, volume strain and shearing strain, respectively
...
(i)
Young’s Modulus: The ratio of the longitudinal stress to the longitudinal strain is
called Young’s modulus for the material of the body
...
Then
Table 8
...
Young’s modulus of
some typical materials
F
Longitudinal stress =
A
and
Name of
substance
∴
M gL
Y =
π r2 ∆ L
(8
...
Y (109Nm–2)
Aluminium
∆L
Longitudinal strain =
L
9
Polystyrene
3
The SI unit of Y in is N m–2
...
8
...
Note that steel is most elastic
...
191
191
MODULE - 2
Physics
Mechanics of Solids
and Fluids
of the material of the body
...
4)
The reciprocal of bulk modulus of a substance is called compressibility :
k =
1
1 ∆V
=
B V ∆P
(8
...
e
...
If a tangential force F acts on an area A and θ is the shearing strain, the modulus of rigidity
Shearing stress
F/A
F
=
η = Shearing strain =
θ
Αθ
(8
...
However, fluids do not have
Young’s modulus and shear modulus because a liquid can not sustain a tensile or shearing
stress
...
3 : Calculate the force required to increase the length of a wire of steel of
cross sectional area 0
...
Given Y = 2 × 1011 N m–2
...
If ∆L is the increase and L is the normal
length of wire then
∴
or
1
∆L
=
L
2
F×L
Y = A × ∆L
Y × A × ∆L
(2 × 1011 Nm – 2 ) (0
...
1 × 10 N = 106 N
Example 8
...
0012 %
...
Calculate the bulk modulus of rubber
...
6 × 106 Nm–2
Notes
0
...
2 × 10-5
100
V
Bulk Modulus B =
3
...
0 × 1011 Nm–2
1
...
3
...
12 : A stretched rubber
tube
...
8
...
(This is also true for a wire but may
not be easily visible
...
The strain perpendicular to the
applied force is called lateral strain
...
e
...
It is denoted by a Greek
letter σ (sigma)
...
If a wire (rod or tube) of length l and diameter d is elongated by applying a stretching
force by an amount ∆ l and its diameter decreases by ∆d, then longitudinal strain
α =
∆l
l
lateral strain
β =
∆d
d
and Possion’s ratio
σ =
∆ d/d
l ∆d
=
∆l/l
d ∆l
(8
...
The value of Poisson’s ratio depends only on the nature of material and for most of the
substances, it lies between 0
...
4
...
e
...
e
...
5
...
5
...
193
193
MODULE - 2
Physics
Mechanics of Solids
and Fluids
Activity 8
...
Make one wire to execute torsional vibrations for some time
...
Observe the rate of decay of
vibrations of the two wires
...
The wire gets tired or fatigued and finds it difficult to continue vibrating
...
Some other facts about elasticity :
1
...
For example, if carbon is added to iron or potassium is added to gold, their elasticity
increases
...
The increase in temperature decreases elasticity of materials
...
Similarly, plastic becomes highly elastic when cooled in liquid
air
...
The value of modulus of elasticity is independent of the magnitude of stress and
strain
...
Example 8
...
Calculate the modulus of rigidity, if top of the cube is displaced by
0
...
with respect to bottom
...
01 cm
Shearing strain = y = 20 cm
Hence,
= 0
...
= 0
...
0005
= 2 × 107 N m–2
Example 8
...
Calculate the extension and lateral strain, if Poisson’s ratio is 0
...
Given Young’s modulus of the wire = 11 × 1010 N m–2
...
05 × 10-3 m, y = 11 × 1010 Nm–2 F = 10 × 9
...
25
...
8ms – 2 ) × (5m)
F
...
14 (0
...
63 × 104
MODULE - 2
Mechanics of Solids
and Fluids
Notes
= 5
...
6 × 10 –3 m
=
5m
= 1
...
125 × 1
...
14 × 10–3
...
Intext Questions 8
...
Is the unit of longitudinal stress same as that of Young’s modulus of elasticity? Give
reason for your answer
...
2
...
Justify
...
The length of a wire is cut to half
...
4
...
The length of the first wire is half that of the
second and its diameter is double that of the second wire
...
5
...
Calculate Young’s modulus of material of the wire
...
3
...
4 Applications of Elastic Behaviour of Materials
Elastic behaviour of materials plays an important role in
our day to day life
...
A uniform beam clamped at one end
and loaded at the other is called a Cantilever [Fig
...
The hanging bridge of Laxman Jhula in Rishkesh and
Vidyasagar Sethu in Kolkata are supported on cantilevers
...
ii)
...
Therefore, by increasing
thickness, we can decrease depression under the same load more effectively
...
iii), the sag in the middle is given by
δ=
(ii)
M g l3
4 b d3γ
(iii)
Thus for a given load, we will select a material with a large Young’s
modulus Y and again a large thickness to keep δ small
...
To avoid this, a large load
bearing surface is provided
...
In cranes, we use a thick metal rope to lift and move heavy loads from
one place to another
...
A single wire of this radius will
practically be a rigid rod
...
This provides ease in
manufacturing, flexibility and strength
...
Cantilever
196
Elastic Properties of Solids
What You Have Learnt
MODULE - 2
Mechanics of Solids
and Fluids
A force which causes deformation in a body is called deforming force
...
The property of matter to restore its original shape and size after withdrawal of
deforming force is called elasticity
...
If a body completely retains its modified form after withdrawal of deforming force, it
is said to be perfectly plastic
...
Its units is Nm–2
The strain equals the change in dimension (e
...
length, volum or shape) per unit
dimension
...
In normal state, the net inter-atomic force on an atom is zero
...
However, for smaller separation, these forces become repulsive
...
A body beyond the elastic limit behaves like a plastic body
...
Young’s modulus is the ratio of longitudinal stress to longitudinal strain
...
Modulus of rigidity is the ratio of the shearing stress to shearing strain
...
Terminal Questions
1
...
Give examples of elastic and plastic objects
...
Explain the terms stress, strain and Hooke’s Law
...
Explain elastic properties of matter on the basis of inter-molecular forces
...
Define Young’s modulus, Bulk modulus and modulus of rigidity
...
Discuss the behaviour of a metallic wire under increasing load with the help of stressstrain graph
...
Why steel is more elastic than rubber
...
Why poission’s ratio has no units
...
In the three states of matter i
...
, solid, liquid and gas, which is more elastic and why?
3
...
A metallic wire 4m in length and 1mm in diameter is stretched by putting a mass 4kg
...
Given that the Young’s modulus of elasticity for
the material of the wire is 13
...
10
...
02% when taken to the bottom of sea 1km deep
...
You make take density of
sea water as 1000 kgm–3 and g = 9
...
11
...
2% in the length of a metallic wire
of radius 0
...
Given Y = 9 × 1010 N m–2
...
What are shearing stress, shearing strain and modulus of rigidity?
13
...
Find out the strain?
14
...
How does the plasticity helps us?
15
...
When the other end of
wire is pulled by a force F, its length increases by x
...
Answers to Intext Questions
8
...
If R > R0 , the nature of force is attractive and if (ii) R < R0 it is repulsive
...
Longitudinal stress and linear strain
...
The ratio will decrease
...
The stress corresponding to breaking point is known as breaking stress
...
0
...
8
...
Both have same units since strain has no unit?
2
...
Therefore solids are more elastic than liquid and gases
...
Half
...
1:8
5
...
Answers To Terminal Problems
9
...
11
...
15
...
15 m
...
9 × 10–10 N m–2
22
...
Properties of Fluids
MODULE - 2
Mechanics of Solids
and Fluids
9
PROPERTIES OF FLUIDS
Notes
In the previous lesson, you have learnt that interatomic forces in solids are responsible for
determining the elastic properties of solids
...
Have
you ever visited the site of a dam on a river in your area / state/ region? If so, you would
have noticed that as we go deeper, the thickness of the walls increases
...
You can explain all these observations on the basis of properties of liquids like
hydrostatic pressure, Pascal’s law and surface tension
...
Have you experienced that you can walk faster on land than under water? If you pour
water and honey in separate funnels you will observe that water comes out more easily
than honey
...
You may have experienced that when the opening of soft plastic or rubber water pipe is
pressed, the stream of water falls at larger distance
...
You will learn about it in this lesson
...
;
explain surface tension and surface energy ;
derive an expression for the rise of water in a capillary tube;
3
...
9
...
If area is large, you will have to apply greater force
...
This effect of force
on unit area is called pressure
...
9
...
It shows the shape of the side wall of a dam
...
Do we use similar shape for the walls of our house
...
Do you know the basic physical characteristic which makes us to
introduce this change?
Fig
...
1 : The structure of side wall of a dam
From the previous lesson you may recall that solids develop
shearing stress when deformed by an external force, because
the magnitude of inter-atomic forces is very large
...
9
...
Also, the fluid exerts a force on the
container normal to its walls at all points
...
We denote it by P :
P=
Thrust
area
(9
...
9
...
9
...
1 Hydrostatic Pressure at a point in
side a liquid
A
MODULE - 2
Mechanics of Solids
and Fluids
P2
h
Consider a liquid in a container and an imaginary right
circular cylinder of cross sectional area A and height h, as
P1
shown in Fig
...
3
...
Therefore, the upward force exerted by the
Fig
...
3 : An imaginary
liquid on the bottom of the cylinder is P1A and the downward
cylinder of height h
force on the top of the cylinder is P2 A
...
∴ The net force in upward direction is (P1A – P2A)
...
A
...
∴ Weight of the liquid in the cylinder = ρ
...
h
...
e
...
2)
So, the pressure P at the bottom of a column of liquid of height h is given by
P = ρgh
That is, hydrostatic pressure due to a fluid increases linearly with depth
...
If we consider the upper face of the cylinder to be at the open surface of the liquid, as
shown in Fig
...
4), then P2 will have to be replaced by Patm (Atmospheric pressure)
...
9
...
3)
3
...
(9
...
5)
...
9
...
Example 9
...
What should be the thickness of the side wall at the bottom of a water dam of
depth 100 m
...
8 ms–2
...
8
= 9
...
8×105 Nm–2
...
8 × 10 5 Nm
–2
10 5
–2
t=
Nm
= 9
...
1
...
The pressure exerted by the atmosphere is known as the atmospheric pressure
...
V
...
He took two hollow hemispheres made of copper,
having diameter 20 inches and tightly joined them with each other
...
When air between them was exhausted with an air pump,
8 horses were required to pull the hemispheres apart
...
6 : Toricelli’s
Barometer
202
Toricelli used the formula for hydrostatic pressure to determine
the magnitude of atmospheric pressure
...
9
...
He observed that the column of
76 cm of mercury above the free surface remained filled in
the tube
...
Therefore,
Properties of Fluids
MODULE - 2
Mechanics of Solids
and Fluids
P atm = h ρ g = 0
...
8 Nm–2
= 1
...
01 × 105 Pa
9
...
It is
because of the difference in the upward forces exerted by these fluids on these object
...
The nature of buoyant force that acts on objects placed inside a fluid was discovered
by
...
It state that when an object is submerged partially or fully in a fluid, the
magnitude of the buoyant force on it is always equal to the weight of the fluid displaced
by the object
...
7
...
Fig
...
7:
(b) : A totally submerged
object of density less than
that of the fluid
experiences a net upward
force
...
Another example of buoyant force is provided by the motion
of hot air balloon shown in Fig
...
8
...
Floating objects
You must have observed a piece of wood floating on the
surface of water
...
However,
the displaced water exerts buoyant force which acts upwards
...
It means that a
floating body displaces the fluid equal to its own weight
...
9
...
203
203
MODULE - 2
Physics
Mechanics of Solids
and Fluids
Archimedes
( 287- 212 B
...
He is well known for
discovering the nature of buoyant forces acting on objects
...
It is an inclined
rotating coiled tube used originally to lift water from the hold
of ships
...
Notes
Once Archimedes was asked by king Hieron of his native city Syracuse to determine
whether his crown was made up of pure gold or alloyed with other metals without
damaging the crown
...
He was so excited about his
discovery that he ran undressed through the streets of city shouting “ Eureka, Eureka’’,
meaning I have found it
...
3 Pascal’s Law
While travelling by a bus, you must have observed that the driver stops the bus by applying
a little force on the brakes by his foot
...
Packing of cotton bales is also done with the help of hydraulic press which works on the
same principle
...
This law is also known as the law of transmission of liquid pressure
...
3
...
The basic arrangement is shown in Fig
...
9
...
On the other side, the piston of large area A2 is attached to a
platform where heavy load may be placed
...
Since the pressure
is same on both the sides, we have
F1
Heavy Load
A1
A2
F2
Fig
...
9 : Hydraulic lift
204
Fig
...
10 : Hydraulic jack
Properties of Fluids
MODULE - 2
Mechanics of Solids
and Fluids
force
F1
Pressure on the smaller piston, P =
=
A1
area
According to Pascal’s law, the same pressure is transmitted to the larger cylinder of area A2
...
4)
A1
2
It is clear from Eqn
...
4) that force F2 > F1 by an amount equal to the ratio (A2/A1)
F = pressure × area =
Notes
With slight modifications, the same arrangement is used in hydraulic press, hydraulic balance,
and hydraulic Jack, etc
...
10)
...
(C) Hydraulic Brakes
While traveling in a bus or a car, we see
how a driver applies a little force by his
foot on the brake paddle to stop the
vehicle
...
The wheels stop
rotating at the same time and the vehicle
comes to stop instantaneously
...
9
...
9
...
1
1
...
2
Calculate the pressure at the bottom of an ocean at a depth of 1500 m
...
024 × 103 kg m–3, atmospheric pressure=1
...
80 ms–2
...
3
...
205
205
MODULE - 2
Physics
Mechanics of Solids
and Fluids
hydraulic lift
...
05m2
balance or lift the elephant?
...
If a pointed needle is pressed against your skin, you are hurt but if the same force is
applied by a rod on your skin nothing may happen
...
Notes
5
...
1m2 of a big hydraulic lift
...
...
4 Surface Tension
It is common experience that in the absence of external forces, drops of liquid are always
spherical in shape
...
The water drops falling from a tap or shower are also spherical
...
But you can not make pure water bubbles with same case? All the above experiences are
due to a characteristic property of liquids, which we call surface tension
...
Activity 9
...
Prepare a soap solution
...
Add a small amount of glycerin to it
...
Take a narrow hard plastic or glass tube
...
4
...
5
...
6
...
To understand as to how surface tension arises, let us refresh our knowledge of
intermolecular forces
...
The intermolecular forces are of two types: cohesive and adhesive
...
It is
the force of adhesion which makes it possible for us to write on this paper
...
show strong adhesion
...
206
Properties of Fluids
MODULE - 2
Mechanics of Solids
and Fluids
Activity 9
...
1
...
Put a few drops of water on it
Notes
3
...
4
...
Glass sheet
Water drops
Fig
...
12 Water drops remain stuck to the glass sheet
The Adhesive forces between glass and water molecules keep the water drops sticking on
the glass sheet, as shown in Fig
...
12
...
4
...
In
Fig
...
13, molecules are shown at different heights
in a liquid
...
However, it is not the case for the molecules at the
surface
...
13 : Resultant force acting on P
Molecules S and R, which lie on the surface layer,
and Q is zero but molecules
experience a net resultant force downward because
R and S experience a net
vertically downward force
...
If we consider the molecules of liquid on the upper half of the
surface of the liquid or liquid-air interface, even then the molecules will experience a net
downward force because of less number of molecules of liquid
...
This means that surface layer possesses an additional
energy, which is termed as surface energy
...
Therefore,
the area of surface must be minimum
...
This produces a tension in the surface, called surface
tension
...
207
207
MODULE - 2
Mechanics of Solids
and Fluids
Notes
Physics
Surface tension is a property of the liquid surface due to which it has the tendency
to decrease its surface area
...
Let us now understand this physically
...
14
...
The surface tension of a liquid can be defined as the force
per unit length in the plane of liquid surface :
B
T = F/L
F
(9
...
14) and tangential to the liquid surface
...
F
A
Fig
...
14 : Direction of
surface tension Let us take a rectangular frame, as shown in Fig
...
15 having
a sliding wire on one of its arms
...
A soap film will be formed on the frame
surface
and have two surfaces
...
Let T be the surface tension of the soap solution and L be the length of the wire
...
9
...
Therefore, the total
force F on the wire = 2TL
...
To keep the wire in equilibrium we
will have to apply an external uniform force equal to F
...
9
...
Let us
denote it by A
...
By rearranging terms, we get the required expresion for
surface tension :
T = W/A
MODULE - 2
Mechanics of Solids
and Fluids
(9
...
We can also say that surface tension
is equal to the surface energy per unit area
...
A simple experiment described below demonstrates the property of surface tension of
liquid surfaces
...
3
Take a thin circular frame of wire and dip it in a soap solution
...
Now take a small circular loop of cotton thread and put it gently on the soap
film
...
9
...
Now take a
needle and touch its tip to the soap film inside the loop
...
16 (a) : A soap film with
closed loop of thread
A
T
Fig
...
16 (b) : The shape of the thread
without inner soap film
You will find that the loop of cotton thread takes a circular shape as shown in Fig 9
...
Initially there was soap film on both sides of the thread
...
When inner side was punctured by the needle,
the outside surface pulled the thread to bring it into the circular shape, so that it may
acquire minimum area
...
4
...
Have you seen mosquitoes sitting on water
surface? They do not sink in water due to surface tension
...
209
209
MODULE - 2
Mechanics of Solids
and Fluids
Physics
the mosquito touch the liquid surface, the surface becomes concave due to the weight of
the mosquito
...
Its vertical component acts upwards
...
17
...
9
...
9
...
If the surface is plane, i
...
θ = 900, the surface tension on the two sides tangential to the
surface balances and the resultant tangential force is zero [Fig
...
18 (a)]
...
(9
...
9
...
Thus, whenever the surface is curved, the surface tension gives rise to a pressure directed
towards the center of curvature of the surface
...
Therefore, there is always an excess pressure on
the concave side of the curved liquid surface [Fig
...
18 b)]
...
9
...
e
...
(The liquid area in contact with
air is called the surface of the liquid
...
Then
P = (Pi – P0)
where Pi and P0 are the inside and outside pressures of the drop, respectively (Fig 9
...
The work done on the drop for this increase in area is given
by
Notes
Fig
...
19 (a) : A spherical
drop
W = Extra surface energy = T∆A = T
...
7)
If the drop is in equilibrium, this extra surface energy is equal to the work done due to
expansion under the pressure difference or excess pressure P:
Work done = P ∆V = P
...
8)
On combining Eqns
...
7) and (9
...
4π r2 ∆r = T
...
9)
(ii) Air Bubble in water
p0
An air bubble also has a single surface, which is the inner surface
(Fig
...
19b)
...
10)
(iii) Soap bubble floating in air
The soap bubble has two surfaces of equal surface area (i
...
Fig
...
19 b : Air Bubble
the outer and inner), as shown in Fig
...
19(c)
...
11)
where T is suface tension of soap solution
...
Now you can understand why a little extra pressure
is needed to form a soap bubble
...
9
...
3: Calculate the difference of pressure between inside and outside of a
(i) spherical soap bubble in air, (ii) air bubble in water, and (iii) spherical drop of water,
each of radius 1 mm
...
2 × 10–2 Nm–1 and surface
tension of soap solution = 2
...
Solution:
(i) Excess pressure inside a soap bubble of radius r is
3
...
5 × 10 –2
Nm–1
1 × 10 –3 m
= 100 Nm–2
Notes
(ii) Excess pressure inside an air bubble in water
= 2T ′/ r
=
2 × 7
...
Water is used as cleaning agent
...
This is desirable for washing and cleaning since high surface tension of
pure water does not allow it to penetrate easily between the fibers of materials, where dirt
particles or oil molecules are held up
...
That is why
detergents are more effective than soap
...
Soap
molecules
water
Soap molecules with head attracted to water
Platter with particles of greasy dirt
Inert ends surround dirt and the platter dirt
can now be dislodged say by moving water
...
Detergent is added the inert waxy ends of its
molecules are attracted to boundary where
water meals dirt
...
20 : Detergent action
The addition of detergent, whose molecules attract water as well as oil, drastically reduces
the surface tension (T) of water-oil
...
e
...
This kind of process using
surface active detergents is important for not only cleaning the clothes but also in recovering
oil, mineral ores etc
...
If
you stick a tablet of camphor to the bottom of a wax-duck and float it on still water
surface, you will observe that it begins to move randomly after a minute or two
...
This creates a net difference of force of
surface tension which makes the duck to move
...
Therefore, answer the following
questions
...
2
1
...
2
...
...
Do solids also show the property of surface tension? Why?
...
Why does mercury collect into globules when poured on plane surface?
...
Which of the following has more excess pressure?
(i) An air bubble in water of radius 2 cm
...
Surface tension of soap solution is 25 × 10–3 Nm–1
...
5 Angle of Contact
You can observe that the free surface of a liquid kept in a container is curved
...
Similarly, when mercury is filled in a
glass jar, its surface become convex
...
To characterize it, we introduce the concept of angle of contact
...
3
...
9
...
