Unit 1
Fundamental Quantities
Contents
1.1 Introduction
1.2 Objectives
1.3 Fundamental Quantities
1.4 Unit of length (meter)
1.5
Mass
1.6 Temperature
1.7 Electric Current
1.8 Luminosity
1.9 Derived Quantities
1.10 Summary
1.11 Keywords
1.12 Exercise
1.1 Introduction
The 'INTERNATIONAL SYSTEM OF UNITS', abbreviated SI, defines the seven quantities
listed in the left box below and the specific methods to build up a physical standard for each,
called the unit, against which any other quantity of the same type can be compared. Fundamental
Quantities are
i.
Length (meter)
ii.
Mass (kilogram)
iii.
Time (second)
iv.
Electric current (ampere)
v.
Thermodynamic temperature (kelvin)
vi.
Amount of substance (mole)
vii.
Luminous intensity (candela)
These quantities, are considered to be the building blocks of physics. They are used to express
the laws of physics. Because physics expresses everything in the Universe in terms of these basic
quantities, one concludes for oneself that in physics there are no other basic quantities than those of
this SI unit system.
1.2 Objectives
At the end of this chapter you will be able to:
•
Explain Fundamental Quantities
•
Know Unit of length, Mass, Temperature
•
Define Electric Current, Luminosity
•
Explain Derived Quantities
1.3 Fundamental Quantities
Fundamental quantities can be considered to be dimensions, but in a loose context. In physics,
when we refer to dimensions, we usually refer to space and time (and theoretically higher similar
dimensions), and not the dimension of the luminosity (see below).
The fundamental quantities:
•
Time
•
Space (or length)
•
Mass
•
Temperature
•
Electrical current
•
Luminosity
•
Amount of matter
Each fundamental quantity has an associated unit in the SI system:
•
Time: seconds (s)
•
Space: meters (m)
•
Mass: kilograms (kg)
•
Temperature: degrees kelvin (K)
•
Electrical current: ampere (A)
•
Luminosity: candela (l)
•
Amount of matter: mole
Time is perhaps the most abstract of the fundamental quantities, possibly because we experience
it in a linear way; we can't get out of it. Space can be experience from a nonlinear perspective,
and is more easily grasped as a distinct concept. The same goes for mass, which is very tangible.
Temperature is experiential but its quantum definition veers into abstract territory (in simplistic
terms, it is the amount of atomic vibration in a system). Like temperature, electrical current is an
everyday experience, but gets stranger the more closely it is analyzed (the "flow" of electrons
through a medium). Luminosity is straightforward (we can see how it changes). The mole is a
stumbling block for students of chemistry, but ultimately makes sense as a measurement of items
(particularly of matter).
Time
In physics, the treatment of time is a central issue. It has been treated as a question of geometry.
(See: philosophy of physics.) One can measure time and treat it as a geometrical dimension, such
as length, and perform mathematical operations on it. It is a scalar quantity and, like length,
mass, and charge, is usually listed in most physics books as a fundamental quantity. Time can be
combined mathematically with other fundamental quantities to derive other concepts such as
motion, energy and fields. Time is largely defined by its measurement in physics. Physicists
measure and use theories to predict measurements of time. What exactly time "is" and how it
works is still largely undefined, except in relation to the other fundamental quantities. Currently,
the standard time interval (called conventional second, or simply second) is defined as 9 192 631
770 oscillations of a hyperfine transition in the 133 caesium atom.
1.4 Unit of length (meter)
The origins of the meter go back to at least the 18th century. At that time, there were two
competing approaches to the definition of a standard unit of length. Some suggested defining the
meter as the length of a pendulum having a half-period of one second; others suggested defining
the meter as one ten-millionth of the length of the earth's meridian along a quadrant (one fourth
the circumference of the earth). In 1791, soon after the French Revolution, the French Academy
of Sciences chose the meridian definition over the pendulum definition because the force of
gravity varies slightly over the surface of the earth, affecting the period of the pendulum.
Thus, the meter was intended to equal 10-7 or one ten-millionth of the length of the meridian
through Paris from pole to the equator. However, the first prototype was short
by 0.2 millimeters because researchers miscalculated the flattening of the
earth due to its rotation. Still this length became the standard. (The engraving
at the right shows the casting of the platinum-iridium alloy called the "1874
Alloy.") In 1889, a new international prototype was made of an alloy of
platinum with 10 percent iridium, to within 0.0001, that was to be measured
at the melting point of ice. In 1927, the meter was more precisely defined as the distance, at 0°,
between the axes of the two central lines marked on the bar of platinum-iridium kept at the
BIPM, and declared Prototype of the meter by the 1st CGPM, this bar being subject to standard
atmospheric pressure and supported on two cylinders of at least one centimeter diameter,
symmetrically placed in the same horizontal plane at a distance of 571 mm from each other.
The 1889 definition of the meter, based upon the artifact international prototype of platinumiridium, was replaced by the CGPM in 1960 using a definition based upon a wavelength of
krypton-86 radiation. This definition was adopted in order to reduce the uncertainty with which
the meter may be realized. In turn, to further reduce the uncertainty, in 1983 the CGPM replaced
this latter definition by the following definition:
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.
Note that the effect of this definition is to fix the speed of light in vacuum at exactly 299 792 458
m·s-1. The original international prototype of the meter, which was sanctioned by the 1st CGPM
in 1889, is still kept at the BIPM under the conditions specified in 1889.
1.5 Mass
Mass, in physics, the quantity of matter in a body regardless of its volume or of any forces
acting on it. The term should not be confused with weight, which is the measure of the force of
gravity (see gravitation) acting on a body. Under ordinary conditions the mass of a body can be
considered to be constant; its weight, however, is not constant, since the force of gravity varies
from place to place. There are two ways of referring to mass, depending on the law of physics
defining it: gravitational mass and inertial mass. The gravitational mass of a body may be
determined by comparing the body on a beam balance with a set of standard masses; in this way
the gravitational factor is eliminated. The inertial mass of a body is a measure of the body's
resistance to acceleration by some external force. One body has twice as much inertial mass as
another body if it offers twice as much force in opposition to the same acceleration. All
evidence seems to indicate that the gravitational and inertial masses of a body are equal, as
demanded by Einstein's equivalence principle of relativity; so that at the same location equal
(inertial) masses have equal weights. Because the numerical value for the mass of a body is the
same anywhere in the world, it is used as a basis of reference for many physical measurements,
such as density and heat capacity. According to the special theory of rel
relativity,
ativity, mass is not
strictly constant but increases with the speed according to the formula m=m0/
,
where m0 is the rest mass of the body, v is its speed, and c is the speed of light in vacuum. This
increase in mass, however, does not become appreciable until very great speeds are reached.
The rest mass of a body is its mass at zero velocity. The special theory of relativity also leads to
the Einstein mass-energy
energy relation, E=mc2, where E is the energy, and m and c are the
(relativistic) mass and the speed of light, respectively. Because of this equivalence of mass
and energy,, the law of conservation of energy was extended to include mass as a form of
energy.
1.6 Temperature
Definition: Temperature is a measurement of the average kinetic energy of the molecules in an
object or system and can be measured with a thermometer or a calorimeter. It is a means of
determining the internal energy contained within the system.
Heat vs. Temperature
Note
ote that temperature is different from heat,, though the two concepts are linked. Temperature is
a measure of the internal energy of the system, while heat is a measure of how energy is
transferred
ed from one system (or body) to another. The greater the heat absorbed by a material,
the more rapidly the atoms within the material begin to move, and thus the greater the rise in
temperature.
Temperature Scales
Several temperature scales exist. In America, the Fahrenheit temperature is most commonly
used, though the SI unit Centrigrade (or Celsius) is used in most of the rest of the world. The
Kelvin scale is used often in physics, and is adjusted so that 0 degrees Kelvin is absolute zero.
zero
1.7 Electric Current
If the two requirements of an electric circuit are met, then charge will flow through the external
circuit. It is said that there is a current – a flow of charge. Using the wordcurrent in this context
is to simply use it to say that something is happening in the wires – charge is moving. Yet current
is a physical quantity that can be measured and expressed numerically. As a physical
quantity, current is the rate at which charge flows past a point on a circuit. As depicted in the
diagram below, the current in a circuit can be determined if the quantity of charge Q passing
through a cross section of a wire in a time t can be measured. The current is simply the ratio of
the quantity of charge and time.
Current is a rate quantity. There are several rate quantities in physics. For instance, velocity is a
rate quantity – the rate at which an object changes its position. Mathematically, velocity is the
position change per time ratio. Accelerationis a rate quantity – the rate at which an object
changes its velocity. Mathematically, acceleration is the velocity change per time ratio.
And power is a rate quantity – the rate at which work is done on an object. Mathematically,
power is the work per time ratio. In every case of a rate quantity, the mathematical equation
involves some quantity over time. Thus, current as a rate quantity would be expressed
mathematically as
Note that the equation above uses the symbol I to represent the quantity current.
As is the usual case, when a quantity is introduced in The Physics Classroom, the standard metric
unit used to express that quantity is introduced as well. The standard metric unit for current is
the ampere. Ampere is often shortened toAmp and is abbreviated by the unit symbol A. A
current of 1 ampere means that there is 1 coulomb of charge passing through a cross section of a
wire every 1 second.
1 ampere = 1 coulomb / 1 second
1.8 Luminosity
The luminosity of a facility is a measure of the rate at which particles collide. Luminosity is
directly related to the intensity of the particle beam (or beams) employed (and, in a collider, to
the size of a spot onto which the beams are focused). Elementary-particle physicists measure
luminosity in units of inverse square centimeters times inverse seconds (cm−2 s−1. This allows
one to calculate an event rate by multiplying luminosity by the effective cross-sectional area of
the particles that are colliding. Typical luminosities are in the range 1030 to 1035 cm−2 s−1. Such
large luminosities are required because the effective areas of the colliding particles are so
small—for example, high-energy electrons have an effective area for producing Z0 particles of
only about 10−32 cm2, leading to an interaction rate of one every 10 seconds in a facility
operating with a luminosity of 1031 cm−2 s−1.
1.9 Derived Quantities
All other quantities in physics can be expressed in terms of the fundamental quantities. Examples
are velocity (space divided by time), acceleration (space divided by time squared), force (mass
times space divided by time squared) or energy (mass times the constant representing the speed
of light squared - aka. space divided by time all squared). Understanding this concept helps in
understanding how all equations work, and how different "things" are related. In teaching
physics this is an important concept to transfer - teaching students not just to work an equation,
but really understand it in terms of how derived quantities relate to fundamental quantities.
Many other quantities can be derived out of the combination of the basic quantities. For instance
a speed is the ratio of a length by a time. An acceleration is the ratio of a speed by a time. And
a force is the multiplication of an acceleration by a mass.
Is Force a Fundamental or a Derived Quantity?
How is this quantum mechanics' fundamental concept of force (a pillar of the theory) getting
along
with
the
SI
quantities,
in
which
it
doesn't
even
appear?
As shown above, in the BOX: "Derived Quantities" a force is only a mental byproduct of a series
of other mental concepts that of speed and acceleration all mathematically derived from length,
time and mass.
Is "force" a fundamental constituent of the universe as asserted in the quantum theory, or is
"force" a derived quantity of length, time and mass?
Note that that quantum theory's force has to be labeled a "constituent" of the universe rather than
a "quantity".
And that brings a second question: Could that quantum theory's force be a "live phenomenon"
rather than an "inert quantity"?
The Future
There may be more fundamental quantities discovered in the future, or it may be found that
fundamental quantities are actually derived from others. From our current perspective, this seems
unlikely, but should not be ruled out entirely. That is part of the immense joy of learning about
the universe; there is inevitably new and fascinating discoveries to be made.
Physical quantities
In a school physics laboratory, there are a host of different quantities we may measure, from the
length of a bench to the voltage supplied by a battery. In physics, seven quantities are seen as
fundamental. You will come across five
1.10 Summary
Fundamental quantities are numbers that we need to describe the world around us, which we
cannot express in terms of "simpler," more basic quantities. I'll give a few examples. My weight
is not a fundamental quantity, because it depends on how much stuff makes up my body. A good
approximation to my weight is just the sum of the weights of all of the protons, neutrons, and
electrons which make me up. If I could count those, I could calculate my weight in terms of the
masses of the proton, the neutron, and the electron (and I'd need the local strength of gravity to
get my weight once I know my mass). The mass of a proton is more fundamental, although we
know now that protons are made up of smaller pieces (quarks and gluons). People are actively
trying to refine calculations of the proton mass in terms of the stuff inside a proton and their
interactions.
1.11 Keywords
•
•
•
•
•
•
•
•
SI
Length
Mass
Time
Electric current
Thermodynamic temperature
Amount of substance
Luminous intensity
1.12 Exercise
1) Explain Fundamental Quantities
2) Give the Unit of length, Mass, Temperature
3) Define Electric Current, Luminosity
4) Explain Derived Quantities
Unit 2
Derived Quantities
Contents
2.1 Introduction
2.2 Objectives
2.3 SI derived units
2.4 Summary
2.5 Keywords
2.6 Exercise
2.1 Introduction
Derived units are units which may be expressed in terms of base units by means of mathematical
symbols of multiplication and division. Certain derived units have been given special names and
symbols, and these special names and symbols may themselves be used in combination with the
SI and other derived units to express the units of other quantities.
2.2 Objectives
At the end of this chapter you will be able to:
•
SI derived units
•
Define Area, Volume, Velocity
•
Explain the terms Acceleration, Momentum, Force
•
Define Impulse, Power, Energy
2.3 SI derived units
Other quantities, called 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 ease of understanding and convenience, 21 SI derived units have been given special names
and symbols, as shown in Table. The special names and symbols of the 21 SI derived units with
special names and symbols given in Table may themselves be included in the names and
symbols of other SI derived units.
Derived quantity
Name
Symbol
Expression
Expression
in terms of in
other
terms
of
SI SI base units
units
plane angle
radian (a)
rad
-
m·m-1 = 1 (b)
solid angle
steradian (a)
sr (c)
-
m2·m-2 = 1 (b)
frequency
hertz
Hz
-
s-1
force
newton
N
-
m·kg·s-2
pressure, stress
pascal
Pa
N/m2
m-1·kg·s-2
J
N·m
m2·kg·s-2
W
J/s
m2·kg·s-3
C
-
energy, work, quantity of joule
heat
power, radiant flux
watt
electric charge, quantity of coulomb
s·A
electricity
electric
potential volt
V
W/A
m2·kg·s-3·A-1
F
C/V
m-2·kg-1·s4·A2
V/A
m2·kg·s-3·A-2
difference,
electromotive force
capacitance
farad
electric resistance
ohm
electric conductance
siemens
S
A/V
m-2·kg-1·s3·A2
magnetic flux
weber
Wb
V·s
m2·kg·s-2·A-1
magnetic flux density
tesla
T
Wb/m2
kg·s-2·A-1
inductance
henry
H
Wb/A
m2·kg·s-2·A-2
Celsius temperature
degree
°C
The image
part with
relationshi
p ID rId5
was not
found in
-
K
Celsius (e)
luminous flux
lumen
lm
cd·sr (c)
m2·m-2·cd = cd
illuminance
lux
lx
lm/m2
m2·m-4·cd = m-2·cd
activity (of a radionuclide)
becquerel
Bq
-
Gy
J/kg
absorbed
dose,
specific gray
energy (imparted), kerma
s-1
m2·s-2
dose equivalent (d)
(a)
sievert
Sv
J/kg
m2·s-2
The radian and steradian may be used advantageously in expressions for derived units to
distinguish between quantities of a different nature but of the same dimension.
Radian is the measure of a central plane angle that subtends an arc that is the same length as the
radius of the circle. Equal to 57.2958°.
Steradian is the measure of a central solid angle that subtends a surface that is the same area as
the square radius of the sphere.
(b)
In practice, the symbols rad and sr are used where appropriate, but the derived unit "1" is
generally omitted.
(c)
In photometry, the unit name steradian and the unit symbol sr are usually retained in
expressions for derived units.
(d)
Other quantities expressed in sieverts are ambient dose equivalent, directional dose equivalent,
personal dose equivalent, and organ equivalent dose.
(e)
The unit of Celsius temperature is the degree Celsius, symbol °C. The numerical value of a
Celsius temperature t expressed in degrees Celsius is given by t/°C = T/K - 273.15.
Derived Units Table
QUANTITY
NAME
SYMBOL
acceleration
metre per second squared
m/s²
angular acceleration
radian per second squared
rad/s²
angular momentum
kilogram metre squared per second
kg•m²/s
angular velocity
radian per second
rad/s
area
square metre
m²
cœfficient of linear expansion
1 per kelvin
K ‾¹
concentration (of amount of substance)
mole per cubic metre
mol/m³
density
kilogram per cubic metre
kg/m³
diffusion cœfficient
metre squared per second
m²/s
electric current density
ampere per square metre
A/m²
exposure rate (ionising radiation)
ampere per kilogram
A/kg
kinematic viscosity
metre squared per second
m²/s
luminance
candela per square metre
cd/m²
magnetic field strength
ampere per metre
A/m
magnetic moment
ampere metre squared
A•m²
mass flow rate
kilogram per second
kg/s
mass per unit area
kilogram per square metre
kg/m²
mass per unit length
kilogram per metre
kg/m
molality
mole per kilogram
mol/kg
molar mass
kilogram per mole
kg/mol
molar volume
cubic metre per mole
m³/mol
moment of inertia
kilogram metre squared
kg•m²
moment of momentum
kilogram metre squared per second
kg•m²/s
momentum
kilogram metre per second
kg•m/s
radioactivity (disintergration rate)
1 per second
s‾¹
rotational frequency
1 per second
s‾¹
specific volume
cubic metre per kilogram
m³/kg
speed
metre per second
m/s
velocity
metre per second
m/s
volume
cubic metre
m³
wave number
1 per metre
m‾¹
Area:
Area is a quantity expressing the two-dimensional size of a defined part of a surface, typically a
region bounded by a closed curve. The term surface area refers to the total area of the exposed
surface of a 3-dimensional solid, such as the sum of the areas of the exposed sides of a
polyhedron . Area is an important invariant in the differential geometry of surfaces.
Volume:
The volume of any solid, liquid, gas, is how much three
three-dimensional
dimensional space it occupies, often
quantified numerically. One-dimension
dimensional figures (such as lines) and two-dimensional
dimensional shapes
(such as squares) are assigned zero volume in the three
three-dimensional
dimensional space. Volume is commonly
presented in units such as cubic meters, cubic centimeters, liters, or milliliters.
