ADVANCED DIPLOMA
ACADEMIC YEAR 2019-2020
FIRST YEAR ,FIRST SEMESTER
MODULE NAME: ENGINEERING PHYSICS
MODULE CODE: PHY 101
DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
OPTION: ELECTRONICS AND TELECOMUNICATION ENGINEERING
Module Leader: Philippe SINDAYIHEBA
October 2019
Contents
PART I: MECHANICS .................................................................................................................. 1
CHAPTER I: MEASUREMENT OF PHYSICAL QUANTITIES ................................................ 1
1.1.
THE CONCEPT OF PHYSICAL QUANTITIES ........................................................... 1
1.2.
FUNDAMENTAL AND DERIVED QUANTITIES ...................................................... 1
1.3.
SCALAR AND VECTOR QUANTITIES ....................................................................... 1
1.4 THE INTERNATIONAL SYSTEM OF UNITS (SI) .......................................................... 2
1.5 DIMENSIONAL ANALYSIS .............................................................................................. 3
CHAPTER II: KINEMATICS ........................................................................................................ 5
2.1 CONCEPT OF FRAME OF REFERENCE ......................................................................... 5
2.2 RECTILINEAR MOTION (MOTION IN A STRAIGHT LINE) ........................................ 5
2.3 PLANAR MOTION............................................................................................................ 16
CHAP III: DYNAMICS OF A POINT ..................................................................................... 32
3.0. INTRODUCTION ......................................................................................................... 32
3.1. NEWTON’S LAW OF MOTION ................................................................................. 32
3.2. APPLICATION OF NEWTON’S LAWS OF NEWTON’S LAWS OF MOTION ...... 34
3.3. ENERGY AND POWER............................................................................................... 58
3.4. LINEAR MOMENTUM ................................................................................................ 63
CHAP IV: ROTATION OF RIGID BODIES ABOUT FIXES AXIS ..................................... 71
4.0. INTRODUCTION ......................................................................................................... 71
4.1. CONCEPT OF ROTATIONAL MOTION ................................................................... 72
4. 2. MOMENT OF FORCE OR TORQUE ......................................................................... 73
4.4. MOMENT OF A COUPLE OF FORCE ....................................................................... 75
4.5. ANGULAR MOMENTUM AND ITS CONSERVATION .......................................... 77
4.5. KINETIC ENERGY OF ROLLING OBJECT .............................................................. 79
CHAP V: STATICS .................................................................................................................. 83
5.0. INTRODUCTION ......................................................................................................... 83
5.1. EQUILIBRIUM CONDITION FOR COPLANAR CONCURRENT FORCES .......... 88
5.2. EQUILIBRIUM CONDITIONS FOR BODIES UNDER COPLANAR NONCONCURRONT FORCES ................................................................................................... 89
CHAP VI: FLUIDS MOTION.................................................................................................. 95
6.0.
i
INTRODUCTION ...................................................................................................... 95
6.1. VISCOSITY ................................................................................................................... 95
6.2. STEADY AND TURBULENT FLOW ......................................................................... 97
6.3. MOTION IN A FLUID .................................................................................................. 98
6.4. BERNOUILLI’S EQUATION AND ITS APPLICATION ........................................ 101
CHAP VII: KINETIC THEORY OF MATTER .................................................................... 107
7.0. INTRODUCTION ....................................................................................................... 107
7.1 BASIC ASSUMPTIONS OF KINETIC THEORY ...................................................... 108
7.2. FORCES ACTING BETWEEN MOLECULES ......................................................... 108
7.3. SOLIDS........................................................................................................................ 108
CHAP VIII: THERMODYNAMICS .......................................................................................... 114
8.0.
INTRODUCTION ........................................................................................................ 114
8.1.
TEMPERATURE AND HEAT ................................................................................... 114
8.2.
THERMAL EXPANSION ........................................................................................... 122
8.3.
HEAT TRANSFER ...................................................................................................... 124
8.4.
HEAT CAPACITY, SPECIFIC HEAT, AND LATENT HEAT................................. 131
LATENT HEAT & CHANGE OF STATE ................................................................................... 135
8.5.
GAS LAWS.................................................................................................................. 138
8.6.
LAWS OF THERMODYNAMICS ............................................................................. 147
8.7.
HEAT ENGINES & HEAT PUMPS ........................................................................... 162
REFERENCE .............................................................................................................................. 166
ii
PART I: MECHANICS
Mechanics is a branch of physics concerning the motions of objects and their response to forces.
Descriptions of such behavior begin with a careful definition of such quantities as displacement,
time, velocity, acceleration, mass, and force.
CHAPTER I: MEASUREMENT OF PHYSICAL QUANTITIES
1.1.
THE CONCEPT OF PHYSICAL QUANTITIES
A physical quantity is a physical property of a phenomenon, body, or substance that can be
quantified by measurement. Hence the value of a physical quantity Q is expressed as the product
of a numerical value Q and a unit of measurement Q .
Q  Q Q
If the temperature T of a body is quantified (measured) as 300 Kelvin this is written T  300 K
where T is the symbol for the physical quantity, and K is the symbol for the unit. The
measurement of physical quantities depends on a number of parameters picked by the observer;
including the coordinate system, and metric.
1.2.
FUNDAMENTAL AND DERIVED QUANTITIES
Fundamental or base quantities are the physical quantities, which are self explanatory. Like
length, mass, time, are known as primary fundamental quantities. The fundamental quantities are
quantities that can be measured using measuring instruments. Derived quantities are physical
quantities, which are explained through fundamental quantities, like area, density, etc. A derived
quantity can be also a combination of both fundamental and derived quantities. Measurement of
derived quantities sometimes requires formulas.
1.3.
SCALAR AND VECTOR QUANTITIES
Most of the physical quantities encountered in physics are either scalar or vector quantities. A
scalar quantity is defined as a quantity that has magnitude only. Typical examples of scalar
quantities are time, speed, temperature, and volume. For example, the units for time (minutes,
days, hours, etc.) represent an amount of time only and tell nothing of direction. A vector
quantity is defined as a quantity that has both magnitude and direction. By definition, a vector
has both magnitude and direction. Direction indicates how the vector is oriented relative to some
reference axis. Examples of vector quantities are displacement, force, electric field, weight, etc.
1
1.4 THE INTERNATIONAL SYSTEM OF UNITS (SI)
1.4.1 SI BASE UNITS
The SI is founded on seven SI base units for seven base quantities assumed to be mutually
independent
Base quantity
SI base unit
Symbol
Definition
m
The meter is the length of the path travelled by light
in vacuum during a time interval of 1/299 792 458 of
a second.
Length
Name
Metre
Mass
Kilogram Kg
The kilogram is the unit of mass; it is equal to the
mass of the international prototype of the kilogram.
Time
Second
s
The second is the duration of 9 192 631 770 periods
of the radiation corresponding to the transition
between the two hyperfine levels of the ground state
of the cesium 133 atom.
Electric current
Ampere
A
The ampere is that constant current which, if
maintained in two straight parallel conductors of
infinite length, of negligible circular cross-section,
and placed 1 meter apart in vacuum, would produce
7
between these conductors a force equal to 2  10
Newton per meter of length.
Thermodynamic
temperature
Kelvin
K
The kelvin, unit of thermodynamic temperature, is
the fraction 1/273.16 of the thermodynamic
temperature of the triple point of water.
Amount of
substance
Mole
mol
Luminous intensity
Candela
cd
The mole is the amount of substance of a system
which contains as many elementary entities as there
are atoms in 0.012 kilogram of carbon 12.
The candela is the luminous intensity, in a given
direction, of a source that emits monochromatic
radiation of frequency 5.40  10 Hertz and that has
a radiant intensity in that direction of 1/683 watt per
steradian.
12
2
1.4.2 SI DIMENSIONS OR DERIVED UNITS
Derived quantities are defined in terms of the seven base quantities via a system of quantity
equations. The SI derived units for these derived quantities are obtained from these equations
and the seven SI base units. For example the speed is expressed in m/s, mass density kg/m3, etc
1.5 DIMENSIONAL ANALYSIS
The ease of switching between units is another feature of the metric system. To convert between
SI units, we multiply or divide by the appropriate power of 10. Prefixes are used to change SI
units by powers of 10, as shown in the following table. You often will encounter these prefixes
in daily life, as in, for example, milligrams, nanoseconds, and gigabytes.
Femto
F
SI prefixes
Scientific
Prefix
notation
15
Deka
10
Pico
P
10 12
Hecto
h
10 2
Nano
10 9
Kilo
k
10 3
micro
N

10 6
Mega
M
10 6
milli
M
10 3
Giga
G
10 9
centi
C
10 2
Tera
T
1012
deci
D
10 1
Peta
P
1015
Prefix
Symbol
Symbol
Scientific
notation
da
101
We can use units to check our work. You often will need to use different versions of a formula,
or use a string of formulas, to solve a physics problem. This method of treating the units as
algebraic quantities, which can be cancelled, is called dimensional analysis. To check that you
have set up a problem correctly, write out the equation or set of equations you plan to use. Before
performing calculations, check that the answer will be in the expected units.
Exercises
1. Convert the speed 5.30 m/s to km/h.
2. How many megahertz is 750 kilohertz?
3. The mass density of a substance is the ratio of its mass in kilograms to its volume in
cubic meters. A piece of lead of mass 11.97g has a volume of 1.05m3. Determine the
density of lead in SI units.
8
4. The radius of the sun is 7.0  10 m and its mass is 2.0  10 30 kg .
a) What is the density of the sun?
3
b) The density of water is 1000kg/m3.What is the ratio of the sun’s density to that of
water?
5. Given the time t, acceleration a, velocity v, mass m, distance x, charge q, Electric field E
and force F, determine if the dimensions in each of the following equations are consistent
2
a) v  at
b) ma  qE
mv 2
 Fx 2
2
6. Knowing that Ampere A is the unit of electric current and Farad F is the unit of
capacitance;
a) establish an expression that may link the two units
b) Why the multiplication of the units Ohm  and Farad F may give a constant of time?
c) xma 
4
CHAPTER II: KINEMATICS
Kinematics is the branch of classical mechanics that describes the motion of bodies (objects) and
systems (groups of objects) without consideration of the forces that cause the motion.
2.1 CONCEPT OF FRAME OF REFERENCE
A frame of reference in physics may refer to a coordinate system or set of axes within which to
measure the position, orientation, and other properties of objects in it, or it may refer to an
observational reference frame tied to the state of motion of an observer.
For example, the driver of a car travelling with uniform velocity has zero velocity relative to a
passenger in the car but has a constant velocity relative to a pedestrian standing at the roadside.
The car is one frame of reference and the roadside is another. Thus it would appear that the
observed motion of a body depends on the motion of the observer as well as those of the body.
The description of the motion of any object must always be given relative to some particular
reference frame. In physics, we often draw a set of coordinate axes to represent a frame of
reference.
Two frames of reference can distinguished according to their motions
- An inertial frame of reference is a non-accelerating frame of reference, for example if a
person is observing a moving car while at rest or while moving at constant velocity.
- A non-inertial frame of reference is an accelerating frame of reference, for example a
rotating frame of reference. (Rotation requires centripetal force and centripetal
acceleration so any rotating object always requires a centripetal acceleration to rotate.)
2.2 RECTILINEAR MOTION (MOTION IN A STRAIGHT LINE)
For a rectilinear motion, we can choose as reference of frame, one coordinate axis as the line
along which the motion takes place. Suppose an object moving along X-axis. The position of the
particle at any moment is given by its x coordinate and the motion becomes one-dimensional.
2.2.1 DISPLACEMENT
The displacement is how far an object has moved from the starting point. To see the difference
between total distance and displacement, imagine a person walking 70m and then turning around
and walking back a distance of 30m.
The total distance travelled is 100m, but the displacement is only 40m since the person is now
only 40m from the starting point. The displacement is a quantity that has both magnitude and
5
direction. The dark arrow in the above figure represents the displacement whose magnitude is
40m and whose direction is to the right.
2.2.2 TIME INTERVAL
Time interval is referred to as how long an event takes place. Then if an event starts at a time t 0
and ends at a time t the time interval is a numerical difference between the two times
t  t  t 0
The values t 0 and t can be read on the stopwatch.
2.2.3 AVERAGE VELOCITY AND INSTANTANEOUS VELOCITY
Velocity or speed may be defined as the rate at which something happens, moves or functions
within a time interval, that is, how fast is a progress, movement or an operation. In general, the
average speed of an object is defined as the covered distance divided by the time it takes to
travel this distance.
dis tan ce
(3.2.1)
average speed 
time
Suppose that an object is moving on X-axis so that at an instant t 0 , the object is at point
(coordinate) x 0 , and at another later instant t the object is at point x .
If during the time interval t  t  t 0 , the displacement of an object is x  x  x0 , then the
average speed according to the equation (3.2.1) is
x x  x0
v

t
t  t0
The instantaneous speed at any moment is
x dx

t 0 t
dt
v  lim
(3.2.2)
In physics we make a distinction between speed and velocity. Speed is simply a positive number,
with units. Velocity, on the other hand, is used to define both the magnitude of how fast an
object is moving and the direction in which it is moving. Namely, the average velocity is defined
in terms of displacement, rather than total distance.




If r0  x0 i and r  x i are position vectors of an object moving along X-axis, respectively at
the instants t 0 and t , the average velocity is
6

 r x  x0 
v

i
t
t  t0
Thus the instantaneous velocity is



r dr dx 
v  lim

 i
t 0 t
dt dt
(3.2.3)
(3.2.4)
2.2.4 AVERAGE ACCELERATION AND INSTANTANEOUS ACCELERATION
Acceleration is the rate at which something increases in velocity within a time interval. The
average acceleration between the instants t 0 and t is given by
  
 v v  v0 v  v0 
a


i
(3.2.5)
t
t  t0
t  t0
The instantaneous acceleration for a particle moving along the OX axis is



v dv dv 
(3.2.6)
a  lim

 i
t 0 t
dt dt
2.2.5 UNIFORM RECTILINEAR MOTION (URM)
The following figure shows how the walker accelerates within equivalent time intervals.
Consider an object moving along X-axis with a constant speed v .
If a particle starts moving from position x0 at an instant t 0 , the v-t graph below shows that the
velocity is constant for the whole period of time t  t  t 0 .
7
The area under the v-t graph is equal to the object’s displacement. Under the graph there is a
rectangular area xRectangle .
x Rectangle  vt
x  x0  vt  t 0 
The position x of an object moving at a constant velocity v at any other time t relatively to x0 is
given by
x  vt  t0   x0
(3.2.7)
Graphs of vt  and xt 
2.2.6 UNIFORMLY ACCELERATED RECTILINEAR MOTION (UARM)
Many practical situations occur in which the acceleration is constant or close enough that we can
assume it is constant. We defined acceleration as the rate at which something increases in
velocity. In this case, both the acceleration and velocity of the particle are either negative or
positive in a same time, meaning that their vectors are always parallel.
Consider a particle moving from x0 to x with a positive constant acceleration a starting with
speed v0 . Assuming the acceleration positive and constant, the velocity keeps changing
monotonously as shown by the v-t graph below.
8
The area under the v-t graph is equal to the object’s displacement. The area under the graph can
be calculated by dividing it into a rectangle and a triangle. The area of the rectangle is
1
xRectangle  v0 t xTriangle  vt
2
As the average velocity according to (2.3.5) is v  at , the area of the triangle can also be
1
at 2
2
The total area x is given by
x  xT riangle  x Rectangle
xTriangle 
1
at 2  v0 t
2
1
2
x  x0  at  t 0   v0 t  t 0 
2
The position x of an accelerated object at any other time t relatively to x0 is given by
x 
1
2
x  x0  at  t0   v0 t  t0 
2
Solving the equation (3.2.5) for t  t 0 , we get
(3.2.8)
v  v0
(3.2.9)
a
Substituting equation (3.2.9) into the equation (3.2.8) we get
(3.2.10)
v 2  v02  2ax  x0 
t  t0 
To find the average speed, we divide the equation (3.2.8) by t  t 0 and we get
v
9
v  v0
2
(3.2.11)
2.2.7 UNIFORMLY DECELERATED RECTILINEAR MOTION (UDRM)
For this case, one of the acceleration or velocity of the particle is considered as negative and
takes a negative value in equations. Acceleration and velocity have antiparallel (parallel but
opposite) vectors
Graphs of xt , vt  and at 
UARM
UDRM
10
Exercises
6
1
1. An electron hits a television screen with a speed 3  10 ms .
accelerated from rest 0.04m, find its average acceleration.
1
Supposing that it is
1
2. An object moves initially with a velocity of 3ms and a uniform acceleration of 4ms .
What is the speed and the distance moved during the next 7 seconds if
a) The acceleration and the velocity are in the same direction?
b) The acceleration and the velocity are in the same direction?
3. Before an airplanes takes off, it accelerates uniformly 600m within 15 s. What are the
speed and the acceleration when it takes off?
4. A race car moving from rest reached 60km / h within 15s
a) Find the distance it ran and the average acceleration in terms of m/min.
b) Suppose that it had a constant acceleration. How long would it take to reach some
more 80km/h and what would be the total distance it ran?
2
5. A car from rest accelerated uniformly 1m / s in 1s after which the engine broke down
2
causing the car to slow down at 5cm / s in 10s due to friction with the road. After they
braked and the car stopped after some more 5s
a) Find the total distance covered by the car
b) Represent graphically distance, speed and acceleration in terms of time
6. An object in the UARM moves 55ft in 2s. The next 2s, it moves 77ft.
a) Determine the initial speed and the acceleration
b) What will be the distance in the next 4s?
7. A car in the accelerated motion along the line OX, moves form position x1 and x 2
respectively on time t1 and t 2 . Prove that the acceleration of the car is given by
a  2x2t1  x1t 2  t1t 2 t1  t 2 
2
8. A car moves from rest accelerating at 4ms within 4seconds. During the 10 next
2
second, the car moves uniformly after what they break and the car slows down at 8ms
11
until it stops. Draw a v-t graph and prove that the area between the curve and the axis can
represent the covered distance.
9. Suppose a car waiting at a red traffic light. When the light turns to green, the car
2
accelerates uniformly at 2ms within 6s then proceeds moving with a uniform speed. At
the moment the car starts at a green traffic light, a truck passes by, heading in the same
1
direction with a uniform speed 10ms . how long and how far from the traffic lights the
car catches up with the truck?
10. A car was moving uniformly at 45km/h when the red traffic light turned on, in the next
crossroads. It took the driver 0.7s to react before slowing down at a constant acceleration
7m / s 2 . Determine the distance the car moved from the time the driver saw the red light
until the time when the car stopped.
11. Two cars A and B, moves uniformly at 72km/h and 45km/h respectively in the opposite
direction of a 200km distance. How long takes car A to meet car B if B started 2h before?
2.2.8 FREE-FALL
One of the most common examples of uniformly accelerated rectilinear motion is that of an
object allowed to fall freely near the Earth’s surface. For free fall, Galileo postulated that: “at a
given location of the Earth and in the absence of air or other resistance all objects fall with the
same constant acceleration”. In the free-fall motion air resistance is negligible and the action can
be considered due to gravity alone.
We call this acceleration the acceleration due to gravity on the Earth, and we give it the symbol
2
g . Its magnitude is approximately g   9.8 m / s . g varies slightly according to latitude and
elevation. The effects of air resistance are often small, and we will neglect them for the most
part. We will also suppose that an object moves along Y-axis.
2.2.8.1 GENERAL CONDITION
Suppose an object released to fall from a height y 0 . The vectors of g and v0 are parallel. The
following figure shows how the ball accelerates within equivalent time intervals.
12
The position, the speed and the acceleration of the object after a time t  t  t 0 are respectively:
1
2
g t  t 0   v0 t  t 0   y0
2
v  g t  t 0   v0
y
ag
(3.2.12)
(3.2.13)
(3.2.14)
The relation between y, v and g is given similarly to the equation (3.2.10):
v 2  v02  2 g  y  y0 
(3.2.15)
2.2.8.2 INITIAL CONDITION
Position, speed and acceleration
Suppose an object released to fall from rest v0  0, t 0  0 . From equations (3.2.12), (3.2.13) and
(3.2.14), the position, the speed and the acceleration of the object after a time t  t are
respectively:
y
1 2
gt  y0
2
(3.2.16)
v  gt
(3.2.17)
ag
(3.2.18)
The duration of the motion
When the object reaches the ground y  0 and it follows that
13
1 2
gt  y 0
2
The duration or final time t f of the motion is given by
0
tf  t  
2 y0
g
(3.2.19)
2.2.8.2.3 THE FINAL SPEED
When the object reaches the ground, the position is y  0 . From equation (3.2.15), the relation
between y, v and g becomes:
v 2  2gy0
The final speed v f of the motion is given by
v f  v  2gy0
(3.2.20)
2.2.9 MOTION OF A PROJECTILE THROWN VERTICALLY UPWARD
An object shot through the air is called a projectile. Consider an object thrown upward from a
tower of height y 0 into the air with an initial speed v0 .
2.2.9.1 GENERAL CONDITION
2.2.9.1.1 POSITION AND SPEED
The object is in the UDRM when it is moving upward. The position and the speed of the object
after a time t  t  t 0 are respectively:
14
1
2
g t  t 0   v0 t  t 0   y0
2
v  g t  t0   v0
y
(3.2.21)
(3.2.22)
The relation between y, v and g is:
v 2  v02  2 g  y  y0 
(3.2.23)
2.2.9.2 INITIAL CONDITION
2.2.9.2.1 POSITION, SPEED
Suppose an object at rest thrown vertically upward from a height y0  t 0  0 . From equations
(3.2.21), (3.2.22) and (3.2.23), the position and the speed of the object after a time t are
respectively:
1
(3.2.24)
y  gt 2  v0 t  y0
2
v  gt  v0
(3.2.25)
2.2.9.2.2 MAXIMUM HEIGHT
When the object reaches its maximum height
v  0  gt  v0  0
The time t m it takes the object to reach the maximum height:
tm  t  
v0
g
(3.2.26)
The maximum height y m can be found affirming that y m  y t m  .
ym 
1
2
g t m   v0 t m   y 0
2
2
 v 
1  v 
 y m  g   0   v0   0   y 0
2  g
 g
 ym 
1 v02 v02
  y0
2 g
g
1 v02
 ym  
 y0
2 g
15
(3.2.27)
2.2.9.2.3 DURATION OF THE MOTION
When the object falls back down it moves as in the free fall motion. Note also that the speed of
the object at any height is the same when going up as coming down (but the directions are
opposite). In addition t m is supposedly the time to reach the initial position. If the object is
launched on the ground level the height y 0  0. the duration of the motion
t f  2t m  
2v0
g
(3.2.28)
2.3 PLANAR MOTION
In this part we will consider the description of motion in two dimensions such as projectile
motion and circular motions in two dimensions near the Earth’s surface. When an object moves
in two dimensions, its motion can be broken into its X- and Y-components.
2.3.1 MOTION OF A PROJECTILE THROWN HORIZONTALLY
Along Y-axis the object moves as in a free fall which means that the motion is accelerated by g in
the opposite direction of the axis. On the X-axis there is no acceleration and the object moves as
in a URM.
2.3.1.1 GENERAL CONDITION
2.3.1.1.1 VELOCITY
At t 0 , the components of the initial velocity are
v0 x  v0 cos
v0 y  v0 sin 
We can write the velocity at t 0
16





v0  v0 x i  v0 y j  v0 cos i  v0 sin  j
(3.2.29)
At every point of the path, the components of the velocity after the time t  t  t 0 are
respectively:
v x  v0 cos
(3.2.30)
vy  g t  t0   v0 sin 
(3.2.31)
The velocity at the moment t





v  v x i  v y j  v0 cos  i  g t  t 0   v0 sin   j
(3.2.32)
The magnitude of the velocity at the moment t is
v  v x2  v y2
v  v02  2 gtv0 sin   g 2 t  t 0 
2
(3.2.33)
At any moment, the angle made by the velocity with the horizontal is deduced from
vy
tan  
vx
(3.2.34)
2.3.1.1.2 POSITION
At every point of the path, the position and the speed of the object relatively to the y direction
after a time t  t  t 0 are respectively:
1
2
g t  t 0   v0 sin  t  t 0   y0 )
2
The relation between y, v and g is:
y
v
2

 v02 sin 2   2 g  y  y 0 
(3.2.35)
(3.2.36)
At every point of the path, the velocity v0 x doesn’t change in the x direction. The range after a
time t  t  t 0 is:
x  v0 cos t  t 0   x0
The vector position is given by