The angle of contact is acute for concave
spherical meniscus, e
...
water with glass
and obtuse (or greater than 900) for
convex spherical meniscus e
...
water
in paraffin or mercury in glass tube
...
As
the liquid is present only in the lower
quadrant, the resultant cohesive force
Fig 9
...
9
...
Similarly due
(b) paraffin wax jar
to symmetry, the resultant adhesive
force Fa acts outwards at right angles to the walls of the container vessel
...
P
Fa
Fc sin θ
Fa
θ
Fc
Fc
Fc cos θ
Fc
Fa
Fa
Fc
F
F
(a)
Fc
(b)
(c)
Fig
...
22 : Different shapes of liquid meniscuses
CASE 1: If Fa > Fc sin θ, the net horizontal force is outward and the resultant of (Fa – Fc
sin θ) and Fc cos θ lies outside the wall
...
Obviously
such a surface at the boundary is concave spherical ( Since radius of a circle is perpendicular
to the circumference at every point
...
Case 2 : If Fa < Fc sin θ the resultant F of (Fc sin θ – Fa) acting horizontally and
Fc cos θ acting vertically down wards is in the lower quadrant acting into the liquid
...
This is true for the case of mercury filled in the glass tube
...
9
...
The ink rises
in the narrow air gaps in the blotting paper
...
Also water given to the fields rises in the innumerable capillaries
in the stems of plants and trees and reaches the branches and leaves
...
Thus, water trapped in the soil is taken up by the plants
...
Such an important phenomenon of the elevation or depression
of a liquid in an open tube of small cross- section (i
...
, capillary tube) is basically due to
surface tension and is known as capillary action
...
9
...
1 Rise of a Liquid in a Capillary Tube
Let us take a capillary tube dipped in a liquid, say water
...
9
...
This is essentially because the forces of adhesion
between glass and water are greater than cohesive forces
...
9
...
9
...
We
know that pressure just below the meniscus is less than the pressure just above it by 2T/R, i
...
P B = PA – 2T/R
(9
...
But pressure at A is equal to the pressure at D and is equal to the atmospheric pressure
3
...
And pressure at D is equal to pressure at C
...
But we know that the pressure at all points at the same level in a liquid must
be same
...
Thus liquid begins to rise in the capillary tube to a certain height h (Fig 9
...
Thereafter, water stops
rising
...
13)
where ρ is the density of the liquid and g is the acceleration due to gravity
...
9
...
13)
h p g = 2T/ r /cos θ
θ
Q
S
Fig
...
24 : Angle of contact
or
h = 2T cosθ / r ρ g
(9
...
e
...
Intext Questions 9
...
Does the value of angle of contact depend on the surface tension of the liquid?
...
The angle of contact for a solid and liquid is less than the 900
...
3
...
...
Calculate the radius of a capillary to have a rise of 3 cm when dipped in a vessel
containing water of surface tension 7
...
The density of water is
1000 kg m–3, angle of contact is zero, and g = 10 m s–2
...
5
...
216
Properties of Fluids
MODULE - 2
Mechanics of Solids
and Fluids
9
...
9
...
Next watch
the flow of two liquids (e
...
glycerin and water) through
identical pipes
...
Drop a steel ball
through each liquid
...
These observations indicate a characteristic
property of the liquid that determines their motion
...
Let us now learn how it
arises
...
9
...
7
...
Similarly, whenever a fluid flows, two adjacent layers of the fluid exert a tangential force
on each other; this force acts as a drag and opposes the relative motion between them
...
Fig
...
26 shows a liquid flowing through a tube
...
Other layers are in motion and have different velocities Let v be the velocity
of the layer at a distance x from the surface and v + dv be the velocity at a distance x +
dx
...
9
...
The
quantity dv/dx is called the velocity gradient
...
217
217
MODULE - 2
Mechanics of Solids
and Fluids
Physics
• velocity gradient (dv/dx) in a direction perpendicular to the flow of liquid : F α dv/dx
On combining these, we can write
F α A dv/dx
or
Notes
F = – η A (dv/dx)
(9
...
The negative
sign indicates that force is frictional in nature and opposes motion
...
In cgs system, the unit of viscosity is poise
...
1 : Viscosity of a
few typical fluids
Name T [0C]
of fluid
Viscosity η
(PR)
Water
20
1
...
3 × 10–3
blood
37
2
...
9 × 10–5
1 poise = 0
...
8 Types of Liquid Flow
Have you ever seen a river in floods? Is it similar to the flow of water in a city water
supply system? If not, how are the two different? To discover answer to such questions,
let as study the flow of liquids
...
8
...
If every particle
passing through a given point of the path
follows the same line of flow as that of
preceding particles, the flow is said to
be streamlined
...
9
...
In steady flow, the streamlines coincide with the line of
flow (Fig
...
27)
...
When the velocity of flow is less than the critical velocity of a given liquid flowing through
a tube, the motion is streamlined
...
e
...
Such a flow is called laminar
flow
...
Such a motion is said to be turbulent
...
8
...
Let A1 and A2
denote the areas of cross section of the tube where the
B
fluid is entering and leaving, as shown in Fig
...
28
...
9
...
So volume of the liquid entering per second= A1 × v1
...
Therefore, we get
A1 v1 ρ = A2 v2 ρ
or
A1 v1 = A2 v2
This expression is called equation of continuity
...
8
...
But when the velocity of flow exceeds the critical
velocity, the flow becomes turbulent
...
e
...
Experiments show that vc α η ; vc α
Hence, we can write
1
1
and vc α
...
η/ρ d
(9
...
It has no dimensions
...
The flow becomes unsteady
when R is between 1000 and 2000 and the flow becomes turbulent for R greater than 2000
...
1: The average speed of blood in the artery (d =2
...
Is the flow laminar or turbulent? Density of blood
1
...
0 × 10–2 poise
...
(9
...
On substituting
3
...
05g cm –3 )
R =
(4
...
9
...
This is known as Stokes’ law
...
It has been found experimentally that K = 6π
...
17)
Stokes’ Law can also be derived using the method of dimensions as follows:
According to Stokes, the viscous force depends on:
coefficient of viscosity (η) of the medium
radius of the spherical body (r)
velocity of the body (v)
Then
F α ηa rb vc
or
F = K ηa rb vc
where K is constant of proportionality
Taking dimensions on both the sides, we get
[MLT –2] = [ML–1T–1]a [L]b [LT–1]c
or
[MLT –2] = [Ma L–a+b+c T–a-c]
Comparing the exponents on both the sides and solving the equations we get a = b = c = 1
...
9
...
The forces acting on the body will be
B
F
v
W
viscous liquid
(i)
Weight of the body W acting downward
...
Fig
...
29 : Force acting on a sphere
(iii)
The buoyant force B acting upward
...
Then, the body falls with
a constant velocity known as terminal velocity
...
W = (4/3) π r3 ρg
B = (4/3) π r3 σg
and
The net force is zero when object attains terminal velocity
...
18)
9
...
2 Applications of Stokes’ Law
A
...
The soldier then descends with constant velocity and opens his parachute close
to the ground at a pre-calculated moment, so that he may land safely near his destination
...
Velocity of rain drops
When raindrops fall under gravity, their motion is opposed by the viscous drag in air
...
That is why rain drops reaching the earth do not have very high kinetic energy
...
2: Determine the radius of a drop of rain falling through air with terminal
velocity 0
...
Given η = 1
...
21 kg m–3, σ = 1
...
8 m s–2
...
221
221
MODULE - 2
Mechanics of Solids
and Fluids
Physics
2r 2 (ρ – σ) g
v0 =
9η
On rearranging terms, we can write
r =
Notes
=
9η v 0
2 (ρ – σ)g
9 × 1
...
12
m
2 (1000 – 1
...
8
= 10–5 m
Intext Questions 9
...
Differentiate between streamline flow and turbulent flow?
...
Can two streamlines cross each other in a flowing liquid?
...
Name the physical quantities on which critical velocity of a viscous liquid depends
...
4
...
01m if the coeflicient of
viscosity of air is 1
...
2 kg m–3
...
Take g = 10 m s–2
...
5
...
Daniel Bernoulli (1700-1782)
Daniel Bernoulli, a Swiss Physicist and mathematician was born
in a family of mathematicians on February 8, 1700
...
His famous work,
Hydrodyanamica was published in 1738
...
He is known as the founder of mathematical physics
...
222
Properties of Fluids
MODULE - 2
Mechanics of Solids
and Fluids
9
...
All these can be
understood on the basis of Bernoulli’s principle
...
Notes
9
...
1 Energy of a Flowing Fluid
Flowing fluids possess three types of energy
...
The third type of energy possessed by the fluid is pressure energy
...
The pressure energy can be taken as the product of pressure difference
and its volume
...
10
...
Three important
assumptions were made to develop this equation:
1
...
e
...
2
...
3
...
P2 A2
U2
P1 A1
U2
h2
U1
h2
Fig
...
30
We consider a tube of varying cross section shown in the Fig
...
30
...
3
...
That is, the sum of pressure energy, kinetic energy and potential energy of a fluid remains
constant in streamline motion
...
4
Notes
1
...
2
...
9
...
3
...
Watch the paper
...
Fig
...
31
9
...
3 Applications of Bernoulli’s Theorem
Bernoulli’s theorem finds many applications in our lives
...
A
...
The device is
inserted in the flow pipe, as shown in the Fig
...
32
P1
H1
Main
pipe
v
v
h
A2
A
P1
P2
h1
h2
v2 A2 A1 r1
B
v
Main
pipe
Venturimeter
Fig
...
32 : A Venturimeter
It consists of a manometer, whose two limbs are connected to a tube having two different
cross-sectional areas say A1 and A2 at A and B, respectively
...
Then applying Bernoulli’s theorem for the steady
flow of liquid through the venturimeter at A and B, we can write
Total Energy at A = Total Energy At B
1
mp1
1
mp2
2
2
m υ1 + mgh +
=
m υ2 + mgh +
2
d
2
d
On rearranging terms we can write,
2
⎤
v12 d ⎡⎛ v 2 ⎞
d
2
2
⎢⎜ ⎟ − 1⎥
( p 1 – p2 ) =
( v2 – v1 ) =
2 ⎣⎝ v1 ⎠
2
⎦
224
(9
...
E
...
This is called Venturi’s Principle
...
Therefore
A 1v 1 = A2v2
(9
...
e
...
Notes
Using this result in Eqn
...
19), we conclude that pressure is lesser at the narrow ends;
p1 – p2 =
v12 d
2
2
⎡ A1 ⎤
⎢ A2 − 1⎥
⎣ 2 ⎦
1 2
= dv1
2
⎡⎛ A ⎞ 2 ⎤
⎢⎜ 1 ⎟ − 1⎥
⎢⎝ A 2 ⎠
⎥
⎣
⎦
2( p1 – p2 )
⎛ A2 ⎞
d ⎜ 1 ⎟ –1
2
⎝ A2 ⎠
v1 =
(9
...
Thus
h since all other parameters are constant for a given
v1 = K
h;
where K is constant
...
Bernaulli’s principle has many applications in the design of many useful appliances like
atomizer, spray gun, Bunsen burner, carburetor, Aerofoil, etc
...
9
...
When the rubber bulb A is squeezed, air
blows through the tube B and comes out of the narrow orifice with larger velocity creating
a region of low pressure in its neighborhood
...
As the liquid reaches the
3
...
(ii) Spray gun : When the piston is moved in, it blows the air out of the narrow hole ‘O’
with large velocity creating a region of low pressure in its neighborhood
...
g
...
The liquid on reaching the end gets sprayed by out blown air from the
piston (Fig
...
34)
...
The air, therefore, rushed in through the side hole
A and gets mixed with the gas
...
9
...
(iv) Carburetor : The carburetor shown in Fig
...
36
...
The
energy is supplied by the explosion of this mixture inside the
cylinders of the engine
...
There is a decrease in the pressure on the side A
due to motion of the piston
...
This causes a low pressure
near the nozzle B (due to constriction, velocity of air sucked
is more near B) and, therefore, petrol comes out of the nozzle
B which gets mixed with the incoming
...
The mixture of
vaporized petrol and air forming the fuel then enters the
cylinder through the tube A
...
9
...
9
...
9
...
9
...
The nozzle has
therefore, to be opened and cleaned
...
The shape of the body
of the aeroplane is designed specially as shown in the Fig
...
37
...
Due to crowding of more
streamlines on the upper side, it becomes a region of more velocity and hence of
comparatively low pressure region than below it
...
226
Properties of Fluids
MODULE - 2
Mechanics of Solids
and Fluids
Notes
Fig
...
37 : Crowding of streamlines on the upper side
...
e
...
(a) Attracted disc paradox : When air is blown through a narrow tube handle into the
space between two cardboard sheets [Fig
...
38] placed one above the other and the upper
disc is lifted with the handle, the lower disc is attracted to stick to the upper disc and is
lifted with it
...
9
...
9
...
When the ball shifts to the lefts , then most of the
jet streams pass by its right side thereby creating a region of
high velocity and hence low pressure on its right side in
comparison to that on the left side and the ball is again pushed
back to the center of the jet stream
...
Fig
...
39 :Dancing Pring
9
...
Water from the tap is allowed to come out of the
narrow jet end of the tube A
...
Air is, therefore,
sucked from the vessel to be evacuated through the tube B; gets mixed with the steam of
water and goes out through the outlet
...
, the pressure of air in the
vessel is decreased to about 1 cm of mercury by such a pump
3
...
This is called swing of the ball
...
9
...
That when a ball is moved forward, the air occupies
the space left by the ball with a velocity v (say)
...
So the resultant velocity of air above the
ball becomes (v – u) and below the ball becomes (v + u)
...
Fig
...
40 : Filter Pump
v
u
Curved path
of the ball
u
v
Fig
...
41 : Swing of a cricket ball
Example 9
...
942)
...
5m?
Solution: Let B be the hole near the bottom
...
We can apply the Bernoulli’s
theorem to the points A and B for the streamline flow of
small mass m
...
9
...
At B, vB = v = ?, pB = p, hB = height of the hole above the ground
...
5m
and d = density of the water
...
8 × 2
...
5
1
...
2
...
3
What are the conditions necessary for the application of Bernoulli’s theorem to solve
the problems of flowing liquid?
...
Water flows along a horizontal pipe having non-uniform cross section
...
20m/s
...
50 m/s?
...
Why do bowlers in a cricket match shine only one side of the ball?
...
According to Pascal’s law, when pressure is applied to any part of an enclosed liquid,
it is transmitted undiminished to every point of the liquid as well as to the walls of the
container
...
The surface tension of a liquid may be defined as force per unit length acting on a
imaginary line drawn in the surface
...
Surface tension of any liquid is the property by virtue of which a liquid surface acts
like a stretched membrane
...
The liquid surface in a capillary tube is either concave or convex
...
229
229
MODULE - 2
Mechanics of Solids
and Fluids
Physics
due to surface tension
...
The property of a fluid by virtue of which it opposes the relative motion between its
adjacent layers is known as viscosity
...
e
...
Coefficient of viscosity of any liquid may be defined as the magnitude of tangential
backward viscus force acting between two successive layers of unit area in contact
with each other moving in a region of unit velocity gradient
...
Bernoulli’s theorem states that the total energy of an element of mass (m) of an
incompressible liquid moving steadily remains constant throughout the motion
...
Derive an expression for hyhostatic pressure due to a liquid column
...
State pascal’s law
...
230
Properties of Fluids
3
...
Find its dimensional formula
...
Describe an experiment to show that liquid surfaces behave like a stretched membrane
...
The hydrostatic pressure due to a liquid filled in a vessel at a depth 0
...
0 N m2
...
8 m
...
In a hydraulic lift, how much weight is needed to lift a heavy stone of mass 1000 kg?
Given the ratio of the areas of cross section of the two pistons is 5
...
Notes
7
...
If Fa is force of adhesion,
Fc is force of cohesion and θ = angle of contact, which of the following relations
should hold good?
(a) Fa > Fc sinθ; (b) Fa < Fc sinθ; (c) Fa cosθ = Fc; (d) Fa sinθ > Fc
8
...
What happens to
the temperature of the water drop? Why?
9
...
Calculate the approximate rise of a liquid of density 103 kg m–3 in a capillary tube of
length 0
...
2 × 10–3 m
...
27 × 10–2 N m–1
...
Why is it difficult to blow water bubbles in air while it is easier to blow soap bubble in
air?
12
...
13
...
What do you expect out of the following
observations?
(i) The air from smaller balloon will rush into the bigger balloon till whole of its air
flows into the later
...
What will be your answer if the balloons are replaced by two soap bubbles of different
sizes
...
Which process involves more pressure to blow a air bubble of radius 3 cm inside a
soap solution or a soap bubble in air? Why?
15
...
16
...
Derive the units and dimensional formula
of coefficient of viscosity
...
What is Reynold’s number? What is its significance? Define critical velocity on the
basis of Reynold’s number
...
State Bernoulli’s principle
...
3
...
Explain Why :
(i) A spinning tennis ball curves during the flight?
(ii) A ping pong ball keeps on dancing on a jet of water without falling on to either
side?
(iii) The velocity of flow increases when the aperture of water pipe is decreased by
squeezing its opening
...
(v) If mercury is poured on a flat glass plate; it breaks up into small spherical droplets
...
Calculate the terminal velocity of an air bubble with 0
...
15 kg m–1 s–1 and density 0
...
What will be the terminal
velocity of the same bubble while rising in water? For water η = 10–2 kg m–1 s–1
...
A pipe line 0
...
1
m
...
2 m pipe-line is 2 m s–1
...
22
...
5 gm cm–3 and 1
...
3 Poise
...
Water at 20ºC flows with a speed of 50 cm s–1 through a pipe of diameter of 3 mm
...
005 × 10–2 Poise; and
Density of water at 20ºC as = 1 g cm–3
...
Modern aeroplane design calls for a lift of about 1000 N m–2 of wing area
...
If the velocity of flow
past the lower wing surface is 100 m s–1, what is the required velocity over the upper
surface to give a desired lift of 1000 N m–2? The density of air is 1
...
25
...
If the pressure of
water equals 5 cm of mercury at a point where the velocity of flow is 28 cm s–1, then
what is the pressure at another point, where the velocity of flow is 70 cm s–1? [Tube
density of water 1 g cm–3]
...
1
1
...
232
Properties of Fluids
2
...
5 × 107 Pa
3
...
5
= 500 N m–2
...
05
5000
= 500 N m–2
...
4
...
w
50
=
, w = 5000 kg wt
...
1 10
5
...
2
1
...
2
...
3
...
4
...
5
...
7 N m–2
...
5 N m–2
...
3
1
...
2
...
3
...
The fall in the
level of mercury in capillary makes it difficult to enter
...
2 × 10 –2
4
...
8 × 10–6m
...
Due to capillary action
...
233
233
MODULE - 2
Mechanics of Solids
and Fluids
Physics
9
...
If every particle passing through a given point of path follows the same line of flow as
that of preceding particle the flow is stream lined, if its zig-zag, the flow is turbulent
...
No, otherwise the same flow will have two directions
...
Critical velocity depends upon the viscous nature of the liquid, the diameter of the tube
and density of the liquid
...
012 ms–1
5
...
9
...
High velocity of air creates low pressure on the upper part
...
Decreasing in the area creates large pressure
...
The fluid should be incompressible and non-viscous on (very less)
...
4
...
So that the stream lines with the two surfaces are different
...
Answers to the Terminal Exercises
5
...
67 N m–2
...
200 N, No
...
2
...
21
...
3 × 10–2 m3 s–1
...
7
...
19 m s–1
...
1500, Unsteady
...
2 cm of mercury
...
Do not send your assignment to NIOS
1
...
Which of the two
will serve as better shock absorber?
(1)
Stress
Stress
Strain
(a)
Strain
(b)
2
...
If
extension in A is twice the extension in B what is the ratio of the radii of A and B
...
Why are the walls of a dam made thicker at the base?
(1)
4
...
Why?
(1)
5
...
Which is more elastic iron or rubber?
(1)
3
...
Is Surface tension dependent on the area of the surface?
(1)
8
...
(1)
9
...
0012%
...
360 km, density of water is 1g cm–3 and acceleration due to gravity is 10 N kg–
1
...
[Ans : 3 × 1011N m–2]
(2)
10
...
section is subjected to an
increasing load
...
Explain why the detergents should have small angle of contact
...
A 40 kg girl, wearing high heel shoes, balances on a single heel which is circular and has a diameter
10 mm
...
(i) Why does a spinning cricket ball in air not follow a parabolic trajectory?
(ii) Discuss the magnus effect
...
State Bernoulli’s principle
...
It is in level flight with a speed
of 960 km h–1
...
Also
estimate the fractional increase in the speed of the air on the upper surface of the wing relative to the
lower surface
...
2 kg m–3
...
08 ⎪
2
⎪∆P' = A = 6
...
A smooth spherical body of density(ρ) and radius(r), falling freely in a highly viscous liquid of density σ
and coefficient of viscosity(η) with a velocity (v), state the law for the magnitude of the tangential
backward viscous force (F) acting on the body
...
(4)
16
...
Discuss the behaviour of molecules in a liquid and hence explain
surface energy
...