Velocity:
Velocity is the rate of change of position. It is a vector physical quantity; both magnitude and
direction are required to define it. In the SI (metric) system, it is measured in meters per second:
(m/s) or ms−1.
−1. The scalar absolute value (magnitude) of velocity is sp
speed.
eed. For example, "5
meters per second" is a scalar and not a vector, whereas "5 meters per second east" is a vector.
Acceleration:
In physics, and more specifically kinematics, acceleration is the change in velocity over time.
Because velocity is a vector, it can change in two ways: a change in magnitude and/or a change
in direction. In one dimension, i.e. a line, acceleration is the rate at which something speeds up or
slows down.
However, as a vector quantity, acceleration is also the rate at which direction changes.
Acceleration has the dimensions L T
T−2.
−2. In SI units, acceleration is measured in metres per
second squared (m/s2).
Momentum:
In classical mechanics, momentum is the product of the mass and velocity of an object (p = mv).
In relativistic mechanics, this quantity is multiplied by the Lorentz factor. Momentum is
sometimes referred to as linear momentum to distinguish it from the related subject of angular
momentum.
Linear momentum is a vector quantity, since it has a direction as well as a magnitude. Angular
momentum is a pseudovector quantity because it gains an additional sign flip under an improper
rotation. The total momentum of any group of objects remains the same unless outside forces act
on the objects (law of conservation of momentum).
Force:
In physics, the concept of force is used to describe an influence that causes a free body to
undergo an acceleration. Force can also be described by intuitive concepts such as a push or pull
that can cause an object with mass to change its velocity (which includes to begin moving from a
state of rest), i.e., to accelerate, or which can cause a flexible object to deform. An applied force
has both magnitude and direction, making it a vector quantity.
Impulse:
In classical mechanics, an impulse is defined as the integral of a force with respect to time. When
a force is applied to a rigid body it changes the momentum of that body. A small force applied
for a long time can produce the same momentum change as a large force applied briefly, because
it is the product of the force and the time for which it is applied that is important. The impulse is
equal to the change of momentum.
The impulse, J, delivered by a constant force, F. Impulse is a vector quantity defined as the
product of the force acting on a body and the time interval during which the force is exerted. If
the force changes during the time interval, F is the average net force over that time interval. The
impulse caused by a force during a specific time interval is equal to the body’s change of
momentum during that time interval: impulse, effectively, is a measure of change in momentum.
J = F delta t = delta p
The unit of impulse is the same as the unit of momentum, kg · m/s.
Power:
In physics, power is the rate at which work is performed or energy is converted. The dimension
of power is energy divided by time. The SI unit of power is the watt (W), which is equal to one
joule per second. Non-SI units of power include ergs per second (erg/s), horsepower (hp), metric
horsepower (Pferdestärke (PS) or cheval vapeur, CV), and foot-pounds per minute.
Energy:
In physics, energy is a quantity that can be assigned to every particle, object, and system of
objects as a consequence of the state of that particle, object or system of objects. Different forms
of energy include kinetic, potential, thermal, gravitational, sound, elastic, light, and
electromagnetic energy. The forms of energy are often named after a related force.
2.4 Summary
The International System of Units (SI) specifies a set of seven base units from which all
other units of measurement are formed, by products of the powers of base units. These other
units are called SI derived units. The number of derived units is unlimited.
The names of SI units are always written in lowercase. The symbols of units named after
persons, however, are always written with an initial capital letter (e.g., the symbol of hertz is Hz;
but metre becomes m).
2.5 Keywords
•
Area
•
Volume
•
Velocity
•
Acceleration
•
Momentum
•
Force
•
Power
•
Energy
•
Impulse
2.6 Exercise
1) List the SI derived units.
2) Define Area, Volume and Velocity.
3) Explain the terms Acceleration, Momentum, Force.
4) Define Impulse, Power, Energy, and Impulse.
Unit 3
Concurrent Forces
Contents
3.1 Introduction
3.2 Objectives
3.3 Concurrent Forces
3.4 What is Equilibrium?
3.5 Equilibrium of Concurrent Forces
3.6 Diagrams
3.7 Number of Forces acting on a Body
3.8 Worked Example
3.9 Summary
3.10 Keywords
3.9 Exercise
3.1 Introduction
A body that is at rest, or is moving with constant velocity, is in a state of equilibrium. The
acceleration of a body in equilibrium is zero in any direction Therefore the resultant force in any
direction is also zero. The converse of this statement is not necessarily true because, although
forces with zero resultant cannot make and object move in a line they can cause an object
to turn .
However a set of concurrent forces ( i.e. all passing through one point ) can never cause turning
so, as at present we will deal only with concurrent forces, the problem of turning will not arise
yet. If the resultant is zero, the collected components in each direction must individually be zero
i.e. X = 0 and Y = 0. Applying this fact to a concurrent system in equilibrium, in which some
forces are unknown, provides a method for finding the unknown quantities.
3.2 Objectives
At the end of this chapter you will be able to:
•
Explain Concurrent Forces
•
Know What is Equilibrium?
•
Explain Equilibrium of Concurrent Forces
•
Know Number of Forces acting on a Body
3.3 Concurrent Forces
In a concurrent force system, all forces pass through a common point. In the previous case
involving the application of two forces to a body, it was necessary for them to be colinear,
opposite in direction, and equal in magnitude for the body to be in equilibrium. If three forces are
applied to a body, as shown in the figure, they must pass through a common point (O), or else the
condition, SMo = 0, will not be satisfied and the body will rotate because of unbalanced
moment. Moreover, the magnitudes of the forces must be such that the force equilibrium
equations,
SFx = 0, SFy= 0, are satisfied.
Concurrent Forces
It is fairly easy to see the reasoning for the first condition. Consider the two forces, F1 and F2,
intersecting at point O in the figure. The sum of moments of these two forces about point 0 is
obviously equal to zero because they both pass through 0. If F3 does not pass through 0, on the
other hand, it will have some nonzero moment about that point. Since this nonzero moment will
cause the body to rotate, the body will not be in equilibrium.
Therefore, not only do three nonparallel forces applied to a body have to be concurrent for the
body to be in an equilibrium state, but their magnitudes and directions must be such that the
force equilibrium con-ditions are satisfied (SFx = SFy = 0). Notice that there is no need for the
moment equilibrium equation in this case since it is automatically satisfied.
3.4 What is Equilibrium?
Equilibrium means "no acceleration". Since a force is a "push" or "pull" exerted on a body,
equilibrium means that the total of all forces acting on a body must be zero.
According to Newton's second law, F = m * a
If a = 0 then F must be zero.
(Remember! In Newton's second law F is the TOTAL force on the body)
Since we are studying STATICS, from now on we assume ever body is in equilibrium.
3.5 Equilibrium of Concurrent Forces
Concurrent means that the forces intersect through a single point. If forces are concurrent, we
can add them together as vectors to get the resultant. If the body is not accelerating, it must be
in equilibrium, so that means the resultant is zero. For concurrent forces, the body is a point. So
for concurrent forces in equilibrium,
ilibrium, the forces should all add up to give zero.
If a body is not accelerating is in equilibrium,, so resultant of
all forces = 0.
A typical concurrent force situation is a lifting eye. The pulling forces in any cables must pass
through the centre of the eye. If there is only one eyebolt (correctly positioned over the centre of
gravity) and the load is suspended, the bolt force must pass through the same centre. Hence all
forces pass through one point (the centre of the eye), so we have concurrent equi
equilibrium.
All forces on a suspended load are concurrent. (Assuming the load remains level when lifted). It
is possible to maintain equilibrium even when the cables are at different angles. In the example
below, Cable B must have less tension than Cable A;
3.6 Diagrams
1. The Space Diagram (SD)
The initial problem is usually sketched. This illustration or picture shows the layout and
dimensions. If this diagram was drawn to scale, the units would be length (mm, m etc). It is nice
to be accurate, but it does not have to be to scale.
2. The Free Body Diagram (FBD)
The Free Body Diagram is a strict diagram that isolates the body for study. See Free Body
Diagrams for more information. The idea of the FBD is to focus on one particular part or group
of parts (called the body) and replace every external member with the force they would apply.
1. Isolate the body. (An outline is best because we are supposed to forget about the inside of the
body)
2. Locate border crossings. Identify the contact points where forces are crossing the boundary.
Gravity acts through centre.
3. Line of Action. Some types of connections have a known direction. E.g. Cables have force
running through the centreline.
4. "To the Body". Since Newton's 3rd law has every action with an opposite reaction, we must
eliminate half the forces. Identify those forces that are applied "to the body", and eliminate those
done "by the body".
If the FBD were drawn to scale, the body might be length (mm, m etc) and the forces might be
another scale (N, kN etc).
Do not get Linear dimensions and Force dimensions mixed up. You cannot add metres and
Newtons together!
3. The Force Polygon (FP)
The force polygon must be drawn strictly to scale, and everything is a Force. The only
information coming from the FBD is;
•
Force magnitudes
•
Force Angles
Do not attempt to bring any FBD Lengths into the FP. There are no metres in the Force Polygon.
In some cases the Free Body diagram does not even look like the original. This is most obvious
for concurrent forces. Since all forces go through one point, we can treat the body as a DOT!
Cable connection in a structure, specially designed to make the centreline of every cables
intersect at one point.
Example Diagrams: These cranes are not accelerating, so they are in equilibrium. Therefore all
the forces on anybody should add up to zero. The body is actually the connection point which is
probably a lifting eye of a hook. The FBD shows as much as we know from the Space diagram in this case angles are known but only one magnitude. The force polygon should form a closed
loop (since resultant = 0), so this defines the lengths (and hence the magnitudes) of F1 and F2.
CAD programs are very helpful when working with force polygons.
Special Contact Points
When drawing the Free Body Diagram we must include all the forces that cross the boundary
(outline) of the body. Some of these contact points have special clues about the direction of the
force and location of the force.
1. Cable Joint: The force must run through the centreline of the cable
2. Frictionless Joint: The force must be perpendicular to the surface.
3. Wheels and Rollers: The force must run through the centre of the axle. Free running
wheels are frictionless so force is perpendicular to the surface and all forces pass through
the centre of the axle.
4. Pulleys: The tension in the cable is the same on each side of the pulley and all forces pass
can be made to go through the centre of the axle.
5. Friction: The force can be in any direction
6. Pin Joint: The force can be in any direction
7. Contact points are also called Support Reactions: Here is a table (Ignore the last one
at this stage).
3.7 Number of Forces acting on a Body
One Force
This is impossible for equilibrium. The forces are supposed to add up to zero (unless the body
is accelerating. E.g. A falling rock).
Two Forces
If a body has only 2 forces, they must be co-linear. E.g. A linkage between 2 pivot pins must
have the force running through the line of the pins. (This assumes gravity force is ignored,
otherwise you have three forces)
Three Forces
If a body has exactly 3 forces, they must be concurrent. This is called the Three Force Principle.
This can be very handy in solving problems because many mechanisms have bodies with 2 or 3
forces.
Four or more Forces...
We cannot assume the forces will be concurrent, unless specially made that way. (Like the fiveway cable connection below). When forces are not concurrent they can create rotations, which
we deal with in a later chapter. (Non Concurrent forces)
Five deliberately concurrent forces
The Equilibrium Equations
Equilibrium simply says the resultant is zero. Mathematically, this can be stated that the Fx and
Fy components are zero.
So, for concurrent forces in 2 dimensions (planar), equilibrium means that...
Very often we know the angle of the forces but not the magnitudes. When solving
mathematically, this means we will need to use simultaneous equations. (See example below)
3.8 Worked Example 1
Example: A Lifting Eye
Two ropes are attached to this lifting eye. Force
A is at 75o, and Force B is at 60o from the
horizontal.
If the load is 240kg, what are the tensions in
cable A and cable B?
The Free Body Diagram
1. Isolate: Take the eye as a dot
2. Border crossings: There are 3 forces
3. Line of action: Must run along cable
centrelines
4. "To the body": Cables always pull
Notes:
o
• You should convert to 360 notation in
the FBD.
• A FBD is almost always compulsory.
Graphical (CAD) method
Start with what you know.
We know the weight (gravity force) of 240kg
load.
Fg = 240 * 9.81 = 2354.4N
See instruction for How to add forces in Solid
Edge.
1. Start somewhere on the page (draw a small
circle to show the start point).
2. Draw the gravity force: 2354.4 at 270o
3. Now draw Fa as a line from the end of the
last one. Use an unknown length (take 1000 for
starters) at 75o.
4. Go back to origin and draw Fb as another
unknown length (say 1000) at 120o.
5. Now trim-corner Fa & Fb, Fb & Fg, into a
triangle.
6. Add dimensions.
Mathematical (components) method
Step 1: Convert angles to 360 Notation:
Fa = Fa N at 75o
Fb = Fb N at 120o
Fg = 240 * 9.81 = 2354.4N at 270o
Step 2: Get X and Y components:
Fax = Fa * cos(75) = 0.2588*Fa
Fay = Fa * sin (75) = 0.9659*Fa
Fbx = Fb * cos(120) = -0.5*Fb
Fby = Fb * sin (120) = 0.8660*Fb
Fgx = 2354.4 * cos(270) = 0 N
Fgy = 2354.4 * sin (270) = -2354.4 N
Step 3: Write Equilibrium equations;
Fx = 0;
Fax + Fbx + Fgx = 0
0.2588*Fa - 0.5*Fb + 0 = 0
1)
Fy = 0;
Fay + Fby + Fgy = 0
0.9659*Fa + 0.8660*Fb - 2354.4 = 0
2)
(eqn
(eqn
Step 4: Solve equations;
These are simultaneous equations that can be
solved by substitution (Or matrices for many
variables)
From eqn 1: Fb = 0.5176*Fa
Substitute this into eqn 2...
0.9659*Fa + 0.8660*0.5176*Fa - 2354.4 = 0
Now we have one variable so we can solve it:
Fa = 2354.4 / 1.4142 = 1664.8 N
Now subs back into eqn 1..
Fb = 0.5176 * Fa = 861.77 N
Using CAD to check the maths.
The Horiz/Vertical dimensions are added to
check the Fx and Fy components of each
force.
Mathematical (Triangle Geometry) Method
Since there are only 2 forces, the Force
Polygon is a triangle. However, it is not a rightangle triangle, so we need the Cosine rule or
the Sine rule.
Find the angles from geometry;
B = 90 - 75 = 15o
C = 75 + (180-120) = 135o
A = 180 - B - C = 30o
So we can use the Sine Rule: (Whew!)
c/sin(C) = 3329.6
Amazing eh? This is the sine number for this
triangle...
OK now, since
b/sin(B) = 3329.6
then
b = 3329.6*sin(15) = 861.77 N
Also
a = 3329.6 * sin(30) = 1664.81 N
Most maths books use capital letters for
angles and lower case for the length of
the opposite sides. Therefore;
2
2
2
• COSINE RULE: a = b + c 2*b*c*Cos (A)
• SINE RULE: a/sin(A) = b/sin(B) =
c/sin(C)
3.9 Summary
In a concurrent force system, all forces pass through a common point. In the previous case
involving the application of two forces to a body, it was necessary for them to be colinear,
opposite in direction, and equal in magnitude for the body to be in equilibrium. If three forces are
applied to a body, as shown in the figure, they must pass through a common point (O), or else the
condition, SMo = 0, will not be satisfied and the body will rotate because of unbalanced
moment. Moreover, the magnitudes of the forces must be such that the force equilibrium
equations, SFx = 0, SFy= 0, are satisfied.
3.10 Keywords
• Equilibrium
• SD
• FBD
• FP
• Pulleys
3.9 Exercise
1) Explain Concurrent Forces.
2) What is Equilibrium?
3) Explain Equilibrium of Concurrent Forces.
4) Explain Forces acting on a Body.
Unit 4
Parallelogram law of forces
Contents
4.1 Introduction
4.2 Objectives
4.3 Parallelogram Law of Forces
4.4 Addition Using Parallelogram Law of Forces.
4.5 Calculation of Resultant Force Using Parallelogram Law of Forces
4.6 In Vector Addition
4.7 Summary
4.8 Keywords
4.9 Exercise
4.1 Introduction
If you have two forces pulling at the same point from different directions, is that any different
from one force pulling halfway between them? The answer is no. As you can see on the right, the
two forces (with the magnitude of the force represented by the length of the line) have been used
as two sides of a parallelogram.
Fig.4.1
The diagonal drawn across is the direction and magnitude of the resultant. A resultant is the
single force that can represent the two original forces. This is a simple, graphical way to add two
forces together and is referred to as the Parallelogram Law.
Here's the part we'll be using again: using the same reasoning, this time backwards, one force can
be broken down into two. This is called breaking a force into its components. This is especially
useful if you place your force in a coordinate plane (like graph paper) and line those two new
forces along the x and y axes (the horizontal and vertical lines). From this, you can make
a vector, a mathematical way to represent a force by its components.
4.2 Objectives
At the end of this chapter you will be able to:
•
Explain Parallelogram Law of Forces
•
Perform Addition Using Parallelogram Law of Forces.
•
Do the Calculation of Resultant Force Using Parallelogram Law of Forces
4.3 Parallelogram Law of Forces
If two forces, acting at a point, are represented in magnitude and direction by the two sides of a
parallelogram drawn from one of its angular points, their resultant is represented both in
magnitude and direction by the diagonal of the parallelogram passing through that angular point.
Resultant Force:
If two or more forces P, Q, S,… act upon a rigid body and if a single force, R, can be found
whose effect upon the body is the same as that of the forces P, Q, S, … this single force R is
called the resultant of the other forces.
Resultant of forces acting in the same direction (same straight line) is equal to their sum.
Magnitude and Direction of the Resultant of Two Forces:
Let OA and OB represent the forces P and Q
acting at a point O and inclined to each other at
an angle a then the resultant R and direction ‘q’
(shown in figure) will be given by R =
√P2+Q2+2PQcosα
Fig.4.2
and tan θ = Qsinα/P+Qcosα
Case
(i):
If
P
=
Q,
then
tanq =
tan
(α/2) => θ = α/2
Case (ii): If the forces act at right angles, so that
α = 90°, we have R = √P2+Q2 and tanθ = Q/P
Solved Example 1:
The resultant of two forces P and Q is R. If Q is doubled, R is doubled and when Q is reversed, R
is again doubled, show that P : Q :R ::√2: √3:√2.