 1


2
r  x i  y j  v0 cos t  t 0   x0  i   g t  t 0   v0 sin  t  t 0   y 0  j
2

17
(3.2.37)
(3.2.38)
2.3.1.2 INITIAL CONDITION
Consider, for example, a projectile thrown horizontally from the top of a cliff of height y 0 , with
the initial velocity v0 .
2.3.1.2.1 VELOCITY
In the beginning the object is thrown horizontally along X-axis   0 . The components of the
initial velocity at t 0  0 are
v0 x  v0 cos  v0
v0 y  v0 sin   0
Using equation (3.2.31) the components of the velocity become at the moment t
v x  v0
v y  gt
This leads to the velocity





v  v x i  v y j  v0 i  gtj
(3.2.39)
We deduce the speed of the projectile at the moment t :
v  v x2  v y2  v02  g 2 t 2
(3.2.40)
2.3.1.2.2 POSITION
The projectile coordinates at the moment t becomes
x  v0 x t
 x  v0 t
18
(3.2.41)
1 2
gt  v0 y t  y0
2
1
(3.2.42)
 y  gt 2  y0
2
The position vector of the projectile at the moment t is


 1


(3.2.43)
r  x i  y j  v0 t i   gt 2  y 0  j
2

The projectile motion is a parabola. Solving (3.2.38) for t t 0 
y
t
x
v0
Replacing in (3.2.39) the equation of the parabola becomes
g
(3.2.44)
yx    2 x 2  y 0
2v0
2.3.1.2.2 DURATION AND FINAL RANGE OF THE MOTION
When the projectile reaches the ground, y  0 and equation (3.2.42) yields
1 2
gt  y0  0
2
We deduce the final time it takes the projectile to reach the ground:
tf  t  
2 y0
g
(3.2.45)
Replacing t f in (1), the final range from the cliff is given by
x f  v0 
2 y0
g
(3.2.46)
2.3.2 MOTION OF A PROJECTILE THROWN OBLIQUELY
Now, we are going to examine the more general motion of object moving through the air in two
dimensions near the Earth’s surface, such as a golf ball, athletes doing the long jump or high
jump. Specifically, in the absence of air resistance, there is no acceleration in the horizontal or x
direction and the object moves as in the URM. In the y direction, the object moves as in the
UDRM until when the object reaches the maximum height y m . After, the motion proceeds as
when the projectile is thrown horizontally.
19

Consider now, a projectile fired obliquely with initial velocity v0 at an angle  above the
horizontal axis.
2.3.2.1 GENERAL CONDITION
When an object is fired obliquely, in the upward motion, it goes decelerating. After reaching the
maximum height, it goes downward accelerating. We still can refer to the general conditions of
the motion of a projectile thrown horizontally
20
2.3.2.2 INITIAL CONDITION
2.3.2.2.1 POSITION
The projectile coordinates at the moment t are
x  v0 cos t
1 2
gt  v0 sin  t  y0
2
The position vector of the projectile at the moment t is


 1


r  x i  y j  v0 cos t  i   gt 2  v0 sin  t  y 0  j
2

y
(3.2.47)
(3.2.48)
(3.2.49)
2.3.2.2.2 MAXIMUM HEIGHT
When the object reaches its maximum height, v y  0 .
v y  gt  v0 sin 
v y  0  gt  v0 sin   0 .
We deduce the time it takes for the projectile to reach the maximum height:
tm  
v0 sin 
g
(3.2.50)
The maximum height y m can be found affirming that y m  y t m  . Replacing (3.2.50) in
equation (3.2.48) we get
v02 sin 2 
ym  
2g
(3.2.51)
Note that the maximum height for a given initial velocity v0 will be reached if the object is
0
launched at   90 , which the motion of a projectile is thrown vertically.
21
2.3.2.2.3 MOTION DURATION
The motion of the object from the maximum height is similar to the motion of a projectile
thrown horizontally from the equivalent height .The time t g it takes the object to reach the
ground from the maximum height is also deduced similarly.
tg  
2 ym
g
Then, the total duration of the motion will be the sum of the time to reach the maximum height
t m and the time to reach the ground from the maximum height t .
t f  tm  t  
v0 sin 
2y
  m
g
g
(3.2.52)
2.3.2.2.4 OBLIQUE MOTION LAUNCHED FROM THE GROUND LEVEL
2.3.2.2.4.1 MOTION DURATION
If the object is launched from the ground  y 0  0 , it takes the same time to reach the maximum
height as to fall back to the ground.
t f  2t m  
22
2v0 sin 
g
(3.2.53)
2.3.2.2.4.1 RANGE
We can find how far away the projectile hits the ground as we know that x f  xt f :
 2v sin  
 2 sin  cos 
  v02 

x f  xt f  v0 cos t f  v0 cos  0
g
g




(3.2.54)
Since 2 sin  cos  sin 2 ,
xm 
v02 sin 2
g
(3.2.55)
The maximum range, for a given initial velocity v0 , is obtained when the sine takes on its
0
maximum value of 1, which occurs for   45 . In this case
xmax
v02

g
(3.2.56)
Exercises
1
1. A stone falls from a balloon descending at a speed of 12ms . Find the distance from the
balloon where the stone is at the end of 10s. (Find the previous distance in the case where
the balloon ascends.)
1
2. A stone is thrown vertically upward with a speed of 20 ms . When will the speed become
6ms 1 and what will be the height then?
3. A stone is thrown upward from the bottom of a well 88ft deep at 240 ft s 1 . what time
does it take the stone to reach the edge ( discuss two possibilities)
4. A man throws a brick vertically upward from the top of a building at 40 ft s 1 and the
brick hits the ground 4.25s later. Disregarding how tall is the man,
a) Find the maximum height at which the brick has possibly reached.
b) How tall is the building?
c) When does the brick hit the ground?
1
5. Two objects are thrown vertically upward at a same speed 100ms but the second object
is thrown 4s after. How long does it take the two objects to meet each other?
1
6. One stone is thrown vertically upward from a tower at 29.4ms . After 4s, the other is
released to fall freely. Prove that the first stone will overtake the second just 4s after
releasing the second stone.
7. A small stone is released from the top of a tower and the sound when the stone hits the
ground is heard 6.5s after. If the sound travels at 1120 ft s 1 , find the height of the tower.
23
8. A projectile is launched 60o at 600m/s upward from the horizontal find
a) The maximum range
b) The maximum height
c) The speed and the height after 30s
d) At what time and speed is the projectile 10km high?
9. A bomber flies horizontally 1.2km of altitude with a speed of 900km/h
a) How long before overtaking the target, must the plane release the bomb?
b) What is the speed of the bomb when it reaches the ground?
c) What is the speed of the bomb 10s after being released?
d) What is the speed of the bomb 200m high?
e) What is the angle of velocity when the bomb hits the ground?
f) What is the maximum range?
10. An artillery shell was launched 35o high and hit the target at a horizontal distance 4km
from the gun.
a) Find the initial speed of the shell
b) Find The total duration of the shelling
c) How much high the shell reached and what was the speed then?
11. A train was moving 72km/h when a suspended lantern on the roof fell down in the
corridor because of jolts.
a) Find the distance that the train had run within the time it took the lantern to reach the
floor.
b) How far did the lantern fall if it had been hung 2.3m from the floor?
12. A machine-gun located on the top of a 400ft tall cliff, fired bullets at 786 ft s 1 , 30o high
above the horizontal while a car was heading towards the cliff 60mi/h on a horizontal
road
a) If the bullets hit the car, how far from the cliff the car was shot?
b) At what distance from the cliff was the car when the gunner opened fire? (retake the
same question in the case where the car was running away from the cliff)
13. A machine-gun fires bullet at 650 ft s 1 . Determine the angle at which a bullet hits a
distant target located 450ft away and 18ft high.
14. A plane flies horizontally 200km/h at 1km above the sea. The plane was supposed to
release a bomb upon a boat heading in the same direction moving at 20km/h. prove that
the bomb must be released when the horizontal distance between the plane and the boat
is730m. (retake the same problem in the case where the plane and the boat are in the
opposite directions)
24
2.3.3 CIRCULAR MOTION
In physics, circular motion is rotation along a circle: a circular path or a circular orbit. It can be
uniform, that is, with constant angular rate of rotation, or non-uniform, that is, with a changing
rate of rotation. Examples of circular motion include: an artificial satellite orbiting the Earth, a
stone which is tied to a rope and is being swung in circles, etc.
2.3.3.1 ANGULAR VELOCITY
To indicate the angular position of a particle, or how far it has rotated we specify the angle  of a
line joining the centre of the particle and its position with respect to some reference line, such as
the x axis. Consider an object moving in a circle of radius r with a uniform speed v round a
fixed point 0 as centre.
When an object rotates the angular displacement      0 , the average angular velocity is
defined as
    0


t
t  t0
(3.2.57)
The linear displacement or the Arc of the object along the circle is
s  s  s0  r
The linear average velocity
s

v
r
 r
t
t
(3.2.58)
We define the instantaneous angular velocity as the very small angle  , through which the
object turns in the very short time interval t :
 d
(3.2.59)
  lim

t 0 t
dt
25
The instantaneous linear velocity
r
d
v  lim
r
t 0 t
dt
(3.2.60)
The angular velocity is generally specified in radians per second rad / s  whereas the
instantaneous linear velocity is expressed in m / s  .
2.3.3.2 PERIODIC TIME AND FREQUENCY
The period T is the time needed for the object to make one complete revolution. During this time,
the object travels a distance equal to the circumference of the circle. The frequency f is referred
to as the number of revolutions made by an object in one second. The unit of frequency is Hertz.
1
T
The object’s angular speed is then represented by
2r

 2rf
T
f 
(3.2.61)
(3.2.62)
2.3.3.3 THE AVERAGE ANGULAR ACCELERATION
The average angular acceleration is defined as the change in angular velocity divided by the time
required to make this change:

    0

t
t  t0
(3.2.63)
The average linear acceleration
s r
a

 r
t
t
(3.2.64)
The instantaneous angular acceleration is
 d
  lim

t 0 t
dt
(3.2.65)
The instantaneous linear acceleration
v

d
a  lim
r
r
t 0 t
t
dt
(3.2.66)


The angular acceleration is generally specified in radians per second rad / s 2 whereas the


instantaneous linear acceleration is expressed in m / s 2 .
26
2.3.3.4 UNIFORM CIRCULAR MOTION (UCM)
Consider an object moving in a circle at a constant speed, such as a stone being whirled on the
end of a string. Are these objects accelerating? At first, we might think that they are not because
their speeds do not change. However, remember that acceleration is the change in velocity, not
just the change in speed. Because the direction of velocity is changing, the objects must be
accelerating.
2.3.3.4.1 TANGENTIAL AND CENTRIPETAL ACCELERATION
In the circular motion, the tangential acceleration aT always points in the direction tangent to the
circle and changes the rate of velocity in terms of magnitude because their vectors are always in
the same or opposite direction. The tangential acceleration can be considered as the linear
acceleration. The centripetal acceleration (normal acceleration or radial acceleration) a N
changes the velocity in terms of direction and its vector is perpendicularly directed inward the
circle.
Since the velocity is constant, the tangential component of acceleration doesn’t exist in UCM:
dv
aT 
0
dt
Consequently, the tangential component of the acceleration is also zero. Only, the normal
component of the acceleration exists in UCM. Thus, the velocity v changes in direction but not

in magnitude. The figure below shows the representation of the angular velocity  using a

distant reference frame i , j , k .
27
The centripetal component of the acceleration is directed along the radius. From the above figure
we can notice that r  R sin  .
v  r  R sin 

This relation indicates that the vector v can be expressed in the vector form by the equation:
  
v R
It follows that
dv
dR
aN 
  sin 
dt
dt


 dv  dr
a N 

dt
dt
 

aN v
(3.2.68)
Introducing the equation (3.2.67) into (3.2.68)
    

aN    v      R


(3.2.69)
The magnitude of the centripetal acceleration is given by
a N  v sin 


Since   v , its magnitude is
a N  v
As v  r , we also find that
aN 
v2
 r 2
r
(3.2.70)

 
The figure below shows the representation of the vectors a N ,  and v .
28
Graphs of  t  and  t 
2.3.3.5 UNIFORMLY ACCELERATED CIRCULAR MOTION (UACM)
Consider a particle moving along a circle with a positive constant angular acceleration  . The
angular acceleration of the particle, which is assumed constant in time, is given by
d

 d  dt
dt
The angular velocity
   t  t 0    0
(3.2.71)
Similarly to the equation (3.2.8) The angular position of the particle after a time t  t  t 0 can
be found by using the equation
1
   t  t 0 2  0 t  t 0    0
2
Substituting t  t 0 
  0
into the equation (3.2.72)

 2  02  2    0 
29
(3.2.72)
(3.2.73)
To find the average angular velocity, we divide the equation (3.2.72) by t  t 0 .

  0
(3.2.74)
2
2.3.3.6 ACCELERATION IN A NON-UNIFORM CIRCULAR MOTION
Circular motion at a constant speed occurs when the acceleration of the object is directed toward
the centre of the circle (Only the centripetal component is available). If the acceleration is not
directed toward the centre but is at an angle, as shown in figure below, the acceleration has two
components; the centripetal and the tangential component.
The tangential component of the acceleration is the rate that changes the magnitude of the
velocity
aT 
dv d r 
d

r
 r
dt
dt
dt
The centripetal acceleration arises from the change in direction of the velocity and, as we have
seen, is given by
a N  r 2  v 
v2
r

If the tangential acceleration is in the direction of motion i.e. parallel to v , the speed is increasing



and if it is antiparallel to v is the speed is decreasing. In either case, aT and a N are always
perpendicular to each other; and their directions change continually as the object moves along its
circular path. The total vector acceleration is the sum of these two:
  
a  aT  a N
(3.2.75)

The magnitude of a at any moment is
30
a  a N2  aT2
(3.2.76)
Exercises
1. A compact disc rotates uniformly at 13.2rad s 1 .
a) Find the frequency and the period of rotation.
b) What time does it takes the disk to rotate 780o?
c) What time does it take the disk rotate 12 turns?
2. Find the angular speed of the three hands of a watch
3. Find the angular speed, the linear speed and the radial acceleration of the moon provided
that the distance earth-moon is 38.2  10 km and that the moon makes a full rotation
around the earth in 28 days.
4. Find the magnitude of the velocity and the centripetal acceleration of the earth motion
4
11
around the sun provided that the radius of the terrestrial orbit is 1.49  10 m and its
period is 3.16  10 s .
5. Find the magnitude of the velocity and the centripetal acceleration of the sun motion in
7
20
the Milky Way provided that the radius of the solar orbit is 2.4 10 m and its period is
6.3 10 7 s .
6. A 3m-diameter wheel rotates 120 turns/min. find the frequency, the period, both the
angular and the linear speed of one part of the rim.
1
1
7. The angular velocity of a fan increases from 20rad s to 30rad s in 5s. Find the angular
acceleration and the total angle of which it has rotated.
8. A motionless pulley A begins to rotate by increasing uniformly its angular velocity to
0.4rad s 1 within 4s . it transmits its motion to a pulley B using a belt as shown in the
following
a) Establish a relationship between the angular velocities of the two pulleys and another
between their radii
b) How long will take the pulley B to rotate 300 turns/s
31
CHAP III: DYNAMICS OF A POINT
3.0. INTRODUCTION
Dynamics is a branch of physics concerned with the study of forces and the torques and their
effect on the motion, as opposes to kinematics, which deal studies the motion of the object
without reference to its causes.
3.1. NEWTON’S LAW OF MOTION
Newton’s studied and developed Galileo’s idea about motion and subsequently stated the three
laws which now bear his name.
3.1.1. Newton’s first law ''principle of inertia''
The first law stated that the body in its states of rest on moving with a constant velocity
continues to do so unless acted on by an external force.
This law implies that the matter has a built in reluctance to change its state of rest or motion
This property passed by all bodies is called inertia
INERTIA: Is its reluctance of body to state moving and its reluctance to stop moving after it
begun moving
Or
Inertia of matter is its reluctance to change its state of rest or motion.
Mass: the mass of body is a measure of inertia. “A large mass requires a large force to produce
certain acceleration on to stop it moving”.
Note that: this law readily defines a force as something that change the state of rest or form
motion of body.
GALILEAN REFERENCE FRAMES
Definition: Galilean reference frames or an inertial frame of references is one in which the
motion of a particle not subject to forces is in straight line at constant speed. Observers
belonging to different inertial reference frame will came up with the same laws of physics
Ex: An observer at rest on the surface of the earth is an (approximate) inertial reference frame; a
passenger on a train that moves with a constant velocity relative to an observer on the earth is
also an inertial reference frame.
The two observers will discover the same laws of physics by performing experiments in their
respective frames.
32
3.1.1.2. Newton’s second law
Newton’s second law states that the note of change of momentum of a body is proportional to the
resultant force on the body and takes place in the direction of the force.
Momentum is the product of mass and velocity of a moving body
force 
F
changeof momentum
time
dp  mv  mu  mv  u 

 ma

dt 
t
t

F  ma
F  forve
Where: m  mass of moving body
a  accelerati on
Note that: the resultant forces acting on the body are of two kinds, such as


Internal forces
External forces
Internal forces: one the force of the body itself
E.g:- magnetic force
-Electrical force
-Tension force
-Gravity forces
External forces: are forces applied on the body, they are caused by the externed agent
e.g.





applied force
Normal force
Tension force
Air resistance force
Etc….
Special case
The body in a free fall obeys Newton’s second law. (i.e. the weight of the body is the force of
gravity acting on it towards the centre of the earth.
If g is the acceleration of the body towards the center of the earth, then we can substitute
33
F W , a  g
then F  ma
Give us
W= mg
where g = 9.82m/s2
Or g = 9.81N/kg
So that the weight of body is proportional to its mass
3.1.3. Newton’s third law ''principle of action and reaction''
Newton’s third law states that ˝to everybody action there is an equal and opposite reaction''
This means that whenever one object exerts a force on a second object the second
object exerts a force in the opposite direction on the first.
FA= -FB
3.2. APPLICATION OF NEWTON’S LAWS OF NEWTON’S LAWS OF MOTION
 Types of forces
Consider a block sliding on a horizontal table, as in figure below

R

N

F

Ff

W
Forces acting on the object are:
34
Motion





The motional force F which is an external force applied to the object;

The weight W pointing vertically downward. The weight acts at the centre of the object.