A soap bubble has two surfaces of equal surface area i
...
the outer and the inner but pressure inside is
different from the pressure outside
...
(4)
18
...
(4)
19
...
(5)
20
...
Show that pressure energy, kinetic energy,
and potential energy per unit Volume of a fluid remains constant in a stream line motion
...
Also discuss what would happen if the thin tube of uniform bore immersed in water is of
insufficient length
...
Kinetic Theory of Gases
11
...
Heat Transfer and Solar Energy
Kinetic Theory of Gases
MODULE - 3
Thermal Physics
10
KINETIC THEORY OF GASES
Notes
studied in
lessons, at standard temperature and pressure,
As you havethree states –the previous and gas
...
At room temperature, these atoms/
molecules have finite thermal energy
...
This state of matter is said to be the gaseous state
...
Under different conditions of temperature, pressure and volume, gases exhibit different
properties
...
In this lesson you will learn the kinetic theory of gases which is
based on certain simplifying assumptions
...
Why the gases
have two types of heat capacities will also be explained in this lesson
...
3
...
1 Kinetic Theory of Gases
You now know that matter is composed of very large number of atoms and molecules
...
Kinetic theory of gases attempts to relate the macroscopic or bulk properties
such as pressure, volume and temperature of an ideal gas with its microscopic properties
such as speed and mass of its individual molecules
...
(A gas whose molecules can be treated as point masses and there is no
intermolecular force between them is said to be ideal
...
10
...
1 Assumptions of Kinetic Theory of Gases
Clark Maxwell in 1860 showed that the observed properties of a gas can be explained on
the basis of certain assumptions about the nature of molecules, their motion and interaction
between them
...
We now state these
...
The intermolecular forces between them are negligible
...
These
collisions are perfectly elastic
...
(iv) Between collisions, molecules move in straight lines with uniform velocities
...
(vi) Distribution of molecules is uniform throughout the container
...
Moreover, since a molecule moving in space will have velocity components
along x, y and z–directions, in view of assumption (vi)it is enough for us to consider the
motion only along one dimension, say x-axis
...
10
...
Note that if there were
N (= 6 ×1026 molecules m–3), instead of considering 3N paths, the assumptions have reduced
the roblem to only one molecule in one dimension
...
Its x, y and z components are u, v and w, respectively
...
On striking the wall, this molecule will rebound in
the opposite direction with the same speed u, since the collision has been assumed to be
perfectly elastic (Assumption ii)
...
Hence, the change in momentum of a molecule is
mu – (–mu) = 2mu
If the molecule travels from face LMNO to the face ABCD with speed u along x–axis
and rebounds back without striking any other molecule on the way, it covers a distance 2l
in time 2l/u
...
238
Kinetic Theory of Gases
According to Newton’s second law of motion, the rate of change of momentum is equal to
y
the impressed force
...
10
...
Since there are N molecules of the gas, the
in a container
total rate of change of momentum or the total force
exerted on the wall ABCD due to the impact of all the
N molecules moving along x-axis with speeds, u1, u2,
...
+ u N )
l
We know that pressure is force per unit area
...
+ u N
l
P =
l2
)
m
2
2
u 2 + u 2 +
...
1)
–2
If u represents the mean value of the squares of all the speed components along x-axis,
we can write
2
2
2
2
–
u 2 = u 1 + u 2 + u 3 +
...
+ u N
Substituting this result in Eqn
...
1), we get
P =
Nmu 2
l3
(10
...
239
239
MODULE - 3
Thermal Physics
Physics
since u, v and w are components of c along the three orthogonal axes
...
e
...
This means that the mean value of u2, v2, w2 are
equal :
–
–
u 2 = v 2 = w2
–2
–
u2 = c
3
so that
Substituting this result in Eqn
...
2), we get
P
=
1 Nm –
c2
3 l3
But l 3 defines the volume V of the container or the volume of the gas
...
3)
Note that the left hand side has macroscopic properties i
...
pressure and volume and the
right hand side has only microscopic properties i
...
mass and mean square speed of the
molecules
...
3) can be re-written as
P =
If ρ =
mN
is the density of the gas, we can write
V
P =
or
1 Nm –
c2
3 V
1 –
ρ c2
3
3P
–
c2 = ρ
(10
...
(10
...
3a)
The following points about the above derivation should be noted:
(i)
240
From Eqn
...
4) it is clear that the shape of the container does not play any
Kinetic Theory of Gases
role in kinetic theory; only volume is of significance
...
A cube only simplifies our calculations
...
MODULE - 3
Thermal Physics
Notes
–
(iii) The mean square speed c 2 is not the same as the square of the mean speed
...
Suppose we have five molecules and their speeds are 1, 2, 3, 4, 5 units, respectively
...
On the other hand, the mean square speed is
55
12 + 22 + 32 + 42 + 52
=
= 11
5
5
Thus we see that mean square speed is not the same as square of mean speed
...
1 : Calculate the pressure exerted by 1022 molecules of oxygen, each of
mass 5 × 10–2 6 kg, in a hollow cube of side 10 cm where the average translational speed of
molecule is 500 m s–1
...
Time taken to make successive impacts on the same face is equal to the time spent in
travelling a distance of 2 × 10 cm or 2 × 10–1 m
...
25 × 10–19 N
4 × 10−4 s
The force on the side due to one third molecules
and
pressure =
f =
1
× 1
...
7 N
3
417 N
Force
=
100 × 10–4 m 2
Area
= 4
...
241
241
MODULE - 3
Physics
Thermal Physics
Intext Questions 10
...
(i) A gas fills a container of any size but a liquid does not
...
Why?
...
What is an ideal gas?
...
How is pressure related to density of molecules?
...
2 Kinetic Interpretation of Temperature
From Eqn
...
3) we recall that
P V=
1
–
m N c2
3
Also, for n moles of a gas, the equation of state is PV = n RT, where gas constant R is
equal to 8
...
On combining this result with the expression for pressure, we get
nRT =
1
–
m N c2
3
3
Multiplying both sides by 2n we have
3
1 Nmc 2
1
–
RT =
= m NA c 2
2
2 n
2
where
N
= NA is Avogadro’s number
...
Its value is 6
...
In terms of NA, we can
write
3 ⎛ R ⎞T
1 –
⎜
⎟ = m c2
2 ⎝ NA ⎠
2
But
1 –
m c 2 is the mean kinetic energy of a molecule
...
5)
Kinetic Theory of Gases
MODULE - 3
Thermal Physics
where
k =
R
NA
(10
...
The value of k is 1
...
Notes
In terms of k, the mean kinetic energy of a molecule of the gas is given as
ε =
1 –
3
m c2 = k T
2
2
Hence, kinetic energy of a gram mole of a gas is
(10
...
This fact is known as the
kinetic interpretation of temperature
...
In other words, all molecular motion
ceases to exist at absolute zero and the molecules behave as if they are frozen in space
...
The energy at absolute zero is known as zero point energy
...
(10
...
It means that lighter molecule, on an average, move faster than
heavier molecules
...
So according to kinetic theory, the hydrogen molecules should move 4 times
faster then oxygen molecules
...
This observed fact provided an early important evidence for the validity
of kinetic theory
...
3 Deduction of Gas Laws from Kinetic Theory
(i) Boyle’s Law
We know that the pressure P exerted by a gas is given by Eqn
...
3) :
PV =
1 –
M c2
3
When the temperature of a given mass of the gas is constant, the mean square speed is
–
constant
...
(10
...
Thus,
we can write
3
...
9)
This is Boyle’s law, which states that at constant temperature, the pressure of a given
mass of a gas is inversely proportional to the volume of the gas
...
(10
...
e, V ∝ c 2 , if M and P do not vary or M and P are constant
...
9)
This is Charle’s law : The volume of a given mass of a gas at constant pressure is
directly proportional to temperature
...
Using a
vacuum pump designed by Robert Hook, he demonstrated that
sound does not travel in vacuum
...
A founding fellow of Royal Society of London, Robert Boyle remained a bachalor
throughout his life to pursue his scientific interests
...
(iii) Gay Lussac’s Law – According to kinetic theory of gases, for an ideal gas
P =
1 M –
c2
3 V
For a given mass (M constant) and at constant volume (V constant),
–
P ∝ c2
But
–
c2 ∝ T
∴
P∝T
(10:11)
which is Gay Lussac’s law
...
(iv) Avogadro’s Law
Let us consider two different gases 1 and 2
...
(10
...
e
...
(10
...
This statement is Avogadro’s Law
...
Let their
– – –
2
2
densities be ρ1, ρ2, ρ3
...
respectively
...
They all will have the same volume
...
3
3
3
1 –2 1 –2 1 –2
ρ c , ρ c , ρ c
...
If we denote these by P1, P2, P3, respectively we get
P = P1 + P2 + P3 +
...
13)
In other words, the total pressure exerted by a gaseous mixture is the sum of the
partial pressures that would be exerted, if individual gases occupied the space in
turn
...
(vi) Graham’s law of diffusion of gases
Graham investigated the diffusion of gases through porous substances and found that the
rate of diffusion of a gas through a porous partition is inversely proportional to the
square root of its density
...
On the basis of kinetic theory of gases, the rate of diffusion through a fine hole will be
proportional to the average or root mean square velocity crms
...
(10
...
245
245
MODULE - 3
Physics
Thermal Physics
3P
–
c2 = ρ
—
or
Notes
c 2 = crms =
3P
ρ
That is, the root mean square velocities of the molecules of two gases of densities ρ 1 and
ρ 2 respectively at a pressure P are given by
(crms)1 =
3P
ρ1
and
(crms)2=
3P
ρ
2
Thus,
Rate of diffusion of one gas
(crms )1
=
Rate of diffusion of other gas
(crms )2 =
ρ2
ρ1
(10
...
Example 10
...
Take m(H2) as 3
...
38 × 10–23 J mol–1 K–1
Solution : We know that
c rms
=
3kT
=
m
3 ´ (1
...
347 ´ 10 –27 kg
= 1927 m s–1
Intext Questions 10
...
Five gas molecules chosen at random are found to have speeds 500 ms–1, 600 ms–1,
700 ms–1, 800 ms–1, and 900 ms–1
...
...
If equal volumes of two non–reactive gases are mixed, what would be the resultant
pressure of the mixture?
...
When we blow air in a balloon, its volume increases and the pressure inside is also
more than when air was not blown in
...
Example 10
...
T
...
, pressure being constant (STP = Standard temperature and
pressure)
...
(10
...
T
...
be c0
...
4 : Calculate the average kinetic energy of a gas at 300 K
...
38
× 10–23 JK–1
...
38 × 10–23 J K–1 and T= 300 K, we get
∴
E =
3
(1
...
21 × 10–21 J
10
...
1 The Law of Equipartition of Energy
We now know that kinetic energy of a molecule of a gas is given by
1 —2 3
m c = kT
...
e
...
That is to say, for a molecule all the three directions are equivalent :
and
Since
u = v = w
1 –
–
–
–
u 2 = v2 = w 2 = c 2
3
– 2 + w2
2
2
c =u + v
3
...
Therefore,
2
Ex = Ey = Ez
...
Hence, we get an
2
important result :
Ex = Ey = Ez =
1
kT
2
Since three velocity components u, v and w correspond to the three degree of freedom of
the molecule, we can conclude that total kinetic energy of a dynamical system is equally
1
k T for each degree of
2
divided among all its degrees of freedom and it is equal to
freedom
...
Let us apply this law for different types of gases
...
For a monoatomic molecule,
we have only translational motion because they are not capable of rotation (although they
can spin about any one of the three mutually perpendicular axes if it is like a finite sphere)
...
15)
A diatomic molecule can be visualised as if two spheres are joined by a rigid rod
...
However, the
rotational inertia about an axis along the rigid rod is negligible compared to that about an
axis perpendicular to the rod
...
z
2
2
Now the special description of the centre of mass of a diatomic gas molecules will require
three coordinates
...
Hence
⎛1
⎝2
⎞
⎠
⎛1
⎝2
⎞
⎠
E = 3 ⎜ kT ⎟ + 2 ⎜ kT ⎟
=
248
5
kT
2
(10
...
He also
worked with Bunsen, Kirchhoff and Helmholtz
...
The cause behind these attempts, people say, were his differences with
Mach and Ostwald
...
Crater Bolzmann on moon is named in his memory and honour
...
5 Heat Capacities of Gases
We know that the temperature of a gas can be raised under different conditions of volume
and pressure
...
In each of these cases, the amount of
thermal energy required to increase unit rise of temperature in unit mass is different
...
If we supply an amount of heat ∆Q to a gas to raise its temperature through ∆T, the heat
capacity is defined as
Heat capacity =
∆Q
∆T
The heat capacity of a body per unit mass of the body is termed as specific heat capacity
of the substance and is usually denoted by c
...
17)
Eqns
...
16) and (10
...
18)
Thus, specific heat capacity of a material is the heat required to raise the temperature
of its unit mass by 1 ºC (or 1 K)
...
It may also the expressed in joules per kg per K
...
2 × 103 J kg–1 K–1
...
In order to study the heat capacity of
a gas, we keep the pressure or the volume of a gas constant
...
249
249
MODULE - 3
Thermal Physics
Physics
(i)
Specific heat at constant volume, denoted as cV
...
(a)
The specific heat capacity of a gas at constant volume (cv) is defined as the
amount of heat required to raise the temperature of unit mass of a gas through 1K,
when its volume is kept constant :
Notes
⎛ ∆Q ⎞
⎟
⎝ ∆T ⎠ V
cv = ⎜
(b)
(10
...
⎛ ∆Q ⎞
⎟
⎝ ∆T ⎠ P
cp = ⎜
(10
...
We know that when pressure is kept constant, the volume of the gas increases
...
This means the specific heat capacity of a gas at constant pressure is greater than its
specific heat capacity at constant volume by an amount which is thermal equivalent of the
work done in expending the gas against external pressure
...
21)
10
...
10
...
Since the gas has been assumed to be ideal (perfect), there is
no intermolecular force between its molecules
...
A
P
V1
V2
Fig
...
2 : Gas heated at constant pressure
Let P be the external pressure and A be the cross sectional area of the piston
...
Now suppose that the gas is heated at constant pressure by
1K and as a result, the piston moves outward through a distance x, as shown in Fig
...
2
...
Therefore, the
work W done by the gas in pushing the piston through a distance x, against external pressure
P is given by
MODULE - 3
Thermal Physics
W =P×A×x
= P × (Increase in volume)
Notes
= P (V2 – V1)
We know from Eqn
...
22) that cp – cv = Work done (W) against the external pressure in
raising the temperature of 1 mol of a gas through 1 K, i
...
cp – cv = P (V2 – V1)
(10
...
e
...
23)
PV 2 = R (T + 1)
(10
...
(10
...
(10
...
25)
Hence from Eqns
...
19) and (10
...
26)
where R is in J mol–1 K–1
Converting joules into calories, we can write
cp – cv =
R
J
(10
...
18 cal is the mechanical equivalent of heat
...
5 : Calculate the value of cp and cv for a monoatomic, diatomic and triatomic
gas molecules
...
e
...
(i) We know that for monoatomic gas, total energy =
3
RT
2
3
...
2
2
2
cV =
3
5
R + R = R
...
2
2
(iii) You should now find out cV and cp for triatomic gas
...
3
1
...
2
...
3J mol–1 K–1)
...
3
...
Brownian Motion and Mean Free Path
Scottish botanist Robert Brown, while observing the pollen grains of a flower
suspended in water, under his microscope, found that the pollen grains were tumbling
and tossing and moving about in a zigzag random fashion
...
But when motion of pollens of
dead plants and particles of mica and stone were seen to exhibit the same behaviour,
it became clear that the motion of the particles, now called Brownian motion, was
caused by unbalanced forces due to impacts of water molecules
...
The Brownian
displacement was found to depend on
...
(ii)
The Brownian motion also increases with the increase in the temperature and
decreases with the viscosity of the medium
...
The
252
Kinetic Theory of Gases
MODULE - 3
Thermal Physics
average distance between two successive collisions of the molecules is called mean
free path
...
Notes
What You Have Learnt
Kinetic theory assumes the existence of atoms and molecules of a gas and applies
the law of mechanics to large number of them using averaging technique
...
The pressure of a gas is the average impact of its molecules on the unit area of the
walls of the container
...
At absolute zero of temperature, the kinetic energy of a gas is zero and molecular
motion ceases to exist
...
This provided an early evidence
in favour of kinetic theory
...
Hence there are two specific heats of gas :
i) Specific heat capacity at constant volume (cV)
ii) Specific heat capacity at constant pressure (cp)
These are related as
c p = W + cV
cp – cV =
R
J
The law of equipartition of the energy states that the total kinetic energy of a dynamical
system is distributed equally among all its degrees of freedom and it is equal to
1
k T per degree of freedom
...
2
3
...
Can we use Boyle’s law to compare two different ideal gases?
2
...
If the absolute temperature of a gas is raised four times, what will happen to its
kinetic energy, root-mean square velocity and pressure?
4
...
If three molecules have velocities 0
...
6
...
Use the concepts of kinetic theory of gases to derives an expression for the rootmean square velocity of the molecules in term of pressure and density of the gas
...
i) Calculate the average translational kinetic energy of a neon atom at 25 0C
...
A container of volume of 50 cm3 contains hydrogen at a pressure of 1
...
Calculate (a)the number of molecules of the gas in the container, and (b)their root-mean square speed
...
3 J mol–1 K–1 , N = 6 × 1023 mol–1
...
9
...
0 mm Hg at a
temperature of 50 K
...
0 K is
800 m s–1, what will be their root-mean square velocity at this new temperature?
10
...
11
...
12
...
13
...
14
...
Calculate the root-mean square of the molecules of hydrogen at 0 0C and at 100 0C
( Density of hydrogen at 00C and 760 mm of mercury pressure = 0
...
16
...
8 × 1024 and the root-mean square speed of the molecules is
254
Kinetic Theory of Gases
1
...
Avogadro’s number 6
...
02)
...
Define specific heat of a gas at constant pressure
...
18
...
Prove that for a triatomic gas
cV = 3R
Notes
19
...
Given R = 8
...
Answers to Intext Questions
10
...
(i) Because in a gas the cohesive force between the molecules are extremely small
as compared to the molecules in a liquid
...
The bonds between the
molecules are stronger giving a ordered structure
...
The gas which follows the kinetic theory of molecules is called as an ideal gas
...
P=
1 –
ρ c2
3
10
...
–
Average speed c
=
500 + 600 + 700 + 800 + 900
5
= 700 m s–1
–
Average value of c 2
=
5002 + 6002 + 7002 + 8002 + 9002
5
= 510,000 m2 s–2
crms =
c2 =
510, 000 = 714 m s–1
crms and c are not same
2
...
e
...
3
...
3
...
3
1
...
cV for a diatomic molecule =
cV =
5
R
2
5
× 8
...
75 J mol–1 K–1
...
05 J mol–1 C–1
...
zero
3
...
4
...
2
7
...
18 × 10–21 ms–1, – 124 ºC
8
...
9 × 1011 m s–1
9
...
1800 ms–1, 2088 ms–1
16
...
97 × 103 Nm– 2
17
...
45 J mol–1 K–1, 20
...
256
5
k T
...
When you rub your hands
together, you get the feeling of warmth
...
This suggests that there is a relationship between mechanical
work and thermal effect
...
A quantitative description of thermal
phenomena requires a definition of temperature, thermal energy and internal energy
...
In this lesson you will learn three laws of thermodynamics : the zeroth law, the first law
and the second law of thermodynamics
...
As such, the zeroth, first and second law introduce the concept of temperature,
internal energy and entropy, respectively
...
You will also learn that Carnot’s engine has maximum
efficiency for conversion of heat into work
...
3
...
1 Concept of Heat and Temperature
11
...
1 Heat
Notes
Energy has pervaded all facets of human activity ever since man lived in caves
...
The energy that cooks our food,
lights our houses, runs trains and aeroplanes originates in heat released in burning of wood,
coal, gas or oil
...
If you
touch the nozzle, you will observe that pump gets hot
...
You will agree that in these processes heating is
not caused by putting a flame or something hot underneath the pump or the hand
...
These examples, in fact, indicate a
relation between mechanical work and thermal effect
...
But a cup of hot coffee placed on the table cools down
...
This energy transfer continues till thermal equilibrium is reached
...
It also shows that
the direction of energy transfer is always from the body at high temprature to a body at
lower temperature
...
So we can say that
heat is the form of energy transferred between two (or more) systems or a system and
its surroundings because of temperature difference
...
What is the nature of this form of energy? The answer to this question
was provided by Joule through his work on the equivalence of heat and mechanical work :
Mechanical motion of molecules making up the system is associated with heat
...
One calorie is defined as the quantity of heat energy required
to raise the temperature of 1 gram of water from 14
...
5oC
...
Kilocalorie (k cal) is the larger unit of heat energy :
1 kcal = 103 cal
...
18 J
11
...
2 Concept of Temperature
While studying the nature of heat, you learnt that energy exchange between a glass of
cold water and its surroundings continues until thermal equilibrium was reached
...