Solution:
Let a be the angle between the forces P and Q. Now from the given conditions, we have
R2 + P2 + Q2 + 2PQ cos α
and
…(1)
(2R)2 + P2 + (2Q)2 + 2P(2Q)cos α
4R2 = P2 + 4Q2 + 4PQ cosα
and
(2R)2 = P2 + (–Q)2 + 2P (–Q) cosα
=>
4R2 = P2+Q2 - 2PQ cosα
…(2)
…(3)
Adding (1) and (3), we get 2P2 + 2Q2 – 5R2 = 0
…(4)
Eliminating a from (2) and (3), we get
P2 + 2Q2 – 4R2 = 0
…(5)
From (4) and (5), we have
P2/–8+10 = Q2/–5+8 = R2/4–2 => P2/2 = Q2/3 = R2/2 => P:Q:R::√2: √3: √2.
4.4 Addition Using Parallelogram Law of Forces.
This topic is no more difficult than it looks. All you need is draw a parallelogram. Just bear a
few rules in mind when you do it.
1. Shift the Force vectors in such a way that the tail of one vector is touching the tail of
other vector.
2. Complete a parallelogram taking vectors as adjacent sides.
3. The resultant force vector is the diagonal of the parallelogram that originates from the
touching tails of the vectors.
4. You are done !
Fig.4.3
The figure is pictorial representation on how to do this.
4.5 Calculation of Resultant Force Using Parallelogram Law of Forces
Till now we discussed how to represent the resultant force vector pictorially. Trigonometry
however, has also provided us with the formula on calculating the magnitude and direction of
resulting force vector. Suppose the magnitudes of
Let
be the resultant force vector for sum of
these vectors. Then
and
and
are
. Also we let
and
respectively.
be the angle between
is given by the equation
Example Problem on Parallelogram Law of Forces
In this section we shall represent the resultant force using the Parallelogram Law on a graph
paper and calculate the same. After that we shall evaluate it using the formula provided to us by
trigonometry. The tools for graph paper solution are provided via tutor vista portal where you
take sessions online.
Fig.4.4
Applying the Formula that Trigonometry gives us, we find
=
=
The results using the two methods, although close, are not exactly equal due to construction
errors and is within the error of least count of graph which is 0.2 N. (one small div = 0.2 N)
4.6 In Vector Addition
In mechanics there are two kind of quantities
•
scalar quantities with magnitude - time, temperature, mass etc.
•
vector quantities with magnitude and direction - velocity, force etc.
When adding vector quantities both magnitude and direction are important. Common methods
adding coplanar vectors (vectors acting in the same plane) are
•
the parallelogram law
•
the triangle rule
•
trigonometric calculation
The Parallelogram Law
Fig.4.5
The procedure of "the parallelogram of vectors addition method" is
•
draw vector 1 using appropriate scale and in the direction of its action
•
from the tail of vector 1 draw vector 2 using the same scale in the direction of its action
•
complete the parallelogram by using vector 1 and 2 as sides of the parallelogram
•
the resulting vector is represented in both magnitude and direction by the diagonal of the
parallelogram
The Triangle Rule
Fig.4.6
The procedure of "the triangle of vectors addition method" is
•
draw vector 1 using appropriate scale and in the direction of its action
•
from the nose of the vector draw vector 2 using the same scale and in the direction of its
action
•
the resulting vector is represented in both magnitude and direction by the vector drawn
from the tail of vector 1 to the nose of vector 2
Trigonometric Calculation
Fig.4.7
The resulting vector of two coplanar vector can be calculated by trigonometry using "the cosine
rule" for a non-right-angled triangle.
FR = [ F12 + F22 − 2 F1 F2 cos(180o - (α + β)) ] 1/2
(1)
where
F = the vector quantity - force, velocity etc.
α + β = angle between vector 1 and 2
The angle between the vector and the resulting vector can be calculated using "the sine rule" for
a non-right-angled triangle.
α = sin-1 [ F1 sin(180o - (α + β)) / FR ]
(2)
where
α + β = the angle between vector 1 and 2 is known
Example - Calculating Vector Forces
A force 1 of magnitude 3 kN is acting in a direction 80o from a force 2 of magnitude 8 kN.
The resulting force can be calculated as
FR = [ (3 kN)2 + (8 kN)2 - 2 (5 kN)(8 kN) cos(180o - (80o)) ] 1/2
= 9 kN
The angle between vector 1 and the resulting vector can be calculated as
α = sin-1[ (3 kN) sin(180o - (80o)) / (9 kN) ]
= 19.1o
The angle between vector 2 and the resulting vector can be calculated as
α = sin-1[ (8 kN) sin(180o - (80o)) / (9 kN) ]
= 60.9o
4.7 Summary
If you have two forces pulling at the same point from different directions, is that any different
from one force pulling halfway between them? The answer is no. [Parallelogram Example] As
you can see on the right, the two forces (with the magnitude of the force represented by the
length of the line) have been used as two sides of a parallelogram.
The diagonal drawn across is the direction and magnitude of the resultant. A resultant is
the single force that can represent the two original forces. This is a simple, graphical way to add
two forces together and is referred to as the Parallelogram Law.
4.8 Keywords
•
Magnitude
•
Parallelogram Law of Forces
•
Resultant Force
•
Scalar
•
Vector
•
Triangle rule
4.9 Exercise
1) Explain Parallelogram Law of Forces.
2) Perform Addition Using Parallelogram Law of Forces.
3) Do the Calculation of Resultant Force Using Parallelogram Law of Forces
Unit 1
Stress and Strain
Contents
1.1 Introduction
1.2 Objectives
1.3 Stress and Strain
1.4 Types of Stress and Strain
1.5 Summary
1.6 Keywords
1.7 Exercise
1.1 Introduction
When a stretching force (tensile force) is applied to an object, it will extend. We can draw its
force - extension graph to show how it will extend. Note: that this graph is true only for the
object for which it was experimentally obtained. We cannot use it to deduce the behaviour of
another object even if it is made of the same material. This is because extension of an object is
not only dependent on the material but also on other factors like dimensions of the object (e.g.
length, thickness etc.) It is therefore more useful to find out about the characteristic extension
property of the material itself. This can be done if we draw a graph in which deformation is
independent of dimensions of the object under test. This kind of graph is calledstress- strain
curve.
1.2 Objectives
At the end of this chapter you will be able to:
•
Explain Stress and Strain
•
List the types of Stress and Strain
1.3 Stress and Strain
Stress
Stress is defined as the force per unit area of a material.
i.e. Stress = force / cross sectional area:
where,
σ = stress,
F = force applied, and
A= cross sectional area of the object.
Units of s : Nm-2 or Pa.
Strain
Strain is defined as extension per unit length.
Strain = extension / original length
where,
ε = strain,
lo = the original length
e = extension = (l-lo), and
l = stretched length
Strain has no units because it is a ratio of lengths.
We can use the above definitions of stress and strain for forces causing tension or compression.
If we apply tensile force we have tensile stress and tensile strain. If we apply compressive
force we have compressive stress and compressive strain.
A useful tip: In calculations stress expressed in Pa is usually a very large number and strain is
usually a very small number. If it comes out much different then, you've done it wrong!
Consider a lump of clay. We can stretch it, squeeze it or twist it. In terms of physics, we say that
we apply "tensile", "compressional" or "torsional" forces, respectively. In order to quantitatively
describe our fun, we define the "stress" which we apply to the clay across any cross section of it
as the force per unit area. Note that these dimensions are those of pressure, and are equivalent to
energy per unit volume ("energy density"):
N / m 2 = J / m 3.
The resulting "strain" (deformation) which the clay experiences is defined as the fractional
extension perpendicular to the cross section we are considering. For instance, when stretching a
cylindrical piece of clay of radius 1 cm, with a force of 100 dynes, the stress is 100 / π dynes per
square cm. The cross section is a circular cut perpendicular to the force we applied to the clay,
and the strain is parallel to the force. If its initial length was 10 cm, and it stretched an additional
2 cm, the strain which it experienced was 2 cm / 10 cm = 0.2. Note that the strain is
dimensionless. Note also that we did 200 ergs of work (100 dynes times 2 cm extension) to
stretch the clay.
The graph of stress versus strain for a material is a veritable cornecopia of information:
Fig 1.1
The slope of the curve at any point is called the "Young's Modulus", and has dimensions of force
over area. Its numerical value is indicative of the "stiffness" of the material: smaller values
indicate that less stress is required for more strain. Likewise, larger values of Young's Modulus
indicate that more stress is required for a given strain. Note that strain may be tensile,
compressive or torsional; in general, the stress versus strain curve will differ for each material,
and for each type of stress. The "strength" of the material is the maximum value of the stress
before breakage. The "extensibility" is the maximum value of the strain before breakage. The
"toughness" is the area under the curve between the vertical dashed lines, and is equal to the
energy required to break the object. The partial area under the curve up to a given strain (less
than the extensibility) is the "work of extension".
Let us revisit the collision in the first section of this chapter. Assume for a moment that the
driver's head had a mass of 5 kg, and that the area of that portion of the head which hit the
windshield was 25 cm 2. The compressive strength of bone is about 16 x 10 7 Pa (N / m 2). From
the definition of pressure as the force per unit area, we find that the force on the head required to
fracture the skull would be 400,000 N (remember to convert the area from cgs to SI units!). This
implies an acceleration of 80,000 m / s 2 (from F = m a), and if the driver's head came to rest in 3
ms, the inital velocity would have to have been 240 m / s!
As a further example of stress and strain, consider a spring. We will treat the spring as a one
dimensional object, so the only stress will be extensional (or compressional, which will be the
negative of extension), and the units will be force per unit extension. Within a range of
extension, the spring obeys "Hooke's Law", which states that Young's Modulus is a constant: the
stress versus strain curve is a straight line:
F = - k ∆x,
where k (which is positive) is Young's Modulus, here called the "spring constant" (with
dimensions of force over length), and ∆x is the amount of extension. The negative sign indicates
that the force is in the opposite direction of extension: if you extend the spring, the force tries to
restore it to its original length. Likewise, if you compress the spring (∆x < 0), the force attempts
to expand the spring, again to its original length. The area under the curve is
U = 1/2 k (∆x) 2,
which gives the work of extension, or alternatively, the amount of potential energy stored in the
spring. We will return to this model when we deal with arterial walls, which we will treat as
springs.
Finally, we mention that the stress versus strain curve is not necessarily the same during the
relaxation of stress as it was during the loading (application) of the stress. This phenomenon is
called "hysteresis", and the ratio of the area under the relaxation curve to that under the loading
curve (for a given strain) is called the "resiliance" (usually expressed as a percentage).
1.4 Types of Stress and Strain
Up to now we have been studying the dynamics of rigid bodies, that is, idealized objects that
have a definite size and shape, but one in which the particles making up the object are
constrained so that the relative positions of the particles never changes. In other words, the rigid
body does not ever stretch, squeeze or twist. However, we know that in reality this does occur,
and
we need
to
find
a way to
describe it.
This
is
done by the
concepts
of stress, strain and elastic modulus. Stress is a measurement of the strength of a material,
strain is a measure of the change in the shape of the object that is undergoing stress and elastic
modulus is a measurement of the amount of stress needed to change the shape of the object.
There are three main types of stress. If we stretch or compress an object, we are subjecting it
to a tensile stress. If an object is subjected to a force along an entire surface, changing its
volume, then it is said to be experiencing a bulk stress. Finally, if the force is acting tangentially
to the surface, causing it to twist, then we are subjecting it to a shear stress.
Tensile Stress
Consider a bar of cross sectional area A being subjected to equal and opposite forces F pulling
at the ends. If this were a rope, we would say that it is experiencing a tension force. Taking this
concept over, we say that the bar is under tension, and is experiencing a stress that we define to
be the ratio of the force to the cross sectional area
Stress = F/A
This stress is called the tensile stress because every part of the object is subjected to a tension.
The SI unit of stress is the Newton per square meter, which is called the Pascal
1 Pascal = 1 Pa = 1 N/m2
Example:
A 250 kg bob is attached to a steel cable with a diameter of 0.05 m. If we take the cable to be
essentially massless, what is the tensile stress experienced by the cable?
Fig.1.2
The stress is just the force divided by the area
If the bar is being pressed instead of pulled, then we say that it is undergoing compressive
stress instead of tensile stress.
Tensile Strain
The fractional amount that an object stretches when it is subjected to a tensile stress is called
the tensile strain. Mathematically, we write this as
(63)
where l0 is the original unstressed length of the bar.
Elastic Modulus
Robert Hooke found that, when the forces are not too large, the amount of strain experience by
an object was directly proportional to the stress. This is another example of Hooke's law. Define
the elastic modulus to be
Using the definitions of stress and strain, this can be rearranged to yield
For tensile stress, the elastic modulus is called the Young's modulus and is denoted by Y.
When a material is stressed, the dimensions perpendicular to the direction of the stress become
smaller by an amount proportional to the fractional change in length. This can be written as
where r is a dimensionless constant called Poisson's ratio. Like Young's modulus, it is a
property of the material and can be used to characterize it.
Example:
A 10000 kg box hangs by a 20 m long cable which has a cross sectional area of 0.15 m2. When
an additional 250 kg is added to the box, the cable is seen to stretch 0.001 mm. What is the
stress, strain and Young's modulus for the cable? What is the material used in the cable?
Fig.1.3
Comparing this with a standard chart of material characteristics, we see that the cable was
probably made of tungsten.
Shear Stress and Strain
Now consider a force that is applied tangentially to an object
Fig.1.4
The ratio of the shearing force to the area A is called the shear stress
If the object is twisted through an angle q, then the strain is
Shear Strain = tanq
Finally, we can define the shear modulus, MS, as
The shear modulus is also known as the torsion modulus.
1.5 Summary
Stress is a measurement of strength, it is how much pressure a material can withstand without
undergoing physical change. There are a number of different types of stress. Stress can cause a
material to change shape, and the degree of deformation is known as strain. Far from being a
nuisance, strain is actually a critical parameter in the growth and morphological evolution of
nanostructures, and its control is essential for producing the desired optical and electronic
properties in both one- and two-dimensional materials.
1.6 Keywords
•
Stress
•
Strain
•
Tensile force
•
Compressive stress
•
Compressive strain
•
Tensile Stress
•
Elastic Modulus
1.7 Exercise
1) Explain Stress and Strain
2) List the types of Stress and Strain
Unit 2
Young's Modulus
Contents
2.1 Introduction
2.2 Objectives
2.3 Young Modulus
2.5 Stress - strain graph beyond elastic behavior
2.6 Energy in stress-strain graphs
2.7 Worked Example
2.8 Deformation and fracture
2.9 Summary
2.10 Keywords
2.11 Exercise
2.1 Introduction
'Young's Modulus' or modulus of elasticity is a measurement of the rate of change of strain as a
function of stress. It represents the slope of the straight-line portion of a stress-strain curve. With
respect to tensile testing, it may be referred to as Tensile Modulus. This method of testing is used
to determine a sample's behavior under an axial stretching load. Common tensile test results
include elastic limit, tensile strength, yield point, yield strength, elongation, and Young's
Modulus. Young's Modulus is reported commonly as N/mm2 (lbs/in2), MPA (psi).
2.2 Objectives
At the end of this chapter you will be able to:
•
Explain Young Modulus
•
Know Stress - strain graph beyond elastic behavior
•
Explain Energy in stress-strain graphs
•
Know Deformation and fracture
2.3 Young Modulus
Instead of drawing a force - extension graph, if you plot stress against strain for an object
showing (linear) elastic behaviour, you get a straight line.
Fig.2.1
This is because stress is proportional to strain. The gradient of the straight-line graph is the
Young's modulus, E
E is constant and does not change for a given material. It in fact represents 'stiffness' property of
the material. Values of the young modulus of different materials are often listed in the form of a
table in reference books so scientists and engineers can look them up.
Units of the Young modulus E: Nm-2 or Pa.
Note: The value of E in Pa can turn out to be a very large number. Therefore some times the
value of E may be given MNm-2.
2.4 Stress-strain graphs
Young Modulus
If you plot a stress against strain of a material with the (linear) elastic behaviour, you get a
straight line.
Fig.2.2
i.e. stress is proportional to strain. The gradient of the above straight line is the Young's modulus,
E
and
E is constant and does not change for a given material, no matter what the size of the sample we
test. It can be considered as a property of the material. The value of E reflects the stiffness of the
material. Stiffer materials have higher values of E. Young's modulus values of different
materials are often listed in the form of a table in reference books so scientists and engineers can
look them up.
Units of the Young modulus E: Nm-2 or Pa.
Note: The value of E in Pa can turn out to be a very large number. It is for this reason that, some
times the value of E may be given MNm-2.
Note: Because 'stress' and 'strain' are (uniquely) related to force and extension, it is not surprising
that the two graphs, stress v/s strain and force v extension, have similar shapes and
characteristics.
Experimental Determination of stress-strain graph and E
We can experimentally determine the value of E by choosing a specimen of the material in a
convenient shape and form. For example, it is easier to deal with a specimen in the form of a
long, thin wire for determining the value of Young's modulus of a metal. In principle we can
apply different forces to a wire by hanging different weights on it and measure the extension of
the wire for the magnitudes of the force applied to draw a stress strain graph. We have already
noted that strain is a small number so it needs to be measured more accurately. We can do this
by using :
Searle's apparatus
This is a schematic diagram of the apparatus.
Fig.2.3
We actually use two wires of equal lengths attached to a rigid support. Although the support is
rigid it to can 'give' slightly under the forces applied. This can affect results. By using two wires,
spurious strain can be eliminated from the measurements. One wire acts as a control wire. We
can accurately measure extension of the other (test) wire. Both control and test wires are attached
to the other ends by a horizontal bar supporting a spirit level. The bar is hinged to the control
wire so that when the test wire is extended due to the addition of weights on the side of the test
wire, the spirit level is tilted by a small amount. We can remove any tilt of the spirit level and
restore it to the horizontal position by turning the screw of a micrometer, which is positioned on
the test wire side and making the bar mounted spirit level travel in the desired direction.
Caution: It is possible that a wire under tension can snap suddenly and damage eyes. Wear
safety glasses. It is also possible that weights attached to the wires could fall down and land on
your feet or other part of the body.
Experimental determination consists of the following steps:
Step 1: Attach equal weights both wires to make them equally taut.
Step 2: Measure the initial length of the wire several times to obtain the average value of lo
Step 3: Measure the diameter of the wire at several points along the wire and the average value
of the diameter (d) and then calculate the circular cross-sectional area
From the formula:
A=
(πd2)
Step 4:Adjust the spirit level so that it is in the horizontal position by turning the micrometer.
Record the micrometer reading to use it as the reference reading.
Step 5:Load the test wire with a further weight. Wait while the wire is being stretched to the
equilibrium position and the spirit level is maximally tilted.