The normal force N exerted by the table on the block acts upward. The normal force

N is a contact force that arises from the repulsion between the atoms of the object and
the atom of the plane. This repulsion prevents the object from penetrating the plane.
Force of friction: When a body is in motion on a rough surface, or through a viscous
medium such as air or water, there is resistance to motion because of the interaction of
the body with its surroundings. We call such resistance a force of friction.

The friction force Ff acts horizontally, parallel to the tabletop, in direction opposite to the
motion. This force, just like the normal force, is a contact force which acts over the entire bottom
surface of the block.

If we apply an external force F to the block, acting to the right, the block will remain stationary

if F is not too large. The force that keeps the block from moving acts to the left and is called the

frictional force Ff . As long as the block is in equilibrium F f  F . Since the block is stationary,

we call this frictional force, the force of static friction, F fs and the friction between the two
surfaces at rest, static friction. Experiments show that this force arises from the roughness of the
two surfaces, so that contact is made only at a few points. Actually, the frictional force is much
more complicated than presented here since it ultimately involves the electrostatic force between
atoms or molecules where the surfaces are in contact.

If we increase the magnitude of F , the block will eventually slip. When the block is on the verge
of slipping, F fs is a maximum. When F exceeds F fs ,max the block moves and accelerates to the
right. When the block is in motion, the retarding frictional force becomes less than F fs ,max . When

the block is in motion, we call the retarding force the force of kinetic friction, F fk . The
unbalanced force in x direction F  F fk produces acceleration to the right. If F  F fk , the block
moves to the right with constant speed. If the applied force is removed, then the frictional force
is acting to the left decelerates the block and eventually brings it to rest.
In a simplified model, we can imagine that the force of kinetic friction is less than F fs ,max
because of the reduction in roughness of the two surfaces when the object is in motion.
Experimentally, one finds that both F fs and F fk are proportional to the normal force acting on
the block. The experimental observations can be summarized by the following laws of friction:
1. The force of static friction between any two surfaces in contact is opposite the applied force
and can have values given by:
F fs   s N
35
(1.5)
Where the dimensionless constant  s is called the coefficient of static friction and N is the
normal force. The equality in Equation (1.5) holds when the block is on the verge of slipping,
that is, when F fs  F fs ,max   s N . The inequality holds when the applied force is less than this
value.
2. The force of kinetic friction acting on an object is opposite to the direction of motion of the
object and is given by
F fk   k N
(1.6)
Where  k is the coefficient of kinetic friction.
3. The values of  k and  s depend on the nature of the surfaces, but  k is generally less than  s .
The coefficients of friction are nearly independent of the area of contact between the surfaces.
Although the coefficient of kinetic friction varies with speed, we shall neglect any such
variations.


The resultant of the normal force N and the frictional force F f is the reaction force of the plane
  
on the object, R  N  F f . Its magnitude is
R  N 2  F f2
(1.7)
3.2 Applications of Newton’s laws of motion
i) Object sliding on a horizontal plane
Consider a block of mass m sliding on a horizontal plane as shown the figure below:
36

R

N
Motion

F

Ff

W
Applying Newton’s second law to the mass m , we have
     
 

 F  F  R  W  F  F f  N  W  ma
(1.8)
Assuming the motion of m is to the right, after projecting all forces on the direction of motion,
we get
F
x
 F  F f  ma
(1.9)
Projecting all forces of the equation (1.8) on Y axis, we get
F
y
 N W  0
N  W  mg
(1.10)
In this case, the friction force can be written
F f   k N   k mg
(1.11)
Introducing the equation (1.11) into (1.9) we find the expression of the applied force:
F  F f  ma   k mg  ma  m( k g  a)
(1.12)
The reaction force is, in this case, given by
R  F f2  N 2  mg  k2  1
37
(1.13)
ii) Motion along an inclined plane
a) Object sliding up on inclined plane
Consider an object of mass m sliding up on inclined plane at an angle  to the horizontal.
Y

N

F


R
Motion

Ff

W sin 


W cos

W

Applying Newton’s second law to the mass m , we have
     
 

F

F

R

W

F

F

N
 W  ma

f
Projecting all forces on the direction of the motion, we get
F
x
 F  F f  W sin   ma (1.14)
Projecting all forces of the equation above on Y axis, we get
F
y
 N  W cos  0
or
N  W cos  mg cos
(1.15)
In this case, the friction force can be written
F f   k N   k mg cos
(1.16)
Introducing the equation (3.16) into (3.14) we find the expression of the applied force:
F  F f  W sin   ma   k mg cos  mg sin   ma
38
or
F  mg  k cos  sin    a 
(1.17)
The reaction force is, in this case, given by
R  F f2  N 2  mg cos  k2  1
(1.18)
and its direction relative to Y axis is given by
tan  
Ff
N

 k mg cos
 k
mg cos
(1.19)
b) Object sliding down an inclined plane
Consider now, an object of mass m sliding down an inclined plane at an angle  to the
horizontal.

R


N

Ff
Motion

W sin 

W cos

F



W
Applying Newton’s second law to the mass m , we have
     
 

F

F

R

W

F

F

N
 W  ma

f
Projecting all forces on the direction of the motion, we get
F
x
 F  F f  W sin   ma
(1.20)
Projecting all forces of the equation above on Y axis, we get
39
F
y
 N  W cos  0
or N  W cos  mg cos
The friction force can be written F f   k N   k mg cos
The equation of motion is
F  F f  W sin   ma   k mg cos  mg sin   ma
F  mg  k cos  sin    a 
or
(1.21)
Exercises:1.
Two masses are connected as shown here. Friction is negligible. What is the acceleration of each
mass? What is the tension in the string?
iii) Atwood’s machine
When two unequal masses are hung vertically over a light, frictionless pulley, the arrangement is
m1
called Atwood’s machine
40
m2
The device is sometimes used in the laboratory to measure the acceleration of gravity. The freebody diagrams of the masses are shown in figures below, where we assume that m2  m1 . Here
we introduce a new force, the tension, i.e. the force that the string exerts on the object.

T

T

a
m2
m1


W2  m2 g


W1  m1 g
Applying Newton’s law to m1 and m2 , we have
For m1 :
  

F
  T  W1  m1a (1.22)
For m2 :




(1.23)
 m2 a
Projecting all forces on the direction of the motion, we get
For m1 :
 F  T  W1  m1a or
F  T W
2
T  m1 g  a 
For m2 :
(1.24)
 m2 a Or T  m2 g  a  (1.25)
When (1.25) is subtracted from (1.24), T drops out and we get
 F  T  W
2
 m  m1 
 g
a   2
 m1  m2 
If (3.26) is substituting into (3.24), we get
41
(1.26)

a
 2m1 m 2
T  
 m1  m 2

 g

(1.27)
Note that when m1  m2 , a  0 and T  m1 g  m2 g , as we would expect for the balanced case.
Also, if m2  m1 , a  g (a freely falling body) and T  2m1 g .
UNIFORM MOTION IN CIRCLE
Forces in circular motion
Due to inertia, anybody moving in circle experiences an inertial force called centripetal force
away from the centre.
centrifugal force  mar  m
v2
 mw2 r
r
In order the body to continue moving in a circle, the centripetal force must be counter – balanced
by a force towards the centre called centripetal force.
Examples of uniform circular motion
a) The conical pendulum
Consider mass m attached to a string of length l describing a horizontal circle of wines r at
uniform speed v.
Let the string be at angle O the vertical and having a tension T
Resolving vertical component
T cos  mg
mv 2
T sin  
r
2
v
T tan 
rg
42
v  rg tan  
1/ 2
, where r  l sin 
If T is the period of oscillation
T
2r
2r

v
(rg tan  ) 2
 2
l cos
g
 2
h
g
b) Motion of a vehicle on a flat (unbanked) circular track
Consider a vehicle of mass m, taking an unbanked curve of wards r
For not skidding, the friction force must at least exceed the centrifugal force
mv 2
(F1 + F2)   (R1 + R2) 
r
mg 
43
mv 2
r
Where  - is the coefficient of kinetic friction.
The maximum safe v’max at which the vehicle can take the curve without skidding is given by
V ' max  gr
Angle of banking
The maximum safe speed for no skidding
V ' max  gr . Calculated on the assumption that all centripetal force is provided by all friction
between tyres and road surface, would be too small for practical purposes. In order to increase
the maximum speed, road curves are linked outwards so that some extra centripetal force is
provided by the horizontal components of the normal reaction.
Calculation
The angle of banking must be determined before the road is constructed: A curve is banked for a
certain speed and when determining the angle of banking is ignored
Assuming there is no friction, all the centripetal force is provided by the horizontal
component of normal reaction.
Resolving vertically
R1  R2 cos  mg
Resolving horizontal
44
R1  R2 sin   mv
2
r
By division, tan  
v2
r
Angle of banking
 v2 

rg
 
  tan 1 
UNIVERSAL GRAVITATION
Introduction: Besides Newton developing the three law of motion. Sir Isaac Newton also
examined the motion of planets and the moon.
In particular he concerned about the nature of force that must keep the moon in
its neatly circular orbit around the earth. Newton also thinks on the problem of gravity
Gravitation field
Gravitation field is defined as the area where the gravitation force of attraction can have action.
The gravitation field strength is defined as the force acting on unit mass placed in field (its unit is
N/kg)
E.g the gravitation field strength on the earth is 9.81N/kg.
Newton’s law of gravitation
It stated that “every particle of matter in the universal attracts every other particle with a force
which is directly proportional to the product of their masses and inversely proportional to the
square of their distance”.
45
Mathematically
F&
→F = G
Where G- a constant called the universal gravitation constant or Cavendish constant and it is
assumed to leave the same value everywhere for all matter.
From the above relation
so G can be expressed
In
Measurement shows that
G= 6.67 x 10-11
KEPLE’S LAWS
Base on tycho Brahe’s astronomical observation Joharnnes Kepler’s arrived at the complete
disruption of plane from motion.
1st LAW: Each planets move in an ellipse which has the sun as one focus or the planets describe
ellipse about the sun at one focus.
46
2st LAW: The imaginary line joining the planet to the sun (the radius vector) sweeps out equal
area in equal intervals of time .
3. The squares of period of revolution about the sun are directly proportional to the cubes of the
mean distances from the sun.
r3  T 2
NB: Kepler’s laws are geometrical summary of the observations on the motion of planets.
The laws started how the planets orbited the sun but gave no indication as to why they moved in
this way.
47
Newton’s law of gravitation
Newton’s analysis of Kepler’s law lead to an explanatory of the planet’s orbit and formed a
basis his law of universal gravitation.
Newton showed that
1) As consequence of Kepler’s first second law, the only force on a planet was a central one
towards the sun.
2) As consequence of Kepler’s and first and third law, the force between the sun and a planet
obeyed an inverse square law and law proportional to the planet’s mass
Proof
Consider a planet of mass mp moving in a circular orbit of radius round the sun ms with a speed
V.
F  mp
v2
Circular motion
r
But v 
2 r
where T is the period
T
F
mp
But
r

4 2 r 2
T2
r3
 a (from Kepler’s 3rdlaw)
2
T
There for
F&
Ex: Radius of moon’s orbit
- this obeys an inverse square law (F <
r = 384x 108m
Period of revolution of motion T = 27.3 days
Acceleration due to gravity at the earth surface 9.81m/S2
48
)
Re = 64x106
Radius of earth
The centripetal acceleration g’ of the moon using two methods
1st method
F  mg '  mw2 r  m 
4 r
T2
Where m-mass of moon
g' 
4 2 r
4 2  3.84  10 8

T2
27.3  24  3500 2
g’= 0.00272m/s2
2nd method
Assuming the square law
At the earth’ surface:
g
1
R E2
(1)
At the moon’s orbit
g '
1
R E2
(2)
Dividing (2) by (1)
g ' RE
 2
g
r
g' 
6.4  10 6
3.84  10 
8 2
 9.81  0.00273 m s 2
the two values calculates moon’s centripetal acceleration agreed well. Newton extended his
ideas to reality to any two bodies; he proposed that the force between two point masses was
-
Proportional to the product of their masses.
Inversely proportional to the square of their mean distance apart
Newton’s law of gravitation states that every particle in the universe attracts every other particle
with the force directly proportional to the product of their masses and inversely proportional to
the square of their distance apart.
49
Solving problems
Problems in gravitation are solved using
= mw2r (circular motion)
1. F =
(Newton’s law of gravitation)
2. F =
Ex1: The distance between the sun and the earth is 1.5x1011m and the time of revolution of the
earth around the sun is 30x107 . Calculate the mass of the sun.
Ans: Ms =2X1030kg
Ex2: the distance between the earth and the moon is 4x108 and the period of revolution of the
moon around the earth is 2.4x106 s (month). Calculate the mass of the earth.
Ans
Me = 6.0x1024kg
APPLICATION OF UNIVERSAL GRAVITATION
A) Relation between acceleration of free fall and G (universal constant of gravitation)
Consider a body of miss m placed at the earth’s surface
1) Newton’s 2nd law
F= mg
F
GM E m
rE2
2) Newton’s law of gravitation
Combining eg (1) and eg (2)
mg 
g
GM E m
rE2
GM E
Where g acceleration at the surface of the earth
rE2
G- Cavendish constant (6.66x10-11 Nm2/kg2)
Me – the earth’s mass
1. Variation of g with light
If g’ is the acceleration due to gravity at a distance a from the center of the earth
50
Then
g' 
GM E
(1)
a2
Where a = (r+h)
r = the radius of the earth
H-height above the earth
As the quantity GMe = is constant, the acceleration of free fall g decrease with the height
We know that
g
GM E
rE2
g rE  GM E (2)
Use (2) in (1), we get
g' 
rE2
rE  h2
g
2. Variation of g with depth
As g 
g' 
1
GM
r2
Gm1
b2
(2)
m1 is the mass of uniform sphere inside the earth.
The ratio between the masses is given by
(3) (as the earth is sphere)
For (1) and (2)
51
=
.
(4)
Using (3) in (4), we get
g1 
b
g
r
Where b= Re –h
g decreases linearly with the depth
Generally variation of acceleration of free fall on field strength
From the point outside the earth, the gravitation force obeys on inverse square law so the
acceleration of free fall g’ is given by
g’
where a is the distance to the center of the earth
The maximum value of g’ is obtained at earth’s surface where a  rE
Inside the earth, the value of g’ is dot inversely proportional to the square of the distance from
the centre.
Assuming a uniform earth density which is true in practice, theory shows that g’
varies linearly with the density from the centre.
Graphical representation of g
52
ENERGY OF A SATELLIE
A) Satellite has two forms of mechanical energy
i) Kinetic energy is energy possessed by a body by its virtue of its motion
ii) Potential energy is energy possessed by a body by virtue of its position in the gravitation
field
KINETIC ENERGY
We know that
mE m mv 2 r
G 2 
r
r
m m
1
mv 2  G E
2
2r
KE  G
mE m
2r
(1)
Potential Energy
Consider a body of mass m moved from a distance r from the centre earth to infinity
F G
53
mE m
at distance x
r2
Work one in moving a miss m from F to infinity in the gravitational force = gain in PE


r
r
Gain in PE   Fdx   G
mE m
m m
dx  G E
2
r
r
Potential energy is chosen to be zero and maximum at infinity so as to make the total energy
negative.
NB: by conversion, the total energy of a free particle must be positive and that of a bound
particle negative. A Satellite is a bound particle, bound to the earth by gravitation force –
therefore it must have a negative total energy.
PE  G
mE m
r
Potential energy at r
Ep  0  G
mE m
m m
 G E
r
r
(2) Always negative
Comparing e.g. (1) and (2)
E p  2Ek
The potential energy is twice its kinetic energy numerically and apposite sign
NB: The gravitation potential V at point in a field is defined as the work done in bringing a unit
from infinity to the point.
V
Ep
m
 G
mE
r
Total mechanical energy
Etot  PE  KE
mE m
m m
G E
2r
r
m m
 G E always negative
2r
G
Velocity of escape Vesc from the earth is surface
Velocity of escape vesc is the minimum vertical velocity with which a body must be projected
from the earth’s surface so as to escape completely from the earth’s gravitational attraction or for
it never to return.
54
Consider a body of mass m at a distance x (great than the earth’s radius, RE) from the
centre of the earth.
F
GM E m
x2
Work needed to move a body from the earth’s surface to infinity

W
GM E m
GM E m
dx 
2
RE
x
RE

The initial Kinetic energy must be equal to the required work.
½ mv2esc =
Vesc =
GM E m
RE
2GM E
RE
Using G = 6.67x10-11Nm2/kg
, ME= 6.0x1024kg
RE = 6.4 X 106m
Vesc
 2GM E
 
 RE



1
Vesc = 11.2km/s
2
 2  6.67  10 11  6.0  10 24 

 
6.4  10 6


1
2
(from the earth’s surface)
In general the velocity of escape from any planet is given by
Vesc =
Where m = mass of the planet
R = Mean radius of planet
Velocity of escape from the moon’s surface
Mass of moon m = 7.3x1022kg
Radius of moon Rm = 1740km = 1.74x106m
G = 6.67X10-11Nm2/kg2
Vesc=
55
=(
)= 2.4km/s
Velocity of a body around the earth
Gravitation force F 
GM E m
x2
Where x  rE  h
Then F 
GM Em
rE  h 2
From Newton second law
GM Em
mv 2