Thus, we can say that temperature of a body is the property which determines
whether or not it is in thermal equilibrium with other bodies
...
1
...
Thermodynamic Terms
(i)
MODULE - 3
Thermodynamic system : A thermodynamic system refers to a definite quantity of
matter which is considered unique and separated from everything else, which can
influence it
...
The boundary may enclose a solid, a liquid or a gas
...
The region
of space outside the boundary of a system constitutes its surroundings
...
A water heater is an open system
...
A gas enclosed in a cylinder fitted with a piston is a closed
system
...
A filled thermos flank is an ideal example of an isolated
system
...
To describe
a thermodynamic system, we use its physical properties such on temperature (T),
pressure (P), and volume (V)
...
(iii) Indicator diagram : You have learnt about displacement–time and velocity–time
graphs in lesson 2
...
This graph indicates how pressure (P) of a system varies with its volume (V)
during a thermodynamic process and is known as an indicator diagram
...
It is equal
to the area under the P-V diagram (Fig
...
1)
...
Then, work done by the system
...
1)
= Area of a shaded strip ABCD
Now total work done by the system when it
expands from V1 to V2 = Area of P1P2V2V1P1
Note that the area depends upon the shape of
the indicator diagram
...
∆V
D
The indicator diagram is widely used in
P2
A
calculating the work done in the process of
expansion or compression
...
The work done on the
V1
V2
B C
system increases its energy and work done by
Fig
...
1 : Indicater Diagram
the system reduces it
...
You must note that the area enclosed by an isotherm
3
...
We may conclude that
work done by or on a system depends on the path
...
11
...
If it is left
to itself, it is common experience that after some time, the liquid attains the room
temperature
...
If within the system, there are variations in pressure or elastic stress, then parts of the
system may undergo some changes
...
Then we say that it is in mechanical equilibrium
...
Then the system is said to be in chemical equilibrium
...
The macroscopic properties of a system in this state do not
change with time
...
2
...
For example, the
expansion of a gas in a cylinder at constant pressure due to heating is a thermodynamic
process
...
Now we will consider different types of thermodynamic processes
...
A reversible process is executed very slowly and in a
controlled manner
...
You will see that it changes to water
...
•
Consider a spring supported at one end
...
You will note that the spring elongates (increases in length)
...
You will see that spring retraces its initial positions
...
As such, a reversible process can only be idealised and never achieved in practice
...
All natural process are irrerersible
...
It means that for irrerersible process,
the intermediate states are not equilibrium states and hence such process can not be
represented by a path
...
MODULE - 3
Thermal Physics
Notes
(iii) Isothermal process : A thermodynamic process that occurs at constant temperature
is an isothermal process
...
The change in pressure
or volume is carried out very slowly so that any heat developed is transferred into the
surroundings and the temperature of the system remains constant
...
In such a process, ∆Q, ∆U and ∆W are finite
...
For example, the expansion and compression
of a perfect gas in a cylinder made of perfect insulating walls
...
Neither any amount of heat leaves the system nor enters it
from the surroundings
...
The change in the internal energy of the system is equal to the work done on the
system
...
So, ∆U becomes
positive and the internal energy of the system increases
...
It is taken as positive and ∆U becomes negative
...
(v) Isobaric process : A thermodynamic process that occurs at constant pressure is an
isobaric process
...
(vi) Isochoric process : A thermodynamic process that occurs at constant volume is an
isochoric process
...
In this process, volume of the gas remains constant so that no
work is done, i
...
∆W = 0
...
In a Cyclic Process the system returns back to its initial state
...
∆U = 0
...
11
...
2 Zeroth Law of Thermodynamics
Let us consider three metal blocks A, B and C
...
Further suppose that block A is also in thermal equilibrium with block C
...
It follows that the temperatures of blocks B and C are equal
...
3
...
The
different states of matter are called its phases
...
We can discuss these three
phases using a three dimensional diagram drawn in pressure (P), temperature (T)
and volume (V)
...
Thus, we discuss
the three phases of matter by drawing a pressure-temperature diagram
...
B
line
Ice
D
e
lin
am
Ste
A
lin
e
C
P
Ho
ar
fro
st
P
E
F
T
Fig
...
3: Phase diagram of water
Refer to Fig
...
3, which shows phase diagram of water
...
Curve CD shows the variation of melting point of ice with pressure
...
Curve AB shows variation of boiling point of water
with pressure
...
Curve EF shows change of ice
directly to steam
...
This curve is also known as
Hoarfrost Line
...
This point is called triple point
...
When we heat a solid, its temperature increases till it reaches a temperature at
which it starts melting
...
During
this change of state, we supply heat continuously but the temperature does not rise
...
On heating a liquid, its temperature also rises till its boiling point is reached
...
The amount of heat required to convert unit mass of liquid in its gaseous state
at constant temperature is called latent heat of vaporization of the liquid
...
2
...
For this reason, in kelvin’s scale of thermometry, triple
point of water is taken as the upper fixed point
...
It is possible that by adjusting temperature and pressure, we can obtain all the
262
Thermodynamics
three states of matter to co-exist simultaneously
...
MODULE - 3
Thermal Physics
P
Intext Questions 11
...
Fill in the blanks
(i) Zeroth law of thermodynamics provides the
basis for the concept of
...
P2
Notes
C
B
V1
Fig
...
2
V2
V
(iii) The unit of heat is
...
Fig
...
2 is an indicator diagram of a thermodynamic process
...
...
Fill in the blanks
...
in the opposite direction
from its final state to its initial state
...
process is that which cannot be retraced along the same
equilibrium states from final state to the initial state
...
State the basic difference between isothermal and adiabatic processes
...
5
...
...
3 Internal Energy of a System
Have you ever thought about the energy which is released when water freezes into ice ?
Don’t you think that there is some kind of energy stored in water
...
This stored energy is called the internal energy
...
263
263
MODULE - 3
Thermal Physics
Notes
Physics
the energies of individual components/constituents
...
Let us
now discuss these
...
These molecules are in a state of constant
rapid motion and hence possess kinetic energy
...
(b) Internal potential energy : The energy arising due to the inter-molecular forces is
called the internal potential energy
...
The energy of the system may be increased by causing its molecules
to move faster (gain in kinetic energy by adding thermal energy)
...
e
...
Internal energy is denoted by the letter U
...
Suppose
W amount of work is done on the system in going from initial state i to final state f
adiabatically
...
Since work is done on the system, internal energy of final state will be higher
than that of the initial state
...
We may point out here that unlike work, internal energy depends on the initial and final
states, irrespective of the path followed
...
Note that if some work is done by
the system, its internal energy will decrease
...
4 First Law of Thermodynamics
You now know that the zeroth law of thermodynamics tells us about thermal equilibrium
among different systems characterised by same temperature
...
Let us consider two examples : (i) Two
systems at different temperatures are put in thermal contact and (ii) Mechanical rubbing
between two systems
...
To explain such processes, the first law of thermodynamics
was postulated
...
It states that change in internal energy of a system during a
264
Thermodynamics
thermodynamic process is equal to the sum of the heat given to it and the work done
on it
...
Then increase in internal energy of the system, ∆U, according to the first law of
thermodynamics is given by
∆U = ∆Q – ∆W
(11
...
Here ∆Q, ∆U and ∆W
all are in SI units
...
3 b)
The signs of ∆Q, ∆U and ∆W are known from the following sign conventions :
1
...
The work is positive when a system expands
...
The work done does
not depend on the initial and final thermodynamic states; it depends on the path followed
to bring a change
...
Heat gained by (added to) a system is taken as positive, whereas heat lost by a system
is taken as negative
...
The increase in internal energy is taken as positive and a decrease in internal energy
is taken as negative
...
However, the difference (∆Q – ∆W) which represents ∆U, remains
the same for all paths of transformations
...
11
...
1 Limitations of the First Law of Thermodynamics
The first law of thermodynamics asserts the equivalence of heat and other forms of energy
...
The electrical energy that lights our
houses, operates machines and runs trains originates in heat released in burning of fossil or
nuclear fuel
...
It explains the fall in temperature with height; the
adiabatic lapse rate in upper atmosphere
...
However, consider the following processes :
•
You know that heat always flows from a hot body to a cold body
...
It
means that this law fails to indicate the direction of heat flow
...
This law does not indicate as to why heat developed in the
target cannot be changed into the kinetic energy of bullet to make it fly
...
3
...
Now take a pause and answer the following questions :
Intext Questions 11
...
Fill in the blanks
(i)
The total of kinetic energy and potential energy of molecules of a system is called
its
...
the system
...
The first law of thermodynamics states that
...
11
...
So a question
may arise in your mind : Can heat be wholly converted into work? Under what conditions
this conversion occurs? The answers of such questions are contained in the postulate of
Second law of thermodynamics
...
However, here you will study Kelvin-Planck and Clausius statements of second
law of thermodynamics
...
(Heat engine is discussed in next section
...
There is no engine which converts the
whole heat into work, without rejecting some heat to the sink
...
Clausius statement of second law of thermodynamics is based on the performance of
a refrigerator
...
It transfers
heat from a colder body to a hotter body when external work is done on it
...
To do this external work, supply of
energy from some external source is a must
...
It is impossible for any process to have as its sole result to transfer heat from a
colder body to a hotter body without any external work
...
266
Thermodynamics
MODULE - 3
Thermal Physics
11
...
1 Carnot Cycle
You must have noticed that when water is boiled in a vessel having a lid, the steam
generated inside throws off the lid
...
A device which can convert heat into work is called a heat engine
...
These may be categorised in three types : steam engine, internal combustion engine and
gas turbine
...
Let us learn about it now
...
11
...
Such a cycle is represented on the P-V diagram in Fig
...
4
...
mol
...
11
...
Original condition of the substance is represented by point A on the
indicator diagram
...
Fig
...
5 : The cylinder with working substance
(a)
Isothermal expansion : The cylinder is put in thermal contact with the source and
allowed to expand
...
Thus
working substance does work in raising the piston
...
But it is in thermal contact with the source
...
267
267
MODULE - 3
Physics
Thermal Physics
So it will absorb a quantity of heat H1 from the source at temperature T1
...
At B, the values of pressure and volume are P2 and V2
respectively
...
11
...
We
call it isothermal expansion process
...
Then, in accordance with the first law of
thermodynamics, H1 will be equal to the external work done by the gas during
isothermal expansion from A to B at temperature T1
...
Then it will be equal to the
area ABGEA
...
It further decreases the load on the piston to
P3
...
Therefore, the working substance performs external work in
raising the piston at the expense of its internal energy
...
The gas is thus allowed to expand adiabatically until its temperature falls to T2, the
temperature of the sink
...
We call it adiabatic expansion
...
(c)
Isothermal compression : Remove the cylinder from the non-conducting stand
and place it on the sink at temperature T2
...
It is represented by the point D on the indicator diagram (Fig
...
4)
...
Thus, there
is no change in the temperature of the system
...
It is shown by the curve CD (Fig
...
4)
...
Hence
W3 = Area CHFDC
(d)
Adiabatic compression : Once again place the system on the non-conducting stand
...
The substance will under go an adiabatic
compression
...
This is an adiabatic
compression process and represented by the curve DA on the indicator diagram
(Fig
...
4)
...
Then
W4 = Area DFEAD
During the above cycle of operations, the working substance takes H1 amount of heat
from the source and rejects H2 amount of heat to the sink
...
Notes
You have studied that the initial and final states of the substance are the same
...
Hence according to the first law of
thermodynamics
W = H1 – H2
Therefore, heat has been converted into work by the system, and any amount of work can
be obtained by merely repeating the cycle
...
7
...
It is denoted as η:
η =
or
Heat converted into work
Heat taken from source
η =
H1 - H 2
H
= 1- 2
H1
H1
It can be shown that for Carnot’s engine,
H 2 T2
=
H1 T1
Hence,
T2
η =1– T
1
Note that efficiency of carnot engine does not depend on the nature of the working
substance
...
But for H2 to
be zero, T2 must be zero
...
The
entire heat taken from the hot source is converted into work
...
Therefore, a steam engine can operate only between finite temperature
limits and its efficiency will be less than one
...
11
...
3 Limitation of Carnot’s Engine
You have studied about Carnot’s cycle in terms of isothermal and adiabatic processes
...
269
269
MODULE - 3
Thermal Physics
Physics
moves very slowly
...
On the other hand, during the adiabatic process,
the piston moves extremely fast to avoid heat transfer
...
Due to these very reasons, all practical engines have an efficiency
less than that of Carnot’s engine
...
3
1
...
(i)
In a Carnot engine, when heat is taken by a perfect gas from a hot source, the
temperature of the source decreases
...
(ii) In Carnot engine, if temperature of the sink is decreased the efficiency of engine
also decreases
...
2
...
Calculate T
...
(ii) A Carnot engine working between an unknown temperature T and ice point
gives an efficiency of 0
...
Deduce the value of T
...
What You Have Learnt
Heat is a form of energy which produces in us the sensation of warmth
...
The most commonly known unit of heat energy is calorie
...
18 J and
1k cal = 103 cal
...
Work done during expansion or compression of a gas is P∆V = P(Vf – Vi)
...
The sum of kinetic energy and potential energy of the molecules of a body gives the
internal energy
...
270
Thermodynamics
The first law of thermodynamics states that the amount of heat given to a system is
equal to the sum of change in internal energy of the system and the external work
done
...
The process which can be retraced in the opposite direction from its final state to
initial state is called a reversible process
...
A process that occurs at constant
temperature is an isothermal process
...
The different states of matter are called its phase and the pressure and temperature
diagram showing three phases of matter is called a phase diagram
...
It is characterised by a particular temperature and pressure
...
According to Clausius statement of second law, heat can not flow from a colder body
to a hotter body without doing external work on the working substance
...
(iii)working substance which performs mechanical work after being supplied with
heat
...
Such a cycle is called a Carnot cycle
...
1
T2
= 1 – T , T1 = Temperature of the source, and T2 = Temperature of the sink
...
Terminal Exercise
1
...
3
...
What do you mean by an indicator diagram
...
3
...
4
...
Notes
5
...
State the Second law of thermodynamics
...
Discuss reversible and irreversible processes with examples
...
Explain Carnot’s cycle
...
9
...
10
...
(i) What is the
temperature of the sink
...
Answers to Intext Questions
11
...
(i) Temperature (ii) C (iii) Joule or Calorie
2
...
(i) retrace
(ii) irreversible
4
...
5
...
e
...
11
...
(i) Internal energy
(ii) on
2
...
11
...
(i) False
(ii) True
2
...
1K
Answers to Terminal Problems
9
...
10
...
In this lesson you
learn about the processes of heat transfer
...
Before reaching the earth, it passes through vacuum as well as
material medium between the earth and the sun
...
This study enables us to determine the temperatures of stars even though they are very
far away from us
...
When one end of a metal rod is heated, its other end also becomes hot after some
time
...
Heat energy falling on the walls of our homes also enters inside through
conduction
...
They move from the bottom of the pot to the water surface and carry heat
energy
...
These processes are responsible
for various natural phenomena, like monsoon which are crucial for existence of life on the
globe
...
Objectives
After studying this lesson, you should be able to :
distinguish between conduction convection and radiation;
define the coefficient of thermal conductivity;
describe green house effect and its consequenies for life on earth; and
apply laws governing black body radiation
...
273
273
MODULE - 3
Thermal Physics
Notes
Physics
12
...
The second law
postulates that the natural tendency of heat is to flow spontaneously from a body at higher
temperature to a body at lower temperature
...
From kinetic theory, you may recall that
temperature of a gas is related to its average kinetic energy
...
There are three processes by which transfer of heat takes place
...
In conduction and convection, heat transfer takes place through
molecular motion
...
Heat transfer through conduction is more common
in solids
...
When heated, they can not leave their sites;
they are constrained to vibrate about their respective
equilibrium positions
...
12
...
12
...
The atoms near the end A become
metal rod
hot and their kinetic energy increases
...
E
...
These atoms further transfer some K
...
This process continues and kinetic energy is transferred to
atoms at the other end B of the rod
...
Thus, heat is transferred from atom to atom by
conduction
...
Fig
...
2 : Convection
currents are
formed in water
when heated
In convection, molecules of fluids receive thermal energy
and move up bodily
...
Put a bunsen flame under the
flask
...
The density of water decreases and the buoyant force
causes it to move upward (Fig
...
2)
...
Thus, a convection current of
hotter water going up and cooler water coming down is
set up
...
These convection
currents can be seen as KMnO4 colours them red
...
You will learn about the
characteristics of these waves in a later section
...
Heat from
the sun comes to us mostly by radiation
...
274
Heat Transfer and Solar Energy
MODULE - 3
Thermal Physics
12
...
1 Conduction
d
Consider a rectangular slab of area of cross-section A
and thickness d
...
12
...
Let
A
A
us consider all the factors on which the quantity of heat
Tc
Q transferred from one face to another depends
...
Also, greater the thickness,
Fig
...
3 : Heat conduction
lesser will be the heat transfer (Q α 1/d)
...
Finally longer the time t allowed
surface area A, when
the faces are kept at
for heat transfer, greater will be the value of Q
...
t
Notes
Tc
...
1)
d
where K is a constant which depends on the nature of the material of the slab
...
Thermal conductivity of a material is defined as the amount of heat transferred in one
second across a piece of the material having area of cross-section 1m2 and edge 1m when
its opposite faces are maintained at a temperature difference of 1 K
...
The value of K for some materials is given in Table 12
...
1 : Thermal
Conductivity of some
materials
Material
Thermal
conductivity
(Wm–1 K–1)
Example 12
...
has side 30 cm and thickness of
5
...
If outside temperature is 45°C, estimate the amount of ice melted in 6 h
...
01 J s–1 m–1 °C–1 and latent heat of fusion of ice is 335 J g–1
...
(12
...
2
Glass
0
...
60
Body talc
0
...
025
Thermocole
0
...
01 J s–1 m–1 ° C–1) × (900 × 10–4 m2) × (45 ºC)
× (6 × 60 × 60 s) / (5 × 10–2 m)
Q =
= 10496 J
Since the box has six faces, total heat passing into the box
Q = 10496 × 6 J
The mass of ice melted m, can be obtained by dividing Q by L :
m = Q/L
=
10496J
×6
335 Jg –1
= 313 × 6 g = 1878 g
3
...
1 that metals such as copper and aluminium have high thermal
conductivity
...
This is the reason
why cooking vessels and heating pots are made of copper
...
Substances having low value of K are
sometimes called thermal insulators
...
Wool is a good thermal insulator
because air is trapped between its fibres
...
Even if a few cotton clothes are put on one above another, the air trapped in-between
layers stops cold
...
Sometimes we wrap the ice slab in jute bag, which also has
low thermal conductivity
...
1
...
Do you know the
reason? Let us discover it
...
Warm air from the shore rises
and moves upwards (Fig
...
4)
...
The net effect of these convection
Fig
...
4 : Convection currents
...
The convection current shore,which is hotter, to water, which is
from water to the shores is cooler
...
many factors
...
However, the rate of heat transfer by convection
depends on the temperature difference between the surfaces and also on their areas
...
Intext Questions 12
...
Distinguish between conduction and convection
...
2
...
...
Explain why do humans wrap themselves in woollens in winter season?
...
A cubical slab of surface area 1 m2, thickness 1 m, and made of a material of thermal
conductivity K
...
Compute the energy transferred across the surface in one second
...
...
During the summer, the land mass gets very hot
...
This results in the onset of sea breezes
...
MODULE - 3
Thermal Physics
...
1
...
This energy
is called radiant energy and is in the form of electromagnetic waves
...
They can easily be reflected from polished surfaces and focussed using a lens
...
The
sun, at 6000 K emits energy mainly in the visible spectrum
...
The human body also radiates energy in the infra-red region
...
Take a piece of blackened platinum wire in a
dark room
...
You will note that the wire has become
hot
...
After sometime, the wire will begin
to radiate
...
This shows that the wire is just emitting red radiation of sufficient intensity to
affect the human eye
...
With further increase in temperature,
the colour of the emitted rediation will change from dull red to cherry red (at nearly
900°C) to orange (at nearly 1100°C), to yellow (at nearly 1250°C) until at about 1600°C,
it becomes white
...
Secondly, with increase in temperature, waves
of shorter wavelengths (since red light is of longer wavelength than orange
...
)
are also emitted with sufficient intensity
...
12
...
The most intense of these
waves will have a particular wavelength
4000 K
(sayλm)
...
The intensity
3000 K
decreases for wavelengths either greater
or less than this value (Fig
...
5)
...
You
may study the curves shown in Fig
...
5
and verify the following two facts
...
12
...
277
277
MODULE - 3
Physics
area between each curve and the holizontal axis) increases rapidly with temperature
...
e
...
The λm shifts towards shorter wavelengths
with increasing temperature
...
It states that λ m shifts towards shorter wavelengths as the temperature of a
body is increased
...
, strictly valid only for black bodies
...
2)
The constant in Eqn
...
2) has a value 2
...
This law furnishes us with a
simple method of determining the temperature of all radiating bodies including those in
space
...
Using Eqn
...