Step 6: Adjust the micrometer screw to restore the spirit level into the horizontal position.
Step 7: Subtract the first micrometer reading from the second micrometer reading to obtain the
extension (e) of the test wire.
Step 8: Calculate stress and strain from the formulae
and
Step 9: Repeat steps 4,5,6 to obtain more values of stresses and strains
Step 10: Plot the above values on stress strain graph; it should be a straight line. Determine the
value of the gradient E.
Worked example
A wire of length 2m and diameter 0.4mm is hung from the ceiling. Find the extension
caused in the wire when a weight of 100N is hung on it. Young Modulus (E) for the wire is
2.0 x 1011 Pa.
Answer: e ~ 8 mm.
2.5 Stress - strain graph beyond elastic behaviour
In this 'Learn-it' so far, we have drawn stress-strain graphs for the elastic behaviour of a material.
In the elastic region the stress-strain graph is a straight line. We can, however draw a stress strain
graph beyond the elastic region. The graph, then becomes non-linear because Hooke's law is not
obeyed and stress is not proportional to strain.
Here are schematic stress-strain graphs of copper and glass.
Fig.2.4
Note: that both graphs end at points marked X. These points are called breaking points. A
material physically breaks at its breaking point. The stress at the breaking point is called
the breaking stress of the material. Breaking stress of a material, in principle, is related to the
energy required to break internal bonds between the atoms or the molecules of the material. It is
very important for designers and engineers to know the value of the breaking stress for the
materials they use.
Fig.2.5
This diagram schematically shows the stress strain curve of rubber. It is different from the
other two stress-strain graphs, above, in the following respects:
1. Within the range of the stress and strain of the graph, rubber undergoes high strains (extension)
without breaking. For example, one kind of rubber (polyisoprene) can be stretched ~500%
without breaking
2. Although, rubber on loading returns to its original length (zero extension), the stress-strain
stress
graph
has two branches (generated by loading up stress and unloading stress). The loop formed by the
two branches is called hysteresis loop. It actually represents the fact that rubber is not a very
good material for storing energy. In on
onee loading and unloading cycle the strain energy,
represented by the area bound by the hysteresis loop is lost and eventually dissipated as heat.
2.6 Energy in stress-strain
strain graphs
Evaluation of strain energy from stress - strain graph
We know that when a material behaves elastically, the work done on straining it is stored as
energy in it. We call this (elastic) strain energy. We can derive the strain energy density
(ρe) in a material by calculating the area under its stress - strain graph. The definition
definitio of the
density of energy is analogous to the definition of the density of mass. It is the energy stored
per unit volume (how many joules are stored in 1m3 of the material).
Fig.2.6
Jm-3
Where:
F is the applied force,
e is extension obtained at force F
F,
A is the area of the cross section of the object and
l is the length of the object
With the knowledge of ρε we can calculate the total energy stored in an object (i.e. that given by
the area under the force - extension graph) if we know the volume of the object.
We can demonstrate this by calculating the work done per unit volume from the total work (W)
done on the object derived from the force - extension graph.
Work done per unit volume = total work done/ total volume
V=
But V = Al
Work done per unit volume =
(
)(
)
2.7 Worked Example
A mass of 200N is hung from the lower end of a steel wire hanging from the ceiling of the
laboratory. The length of the wire 5m, its diameter is 1mm, Young's modulus is: 2 x 1011 Nm-2.
Calculate the strain energy density of the wire and the total energy stored in it.
Apply the formulae for strain density and total work done
F = 200N
d = 1mm = 1 x 10-3m
A = ¼ (πd2) = 0.785 x 10-6m2
l = 5m
Strain energy density =1.62 x105Jm-3
Total strain energy stored = 0.637 J
Density
We know that some materials are light while some are very heavy. For instance, a cube of
polythene will weigh a lot less than the same size cube of steel.
We can systematically compare amount of matter in different materials by defining a property
called density (r).
Density = mass per unit volume
Where,
M is the mass of an object, and V is its volume.
Units of ρ: kgm-3.
Tip: From the knowledge of density and molar mass of a solid or liquid, it is possible to estimate
the average volume occupied by each molecule in a substance (Vmolecule) by using the following
formula.
Where, Mmole its molar mass, and
A = Avagadro's number = 6.023 x 1023.
Approximate molecular spacing (i.e. distance D between adjacent molecules) in the material is
given by the cube root of the volume of the molecule.
D = (Vmolecule)1/3.
Additional Definitions: Molar mass of a material = mass of one mole of the material.
Mole of a material = amount of material containing 6.023 x 1023 molecules.
2.8 Deformation and fracture
When looking at different materials for mechanical purposes we use 'stress-strain' curves. We
saw that for materials obeying Hooke's law the stress strain graph is a straight line. However, this
straight line forms just a part of the stress strain curve. The whole of the stress-strain curve of a
material is an invaluable aid to describing its mechanical behaviour.
For instance, engineers and scientists can easily compare the mechanical properties of different
materials by comparing their stress strain curves and decide which material would be better
suited for a particular use. Typically, the whole (tensile) stress strain curve of a material is made
up of different regions.
Fig.2.7
1. Linear elastic region (region I): At relatively low strains (region I) the material obeys Hookes'
Law and stress is proportional to strain. This part of the curve is a straight line. The constant of
proportionality is the Young modulus and its value is given by the gradient of the straight line.
2. Non-linear elastic region (region II): If we stretch the material beyond its elastic region, we
soon reach the elastic limit of the material. This part of the curve is not a straight line i.e.
Hooke's law is not obeyed. However, if the force (load) is removed the material can go back to
its original shape
3. Yield region (region III): the material suddenly experiences increased deformation. This point
of the graph is known as the upper yield point (Y1). Interestingly, the stress begins to decrease
with the increasing strain until another point the lower yield point (Y2) is reached.
4. Beyond the lower yield point (region IV): Stress increases again with the increasing strain.
However, this increase is not elastic. The material begins to change its cross sectional area
uniformly. At some point, the value of the stress reaches its maximum value. This value of the
stress is known as the ultimate tensile strength (UTS) of the material
5. If the material is stretched further beyond the UTS (region V): the material shows 'necking'
i.e. one part of it narrows considerably. Although the stress dramatically increases locally in the
necking region; the overall stress decreases again with the increasing strain until the breaking
point is reached and the material fractures in the necked region The stress at this point is
called the breaking stress. We know that in the necked region the material can develop
microscopic holes, reducing further, the effective cross sectional area which causes increased
local stresses.
Different materials show different stress strain curves, and the size of each of the above regions
may differ considerably from material to material. Here are some examples:
Fig.2.8
Some materials, called brittle materials, are very stiff (e.g. glass, cast iron) and exhibit elastic
behaviour up to relatively very small values of strains. They hardly change their shape in the
elastic region. Under greater stresses they do not show any yielding but develop cracks at the
surface, which open up as the stress increases and snap to fracture at the breaking point. This
kind of fracture is called brittle fracture.
Fig.2.9
Many metallic materials (e.g. tin copper, silver) on the other hand are ductile . They tend to
plastically elongate under increasing tensile stress. Any cracks on the surface in ductile materials
do not become large under stress because the constituent atoms slide over each other. Before
breaking point, a ductile material tends to 'neck' decreasing in cross sectional area and eventually
breaking.. This kind of fracture is called ductile yielding.
Fig.2.10
Materials like rubber show hysteresis but are resilient. These materials can be stressed
repeatedly through hysteresis cycles without losing their strength and plastic deformation.
2.9 Summary
Young's Modulus also known as the Elasticity Modulus of a material is the ratio of the stress
versus the strain within the Elastic region of the Stress-Strain diagram. This is usually found
from the slope of the stress vs. strain curve.
Elasticity Modulus = Stress / Strain
This mechanical property for a given material was named after Thomas Young, usually denoted
by the symbol E and have the units of Pascals (Pa) or GigaPascals (GPa). This is a more
generalized version of the Hook's law, similar to the spring constant. A high Young's modulus
means that the material is stiffer.
2.10 Keywords
•
Young Modulus
•
Stress
•
Strain
•
Density
2.11 Exercise
1) Explain Young Modulus
2) Define Stress - strain graph beyond elastic behavior
3) Explain Energy in stress-strain graphs
4) Define Deformation and fracture
Unit 3
Viscosity
Contents
3.1 Introduction
3.2 Objectives
3.3 Viscosity
3.4 Types of Viscosity
3.5 Navier stokes equation
3.6 Newtonian Fluid
3.7 Second viscosity coefficient
3.8 Experiment
3.9 Summary
3.10 Keywords
3.11 Exercise
3.1 Introduction
The viscosity of a fluid is a measure of its resistance to continuous deformation caused by sliding
or shearing forces. Imagine a fluid between two flat plates; one plate is stationary and the other is
being moved by a force at a constant velocity parallel to the first plate. The applied force per
unit area of the plate is called the shear stress. The applied shear stress keeps the plate
in motion and, when the plate velocity is steady, this shear stress is in equilibrium with the
frictional and drag forces within the fluid. The shear stress is proportional to the speed of the
plate and inversely proportional to the distance between the plates. The proportionality factor
between the shear stress and the velocity difference between the plates is defined as the
coefficient of viscosity or simply the viscosity of the fluid. Thick fluids such as tar or honey
have a high viscosity; thin fluids such as water or alcohol have a low viscosity.
In
general, viscosity is
a
function
of temperature and pressure;
however,
in
some
fluids viscosity is dependent on the rate of shear and time. When brushed on (sheared) quickly,
fluids such as paint have a low viscosity and flow easily. After paint is applied, only the slow and
steady pull of its weight causes it to flow; at this slow shear rate the viscosity of paint is high and
its resists the tendency to flow or sag. Fluids that behave in this manner are called nonNewtonian fluids. Other examples are liquid plastics and mud. For gases and non-polymeric
liquids like water, viscosity is independent of the fluid's shear stress and history. These are called
Newtonian fluids. In the case of gases, the viscosity increases with temperature because of the
increased molecular activity at higher temperatures. Liquids, conversely, generally show
decreasing viscosity with increasing temperature. The flow of liquids in pipes, the performance
of oil-lubricated bearings in engines or oil-filled automotive shock absorbers, and the air
resistance on a moving car or airplane are all dependent on the viscosity of the fluids involved
3.2 Objectives
At the end of this chapter you will be able to:
•
Define Viscosity
•
List the Types of Viscosity
•
Explain Navier stokes equation
•
Explain Newtonian Fluid
3.3 Viscosity
Definition:
Viscosity is a measurement of how resistant a fluid is to attempts to move through it. A fluid
with a low viscosity is said to be "thin," while a high viscosity fluid is said to be "thick." It is
easier to move through a low viscosity fluid (like water) than a high viscosity fluid (like honey).
Most common fluids, called Newtonian fluids (yes, another thing named after that Newton),
have a constant viscosity. There is a greater resistance as you increase the force, but it's a
constant proportional increase. In short, a Newtonian fluid keeps acting like a fluid, no matter
how much force is put into it. In contrast, the viscosity of non-Newtonian fluids is not constant,
but rather varies greatly depending on the force applied. A classic example of a non-Newtonian
is Oobleck, which exhibits solid-like behavior when a large amount of force is used on it.
Another type of non-Newtonian fluid are known as magnetorheological fluids, which respond to
magnetic fields by becoming nearly solid but reverting to their fluid state when removed from
the magnetic field.
The viscosity of a fluid can be defined as the measure of how resistive the fluid is to flow. It is
analogous to the friction of solid bodies in that it also serves as a mechanics for
transforming kinetic energy into thermal energy.
Given two plane parallel plates separated by a distance d and with a fluid between them, keep
one stationary while moving the other at a slow speed
. A common situation is that the
force F required to keep the second plate moving is proportional to its area A and to
.A
fluid in which these quantities are proportional, called a Newtonian fluid, therefore exhibits shear
stress
proportional to
, giving
(1)
Here,
is a constant for the given fluid called the dynamic viscosity.
3.4 Types of Viscosity
Dynamic Viscosity
A parameter
defined such that
(1)
Written explicitly,
(2)
where l is the length scale and u is the velocity scale. In cgs,
has units of g cm-1 s-1, or poise.
1 poise = 1/10 Pa s. Dynamic viscosity is related to kinematic viscosity
by
(3)
where
is the density.
Bulk Viscosity
The bulk viscosity
where
of a fluid is defined as
is the second viscosity coefficient and
is the shear viscosity.
Eddy Viscosity
A parameter which reproduces the effects of turbulent diffusion (mixing) by modeling it with a
fictitious diffusion coefficient. The viscosity operates on a scale known as the mixing length.
Kinematics Viscosity
A coefficient which describes the diffusion of momentum. Let
The unit of kinematic viscosity is the Stoke, equal to 1 cm2 s-1.
be the dynamic viscosity, then
3.5 Navier stokes equation
The Navier-Stokes equations are the fundamental partial differentials equations that describe the
flow of incompressible fluids. Using the rate of stress and rate of strain tensors, it can be shown
that the components
of a viscous force F in a nonrotating frame are given by
(1)
(2)
(Tritton 1988, Faber 1995), where
coefficient,
is the Kronecker delta,
and Einstein summation
is the dynamic viscosity,
is the second viscosity
is the divergence,
is the bulk viscosity,
has been used to sum over j = 1, 2, and 3.
3.6 Newtonian Fluid
Let a fluid be traveling in the
direction with velocity u(y). A Newtonian fluid is one for which
the strain rate
(1)
is equal to the vertical velocity gradient
(2)
As shown in strain rate, this also implies
(3)
so
(4)
3.7 Second viscosity coefficient
For a compressible fluid, i.e., one for which
, where
is the divergence
of the
velocity field, the stress tensor of the fluid can be written
where
is the Kronecker delta
of viscosity (Tritton 1988).
fluid, the term involving
,
is the dynamic viscosity, and
is the second coefficient
is analogous to the first Lamé constant. For an incompressible
drops out from the equation, so
can be ignored.
3.8 Experiment
Falling through a high viscosity liquid
Demonstration
The higher the viscosity of a liquid the more it resists motion of a body through it. The result can
be very low terminal velocity.
Apparatus and materials
•
Measuring cylinder or tall and fairly wide glass tube, 1,000 ml, with firm stopper
•
Glycerine, heavy oil or liquid detergent
•
Ball bearings (approximately 3 mm and 1.5 mm)
•
Chinagraph pencil, water-based pen, or elastic bands
•
Eye protection
Technical notes
A tall glass tube allows a greater distance of fall than a measuring cylinder. Seal the bottom end
firmly with a stopper and rest this on a surface so that it cannot fall out. Do not over-tighten any
clamp that you use to hold such a tube.
Use the pencil, pen or elastic bands to provide equally spaced markers on the measuring cylinder
or glass tube. Do this before the lesson.
Place the ball-bearings in a dish of the same liquid before use. This reduces the occurrence of air
bubbles, which will affect the motion of the ball bearings.
Retrieve ball-bearings from the liquid with a magnet outside the jar. This is a messy activity to
clear away, especially if many ball bearings are allowed to fall and must then be retrieved.
Safety
Glycerine (glycerol or propane -1, 2, 3-triol) will irritate eyes, so eye protection should be worn.
Its properties will change if it is allowed to absorb water vapour from the atmosphere so it must
be kept in a closed container. Waste engine oil is carcinogenic and must not be used.
Procedure
a Set up the measuring cylinder or tall glass tube, filled with the viscous liquid, so that it is
illuminated from above by a bright source. In an otherwise darkened room (full blackout is not
necessary) the ball-bearings then appear as bright points of light.
Fig.3.1
b Release a ball-bearing from just above the liquid surface.
c Ask students to clap as the ball-bearing passes each marker. This is sufficient to show that the
time intervals become the same, and thus that the ball-bearings quickly reach their terminal
velocity.
Teaching notes
1 You could use a more sophisticated timing system, but the point here is to demonstrate
terminal velocity rather than to make precise measurements.
2 Advanced level students could determine the viscosity of the liquid, using Stokes' law. Or they
could investigate the relationship between the radius of a falling ball and its terminal velocity.
When a ball bearing is moving at terminal velocity, the forces acting on it are balanced.
Frictional force acting upwards = weight - upthrust
where η = viscosity
α = radius of the ball bearing
ν ο = terminal velocity
γ = gravitational field strength
ρ = density of the bearing material
σ = density of the liquid
3.9 Summary
Viscosity, resistance of a fluid (liquid or gas) to a change in shape, or movement of
neighbouring portions relative to one another. Viscosity denotes opposition to flow. The
reciprocal of the viscosity is called the fluidity, a measure of the ease of flow. Molasses, for
example, has a greater viscosity than water. Because part of a fluid that is forced to move carries
along to some extent adjacent parts, viscosity may be thought of as internal friction between the
molecules; such friction opposes the development of velocity differences within a
fluid. Viscosity is a major factor in determining the forces that must be overcome when fluids are
used in lubrication and transported in pipelines. It controls the liquid flow in such processes as
spraying, injection molding, and surface coating.
For many fluids the tangential, or shearing, stress that causes flow is directly proportional to the
rate of shear strain, or rate of deformation, those results. In other words, the shear stress divided
by the rate of shear strain is constant for a given fluid at a fixed temperature. This constant is
called the dynamic, or absolute, viscosity and often simply the viscosity. Fluids that behave in
this way are called Newtonian fluids in honour of Sir Isaac Newton, who first formulated this
mathematical description of viscosity.
The viscosity of liquids decreases rapidly with an increase in temperature; the viscosity of gases
increases with an increase in temperature. Thus, upon heating, liquids flow more easily, whereas
gases flow more sluggishly.
The dimensions of dynamic viscosity are force times time divided by area. The unit of viscosity,
accordingly, is newton-second per square metre. For some applications the kinematic viscosity is
more
useful
than
the
absolute,
or
dynamic, viscosity.
Kinematic viscosity is
the
absolute viscosity of a fluid divided by its mass density. (Mass density is the mass of a substance
divided by its volume.) The dimensions of kinematic viscosity are area divided by time; the
appropriate units are metre squared per second. The unit of kinematic viscosity in the centimetregram-second (CGS) system, called the stokes in Britain and the stoke in the U.S., is named for
the British physicist Sir George Gabriel Stokes. The stoke is defined as 1 cm squared per
second.