rE  h rE  h2
v2 
GM E
(1)
rE  h
As we know that
g
GM E
rE2
Then grE2  GM E
(2)
Use (2) in (1)
grE2
v 
rE  h
2
v 2  rE
g
Where g= acceleration of free fall rE radius of the earth.
rE  h
If the satellite is close to the earth then rE  h
V2= g rE
Substituting values
r= 6.4x106m
g= 98m/s2
V=
=
V=8CM/S
56
=7.9 x
m/s2
rE
This is the maximum speed that helps the satellites to remain in their orbit around the earth
This orbit is known as Parking orbits
Exercises
1. A satellite is to be put into orbit 500 km above the earth's surface. If its vertical velocity
after launching is 2000 m/s at this height, calculate the magnitude and direction of the
impulse required to put the satellite directly into orbit, if its mass is 50 kg. Assume radius
of earth, rE = 6400 km. [Answer: 4.0 x 105 kg m/s, 14.6°]
2. A satellite of mass 1000 kg moves in a circular orbit of radius 7000 km round the earth,
assumed to be a sphere of radius 6400 km. Calculate the total energy needed to place the
satellite in orbit from the earth. [Answer: 3.5 x 1010 J]
3. A satellite in a stable orbit contains two closed vessels: one of these is filled with water
while the other is filled with hot steam. Explain why the water exerts very little pressure
on its container but the steam exerts almost the same pressure as it would on earth when
at the same temperature.
4. A satellite of mass 1000 kg moves in a circular orbit of radius 7000 km round the earth,
assumed to be a sphere of radius 6400 km. Calculate the total energy needed to place the
satellite in orbit from the earth. [Answer: 3.5 x 1010 J]
5. Assuming the earth is a uniform sphere of mass M and radius R, show that the
acceleration of free-fall at the earth's surface is given by g = GM / R2. What is the
acceleration of a satellite moving in a circular orbit round the earth of radius
2R? [Answer: 0.25g]
6. A planet of mass m moves round the sun of mass M in a circular orbit of radius r with an
angular speed w. Show (i) that w is independent of the mass of the planet, (ii) that in a
circular orbit of radius 4r round the sun, the angular speed decreases to w/8.
7. A satellite X moves round the earth in a circular orbit of radius R. Another satellite Y of
the same mass moves round the earth in a circular orbit of radius 4R. Show that (i) the
speed of X is twice that of Y, (ii) the kinetic energy of X is greater than that of Y, (iii) the
potential energy of X is less than that of Y. Has X or Y the greater total energy (kinetic
plus potential energy)? [Answer: Y]
8. Find the period of revolution of a satellite moving in a circular orbit round the earth at a
height of 3.6 x 106 m above the earth's surface. Assuming the earth is a uniform sphere of
radius 6.4 x 106m, the earth's mass is 6 x1024 kg and G is 6.7 x 10-11 Nm2kg-2. [Answer:
9910 s]
9. If the acceleration of the free fall at the earth's surface is 9.8 m/s2, and the radius of the
earth is 6400 km, calculate a value of for the mass of the earth (G = 6.7 x 10-11 Nm2kg-2).
Give the theory. [Answer: 6 x 1024 kg]
57
10. Two stars, masses 1020 kg and 2 x 1020 kg respectively, rotate about their common centre
of mass with an angular speed w. Assuming that the only force on a star is the mutual
gravitational force between them, calculate w. Assume that the distance between the stars
is 106 km and that G is 6.7 x 10-11 Nm2kg-2. [Answer: 4.5 x 10-9 rad/s]
11. A preliminary stage of spacecraft Apollo 11's journey to the moon was to place it in an
earth parking orbit. This orbit was circular, maintaining an almost constant distance 189
km from the earth's surface. Assuming the gravitational field strength in this orbit is 9.4
N kg-1, calculate (a) the speed of the spacecraft in this orbit and (b) the time to complete
one orbit. (Radius of the earth = 6370 km) [Answer: (a) 7852 m/s (b) 5250 s
12. Explorer 38, a radio-astronomy research satellite of mass 200 kg, circles the earth in an
orbit of average radius 3R/2, where R is the radius of the earth. Assuming the
gravitational pull on a mass of 1 kg at the earth's surface to be 10 N, calculate the pull on
the satellite. [Answer: 889 N]
13. A satellite of mass 66 kg is in orbit round the earth at a distance of 5.7 R above its
surface, where R is the value of the mean radius of the earth. If the gravitational field
strength at the earth's surface is 9.8 N kg-1, calculate the centripetal force acting on the
satellite. Assuming the earth's mean radius to be 6400 km, calculate the period of the
satellite in orbit in hours. [Answer: 14.4 N, 24.5 h]
14. (a) Explain what is meant by gravitational field strength. In what units is it measured?
Starting with Newton's law of gravitation, derive an expression for g, the acceleration of
free-fall on the earth's surface, stating clearly the meaning of each symbol used. (Assume
that the earth may be considered as a point mass located at its centre.)
(b) g may be found by measuring the acceleration of a free-falling body. Outline how you
would measure g in this way, indicating the measurements needed and how you would
calculate a value of g from them.
(c) At one point on the line between the earth and the moon, the gravitational field caused
by the two bodies is zero. Briefly explain why is it so. If this point is 4 x 104 km from the
moon, calculate the ration of mass of the moon to the mass of earth. (Distance from earth
to moon = 4.0 x 105 km) [Answer: (c) 0.012]
3.3. ENERGY AND POWER
3.3.1 Concept of work and energy
A. Work
1. Definition: Is the product of the force applied and the distance moved in the direction of the
Types of work
Positive work: The work is positive when the direction of motion is in same direction of that
force.
58
Negative work: The work is negative when the direction of motion is appositive to the direction
of the force.
e.g.: work done against friction force
Zero work: the work is zero when the displacement is zero despite of the action of force
The SI unit of work is the joule work is a scalar quantity it has no property of direction but only
size.
Formula of mechanical work
Suppose that a force F pulls an object at a distance d along a line OA acting at angle O to H
F sin  – has n effect along OA
Then work done = F cos  dis tan ce
W  Fd cos
When θ = 90 0 this means that the force is normal to the displacement, then we conclude that
there is no work done by such force.
ENERGY
Def: Energy of body is its capacity to do the work. It is the total work that a body can do.
59
The SI unit of energy is the joule (j) the same as that of work.
Mechanical energy
There are 2 types of mechanical energy
-
Kinetic energy
Potential energy
Kinetic energy
Kinetic energy KE- is the energy possessed by a body due to its motion
The change of KE is equal to the work produced
KE = W = F d
From Kinematic, we know period
where a 
F= ma
v 2  v02
2d
v 2  v02
d
2a
KE = ½ m (v2-vo2) this is the work energy principle
“The net work done on an object is equal to the change in its kinetic energy''
Potential energy
Def: Potential is the energy of body has because of the relative position
Potential energy can either be gravitation potential energy or elastic potential energy
Gravitation potential energy
Is the energy possessed by a body by virtue of its position; it is equal to the work done against
the force of gravity.
Gravitation potential energy = mgh
Elastic potential energy
60
A compressed or stretched spring possesses elastic potential or strain energy.
This energy is due to the state of strain of the object.
pull
F
Extension
F
Work done in stretching the spring
= Average force x extension
0 F 
= 
e
 2 
= Fe
Work done is stored as elastic potential energy
Application exercise
Calculate the elastic gravitation PE stored in a spring when stretched through 4cm by a force of
2N
Ans
Elastic PE = Fe
= x2 (0.04) = 0.04j
Conservation of mechanical energy
Law of conservation of energy
61
The law of conservation of energy states that energy can neither be created nor destroyed but can
only is converted from one form to another.
Example of energy transformation
1. When a ball a thrown vertically upward the KE at the bottom is converted into gravitation
potential energy at the top most position.
2. An electrical motor converts electrical energy into mechanical energy
3. Bicycle dynamo converts mechanical energy into electrical energy
4. Etc A device which converts one form of energy to another is called a transducer
E.g:
Transducer
-
Transformation of energy
Microphone
Loudspeaker
Battery
Electrical water
- Sound to electrical
- Electrical to sound
- Chemical to light
- Electrical to heat
Statement of principle of conservation o mechanical energy
ME = KE+ PE
Summary
62
POWER
The power is defined as the time rate at which work is done. The average power delivered by an
agent is the total work done by the agent divided by the total time interval
Pav 
W
t
The instantaneous power delivered by the agent is
P
dW
dt
If the power is constant P  Pav 
W
and W  Pt
t
The international system of unit, the unit of power is 1 js 1 , which is called 1 watt
Work can also be expressed in the units of power  time . This is the origin of the term kilowatthour
We can also express the power (P) delivered to a body by a force F that acts on it by P  F  v
power  force  velocity
3.4. LINEAR MOMENTUM
10) Definition: momentum is a Physics term which refers to the quantity of motion that an
object has. If an object is in motion then it has the momentum. Momentum can be defined as the
mass in motion or the momentum of a body is the mass of the body multiplied by its velocity
63
According to Newton’s second law of mechanics, the net force acting on the body equals to the
rate of change of momentum.
This momentum is a vector quantity whose direction is the same as that of the velocity of the
body.
Its units is kg m/s or NS
Conservation of linear momentum
Consider a body A of mass m, moving with the speed u, colliding with a body B of mass m2
moving with speed u2 in the same direction.
Let v1 and v2 be the speed of A and B respectively after collision
u1
u2
v1
m1 A
m2 B
m1 A
Before collision
v2
B m2
after collision
Let FAB = force exerted by A on B
FBA = force exerted by Bon A
Using Newton’s 3rd law, FAB is equal but opposite to FBA
FAB= -FBA
(1)
Using Newton’s 2nd law
v u 
FAB  m1  1 1  (2)
 t 
 v  u2 
FAB  m2  2
 (3)
 t 
Where t is the duration of impact combining eq (1), eq (2) and eq(3)
v u 
 v  u2 
m1  1 1   m2  2

 t 
 t 
(4)
Rearrange the equation (4)
m1u1  m2u2  m1v1  m2 v2
total mometum before collision  total momentum after collision
64
The principle of conservation of momentum states that in a system of colliding bodies the total
linear momentum is conserved provided there are no external forces.
IMPULSE
Definition of impulse
The impulse is the total momentum change of the mass
From Newton’s second law
F
mv  mu
t
Fdt  mv  mu
Where Fdt – impulse
mv  mu -change of momentum
From the definition, the impulse is vector quantity and it has the same unit as the momentum
Application
1. Propulsion by reaction
Consider a jet propulsion which is illustrated by the behavior of in inflated ballon when released
with neck open.
The neck is close; there is a state of balance inside ballon with equal pressure
The neck is open the pressure on the surface opposite The neck is now
unbalanced and the ballon is force to move in the opposite direction to that of escaping to that of
escaping air.
65
According to the principle of conservation of momentum, the air and ballon have equal and
opposite amount of momentum.
M air  Vair  mballon  Vballon
2. Recoiling gun
Consider the gun which is held horizontally and when the short is from the gun an explosion
occurs.
Before explosion
after explosion
Both gun and short are at rest
M  0  m  0
mV  M   v
by conservation
0  mV  Mv
 mV  Mv
Jet engine
A jet engine is a reaction engine that discharges a fast moving jet which generates thrust by jet
propulsion in accordance with Newton's laws of motion. This broad definition of jet engines
includes turbojets, turbofans, rockets, ramjets, and pulse jets. In general, most jet engines are
internal combustion engines but non-combusting forms also exist.
66
Collision
Collision is an interacting between bodies in which the time interval during in which the bodies
interact is small relative to the time for which we can observe them.
To distinguish between collisions of bodies, we use the principle of the conservation of the
momentum and the kinetic energies.
In using the principle of conservation of the momentum, whenever objects collide in the absence
of external forces, the total momentum never changes.
Further, we may also apply the principle of the conservation of the kinetic energy to help
distinguish collisions:
The sum of the kinetic energies before collisions is equal to the sum of the kinetic energies after
collision. These two principles are useful in distinguishing collisions as explained.
So; there are two types of collision
-
Elastic collision
Inelastic collision
1. Elastic collision: is the collision where total Kinetic energy and total momentum is conserved.
a) Elastic collision (heat on)
Suppose two masses m1 and m2, moving with different velocity u1and u2 in the same direction
collide had-on with each other.
67
After collision, their velocities change to v1 and v2 respectively.
Then by application of conservation kinetic energy and momentum we find the following
deduction.
m1u1  m2 u 2  m1v1  m2 v 2

1
1
1
1
2
2
2
2
 2 m1u 1  2 m2 u 2  2 m1v 1  2 m2 v 2
(1)
m1u1  m2 u 2  m1v1  m2 v2

2
2
2
2
m1u 1  m2 u 2  m1v 1  m2 v2
(3)
m1 u1  v1   m2 v2  u 2 

2
2
2
2
 m1 u 1  v 1  m2 v 2  u 2



(2)
(4)
(5)

(6)
Divide (6) by (5)

m1 u 12  v 12
  m v
m1 u1  v1 
As
2
2
2
 u 22

m2 v 2  u 2 
a 2  b 2  (a  b)(a  b)
v2  u1  v1  u2
Use (7)
v1 
in (3), we get
2m2 u 2  u1 m2  m1 
m1  m2
In the same may we can find V2
v2 
68
2m1u1  u 2 m2  m1 
m1  m2
(7)
No –head –on
Collision of equal masses
Consider a sphere A of mass m and velocity u1 incident on a stationary sphere of equal mass
Suppose the collision is elastic and after collision A moves with V1 at the angle of its original
direction and B moves with the velocity V2 at angle B.
V = V1+V2
V1  V cos
V2  V cos
V2  V sin 
V1  V sin 
By using conservation of momentum
m1u1  mv1  v2 
u1  v1  v2
Worked example
A billiard ball moving with speed u1 = 3m/s is in the positive direction strikes an equal mass of
ball initially at rest the two balls are observed to more off at 450 ball above the x axis and ball 2
below. What are the speeds of two balls after collision
Collision of different masses
Ex: A snooker ball x of mass 0.3kg moving with velocity 5m/s hits a stationary ball m1=
0.4kg y moves off with a velocity of 2m/s at 300 to the initial direction. Find the velocity v of x
and its direction after hitting y.
V2 = 3m/s
69
Ex: A 7500kg truck travelling at 5m/s cost collides with a 1500kg can moving at 20m/s
In a direction 300 south west, the two vehicles remain tangled together with what speed and what
direction does he wastage begin to move
Explosion and defragmentation
When a body is exploded it is dived immediately in many particles which can moves at different
velocity. This process is called defragmentation.
The explosion can be accused on the body at rest or in motion to study this
we use the conservation of momentum.
Example of explosion and defragmentation
Rocket engine
Let us consider the rocket that is continuously losing mass due to the ejection of combustion gas
as shown in the figure below.
At time
at time
A
B
A
B
dm
t  dt
v  dv
m  dm
v  dv
Let: m- mass of rocket
v- Velocity of rocket at t
dm - The mass of gases ejected at dt
m  dm – Mass of rocket at t  dt
dv – The change of velocity at dt
V’ – velocity of ejected gases
Vr  v ,  v the absolute speed of gases relative to observer.
The acceleration of rocket will be conservation of momentum.
P before
Mv  (m  dm)(v  dv)  dm(v  v , )
–
=a
70
p after
ma  
dm v ,
dt m
Worked example
The mass of gases emitted from real of toy rocket is 0.1kg/s if the speed of gas relative to the
rocket is 2kg. What is initial acceleration of rocket?
Inelastic collision
Inelastic collision is the collision where KE is not conserved.
Conservation of momentum
V
m2 u 2  m1u1
m1  m2
Example
A bullet of mass 10 g travelling horizontally at speed of 1 x 102 m/s end ends itself in a block of
wood of mass 9.9 x 102 suspended by string so that it can suing
i) The vertical height through which the block rises
ii) How much of the bullet’s energy becomes internal energy (g = 10m/s2)
CHAP IV: ROTATION OF RIGID BODIES ABOUT FIXES AXIS
4.0. INTRODUCTION
A Rigid body is a system of particles whose relative position is fixed (i.e. the distance between
particles is fixed).
The rigid body motion can be
-
71
Linear motion or translation
Circular or rotation motion
4.1. CONCEPT OF ROTATIONAL MOTION
Def: This is the motion where its entire particle moves on circle about an axis with a common
angular velocity
In this case each particle is considered as a body which travels equals distance in equal time
along a circular path.
This body has a constant speed but not a constant velocity because of that
-
Speed is solar quantity
Velocity is a vector quantity
Angular displacement and angular velocity
The body a moving around a circular path
The distance travelled by a notating body on circular path is fire by
S=r
where r = radius
= Angular displacement
S = linear displacement
Then
=
Angular velocity
w
2
T
or
w  2f
where f 
1
T
LINEAR VELOCITY
As the definition, linear velocity is the rate of change of linear displacement by the time
=v
As v  r
V=
72
=
(r )
=
d
d
where
w
dt
dt
Therefore v  rw
And w =
Angular acceleration
Angular acceleration

of a rotating boauf is its rate of change of angular velocity
w  w0
The unit is rad/s2
t
So w  w0  t
Relation between w and θ
1
from   w0 t   t 2
2
w  w0
and t 

We obtain  

1
w 2  w02
2

4. 2. MOMENT OF FORCE OR TORQUE
We know that in linear dynamics a force F produces an acceleration in given by F=ma where m
is there mass of the object.
In an analogous way we soon see that a torque T applied to a rotating wheel gives it an angular
acceleration.
Definition
Torque or moment of force is the effectiveness of force in producing rotation about an axis .
It is measured by the product of the force and the perpendicular distance from the axis of rotation
to the line of action of the force.
Tongue = Force perpenducular distance from axis to line of action force
  F r
or   Fr sin 
The unit of tongue is Nm where
r-is the perpendicular project
73
is the angle of projection of the force F to the axis of rotation
4.3. MOMENT OF INERTIA
The mass of body is a measure of its built position to any change of linear momentum i.e mass
measures inertia
The corresponding property for rotational motion is called the moment of inertia.
Generally: the moment of inertia is a property of a body rotating about a particular axis.
Mathematically
n
I   mi r22 . 
i 1
n
I  r2
i 1
This is the same as the mass
In generally
unit: kg
This quantity depends on the mass and is taken as a measure of the moment of inertia of the body
about the axis.
Now let us define the term angular momentum of a rotating body similar to linear momentum
in the translation motion.
The angular momentum L is the product of the moment of inertia and angular speed of a rotating
body
L  I .w
Where I is the moment of inertia W is the angular velocity,
In this case the moment of force or torque is defined as the rate of change of the angular
momentum.
τ=
Because
= is the angular acceleration
This
τ=
Note that: if two torques are applied to a body, producing rotation in opposite direction, then one
of the torque is arbitral considered positive, and the other negative.
74
4.4. MOMENT OF A COUPLE OF FORCE
a. Concept of a couple
Every time we open a door, turn on a tap tighten up a nut with a spanner, we exert a turning
force.
Two factors are involved here, the magnitude of the force applied and the distance of its line of
action from the axis or fulcrum about which turning effect takes place.
A very large turning effect can be produced with comparatively small force provided the distance
from the furculum a large. For this reason it is easier to a right nut with a long spanner than with
a short one
The combined effect of the force and distance which determines the magnitude of the turning
force is called the moment of the force.
Application exercises
1. A known fly wheel of moment of inertia 0.3kg m2 is mounted on horizontal axle of
radius 0.01m and negligible mass compared with the fly wheel. Neglecting fraction, find
i. the angular acceleration if a force of 40N is applied tangentially to the axle
ii. the angular velocity of the flywheel after 10 second from rest
2. Two weights of mass 5.0kg and 7.0kg are mounted 4.0m apart on a light 100 as shown
below; calculate the moment of inertia of the system.
i.
When the system rotated about an axis halfway between the weights
ii.
When the system rotates about an axis 0.50m to the left of the 5.0kg mass.
N.B: moment of inertia for some important special cases about specified axes
75
Application
1. Four solid spheres a,b,c and d each of mass m and radius a are placed with their centre of the
four canners of square of side b
a) calculate the moment of inertia of the system about one side of the sphere if the moment of
about one side of the square if the moment of inertia of each sphere about its diameter is 2/5 ma2
76
b) Calculate the moment of inertia of the system about a diagonal of the sphere
4.5. ANGULAR MOMENTUM AND ITS CONSERVATION
Angular momentum and its relation to torque
In linear or straight an important property of moving object its linear momentum, when the
object spins or rotates about an axis its angular momentum plays an important parts in its motion.
Consider a particle A of rigid object rotating about an axis 0. As in the figure bellows
The momentum of A = mass x velocity
 m1v
 m1 r1 w
as v  rw
The angular momentum of A about O is defined as the moment of the momentum about O
Mathematically
Angular momentum  m1v1  r1  m  r1  r1  w  m1r12 w
total angular momentum   m1 r12 w
 w m1r12
total angular momentum  Iw
Where I is the moment of inertia of the body about O
Relation of angular momentum and torque
77
In linear dynamics a force F acting on an object for a time t produces a momentum change given
by F  t (impulse) = momentum change
In an analogous way a torque z acing for a time t on rotating object produces an angular
momentum change given by
  t (impulse)  angular momentum change
  t  Iw  Iw0
Where I – moment of inertia about the axis concerned
w Initial angular velocity
w0 – Final angular velocity
Conservation of angular momentum
Similar to the principle of conservation of linear momentum also Newton’s third law can be
used to derive the principle of conservation of the angular momentum.
Thus conservation of angular momentum correspond to the conservation of linear momentum
and starter that.
The angular momentum about an axis of a given rotating body or system of bodies is constant if
no external torque acts about the axis.
Work done by force acting on rotating body
a) Rotational Kinetic energy
If we consider the Kinetic energy of rotating object
The rotational KE of the object about an axis is equal to the sum of
Kinetic energy of all its individual masses.
KE 
1
1
m1v12  m1v22  ...........
2
2
KE 
1
1
m1r12 w 2  m1r22 w 2  ...........
2
2
1
1
 2
2
2
 m1 r1  m1 r2  .......... w
2
2

1
 Iw 2
2
KE 
78
rotational KE 
1 2
Iw
2
Work done by a TORQUE
Consider the wheel and opposite forces act tangentially on it; and rotating through an angle 
Work done by each force
 force  dis tan ce
 F  arc AB
 F  r 
Total work done by a couple = torque x angle of rotation
W  orque  
Its unit joule
Application exercice
1. A wheel turns about its centre with a uniform angular speed of 10 rad /s its moment of inertia
about the centre is 2kg m2. Calculate its KE.
4.5. KINETIC ENERGY OF ROLLING OBJECT
Consider a uniform cylinder rolling down on inclined plane without slipping. The object is
assumed to have two types of motion
- Motional motion
- Translation motion
At the same time
The total Kinetic energy is equal to the summation of translation and rotational Kinetic energy
Total Kinetic = ½ mv2 + ½ Iw2
79
If the cylinder does not slip then V  rw
Total KE = ½ mv2 + ½ I ( ) 2
= ½ v2 (m+ )
(1)
Suppose that the cylinder rolls from rest through a distance, s, along the plane the loss of
potential energy = gain in KE.
mgd sin  
v2 
1 2
I 
v m  2 
2 
r 
2mgd sin 
I
m 2
r
(2)
but v 2  2ad
a
2mg sin 
I
m 2
r
(3)
For uniform cylinder
mr 2
I
2
Using (4) in (2)
80
(4)
v2 
v2 
2mgd sin 
m
m
2
4 gd sin 
3
(5)
For Kinematics of rectilinear motion
v 2  2ad
Then equation 5 becomes
a
2 gd sin 
3
(5)
Exercises
1.
A bicycle wheel has a moment of inertia of 0.25 kg m 2 about its axle.
The wheel rotates at 120 rpm. Calculate:
(a) the angular momentum of the wheel
(b) the rotational kinetic energy of the wheel.
2.
A turntable of moment of inertia 5.8 × 10 –2 kg m 2 rotates freely at
3.5 rad s –1 with no external torques. A small mass of 0.18 kg falls vertically onto
the turntable at a distance of 0.16 m from the axis of rotation.
0.16 m
Calculate the new angular speed of the turntable.
3.
4.
A turntable rotates freely at 40 rpm about its vertical axis. A small mass of 50 g
falls vertically onto the turntable at a distance of 80 mm from the central axis.
The rotation of the turntable is reduced to 33 rpm.
Calculate the moment of inertia of the turntable.
A CD of mass 0.020 kg and diameter 120 mm is dropped onto a turntable rotating
freely at 3.0 rad s –1 .
The turntable has a moment of inertia of 5.0 × 10 –4 kg m 2 about its rotation axis.
(a) Calculate the angular speed of the turntable after the CD lands on it. Assume
the CD is a uniform disc with no hole in the centre.
81
(b)
5.
Will your answer to part (a) be bigger, smaller or unchanged if the hole in
the centre of the CD is taken into account? Explain your answer.
A turntable rotates freely at 100 rpm about its central axis. The moment of inertia
of the turntable is 1.5 × 10 –4 kg m 2 about this axis.
A mass of plasticine is dropped vertically onto the turntable and sticks at a
distance of 50 mm from the centre of the turntable.
50 mm
The turntable slows to 75 rpm after the plasticine lands on it.
Calculate the mass of the plasticine.
6.
An ice skater is spinning with an angular velocity of 3.0 rad s –1 with her arms
outstretched.
The skater draws in her arms and her angular velocity increases to
5.0 rad s –1 .
(a) Explain why the angular velocity increases.
(b) When the skater’s arms are outstretched her moment of inertia about the spin
axis is 4.8 kg m 2 .
Calculate her moment of inertia when her arms are drawn in.
(c) Calculate the skater’s change in rotational kinetic energy.
(d) Explain why there is a change in kinetic energy.
7.
A solid sphere of mass 5.0 kg and radius 0.40 m rolls along a horizontal surface
without slipping. The linear speed of the sphere as it passes point A is 1.2 m s –1 .
A
8.
82
As the sphere passes point A calculate:
(a) the linear kinetic energy of the sphere
(b) the angular velocity of the sphere
(c) the rotational kinetic energy of the sphere
(d) the total kinetic energy of the sphere.
A solid cylinder of mass 3.0 kg and radius 50 mm rolls down a slope without
slipping.
0.60 m
40°
The slope has a length of 0.60 m and is inclined at 40° to the horizontal.
(a)
(b)
Calculate the loss in gravitational potential energy as the cylinder rolls from
the top to the bottom of the slope.
Calculate the linear speed of the cylinder as it reaches the bottom of the
slope.
CHAP V: STATICS
5.0. INTRODUCTION
Statics is a part of physics mainly subpart of dynamics which dials with the forces in equilibrium
Those forces would be coplanar or non-coplanar
They must be parallel or non-parallel and the must be acted on the same body
1. Important definition
1. Coplanar concurrent collinear force system:
Is the simplest force system and includes those forces whose vectors lie along the same straight
line.
P1
P2
i.e. Force system : is a collection of forces acting on a body in one or more planes
2. Coplanar concurrent non- parallel force system
Forces whose lines of action pass through a common point are called concurrent forces in this
system lines of action of all the forces meet at a point but have different directions in the same
plane.
83
P3
P2
P1
P4
Coplanar non- concurrent parallel force system
In this system the lines of action of all the forces lie in the same plane and are parallel to each
other but may not have same direction.
P1
p2
p3
p4
P1
p2
4. Coplanar non- concurrent free system
Such system exists where the lines in the same plane but do not pass through a common point.
P1
P2
R
W
Addition of forces ‘resultant force’
A resultant force is a single force which can replace two or more forces and produce the same
effect on the body as the forces.
Laws of forces
The method of determination of the resultant of same forces acting simultaneously on a particle
is called composition of forces the various laws used for the composition of forces are given as
under.
 Parallelogram law forces
84
 Triangle law of forces
 Polygon law of forces
1. Parallelogram law of forces
It states that if two forces, acting simultaneously on a particle, be represented in magnitude
parallelogram then their resultant may be represented in magnitude and the direction by the
diagonal of the parallelogram which passes through their point of intersection
Analytical method
From the right – angled triangle odc
=
R=
=
=
R=
Let the resultant makes makes an angle
Then
tan
=
with P as show above
=
=
1st case: if O= o ie p and Q act along the same straight line
Then R = P+Q
2nd case: O = 900 – p and Q act right angle each other, then
R=
85
3rd case: if O = 1800, P and Q act along the same straight line but in opposite direction
R PQ
The resultant will act in the direction of the greater force
2. Triangle law of forces
It states that if two forces acting simultaneously on a body are represented in magnitude and the
direction by the two sides of triangle taken in order then their resultant may be represented in
magnitude and the direction by the third side taken in opposite order.
Analytical method
 0CA    
 0CA  
 0CA  180 0  
0a
ac
oc