2), we get
T=
2884 micron K
14 micron = 206K
That is, the temperature of the lunar surface is 206K
Wilhelm Wien
(1864 – 1928)
The 1911 Nobel Leureate in physics, Wilhelm Wien, was son of a
land owner in East Prussia
...
At the University of Berlin, he studied under
great physicist Helmholtz and got his doctorate on diffraction of light from metal
surfaces in 1886
...
In 1896, he succeeded Philip Lenard as
Professor of Physics at Aix-la-chappelle
...
C
...
In
1902, he was invited to succeed Ludwig Boltzmann at University of Leipzig and in
1906 to succeed Drude at University of Berlin
...
In
1920, he was appointed Professor of Physics at munich and he remained there till his
last
...
2
...
If for a particular wavelength λ and a given surface,
rλ , aλ and tλ , respectively denote the fraction of total incident energy reflected, absorbed
and transmitted, we can write
1 = rλ + aλ + tλ
278
(12
...
It means that radiations
incident on black bodies will be completely absorbed
...
Lamp black is the nearest approximation to a black body
...
It is found to transmit
light of long wavelength
...
A piece
of white chalk approximates to a perfectly white body
...
But each body must either absorb
or reflect the radiant energy reaching it
...
Designing a Black Body
Kirchoff’s law also enables us to design a perfectly black body for experimental
purposes
...
Now let us make a small hole in the
enclosure and examine the radiation escaping out of it
...
Thus, if the reflecting power of the surface of the wall is
r, and emissive power is eλ , the total radiation escaping out is given by
Eλ = eλ + eλ rλ + eλ r2 + eλ rλ3 +
...
)
eλ
= 1– r
λ
But from Kirchoffs Law
(12
...
5)
where Eλ is the emission from a black body
...
e
...
(12
...
6)
Substituting this result in Eqn
...
5), we get
eλ = Eλ (1 – rλ )
or
Eλ =
eλ
1 – rλ
(12
...
(12
...
7), we note that the radiation emerging out of the
hole will be identical to the radiation from a perfectly black emissive surface
...
So we see that the
uniformly heated enclosure with a small cavity behaves as a black body for
emission
...
Any radiation passing into the hole will undergo multiple reflections internally within
3
...
This may be further improved by
blackening the inside
...
Fig
...
6 shows a black body due to Fery
...
It has a small
conical opening O
...
12
...
This is to avoid direct
radiation from the surface opposite the hole which
would otherwise render the body not perfectly black
...
1
You have studied that black surface absorbs heat radiations more quickly than a shiny
white surface
...
Take two metal plates A and B
...
Take an electric heater
...
Ensure that coated plates are equidstant
from the heater
...
Polished or white
coated metal surface
—
er
at
he
c
tri
ec
El
te
pla
t a l black
Me
— oated
c
Cork
Cork
B
A
Fig
...
7 : Showing the difference in heat absorption of a black and a shining surface
Switch on the electric heater
...
You will
observe that the cork on the blackened plate falls first
...
This proves that black surfaces are good
aborbers of heat radiations
...
2
...
8)
where σ is Stefan-Boltzmann constant and has the value 5
...
The
temperature is expressed is kelvin, e is emissivity or relative emittance
...
The value of e lies between 0 and 1; being small
for polished metals and 1 for perfectly black materials
...
(12
...
They would have done so if energy were not supplied to them in some way
...
If a body is at the same
temperaturture as its surroundings, the rate of emission is same as the rate of absorption;
there is no net gain or loss of energy and no change in temperature
...
Its temperature will rise till it is equal to the room temperature
...
There will be a net energy loss
...
5)
Example 12
...
Given σ = 5
...
3
...
Hence we can rewrite it as
A=
E
eσ T 4
On substituting the given data, we get
100 W
A = 0
...
7 × 10–8 Wm−2 K−4 × (3000K)4
= 7
...
3
...
2
1
...
Notes
2
...
3
...
...
A person with skin temperature 28 oC is present in a room at temperature
22oC
...
9m2, compute the radiant power of this person
...
12
...
The sun is radiating tremendous amount of energy in the form of
light and heat and even the small fraction of that radiation received by earth is more than
enough to meet the needs of living beings on its surface
...
Some basic issues related with solar radiations are discussed below
...
Solar Constant
To calculate the total solar energy reaching the earth, we first determine the amount of
energy received per unit area in one second
...
Solar
constant for earth is found to be 1
...
Solar constant multiplied by the surface
area of earth gives us the total energy received by earth per second
...
Therefore,
Q = 2 × 3
...
4 × 106 m)2 × (1
...
5 × 1017 W
~ 3
...
If R is radius of the orbit of the planet, then
2
∈
⎛r⎞
=
σ T4
E=
2 ⎜R⎟
⎝ ⎠
4πR
(12
...
7)
2
E′
⎛R⎞
=⎜ ⎟
E
⎝ R′ ⎠
(12
...
52 times the distance of earth from the sun
...
52 ⎟
⎠
2
= 6 × 102 W m–2
2
...
The atmosphere of earth plays an
important role to provide a comfortable temperature for the
living organisms
...
In a greenhouse, plants, flowers, grass etc
...
The glass allows short wavelength radiation
of light to enter
...
It is
subsequently re-radiated in the form of longer wavelength
heat radiations – the infrared
...
These heat radiations are
thus trapped in the greenhouse keeping it warm
...
12
...
The atmosphere, which contains a
trace of carbon dioxide, is transparent to visible light
...
The earth absorbs this light and
subsequently emits it as infrared radiation
...
CO2 reflects these radiations back rather than allowing them to escape into the
atmosphere
...
This effect is referred to as
the greenhouse effect
...
283
283
MODULE - 3
Thermal Physics
Notes
Physics
Due to emission of huge quantities of CO2 in our atmosphere by the developed as well as
developing countries, the greenhouse effect is adding to global warming and likely to pose
serious problems to the existence of life on the earth
...
In the foreseable future, these can cause disasters beyond imagination
beginning with flooding of major rivers and rise in the sea level
...
There is a lurking fear that
these together will create problems of food security
...
In Indian context, it has been estimated that lack of positive action can lead to serious
problems in Gangetic plains by 2030
...
How can
you contribute in this historical event?
12
...
The law can be deduced from stefanBoltzmann law
...
The rate at which heat
is lost per unit area per second by the hot body is
E = eσ (T4 – T04 )A
As T4 – T04 = (T 2 − T02 ) (T 2 + T02 ) = (T − T0 ) (T − T0 ) (T 2 + T02 )
...
9)
(12
...
Hence
∴
E = eσ (T – T0) 4 T03 A
= k (T – T0)
where k = 4eσ T03 A
...
11)
This is Newton’s law of cooling
...
3
1
...
2
...
284
Heat Transfer and Solar Energy
3
...
What You Have Learnt
Heat flows from a body at higher temperature to a body at lower temperature
...
In conduction, heat is transferred from one atom/ molecule to another atom/molecule
which vibrate about their fixed positions
...
In radiation, heat is
transferred through electromagnetic waves
...
The spectrum of energy radiated by a body at temperature T(K) has a
maxima at wavelength λ m’ such that λ mT = constant ( = 2880 µK)
Q=
Stefan-Boltzmann Law
...
The solar constant for the earth is 1
...
Terminal Exercise
1
...
12
...
The bottle contains
some liquid whose temperature we
want to maintain, Look at the diagram
carefully and explain how the
construction of the flask helps in
minimizing heat transfer due to
conduction convection and radiation
...
The wavelength corresponding, to
Fig
...
9
emission of energy maxima of a star is
4000 Aº
...
(1Aº = 10–8 cm)
...
A blackened solid copper sphere of radius 2cm is placed in an evacuated enclosure
3
...
At what rate must energy be supplied to the sphere
to keep its temperature constant at 127° C
...
Comment on the statement “A good absorber must be a good emitter”
Notes
5
...
5cm thick and 50 cm in diameter rests on a
burner which maintains the bottom surface of the pot at 110°C
...
The actual
temperature of the inside surface of the bottom of the pot is 105°C
...
Define the coefficient of thermal conductivity
...
7
...
8
...
[Note : Thermal resistance is reciprocal of thermal conductivity]
9
...
To have
the same thermal resistance of the two rods of these materials of equal thickness
...
Why do we feel warmer on a winter night when clouds cover the sky than when the
sky is clear?
11
...
Why is it more difficult to sip hot tea from a metal cup than from a china-clay cup?
13
...
Why do two layers of cloth of equal thickness provide warmer covering than a single
layer of cloth of double the thickness?
15
...
A
...
We keep our one hand 5 cm above it and other 5 cm below
it
...
Two vessels of different materials are identical in size and in dimensions
...
If ice in both vessles metls completely in 25
minutes and in 20 minutes respectively compare the (thermal conductivities) of metals
of both vessels
...
Calculate the thermal resistivity of a copper rod 20
...
length and 4
...
in diamter
...
2 x 10–2 temperature different acrosss the ends of
the rod be 50°C
...
286
Heat Transfer and Solar Energy
MODULE - 3
Thermal Physics
Answers to Intext Questions
12
...
Conduction is the principal mode of transfer of heat in solids in which the particles
transfer energy to the adjoining molecules
...
Notes
Qd
2
...
The trapped air in wool fibres prevents body heat from escaping out and thus keeps
the wearer warm
...
The coefficient of thermal conductivity is numerically equal to the amount of heat
energy transferred in one second across the faces of a cubical slab of surface area
1m2 and thickness 1m, when they are kept at a temperature difference of 1°C
...
During the day, land becomes hotter than water and air over the ocean is cooler than
the air near the land
...
This causes see breeze because the moist air from the ocean moves to the
land
...
12
...
λ m =
=
Wien 's constant
Temperature
2880µK
300Κ
= 9
...
Hint: Because light colours absorb less heat
...
Hint: (a) λmT = S (b) t = σ T4
4
...
4 W
...
3
1
...
area
= 2
...
Constant addition of CO2 in air will increase greenhouse effect causing global warming
due to which glaciers are likely to melt and flood the land mass of the earth
...
287
287
MODULE - 3
Thermal Physics
Physics
3
...
7210 K
3
...
6 × 50–11 W
Notes
5
...
7 × 105 kg
9
...
4 : 5
18
...
9 m ºC–1 W–1, 0
...
Do not send your assignment to NIOS
1
...
(1)
2
...
Why change in temperature of water from 14
...
5° C is specified in defining one calorie? (1)
4
...
What is indicated by the statement
...
State two reasons due to which all practical engines have an efficiency less than the carnot’s engine
...
Which diagram plays important role to explain the theory of heat engine?
(1)
8
...
(1)
9
...
(2)
10
...
A refrigerator transfers heat from the cooling coil at low temperature to the warm surroundings
...
(2)
12
...
Each rod has its ends at temperature T1 and T2 respectively
(T1 > T2)
...
289
289
13
...
Figure shows three paths through which a gas can be taken from the
state 1 to state 2
...
(4)
1
2
3
−6
Path1 → 2 w12 = (10 + 3) × 10 × 15 × 10 = 0
...
15t]
Volume →
[Hint : Path 1 → 3 → 2 w13 + w32 = 0 + p∆v = 0
...
The P - V diagram of a certain process (carnot cycle) is reflected in figure a
...
(4)
B
A
C
D
O
→ Volume (V)
(a)
B
A
→ Entropy (s)
B
A
C
D
C
D
O
Pressure (P)
Temperature (T)
Hint :
O
→ Volume (V)
15
...
(4)
16
...
Discuss the limitations of first law of thermodynamics
...
State and explain second law of thermodynamics
...
What do you mean by the following terms :
(i) thermal conductivity of a solid (ii) variable state of a metallic rod (iii) steady state of a matallic rod (iv)
coefficient of thermal conductivity
...
Briefly describe a carnot cycle and derive an expression for efficiency of this cycle
...
What is a heat engine? Obtain an expression for its efficiency
...
Obtain an expression for its coefficient of performance
...
2 + 2 + 1 = (5)
290
MODULE - IV
OSCILLATIONS AND WAVES
13
...
Wave Phenomena
Simple Harmonic Motion
MODULE - 4
Oscillations and Waves
13
SIMPLE HARMONIC MOTION
Notes
You are now familiar with motion in a straight line, projectile motion and circular motion
...
But some objects execute
motion which are repeated after a certain interval of time
...
Such a motion is called periodic
motion
...
In this lesson, you will study about the periodic motion, particularly the oscillatory motion
which we come across in daily life
...
Wave phenomena – types of waves and their characteristics–form the subject matter of
the next lesson
...
13
...
The seconds
hand completes its journey around the dial in one minute but the minutes hand takes one
hour to complete one round trip
...
291
291
MODULE - 4
Oscillations and Waves
Notes
Physics
position and completes its motion from one end to the other and back to its initial position
in a fixed time
...
There are two types of periodic motion : (i) non–oscillatory, and (ii) oscillatory
...
However, both the motions are periodic
...
Remember that a motion which repeats itself in equal intervals of time is periodic
and if it is about a mean position, it is oscillatory
...
It also revolves around the sun and completes its revolution in 365 days
...
Similarly all the planets move around the
Sun in elliptical orbits and each completes its revolution in a fixed interval of time
...
Jean Baptiste Joseph Fourier
(1768 – 1830)
French Mathematician, best known for his Fourier series to analyse a
complex oscillation in the form of series of sine and consine functions
...
He
established the partial differential equation governing heat diffusion
and solved it by using infinite series of trigonometric functions
...
From the training as a priest, to a teacher, a revolutionary, a mathematician
and an advisor to Nepolean Bonapart, his life had many shades
...
Lunar crator Fourier and his name on Eiffel tower are tributes to
his contributions
...
1
Suppose that the displacement y of a particle, executing simple harmonic motion, is
represented by the equation :
y = a sin θ
or
(13
...
2)
From your book of mathematics, obtain the values of sin θ and cos θ for θ = 0, 300, 600,
900, 1200, 1500, 1800, 2400, 3000, 3300 and 3600
...
5cm, determine
the values of y corresponding to each angle using the relation y = a sin θ
...
Similarly, using the relation y = a cosθ,
plot another graph between y and θ
...
It shows that a certain type of oscillatory motion can be represented
by an expression containing sine or cosine of an angle or by a combination of such
expressions
...
Intext Questions 13
...
What is the difference between a periodic motion and an oscillatory motion?
...
Notes
Which of the following examples represent a periodic motion?
(i) A bullet fired from a gun, (ii) An electron revolving round the nucleus in an atom,
(iii) A vehicle moving with a uniform speed on a road, (iv) A comet moving around the
Sun, and (v) Motion of an oscillating mercury column in a U-tube
...
3
...
...
2 Simple Harmonic Motion : Circle of Reference
The oscillations of a harmonic oscillator can be represented by terms containing sine and
cosine of an angle
...
We define simple
harmonic motion as under :
A particle is said to execute simple harmonic motion if it moves to and fro about a
fixed point periodically, under the action of a force F which is directly proportional
to its displacement x from the fixed point and the direction of the force is opposite to
that of the displacement
...
Mathematically, we express it as
F = – kx
where k is constant of proportionality
...
13
...
293
293
MODULE - 4
Oscillations and Waves
Notes
Physics
To derive the equation of simple harmonic motion, let us consider a point M moving with a
constant speed v in a circle of radius a (Fig
...
1) with centre O
...
The position vector OM specifies the position of the moving point at time t,
...
The acceleration of the point M is v2/a = a ω2 towards the centre O
...
Let us draw MP
perpendicular to YOY′
...
The force on the particle P towards O is therefore given by
F = maω2 sin ωt
But sin ωt = y/a
...
3)
F = mω2y
The displacement is measured from O towards P and force is directed towards O
...
we can say that the particle P is executing simple harmonic motion
...
Then Eqn
...
3) takes the form
F =–ky
(13
...
The angular
frequency of oscillations is given by
ω2 = k / m
(13
...
Hence
ω = 2π/T
(13
...
(13
...
6), we get an expression for time period :
T = 2π k / m
(13
...
During this time, the particle moves once on the circle and the foot of perpendicular from
its position is said to make an oscillation about O as shown in Fig
...
1
...
13
...
1 Basic Terms Associated with SHM
Displacement is the distance of the harmonic oscillator from its mean (or equilibrium)
position at a given instant
...
Time period is the time taken by the oscillator to complete one oscillation
...
13
...
Frequency is the number of oscillations completed by an oscillator in one second
...
The SI unit of frequency is hertz (symbol Hz)
...
Hence T =1/v or
v = (1/T) s–1
...
MODULE - 4
Oscillations and Waves
Phase φ is the angle whose sine or cosine at a given instant indicates the position and
direction of motion of the oscillator
...
Angular Frequency ω describes the rate of change of phase angle
...
Since phase angle φ changes from 0 to 2π radians in one complete
oscillation, the rate of change of phase angle is ω = 2π/T = 2π v or ω = 2πv
...
1 : A tray of mass 9 kg is supported by a spring of force constant k as
shown in Fig
...
2
...
It begins to
execute SHM of period 1
...
When a block of mass M is placed on the tray, the period
increases to 2
...
Calculate the mass of the block
...
Since ω = 2π/T, from Eqn
...
7) we get
4π2/T2 =
k
m
M
kT 2
m =
4π2
or
When the tray is empty, m = 9kg and T = 1s
...
13
...
Therefore, 9 + M = k × (2)2/4π2
From the above two equations we get
(9 + M )
=4
9
Therefore, M = 27kg
...
2 : A spring of force constant 1600 N m–1 is mounted on a horizontal table
as shown in Fig
...
3
...
0 kg attached to the free end of the spring is pulled
horizontally towards the right through a distance of 4
...
Calculate (i)
the frequency (ii) maximum acceleration and (iii) maximum speed of the mass
...
13
...
Therefore v = 20/2π = 3
...
Maximum acceleration = a ω2 = 0
...
04 × 20 = 0
...
3
...
3 Examples of SHM
In order to clarify the concept of SHM, some very common examples are given below
...
3
...
The other end of the spring is attached to a rigid wall (Fig
...
4))
...
m
P
(i)
P
(ii)
kx
P
(iii)
(iv)
x
P
P
(v)
kx
Fig
...
4 : Oscillations of a spring-mass system
Let us suppose that there is no loss of energy due to air resistance and friction
...
Initially, that is, at t = 0, the block is at
rest and the spring is in relaxed condition [Fig
...
4(i)]
...
13
...
As the spring undergoes an extension x, it
exerts a force kx on the block
...
As the block returns to its initial position
[Fig
...
4 (iii)], it acquires a velocity v and hence a kinetic energy K = (1/2) m v2
...
13
...
In this
position, the block again experiences a force kx which tries to bring it back to the
initial position [Fig
...
4 v]
...
The time period of oscillation is 2π m / k , where k is the force per
unit extension of the spring
...
3
...
13
...
Then let us attach a block of mass
m to the free end of the spring
...
13
...
Obviously, the force constant of the spring is k = mg/l
...
13
...
A force ky acts on the block vertically
296
l
y
(a
)
(b) (c)
Fig
...
5: Vertical oscillations
of a a spring–mass
system
Simple Harmonic Motion
upwards
...
As the block
returns to its initial position, it continues moving upwards on account of the velocity it has
gained
...
The compressed spring now
applies on it a restoring force downwards
...
Thus, the system
continues to execute vertical oscillations
...
8)
This result shows that acceleration due to gravity does not influence vertical oscillations
of a spring–mass system
...
As a child, he was interested in music, art and toy making
...
To pursue the study
of medicine, he entered the University of Pisa
...
For lack of money, Galileo could not complete his studies, but through his efforts, he
learnt and developed the subject of mechanics to a level that the Grand Duke of
Tuscany appointed him professor of mathematics at the University of Pisa
...
Through his
observations, he became convinced that Copernican theory of heliocentric universe
was correct
...
The proposition
being at variance with the Aristotelian theory of geocentric universe, supported by
the Church, Galileo was prosecuted and had to apologize
...
Because sophisticated measuring devices were not available in Galileo’s time, he
had to apply his ingenuity to perform his experiments
...
3
...
3
...
13
...
The
bob is considered a point mass and the string
is taken to be inextensible
...
θ
Notes
T
When the pendulum is displaced through a
small distance from its equilibrium position
mg sinθ
mg cosθ
and then let free, it executes angular
mg
oscillations in a vertical plane about its
equilibrium position
...
13
...
The forces acting on the bob of the pendulum
in the displaced position shown in Fig
...
6 are : (i) the weight of the bob mg vertically
downwards, and (ii) tension in the string T acting upwards along the string
...
The component mg cosθ balances the
tension T and the component mg sinθ produces acceleration in the bob in the direction of
the mean position
...
For small displacement x of
the bob, the restoring force is F = mgθ = mg x/l
...
9)
Measuring Weight using a Spring
We use a spring balance to measure weight of a body
...
e
...
Therefore, extension varies linearly
with load
...
The balance so prepared can be used to measure unknown
weights
...
Then how do they
measure mass of astronauts during regular health
check up? It is again a spring balance based on a
Fig
...
7 : Spring balance for
different principle
...
13
...
of an astronaut
The time period of oscillations of the chair with and
without the astronaut is determined with the help of an electronic clock :
2
1
T
Notes
4π2m
=
k
where m is mass of the astronaut
...