3.10 Keywords
•
Velocity
•
Equilibrium
•
Temperature
•
Dynamic Viscosity
•
Bulk Viscosity
•
Eddy Viscosity
•
Kinematics Viscosity
3.11 Exercise
1. Define Viscosity.
2. List the Types of Viscosity.
3. Explain Navier stokes equation.
4. Explain Newtonian Fluid.
Unit 4
Surface tension
Contents
4.1 Introduction
4.2 Objectives
4.3 Surface Tension
4.4 Pressure Inside a Bubble
4.5 Contact Angle & Capillarity - Liquid in a Vertical Tube
4.6 Surface Tension Physics Experiments & Tricks
4.7 Experimental Surface Tension Values
4.8 Summary
4.9 Keywords
4.10 Exercise
4.1 Introduction
Surface tension, property of a liquid surface displayed by its acting as if it were a stretched
elastic membrane. This phenomenon can be observed in the nearly spherical shape of small
drops of liquids and of soap bubbles. Because of this property, certain insects can stand on the
surface of water. A razor blade also can be su
supported
pported by the surface tension of water. The razor
blade is not floating: if pushed through the surface, it sinks through the water.
4.2 Objectives
At the end of this chapter you will be able to:
•
Explain Surface Tension
•
Give the Pressure Inside a Bubble
•
Explain Contact Angle & Capillarity - Liquid in a Vertical Tube
•
List the Experimental Surface Tension Values
•
4.3 Surface Tension
Surface tension is a phenomenon in which the surface of a liquid, where the liquid is in contact
with gas, acts like a thin elastic sheet. This term is typically used only when the liquid surface is
in contact with gas (such as the air). If the surface is between two liquids (such as water and oil),
it is called "interface tension."
Fig.4.1: The forces acting on a liquid that cause surface tension.
Causes of Surface Tension
Various intermolecular forces, such as Van der Waals forces, draw the liquid particles together.
Along the surface, the particles are pulled toward the rest of the liquid, as shown in the picture to
the right.
Surface tension (denoted with the Greek variable gamma) is defined as the ratio of the surface
force F to the length d along which the force acts:
gamma = F / d
Units of Surface Tension
Surface tension is measured in SI units of N/m (newton per meter), although the more common
unit is the cgs unit dyn/cm (dyne per centimeter).
In order to consider the thermodynamics of the situation, it is sometimes useful to consider it in
terms of work per unit area. The SI unit in that case is the J/m2 (joules per meter squared). The
cgs unit is erg/cm2.
These forces bind the surface particles together. Though this binding is weak - it's pretty easy to
break the surface of a liquid after all - it does manifest in many ways.
Examples of Surface Tension
Drops of water: When using a water dropper, the water does not flow in a continuous stream,
but rather in a series of drops. The shape of the drops is caused by the surface tension of the
water. The only reason the drop of water isn't completely spherical is because of the force of
gravity pulling down on it. In the absence of gravity, the drop would minimize the surface area in
order to minimize tension, which would result in a perfectly spherical shape.
Insects walking on water: Several insects are able to walk on water, such as the water strider.
Their legs are formed to distribute their weight, causing the surface of the liquid to become
depressed, minimizing the potential energy to create a balance of forces so that the strider can
move across the surface of the water without breaking through the surface. This is similar in
concept to wearing snow shoes to walk across deep snowdrifts without your feet sinking.
Needle (or paper clip) floating on water: Even though the density of these objects are greater
than water, the surface tension along the depression is enough to counteract the force of gravity
pulling down on the metal object. Click on the picture to the right, then click "Next," to view a
force diagram of this situation or try out the Floating Needle trick for yourself.
4.4 Pressure Inside a Bubble
Anatomy of a Soap Bubble
When you blow a soap bubble, you are creating a pressurized bubble of air which is contained
within a thin, elastic surface of liquid. Most liquids cannot maintain a stable surface tension to
create a bubble, which is why soap is generally used in the process ... it stabilizes the surface
tension through something called the Marangoni effect.
When the bubble is blown, the surface film tends to contract. This causes the pressure inside the
bubble to increase. The size of the bubble stabilizes at a size where the gas inside the bubble
won't contract any further, at least without popping the bubble.
In fact, there are two liquid-gas interfaces on a soap bubble - the one on the inside of the bubble
and the one on the outside of the bubble. In between the two surfaces is a thin film of liquid.
The spherical shape of a soap bubble is caused by the minimization of the surface area - for a
given volume, a sphere is always the form which has the least surface area.
Pressure Inside a Soap Bubble
To consider the pressure inside the soap bubble, we consider the radius R of the bubble and also
the surface tension, gamma, of the liquid (soap in this case - about 25 dyn/cm).
We begin by assuming no external pressure (which is, of course, not true, but we'll take care of
that in a bit). You then consider a cross-section through the center of the bubble.
Along this cross section, ignoring the very slight difference in inner and outer radius, we know
the circumference will be 2pi R. Each inner and outer surface will have a pressure of
gamma along the entire length, so the total. The total force from the surface tension (from both
the inner and outer film) is, therefore, 2gamma (2pi R).
Inside the bubble, however, we have a pressure p which is acting over the entire cross-section pi
R2, resulting in a total force of p(pi R2).
Since the bubble is stable, the sum of these forces must be zero so we get:
2gamma (2pi R) = p(pi R2)
or
p = 4 gamma / R
Obviously, this was a simplified analysis where the pressure outside the bubble was 0, but this is
easily expanded to obtain the difference between the interior pressure p and the exterior
pressure pe:
p - pe = 4 gamma / R
Pressure in a Liquid Drop
Analyzing a drop of liquid, as opposed to a soap bubble, is simpler. Instead of two surfaces, there
is only the exterior surface to consider, so a factor of 2 drops out of the earlier equation
(remember where we doubled the surface tension to account for two surfaces?) to yield:
p - pe = 2 gamma / R
4.5 Contact Angle & Capillarity - Liquid in a Vertical Tube
Contact Angle
Surface tension occurs during a gas-liquid interface, but if that interface comes in contact with a
solid surface - such as the walls of a container - the interface usually curves up or down near that
surface. Such a concave or convex surface shape is known as a meniscus
Fig.4.2 Determining the contact angle, theta, of liquid in a vertical tube.
The contact angle, theta, is determined as shown in the picture to the right.
The contact angle can be used to determine a relationship between the liquid-solid surface
tension and the liquid-gas surface tension, as follows:
gammals = - gammalg cos theta
where
•
gammals is the liquid-solid surface tension
•
gammalg is the liquid-gas surface tension
•
theta is the contact angle
One thing to consider in this equation is that in cases where the meniscus is convex (i.e. the
contact angle is greater than 90 degrees), the cosine component of this equation will be negative
which means that the liquid-solid surface tension will be positive.
If, on the other hand, the meniscus is concave (i.e. dips down, so the contact angle is less than 90
degrees), then the cos theta term is positive, in which case the relationship would result in
a negative liquid-solid surface tension!
What this means, essentially, is that the liquid is adhering to the walls of the container and is
working to maximize the area in contact with solid surface, so as to minimize the overall
potential energy.
Capillarity
Another effect related to water in vertical tubes is the property of capillarity, in which the surface
of liquid becomes elevated or depressed within the tube in relation to the surrounding liquid.
This, too, is related to the contact angle observed.
If you have a liquid in a container, and place a narrow tube (or capillary) of radius r into the
container, the vertical displacement y that will take place within the capillary is given by the
following equation:
y = (2 gammalg cos theta) / (dgr)
where
•
y is the vertical displacement (up if positive, down if negative)
•
gammalg is the liquid-gas surface tension
•
theta is the contact angle
•
d is the density of the liquid
•
g is the acceleration of gravity
•
r is the radius of the capillary
NOTE: Once again, if theta is greater than 90 degrees (a convex meniscus), resulting in a
negative liquid-solid surface tension, the liquid level will go down compared to the surrounding
level, as opposed to rising in relation to it.
Capillarity manifests in many ways in the everyday world. Paper towels absorb through
capillarity. When burning a candle, the melted wax rises up the wick due to capillarity. In
biology, though blood is pumped throughout the body, it is this process which distributes blood
in the smallest blood vessels which are called, appropriately, capillaries.
4.6 Surface Tension Physics Experiments & Tricks
Quarters in a Full Glass of Water
This is a neat trick! Ask friends how many quarters can go in a completely full glass of water
before it overflows. The answer will generally be one or two. Then follow the steps below to
prove them wrong.
Needed materials:
•
10 to 12 Quarters
•
glass full of water
The glass should be filled to the very rim, with a slightly convex shape to the surface of the
liquid.
Slowly, and with a steady hand, bring the quarters one at a time to the center of the glass. Place
the narrow edge of the quarter in the water and let go. (This minimizes disruption to the surface,
and avoids forming unnecessary waves that can cause overflow.)
As you continue with more quarters, you will be astonished how convex the water becomes on
top of the glass without overflowing!
Possible Variant: Perform this experiment with identical glasses, but use different types of coins
in each glass. Use the results of how many can go in to determine a ratio of the volumes of
different coins.
Floating Needle
Another nice surface tension trick, this one makes it so that a needle will float on the surface of a
glass of water. There are two variants of this trick, both impressive in their own right.
Needed materials:
•
fork (variant 1)
•
piece of tissue paper (variant 2)
•
sewing needle
•
glass full of water
Variant 1 Trick
Place the needle on the fork, gently lowering it into the glass of water. Carefully pull the fork
out, and it is possible to leave the needle floating on the surface of the water.
This trick requires a real steady hand and some practice, because you must remove the fork in
such a way that portions of the needle do not get wet ... or the needle will sink. You can rub the
needle between your fingers beforehand to "oil" it increase your success chances.
Variant 2 Trick
Place the sewing needle on a small piece of tissue paper (large enough to hold the needle). The
needle is placed on the tissue paper. The tissue paper will become soaked with water and sink to
the bottom of the glass, leaving the needle floating on the surface.
Put Out Candle with a Soap Bubble
This trick demonstrates how much force is caused by the surface tension in a soap bubble.
Needed materials:
•
lit candle (NOTE: Do not play with matches without parental approval and supervision!)
•
funnel
•
detergent or soap-bubble solution
Coat the funnel mouth (the large end) with the detergent or bubble solution, then blow a bubble
using the small end of the funnel. With practice, you should be able to get a nice big bubble,
about 12 inches in diameter.
Place your thumb over the small end of the funnel. Carefully bring it toward the candle. Remove
your thumb, and the surface tension of the soap bubble will cause it to contract, forcing air out
through the funnel. The air forced out by the bubble should be enough to put out the candle.
For a somewhat related experiment, see the Rocket Balloon.
Motorized Paper Fish
This experiment from the 1800's was quite popular, as it shows what seems to be sudden
movement caused by no actual observable forces.
Needed materials:
•
piece of paper
•
scissors
•
vegetable oil or liquid dishwasher detergent
•
a large bowl or loaf cake pan full of water
In addition, you will need a pattern for the Paper Fish. To spare you my attempt at artistry, check
out this example of how the fish should look. Print it out - the key feature is the hole in the center
and the narrow opening from the hole to the back of the fish.
Once you have your Paper Fish pattern cut out, place it on the water container so it floats on the
surface. Put a drop of the oil or detergent in the hole in the middle of the fish.
The detergent or oil will cause the surface tension in that hole to drop. This will cause the fish to
propel forward, leaving a trail of the oil as it moves across the water, not stopping until the oil
has lowered the surface tension of the entire bowl.
Experimental Surface Tension Values
The table below demonstrates values of surface tension obtained for different liquids at various
temperatures.
4.7 Experimental Surface Tension Values
Liquid in contact with air Temperature (degrees C) Surface Tension (mN/m, or dyn/cm)
Benzene
20
28.9
Carbon tetrachloride
20
26.8
Ethanol
20
22.3
Glycerin
20
63.1
Mercury
20
465.0
Olive oil
20
32.0
Soap solution
20
25.0
Water
0
75.6
Water
20
72.8
Water
60
66.2
Water
100
58.9
Oxygen
-193
15.7
Neon
-247
5.15
Helium
-269
0.12
4.8 Summary
For surface tension, something different is going on. There is an interaction between the
molecules in the water and the aluminum (and between other water molecules). This attractive
force makes the surface of the water kind of like the skin on pudding (kind of). For the video
above, the effect of the water molecules attracting to the aluminum and to other water molecules
exerts an upward force on the disk. Oh, in this case, there is also some buoyancy force - but
clearly that is not all that keeps it up. If you push that disk below the surface, it will sink. Under
the water, there is no surface tension.
4.9 Keywords
•
Surface tension
•
Anatomy of a Soap Bubble
•
Contact Angle
•
Capillarity
•
Floating Needle
4.10 Exercise
1) Explain Surface Tension
2) Give the Pressure Inside a Bubble
3) Explain Contact Angle & Capillarity - Liquid in a Vertical Tube
4) List the Experimental Surface Tension Values
Unit 1
Projectile Motion
Contents
1.1 Introduction
1.2 Objectives
1.3 What is a Projectile?
1.4 Projectile Motion and Inertia
1.5 Projectile Motion, General Solution
1.6 Summary
1.7 Keywords
1.8 Exercise
1.1 Introduction
Projectile motion refers to the motion of an object projected into the air at an angle. A few
examples of this include a soccer ball begin kicked, a baseball begin thrown, or an athlete
long jumping. Even fireworks and water fountains are examples of projectile motion. In this
lesson you will learn the fundamentals of projectile motion. You will be given examples and
interesting facts.
1.2 Objectives
At the end of this chapter you will be able to:
•
Know What is a Projectile?
•
Explain Projectile Motion and Inertia
•
Know Projectile Motion, General Solution
1.3 What is a Projectile?
A projectile is an object upon which the only force acting is gravity. There are a variety of
examples of projectiles. An object dropped from rest is a projectile (provided that the influence
of air resistance is negligible). An object that is thrown vertically upward is also a projectile
(provided that the influence of air resistance is negligible). And an object which is thrown
upward at an angle to the horizontal is also a projectile (provided that the influence of air
resistance is negligible). A projectile is any object that once projected or dropped continues in
motion by its own inertia and is influenced only by the downward force of gravity.
Fig.1.1
By definition, a projectile has a single force that acts upon it - the force of gravity. If there were
any other force acting upon an object, then that object would not be a projectile. Thus, the freebody diagram of a projectile would show a single force acting downwards and labeled force of
gravity (or simply Fgrav). Regardless of whether a projectile is moving downwards, upwards,
upwards and rightwards, or downwards and leftwards, the free-body diagram of the projectile is
still as depicted in the diagram at the right. By definition, a projectile is any object upon which
the only force is gravity.
1.4 Projectile Motion and Inertia
Many students have difficulty with the concept that the only force acting upon an upward
moving projectile is gravity. Their conception of motion prompts them to think that if an object
is moving upward, then there must be an upward force. And if an object is moving upward and
rightward, there must be both an upward and rightward force. Their belief is that forces cause
motion; and if there is an upward motion then there must be an upward force. They reason, "How
in the world can an object be moving upward if the only force acting upon it is gravity?" Such
students do not believe in Newtonian physics (or at least do not believe strongly in Newtonian
physics). Newton's laws suggest that forces are only required to cause an acceleration (not a
motion). Recall from the Unit 2 that Newton's laws stood in direct opposition to the common
misconception that a force is required to keep an object in motion. This idea is simply not true! A
force is not required to keep an object in motion. A force is only required to maintain an
acceleration. And in the case of a projectile that is moving upward, there is a downward force
and a downward acceleration. That is, the object is moving upward and slowing down.
To further ponder this concept of the downward force and a downward acceleration for a
projectile, consider a cannonball shot horizontally from a very high cliff at a high speed. And
suppose for a moment that the gravity switchcould be turned off such that the cannonball would
travel in the absence of gravity? What would the motion of such a cannonball be like? How
could its motion be described? According to Newton's first law of motion, such a cannonball
would continue in motion in a straight line at constant speed. If not acted upon by an unbalanced
force, "an object in motion will ...". This is Newton's law of inertia.
Fig.1.2
Now suppose that the gravity switch is turned on and that the cannonball is projected
horizontally from the top of the same cliff. What effect will gravity have upon the motion of the
cannonball? Will gravity affect the cannonball's horizontal motion? Will the cannonball travel a
greater (or shorter) horizontal distance due to the influence of gravity? The answer to both of
these questions is "No!" Gravity will act downwards upon the cannonball to affect its vertical
motion. Gravity causes a vertical acceleration. The ball will drop vertically below its otherwise
straight-line, inertial path. Gravity is the downward force upon a projectile that influences its
vertical motion and causes the parabolic trajectory that is characteristic of projectiles.
Fig.1.3
A projectile is an object upon which the only force is gravity. Gravity acts to influence the
vertical motion of the projectile, thus causing a vertical acceleration. The horizontal motion of
the projectile is the result of the tendency of any object in motion to remain in motion at constant
velocity. Due to the absence of horizontal forces, a projectile remains in motion with a constant
horizontal velocity. Horizontal forces are not required to keep a projectile moving horizontally.
The only force acting upon a projectile is gravity!
1.5 Projectile Motion, General Solution
What follows is a general solution for the two dimensional motion of an object thrown in a
gravitational field. This is usually termed a projectile motion problem. The thrown object is
called the projectile. Its path is called the trajectory. We will answer all the usual questions that
arise in a first year physics class regarding this motion. We will not consider air resistance.
Without air resistance, the projectile will follow a parabolic trajectory. We will be throwing the
projectile on level ground on planet Earth. It will leave the point of release, arc through the air
along a path shaped like a parabola, and then hit ground a certain distance from where it was
thrown.
Fig.1.4
As mentioned above, this is a two dimensional problem. Therefore, we will consider x and y
directed displacements, velocities, and accelerations. The projectile will accelerate under the
influence of gravity, so its y acceleration will be downward, or negative, and will be equal in size
to the acceleration due to gravity on Earth. There will be no acceleration in the x direction since
the force of gravity does not act along this axis.
On Earth the acceleration due to gravity is 9.8 m/s2 directed downward. So, for this presentation
acceleration in the y direction, or ay, will be -9.8 m/s2, and acceleration in the x direction, or ax,
will be 0.0 m/s2.
Given the original conditions with which the projectile is thrown we will proceed to find the
components of the original velocity and then move on to answer the following questions:
•
How much time passes till the projectile is at the top of its flight?
•
How high does the projectile rise?
•
How much time passes till the projectile strikes the ground?
•
How far away does the projectile land from its starting point?
Original, or initial, conditions:
The original conditions are the size of the velocity and the angle above the horizontal with which
the projectile is thrown.
Fig.1.5
General:
Original size of velocity: vo
Original angle: theta
Example:
vo = 40.0 m/s
theta = 35 degrees
Components of original velocity:
The usual first step in this investigation is to find the x and y components for the original
velocity.