sin(   ) sin  sin 180 0  


or
Q
P
R


sin(   ) sin  sin 180 0  


or
Q
P
R


sin(   ) sin  sin 
Lami’s theorem
It states that: If three coplanar force acting on a point in a body keep in it equilibrium, then each
is proportional to the sine of the angle between the other two forces.
86
or
Q
P
R


sin  sin  sin 
Polygon law of forces
It states that:
If a number of coplanar concurrent forces, acting simultaneously on a body are represented in
magnitude and direction by the sides of a polygon taken in order, then their resultant may be
represented in magnitude and the direction by the closing side of a polygon taken in the opposite
order.
Resultant by analytical method
The resolved parts in the direction ox and oy of
P1
87
are
P1 cos1
and
P1 sin 1 respectively
P2
are
P2 cos 2
and
P2 sin  2 respectively
P3
are
P3 cos 3
and
P3 sin  3 respective ly
If the resultant R makes angle  with ox then by the principle of resolved parts
R cos  P1 cos 1  P2 cos 2  P3 cos 3
H
And
R sin   P1 sin  1  P2 sin  2  P3 sin  3
 V
R
 H   V 
2
2
Angle made by the resultant
R sin 
 tan   tan  
R cos
  tan 1
H
V
H
V
Note that :
While solving problems, proper care must be taken about signs (+ve or –ve) of resolved parts
Vertical components


Vertical direction
Downward direction
positive (+)
negative (-)
Horizontal components
From left to right: positive (+)
Direction
From right to left: negative (-)
5.1. EQUILIBRIUM CONDITION FOR COPLANAR CONCURRENT FORCES
When several forces act on a particle, the particle is said to be in equilibrium if there is no
unbalanced forces acting on it.
The resultant of all forces acting on the particle is zero
88
Analytical method
1. The algebraic sum of component of all forces in a direction perpendicular to the first
direction, which may be taken as vertical in their plane, must be zero

 Fy  0
2. The algebraic sum of components of all the forces in a direction perpendicular to the first
direction, which may be taken as vertical in their plan, must be zero

F
x
0
5.2. EQUILIBRIUM CONDITIONS FOR BODIES UNDER COPLANAR NONCONCURRONT FORCES
When a body is under the action of a coplanar non-concurrent force system it may rotate due to
resultant moment of the force system or it may set in a horizontal or vertical due to horizontal
and vertical components of forces.
The body can be in equilibrium if the algebraic sum of all external forces and their moment
about any point in their plane is zero.
Mathematically

 Fy  0

F
0
Law of moment
x

M y  0

M
0
N.B All the three conditions of equilibrium spoken above have to be fulfilled.
x
Centre of the gravity of a body
Definition: The centre of gravity of a body is the point through which the whole weigh of a body
may be assumed to act
It is usually denoted by c.g Or simply a, the position of G and this may or may not necessarily
be within the boundary of the body.
Determination of centre of gravity
The centre of gravity of some object may be found by balancing the object on a point.
89
The C.g of some object may be found by suspension take an object the section of which a shown
below, suspend it from point L. The body will not come to rest until its resultant weight is
vertical downward from A. Through L, draw a vertical. line LN then suspend the body from a
point M, and let come to rest, through M draw a vertical line MT the point of LN and MT is the
position of the centre of interaction of gravity.
POSITION OF CENTRE OF GRAVITY OF REGULAR SOLIDS
If the body has no geometrical form, we use the experiment method, as follows:
- Suspend the cardboard by a string. The cardboard is attached to a retort stand and the
string is made in lath.
- If the cardboard is in equilibrium, you draw on it a vertical line.
- suspend the cardboard by another point and if it is in equilibrium state, draw a vertical
line.
- Then, the center of gravity is located at the common point of the two lines
90
The gravitational force can make some object stable and others unstable. 3 states of equilibrium
can rise.
1. Stable equilibrium: an object is in stable equilibrium when if deviated from the equilibrium
position it raises its center of gravity. For example: a cone on its base.
2. Unstable equilibrium: an object is in unstable equilibrium if when deviated from the
equilibrium position; its center of gravity is at its lowest position compared to its position before
deviation. For example: a cone on its top.
3. Neutral equilibrium: an object is in neutral equilibrium if when its center of gravity deviated
from the equilibrium position remains where it was before deviation. For example: a geometric
board.
For having a stability of a body you can make its center of gravity lower (at low position) or
make its base wider.
NB: The center of mass is the point of application of resultant force of all forces acting on a
body.
Exercises
1. a) what is meant statics .
b) Summarize the various conditions which are being satisfied when the body is in
equilibrium.
c) A 3.00m long, 240N, uniform rod at zoo is held in a horizontal position by two ropes
0
at its end as shown below. The left rope makes an angle of 150 with the rod and the
right rope makes an angle θ with the horizontal. A 90N bowler monkey (Alouatta
seniculus) hangs motionless 0.50m from the right end of the rod as he carefully studies
you. Calculate the tension in two ropes and the angle θ. Fist make a free-body diagram of
the rope ./ 7marks
91
2. A 172 cm tall person lies on a light (massless) board which is supported by two scales,
one under the top of her head and one beneath the bottom of her feet as shown. The two
scales read, respectively, 35.1kg and 31.6 kg. What distance is the center of gravity of
this person from the bottom of her feet?
3. A 350 N, uniform, 1.5 m bar is suspended horizontally by two vertical cables at each end.
Cable A can support a maximum tension of 500 N without breaking, and cable B can
support up to 400 N. You want to place a small weight on this bar.
a) What is the heaviest weight that can put on without breaking either cable?
b) Where this weight should be put?
4. A uniform 225 N ladder leans against a smooth vertical wall. The ladder is 9.00 m in
length. It makes an angle of 78.0 with the deck. A 65.0 kg Bengal tiger rests on one of
the rungs 3.00 m from the top end of the ladder. (a) Find the force of friction exerted by
the deck on the bottom of the ladder. (b) What is the upward force exerted by the ladder?
5. The object in the figure below is in equilibrium and has a weight Fw  80 N . Find
FT 1 , FT 2, FT 3 and FT 4 . Give answers to two significant figures
92
6. The figure at the right is for the next three questions. The sign shown in the figure
weighs 200 N. The boom is of uniform construction (the center of mass of the boom is in
its center). The force exerted by the hinge on the boom is 300 N and acts at an angle of
20 above the horizontal. Find the tension in the cable, the weight of the boom, and the
angle the cable makes with the positive x-axis:
7. A sphere of weight 20N and radius 15cm rests against a smooth vertically wall. The
sphere is supported in this position by a string of length 10cm attached to a point on the
sphere and to a point on the wall as shown below
a) Copy the diagram and show the forces acting on the sphere
b) Calculate the reaction on the sphere due to the wall
c) Find the tension in the string.
8. Forces of 2.83 N,4.00N and 6.00 N act on an object O as shown in figure below
93
Find the resultant force on the object
9. Name the type of equilibrium for each position of the ball.
10. Which the configurations of brick a) or b) is more likely to be stable
11. Find the tension in two cords shown below neglect the mass of the cords and assume that
the angle  is 33 0 and the mass is 200kg
12. A 20kg sphere rests between two smooth planes as shown below. Determine the
magnitude of the force acting on the sphere exerted by each plane
13. A shop sign weighing 245N is supported by a uniform 155N beam as shown in figure
below. Find the tension in the guy wire and the horizontal and vertical forces exerted by
the hinge on the beam
94
CHAP VI: FLUIDS MOTION
6.0.INTRODUCTION
A stream or river flows slowly when it runs through open country and faster tough narrow
openings or constrictions. This is due to the fact that water is practically an incompressible
fluid that is changes of pressure cause practically no change in fluid density at various parts.
Fluid flow can be described in terms of two main types.
Streamline flow and turbulent flow or laminar flow
Streamline flow/ steady flow
When the particles of fluid passing successively through a fluid, follow the same path
Turbulent flow: Is an irregular flow characterized by small whirlpool
6.1. VISCOSITY
The term viscosity is used in fluid to characterize the degree of internal friction in the fluid.
This internal friction is associated with the resistance to two adjacent layers of the fluid to more
relative to each other.
95
Note that: steady flow thus involves parallel layers of fluid sliding over each other with different
velocities this creates viscous forces acting between tangentially (as shear forces do between the
layers and impending their motion.
In our model of an ideal fluid, we can make the following assumptions
1. Non viscous fluid
in a non viscous fluid, internal friction is neglected. An object moving through a non
viscous fluid would experience no retarding viscous force.
2. Steady flow
In steady flow we assume that the velocity of the fluid at each point remains constant in
time.
3. Incompressible fluid
The density of the fluid is assumed to remain constant in time.
4. Irrotational flow
Fluid flow is Irrotational if there is no angular momentum of the fluid about any point. If a small
does not rotate about its center of mass, the flow is Irrotational. If flow wheel were to rotate, as it
would if turbulence were present, the flow could be rotational.
If adjacent layers of a material are displaced laterally over each other as in the following figure.
The deformation of the material is called a shear
The simplest type of fluid flow basically involve shear . A fluid is therefore sheared when it
flows pass a solid surface and the opposition set up by the fluid to shear is called Viscosity.
6.1.1. COEFFICIENT OF VISCOSITY
To obtain a definition of viscosity, we consider two layers moving each organist other.
The velocity gradient (change of velocity /distance) in a direction perpendicular to the velocities
is δV/δY
The tangential stress between the layers is therefore F/A where A is the area of contact. Thus the
coefficient of the viscosity n is defined by the equation.
=
96
Is small and F/A large, n is large and the fluid very viscous.
For many pure fluids (water), η is independent to the velocity gradient at a particular
temperature.
η is constant and so the tangential stress is directly proportional to the velocity gradient . Fluids
for which this is true the called Newtonian fluids.
Then, for some liquids such as paints glues and liquid cements, η decreases as the tangential
stress increases and these are sound to be Thixotropic
6.2. STEADY AND TURBULENT FLOW
6.2.1. STEADY FLOW
Take the streamlines for steady flow in a circular pipe
Everywhere there are parallel to the axis of the pipe and represent velocities varying from zero at
the wall of the pipe to the maximum at its axis. The surfaces of equal velocity are the surfaces of
concentric cylinders.
An expression for the volume of liquid passing per second Q through a pipe when the flow is
steady, can be obtained by the method of dimensions.
Q depends on: i) the coefficient of viscosity of the liquid
ii) the radius of the pipe
i)
The pressure gradient (p/l) causing the flow where p is the pressure difference
between the ends of the pipe and l is its length.
Q  k x r y ( p / l ) z Where x, y and z are indices to be found and k is a dimensionless constant.





 

The dimensions of Q is L3T 1 , of  is ML1T 1 , of r L , of p MLT 2 / L2  MT 2 L1 and of
l L .Hence the dimensions of p/l are ML2T 2

 

Equating dimensions L3T 1 = ML1T 1

 L ML
x
y
Equating indices of M, L and T on both sides
0=x+z
97
2
T 2

z
3=-x+y-2z
-1=-x-2z
Solving, we get x=-1, y=4 and z=1
kpr 4
Hence Q 
the value of k cannot be obtained by the method of dimensions but a fuller
l

analysis shows that it equals
and so the complete expression is
8
pr 4
Poiseuille’s formula
Q
8l
6.2.2. TURBULENT FLOW
When the velocity of flow exceeds a particular critical value the motion become turbulent, the
liquid is churned up and the streamlines are no longer parallel and straight.
To study the stability of fluid (liquid) flow, we can use the Reynolds’s number (Re) it is defined
by the equation.
Re=
Where η and one the viscosity and the density respectively of the fluid, is the speed of the
bulk of the fluid and l is characteristic dimension of the solid body concerned.
For cylindrical pipe l is usually the diameter (2r) of the pipe.
Experiment shows that for cylindrical pipes,
When
Re 2200 flow is steady
Re 2200 flow is unstable (critical velocity )
Re 2200 flow is usually turbulent
6.3. MOTION IN A FLUID
6.3.1. STOKE’S THEOREM
The streamlines for a fluid flowing slowly past a stationary solid sphere are shown here
98
When the sphere moves slowly rather than the fluid, the pattern is similar but the streamlines
then show the apparent motion of the fluid particles as seen by someone on the moving sphere.
Viscous forces are thereby brought into play and constitute the resistance experienced by the
moving sphere. If we make the plausible assumption that the viscous retarding force F depends
on : i)the viscosity
ii)
The velocity v and the radius of the sphere, then the expression can be derived for F
by the method of dimensions. Thus
; x,y and z are indices to be found

and k is a dimensionless constant.

 
The dimensional equation is MLT 2  ML1T 1
 LT  L
x
1 y
z
Equating indices of M, L and T on both sides
1= x
1=-x+y+z
-2=-x-y
Solving we get x=1, y=1 and z=1 so F  kvr
A detail treatment, first done by Stokes gives k  6 and so F  6vr ; this is a Stokes’ law
only holds for steady flow in fluid of infinite extent.
Terminal velocity
Now consider the sphere falling vertically under gravity in a viscous fluid. three forces act on it .
-
99
Its weight , W acting down wards ,
The up thrust , U due to the weight of fluid displaced acting upwards , and
The viscous drag F, acting upwards
The resultant downward force equals (W-U-F) and causes the sphere to accelerate until its
velocity and the viscous drag reach values such that
W-U-F = 0
The sphere then continues to fall with a constant velocity, known as its terminal velocity velocity
of say v t now
4
w  r 3 g
3
Where  is the density of the sphere and
4
u  r 3  g
3
Where ρ is the density of the fluid
Also, if stead condition still hold, when velocity v t is reached then , by stokes’s
F  6 r vt
Hence
4 3
4
r  g  r 3  g  6 r vt  0
3
3
2
2r (   ) g
vt 
9
If the velocity of the moving body into fluid reaches a critical velocity, the motion becomes
turbulent. At velocity greater than vc the resistance to motion (drag) increases and is proportion
al to square of the velocity
Where A- is the cross section area of the body perpendicular to its velocity
ρ-is the fluid density
C-is the drag coefficient it has a numerical value between 0 and1
6.3.2. EQUATION OF CONTINUITY
Consider a fluid flowing through a pipe of non uniform site size as in the following figure.
The particles in the fluids move along the streamlines in steady flow; at all point the velocity of
the particle is tangent to the streamline along which it moves
100
In small time interval t, the fluid at the bottom end of the pipe moves at a distance x1  v1 .t .
If A1 is the cross section area in this region, then mass contained in shaded region is
m1  1 A1 x
 1 A1V1 t
Similarly, m2   2 A2V2 t
However, since mass is conserved and because the flow is steady , the mass that cross A1 in a
time t must equal the mass that cross A2 in time t
Therefore m2  m1
=
This expression is called the equation of continuity
Since
is constant for an incompressible fluid the above equation becomes.
=
=contant
That is the product of the area and the fluid speed at all points along the pipe is constant.
6.4. BERNOUILLI’S EQUATION AND ITS APPLICATION
As fluid moves through a pipe or hallow of varying was section and elevation. the pressure will
change along the pipe.
In 1740 Bernoulli obtained a relation between the pressure and velocity at different parts of a
moving incompressible fluid.
If the viscosity of the fluid is negligibly small. There is no frictional force to overcome; in this
case the work done by the pressure difference per unit volume of a fluid following along a pipe
steadily is equal to the gain in kinetic energy per unit volume plus the gain in potential energy
per unit volume.
Consider the flow through a non uniform pipe in a time t. The force on the lower end of force
fluid is given by.
F1  P1 A1
Where P1 is the pressure a point 1,
101
The work done by this force described on the lower end is given by.
w1  F1  x1  P1 A1x1
Where: v1 is the volume of the lower shaded region
In similar manner, the work did on the fluid at the upper end in the time t is given by.
w2  F2  x2  P2 A2 x2
Because the volume that passes through (1) in a time t equals the volume that passes through
(2) in the same time internal is
Thus the net work done by these forces in the time t is W  P1  P1 v
Part on this work goes into changing the KE of the fluid and part goes into changing the
gravitational energy.
If m is the mass passing through the pipe in the time t , then the change in its kinetic energy is
1
1
KE  mv22  mv12
2
2
The change in its potential energy is
1
1
PE  m gy2  m gy1
2
2
We can apply the work done theorem in the form.
W= KE+ PE
Then
P1  P1 v  1 mv22  1 mv12  1 m gy2  1 m gy1
2
2
2
2
If we divide each term by V, and recall that
m