On subtracting one from another, we get
2
1
2
0
T –T
⇒
4π2
=
(m – mo)
k
m =
k
( T 2 – T02 ) + mo
4π2 1
Because the values of T0 and k are fixed and known, a measure of T1 itself shows
the variation in mass
...
3 : Fig
...
8 shows an oscillatory system comprising two blocks of masses
m1 and m2 joined by a massless spring of spring constant k
...
Calculate the angular frequency of
each mass
...
m1
m2
Solution : Let x1 and x2 be the displacements
of the blocks when pulled apart
...
Thus the
acceleration of m 1 is k (x 1 + x 2)/m 1 and Fig
...
8 : Oscillatory system of masses
attached to a spring
acceleration of m2 is k(x1 + x2)/m2
...
Thus the acceleration of the system is
a=
kx
k ( x1 + x2 )
=
µ
⎛ 1
1 ⎞
+
⎜
⎟
⎝ m1 m2 ⎠
where x = x1 + x2 is the extension of the spring and µ is the reduced mass of the system
...
299
299
MODULE - 4
Physics
Oscillations and Waves
ω =
k /µ
(13
...
Intext Questions 13
...
A small spherical ball of mass m is placed in contact with the sunface on a smooth
spherical bowl of radius r a little away from the bottom point
...
13
...
...
13
...
m
Fig
...
10
Fig
...
11
A cylinder of mass m floats vertically in a liquid of density ρ
...
Obtain an expression for the time period of its oscillations
(Fig
...
10)
...
3
...
13
...
The force constant of each band is k
...
13
...
13
...
11)
When t changes to t + ∆ t, y changes to y + ∆y
...
Then
y + ∆y = a sin ωt + a ω∆t cos ωt
...
(13
...
(13
...
12)
Simple Harmonic Motion
so that
v
Oscillations and Waves
∆y/∆t = ωa cos ωt
or
= ωa cost ωt
MODULE - 4
(13
...
Hence, the kinetic energy of the
oscillator at that instant of time is K = (1/2) mv2 = (1/2) ω2a2 cos2 ωt
(13
...
When the displacement is
y, the restoring force is ky, where k is the force
C
constant
...
PQ
We get a straight line graph as shown in Fig
...
12
...
As points
P and Q are close to each other, trapezium
y
O
B
MN
PQNM can be regarded as a rectangle
...
This area
Fig
...
12 : Graph between the
equals the work done against the restoring force
displacement y and the
restoring force ky
ky when the displacement changes by a small
amount ∆y
...
This work done against the conservative force is the potential energy U of the
2
oscillator
...
Therefore, substituting k = mω2 in above expression we get
U =
1
mω2y2
2
Further as y = a sin ωt, we can write
U =
1
mω2a2sin2ωt
2
(13
...
(13
...
16)
3
...
13
...
13
...
From
the graph it is evident that for y = 0, K = E
and U = 0
...
At the mean position, the
potential energy is zero but kinetic energy
is maximum
...
However, the
sum K + U = E is constant
...
3
1
...
2
...
13
...
As a result in each
oscillation some of its energy is dissipated as heat
...
The amplitude of oscillations of a pendulum in
air decreases continuously
...
To understand
damped oscillations perform activity 13
...
S
y (t)
G
B
(a)
t (s)
(b)
Fig
...
14 : Damped vibrations : (a) experimental setup; (b) graphical representation
302
Simple Harmonic Motion
MODULE - 4
Oscillations and Waves
Activity 13
...
(Fig
...
14(a)
...
Paste a millimetre scale (vertically) on the side
of the cylinder just opposite the pointer attached to the block
...
After each oscillation, note down the uppermost
position of the pointer on the millimetre scale and the time
...
Does the graph [Fig
...
14 (b)] show that the amplitude
decreases with time
...
Notes
13
...
3
Take a rigid horizontal rod fixed at both ends
...
13
...
The pendulums A and B are of equal lengths,
C
whereas C has a shorter and D has a longer
B
A
length than A and B
...
13
...
heavy bob
...
You will observe that after a few minutes, the
other three pendulums also begin to oscillate
...
of oscillators are
coupled, they transfer their energy
...
) You will note that the amplitude of A is larger
...
The pendulum B, which has a heavy
bob, transmits its vibrations to each of the pendulums A, C and D
...
The phenomenon is called forced oscillation
...
Both C and D are forced to oscillate with the frequency of B
...
This phenomenon
is known as resonance
...
Such oscillations are known as free
vibrations
...
When a body oscillates under the influence of an external periodic force, the oscillations
are called forced oscillations
...
303
303
MODULE - 4
Oscillations and Waves
Physics
with the frequency of the external force
...
The particular case of forced oscillations in which natural frequencies of the
driver and the driven are equal is known as resonance
...
Notes
Intext Questions 13
...
When the stem of a vibrating tuning fork is pressed against the top of a table, a loud
sound is heard
...
What is the cause of the loud sound
produced?
...
Why are certain musical instruments provided with hollow sound boards or sound
boxes?
...
Tacoma Narrows Suspension Bridge, Washington, USA collapsed during a storm
within six months of its opening in 1940
...
So it swayed the bridge with increasing
amplitude
...
The events of suspension bridge collapse also happened when groups of marching
soldiers crossed them
...
The factory chimneys and cooling towers set into oscillations by the wind and
sometimes get collapsed
...
You might have heard about some singers with mysterious powers
...
When they sing, the glasses of the window panes in the auditorium
are broken
...
3
...
When the frequency of the tuner matches
the frequency transmitted by the specific station, resonance occurs and the antenna
catches the programme broadcasted by that station
...
Oscillatory motion is to and fro motion about a mean position
...
Notes
Simple harmonic motion is to and fro motion under the action of a restoring force,
which is proportional to the displacement of the particle from its equilibrium position
and is always directed towards the mean position
...
Frequency is the number of vibrations completed by the oscillator in one second
...
Angular frequency is the rate of change of phase angle
...
Equation of simple harmonic motion is
y = a sin (ωt + φ0)
y = a cos (ωt + φ0)
or
where y is the displacement from the mean position at a time, φ0 is the initial phase
angle (at t = 0)
...
If,
however, an oscillatory system is driven by an external system, its vibrations are said
to be forced vibrations
...
Terminal Exercise
1
...
2
...
Which of the following functions represent (i) a simple harmonic motion (ii) a periodic
but not simple harmonic (iii) a non periodic motion? Give the period of each periodic
motion
...
The time period of oscillations of mass 0
...
Calculate the time period of oscillation of mass 0
...
3
...
What is phase angle? How is it related to angular frequency?
6
...
When is the magnitude of acceleration of a particle executing simple harmonic motion
maximum? When is the restoring force maximum?
8
...
Obtain an expression for the time period of a simple harmonic
oscillator in terms of mass and force constant
...
Obtain expressions for the instantaneous kinetic energy potential energy and the total
energy of a simple harmonic oscillator
...
Show graphically how the potential energy U, the kinetic energy K and the total energy
E of a simple harmonic oscillator vary with the displacement from equilibrium position
...
The displacement of a moving particle from a fixed point at any instant is given by
x = a cos ωt + b sin ωt
...
12
...
04m
...
13
...
Obtain an expression for its time period in terms of radius of the earth and the
acceleration due to gravity
...
Fig
...
16 shows a block of mass m = 2kg connected
to two springs, each of force constant k = 400Nm–1
...
05m from equilibrium
position and then released
...
13
...
Answers to Intext Questions
13
...
A motion which repeats itself after some fixed interval of time is a periodic motion
...
A periodic motion may or
may not be oscillatory but oscillation motion is perodic
...
(ii), (iv), (v);
3
...
(ii) Motion of a planet in its orbit
...
2eck
MODULE - 4
Oscillations and Waves
Your Progress 13
...
Return force on the ball when displaced a distance x from the equilibrium position is
mg sin θ = mg θ = mg x/r
...
2
...
Therefore ω =
αρg
and m = αpρ
...
Hence, ω2 = g/l or T = 2π
Notes
l/g
...
ω2 = k/m and hence v = 1/2π k / m
...
13
...
K
...
2
...
This work done is dissipated as
heat
...
13
...
When an oscillatory system called the driver applies is periodic of force on another
oscillatory system called the driven and the second system is forced to oscillate with
the frequency of the first, the phenomenon is known as forced vibrations
...
2
...
Therefore, this observation demonstrates forced vibrations
...
3
...
Since a large area is set into vibrations, the intensity of the note
produced increases and its duration decreases
...
3s
11
...
× 10 ms
π
14
...
14 s– 1 (b) 0
...
3 ms– 2 (d) 0
...
307
307
MODULE - 4
Physics
Oscillations and Waves
14
Notes
WAVE PHENOMENA
a stone is dropped
concentric
You would have noticed that whendepressions emergeinto still water in a pond,impact and
rings of alternate elevations and
out from the point of
spread out on the surface of water
...
Here the particles of water are
moving up and down at their places
...
We call it a wave
...
These can also be classified as longitudinal and trnsverse
depending on the direction of motion of the material particles with respect to the direction
of propagation of wave in case of mechnical waves and electric and magnetic vectors in
case of e
...
waves
...
Sound waves travelling through air make it possible for us to listen
...
In fact, wave phenomena is universal
...
V require us to understand wave
phenomena
...
Objectives
After studying this lesson, you should be able to :
explain propagation of transverse and longitudinal waves and establish the
relation v = vλ ;
write Newton’s formula for velocity of longitudinal waves in a gas and explain
Laplace’s correction;
discuss the factors on which velocity of longitudional waves in a gas depends;
explain formation of transverse waves on stretched strings;
derive the equation of a simple harmonic wave;
308
Wave Phenomena
explain the phenomena of beats, interference and phase change of waves on
the basis of principle of superposition
MODULE - 4
Oscillations and Waves
explain formation of stationary waves and discuss harmonics of organ pipes
and stretched strings;
discuss Doppler effect and apply it to mechanical and optical systems;
Notes
explain the properties of em waves, and
state wavelength range of different parts of em spectrum and their applications
...
1 Wave Propagation
From the motion of a piece of straw, you may think that waves carry energy; these do not
transport mass
...
Do you recall
the devastation caused by Tsunami, waves which hit Indonesia, Thailand, Sri lanka and
India caused by a deep sea quake waves of, 20m height were generated and were
responsible for huge loss of life
...
Activity 14
...
Hold
the free end in your hand and give it a jerk side–
ways
...
1(a)]
...
This kink is a wave of short duration
...
You will observe a train of pulses ravelling towards
(c)
the fixed end
...
14
...
14
...
(a) pulse on a slinky,
(b) transverse wave, and
(c) longitudinal Wave
There is another type of wave that you can generate
in the slinky
...
A pulse of compression
thus moves on the spring
...
These
are called longitudinal waves [Fig
...
1(c)]
...
1
...
2
...
It comprises a row
of spherical balls of equal masses, evenly spaced and connected together by identical
springs
...
H
...
in a direction perpendicular to the row of balls with a period T
...
The motion is
3
...
14
...
passed on from one ball to the next one by one
...
This means that in the interval T/8s,
the disturbance propagates from the particle at mark 1 to the particale at mark 2
...
In parts (a)—(i) in Fig
...
2 we have shown the instantaneous positions
of particles at all nine marked positions at intervals of T/8
...
) You will observe that
(i) At t = 0, all the particles are at their respective mean positions
...
The
first and ninth particles are about to move upward whereas the fifth particle is about to
move downward
...
The envelop joining the instantaneous positions
of all the particles at marked positions in Fig
...
2(a) are similar to those in Fig
...
2(i) and
310
Wave Phenomena
represents a transverse wave
...
The important point to note here is that while the wave moves along the string, all
particles of the string are oscillating up and down about their respective equilibrium
positions with the same period (T) and amplitude (A)
...
MODULE - 4
Oscillations and Waves
Notes
In a wave motion, the distance between the two nearest particles vibrating in the
same phase is called a wavelength
...
It is evident that time taken by the wave to travel a distance λ is T
...
14
...
Therefore, the velocity of the wave is
v =
λ
Distance
=
T
Time
(14
...
Therefore,
v = vλ
(14
...
Therefore, the phase change per unit distance
k =
2π
λ
(14
...
4)
We call k the propagation constant
...
But the phase change in time T is 2π hence
ω=
Dividing Eqn
...
3) by Eqn
...
4), we get an expression for the wave velocity :
v =
ω
2πv
=
k
2π/λ
or
v = vλ
Let us now explain how the logitudinal waves propagate
...
5)
14
...
2 Propagation of a Longitudinal Wave
(a)
Fig
...
3 : Graphical representation of a longtudinal wave
...
311
311
MODULE - 4
Oscillations and Waves
Notes
Physics
In longitudinal waves,
the displacement of
particles is along the
(b)
direction of wave
propagation In Fig
...
3,
the hollow circles
represent the mean
(c)
positions of equidistant
Fig
...
4 : Longitudinal waves on a spring are analogous
particles in a medium
...
The arrows show their
(rather magnified) longitudinal displacements at a given time
...
This is clear from the position
of solid circles, which describe instantaneous positions of the particles corresponding to
the heads of the arrows
...
For every arrow directed to the right, we draw a proportionate line upward
...
On drawing a
smooth curve through the heads of these lines, we find that the graph resembles the
displacement-time curve for a transverse wave
...
These represent regions of compression and
rarefaction
...
A sound wave propagating in air is very similar to the longitudinal
waves that you can generate on your spring (Fig
...
4)
...
14
...
3 Equation of a Simple Harmonic Wave in One Dimension
Y
O
–A
Y′
x
X
P
λ
Fig
...
5 : Simple harmonic wave travelling along x-direction
Let us consider a simple harmonic wave propagating along OX (Fig
...
5)
...
Let us
represent the displacement at t = 0 by the equation
y = a sin ωt
(14
...
Then
y = a sin (ωt – φ)
(14
...
Since phase change per unit distance is k, we can write φ = kx
...
(14
...
8)
Wave Phenomena
Oscillations and Waves
Further as ω = 2π/t and k = 2π/λ, we can rewrite Eqn (14
...
9)
In terms of wave velocity (v = λ/T), this equation can be expressed as
2π
(v t – x)
y = a sin
λ
Notes
(14
...
(14
...
However, if
the initial phase angle at O is φ0 , the equation of the wave would be
y (x,t) = a sin [(ωt – kx) + φ0]
(14
...
8)
(14
...
∆x = –
(x – x )
λ
λ 2 1
(14
...
Here the negative sign
indicates that a point positioned later will acquire the same phase at a later time
...
For the first wave,
phase φ , is given by
φ1 =
2π
2π
t1 – x
T
λ
φ2 =
2π
2π
t2 –
x
...
13(a)]
= 2π v ( ∆t)
3
...
1 : A progressive harmonic wave is given by y = 10–4 sin (100πt – 0
...
Calculate its (i) frequency, (ii) wavelength and (iii) velocity y and x are in metre
...
1 π ⇒ λ = 20 m
λ
(iii) v = vλ = 1000 ms–1
14
...
4 Transverse and Longitudinal Waves
We now consider transverse and longitudinal waves and summerise the difference between
them
...
(i)
Displacements of the particles are along
the direction of propagation
of the wave
...
Transverse waves can only be
transmitted in solids or on the
surface of the liquids
...
(ii)
Longitudinal waves give the
appearance of alternate compressions
and rarefaction moving forward
...
(iii)
(iv)
(iii)
(iv)
In case of longitudinal waves, the
graph only represents the
displacement of the particles at
different points at a given time
...
Longitudinal waves for propagation in a medium require volume
elasticity but transverse waves need modulus of rigidity
...
Intext Questions 14
...
State the differences between longitudinal and transverse waves?
...
Write the relation between phase difference and path difference
...
3
...
What is the phase difference between these two waves?
MODULE - 4
Oscillations and Waves
...
2 Velocity of Longitudinal and Transverse Waves
in an Elastic Medium
Notes
14
...
1 Newton’s Formula for Velocity of Sound in a Gas
Newton to derive a relation for the velocity of sound in a gaseous medium, assumed that
compression and rarefaction caused by the sound waves during their passage through the
gas take place under isothermal condition
...
Under such conditions, Newton agreed that
the velocity of sound wave in a gas is given by
v =
P
ρ
(14
...
01 × 105 Nm–2 and ρ = 1
...
On substituting these values in Eqn
...
15) we get
v = 1
...
29 = 280 ms–1
Clouds collide producing thunder and lightening, we hear sound of thunder after the lightening
...
By measuring the time interval between observing the lightening and hearing the sound,
the velocity of sound in air can be determined
...
The percent error in the value
predicted by Newton’s formula and that determined experimentally is
333 – 280
× 100%
333
= 16%
...
Obviously there is
something wrong with Newton’s assumption that during the passage of sound, the
compression and the rarefaction of air take place isothermally
...
2
...
(i)
Air is bad conductor of heat and
(ii)
Compression and rarefactions caused by the sound are too rapid to permit heat to
flow out during compression and flow in during rarefaction
...
315
315
MODULE - 4
Physics
Oscillations and Waves
v =
Hence,
γP
ρ
(14
...
4
...
4 ×1
...
29
= 333ms–1
This value is very close to the experimentally observed value
...
2
...
This result shows that
v α T
=
⇒
(14
...
⎝ 2 × 273 ⎠
~ 333 +
333
t
546
~ 333 + 0
...
17b)
Note that for small temperature variations, velocity of sound in air increases by
0
...
(ii) Effect of pressure
When we increase pressure on a gas, it gets compressed but its density increases in the
same proportion as the pressure i
...
P/ρ remains constant
...
316
Wave Phenomena
MODULE - 4
Oscillations and Waves
(iii) Effect of density
If we consider two gases under identical conditions of temperature and pressure, then
v α
1
ρ
If we, compare the velocities of sound in oxygen and hydrogen, we get
ρhydrogen
voxygen
v hydrogen
=
ρoxygen
=
M hydrogen
M oxygen
=
Notes
2
1
=
4
32
This shows that velocity of sound in hydrogen is 4 times the velocity of sound in oxygen
under identical conditions of temperature and pressure
...
You will discover answer to this question in the next sub–section
...
Example 14
...
T
...
Solution : We know that
v
=
v0
T
=2=
m
T
273
On squaring both sides and rearranging terms, we get
∴
T = 273 × 4 = 1092k
14
...
4 Velocity of Waves in Stretched Strings
The velocity of a transverse wave in a stretched string is given by
v=
F
m
(14
...
The velocity of
longtudinal waves in an elastic medium is given by
v=
E/ρ
(14
...
It may be pointed out here that since the value of elasticity is more in
solids, the velocity of longitudinal waves in solids is greater than that in gases and liquids
...
Intext Questions 14
...
What was the assumption made by Newton in deriving his formula?
...
What was wrong with Newton’s formula?
...
317
317
MODULE - 4
Oscillations and Waves
Physics
3
...
61 ms–1
...
4
...
Notes
5
...
6
...
Write the relation between n, λ, F and m? Further if λ = 2l,
what would be the relation between n, l, F and m?
...
3 Superposition of Waves
Suppose two wave pulses travel in opposite directions on a slinky
...
Activity 14
...
14
...
The crests are moving in the opposite directions
...
14
...
Thereafter,
they continue to move in the same direction in which they were moving before crossing
each other
...
14
...
Now produce one crest and one trough on the slinky as
shown in Fig
...
6(d)
...
They meet [Fig
...
6(e)], overlap and then
separate out
...
Repeat
the experiment again and observe carefully what happens
at the spot of overlapping of the two pulses [(Fig
...
6(b)
and (e)]
...
We
may summarize this result as : At the points where the
two pulses overlap, the resultant displacement is the
vector sum of the displacements due to each of the
two wave pulses
...
This activity demonstrates not only the principle of
superposition but also shows that two or more waves
318
(a)
(b)
(c)
(d)
(e)
(f)
Fig
...
6 : Illustrating principle
of superpositionof
waves
Wave Phenomena
can traverse the same space independent of each other
...
This important property of the waves enable us to tune to a particular
radio station even though the waves broadcast by a number of radio stations exist in space
at the same time
...
14
...
1 Reflection and Transmission of Waves
MODULE - 4
Oscillations and Waves
Notes
We shall confine our discussion in respect of mechanical waves produced on strings and
springs
...
Activity 14
...
14
...
Keeping the slinky
horizontal, give a jerk to its free end so as to produce a transverse wave pulse which
travells towards the fixed end of the slinky (Fig
...
7(a))
...
As it bounces back, the crest becomes a trough travels
back in the opposite direction
...
The equal and opposite reaction not only reverses the
direction of propagation of the wave pulse but also reverses the direction of the displacement
of the wave pulse (Fig
...
7(b))
...
The wave pulse moving in the opposite direction is called
the reflected wave pulse
...
Fig
...
7 :
Reflection from a denser
medium : a phase reversal
...
14
...
For this we perform
another activity
...
4
Suspend a fixed rubber tube from a rigid support
(Fig
...
8 a)
...
On reflection from the
Fig
...
9 : Longitudinal waves are free end, the wave pulse travels upward but
reflected from a denser without any change in the direction of its
medium without change of displacements i
...
crest returns as crest
...