Fig.1.6
General:
X component of original velocity: vox = vocos(theta)
Y component of original velocity: voy = vosin(theta)
Example:
In the x direction:
vox = vocos(theta)
vox = (40.0 m/s)(cos(35 degrees))
vox = (40.0)(0.8191)
vox = 32.76
vox = 32.8 m/s
In the y direction:
voy = vosin(theta)
voy = (40.0 m/s)(sin(35 degrees))
voy = (40.0)(0.5735)
voy = 22.94
voy = 22.9 m/s
How much time passes until the projectile is at the top of its trajectory?
At the top of the trajectory the y, or upward, velocity of the projectile will be 0.0 m/s. The object
is still moving at this moment, but its velocity is purely horizontal. At the top it is not moving up
or down, only across.
Notice that the object is still in motion at the top of the
trajectory; however, its velocity is completely horizontal. It
has stopped going up and is about to begin going down.
Therefore, its y velocity is 0.0 m/s.
Fig.1.7
We need to find out how much time passes from the time of the throw until the time when the y
velocity of the projectile becomes 0.0 m/s. This y velocity at the top of the trajectory can be
thought of as the final y velocity for the projectile for the portion of its flight that starts at the
throw and ends at the top of the trajectory.
We will call this amount of time 'the half time of flight', since the projectile will spend one half
of its time of flight rising to the top of its trajectory. It will spend the second half of its time of
flight moving downward.
General:
We can use the following kinematics equation:
vf = vo + at
Subscript it for y:
vfy = voy + ayt
Solve it for t:
t = (vfy - voy) / ay
Plug in 0.0 m/s for vfy:
t = (0.0 m/s - voy) / ay
If the original y velocity and the y acceleration, i. e., the acceleration due to gravity, are plugged
into the above equation, it will solve for the amount of time that passes from the moment of
release to the moment when the projectile is at the top of its flight.
Example:
Start with:
t = (vfy - voy) / ay
Plug in 0.0 m/s for vfy:
t = (0.0 m/s - voy) / ay
Plug in values for voy and ay:
t = (0.0 m/s - 22.9 m/s) / - 9.8 m/s2
t = -22.9 / -9.8
t = 2.33
t = 2.3 s
In this example 2.3s of time passes while the projectile is rising to the top of the trajectory.
How high does the projectile rise?
Here you need to find the displacement in the y
direction at the time when the projectile is at the
top of its flight. We have just found the time at
which the projectile is at the top of its flight. If we
plug this time into a kinematics formula that will
return the displacement, then we will know how
high above ground the projectile is at when it is at
the top of its trajectory.
Fig.1.8
General:
Here is the displacement formula:
d = vot + 0.5at2
We must think of this displacement in the y direction, so we will subscript this formula for y:
dy = voyt + 0.5ayt2
If now we plug in the half time of flight, which was found above, we will solve for the height of
the trajectory, since the projectile is at its maximum height at this time.
Example:
Starting with:
dy = voyt + 0.5ayt2
Then plugging in known values:
dy = (22.9 m/s)(2.33 s) + (0.5)(-9.8 m/s2)(2.33 s)2
dy = 53.35 - 26.60
dy = 26.75
dy = 27 m
How much time passes until the projectile strikes the ground?
General:
With no air resistance, the projectile will spend an equal amount of time rising to the top of its
projectile as it spends falling from the top to the ground. Since we have already found the half
time of flight, we need only to double that value to get the total time of flight.
Example:
t = 2(2.33 s)
t = 4.66
t = 4.7 s
This is the total time of flight.
How far away does the projectile land from its starting point?
Fig.1.9
The distance from the starting point on the ground to the landing point on the ground is called the
range of the trajectory. This range is a displacement in the x direction. It is governed by the x
velocity of the projectile. This x velocity does not change during the flight of the projectile. That
is, whatever is the value of the x velocity at the start of the trajectory will be the value of the x
velocity throughout the flight of the projectile. The x velocity remains constant because there are
no accelerations in the x direction. The only acceleration is in the y direction, and this is due to
the vertical pull of gravity. Gravity does not pull horizontally. Therefore, the calculation for the
range is simplified.
General:
Let us start with the general displacement formula:
d = vot + 0.5at2
Since we are working in the x direction, we should subscript this equation for x:
dx = voxt + 0.5axt2
Now, since the acceleration in the x direction is 0.0 m/s2, the second term in the above equation
drops out, and we are left with:
dx = voxt
The velocity in the x direction does not change. The projectile maintains its original x velocity
throughout its entire flight. So, the original x velocity is the only x velocity the projectile will
have. We could, therefore, think of the last equation as:
dx = vxt
If we plug in the original x velocity for vx and the total time of flight for t, we will solve for the
horizontal displacement, or range, of the trajectory.
Example:
As shown in the general section above, start with:
dx = vxt
Plug in values. Remember that the x velocity is constant and always equal to its original value
and that the time here is the total time of flight.
dx = (32.8 m/s)(4.66 s)
dx = 152.84
dx = 150 m
1.6 Summary
When a projectile is shot in a uniform gravitational the horizontal and vertical components of
motion are independent, that is the horizontal motion does not affect the vertical motion. No
matter how fast the projectile moves vertically it can not affect the horizontal motion. Projectile
motion can be seen as the superposition of horizontal and vertical motion. Projectile motion can
be described using the uniformly accelerated motion equations.
1.7 Keywords
•
Projectile Motion
•
Inertia
•
Velocity
•
Trajectory
1.8 Exercise
1) What is a Projectile?
2) Explain Projectile Motion and Inertia
3) How much time passes until the projectile is at the top of its trajectory?
4) How high does the projectile rise?
5) How much time passes until the projectile strikes the ground?
Unit 2
Angle of projection
Contents
2.1 Introduction
2.2 Objectives
2.3 Optimum Projection Angle
2.4 Optimum Projection Angles in Undergraduate Teaching
2.5 Summary
2.6 Keywords
2.7 Exercise
2.1 Introduction
One of the best known 'results' of the science of mechanics is that the optimum projection angle
for achieving maximum horizontal range is 45°. However, it is also well known that actual
performers in projectile-related sports seldom use an angle of 45°. For example, typical
projection angles of world-class shot-putters are around 37°. Some researchers have noted that in
shot-putting the landing is about 2 m lower than the launch. Even so, this produces only a small
reduction in the calculated optimum projection angle (to about 42°)
Fig.2.1
The reason for the discrepancy between theory and practice is that the projection speed and
launch height attained by the athlete are not independent of the projection angle, as is assumed in
the conventional calculation of the optimum projection angle. Experiments have shown that the
projection speed an athlete can generate decreases with increasing projection angle, and that this
substantially reduces the optimum projection angle.
2.2 Objectives
At the end of this chapter you will be able to:
•
Explain Optimum Projection Angle
2.3 Optimum Projection Angle
The standard view that the optimum projection angle in shot-putting is about 42° may be
understood by using the well-known formula for the range of a projectile in free flight.
A series of distance versus projection angle curves may be plotted for selected projection speeds.
These curves suggest that the optimum projection angle is just under 45°.
Fig.2.2
This set of calculations contain a serious error. The calculations do not include the fact that an
athlete cannot throw with the same speed at all projection angles. The projection speed an athlete
can generate steadily decreases as the athlete tries to throw with a higher and higher projection
angle.
Fig.2.3
The decrease in projection speed with increasing projection angle is a result of two factors.
When throwing with a high projection angle, the shot-putter must expend a greater effort during
the delivery phase to overcome the weight of the shot, and so less effort is available to accelerate
the shot (i.e. produce projection speed).
The structure of the human body favours the production of putting force in the horizontal
direction more than in the vertical direction. Considering just upper body strength, most athletes
can lift more weight in a bench press exercise than in a shoulder press exercise.
The optimum projection angle for the athlete is obtained by combining the speed-angle relation
for the athlete with the equation for the range of a projectile in free flight. The optimum
projection angle for the athlete is not just under 45°, but about 34°.
Fig.2.4
The optimum projection angle calculated above applies only to the athlete in question. Each
athlete has a unique speed-angle relation that depends on their size, strength, and throwing
technique. This means that each athlete has their own specific optimum projection angle. The
optimum projection angle for a world-class shot-putter may be anywhere from 26° to 38°.
2.4 Optimum Projection Angles in Undergraduate Teaching
The work on optimum projection angles has been incorporated into my biomechanics classes. I
have produced a Microsoft Excel spreadsheet and graphing tutorial to examine the optimum
projection angle in shot-putting. This tutorial highlights to the student the fact that the optimum
projection angle in sports is not 45°.
2.5 Summary
Although it is possible to calculate the optimum projection angle for any athlete, it is not always easy to
do so in practice. The calculation requires knowledge of how the athlete's projection speed changes
with projection angle . You need to take very careful measurements using a high-speed video camera of
many throws (or jumps) by the athlete over a wide range of projection angles. Such measurements
usually require the services of a sports scientist, and the process is time-consuming and expensive. For
the coach without the necessary resources, a process of trial-and-error at the training track is a good
method of identifying your athlete's optimum projection angle. Most athletes will usually 'home-in' on
their optimum projection angle relatively quickly with the aid of a tape measure (to measure their
performance) and with the guidance of a coach (to help the athlete maintain good technique while
experimenting with different projection angles). A high-speed video analysis is beneficial only to the very
best athletes. Elite athletes are closer to their ultimate potential, so expending the time and resources
to obtain an extra small improvement in performance may be justified.
2.6 Keywords
•
Optimum Projection Angle
2.7 Exercise
1. Explain Optimum Projection Angle
Unit 3
Circular Motion
Contents
3.1 Introduction
3.2 Objectives
3.3 Uniform Circular Motion
3.4 Centripetal Force
3.5 Direction of centripetal force and circular trajectory
3.6 Force analysis of uniform circular motion
3.7 Force analysis of non-uniform circular motion
3.8 Problems
3.9 Summary
3.10 Keywords
3.11 Exercise
3.1 Introduction
An object which is moving in a circular path with a constant speed is said to be in uniform
circular motion. For an object to move in a circular path, there must be a force exerted on the
object which is directed toward the center of the circular path called the centripetal force. This
centripetal force gives rise to centripetal acceleration. This centripetal force can be provided by
tension in string, friction acting between tires and the road, or the gravitational force holding a
satellite in orbit around a planet or star.
3.2 Objectives
At the end of this chapter you will be able to:
•
Explain Uniform Circular Motion
•
Define Centripetal Force
•
Know Direction of centripetal force and circular trajectory
•
Explain Force analysis of uniform circular motion and non-uniform circular motion
•
Solve the Problems
3.3 Uniform Circular Motion
Centripetal Acceleration
Before discussing the dynamics of uniform circular motion, we must explore its kinematics.
Because the direction of a particle moving in a circle changes at a constant rate, it must
experience uniform acceleration. But in what direction is the particle accelerated? To find this
direction, we need only look at the change in velocity over a short period of time:
Fig.3.1
Figure %: A particle in Uniform Circular Motion
The diagram above shows the velocity vector of a particle in uniform circular motion at two
instants of time. By vector addition we can see that the change in velocity, ∆v , points toward the
center of the circle. Since acceleration is the change in velocity over a given period of time, the
consequent acceleration points in the same direction. Thus we define centripetal acceleration as
an acceleration towards the center of a circular path. All objects in uniform circular motion must
experience some form of uniform centripetal acceleration.
We find the magnitude of this acceleration by comparing ratios of velocity and position around
the circle. Since the particle is traveling in a circular path, the ratio of the change in velocity to
velocity will be the same as the ratio of the change in position to position. Thus:
=
=
Rearranging the equation,
=
Thus
a=
We now have a definition for both the magnitude and direction of centripetal acceleration: it
always points towards the center of the circle, and has a magnitude of v 2/r .
Let us examine the equation for the magnitude of centripetal acceleration more practically.
Consider a ball on the end of a string, being rotated about an axis. The ball experiences uniform
circular motion, and is accelerated by the tension in the string, which always points toward the
axis of rotation. The magnitude of the tension of the string (and therefore the acceleration of the
ball) varies according to velocity and radius. If the ball is moving at a high velocity, the equation
implies, a large amount of tension is required and the ball will experience a large acceleration. If
the radius is very small, the equation shows, the ball will also be accelerated more rapidly.
3.4 Centripetal Force
Centripetal force is the force that causes centripetal acceleration. By using Newton's Second Law
in conjunction with the equation for centripetal acceleration, we can easily generate an
expression for centripetal force.
F c = ma =
Remember also that force and acceleration will always point in the same direction. Centripetal
force therefore points toward the center of the circle.
There are many physical examples of centripetal force, and we cannot completely explore each
one. In the case of a car moving around a curve, the centripetal force is provided by
the static frictional force of the tires of the car on the road. Even though the car is moving, the
force is actually perpendicular to its motion, and is a static frictional force. In the case of an
airplane turning in the air, the centripetal force is given by the lift provided by its banked wings.
Finally, in the case of a planet rotating around the sun, the centripetal force is given by the
gravitational attraction between the two bodies.
With a knowledge of physical forces such as tension, gravity and friction, centripetal force
becomes merely an extension of Newton's Laws. It is special, however, because it is uniquely
defined by the velocity and radius of the uniform circular motion. All of Newton's Laws still
apply, free body diagrams are still a valid method for solving problems, and forces can still be
resolved into components. Thus the most important thing to remember regarding uniform
circular motion is that it is merely a subset of the larger topic of dynamics.
3.5 Direction of centripetal force and circular trajectory
There is a subtle point about circular motion with regard to the direction of force as applied on
the particle in circular motion. If we apply force on a particle at rest, then it moves in the
direction of applied force and not perpendicular to it. In circular motion, the situation is different.
We apply force (centripetal) to a particle, which is already moving in a direction perpendicular to
the force. As such, the resulting motion from the interaction of motion with external force is not
in radial direction, but in tangential direction.
In accordance with Newton's second law of motion, the particle accelerates along the direction of
centripetal force i.e. towards center. As such, the particle actually transverses a downward
displacement (∆y) with centripetal acceleration; but in the same time, the particle moves
sideways (∆x) with constant speed, as the component of centripetal force in the perpendicular
direction is zero.
It may sound bizzare, but the fact is that the particle is continuously falling towards the center in
the direction of centripetal force and at the same is able to maintain its linear distance from the
center, owing to constant side way motion.
Centripetal force
Figure 3.2: Direction of centripetal force and
circular trajectory
In the figure shown, the particle moves towards center by ∆y, but in the same period the particle
moves left by ∆x. In the given period, the vertical and horizontal displacements are such that
resultant displacement finds the particle always on the circle.
∆x=v∆t∆y=12ar∆t2
3.6 Force analysis of uniform circular motion
As pointed our earlier, we come across large numbers of motion, where natural setting enables
continuous change of force direction with the moving particle. We find that a force meeting the
requirement of centripetal force can be any force type like friction force, gravitational force,
tension in the string or electromagnetic force. Here, we consider some of the important examples
of uniform circular motion drawn from our life experience.
Uniform circular motion in horizontal plane
A particle tied to a string is rotated in horizontal plane by virtue of the tension in the string. The
tension in the string provides the centripetal force for uniform circular motion.
We should, however, understand that this force description is actually an approximation, because
it does not take into account the downward force due to gravity. As a matter of fact, it is not
possible to have a horizontal uniform circular motion (except in the region of zero gravity) by
keeping the string in horizontal plane. It is so because, gravitational pull will change the plane of
string and the tension in it.
In order that there is horizontal uniform circular motion, the string should be slanted such that
the tension as applied to the particle forms an angle with the horizontal plane. Horizontal
component of the tension provides the needed centripetal force, whereas vertical component
balances the weight of the particle.
Uniform circular motion in horizontal plane
Figure 3.3: String is not in the plane of circular
motion.
∑Fx⇒Tsinθ=mar=mv2r
and
∑Fy⇒Tcosθ=mg
Taking ratio,
⇒tanθ=mv2rg
EXAMPLE 1
Problem : A small boy sits on a horizontal platform of a joy wheel at a linear distance of 10 m
from the center. When the wheel exceeds 1 rad/s, the boy starts slipping. Find the coefficient of
friction between boy and the platform.
Solution : For boy to be stationary with respect to platform, forces in both vertical and
horizontal directions are equal. However, requirement of centripetal force increases with
increasing rotational speed. If centripetal force exceeds the maximum static friction, then boy
begins to slip towards the center of the rotating platform.
Horizontal circular motion
Figure 3.4
In vertical direction,
N=mg
In horizontal direction,
⇒mω2r=µsN=µsmg⇒µs=rω2g=5x1210=0.5
Motion of a space shuttle
A space shuttle moves in a circular path around Earth. The gravitational force between earth and
shuttle provides for the centripetal force.
mg‘=mv2r
where g' is the acceleration due to gravity (acceleration arising from the gravitational pull of
Earth) on the satellite.
Here, we need to point out an interesting aspect of centripetal force. A person is subjected to
centripetal force, while moving in a car and as well when moving in a space shuttle. But the
experience of the person in two cases are different. In car, the person experiences (feels) a
normal force in the radial direction as applied to a part of the body. On the other hand, a person
in the shuttle experiences the "feeling" of weightlessness. Why this difference when body
experiences centripetal force in either case?
In the space shuttle, gravity acts on each of the atoms constituting our body and this gravity itself
is the provider of centripetal force. There is no push on the body as in the case of car. The body
experiences the "feeling" of weightlessness as both space shuttle and the person are continuously
falling towards center of Earth. The person is not able to push other bodies. Importantly,
gravitational pull or weight of the person is equal to mg’ and not equal to zero.
Horizontal circular motion in a rotor
Horizontal rotor holds an object against the wall of a rotating cylinder at a certain angular speed.
The object (which could be a person in a fun game arrangement) is held by friction between the
surfaces of the object and the cylinder's inside wall. For a given weight of the object, there is a
threshold minimum velocity of the rotor (cylinder); otherwise the object will fall down.
The object has a tendency to move straight. As the object is forced to move in a circle, it tends to
move away from the center. This means that the object presses the wall of the rotor. The rotor, in
turn, applies normal force on the object towards the center of circular path.
∑Fx=N=mar=mv2r
Since friction is linearly related to normal force for a given pair of surfaces ( µs ), it is possible to
adjust speed of the rotor such that maximum friction is equal to the weight of the object. In the
vertical direction, we have :
Horizontal circular motion in a rotor
Figure 3.5: As a limiting case, the maximum
friction is equal to the weight of the object.