V
The above equation
Reduces to
P1  P2   1 v22  1 v12  1 gy2  1  gy1
2
2
Rearranging terms, we get
102
2
2
P1 
1 2 1
1
1
v1   gy1  P2  v22  gy2
2
2
2
2
This is Bernoulli’s equation and it is usually stated by saying that along a streamline in an
incompressible inviscid fluid.
It is also often expressed as
P+
+
Bernoulli’s principle says that sum of the pressure; the kinetic energy per unit volume, the
potential energy per unit volume has the same value at all points along a streamline.
Application of Bernoulli’s principle
1. A sanctions effect is experienced by a person standing close to the platform at a station
when a fast train passes the fast moving air between the person and train produces a
decrease in pressure air. The excess air pressure on the other side pushes the person
toward the train .
2. Filter pump a filter pump has a narrow section in the middle, so that a jet of water from
the tap flows faster here. This cause a drop in pressure near it and air therefore flows in
from the side tube to which a vessel is connected. The air and water together are forced
through the bottom of the filter pump . a similar principle is used in the engine carburetor
for vehicles.
3. Airfoil lift: the curved shape of an aerofoil creates a faster flow of air over its top surface
than the lower one. This is shows by the closeness of the streamlines above the aerofoil
compared with those below.
From Bernoulli’s principles, the pressure of the air below is greater than that above and
this produces the lift on the aerofoil.
4. Venturi meter
This meter measures the volume of gas or liquid per second flowing through gas pipes or
oil pipes.
103
The meter consists of a horizontal tube with a constriction and replaces part of the piping
of a system. The two vertical tubes record the pressures (above atmospheric) in the fluid
flowing in the normal part of the tube and in the constriction
The venturi meter
If and p2 are the preserve and v1 and v2 the velocities of the fluid of density  and L
and M are at the same horizontal level. Then assuming Bernoulli’s equation holds.
1 2
1
2
v1  gh1  P2   v2  gh2
2
2
Since =
Then
1
1
2
P1  v12  P2   v2
2
2
Q  The quantity of the volume per unit time
Q  A1v1  A2 v2
Q
Q
v1 
, v2 
A1
A2
Therefore
P1 
2 A12 A22 P1  P2 
A12  A22
5) Spinning ball
Q
If a tennis ball is cut, it spins as it travels through the air and experiences a sideways force which
causes it to curve in flight. This is due to air being dragged round by the spinning ball, thereby
increasing the air flow on one side and decreasing it on the other. A pressure difference is thus
created.
Measurement of fluid velocity “Pitot –static tube”
The velocity of a point in a fluid flowing through a horizontal tube can be measured by the
application of the Bernoulli’s equation.
104
In this case is zero and so
P
1 2
v  Constant t
2
Here P is the static pressure at a point in the fluid, which is the pressure unaffected by its
1
velocity. The pressure P  v 2 is the total or dynamic pressure that is the pressure which
2
then fluid would exert if it were bought to rest by striking a surface place normally to the
velocity at the point concerned.
FIGURE
Principe of Pitot –static to be the inner or Pitot tube P. was an opening at one end normal to
the fluid velocicity. a monometer connected to T would measure
. The outer or static
tube has hole Q in its side which are parallel to the fluid velocity. Monometer connected to s
would measure the static pressure P.
In Bernoulli’s equation
1
1
P  gh  v 2  Constant
2
2
The static component is given by P 
1
gh if the flow is horizontal the dynamic component by
2
1 2
1
v and the total pressure by v 2 .
2
2
1
Total pressure –static pressure = P  v 2  P
2
V=
total pressure –static
EXERCISES
1. a) Explain the
i) Terms laminar flow and turbulent flow
ii) Effects of temperature on the viscosity of liquids
b) Water flows steadily along a uniform flow tube of cross- section 30cm2. The static pressure is 1.20
x 105 Pa and the total pressure is 1.28 x 105 Pa. calculate the flow velocity and the mass of water
per second flowing past a section of the tube
2. Water flows along a horizontal pipe of cross- sectional area 48 cm2 which has a constriction of
cross sectional area 12 cm2 at one place. If the speed of water at the constriction is 4m/s. calculate
the speed in wide section. The pressure in the wider section is 1.0 x105 Pa. calculate the pressure at
the constriction. (Density of water 1000kg/m3).
105
3. A venturi meter equipped with a differential mercury manometer is shown below. At the inlet,
point 1, the diameter is 12cm. while at the throat, point 2. The diameter is 6.0 cm. what is the flow
Q of water through the meter if the mercury reading is 22 cm? the density of mercury is 13.6g/cm3.
3
4. A metal sphere of radius 2.0 10 m and mass 3.0 10 4 kg falls under gravity, centrally down a
0
wide tube filled with the liquid at 35 c. the density of the liquid is 700 kg m 3 . The sphere attains a
2
1
terminal velocity of the magnitude 40.0 10 ms . The tube is emptied and filled with another
liquid at the same temperature and of the density 900 kg m 3 . When the metal sphere falls centrally
down the tube, it is found to attain a terminal velocity of the magnitude 25.0 10 2 m s . Determine
0
at 35 c , the ratio of the coefficients of viscosity of the second liquid to that of the first.
5. Water at a gauge pressure of 3.8 atm at street level flows into an office building at speed of 0.60m/s
through a pipe 5.0cm in the diameter. The pipe tapers down to 2.6cm in the diameter by the top
floor, 18m above as the figure below, where the faucet has been left open. Calculate the flow
velocity and gauge pressure in such a pipe on the top floor.( ignore the viscosity)
6. Water runs into a fountain, filling all the pipes, at a steady rate of 0/750m/s3. a) How fast will shoot
out of a hole 4.50cm in diameter? b) At what speed will it shoot out if the diameter of the hole is
three times as large?
7. Water is flowing in a pipe with a varying cross sectional area, and at all point the water completely
fills the pipe. At point 1 the cross section area of the pipe is 0.070m2, and the magnitude of the fluid
velocity is 3.50m/s.
a) What the fluid speed at points in the pipe at points in the pipe where the cross sectional area
is a) 0.105m2 and 0.047m2?
b) Calculate the volume of water discharged from the open end of the pipe in 1.00hour
8. Water is flowing in a pipe with a varying cross sectional area, and at all point the water completely
fills the pipe.
a) At one point in the pipe the radius is 0.150m. What is the speed of water at this point if the
water is flowing into this pipe at a steady rate of 1.20m3/s?
b) At second point in the pipe the water speed is 3.80m/s. what is the radius of the pipe at this
point?
106
9. What gauge pressure is required in the city water mains for a stream from a fire hose connected to
the mains to reach a vertical height of 15.0m? (Assume that the mains have a much larger diameter
than the fire hose.)
10. At one point in the pipeline the water’s speed is 3.00m/s and gauge pressure is 5.00× 104Pa. find
the gauge pressure at a second point in the line, 11.0m lower than the first, if the pipe diameter at
the second point twice that at the first.
11. Air streams horizontally past small airplane’s wings such that the speed is 70.0m/s over the top
surface and 60.0m/s past the bottom surface. If the plane has a wing area of 16.2m2 on the top and
on the bottom, what is the net vertical force that the air exerts on the airplane? The density of the
air is 1.20kg/m3.
12. Water enters a house through a pipe 2.0 cm I.D. (1 cm radius) at an absolute pressure of 4x105 Pa.
The pipe leading to the 2nd floor bathroom 5m above is 1.0 cm in diameter (0.5 cm radius).
Velocity at the inlet is 4m/s. Find v and P at the bathroom
13. Water at a pressure of 3.8atm flows into the street level of an office building at a speed of 0.60m/s
through a pipe 5.0 cm in radius. The pipe tapers to 2.6 cm in radius by the top floor 20 m above.
Compute the flow velocity and pressure at the top floor. All pressures are gauge pressures
14. A garden sprinkler has 150 small holes each 2.0mm2 in area. If water is supplied at the rate of
3.0x10-3m3s-2, what is the average velocity of the spray?
15. A horizontal pipe 10cm diameter has a smooth reduction to a pipe 5cm in diameter. If the pressure
of the water in the larger pipe is 8.0x104Pa and the pressure in the smaller is 6x104 Pa, at what rate
does water flow through the pipe?
16. A water hose 2.50 cm in diameter is used by a gardener to fill a 30.0L bucket. The gardener notes
that it takes 1.00 min to fill the bucket. A nozzle with an opening of cross-section area 0.500 cm2 is
then attached to the hose. The nozzle is held so that water is projected horizontally from a point
1.00 m above the ground. Over what horizontal distance can the water are projected
CHAP VII: KINETIC THEORY OF MATTER
7.0. INTRODUCTION
The Kinetic Theory explains the differences between the three states of matter. It states that all
matter is made up of moving particles which are molecules or atoms. In solids, the particles are
so tightly bound to each another that they can only vibrate but not move to another location.
In liquids, the particles have enough free space to move about, but they still attract one another.
In gases, the particles are far apart and can move about freely since there is much free space.
Solids change into liquids, and liquids into gases, when the particles gain more kinetic energy,
like when being heated and are able to move apart from one another. When the molecules vibrate
more quickly upon heating, some of it escapes from the matter.
107
7.1 BASIC ASSUMPTIONS OF KINETIC THEORY.
1. Matter consists of small particles “the matter consists of large number of a very small particles
either individual atoms or molecules”.
2. Between these small particles there are small spaces (called intermolecular spaces).
3. The particles act on all neighboring particles with the forces of attraction (called
intermolecular forces).
4. All particles of any substance are in more or less rapid motion (molecular motion).
7.2. FORCES ACTING BETWEEN MOLECULES
The force between molecules of a certain substance depended: on the arrangement of the
molecules throughout the bulk of substance (material), the separation of the molecules and the
motion of molecules now the forces are attractive or repulsive.
The force at any point is found from taking the gradient of the potential energy curve, in other
words F = -dV/dr, where V is the potential energy.
Two forces act between the molecules:
(a) the repulsive force which predominates at
short distances
(b) the attractive force which predominates
at long distances
You can see from the graph that when the
molecules are close to each other the
repulsive force predominates, while at
greater distances the attractive force is
larger. The resultant force Is:
(a) repulsive from 0 to M
(b) attractive from M to B but increasing
with distance, and
(c) attractive from B to infinity but
decreasing with distance.
There is a position where the two forces
balance, shown by M on the graph. This is
the equilibrium position for molecules in the solid.
7.3. SOLIDS
a) Nature of solids
The theory states that in solid the molecules ore close together and the attractive and repulsive
forces between neighboring molecules balance.
Also each molecule vibrates to end fro about a fixed position.
108
The molecule in solid could be arranged in a regular repeating pattern like these formed by
crystalline substance.
Thus, they are crystalline and their shape is definite.
B Adhesion and cohesion
The force of attraction between molecules of the some kind is known as cohesion. In solids they
exist a force of cohesion which is strong.
Adhesion is the force between molecules of different kinds
“The force of adhesion between solid and liquid is greater than that between water and gas”.
C Ductility and malleability
Ductility: is the physical property of a material where it is capable of sustaining large permanent
changes in shapes without breaking
An example is when a piece of metal is drawn into wire it is characterized by the internal
structure of material flowing under shear stress.
Gold, copper aluminum and steel have high ductility.
Malleability:
Is a similar property is a materials ability to deform under compressive stress; this is often
characterized by the material’s ability to form a thin sheet by hammering or rolling.
Ductility and malleability are not always coextensive – for instance, while gold has high ductility
and malleability, lead has low ductility but high malleability. The word ductility is sometimes
used to embrace both types of plasticity.
a) Elasticity
An elastic material such a rubber recovers its original shape and size after the force deforming it
has been removed.
A material which does not recover but is deformed permanently like plasticine is plastic.
Elasticity: is a physical property of material which return to their original shape after the stress
that caused their deformation is no longer applied.
Stress and strain
Stress: is the force (in N) acting on unit cross-section area (1m2) or stress is the ratio of the
internal force, F, that occurs when the shape of substance is changed to the area, A, over which
the force acts.
stress 
109
force
F
 
area
A
The unit of stress is the Pascal (Pa)
Strain is the relative amount of deformation produced in a body under stress or is the extension
of unit length (1m)
If e= extension and l-original length then
strain 

change in dimension
extension

oroginal lenght
original length
e
l0
Strain has no unit because it is a ratio of like quantities.
Stress – strain graph
Along 0A stress is proportional to strain and we wire regains to original length when stress in
removed.
Along AL the material is elastic but does not abbey Hooke’s law.
Along regain LB the material is plastic – it does not regain its original length when the load is
removed.
Hooke’s law and elastic modulus
Hooke’s law states that provided the elastic limit is not exceeded, the extension is directly
proportional to the applied force or tension in wire.
110
Graph of force against extension
In the section OE the graph is straight line passing through the origin , obeying Hooke ‘s law up
to the point E, each time the stretching of a simple is unloaded, it return to its original position.
Beyond the point E, the extension is no longer proportional to the applied force. Point E is called
the elastic limit .Beyond this point the graph is no longer a straight line, showing that Hooke’s
law doesn’t apply any more. The stretching of a sample has more permanently deformed
b) Elastic modulus
All deformations of a body whether stretches, compressions, bends or twists, can be regarded as
consisting of one or more of three basic strains. For many materials experiment shows that
provided the elastic limit is not exceeded,
Stress
 cons tan t
strain
The constant is called an “elastic modulus”
1. Young modulus: is considered with change of length strains
It is defined by
Y
tensile stress
tensile strain
Where stress is force per unit area
E
111
Fl0
Ae
Unit Kg
ms 2
E
e
and strain is change of length per unit length
A
l0
2. Shear modulus (G)
In this use the strain involves a change of shape without change of volume. Thus if a tangential
force F is applied along the top surface of area A of a rectangular slock of material fixed to the
bench, the block suffers a change of shape and is deformed so that the front and rear faces
becomes parallelograms.
Shear stres 
tangential force
surface being sheined
shear straim 
dis tan ce sheared
dis tan ce between surfaces
Since �l is usually very small, the ratio
radians
is equal approximatels to shear angle
in
Therefore
G
F
A
3) Bulk modulus k
If a body of volume V is subjected to a small increase of external pressure �P which changes its
volume by the small amount �V. The deformation is a change of volume without a change of
shape.
112
The bulk stress is �P (i.e. increase in force per unit area) and bulk strain is �V/V (i.e. change of
volume per original volume; the bulk modulus K is defined
K=
The negative sign is introduced to make K positive since �V, being a decrease, is negative.
N.B: Solids have all three module, liquids and gases only K. All modulus have the same unit –
Pascal
Bulk modulus of the gas
If the pressure P and the volume of the gas change under condition such that PV is constant
which is Boyle’ s law; the change is said to be isothermal
PV  Constant
pdv  vdp  0
pdv  vdp
PK
“The isothermal bulk modulus is equal to the pressure for isothermal process”
PV   constant
Differentiating the above equation
PdV   V  dP  0
P V  1dV  V  dP
Therefore K  P
113
CHAP VIII: THERMODYNAMICS
8.0.INTRODUCTION
Thermodynamics is the branch of physical science that deals with heat and related processes. A
part of thermodynamics that is especially relevant to power plants focuses on the laws governing
heat transfer from one location to another and transformation of energy from one form into
another. Examples of such processes would be the heat Transfer in a steam generator or the
conversion of heat into work in a steam turbine.
In thermodynamics, a clear definition of a thermodynamic system is necessary. Parameters such
as volume, temperature, pressure, internal energy, are used to define the state of a system.
8.1.TEMPERATURE AND HEAT
In everyday language we use the terms heat and temperature loosely as if they had the same
meaning. In physics they have different meaning.
Example:
Take a beaker half filled with water and place some ice in it. Put a thermometer in the water and
wait till the temperature of the water becomes stable so that the temperature of the water and the
ice are the same. Now place the beaker over a Bunsen burner and start heating it. You'll notice
that the temperature of the water stays the same as long as there is ice left. We all agree that the
flame is heating the water but the thermometer says that the temperature does not change.
Once all the ice melts, the temperature of the water starts to rise. From this we can see that we'll
need to closely examine our ideas about the meanings of heat and temperature.
We'll examine these concepts in more detail later but for the moment, in a nutshell:
(1) Temperature is related to the average kinetic energy of the particles (atoms or molecules).
(2) Heat is the amount of energy transferred to a system of particles
In the above examples, we were transferred heat to the system, which in turn melted the ice but
the temperature did not change!
114
Heat is defined as the thermal energy that flows from one object to another due to a difference in
their temperatures. Heat is a form of energy, is a derived quantity and SI unit is Joule, J. It is
being transferred from a region of higher temperature to low temperature.
The symbol for heat is (Q) and its unit of measurement is also kilojoules (kJ). When two parts of
a thermodynamic system are in thermal contact, heat is transferred between the parts. In
everyday life we interchange the words heat and temperature. In science we define both words
more accurately.
Temperature is a physical quantity characterizing the degree of hotness or coldness of a body.
We say that heat is transferred spontaneously from the hotter part toward the cooler part. This
continues until we reach an equilibrium state where the contacting parts.
Temperature is the measure of the average molecular motions in a system and simply has units of
(degrees F, degrees C, or K). Notice that one primary difference between heat and temperature is
that heat has units of Joules and temperature has units of (degrees F, degrees C, or K). Another
primary difference is that energy can be transported without the temperature of a substance
changing (e.g. latent heat, ice water remains at the freezing point even as energy is brought into
the ice water to melt more ice). But, as a general statement (ignoring latent heat), as heat energy
increases, the temperature will increase. If molecules increase in vibration, rotation or forward
motion and pass that energy to neighboring molecules, the measured temperature of the system
will increase. Heat cannot be measured with a thermometer, like temperature. Heat must be
calculated.
Example: You are sitting in a bathtub full of hot water that is 42°C while your body temperature
is 37°C. Heat flows from the hot water into you (since the temperature of the water is higher).
Example: You are sitting in a tub full of water that is 20°C while your body temperature is
37°C. Heat flows from you into the water (since your temperature is higher). Because heat leaves
your body, you feel cold.
115
ABSOLUTE TEMPERATURE SCALE
There is a physical lower temperature limit of matter. Nothing can be cooled below -273.15 ºC.
So for convenience, scientists have devised the absolute temperature scale which starts with 273.15 ºC and called it 0 Kelvin (not degrees Kelvin!). So the relationship between Celsius and
Kelvin is:
TK=TC+273.15
Where T k is temperature in Kelvin and Tcis temperature in Celsius.
Ice freezes at 0 ºC or TK= 0 +273.15= 273.15 Kelvin. Normal room temperature is at 20 ºC or
TK = 20 + 273.15 = 293.15 Kelvin
TEMPERATURE SCALES
When an object receives heat and its temperature increases, several of its physical properties can
change. A property which changes with temperature is called a thermometric property. These can
include volume, pressure, electrical resistance, emf and colour. With one of these properties, a
thermometer can be made. Two fixed points will also be needed to create a temperature scale to
display numbers.
Gabriel Fahrenheit devised a temperature scale in 1709 or 1714 in Holland. He chose two fixed
points which were easily identifiable and reproducible and a linear thermometric property (linear
expansion).
He set 00 as the coldest temperature that he could get. To do this he used a mixture of ice and
ordinary salt.
He set the normal blood temperature of a person as 1000. He chose the assumed linear expansion
of Mercury along a capillary tube in glass as the thermometric property thus creating the
"Mercury in glass thermometer". Unfortunately he didn't get blood temperature quite right (it is
98.60F) but the freezing point of water was officially set at 320F and its boiling point at 2120F.
Anders Celsius (1701-1744) in 1742 also devised a temperature scale that had 1000 separating
the boiling and freezing points of water. His original scale had the boiling (or steam) point as 00
116
and the freezing (or ice) point as 1000 but this was reversed in 1948 when it was adopted
internationally.
Conversion between Fahrenheit and Celsius Temperatures
You can convert between these scales by comparing the size of the degrees and the starting
points.
100 C0 corresponds to 180 F0
 212  32  180 F 0
9
Then the zero points have to be aligned.  F 0
5
5
 ( F  32)
9
Neither of these temperature scales have a proper zero.
For example, if it was 100C yesterday and 200C today, it is not twice as hot.
TYPES OF THERMOMETER
Mercury thermometer.
117
The physical quantity that is used to determine the temperature of a body by means of a mercury
thermometer is the length of the thread mercury, or to be more exact, the volume of mercury.
When the temperature increases, the volume of the mercury increases too.
The sensitivity of a mercury thermometer can be increased by:
a. reducing the diameter of the capillary tube.
b .Increasing the size of the bulb.
c .using a thinner-walled glass bulb.
Normally mercury is used in a thermometer because it:
a. Expands uniformly.
b. has a higher boiling limit.
c. is opaque and therefore it is easier to read off the temperature.
d. is a good conductor of heat.
One weakness of the mercury thermometer in the measurement of an accurate temperature is that
the glass of the capillary tube also expands when the temperature expands.
In addition to that, it is extremely dangerous if the glass tube breaks because mercury is very
poisonous. Mercury thermometer is suitable to measure temperature between -300Celsius to 300
degree Celsius.
2. Resistance thermometer.
The property of metals that their resistance is temperature-dependent makes them ideal as
thermometers. The metal of choice is platinum as a result of its high melting point (1773oC) and
large resistance.
118
Thermometers which use liquids inside the glass are not suitable to be used for measuring a wide
range of temperature. e.g. temperature ranging from -250 degree Celsius to about 700 degree
Celsius. A suitable thermometer which is used for the above range of temperatures is a resistance
thermometer.
A resistance thermometer uses the property of the change in the platinum wire with a change in
temperature. The current flowing in the wire experiences more resistance when the wire becomes
hot. The change in the resistance of the wire is directly proportional to the change in temperature.
Thermocouple thermometer.
119
An electromotive force (e.m.f) will be produced in a thermocouple when there is a temperature
difference between the hot junction and the cold junction. Once this happens, a current will flow.
This thermometer is very sensitive and responds towards slight change in temperature. Since the
physical quantity which is used to measure the temperature is the e.m.f, this thermometer can be
connected to other electrical circuits to control or record the surrounding temperature. A
thermocouple thermometer is a very sensitive thermometer which is suitable for measuring
temperatures ranging from -250 degree Celsius to 1600 degree Celsius.
The Constant-Volume Gas Thermometer.
120
By international agreement, the standard thermometric property is the pressure of a gas at
constant volume.
A bulb is immersed in a liquid at the temperature to be measured.
By raising or lowering the mercury reservoir the column of Mercury on the left can be set to the
zero mark and the volume of gas thus kept constant.
The pressure is then obtained from the difference between Mercury heights.
The temperature is defined by being proportional to pressure.
Comparing the three scales
121
8.2.THERMAL EXPANSION
When a material is heated or cooled, it changes its dimensions. Generally, it expands when
heated and contracts when cooled although there can be exceptions to this rule.When cooled, it
decreases in size. Matter in its basic state expands when heated but the expansion is most
noticeable in gases.
Example: The Eiffel Tower can extend by 12 centimetres if the temperature increases by 40 ºC.
So if you go to Paris in the summer and stand on top of the Eiffel Tower, you will get a little
more height for your money than if you had gone in winter.
Coefficients of Expansion
Almost all materials expand on heating—the most famous exception being water, which
contracts as it is warmed from 0 degrees Celsius to 4 degrees. This is actually a good thing,
because as freezing weather sets in, the coldest water, which is about to freeze, is less dense than
slightly warmer water, so rises to the top of a lake and the ice begins to form there.
For almost all other liquids, solidification on cooling begins at the bottom of the container. So,
since water behaves in this weird way, ice skating is possible! Also, as a matter of fact, life in
lakes is possible—the ice layer that forms insulates the rest of the lake water from very cold air,
so fish can make it through the winter.
Linear Expansion
The coefficient of linear thermal expansion (CTE, α, or α1) is a material property that is
indicative of the extent to which a material expands upon heating.
Different substances expand by different amounts. Over small temperature ranges, the thermal
expansion of uniform linear objects is proportional to temperature change. Thermal expansion
finds useful application in bimetallic strips for the construction of thermometers but can generate
detrimental internal stress when a structural part is heated and kept at constant length.
The coefficient of linear expansion α of a given material, for example a bar of copper, at a given
temperature is defined as the fractional increase in length that takes place on heating through one
degree:
L  L  L  (1   ) L When T  T  10 c
122
Of course, α might vary with temperature (it does for water, as we just mentioned) but in fact for
most materials it stays close to constant over wide temperature ranges.
For copper,   17 10 6
Where  L is the change in length, L0 is the original length before the change, ∆T is the change in
temperature, and  is the linear thermal coefficient of expansion, which is different for different
materials.
Example: Two strips of metal are riveted together. They both have different thermal expansion
coefficients. This means that if they are heated to the same temperature they will expand to
different lengths. But since they are riveted together the result will be that they bend. This is a
useful property and is used in many applications that involve electricity and heat such as electric
stoves where the current is switched off at a certain temperature. Another example is the
automatic hot water kettle where it switches itself off when the water boils.
Volume Expansion
In reality, materials expand in 3 dimensions so equation 2 is modified to the following:
For liquids and gases, the natural measure of expansion is the coefficient of volume expansion, β.
V  V  V  (1   )V When T  T  10 c
Of course, on heating a bar of copper, clearly the volume as well as the length increases—the bar
expands by an equal fraction in all directions (this could be experimentally verified, or you could
just imagine a cube of copper, in which case all directions look the same).
3
The volume of a cube of copper of side L is V = L . Suppose we heat it through one degree.
Putting together the definitions of , α β above,
V  (1   ), L  (1   ) L, L3  (1   )3 L3orV  (1   )3V
So
(1+ 1   )  (1   ) .
3
But
remember
α
is
very,
very
small—so
even
though
(1   )3  1  3  3 2   3 , the last two terms are completely negligible, so to a fantastically
good approximation:
The coefficient of volume expansion is just three times the coefficient of linear expansion.
123
  3
When the temperature of solid or liquid increased, the atoms vibrate more vigorously, and they
tend to push each other farther apart. There are some exceptions to this behavior; for example.
Because of its unique structure. Water contracts between 00c and 40c. This has important
consequences for the temperature distribution in lakes. If at an initial temperature T0 an object
has length Lo along some dimension, its length will change by amount L  L  L when the e
temperature is changed to T, where T  T  T0 . If TT0 , the object becomes smaller.
Experimentally it is found that if T
Is small, then L  L0 T
 is the coefficient of linear expansion, with unit 0 c 1 .
Area and volume V also change with temperature. Thus for a small square of material originally
of area AA0  L0  L0 at temperature T0, the area at T is
A  L  L  L0 (1  T ) L0 (1  T )  L0 (1  2T   2 (T ) 2 )
2
Since 1 , we neglect the last term, and A  L0 (1  2T )  A0 (1  2T ) .Thus
2
A  2A0 T
In similar fashion a small cube initially of side L0 and volume V0 Changes to volume V when the
temperature is changed. V  L0 (1  T )3  V0 (1  3T ) .Thus
3
V  3VT  VT
Here  is the coefficient of volume expansion WhereV is the change in volume, V0 is the
original volume,  T is the change in temperature.
8.3.HEAT TRANSFER
Heat is a form of energy that flows between points which are at different temperatures. It flows
from an area of high temperature to one of low temperature.
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Heat continues to pass from a hot place to a cold place until both places reach the same
temperature Heat transfer occurs by any of four methods:

Conduction

Convection

Radiation
In terms of heat transfer we defined temperature as the parameter that characterizes the state of
equilibrium. Heat can only be transferred between parts of a system or between systems when
there is a temperature difference.
When heat is transferred to a substance, the substance may change its temperature, its phase, or
both. When the substance changes its temperature, the following equation applies:
Q  mcT
The mechanisms and the rate of heat transfer will be explained in this section.
Conduction
Heat is the energy transferred to an object and is measured in joules. If two objects at different
temperatures are placed in contact, heat will flow from the higher to the lower temperature
object. This is called Conduction.
This is sometimes not obvious: Like when you shake hands with a person with cold hands. The
conclusion that many people make is that cold has travelled from that person to you. It is only
heat that travels. The coldness that you feel is simply the heat leaving your hand.

Materials which conduct heat well are called conductors of heat. Electrical conductors
(such as metals) are good conductors of heat.

Materials which do not conduct heat well are called insulators. Electrical insulators (for
example, wood or glass) are usually good insulators of heat. Materials with low density, such as
air or foamed plastic, are normally also good insulators unless they happen to be electrical
conductors. To prevent heat from moving from one place to another, we usually place an
insulator between.
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Once a good insulator becomes hot, however, it stays that way for a long time, because it is
difficult for the material to lose heat by conduction. Think of a hot ceramic pan and a hot metal
pan: which cools faster?
Conduction involves heat transfer with no transfer of mass. Heat is transferred from one part of a
body to another part, or to another body in physical contact with it.
So how fast does heat travel?
Heat travels at different rates in different materials. The rate of heat transfer by conduction is
dependent on the ability of the material to conduct heat, the area through which heat is flowing,
A, the temperature differential from the hot side to the cold side, ΔΤ, and the wall thickness, x.
Consider a solid material having an area of cross section A and thickness x.
Hot
face
T
T
1T
A
1
11
T
2
Cold
face
x
The amount of heat Q flowing from one face to another depends upon the following factors:
Directly proportional to A, i.eQ α A

Directly proportional to the time of conduction, Q α t

Directly proportional to the temperature difference, Q α (T1 – T2)

Inversely proportional the thickness of the slab, Q α 1/x
>>

Qα
Q=K
K is called the co-efficient of thermal conductivity
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K=
The co-efficient of thermal conductivity is equal to the quantity of heat which flows in one
second through a unit cube of material when its opposite faces are maintained at a temperature
difference of 1oC.
Units of K are cal/cm.oC.s or Kcal/m.oC.s
The rate of heat flux qx = Q/t
=
Or the rate of heat transfer
qx =
… Fourier’s Law
Convection
Convection is the transport of heat by the movement of liquids or gases. People make use of this
when they go hot air ballooning. Hot air rises because it expands when heated and therefore
becomes less dense. The hot air is then captured by the balloon. The volume of the balloon is
chosen so that the buoyancy force on it is larger than the weight of the balloon and the weights
attached to it (that includes people). So the balloon rises.
Rising hot air eventually cools, which means now it is denser and can start falling again. But it
can't go straight down since there is rising hot air below it. So it shifts sideways then starts to
fall. Air circulating in this way is called convection current. This is only one special case. The
same phenomenon occurs in liquids. This causes circular motion of the fluid away from a source
of heat.
Convection in water drives ocean currents; convection in air drives weather patterns; and
convection of molten rock inside the earth is thought to drive plate tectonics.
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Convection leads to the counterintuitive fact that good insulators (like air) can transfer heat
efficiently -- as long as the air is allowed to move freely. Trapped air, as between panes of a
double window, cannot transfer heat well because it cannot mix with air of a different
temperature.
Convection involves heat transfer that is accomplished
by the movement of a fluid. As the fluid moves, it
carries heat with it. Because convection relies on
transfer of molecules from one place to another, it only
occurs in fluids that are free to move. It does not occur
in solids and is usually negligible in fluids that are
trapped.
TYPES OF CONVECTION
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1.Forced Convection
This type of convection uses pressure differences to force the fluid to move. The pressure
differences are generated by equipment, such as pumps, fans, and compressors. This type of
convection is very common in our plants.
2.Natural Convection
In natural convection, there are no external means forcing the fluid to flow, e.g., no running
pump. The movement of the fluid is caused by density and/or pressure differences produced in
the fluid due to heat transfer. The movement of the fluid is caused by density differences coupled
with gravity. The denser hence, heavier fluid descends, forcing the lighter fluid upwards.
Radiation
Energy is transferred by electromagnetic radiation. All of the earth's energy is transferred from
the Sun by radiation. Our bodies radiate electromagnetic waves in a part of the spectrum that we
can't see called the infra-red. However, there are some cameras that can actually see this
radiation. The colour and texture of different surfaces determines how well they absorb the
radiation.
Black body
A black body is one which absorbs all the heat radiations incident to it. When radiations are
allowed to fall on such a body, there neither transmitted nor reflected radiations.
Laws of black body radiations
1. Stefan – Boltzmann Law
The total power per unit area from black body radiation is given by:
E = = T4
Where:
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P = net radiation
A = radiation area
T = temperature of radiation
E = total power per unit = energy radiated by radiator per second per unit area
= 5.67 x 10-8W/m2K4 …. Stefan Boltzmann Constant
2. Wien’s Law
The amount of energy radiated by a black body is not uniformly distributed over all the
wavelengths emitted by the body but is maximum for a particular value of wavelength ( max).
max

maxmax max
The value of these wavelengths is different for different temperature. max is the wavelength of
the most intense radiation and T is the temperature of the black body.
max 1/T
>>max x T = constant
max. T = 0.2892 cm.k
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8.4.HEAT CAPACITY, SPECIFIC HEAT, AND LATENT HEAT
HEAT CAPACITY
If you light a match and put the flame to a small object such as a needle or a pin then you know
that after a few seconds you will not be able to hold the needle because it has become too hot.
On the other hand, if you put the flame of the match to a massive object such as a truck , then
instinctively you know it will not make one bit of difference to its temperature. Although in both
cases the temperature of the flame is the same. So somehow the mass of an object is related to
the temperature it will reach in a certain time.
In the olden days people defined the concept of heat capacity because they thought that
somehow an object can have a capacity for being "filled" with heat just as a bucket has a
capacity for holding water. This is not a correct point of view since it would be theoretically
possible to keep transferring heat to the object without limit. Although in practice the object
might eventually vaporise.
However, we will define it here anyway:
The ratio of the energy transferred and the change in temperature is called the heat capacity
c
Q
T
Where T  T final  Tinitial .Where Q is the amount of heat transferred.
SPECIFIC HEAT
It's a little cumbersome to use heat capacity since this "constant" keeps changing as the mass of
the object changes. So people went further and defined something slightly better. The specific
heat. This is the number of joules required to raise 1kg for 1 Kelvin. Specifically:
C  cm
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Where c is the specific heat and m is the mass. So now the amount of heat transferred Q is given
by
Q  mcT
Where the c is expressed in units of J .kg-1 K-1
For 1 litre of water the mass is 1 kilogram, specific heat is c = 4190 J/kg.K.
Calorimeters.
Calorimeters are special containers used to measure the exchange of heat when substances are
mixed. The name comes from the old unit of energy, the calorie.
Calorimeters are designed so that virtually all of the heat is transferred from one substance to
another inside the container with no heat loss to the surroundings. The insulated lid, ring, and
dead air space are central to preventing heat loss to the surroundings.
DETERMINATION OF SPECIFIC HEAT CAPACITY BY EXPERIMENT
These two methods concern the heating up a known mass and measuring the temperature rise for
a known amount of electrical energy used.
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Specific heat capacity of a liquid by an electrical method.
The heat energy supplied by the electrical element is given to the liquid and its container,
producing a temperature rise Δθ.
The heater current (I) and voltage (V) are monitored for a time (t). Energy supplied by heater
 VIt . Energy absorbed by liquid and container  mLcL T  mccc T
where, mL mass of liquid
mc mass of container
cL specific heat capacity of liquid
cc specific heat capacity of container
Equating the two quantities, VIt  mL cL   mc cc 
mL,mc,cc Are known and V, I, t, Δθ are all measured. So the specific heat capacity of the liquid (
cL ) can be calculated.
133
Specific Heat Capacity of a solid by an electrical method
The method is very similar to that for a liquid except that there is no container. The solid under
test is a lagged cylinder with holes drilled for the thermometer and the heater element. A little
glycerin is added to the thermometer hole to improve thermal contact.
Heat energy supplied by the electrical element is given directly to the solid, producing a
temperature rise ΔT.
VIt  ms cs T
Where, ms is mass of solid
cc is specific heat capacity of solid
ms Is known and V , I , t , T are measured. So the specific heat capacity of the solid ( cs ) can
be calculated.
HEAT EXCHANGE & MIXTURES
The Principle of Heat Exchange (a special case of the Law of Conservation of Energy) states that
when two substances at different temperatures are mixed, the amount of heat lost by the
warmer substance equals the amount of heat gained by the cooler substance - assuming no
heat is lost to the surroundings.
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Heat lost = Heat gained
The heat loss from the 100g of 80°C water is gained by the 100g of 20°C water resulting in a
final temperature of 50°C.
LATENT HEAT & CHANGE OF STATE
Energy is acquired or released when a material changes phase. For example, energy is required
to melt ice and vaporise water. However, energy is given out if water vapour condenses or water
freezes. The heat acquired or released is called the latent heat.
During a phase change there is no change in temperature so we cannot use the equation
containing the specific heat to determine the amount of heat transferred. The formal definition of
latent heat is the energy given out or absorbed without a change in temperature and is given by
Q  mL
Q is the heat, m is the mass, L is a constant for a certain material and is called the Latent heat of
fusion (for melting) or vaporisation (for boiling).Latent heat is the energy involved when a
substance changes state.
Latent heat energy (L) has two components:
ΔU - the increase/decrease in internal PE
ΔW - the external work involved in expansion (+ΔW) and contraction (-ΔW)
135
This can be summarized as:
L  U  W
The phase changes involving latent heat energy are:
phase change
action
symbol
solid to liquid
melting
LF
liquid to solid
fusion
LF
liquid to vapor
vaporization
LV
vapor to liquid
condensation
LV
solid to vapor
sublimation
LS
vapor to solid
sublimation
LS
The graph illustrates the temperature changes when a solid (eg ice) is heated from below its
melting point, to above boiling. The graph illustrates the temperature changes when a solid (eg
ice) is heated from below its melting point, to above boiling.
Note that the changes of state occur in the flat areas. There is no temperature rise here and hence
any increase in KE.
136
Latent heat must be absorbed from the surroundings (and given to the substance) for the
substance to melt or boil. Latent heat is given out to the surroundings (from the substance) when
the substance condenses or freezes.
Adding or removing heat does not always result in a change of temperature. During a change of
state, the heat added is called latent heat because there is no change in temperature. Latent
means "hidden".
Notice in the graph above that while ice is melting (change of state) the temperature stays
constant at 0°C. The temperature also is constant when water boils and changes to steam or
vapor.
When a solid is melting the heat energy added is building up the potential energy of the
molecules to break the electrical forces holding them together. Similarly, when liquids are
turning to gases the heat energy increases the energy of the molecules so they get further apart
and become gas molecules.
137
LANTENT HEAT OF FUSION
Latent heat of fusion is the amount of heat required to melt 1 kg of a substance without
changing its temperature. The latent heat of fusion for water is 3.3 x 105 J/kg, which means that
3.3 x 105 J of energy are needed to change 1 kg of ice at 0°C into water at 0°C.Latent heat of
fusion is the quantity of heat energy absorbed when a solid melts to become a liquid without
changing its temperature. The quantity of heat Q, required to melts a given mass, m, of substance
is given by:
Q=m Lf
Latent heat of fusion is the energy required to melt 1 kg of a solid. The units are J.kgLatent heat of vaporisation is the energy required to evaporate 1 kg of a liquid. Specific latent
heat of vaporization is the quantity of heat required to change 1 kg of a liquid at its normal
boiling point to vapor without changing its temperature.
Boiling point
The boiling point of a substance is the temperature at which its saturated vapor pressure becomes
equal to the external atmospheric pressure. Boiling takes place throughout the entire volume of a
liquid and takes place at one temperature that is the boiling point.
8.5.GAS LAWS
How do gases behave? The behaviors of gases are described mathematically by the gas laws.
These gas laws do not explain gas behavior – rather they just predict how most gases act under
usual conditions.
Chemists found that there were relationships among temperature, volume, pressure, and quantity
of a gas that could be described mathematically. This chapter deals with Boyle’s law, Charles’s
law, Gay-Lussac’s law, the combined gas law, and Dalton’s law of partial pressures. These laws
have one condition in common. They all assume that the molar amount of gas does not change.
In other words, these laws work correctly only when no additional gas enters a system and when
no gas leaks out of it.
138
1. Pressure-volume: Boyle's Law
Robert Boyle, a British chemist who lived from 1627 to 1691 formulated the first gas law, now
known as Boyle’s law.
This law describes the relationship between the pressure and volume of a sample of gas confined
in a container. Boyle found that gases compress, much like a spring, when the pressure on the
gas is increased. He also found that they “spring back” when the pressure is lowered. By
“springing back” he meant that the volume increases when pressure is lowered. It’s important to
note that Boyle’s law is true only if the temperature of the gas does not change and no additional
gas is added to the container or leaks out of the container.
Boyle’s law states that the volume and pressure of a sample of gas are inversely proportional to
each other at constant temperature. For a fixed mass of gas at constant temperature and
pressure, the pressure is inversely proportional to the volume.
This statement can be expressed as follows.
P
1
V
Making the proportionality into an equality,
PV  k
Where k is a constant the expression PV = k means that the product of the pressure and volume
of any sample of gas is a constant, k. If this is true, then P  V under one set of conditions is
equal to P  V for the same sample of gas under a second set of conditions, as long as the
temperature remains constant. Boyle’s law can be expressed by the following mathematical
equation.
Now, consider a fixed mass of gas at one temperature at different pressures and volumes,
P1V1  k P2V2  k
Eliminating the constant k
P1V1  P2V2
139
Although the product of pressure and volume gives the work done on or by a gas, in practice this
may not be a constant quantity throughout a particular experiment. For example a plot of
pressure versus volume may look something like this
According to Boyle’s law, when the pressure on a gas is increased, the volume of the gas
decreases. For example, if the pressure is doubled, the volume decreases by half. If the volume
quadruples, the pressure decreases to one-fourth of its original value.
2. Charles' Law
The French physicist Jacques Charles carried out experiments in 1786 and 1787 that showed a
relationship between the temperature and volume of gases at constant pressure. You know that
most matter expands as its temperature rises. Gases are no different. When Benjamin Thomson
and Lord Kelvin proposed an absolute temperature scale in 1848; it was possible to set up the
mathematical expression of Charles’s law.
Charles’s law states that the volume of a sample of gas is directly proportional to the absolute
temperature when pressure remains constant. For a fixed mass of gas at constant temperature
and pressure, the volume is directly proportional to the temperature (K).Charles’s law can
be expressed as follows.
V
m
T
140
Making the proportionality into an equality,
V  mT
Where m is a constant .Now, consider a fixed mass of gas at one pressure at two different
temperatures and volumes,
V1
 m and
T1
V2
m
T2
Eliminating the constant m,
V1 V2

T1 T2
A graph of volume versus temperature is shown below.
According to Charles’s law, when the temperature of a sample of gas increases, the volume of
the gas increases by the same factor. The expression V/T =k means that the result of volume
divided by temperature is a constant, k, for any sample of gas. If this is true, then V/T under one
set of conditions is equal to V/T for the same sample of gas under another set of conditions, as
long as the pressure remains constant.
3. Guy-Lussac’s law (Pressure Law)
The relationship between the pressure and temperature of a gas is described by Gay-Lussac’s
law. Gay-Lussac’s law states that the pressure of a sample of gas is directly proportional to the
141
absolute temperature when volume remains constant. For a fixed mass of gas at constant
temperature and pressure, the pressure is directly proportional to the temperature (K).
Gay-Lussac’s law can be expressed as follows:
P  nT
P T
Making the proportionality into equality, P  nT where n is a constant
Now, consider a fixed mass of gas at one volume at two different temperatures and pressures,
P1
P
n 2  n.
T1
T2
Eliminating the constant n,
P1 P2