(Note that air is rarer than the rubber tube
...
319
319
MODULE - 4
Oscillations and Waves
Notes
Physics
displacement of the wave pulse
...
You may now raise the question : Do longitudinal waves also behave similarly? Refer to
Fig
...
9, which shows a row of bogies
...
The buffer spring between the engine E and the first bogie gets
compressed and pushes bogie B1 towards the right
...
As this compressed spring expands, the spring between the 1st and the 2nd bogie
gets compressed
...
In this manner the compression arrives at the last buffer spring in contact with
the fixed stand D
...
As a result of this, the buffer spring between the
next two bogies on left is compressed
...
Thus, a compression returns as a
compression
...
In this mechanical model, the
buffer spring and the bogies form a medium
...
Thus, when reflection takes place from a denser medium, the longitudinal waves
are reflected without change of type but with change in sign
...
By ‘change of type’ we mean that rarefaction is reflected back as
compression and a compression is reflected back as rarefaction
...
3
1
...
2
...
3
...
What
happens (i) when the waves are in the same phase? (ii) the waves are in the opposite
phases?
...
What happens when a transverse wave pulse travelling along a string meets the
fixed end of the string?
...
What happens when a wave pulse travelling along a string meets the free end of the
string?
...
What happens when a wave of compression is reflected from (i) a rarer medium (ii)
a denser medium?
...
4 Superposition of Waves Travelling in the Same Direction
Superposition of waves travelling in the same direction gives rise to two different phenomena
(i) interference and (ii) beats depending on their phases and frequencies
...
14
...
1 Interference of waves
Notes
Let us compute the ratio of maximum and minimum intensities in an interference pattern
obtained due to superposition of waves
...
These waves are
represented by the equations
y 1 = a1 sin (ωt – kx)
y 2 = a2 sin [(ωt – kx) + φ]
and
where ω= 2π/T is angular frequency and k =
2π
is wave number
...
According to the principle of superposition, the resultant
displacement at the given location at the given time is
y = y1 + y2 = a1 sin (ωt – kx) + a2 sin [(ωt – kx) + φ]
If we put (ωt – kx) = θ, then
y = a1 sinθ + a2 sin (θ + φ)
= a1 sinθ + a2 sinθ cosφ + a2 sinφ cosθ
Let us put
a2 sinφ = A sinα
a1 + a2 cosφ = A cos α
and
Then
y = A cosα sinθ + A sin αcosθ
= A sin (θ + α)
Substituting for θ we get
y = A sin [(ωt – kx) + α]
Fig
...
10 : Calculating
resultant
amplitude A
Thus, the resultant wave is of angular frequency ω and has an amplitude A given by
A 2 = (a1 + a2 cosφ)2 + (a2 sinφ)2
2
2
= a12 + a2 cos2φ + 2a1a2 cosφ + a2 sin2φ
2
A 2 = a12 + a2 + 2 a1a2cosφ
(14
...
(14
...
If path
difference, between the two waves corresponds to phase difference φ , then
φ=
2πp
2π
, where
is the phase change per unit distance
...
321
321
MODULE - 4
Physics
Oscillations and Waves
When the path difference is an even multiple of
λ
λ
, i
...
, p = 2m , then phase difference
2
2
is given by φ = (2π/λ) × (2m λ/2) = 2mπ
...
(14
...
As intensity of wave at a given position is directly proportional to the square of its amplitude,
we have
Imax α ( a1 + a2)2
When p = (2m + 1) λ/2, then φ = (2m + 1) π and cosφ = –1
...
(14
...
Then Imin α ( a1 – a2)2
Imax
(a1 + a2 )2
Imin = (a1 − a2 )2
Thus
(14
...
These results show that interference is
essentially redistribution of energy in space due to superposition of waves
...
4
...
Let us now investigate what would be the outcome of
superposition of waves of nearly the same frequency
...
Activity 14
...
Let us name them as A and B
...
Now sound them together by a rubber hammer
...
You will observe that the
intensity of sound alternately becomes maximum and minimum
...
One alternation of a maximum and a
minimum is one beat
...
of beats increase
...
The
reason is that our ear is unable to hear two sounds as separate produced in an interval less
than one tenths of a second
...
(a) Production of beats : Suppose we have two tuning forks A and B of frequencies N
and N + n respectively; n is smaller than 10
...
That is, B completes n more vibrations in one second than
the tuning fork A
...
in (1/n) s
...
Suppose at t = 0, i
...
initially, both the tuning
322
Wave Phenomena
forks were vibrating in the same phase
...
If A sends a wave of compression then
2n
B sends a wave of rarefaction to the observer
...
After (1/n)s, B would gain one complete vibration
...
The
intensity observed would become maximum
...
This process
would continue
...
Thus, the number of beats heard in one second equals the difference in the
frequencies of the two tuning forks
...
The beat frequency is n and beat period is 1/n
...
Thus after
(a)
Notes
v1 = 12Hz
(b)
v2 = 10Hz
(c)
Beats
Fig
...
11 : (a) Displacement time graph of frequency 12 Hz
...
Superposition of the two waves produces 2 beats per second
...
Divide it into 12 equal parts of 1 cm
...
5 cm
...
On the line (b) draw 10 wavelengths each of length 1
...
5 cm
...
(c) represents the resultant wave
...
11 is not actual waves but the displacement time graphs
...
The number of beats produced in one second
is ∆v
...
Example 14
...
Determine frequency of the unknown fork
...
Example 14
...
What is the amplitude ratio of the superposing waves?
Solution :
Imax ⎛ a1 + a2 ⎞
=
Imin ⎜ a1 − a2 ⎟
⎝
⎠
2
2
a2
⎛1+ r ⎞
⇒ 9=⎜
⎟ , where r =
a1
...
323
323
MODULE - 4
Physics
Oscillations and Waves
You can easily solve it to get r =
1
, i
...
, amplitude of one wave is twice that of the other
...
4
Notes
1
...
2
...
What will you observe?
...
Two waves of frequencies v and v + ∆v are supperposed, what would be the
frequency of beats?
...
Two tuning forks A and B produce 5 beats per second
...
What was the frequency of
A before loading if that of B is 512 Hz
...
...
5 Superposition of Waves of Same Frequency Travelling
in the Opposite Directions
So far we have discussed superposition of collinear waves travelling in the same direction
...
Superposition of
progressive waves of same wavelength and same amplitude travelling with the same
speed along the same line in a medium in opposite direction gives rise to stationary or
standing waves
...
14
...
1 Formation of Stationary (Standing) Waves
To understand the formation of stationary waves, refer to Fig
...
12 where we have
shown the positions of the incident, reflected and resultant waves, each after T/4s, that is,
after quarter of a period of vibration
...
14
...
Hence the
resultant displacement at each point is zero
...
At t = T/4s [Fig
...
12(ii)], the incident wave has advanced to the right by λ/4, as
shown by the shift of the point P and the reflected wave has advanced to the left by
λ/4 as shown by the shift of the point P′
...
14
...
The resultant displacement
at each point is maximum
...
(v)
N2
N3
A2
Oscillations and Waves
A3
N4
x (i)
P
P′
x (ii)
Notes
(3) t=T/2P
(2) t=T/4 (1) t = 0
A1
P
P′
x (iii)
(4) t=3T/4
(iii) At t = T/2s [Fig
...
12(iii)], the
incident wave advances a
distance λ/2 to the right as shown
by the shift of the point P and the
reflected wave advances a
distance λ/2 to the left as shown
by the shift of the point P′
...
N1
P′
P
x (iv)
P P′
(5) t=4T/4
by the thick continuous curve
...
Hence the particle
velocity at each point is zero and
the strain is maximum
MODULE - 4
x (v)
λ/4
N1
λ/4 P′ λ/4
A1
N2
λ/4 P λ/4
A2
N3
λ/4
A3
N4
Fig
...
12 : Showing formation of stationary
waves due to superposition of two
waves of same wave length, same
amplitude travelling in opposite
direction along the same line
...
14
...
The resultant is a straight line (shown by an
unbroken thick line)
...
Note that
at points N1, N2, N3 and N4, the amplitude is zero but the strain is maximum
...
These
points are called antinodes;
the distance between two successive nodes or between two, successive antinode is
λ/2;
the distance between a node and next antinode is λ/4;
the time period of oscillation of a stationary wave equals the time period of the two
travelling waves whose superposition has resulted in the formation of the stationary
wave; and
the energy alternately surges back and forth about a point but on an average, the
energy flow past a point is zero
...
They are called stationary
waves, because the wave form does not move forward, but alternately shrinks and
3
...
The energy merely surges back and forth and on an average, there is no net
flow of energy past a point
...
5
...
The equation of the reflected
wave is therefore,
y 2 = a sin (ωt – kx)
Thus, owing to the superposition of the two waves, the resultant displacement at a given
point and time is
y = y1 + y2
= a sin (ωt – kx) – a sin (ωt – kx)
Using the trigonometric identity
...
20)
Let us put –2a sin kx = A
...
(14
...
The amplitude of the resultant wave, oscillates in
space with an angular frequency ω, which is the phase change per metre
...
Hence A = 0,
The points where the amplitude is zero are referred to as nodes
...
Obviously the spacing between two nearest points is
λ/2
...
Hence A is maximum
...
Obviously the spacing
between two such neighbouring points is λ/2
...
It may be pointed out here that at nodes, the particle velocity is zero and at antinodes,
particle velocity ∆y/∆t is maximum
...
The energy merely surges back and forth
...
326
Wave Phenomena
14
...
3 Distinction between Travelling and Standing Waves
MODULE - 4
Oscillations and Waves
Let us summarise the main differences between travelling and standing waves
...
Particular conditions of the medium
namely crests and troughs or
compressions and rarefactions appear
to travel with a definite spped depending
on density and elasticity (or tension) of
the medium
...
Each particule or element
of the medium vibrates to and fro like
a pendulum
...
The amplitude of vibration of all the
particles is the same
...
3
...
At nodes the particle velocity is zero
and at antinodes it is maximum
...
Energy is transferred from particle to
particle with a definite speed
...
5
...
In a stationary wave the maximum
velocity is different at different points
...
But all the particles attain
their respective maximum velocity
simultaneously
...
6
...
7
...
Notes
Antinodes are points of no change of
density but at nodes there is maximum
change of density
...
5
1
...
...
What is the distance between two successive nodes, and between a node and next
antinode?
...
Pressure nodes are displacement antinodes and pressure antinodes are displacement
nodes
...
...
Stationary waves of frequency 170Hz are formed in air
...
3
...
6 Characteristics of Musical Sound
The characteristics of musical sounds help us to distinguish one musical sound from another
...
We will now discuss these briefly
...
6
...
It is a subjective quantity which cannot be measured by an instrument
...
However, there does not exist any one-to-one correspondence
between the two
...
But a dull, grave
and flat note is said to be of low pitch
...
On the other hand, the buzzing of mosquito, though of low intensity, is of high pitch
...
6
...
The intensity of waves is the average amount of energy transported by the wave per
unit area per second normally across a surface at a given point
...
As such, logarithmic scale rather than
arithmetic intensity scale is more convenient
...
β = 10 log I/I0
(14
...
This value
corresponds to the faintest sound that can be heard
...
If the intensity of a sound wave equals I0 or 10–12 Wm–2, its
intensity level is then I0 = 0 db
...
The threshold audibility at any frequency is the minimum
intensity of sound at that frequency, which can be detected
...
A sone is the loudness experienced
by a listener with normal hearing when 1kilo hertz
tone of intensity 40db is presented to both ears
...
14
...
This is a graph of auditory area of
good hearing
...
•
•
328
The lower part of the curve shows that the
ear is most sensitive for frequencies
between 2000 Hz to 3000 Hz, where the
threshold of hearing is about 5db
...
14
...
At intensities above those corresponding to
and threshold of feeling
Wave Phenomena
MODULE - 4
Oscillations and Waves
the upper part of the curve, the sensation changes from one of hearing to
discomfort and even pain
...
•
Loudness increases with intensity, but there is no definite relation between the
two
...
•
The height of the upper curve is constant at a level of 120 db for all frequencies
...
•
Distance between the observer and the Source : I α 1/r2 where r is the
distance of the observer from the source (provided it is a point source)
...
Intensity is directly proportional to the square of frequency of the wave (I
α v2)
...
Intensity is directly proportional to the density of the medium (I α ρ)
...
6
...
No
instrument, except a tuning fork, can emit a pure note; a note of one particular frequency
...
may also be produced
...
The resultant wave form of the emitted waves determines the quality
of the note emitted
...
It depend on
the resultant wave form
...
6
...
A wooden or metal pipe producing musical
sound is known as organ pipe
...
If both the ends of the
pipe are open, we call it an open pipe
...
When we blow in gently, almost a pure tone is heard
...
But, when we blow hard, we also hear notes of frequencies which
are integral multiple of the frequency of the fundamental note
...
These frequencies are called
overtones
...
•
At the open end of the pipe, the change in density must be zero since this end is in
communication with atmosphere
...
(a) Open pipe : The simplest mode of vibrations of the air column called fundamental
3
...
14
...
At each end, there is
an antinode and between two antinodes, there is a
node
...
14
...
One more node and one more
antinode has been produced
...
Notes
A
λ/2
N
A
N
λ/4
A
(b)
N
A
(c)
Fig
...
14 : Harmoniscs of an open
Organ pipe
...
To get the second harmonic
you have to blow harder
...
14
...
Thus, in this case
l =
λ λ λ λ
+
+
+
2
4
2
4
λ=
A
2l
3
A
A
N
λ/4
λ/4
λ/2
L
λ/2
A
N
A
N
N
N
(a)
(b)
λ/2
(c)
Fig
...
15 : Harmonics of a closed organ pipe
...
Therefore, the frequency of the note emitted is
n3 =
330
v
λ
=
3v
= 3n1
2l
Wave Phenomena
The note produced is called the 3rd harmonic or 2nd overtone
...
14
...
There is an antinode at the open end and a node at the
closed end
...
On blowing harder one more node and
antinode will be produced (Fig
...
15(b))
...
14
...
The wavelength of the note produced is then given by
l =
λ λ λ 5λ
4l
+
+
+
or λ =
2
2
4
4
5
The frequency of the note emitted then will be
n3 =
v
λ
=
5v
= 5n1
4l
The note produced is called the second overtone or the 5th harmonic of the fundamental
...
In closed pipe, the even order harmonics are missing
...
5 : Two organ pipes – one open and the other closed – are of the same
length
...
Frequency of open pipe v / 2l
Solution : Frequency of closed pipe = v /4l = 2
∴ Frequency of note produced by open pipe = 2 × frequency of fundamental note produced
by closed pipe
...
6
1
...
3
...
What is that characteristic of musical sounds which enables you distinguis between
two notes of the same frequency, and same intensity but sounded by two different
instruments?
...
Notes
Name the characteristic of sound which helps you identify the voice of your friend
...
4
...
5
...
...
What will be the effect of temperature, if any, on the frequency of the fundamental
note of an open pipe?
...
Noise is also one of the by-products of industrialisation and misuse
of modern amentities provided by science to human beings
...
Table 14
...
It increases the rate of heart beat and causes dilation of the pupil of eye
...
It causes emotional disturbance, anxiety and nervousness
...
Oscillations and Waves
It causes impairment of hearing
...
2
...
Notes
(b) Methods of Reducing Noise Pollution
1
...
2
...
3
...
4
...
5
...
6
...
7
...
8
...
Shock Waves
When a source of waves is travelling faster than the sound waves, shock waves are
produced
...
It may be pointed out that
the object which moves with a speed greater than the speed of sound is itself a
source of sound
...
7 Electromagnetic Waves
You know that light is an e
...
wave
...
A
brief description of em waves is given below
...
7
...
m
...
m
...
(i)
e
...
waves are transverse in nature
(ii) They consist of electric (E) and magnetic fields (B) oscillating at right angles to each
other and perpendicular to the direction of propagation (k)
...
[see figures
14
...
333
333
MODULE - 4
Physics
Oscillations and Waves
Notes
Fig
...
16 : Electrical and Magnetic fields in em waves
(iii) They propagate through free space (in vacuum) with a uniform velocity =
1
µ0 ε0
= 3 × 108 ms–1 = c (velocity of light)
...
µr) and
permittivity ε (= ε0
...
Maxwell’s theory placed no restriction on possible wavelengths for e
...
waves and
hence e
...
waves of wavelengths ranging from 6 × 10–13 m have been successfully
produced
...
The whole range of e
...
waves from very long to very short
wavelengths constitutes the electromagnetic spectrum
...
Through his equations of
electromagnetic principles he showed that they implicitly indicated
the existence of em waves which travelled with the speed of
light, thus relating light and electromagnetism
...
He developed a statistical
theory of heat
...
14
...
2 Electromagnetic Spectrum
Maxwell gave the idea of e
...
waves while Hertz, J
...
Bose, Marconi and others
successfully produced such waves of different wavelengths experimentally
...
m
...
334
Wave Phenomena
Electromagnetic waves are classified according to the method of their generation and are
named accordingly
...
m waves is also observed
...
m
...
It is important to remember that the
physical properties of e
...
waves are determined by the frequencies or wavelengths
and not by the method of their generation
...
m
...
MODULE - 4
Oscillations and Waves
Notes
There is no sharp dividing point between one class of e
...
waves and the next
...
c
...
m
...
These weaves have the lowest frequency
...
3m to 106 m ⎫
⎪
⎪
⎬ : Radio waves are generated when charges
Radio Waves ⎨
9
⎪v = 10 Hz to 300Hz ⎭
⎪
⎩
are accelerated through conducting wires
...
⎧λ = 10 –3 m to 0
...
Because of their short wavelengths, they are well suited for
the radar system used in aircraft navigation, T
...
communication and for studying the
atomic and molecular properties of matter
...
It is suggested that solar energy could be harnessed by beaming
microwaves to Earth from a solar collector in space
...
3 × 10 Hz to 3 × 10 Hz ⎪
⎩
⎭
waves, are produced by hot bodies and molecules
...
The temperature of the body, which absorbs these radiations, rises
...
These are detected by a thermopile
...
m
...
5 × 10 Hz to 4
...
It forms a very small
portion of the whole electromagnetic spectrum
...
When an electron-jumps from
outer orbit to inner orbit of lower energy, the balance of energy is radiated in the form
of visible radiation
...
Human eye is most sensitive
3
...
Light is the basis of our communitation with the
world around us
...
5 × 10 Hz ⎭
⎪
⎩
Notes
radiations, which is the main cause of suntans
...
e
...
This ozone layer then radiates out the absorbed energy as heat radiations
...
These ultraviolet rays are used in
killing the bacteria in drinking water, in sterilisation of operation theatres and also in
checking the forgery of documents
...
5 × 10 Hz to 7
...
X-rays
find their important applications in medical diagnostics and as a treatment for certain
forms of cancer
...
X-rays are also used in study of crystal-structure
...
⎧λ = 6 × 10 –17 m to 10 –10 m ⎪
⎫
⎪
⎬ : These are emitted by radioactive
(viii) Gamma rays ⎨
24
18
⎪v = 5 × 10 Hz to 3 × 10 Hz ⎪
⎩
⎭
nuclei such as cobalt (60) and ceasium (137) and also during certain nuclear reactions
in nuclear reactors
...
Thick sheets of lead are used to shield the objects from
the lethal effects of gamma rays
...
m
...
Thus gamma
λ ⎠
⎝
rays are the most energetic and penetrating e
...
waves, while the power frequencies,
and the A
...
radio waves are the weakest radiations
...
They are detected by Geiger tube or scintillation
counter
...
For example, while whole of the human body is opaque to visible
light, human tissues are transparent to X-rays but the bones are relatively opaque
...
336
Wave Phenomena
MODULE - 4
Oscillations and Waves
Notes
Fig
...
17 : Electromagnetic spectrum
Intext Questions 14
...
Fill in the blanks:
(i)
...
(ii) Human eye is most sensitive to
...
(iii)
...
(iv)
...
...
Which of the e
...
waves are more energetic?
(i) Ultraviolet or infrared
...
3
...
m
...
4
...
5
...
m
...
14
...
You will note that the pitch is higher when the engine approaches
but is lower when the engine moves away from you
...
3
...
Let v be velocity of the sound waves relative to the medium, (air), vs velocity of the
source; and vo velocity of the observer
...
J
...
, 29, 1803 in a family of stone mesons
...
So on recommendation of the professor of
mathematics at Salzburg Lycousin, he was sent to Vienna
Polytechnic from where he graduated in 1825
...
He could think of marrying in 1836 only when he got a
permanent post at the technical secondary school at prague
...
But he pushed his
way through all odds and finally got succes in getting the position of the first director
of the new Institute of Physics at Vienna University
...
The RADAR, the SONAR,
the idea of expanding universe there are so many developments in science and technology
which owe a lot to Doppler effect
...
It is important to note that the wave originated at a moving source does not affect the
speed of the sound
...
The wave forgets the
source as it leaves the source
...
Let us first consider the effect of motion of the source
...
In this time, the source moves a distance vs
...
Thus length of each wave decreased to
v – vs
n
λ′ =
...