∑Fy=µsN−mg=0⇒N=mgµs
Combining two equations, we have :
⇒mv2r=mgµs⇒v=√(rgµs)
This is threshold value of speed for the person to remain stuck with the rotor.
We note following points about the horizontal rotor :
1. The object tends to move away from the center owing to its tendency to move straight.
2. A normal force acts towards center, providing centripetal force
3. Normal force contributes to maximum friction as Fs=µsN.
4. Velocity of the rotor is independent of the mass of the object.
3.7 Force analysis of non-uniform circular motion
Motion in vertical loop involves non-uniform circular motion. To illustrate the force analysis, we
consider the motion of a cyclist, who makes circular rounds in vertical plane within a cylindrical
surface by maintaining a certain speed.
Vertical circular motion
In the vertical loop within a hallow cylindrical surface, the cyclist tends to move straight in
accordance with its natural tendency. The curvature of cylinder, however, forces the cyclist to
move along circular path (by changing direction). As such, the body has the tendency to press the
surface of the cylindrical surface. In turn, cylindrical surface presses the body towards the center
of the circular path.
Vertical circular motion
Figure 3.6: The cyclist executes vertical circular
motion along the cylindrical surface.
The free body diagram of the cyclist at an angle “θ” is shown in the figure. We see that the
resultant of normal force and component of weight in the radial direction meets the requirement
of centripetal force in radial direction,
Vertical circular motion
Figure 3.7: Force diagram
N−mgcosθ=mv2r
The distinguishing aspects of circular motion in vertical plane are listed here :
1. Motion in a vertical loop is a circular motion – not uniform circular motion. It is so
because there are both radial force (N – mg cosθ) and tangential force (mg sin θ). Radial
force meets the requirement of centripetal force, whereas tangential force accelerates the
particle in the tangential direction. As a result, the speed of the cyclist decreases while
traveling up and increases while traveling down.
2. Centripetal force is not constant, but changing in magnitude as the speed of the cyclist is
changing and is dependent on the angle “θ”.
The cyclist is required to maintain a minimum speed to avoid free fall. The possibility of free fall
is most stringent at the highest point of the loop. We, therefore, analyze the motion at the highest
point with the help of the free body diagram as shown in the figure.
Vertical circular motion
Figure 3.8: Force diagram at the top
N+mg=mv2r
NOTE:
We can also achieve the result as above by putting the value θ=180° in the equation obtained
earlier.
Th minimum speed of the cyclist corresponds to the situation when normal force is zero. For this
condition,
mg=mv2rv=√(rg)
Vertical motion of a particle attached to a string
This motion is same as discussed above. Only difference is that tension of the string replaces
normal force in this case. The force at the highest point is given as :
T+mg=mv2r
Also, the minimum speed for the string not to slack at the highest point (T = 0),
mg=mv2rv=√(rg)
The complete analysis of circular motion in vertical plane involves considering forces on the
body at different positions. However, external forces depend on the position of the body in the
circular trajectory. The forces are not constant forces as in the case of circular motion in
horizontal plane.
We shall learn subsequently that situation involving variable force is best analyzed in terms of
energy concept. As such, we will revisit vertical circular motion again after studying different
forms of mechanical energy.
3.8 Problems
Problem 1:
A 2 kg ball on a string is rotated about a circle of radius 10 m. The maximum tension allowed in
the string is 50 N. What is the maximum speed of the ball?
Solution for Problem 1 >>
The centripetal force in this case is provided entirely by the tension in the string. If the maximum
value of the tension is 50 N, and the radius is set at 10 m we only need to plug these two values
into the equation for centripetal force:
T=Fc=
implies that v =
thus
v=
= 15.8 m/s
Problem 2:
During the course of a turn, an automobile doubles its speed. How much additional frictional
force must the tires provide if the car safely makes around the curve?
Solution for Problem 2 >>
Since F c varies with v 2 , an increase in velocity by a factor of two must be accompanied by an
increase in centripetal force by a factor of four.
Problem 3:
A satellite is said to be in geosynchronous orbit if it rotates around the earth once every day. For
the earth, all satellites in geosynchronous orbit must rotate at a distance of 4.23×107 meters from
the earth's center. What is the magnitude of the acceleration felt by a geosynchronous satellite?
Solution for Problem 3 >>
The acceleration felt by any object in uniform circular motion is given by a =
. We are given
the radius but must find the velocity of the satellite. We know that in one day, or 86400 seconds,
the satellite travels around the earth once. Thus:
v=
=
=
= 3076 m/s
thus
a=
=
= .224 m/s2
Problem 4:
The maximum lift provided by a 500 kg airplane is 10000 N. If the plane travels at 100 m/s, what
is its shortest possible turning radius?
Solution for Problem 4 >>
Again, we use the equation F c =
. Rearranging, we find that r =
maximum value for the lift of the plane, we find that
r min =
= 500m
. Plugging in the
3.9 Summary
The specific requirement of a continuously changing radial force is not easy to meet by
mechanical arrangement. The requirement means that force should continuously change its
direction along with particle. It is a tall order. Particularly, if we think of managing force by
physically changing the mechanism that applies force. Fortunately, natural and many craftily
thought out arrangements create situations, in which the force on the body changes direction with
the change in the position of the particle - by the very act of motion. One such arrangement is
solar system in which gravitational force on the planet is always radial. Centripetal force is a
name given to the force required for circular motion. The net component of external forces which
meet this requirement is called centripetal force. In this sense, centripetal force is not a separately
existing force. Rather, we should look at this force as component of the external forces on the
body in radial direction.
3.10 Keywords
•
Centripetal Acceleration
•
Centripetal Force
•
Circular trajectory
•
Space shuttle
3.11 Exercise
1) Explain Uniform Circular Motion
2) Define Centripetal Force
3) A popular daredevil trick is to complete a vertical loop on a motorcycle. This trick is
dangerous, however, because if the motorcycle does not travel with enough speed, the
rider falls off the track before reaching the top of the loop. What is the minimum speed
necessary for a rider to successfully go around a vertical loop of 10 meters?
Unit 4
Application of Circular Motion
Contents
4.1 Introduction
4.2 Objectives
4.3 Applications of Circular Motion
4.4 Sample Roller Coaster Problem
4.5 Summary
4.6 Keywords
4.7 Exercise
4.1 Introduction
Curved track is to be considered as a part of the circular path. Consider a motor car moving
along a curved track. Now two forces are acting on the car, one is the weight of the car acting
down ward and second is the normal reaction acting upward. In addition to these two forces, a
horizontal centripetal force should act on the car for executing curved track on the tyres of the
car towards inner side.
4.2 Objectives
At the end of this chapter you will be able to:
•
Know the Applications of Circular Motion
•
Solve Sample Roller Coaster Problem
4.3 Applications of Circular Motion
Roller Coasters and Amusement Park Physics
Americans are wild about amusement parks. Each day, we flock by the millions to the nearest
park, paying a sizable hunk of money to wait in long lines for a short 60-second ride on our
favorite roller coaster. The thought prompts one to consider what is it about a roller coaster ride
that provides such widespread excitement among so many of us and such dreadful fear in the
rest? Is our excitement about coasters due to their high speeds? Absolutely not! In fact, it would
be foolish to spend so much time and money to ride a selection of roller coasters if it were for
reasons of speed. It is more than likely that most of us sustain higher speeds on our ride along the
interstate highway on the way to the amusement park than we do once we enter the park. The
thrill of roller coasters is not due to their speed, but rather due to their accelerations and to the
feelings of weightlessness and weightiness that they produce. Roller coasters thrill us because of
their ability to accelerate us downward one moment and upwards the next; leftwards one moment
and rightwards the next. Roller coasters are about acceleration; that's what makes them thrilling.
And in this part of Lesson 2, we will focus on the centripetal acceleration experienced by riders
within the circular-shaped sections of a roller coaster track. These sections include the clothoid
loops (that we will approximate as a circle), the sharp 180-degree banked turns, and the small
dips and hills found along otherwise straight sections of the track.
Fig 4.1
The most obvious section on a roller coaster where centripetal acceleration occurs is within the
so-called clothoid loops. Roller coaster loops assume a tear-dropped shape that is geometrically
referred to as a clothoid. A clothoid is a section of a spiral in which the radius is constantly
changing. Unlike a circular loop in which the radius is a constant value, the radius at the bottom
of a clothoid loop is much larger than the radius at the top of the clothoid loop. A mere
inspection of a clothoid reveals that the amount of curvature at the bottom of the loop is less than
the amount of curvature at the top of the loop. To simplify our analysis of the physics of clothoid
loops, we will approximate a clothoid loop as being a series of overlapping or adjoining circular
sections. The radius of these circular sections is decreasing as one approaches the top of the loop.
Furthermore, we will limit our analysis to two points on the clothoid loop - the top of the loop
and the bottom of the loop. For this reason, our analysis will focus on the two circles that can be
matched to the curvature of these two sections of the clothoid. The diagram at the right shows a
clothoid loop with two circles of different radius inscribed into the top and the bottom of the
loop. Note that the radius at the bottom of the loop is significantly larger than the radius at the
top of the loop.
Fig.4.2
As a roller coaster rider travels through a clothoid loop, she experiences an acceleration due to
both a change in speed and a change in direction. A rightward moving rider gradually becomes
an upward moving rider, then a leftward moving rider, then a downward moving rider, before
finally becoming a rightward-moving rider once again. There is a continuous change in the
direction of the rider as she moves through the clothoid loop. And as learned in Lesson 1, a
change in direction is one characteristic of an accelerating object. In addition to changing
directions, the rider also changes speed. As the rider begins to ascend (climb upward) the loop,
she begins to slow down. As energy principles would suggest, an increase in height (and in turn
an increase in potential energy) results in a decrease in kinetic energy and speed. And
conversely, a decrease in height (and in turn a decrease in potential energy) results in an increase
in kinetic energy and speed. So the rider experiences the greatest speeds at the bottom of the loop
- both upon entering and leaving the loop - and the lowest speeds at the top of the loop.
This change in speed as the rider moves through the loop is the second aspect of the acceleration
that a rider experiences. For a rider moving through a circular loop with a constant speed, the
acceleration can be described as being centripetal or towards the center of the circle. In the case
of a rider moving through a noncircular loop at non-constant speed, the acceleration of the rider
has two components. There is a component that is directed towards the center of the circle (ac)
and attributes itself to the direction change; and there is a component that is directed tangent (at)
to the track (either in the opposite or in the same direction as the car's direction of motion) and
attributes itself to the car's change in speed. This tangential component would be directed
opposite the direction of the car's motion as its speed decreases (on the ascent towards the top)
and in the same direction as the car's motion as its speed increases (on the descent from the top).
At the very top and the very bottom of the loop, the acceleration is primarily directed towards the
center of the circle. At the top, this would be in the downward direction and at the bottom of the
loop it would be in the upward direction.
Fig.4.3
We learned in Lesson 1 that the inwards acceleration of an object is caused by an inwards net
force. Circular motion (or merely motion along a curved path) requires an inwards component of
net force. If all the forces that act upon the object were added together as vectors, then the net
force would be directed inwards. Neglecting friction and air resistance, a roller coaster car will
experience two forces: the force of gravity (Fgrav) and the normal force (Fnorm). The normal force
is directed in a direction perpendicular to the track and the
gravitational force is always directed downwards. We will
concern ourselves with the relative magnitude and
direction of these two forces for the top and the bottom of
the loop. At the bottom of the loop, the track pushes
upwards upon the car with a normal force. However, at the
top of the loop the normal force is directed downwards;
since the track (the supplier of the normal force) is above the
Fig. 4.4
car, it pushes downwards upon the car. The free-body diagrams for these two positions are
shown in the diagrams at the right.
The magnitude of the force of gravity acting upon the passenger (or car) can easily be found
using the equation Fgrav = m•g where g = acceleration of gravity (9.8 m/s2). The magnitude of
the normal force depends on two factors - the speed of the car, the radius of the loop and the
mass of the rider. As depicted in the free body diagram, the magnitude of Fnorm is always greater
at the bottom of the loop than it is at the top. The normal force must always be of the appropriate
size to combine with the Fgrav in such a way to produce the required inward or centripetal net
force. At the bottom of the loop, the Fgrav points outwards away from the center of the loop. The
normal force must be sufficiently large to overcome this Fgrav and supply some excess force to
result in a net inward force. In a sense, Fgravand Fnorm are in a tug-of-war; and Fnorm must win by
an amount equal to the net force. At the top of the loop, both Fgrav and Fnorm are directed inwards.
The Fgrav is found in the usual way (using the equation Fgrav = m•g). Once more the Fnorm must
provide sufficient force to produce the required inward or centripetal net force.
4.4 Sample Roller Coaster Problem
Sample Roller Coaster Problem 1
Anna Litical is riding on The Demon at Great America. Anna experiences a downward
acceleration of 15.6 m/s2 at the top of the loop and an upward acceleration of 26.3
m/s2 at the bottom of the loop. Use Newton's second law to determine the normal force
acting upon Anna's 864 kg roller coaster car.
Steps 1 and 2 involve the construction of a free body diagram and the identification of known
and unknown quantities. This is shown in below.
Given Info:
m = 864 kg
atop = 15.6 m/s2 , down
bottom = 26.3 m/s2 , up
Find:
Fnorm at top and bottom
Fig.4.5
Step 3 of the suggested method would not apply to this problem since there are no forces
directed "at angles" (that is, all the forces are either horizontally or vertically directed). Step 4 of
the suggested method involves the determination of any known forces. In this case, the force of
gravity can be determined from the equation Fgrav = m • g. Using a g value of 9.8 m/s2, the force
of gravity acting upon the 864-kg car is approximately 8467 N. Step 5 of the suggested method
would be used if the acceleration were not given. In this instance, the acceleration is known. If
the acceleration were not known, then it would have to be calculated from speed and radius
information.
Step 6 of the suggested method involves the determination of an individual force - the normal
force. This will involve a two-step process: first the net force (magnitude and direction) must be
determined; then the net force must be used with the free body diagram to determine the normal
force. This two-step process is shown below for the top and the bottom of the loop.
Bottom of Loop
Top of Loop
Fnet = m * a
Fnet = m * a
Fnet = (864 kg) * (26.3 m/s2, up)
Fnet = (864 kg) * (15.6 m/s2, down)
Fnet = 22 723 N, up
Fnet = 13478 N, down
From FBD:
From FBD:
Fnorm and Fgrav together must combine
Fnorm must be greater than the Fgrav by
together (i.e., add up) to supply the
22723 N in order to supply a net
required inwards net force of 13478 N.
upwards force of 22723 N. Thus,
Thus,
Fnorm = Fgrav + Fnet
Fnorm = 31190 N
Fnorm = Fnet - Fgrav
Fnorm = 5011 N
Observe that the normal force is greater at the bottom of the loop than it is at the top of the loop.
This becomes a reasonable fact when circular motion principles are considered. At all points
along the loop - which we will refer to as circular in shape - there must be some inward
component of net force. When at the top of the loop, the gravitational force is directed inwards
(down) and so there is less of a need for a normal force in order to meet the net centripetal force
requirement. When at the bottom of the loop, the gravitational force is directed outwards (down)
and so now there is a need for a large upwards normal force in order to meet the centripetal force
requirement. This principle is often demonstrated in a physics class using a bucket of water tied
to a string. The water is spun in a vertical circle. As the water traces out its circular path, the
tension in the string is continuously changing. The tension force in this demonstration is
analogous to the normal force for a roller coaster rider. At the top of the vertical circle, the
tension force is very small; and at the bottom of the vertical circle, the tension force is very large.
(You might try this activity yourself outside with a small plastic bucket half-filled with water.
Give extra caution to stay clear of all people, windows, trees and overhead power lines. Repeat
enough cycles to observe the noticeable difference in tension force when the bucket is at the top
and the bottom of the circle.)
If you have ever been on a roller coaster ride and traveled through a loop, then you have likely
experienced this small normal force at the top of the loop and the
large normal force at the bottom of the loop. The normal force
provides a feel for a person's weight. (As will be discussed later in
Lesson 4, we can never feel our weight; we can only feel other forces
that act as a result of contact with other objects.) The more you weigh,
Fig.4.6
the more normal force that you will experience when at rest in your seat. But if you board a
roller coaster ride and accelerate through circles (or clothoid loops), then you will feel a normal
force that is constantly changing and different from that which you are accustomed to. This
normal force provides a sensation or feeling of weightlessness or weightiness. When at the top of
the loop, a rider will feel partially weightless if the normal forces become less than the person's
weight. And at the bottom of the loop, a rider will feel very "weighty" due to the increased
normal forces. It is important to realize that the force of gravity and the weight of your body are
not changing. Only the magnitude of the supporting normal force is changing! (The phenomenon
of weightlessness will be discussed in much more detail later in Lesson 4.)
There is some interesting history (and physics) behind the gradual usage of clothoid loops in
roller coaster rides. In the early days of roller coaster loops, circular loops were used. There were
a variety of problems, some of which resulted in fatalities, as the result of the use of these
circular loops. Coaster cars entering circular loops at high speeds encountered excessive normal
forces that were capable of causing whiplash and broken bones. Efforts to correct the problem by
lowering entry speeds resulted in the inability of cars to make it around the entire loop without
falling out of the loop when reaching the top. The decrease in speeds as the cars ascended the
large circular loop resulted in coaster cars turning into projectile cars (a situation known to be not
good for business). The solution to the problem involved using low entry speeds and a loop with
a sharper curvature at the top than at the bottom. Since clothoid loops have a continually
changing radius, the radius is large at the bottom of the loop and shortened at the top of the loop.
The result is that coaster cars can enter the loops at high speeds; yet due to the large radius, the
normal forces do not exceed 3.5 G's. At the top of the loop, the radius is small thus allowing a
lower speed car to still maintain contact with the track and successfully make it through the loop.
The clothoid loop is a testimony to an engineer's application of the centripetal acceleration
equation - a = v2/R. Now that's physics for better living!
The above discussion and force analysis applies to the circular-like motion of a roller coaster car
in a clothoid loop. The second section along a roller coaster track where circular motion is
experienced is along the small dips and hills. These sections of track are often found near the end
of a roller coaster ride and involve a series of small hills followed by a sharp drop. Riders often
feel heavy as they ascend the hill (along regions A and E in the diagram below). Then near the
crest of the hill (regions B and F), their upward motion makes them feel as though they will fly
out of the car; often times, it is only the safety belt that prevents such a mishap. As the car begins
to descend the sharp drop, riders are momentarily in a state of free fall (along regions C and G in
the diagram below). And finally as they reach the bottom of the sharp dip (regions D and H),
there is a large upwards force that slows their downward motion. The cycle is often repeated
mercilessly, churning the riders' stomachs and mixing the afternoon's cotton candy into a slurry
of ... . These small dips and hills combine the physics of circular motion with the physics of
projectiles in order to produce the ultimate thrill of acceleration - rapidly changing magnitudes
and directions of acceleration. The diagram below shows the various directions of accelerations
that riders would experience along these hills and dips.