T1 T2
According to Gay-Lussac’s law, when the temperature of a sample of gas increases, the pressure
of the gas increases by the same factor.
3. Avogadro’s Law
This law looks at the relationship between volume and number of moles, where the temperature
and pressure must be held constant. A graph of volume versus number of moles is shown below.
142
The mathematical relationship can be determined to be
V V
V
 constant or 1  2 , or in words we would say that
n
n1 n2
Volume and number of moles are directly proportional at constant pressure and temperature
5. Dalton’s law of partial pressures
Air is a mixture of approximately 78% N2 , 20% O2 , 1% Ar, and 1% other gases by volume, so
at any barometric pressure 78% of that pressure is exerted by nitrogen, 20% by oxygen, and so
on.
Ptotal  PGas1  PGas2  PGas3  PGas4  .........PGasn
The ideal gas equation of state applies to mixtures just as to pure gases. It was in fact with a gas
mixture, ordinary air, that Boyle, Gay-Lussac and Charles did their early experiments. The only
new concept we need in order to deal with gas mixtures is the partial pressure.
The pressure exerted by a gas depends on the force exerted by each molecular collision with the
walls of the container, and on the number of such collisions in a unit of area per unit time. If a
gas contains two kinds of molecules, each species will engage in such collisions, and thus make a
contribution to the total pressure in exact proportion to its abundance in the mixture. The
contribution that each species makes to the total pressure of the gas is known as the partial
pressure of that species.
The above is essentially a statement of Dalton's law of partial pressures. This phenomenon is
described by Dalton’s law of partial pressures, which says that the total pressure of a mixture of
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gases is equal to the sum of the partial pressures of the component gases. It can be stated
mathematically as follows.
Algebraically, we can express this law by
PTotal  P1  P2  ...   Pi
i
Dalton himself stated this law in a simple:
Every gas is a vacuum to every other gas. The partial pressure of any one gas is directly
proportional to its abundance in the mixture, and is just the total pressure multiplied by the mole
fraction of that gas in the mixture.
Dalton’s law applies to mixtures of gases. The partial pressure is the pressure that each
component of a gaseous mixture would have if that component were isolated in a vessel of the
same size and at the same temperature as the vessel containing the gaseous mixture. The diagram
below can help in understanding this. The gases in the vessels below are all at the same
temperature, and the vessels are all of the same volume…
In words, Dalton’s Law becomes:
The sum of the partial pressures is equal to the total pressure of a gaseous mixture.
Note that the identities of the gases are not important here, just as this was not important with
any of the other gas laws. Next we will show how the total pressure is related to the number of
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moles. Solve the ideal gas law for P, and then replace each term on the right side of the equation
of Dalton’s law. Remember that each gas has the same temperature and volume….
Ptotal 
n1 RT n2 RT n3 RT
 RT 
 RT 


 ....  (n1  n2  n3 )
  ntotal 

V
V
V
V 
V 
This tells us that the total pressure of the gas is only depends on the total moles of gas, as well as
the temperature and volume. Again, the specific identities of the gas components are not
important to the total pressure. Dalton’s law can be extended further, to look at the relationship
between composition and pressure. Composition in a gaseous mixture is often described with a
quantity called the mole fraction. For component 1 in a gaseous mixture, its mole fraction, X1, is
given by…
X1 
n1
ntotal
The mole fraction is used instead of concentration because of the difficulty in identifying the
solvent and solutes in a gaseous mixture. Also, the sum of all the mole fractions for a mixture
must equal one.
By comparison with the terms in Equation above, you can show that:
P1
n RT / V
 1
 X 1 or
Ptotal ntotal RT / V
P1  X 1 Ptotal
6. Graham's Law
Discovered by Thomas Graham of Scotland in 1830, he considered a sample of two different
gases at the same Kelvin temperature. Since temperature is proportional to the kinetic energy of
the gas molecules, the kinetic energy of the two gas samples is also the same.
In equation form, we can write: K.E1  K.E2
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1
K .E  mv2
2
Since
, we can write the following equation:
m1v1  m2v2
2
2
Note that the value of one-half cancels. The equation above can be rearranged algebraically into
the following:
 m1 
 m 
2

1
2
 V2
V1
This last equation is the modern way of stating Graham’s law.
7. Ideal Gas Law
An ideal gas is defined as a hypothetical substance that obeys the ideal gas equation of state. We
will see later that all real gases behave more and more like an ideal gas as the pressure
approaches zero. A pressure of only 1 atm is sufficiently close to zero to make this relation
useful for most gases at this pressure. If the variables n, P, V , and T have known values, then a
gas is said to be in a definite state, meaning that all other physical properties of the gas are also
defined. The relation between these state variables is known as an equation of state. The ideal
gas equation of state can be derived by combining the expressions of Boyle's, Charles', and
Avogadro's laws.
The above three proportionalities can be all solved for V and written as shown below:
Boyle’s Law
V
Charles’ Law
Avogadro’s Law
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1
p
V T
V n
nT
or
P
nT
V R
P
V
If these three equations are combined we can write
Where R is proportionality constant called the gas constant
The equation V  R
nT
is typically rewritten as PV  nRT , and is called the Ideal Gas Law.
P
What is an "ideal gas"? It is a gas that follows this law at all temperatures and pressures.
8.6. LAWS OF THERMODYNAMICS
Thermodynamic systems
Energy transfer is studied in three types of systems:
Open
Open systems can exchange both matter and energy with an outside system. They
systems:
are portions of larger systems and in intimate contact with the larger system. Your
body is an open system.
Closed
Closed systems exchange energy but not matter with an outside system. Though they
systems:
are typically portions of larger systems, they are not in complete contact. The Earth
is essentially a closed system; it obtains lots of energy from the Sun but the
exchange of matter with the outside is almost zero.
Isolated
Isolated systems can exchange neither energy nor matter with an outside system.
systems:
While they may be portions of larger systems, they do not communicate with the
outside in any way. The physical universe is an isolated system; a closed thermos
bottle is an isolated system.
Basic definitions and units
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Internal Energy
A body is said to possess energy if it is capable of doing work. Such a definition used in
mechanics arises from the fact that most forms of energy can be transformed into work.
Internal energy of a thermodynamic system is the total amount of energy contained in the
system. In general, the internal energy of a substance is made up of two parts: kinetic energy due
to molecular motion and potential energy arising from molecular dispersion.
The symbol for internal energy is U and in the International System (SI) it is measured in joules
(J) or kilojoules (kJ). 1kJ=1000J.
Equilibrium Thermodynamics
Thermodynamic equilibrium occurs in a system when
• the resultant force is zero throughout the system,
• the temperature is constant throughout the system, and
• the internal structure and chemistry do not change.
When one or more of these variables change, then the system is said to undergo a change of state
.In equilibrium thermodynamics any change of state that takes place is performed very, very,
slowly. This is to keep the system always infinitesimally near an equilibrium state.
Work
Work (W) is the product between force and displacement.
If we look at the common bicycle pump, we can easily understand the way work is used in
thermodynamics. When the piston is displaced, work (W) is simply the product of force (F)
multiplied by distance (d):
W = F. d
In mechanics, the definition of pressure (p) is force divided by the surface area (S) of the piston:
p
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F
and F= P.S
S
The change in volume is equal to:
ΔV = S. d and d 
V
S
Therefore, from the above, we get the relationship used for defining work in thermodynamics:
W = p. ΔV
Work is connected with changes of internal energy or enthalpy, and is also measured in
kilojoules (kJ).
The Work done by a system in changing volume.
External work is done on or by a system, when the system as a whole exerts a force on its
surroundings and produces a displacement. Only external work is considered here. For example,
a gas confined to a cylinder can expand its volume by displacing a piston.
Take the situation where an insulated cylinder has a piston weighed down by a container of Lead
shot, and a base that has a heat source, which can change the temperature of the system.
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The system in the diagram is initially in equilibrium. A tiny amount of Lead shot is removed and
the reduced weight provides a small constant force, F, to move the piston a small distance, ds.
The differential work done by the system is:
dW  F .ds  ( P. A).ds  P.dV Joule
Where P is the small constant pressure and dV is the differential volume change. Since the Joule
is a work or energy unit, the product of pressure times volume gives a quantity of work.
When the volume increases from an initial volume, Vi, to a final volume, Vf, the work done by
the system is the area under the P vsV curve:
Vf
W   PdV
Vi
The work done by the system is positive because the volume has increased. In evaluating the
work done by the system, the actual way that the pressure changes will determine the result.
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Case 1:
The first change, from "i", to "a", occurs at constant pressure (an isobaric change).
To increase the volume at constant pressure, the temperature has to be turned up by adding heat
(+Q1) while the piston is allowed to move freely against the weight.
The second change, from "a", to "f", occurs at constant volume (an isochoric change).
To decrease the pressure at constant volume, the piston has to be fixed in place and the
temperature turned down by subtracting heat ( -Q2).
Case 2:
The first change is at constant volume: the piston fixed in place and the temperature reduced.
The second change is at constant pressure: the piston is free and the temperature increased.
Considering the work done by the system in each case:
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The system moves from identical initial states to identical final states. The work done by the
system is larger along the first path than the second (as see by the colored areas). The work can
be made as small as you like or as large as you like by choosing a suitable path.
Cyclic Processes
Since the work done by the system and the heat added to the system are dependent on the path
taken, there will be net work done by the system in cycling around a closed path as shown below.
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The Zeros Law of Thermodynamics
This states:
If two systems are in thermal equilibrium with a third system, they are also in thermal
equilibrium with each other.
To understand this concept we must first appreciate what thermal equilibrium is. Consider a
body at a high temperature in contact with a body at a low temperature. Heat is transferred from
the high temperature body to the lower temperature body until the temperatures are equalized.
When this equal, constant temperature is reached and maintained, the two bodies are said to be in
thermal equilibrium.
Consider three bodies, X, Y and Z.
Z is in thermal equilibrium with Y, X is in thermal equilibrium with Y .Then Z is in thermal
equilibrium with X.
The First Law of Thermodynamics
From the diagrams above, the work done by the system and the heat added to the system vary
with the path taken. However the difference between the heat added and the work done by the
system is the same for all paths. This constant difference represents the change in internal energy
between
the
two
states.
The first law of thermodynamics says that when a system changes from one state to another, the
change in internal energy equals the heat added to the system minus the work done by the
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system.
In symbolic form:
U  Q  W
Note that a body does not "contain heat" or "contain work". Heat and work are energy in
transition that changes the internal energy. When heat is added to the system the internal energy
increases. When work is done by the system the internal energy decreases.
Special Cases
• Adiabatic changes.
In an adiabatic change there is no heat transfer between the system and its surroundings.
From the First Law:
U  Q  0  W
U  W
When no heat is transferred,
• work done by the system decreases the internal energy so the tempe
drops, and
• work done on the system increases the internal energy so the tempe
rises.
One way of ensuring an adiabatic change is to carry it out rapidly. This happens with sound
waves and internal combustion engines.
• Constant Volume changes.
Since the volume doesn't change no work is done.
From the First Law:
U  Q  W  0
U  Q
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Heat added to the system increases the internal energy. Heat removed from the system decreases
the internal energy.
• Cyclic Changes.
In a cyclic process the system returns to its initial state and so the internal energy is not changed.
From the First Law:
U  0  Q  W
Q  W
The heat added to the system equals the work done by the system.
• Free Expansions.
A free expansion is an adiabatic change that does no work.
From the First Law:
U  Q  0  W  0
U  0
155
From the diagram, the insulation means no heat transfer and there is no change in the volume of
the system. As the name implies a free expansion is not a quasi-static process, it can only happen
rapidly and we cannot describe it on a P Vs V diagram.
THERMAL PROCESSES
A system can interact with its surroundings in many ways, and the heat and work that come into
play always obey the first law of thermodynamics. This section introduces four common thermal
processes. In each case, the process is assumed to be quasi-static, which means that it occurs
slowly enough that a uniform pressure and temperature exist throughout a11 regions of the
system at all times.
An isobaric process is one that occurs at constant pressure. For instance, Figure below shows a
substance (solid, liquid, or gas) contained in a chamber fitted with a frictionless piston.
The pressure P experienced by the substance is always the same, because it is determined by the
external atmosphere and the weight of the piston and the block resting on it.
Heating the substance makes it expand and do work W in lifting the piston and block through the

displacement S .The work can be calculated from W = Fs, where F is the magnitude of the force
and s is the magnitude of the displacement. The
Force is generated by the pressure P acting on the bottom surface of the piston (area = A),
according to F = PA. With this substitution for F, the work becomes
W = (PA) s. But the product A.S is the change in volume of the material, V  V f  Vi ,
Where V f and Vi are the final and initial volumes, respectively.
Isobaric process
W  pV  p(V f  Vi ) (1)
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Figure: The substance in the chamber is expanding isobarically because the pressure is held
constant by the external atmosphere and the weight of the piston and the block.
Consistent with our sign convention, this result predicts a positive value for the work done by a
system when it expands isobarically ( V f exceeds Vi . Equation (1) also applies to an isobaric
compression ( V f less than Vi . Then , the work is negative, since work must be done on the
system to compress it.
1t is often convenient to display thermal processes graphically.
For instance, Figure below shows a plot of Pressure versus volume for an isobaric expansion.
Since the pressure is constant, the graph is a horizontal straight line, beginning at the initial
volume and ending at the final volume V f . 1n terms of such a plot, the work W  p(V f  Vi ) is
the area under the graph, which is the shaded rectangle of height P and width V f  Vi .
157
Figure: For an isobaric process, a pressure-versus-volume plot is a horizontal straight line, and


the work done W  p(V f  Vi is the colored rectangular area under the graph.
Another common thermal process is an isochoric process, one that occurs at constant volume.
Figure (a) illustrates an isochoric process in which a substance (solid, liquid,or gas) is heated.
The substance would expand if it could, but the rigid container keeps the volume constant, so the
pressure-volume plot shown in Figure (b) is a vertical straight line. Because the volume is
constant, the pressure inside rises and the substance exerts more and more force on the walls.
Although enormous forces can be generated in the closed container, no work is done (W = 0 J),
since the walls do not move. Consistent with zero work being done, the area under the vertical
straight line in Figure (b) is zero.
Since no work is done, the first law of thermodynamics indicates that the heat in an isochoric
process serves only to change the internal energy: U  Q  W  Q .
158
Isotherms
An ideal gas in a box has three thermodynamic variables: P, V, T. But if there is a fixed mass of
gas, fixing two of these variables fixes the third from PV  nRT (for n moles). In a heat engine,
heat can enter the gas, then leave at a different stage. The gas can expand doing work, or contract
as work is done on it. To track what’s going on as a gas engine transfers heat to work, say, we
must follow the varying state of the gas. We do that by tracing a curve in the (P, V) plane.
Supplying heat to a gas which consequently expands and does mechanical work is the key to the
heat engine. But just knowing that a gas is expanding and doing work is not enough information
to follow its path in the (P, V) plane. The route it follows will depend on whether or not heat is
being supplied (or taken away) at the same time. There are, however, two particular ways a gas
can expand reversibly—meaning that a tiny change in the external conditions would be sufficient
for the gas to retrace its path in the (P, V) plane backwards. It’s important to concentrate on
reversible paths, because as Carnot proved and we shall discuss later, they correspond to the
most efficient engines. The two sets of reversible paths are the isotherms and the adiabatic.
Isothermal behavior: the gas is kept at constant temperature by allowing heat flow back and
forth with a very large object (a “heat reservoir”) at temperature T. From , it is PV  nRT , it is
evident that for a fixed mass of gas, held at constant T but subject to (slowly) varying pressure,
the variables P, V will trace a hyperbolic path in the (P, V) plane.
This path, PV  nRT1, say is called the isotherm at temperature T1. Here are two examples of
isotherms:
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Isotherms PV=RT for one mole at 273K,373k
The Adiabatic Expansion of an Ideal Gas
In an adiabatic expansion, no heat is transferred between the system and the environment, i.e. ΔQ
goes to zero in the 1st Law of Thermodynamics. This happens when the change occurs very
quickly (as with sound waves) or slowly with a system completely insulated from its
environment.
On a PV graph, an adiabatic process has a steeper curve than isothermal processes. The drop in
temperature that occurs in an adiabatic expansion is due to small but non-zero attractive forces
160
between molecules (despite the kinetic theory assumptions). Kinetic Energy is used up in
overcoming the attractive forces.
To find an expression that characterizes adiabatic processes, the starting point is the ideal gas
equation. This describes the state of the gas. When the state changes, there are changes in
pressure (Δp), volume (ΔV) and temperature (ΔT).
The Second Law of Thermodynamics
Ice-cream melts when left out on a warm day. A cold can of soda warms up on a hot day at a
picnic. Ice cream and soda never become colder when left in a hot environment, for heat always
flows spontaneously from hot to cold and never from cold to hot.
The spontaneous flow of heat is the focus of one of the most profound laws in al1 of science, the
second law of thermodynamics.
THEN, THE SECOND LAW OF THERMODYNAMICS: THE HEAT FLOW STATEMENT.
Heat flows spontaneously from a substance at a higher temperature to a substance at a lower
temperature and does not flow spontaneously in the reverse direction.
It is important to realize that the second law of thermodynamics deals with a different aspect of
nature than does the first law of thermodynamics. The second law is a statement about the
natural tendency of heat to f10w from hot to cold, whereas the first law deals with energy
conservation and focuses on both heat and work. A number of important devices depend on heat
and work in their operation, and to understand such devices both laws are needed. For instance,
an automobile engine is a type of heat engine because it uses heat to produce work.
There are two formulations used for the Second Law of Thermodynamics.
The formulations are equivalent, but apply in different processes. One is connected with heat
engines and the other with heat transfer.
It is not possible to convert all the heat supplied to a heat engine into useful work.
Heat cannot be naturally transferred from a colder body to a hotter body.
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Carnot, proved that Q2 cannot be zero and that, the ratio Q2 / Q1 is equal to the ratio T2/T1, where T1 represents the absolute temperature of the hot source and T2 the absolute temperature
of the cold source. Therefore, the maximum thermal efficiency of a heat engine is:
 1
T2
T1
There are a number of statements outlining this law:
The Clausius Statement
No process is possible whose sole result is the transfer of heat from a body of lower
temperature to a body of higher temperature.
The Kelvin Statement
No process is possible in which the sole result is the absorption of heat from a reservoir and
its complete conversion into work.
A general statement
It is not possible to convert heat continuously into work without at the same time
transferring some heat from a warmer body to a colder one.
8.7.HEAT ENGINES & HEAT PUMPS
A heat engine is any device that uses heat to perform work. It has three essential features: a heat
engine. Heat is supplied to the engine at a relatively high input temperature from a place called
the hot reservoir. Part of the input heat is used to perform work by the working substance of the
engine which is the material within the engine that actually does the work (e.g. ,thegasoline- air
mixture in an automobile engine).The remainder of the input heat is rejected to a place called the
cold reservoir; which has a temperature lower than the input temperature.
162
Heat engines: are devices that use the flow of energy from a hot source to a cold sink to produce
work.
Figure above illustrates these features. The symbol QH refers to the input heat, and the subscript
H indicates the hot reservoir. Similarly, the symbol Qc stands for the rejected heat and the
subscript C denote the cold reservoir". The symbol W refers to the work done. The vertical bars
enclosing each of these thee symbols in the drawing are included to emphasize that we are
concerned here with the absolute values, or magnitudes, of the symbols. Thus QH indicates the
magnitude of the input heat, Qc denotes the magnitude of the rejected heat, and W stands for
the magnitude of the work done.
Since QH , Qc , and W ，refer to magnitudes only, they never have negative values assigned to
them when they appear in equations.
To be highly efficient, a heat engine must produce a relatively large amount of work from as
little input heat as possible. Therefore the efficiency e of a heat engine is defined as the ratio of
the magnitude of the work W done by the engine to the magnitude of the input heat QH
e
W
QH
If the input heat were converted entirely into work, the engine would have an efficiency of 1.00,
since W  QH , such an engine would be 100% efficient. Efficiencies are often quoted as
percentages obtained by multiplying the ratio
163
W
by a factor 100.
QH
An engine, like any device, must obey the principle of conservation of energy. Some of the
engine's input heat QH is converted into work W , and the remainder is Qc rejected to the cold
reservoir. If there are no other losses in the engine, the principle of energy conservation requires
that
QH  W  Qc
Solving this equation for W and substituting the result into Equation above leads to the
following alternative expression for the efficiency e of a heat engine:
e
QH  Qc
Q
1 c
QH
QH
Heat pumps: are devices that use work to alter the energy flow from cold to hot (eg a
refrigerator).
CARNOT'S PRINCIPLE AND THE CARNOT ENGINE
What is it that allows a heat engine to operate with maximum efficiency?
The French engineer Sadi Carnot (1796-1832) proposed that a heat engine has maximum
efficiency when the processes within the engine are reversible. A reversible process is one in
which both the system and its environment can be returned to exactly the states they were in
before the process occurred.
164
In a reversible process, both the system and its environment can be returned to their initial states.
Therefore, a process that involves an energy-dissipating mechanism, such as friction, cannot be
reversible because the energy wasted due to friction would alter the system or the environment or
both.
There are also reasons other than friction why process may not be reversible. For instance, the
spontaneous flow of heat from a hot substance to cold substance is reversible, even though
friction is not present.
The Thermal Efficiency (η - eta) is given by:

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TH  TC
 100
TH
REFERENCE
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Fundamentals of Physics by David Halliday, Robert Resnick and Jearl Walker
Advanced Level Physics, by Nelkon, Michael, Parker, Philip
Advanced Physics by Tom Duncan by John Murray publications
Advanced Physics by Tom Duncan by John Murray publications
Fundamental Physics by K.L. Gomber and K.L.Gogia, Pradeep Publications
Fundamentals of Physics by David Halliday, Robert Resnick and Jearl Walker
Mechanics by Brijlal and Subramaniam, S. Chand & Co
Waves and Oscillations by Brijlal and Subramaniam, S. Chand & Co
Heat and Thermodynamics by Brijlal and Subramaniam, S. Chand & Co
Applied Physics by Dale Ewen, Neill Schurter and Erik Gundersen
Engineering Science, Fifth Edition by W. Bolton
Schaum's Outline of Applied Physics, 4th ed., by Arthur Beiser
Advanced University Physics, by Rogalski, Mircea Serban
Advanced Level Physics, by Nelkon, Michael, Parker, Philip
Journals
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Physics Education, Institute of Physics Publishing, UK.
American Journal of Physics, American Association of Physics Teachers, USA.
Physics Today, American Institute of Physics, USA.
The Physics Teacher, American Association of Physics Teachers, USA.
Key websites and on-line resources
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www.en.wakipedia.org
http://www.aaas.org
www.Intute.ac.uk/sciences/physics
http://www.phys.vt.edu/PhysNet