22)
S
A
v
(a)
vs
S
A
(v – vs)
(b)
Fig
...
18 : Crowding of waves when source is moving
338
Wave Phenomena
Now let us consider the effect of motion of the observer
...
But in the mean
time, the observer moves from O to O′
...
The number of the waves passing across
the observer in one second is therefore,
n′ = (v – v0)/λ′
(14
...
14
...
(14
...
24)
where n′ is the observed frequency when both observer and source are moving in the
direction from the source to the observer
...
(14
...
Similarly , v0 and vs are taken positive if these are in the direction
of v and vice versa
...
In particular, it led us to the concept of expansion of the universe
...
Example 14
...
If this shift, called the red shift, is 0
...
Solution : In this case, the source of waves is the star
...
We have shown that in such a case
λ′ =
But n = v /λ Therefore, λ′
=
v – vs
n
v – vs
v/λ
=λ
(v – v s )
v
⎛
= λ ⎜1 –
⎝
vs ⎞
⎟
v⎠
On rearranging terms, we can write
3
...
032/100
...
032/100) = – 9
...
λ
The negative sign shows that the star is receding away
...
8
1
...
0kHz
...
Calculate the
frequency of the sound reflected by the sonar
...
...
An engine, blowing a whistle of frequency 200Hz moves with a velocity 16ms–1
towards a hill from which a well defined echo is heard
...
Velocity of sound in air is 340ms–1
...
Constancy of Speed of Light
Aristotle, believed that light travels with infinite velocity
...
Feazeu, Focult, Michelson
and many other scientists carried out experiments to determine the speed of light in
air with more and more precision
...
One of the postulates was the constancy of speed of light in
vacuum, irrespective of the wavelength of light, the velocity of the source or the
observer
...
However, the Autralian researcher Barry Setterfield and Trevn Norwah have studied,
the data of 16 different experiments on the speed of light in vacuum, carried out over
the last 300 years, by different scientists at different places
...
If this hypothesis is sustained and
coroborated by experiements, it will bring in thorough change in our world view
...
340
Wave Phenomena
MODULE - 4
Oscillations and Waves
What You Have Learnt
The distance between two nearest points in a wave motion which are in the same
phase is called wavelength
...
Notes
The energy transmitted per second across a unit area normal to it is called intensity
...
Velocities of transverse wave and
longitudinal waves is given by v =
T / m and v = E / ρ respectively
...
But there is no phase
reversal on reflection from a rarer medium
...
Superposition of two colliner waves of
same frequency but differing phases, when moving in the same direction results in
redistribution of energy giving rise to interference pattern
...
In such waves, waveform does not move
...
It is, therefore, obvious that between two nodes, there is an antinode
and between two antinodes there is a node
...
Intensity level is defined by the equation β = 10log (I/I0), where I0 is an arbitrarily
chosen reference intensity of 10–12 W m–2
...
db)
Quality of a note is the characteristic of musical sounds which enable us to distinguish
two notes of the same pitch and same loudness but sounded by two different
instruments
...
Light is an e
...
wave with wavelength in the range 4000 Å – 7500 Å
...
m
...
e
...
waves are used for wireless radio communication, TV transmission, satellite
communication etc
...
How will you define a wave in the most general form?
2
...
341
341
MODULE - 4
Physics
Oscillations and Waves
(ii) longitudinal wave
...
3
...
Derive the equation of a simple harmonic wave of angular frequency of (i) transverse
(ii) longitudinal waves
...
What are the essential properties of the medium for propagation of (i) transverse
waves (ii) longitudinal waves
...
Derive an expression for the intensity of the wave in terms of density of the medium,
velocity of the wave, the amplitude of the wave and the frequency of the wave
...
Write Newton’s formula for the velocity of sound in a gas and explain Laplace’s
correction
...
When do two waves interfere (i) constructively (ii) destructively?
9
...
Show using trigonometry that when two simple harmonic waves of the same angular
frequency ω and same wavelenght λ but of amplitudes a1 and a2 are superposed, the
resultant amplitude is A =
2
2
a2 + a2 + 2a1 a2 cos θ , where θ is the phase difference
between them
...
What are beats? How are they formed? Explain graphically
...
Discuss graphically the formation of stationary waves
...
12
...
13
...
What are the characteristics of musical sounds
...
15
...
What is meant by quality of sound? Explain with examples?
17
...
Show that an open pipe is richer in harmonics
...
Show that (i) the frequency of open organ pipes
...
19
...
State the causes of noise pollution, its harmful effects and methods of minimising it
...
Explain Doppler’s effect and derive an expression for apparant frequency
...
22
...
Calculate the velocity of sound in a gas in which two waves of wavelengths 1
...
01m produce 10 beats in 3 seconds
...
What will be the length of a closed pipe if the lowest note has a frequency 256Hz at
20C
...
25
...
Calculate the
frequency of the waves as perceived by the observer when (a) the source and the
observer are stationary, (b) the source is moving with a velocity of 50ms–1 towards
the observer, and (c) the source is moving with a velocity of 50ms–1 away from the
observer
...
Notes
26
...
m
...
27
...
m
...
Give the range of wavelengths of the following e
...
waves:
(i) Radio Waves (ii) Microwaves : (iii) Ultraviolet; (iv) x-rays
...
How are x-rays produced?
30
...
m
...
Fill in the blanks
...
(ii) ___________ are more harmful to our eyes than x-rays
...
(iv) Infra red rays are less energies than_______________
(v) In an e
...
wave propagating along z-direction, if the E field oscillates in the X,Z
plane then the B field will oscillate in the _______________ plane
...
m
...
H
(vii) The frequency range of F
...
band is ________________
...
V
...
Answers to Intext Questions
14
...
See section 14
...
4
...
If p be the path difference, then the phase difference is θ =
3
...
3
...
2
1
...
Newton assumed that isothermal conditions instead of adiabatic conditions for sound
propagation
...
3570C
...
v=
6
...
Therefore, n =
T
m
1
λ
T
m
Further, for the simplest mode of vibration, at the two ends of the string, there are nodes
and in between the two nodes is an antinode
...
If the string vibrates in p segments, the λ = p l/2 or λ = 2l/p
...
m
14
...
14
...
25/9
...
Beats with frequency 4Hz are produced
...
Frequency of beat is ∆v
...
517, on loading the frequency of A decreases from 517 to 507
...
5
1
...
2
...
4
...
14
...
2
...
Timbre
4
...
344
Pitch increases with increase in frequency
...
Wave Phenomena
For an open pipe l = λ/2
...
MODULE - 4
Oscillations and Waves
Comparing (i) and (ii) we find that n′ = 2n
6
...
As v increases with increase in temperature n also increases
...
7
(i) microwaves
...
(iv) X – rays
...
2
...
3
...
Ozone
...
Perpendicular to each other
...
8
1
...
1450 – 100
1450
135
× 10 = 37
...
145
n′ = 200 ×
= 200 ×
340 + 16
340 – 16
356
= 220 Hz
...
337 ms–1
24
...
25
...
3
...
Do not send your assignment to NIOS
1
...
(b) y = sin
...
ωt
...
ωt + cos ωt
...
Four simple pendulum A, B, C and D are suspended from the same support
...
Which two of these pendulums will oscillate with the same frequency
...
A mass m when made to oscillate on a spring of force constant k oscillates
with a frequency v
...
What is the new frequency of oscillation of mass m
...
Give an example of a motion which is periodic but not oscillatory
...
Draw a graph showing the variation of velocity of sound in air with pressure
...
Is there a deviation in the direction of propagation of a sound wave in passing from air to water?
Explain
...
What happens when a transverse wave pulse travelling on a string meets the fixed end of the string?(1)
8
...
9
...
For simplicity, the sense of rotation may be taken to be anticlockwise
...
(2)
10
...
Find
the ratio of maximum and minimum intensities in the interference pattern
...
Two tuning forks A and B are marked 480 hz each
...
What can you say about the frequency marked on the tuning forks
...
(a) Name the em waves used in aircraft navigation by radar?
(b) Which gas in atmosphere absorbs u-v radiation?
(2)
13
...
Using the formula explain why the speed of sound
in air (a) increases with temperature (b) increases with humidity
...
A transverse harmonic wave on a string is described by
y(x1t) = 3
...
018x)
find (i) amplitude of particle velocity
...
(4)
15
...
If the waves strike a water surface, find the
difference in the wave lengths of transmilted sound and reflected sound
...
(4)
16
...
The mass of the wire is 3
...
0 × 10-2 kg m-1
...
A pipe 20 cm long is closed at one end
...
(4)
18
...
(ii) the shape of a pulse get distorted during propagation in a dispersive medium
...
(iv) a note played on voilin and sitar has the same frequency but the two may still be distinguished from
each other
...
Discuss the applications of doppler effect in measuring
(i) the velocity of recession of stars (ii) velocity of enemy boat by SONAR
...
The transverse displacement of a string of length 1
...
03 kg which is clamped at both ends, is
given by
(5)
⎛ 2 πx ⎞
y = 0
...
(i) Does it represent travelling wave or stationary wave?
(ii) Interpret the wave as a result of superposition of two waves
...
3
...
Every event that occurs in the natural world has some features that can
be viewed in these terms
...
Physics lies behind all technological advancements, such as,
computer, internet, launching of rockets and satellites, radio and T
...
It also finds
applications in such simple activities of men as lifting a heavy weight or making a long jump
...
Keeping in view the issues highlighted in the National Curriculum Framework (NCF) for School Education,
present Physics curriculum has been so designed that it not only focusses on the basic concepts of Physics but
relates them to the daily life activities
...
The basic themes of Physics which would be of interest to all, particularly to
those who are interested in pursuing Physics as a career in life have been selected to form core content of the
curriculum
...
Though mathematics is basic to the understanding of most of the problems of physics, in the present course,
stress has been given to avoid rigour of mathematics like intergration and differentiation
...
Course Objectives
The basic objectives of the sr
...
As a part of this process, the course also aims at developing the following abilities in the learner:
•
experimental skills like taking observations, manipulation of equipment, and communicative skills such as
reporting of observations and experimental results;
•
problem solving ability e
...
Course Structure
The physics curriculum at sr
...
(i)
The theoretical part of the Physics curriculum includes two parts – core modules and optional modules
...
Core modules : The core modules comprise of the essential concepts and phenomena of physics which
348
a student at this level should know
...
Core Modules
Marks
Minimum Study
Time (hours)
...
Motion, Force and Energy
45
2
...
Thermal Physics
08
25
4
...
Electricity and Magnetism
14
45
6
...
Atoms and Nuclei
07
25
8
...
14
68
230 hours
Optional Modules : The optional modules are in the application oriented specific fields like Electronics
and Communication and Photography and Audi– Videography
...
Each modules carries a weightage of 12 makrs which makes 15%
of total theory marks
...
Electronics and Communication
12
30
2
...
It carries a weightage of 20% marks in the
term end examination
...
Module 1 : Motion, Force and Energy
Approach : Besides highlighting the importance of universal standard units of measurement, applications of
dimensions and vectors in the study of physics to be described in this module
...
Significance of gravitation, concept of work and energy are to be highlighted
...
Unit 1
...
: Units, Dimensions and Vectors
•
Units of measurement – fundamental and derived units
Supprotive Video programme
•
Dimensions of physical quantities
1
...
349
349
•
Vectors and their graphical representation
•
Addition and subtraction of vectors
•
Resolution of vectors into rectangular components
(two dimensions)
•
Unit vector
•
Scalar and Vector products
2
...
2 motion in a Straight line
•
Distance and displacement
Supportive Video Programme
•
Speed, velocity and acceleration
1
...
•
Uniformly accelerated motion
•
Position – time and velocity – time graphs
•
Equations of motion with constant acceleration including
motion under gravity
•
Relative motion
3
...
3: Newton’s laws of motion
Supprotive Video Programme
•
Concept of force and inertia
1
...
Frictional force
•
Concepts of momentum
•
Second law of motion
•
Third law of motion
•
Impulse
•
Conservation of linear momentum
•
Friction – static and kinetc, factors affecting friction
•
Importance of friction and methods of reducing fiction
•
Free body diagram technique
•
Elementary idea of inertial and non – inertial frames of references
...
4: Motion in a Plane
•
Supportive Video Programme
Projectile motion (time of fligtht, range and
maximum height)
1
...
Circular Motion
•
Uniform circular motion
•
Centripetal acceleration
•
Circular motion in daily life
350
Unit 1
...
Planetary Motion
• Acceleration dute to gravity and its variation with
2
...
Unit 1
...
Work and Power
•
Work done by a varying force (graphical method)
2
...
•
Conservation of energy (spring pendulum, etc)
•
Elastic and inelastic collisions
•
Power and its units
...
Unit 1
...
Rotational Motion
•
Moment of inertia, radius of gyration and its
significance
•
Theorems of parallel and perpendicular axes
concerning moment of inertia and their uses in simple
cases (no derivation)
•
Equations of motion for a uniformly rotating rigid
body (no derivation)
•
Angular momentum and law of conseration of angular
momentum with simple applications
•
Rotational and transnational motions with examples
(motion of ball, cylinder, flywheel on an incline plane)
•
Rotational energy
3
...
This module explains the elastic behaviour of the solids and highlights source of elastic behaviour of solids
...
have been explained
with the help of daily like examples and their applications have been highlighted
...
Unit 2
...
Elastic Behaviour of solids
• Inter –molecular forces
• Young’s modulous, bulk modulous, modulous of rigidity
and compressibility
• Some applications of elastic behaviour of solids like
cantilever, girder etc
...
Unit 2
...
Hydrostatic Pressure
•
Pascal’s law and its applications
...
Surface Tension
•
Forces of cohesion and adhesion
3
...
The concept of temperature is to be explained by thermal equilibrium
...
Working of heat engines and refrigerators will be explained
...
The
concept of thermal pollution and the issue of green house effect will also be dealt with in this module
...
Unit 1
...
E
...
Unit 3
...
Thermodynamic Processes
Thermal equilibrium – Zeroth law of thermo dynamics
and concept of temperature
•
Thermodynamic variables and thermodynamic
equilibrium
•
2
...
•
First law of thermodynamics – internal energy
•
Phase change, Phase diagram, latent heat and triple
Point carnot’s cycle and its efficiency – second
law of thermodynamics, heat engine and refrigerator
•
Limitations of Carnot’s engine
12
...
3 : Heat Transfer and Solar Energy
Supportive Video Programme
•
1
...
•
Green house effect
•
Solar energy
Module 4 : Oscilations and Waves
Approach : Besides explaining the terms associated with periodic motion, the harmonic motion will be described
with the help of common examples
...
13
...
1: Simple Harmonic Motion
Supportive Video Programme
•
1
...
Unit 4
...
Formation and Propagation of Waves
3
...
2
...
(qualitative only)
• Electromagnetic waves and their properties
• Em – waves spectra
• Constancy of speed of light (non – evaluative in a box)
Module 5 Electricity and Magnetism
Approach : The basic concept of electrostatics and frictional electricity will be described in the module
...
Different types of capacitors, their
compbinations and applications will be explained
...
Significance of maganetic effedct of current and electromagnetic induction has been
emphasized
...
15
...
1 : Electric Charge and Electric Field
Supportive Video Programme
•
Frictional electricity – electric charges and their
conservation
1
...
16
...
2 : Electric Potential and Capacitors
• Electric potential due to a point charge
354
Supportive Video Programme
• Electric potential at a point due to a dipole
(axial and equatorial)
...
• Different type of capacitors and their applications
• Capacitors in series and parallel combinations
• Energy stored in a capacitor
• Dielectrics and their polarization
• Effects of dialectics on capacitance
17
...
3 : Electric Current
Supportive Video Programme
• Electric current in a conductor
1
...
Heating Effect of Electric Current
• Ohm’s law, ohmic and non – ohmic resistances –
• Colour coding of resistors
...
• Heating effect of electric current – Joule’s law of
heating
18
...
4 Magnectism and Magnetic Effect of Electric Current
• Bar Magnet and its magnetc field
Supportive Video Programme
• Magnetic effect of electric current
1
...
3
...
Unit 5
...
Generation and Transmission of
• Alternating current and voltage illustrating with
of Electric Current
Phase diagram – peak and rms values
• Circuits containing only R, L or C separately –
phase relationship between I & V
• LCR series combination (using phaser diagram only)
and resonance
• Generators – AC and DC
• Transformers and their applications
• Transmission of electric power
• Problem of low voltage and load shedding
(concepts of stabilizer and inverters )
Module 6 : Optics and Optical Instruments
Approach : After giving a brief introduction of reflection of light, the basic concepts like refraction, total internal
reflection, dispersion, scattering, of light will be described in the module
...
Further applications of the properties
of light have been described to construct various types of optical instruments
...
Unit 6
...
Reflection of light
• Refraction of light, Snell’s law of refraction
2
...
Unit 6
...
Rainbow
• Defects of image formation–spherical and chromatic
aberration (qualitative only)
• Scattering of light in atmosphere
...
Unit 6
...
Supportive Video Programme
• Interference–Young’s double slit experiment
• Diffraction of light at a single slit (qualitative)
• Polarization-Brewster’s law and its application in daily life
23
...
4 : Optical Instruments
• Simple and Compound microscopes and their
magnifying power
Supportive Video Programme
1
...
Nuclei and radio activity
have been explained along with their applications
...
24
...
1 : Structure of Atom
• Alpha-Particle scattering and Rutherford’s atomic model
Supportive Video Programme
• Bohr’s model of hydrogen atom and energy levels
1
...
Unit 7
...
Photo electric Effect and its Applications
3
...
Unit 7
...
Radioactivity and its Applications
• Nuclear forces, mass-energy equivalence
• Mass defect and binding-energy curve
• Radioactivity-alpha, beta decay and gamma emission
• Half life and decay constant of nuclei
• Applications of radioactivity
27
...
4 : Nuclear Fission and Fusion
• Nuclear reactions
Supportive Video Programme
• Nuclear fission and chain reaction
1
...
Besides highlighting
the basis of semiconductors, different types of semiconductor devices and their applications have been explained
in the module
...
Unit 8
...
Unit 8
...
Working principles of different electronic
devices used in daily life have been explained
...
30
...
Electronics in Daily Life
• Timer – digital clock
• Processor – calculator
• LCD
• transducers and control system – Burglar alarm/fire alarm
31
...
Communication systems
receiver media of communication and antenna
• Types of signals – analogue & digital
• Electromagnetic waves in communication
32
...
Modulation and Demodulation
• Demodulation
• Role of tuner
• Common communication devices–radio/TV/Fax/Modern etc
...
Unit 4 : Communication Media
• Guided Media – transmission lines and optical fibre
Supportive Video Programme
• Unguided Media and antenae–ground wave Communication, Communication media
sky wave communication, space wave communication and
satellite communication
...
359
359
Optional Module – 2 : Photography and Audio–Video–Graphy
APPROACH : The basic principles of physics used in the field of photography and audio–videography have
been described in different units of the module
...
30
...
Supportive Video Programme
1
...
• Choosing a camera, picture size
...
31
...
Supportive Video Programme
• Characteristics of film
1
...
Unit 3 : Audio–Video Recording
• Basic principle of recording
Supportive Video Programme
• Conversion of audio signal into electrical signals,
1
...
• Storage of audio–video signals on tapes
...
• Tape characteristics, structure and composition, tape
format, tape speeds, important tape parameters,
• Presentation of tapes, storage techniques, precautions
during handling and transportation
...
Unit 4 : Compact Disc for Audio–Video Recording
• Limitations of traditional audio–video recording systems,
Supportive Video Programme
• Compact Disc
• Need for compact disc, advantages of compact disc
...
Compact Disc for Audio–Video
Recording
• CD for audio recording,
• Basic principle of audio recordings,
• Methods of CD – audio-recording,
360
• CD for video-recording,
• Basic principle for video recording
• Methods of CD – video recording
• General operating and installation precautions,
• CD – players, operating principle,
• Quality of reproduction
...
Measurement of physical quantities using single scales like metre scale, graduated cylinder,
thermometer, spring balance, stopwatch, ammeter, voltmeter,
2
...
3
...
Section-B
1
...
Plot the L-T2 graph and
use it to determine (a) the length of a second’s pendulum, (b) the value of acceleration due to gravity
...
Determine the weight of a given body using parallelogram law of forces
...
3
...
4
...
5
...
6
...
7
...
8
...
Determine the terminal
velocity of the body in a viscous liquid and determine the coefficient of viscosity of that liquid
...
Study the formation of stationary waves in (a) stretched strings and (b) air columns
...
2
...
Determine the focal length of (a) convex
lens, and (b) concave mirror
...
4
...
Determine the internal resistance of a cell using a potentiometer
...
Convert the
galvanometer into a voltmeter of suitable range and verify it
...
3
...
7
...
Study the characteristics of an npn transistor in common emitter configuration
...
Home Activities (Suggestive)
1
...
2
...
Determine the
refractive index of the glass using the graph
...
Study the relationship between the angle of rotation of a plane mirror and the change in angle of reflection
...
Draw magnetic filed line due to a bar magnet keeping (i) North pole pointing north, and (ii) North pole
pointing south
...
362
Final fold and seal
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