Fig.4.7
At various locations along these hills and dips, riders are momentarily traveling along a circular
shaped arc. The arc is part of a circle - these circles have been inscribed on the above diagram in
blue. In each of these regions there is an inward component of acceleration (as depicted by the
black arrows). This inward acceleration demands that there also be a force directed towards the
center of the circle. In region A, the centripetal force is supplied by the track pushing normal to
the track surface. Along region B, the centripetal force is supplied by the force of gravity and
possibly even the safety mechanism/bar. At especially high speeds, a safety bar must supply even
extra downward force in order to pull the riders downward and supply the remaining centripetal
force required for circular motion. There are also wheels on the car that are usually tucked under
the track and pulled downward by the track. Along region D, the centripetal force is once more
supplied by the normal force of the track pushing upwards upon the car.
Fig.4.8
The magnitude of the normal forces along these various regions is dependent upon how sharply
the track is curved along that region (the radius of the circle) and the speed of the car. These two
variables affect the acceleration according to the equation
a = v2 / R
and in turn affect the net force. As suggested by the equation, a large speed results in a large
acceleration and thus increases the demand for a large net force. And a large radius (gradually
curved) results in a small acceleration and thus lessens the demand for a large net force. The
relationship between speed, radius, acceleration, mass and net force can be used to determine the
magnitude of the seat force (i.e., normal force) upon a roller coaster rider at various sections of
the track. The sample problem below illustrates these relationships. In the process of solving the
problem, the same problem-solving strategy enumerated above will be utilized.
Sample Roller Coaster Problem 2
Anna Litical is riding on The American Eagle at Great America. Anna is moving at
18.9 m/s over the top of a hill that has a radius of curvature of 12.7 m. Use Newton's
second law to determine the magnitude of the applied force of the track pulling down
upon Anna's 621 kg roller coaster car.
Steps 1 and 2 involve the construction of a free body diagram and the identification of known
and unknown quantities. This is shown in below.
Given Info:
m = 621 kgv = 18.9 m/s
R = 12.7 m
Find:
Fapp at top of hill
Step 3 of the suggested method would not apply to this problem since there are no forces
directed "at angles" (that is, all the forces are either horizontally or vertically directed). Step 4 of
the suggested method involves the determination of any known forces. In this case, the force of
gravity can be determined from the equation Fgrav = m * g. So the force of gravity acting upon
the 621-kg car is approximately 6086 N. Step 5 of the suggested method involves the calculation
of the acceleration from the given values of the speed and the radius. Using the equation given in
Lesson 1, the acceleration can be calculated as follows
a = v2 / R
a = (18.9 m/s)2 / (12.7 m)
a = 28.1 m/s2
Step 6 of the suggested method involves the determination of an individual force - the applied
force. This will involve a two-step process: first the net force (magnitude and direction) must be
determined; then the net force must be used with the free body diagram to determine the applied
force. This two-step process is shown below.
Fnet = m • a
Fnet = (621 kg) •
(28.1 m/s2, down)
Fnet =
17467
N,
down
Fapp and Fgrav must combine together (i.e., add up) to supply
the required downwards net force of 17467 N.
As shown in FBD
at right:
Fapp = Fnet - Fgrav
Fnorm = 11381 N
This same method could be applied for any region of the track in which roller coaster riders
momentarily experience circular motion.
4.5 Summary
Curved track is to be considered as a part of the circular path. Consider a motor car moving
along a curved track. Now two forces are acting on the car, one is the weight of the car acting
down ward and second is the normal reaction acting upward. In addition to these two forces, a
horizontal centripetal force should act on the car for executing curved track on the tyres of the
car towards inner side.
According to Newton’s third law of motion, the tyres exert and equal and opposite
pressure on the track towards the outer side. Hence there must be a frictional force between the
track and the tyres. This frictional force tries to balance the centrifugal force.
If the track is in horizontal level, the friction may not be sufficient to balance the
centrifugal force. Then the car may slip away from the track. To avoid this, the horizontal level
of the curved track is formed (banked) as to tilt inwards. This formation is called as banking of
curved tracks.
In the case of train moving along a curved track, the necessary centripetal force is supplied
by the pressure exerted by the rails on the flanges of the wheels. Also the flanges of the wheels
exert an equal and opposite pressure on the rails. Hence there must be a large amount of friction
and it leads to damage the rails also to derailment of the train.
To avoid this damage and derailment, the outer rail is suitably raised over the inner rail to
eliminate the flange pressure on the rails. This formation is called as banking of railway
tracks.
4.6 Keywords
•
Curved track
•
Circular Motion
•
Roller Coaster
•
Clothoid loops
4.7 Exercise
1) Anna Litical is riding on The Shock Wave at Great America. Anna experiences a
downward acceleration of 12.5 m/s2at the top of the loop and an upward acceleration of
24.0 m/s2 at the bottom of the loop. Use Newton's second law to determine the normal
force acting upon Anna's 50-kg body at the top and at the bottom of the loop.
2) Noah Formula is riding a roller coaster and encounters a loop. Noah is traveling 6 m/s at
the top of the loop and 18.0 m/s at the bottom of the loop. The top of the loop has a radius
of curvature of 3.2 m and the bottom of the loop has a radius of curvature of 16.0 m. Use
Newton's second law to determine the normal force acting upon Noah's 80-kg body at the
top and at the bottom of the loop.
3) Noah Formula is riding an old-fashioned roller coaster. Noah encounters a small hill
having a radius of curvature of 12.0 m. At the crest of the hill, Noah is lifted off his seat
and held in the car by the safety bar. If Noah is traveling with a speed of 14.0 m/s, then
use Newton's second law to determine the force applied by the safety bar upon Noah's 80kg body.
4) Anna Litical is riding a "woody" roller coaster. Anna encounters the bottom of a small
dip having a radius of curvature of 15.0 m. At the bottom of this dip Anna is traveling
with a speed of 16.0 m/s and experiencing a much larger than usual normal force. Use
Newton's second law to determine the normal force acting upon Anna's 50-kg body.
Unit 1
Moment of Inertia
Contents
1.1 Introduction
1.2 Objectives
1.3 Moment Of Inertia
1.4 Polar Moment Of Inertia
1.5 Moment of Inertia Example
1.6 Mass Moment of Inertia
1.7 Area Radius of Gyration
1.8 Mass Radius of Gyration
1.9 Summary
1.10 Keywords
1.11 Exercise
1.1 Introduction
"Moment of inertia," or MOI, is a property of physics that indicates the relative difference in
how easy or difficult it will be to set any object in motion about a defined axis of rotation. The
higher the MOI of an object, the more force will have to be applied to set that object in a
rotational motion. Conversely, the lower the MOI, the less force needed to make the object rotate
about an axis.
To understand MOI, think of a spinning ice skater. At the beginning of the spin, the skater
extends her arms and the rotation speed is slow. As the skater pulls her arms in closer to her
body, the speed of the spin greatly increases. Thus when the arms are extended, the skater's
Moment of Inertia is very high, and the result is a slower spin because the high MOI of the skater
is resisting the speed of rotation. Conversely, the reason the spin speed increases when the skater
pulls in her arms is that as the arms get closer to her body, the skater's MOI falls lower and
lower, creating less resistance to the rotation.
There are several different moments of inertia that are factors in the performance of a golf club.
Remember, MOI has to first be defined by identifying what axis the object is rotating around.
There is an MOI for the whole golf club which, when swung, is "rotated" around the golfer
during the swing.
There are also three different MOIs which can be measured for the clubhead itself. Two of these
MOIs are important in the design of any clubhead. First, when you hit a shot off the center of the
face, even though the head is secured to a shaft, the head will try to rotate around the vertical
axis going through the clubhead'scenter of gravity. Second, and at the same time, when the golfer
swings the club on the downswing, the clubhead is rotating around the axis through the center of
the shaft. The first example refers to the MOI of the clubhead about its center of gravity. In
marketing terms, this is the head design property that has a bearing on the amount of
"forgiveness" a clubhead offers for off-center strikes. The larger the clubhead, and/or the more
the designer incorporates perimeter weighting, the higher the MOI of the clubhead about its
center of gravity vertical axis will be. The higher the MOI of the head about its vertical CG axis,
the less the head will twist in response to an off-center hit, and the less distance will be lost from
that off-center hit.
The smaller the head and the more head weight is positioned close to the center of the head, the
lower the MOI of the head will be around its vertical CG axis, and the more distance will be lost
when the ball is hit off center.
Again, higher MOI equals more resistance to the object being rotated around an axis; lower MOI
equals less resistance to the object rotating around an axis.
The second example refers to the MOI of the clubhead about the shaft axis. Little is spoken about
this MOI in equipment marketing, but it is an important head design factor that can affect the
accuracy of the shot, not the distance. The bigger the head or the more weight that is placed far
out on the toe of the clubhead, the higher the MOI of the head will be about the shaft's axis. The
smaller the head or the more weight that is positioned in the heel area of the head, the lower the
MOI of the head will be about the shaft's axis. The higher the clubhead MOI around the shaft,
the more tendency there is for a golfer to leave the face open at impact. The lower the clubhead
MOI around the shaft, the more tendency there is for a golfer to rotate the face more closed at
impact.
As stated earlier, the whole golf club also has an MOI. The longer the club, the heavier the head,
the heavier the total weight of the head, shaft and grip added together, the higher the MOI will be
for the whole club. Conversely, the shorter the club, the lighter the head, the lower the weight of
the head, shaft and grip, then the lower the MOI will be for the club.
The MOI of the club is important to matching the swing feel of all the clubs in the bag.
Clubfitting theory states that if all clubs in a set are made to have the same, identical MOI, the
golfer will be more consistent because each club will require the same effort to swing.
The
current
method
for
matching
clubs
in
swing
feel
is
called
swingweight
matching.Swingweight is an expression of the ratio of the weight in the grip end of the club to
the weight in the rest of the club on down to the clubhead. Swingweight-matched golf clubs
are not matched for MOI, but come relatively close to MOI matching. MOI matching of clubs is
a swing matching system currently offered only by more advanced custom clubmakers.
1.2 Objectives
At the end of this chapter you will be able to:
•
Explain Moment Of Inertia
•
Know Polar Moment Of Inertia
•
Give example of Moment of Inertia
•
Define Mass Moment of Inertia
•
Know Area Radius of Gyration
1.3 Moment Of Inertia
The Area Moment Of Inertia of a beams cross-sectional area measures the beams ability to resist
bending. The larger the Moment of Inertia the less the beam will bend.
The moment of inertia is a geometrical property of a beam and depends on a reference axis. The
smallest Moment of Inertia about any axis passes throught the centroid.
The following are the mathematical equations to calculate the Moment of Inertia:
Ix
equ. (1)
Iy
equ. (2)
y is the distance from the x axis to an infinetsimal area dA.
x is the distance from the y axis to an infinetsimal area dA.
1.4 Polar Moment Of Inertia
The Polar Area Moment Of Inertia of a beams cross-sectional area measures the beams ability to
resist torsion. The larger the Polar Moment of Inertia the less the beam will twist.
The following are the mathematical equations to calculate the Polar Moment of Inertia:
equ. (3)
Jz
x is the distance from the y axis to an infinetsimal area dA.
y is the distance from the x axis to an infinetsimal area dA.
PERPENDICULAR AXIS THEOREM:
The moment of inertia of a plane area about an axis normal to the plane is
equal to the sum of the moments of inertia about any two mutually
perpendicular axes lying in the plane and passing through the given axis.
Using the PERPENDICULAR AXIS THEOREM yeilds the following equations for the Polar
Moment of Inertia:
Jz = Ix+Iy
equ. (4)
1.5 Moment of Inertia Example
Find the moment of inertia about the x axis. Of the following figure:
Fig.1.1
Calculating Ix:
Equ. 1
From Equ. 1
Moment of Inertia about other axes
fig.1.2
Cx
Cy
Area
Moment of Inertia
Moment of Inertia
Polar Moment of Inertia
about the x axis
about the y axis
about the z axis
Ix
Iy
Jz
Radius of Gyration
Radius of Gyration
Radius of Gyration
about the x axis
about the y axis
about the z axis
kx
ky
rz
Moment of Inertia
Moment of Inertia
Polar Moment of Inertia
about the xc axis
about the yc axis
about the zc axis
Ixc
Iyc
Jzc
Radius of Gyration
Radius of Gyration
Radius of Gyration
about the xc axis
about the yc axis
about the zc axis
kxc
kyc
rzc
1.6 Mass Moment of Inertia
The Mass Moment of Inertia of a solid measures the solid's ability to resist changes in rotational
speed about a specific axis. The larger the Mass Moment of Inertia the smaller the angular
acceleration about that axis for a given torque.
The mass moment of inertia depends on a reference axis, and is usually specified with two
subscripts. This helps to provide clarity during three-dimensional motion where rotation can
occur about multiple axes.
Following are the mathematical equations to calculate the Mass Moment of Inertia:
Fig.1.3
where x is the distance from the yz-plane to an infinitesimal area dA.
y is the distance from the zx-plane to an infinitesimal area dA.
z is the distance from the xy-plane to an infinitesimal area dA.
1.7 Area Radius of Gyration
The Radius of Gyration kx of an Area (A) about an axis (x) is defined as:
equ. (1)
equ. (2)
kx
Where Ix is the Moment of Inertia about the axis (x), and A is the area. If no axis is specified the
centroidal axis is assumed.
Using the Perpendicular Axis Theorem and equ. (1) from above it can be shown that:
equ. (3)
1.8 Mass Radius of Gyration
The Radius of Gyration kxx of a Mass (m) about an axis (x) is defined as:
equ. (4)
equ. (5)
k
Where I is the Moment of Inertia about the axis (x), and m is the mass.
If no axis is specified the centroidal axis is assumed.
Area Radius of Gyration Definition
The Radius of Gyration of an Area about a given axis is a distance k from the axis. At this
distance k an equivalent area is thought of as a line Area parallel to the original axis. The
moment of inertia of this Line Area about the original axis is unchanged.
Mass Radius of Gyration Definition
The Radius of Gyration of a Mass about a given axis is a distance k from the axis. At this
distance k an equivalent mass is thought of as a Point Mass. The moment of inertia of this Point
Mass about the original axis is unchanged.
1.9 Radius of Gyration Concept
Figure 1.4
It is possible for two EQUAL areas to have equal MOMENT'S OF INERTIA about the same
axis while having different dimensions.
Figure 1 contains two rectangles. Setting the Area's and Moment's of Inertia equal, the following
relationships can be found.
equ.
(1)
equ.
(2)
h1 and b1 are constants.
b2 depends on h2
d(h2) is:
equ.
(3)
In equ. (3) there is a limited range of h2 that yields a real solution.
d is maximized when h2 goes to zero.
This corresponds to an infinitely long, infinitely thin rectangle.
The maximum value of d is the Radius of Gyration.
equ.
(4)
This is equivalent to using the general Radius of Gyration equation on a rectangle.
The mass Radius of Gyration is the same concept just applied to a mass.
Parallel Axis Theorem
Transfer of Axis Theorem
For Area Moments of Inertia
: is the cross-sectional area.
: is the perpendicuar distance between the centroidal axis and the parallel axis.
For Area Radius of Gyration
: is the Radius of Gyration about an axis Parallel to the Centroidal axis.
: is the Radius of Gyration about the Centroidal axis.
: is the perpendicuar distance between the centroidal axis and the parallel axis.
For Mass Moments of Inertia
: is the mass of the body.
: is the perpendicuar distance between the centroidal axis and the parallel axis.
For Mass Radius of Gyration
: is the Radius of Gyration about an axis Parallel to the Centroidal axis.
: is the Radius of Gyration about the Centroidal axis.
: is the perpendicuar distance between the centroidal axis and the parallel axis.
Problem
The Moment of Inertia (I) is a term used to describe the capacity of a cross-section to resist
bending. It is always considered with respect to a reference axis such as X-X or Y-Y. It is a
mathematical property of a section concerned with a surface area and how that area is distributed
about the reference axis. The reference axis is usually a centroidal axis.
The moment of inertia is also known as the Second Moment of the Area and is expressed
mathematically as:
Ixx = Sum (A)(y2)
In which:
Ixx = the moment of inertia around the x axis
A = the area of the plane of the object
y = the distance between the centroid of the object and the x axis
Fig.1.5
The Moment of Inertia is an important value which is used to determine the state of stress in a
section, to calculate the resistance to buckling, and to determine the amount of deflection in a
beam. For example, if a designer is given a certain set of constraints on a structural problem (i.e.
loads, spans and end conditions) a "required" value of the moment of inertia can be determined.
Then, any structural element which has at least that specific moment of inertia will be able to be
utilized in the design. Another example could be in the inverse were true: a specific element is
given in a design. Then the load bearing capacity of the element could be determined.
Let us look at two boards to intuitively determine which will deflect more and why. If two
boards with actual dimensions of 2 inches by 8 inches were laid side by side - one on the two
inch side and the other on the eight inch side, the board which is supported on its 2" edge is
considerably stiffer than that supported along its 8" edge. Both boards have the same crosssectional area, but the area is distributed differently about the horizontal centroidal axis.
Fig.1.6
Calculus is ordinarily used to find the moment of inertia of an irregular section. However, a
simple formula has been derived for a rectangular section which will be the most important
section for this course.
Ixx = (1/12 ) (b)(h3)
In which the value b is always taken to be the side parallel to the reference axis and h the height
of the section. This is very important to note! If the wrong value is assumed for the value of b,
the calculations will be totally wrong.
1.9 Summary
Moment of inertia is the name given to rotational inertia, the rotational analog of massfor linear
motion. It appears in the relationships for the dynamics of rotational motion. The moment of
inertia must be specified with respect to a chosen axis of rotation. For a point mass the moment
of inertia is just the mass times the square of perpendicular distance to the rotation axis, I = mr2.
That point mass relationship becomes the basis for all other moments of inertia since any object
can be built up from a collection of point masses.
1.10 Keywords
•
Inertia
•
•
•
•
•
Polar Moment
Mass Moment
Radius of Gyration
Mass Radius
MOI
1.11 Exercise
1) Explain Moment Of Inertia.
2) Explain Polar Moment Of Inertia.
3) Give example of Moment of Inertia.
4) Define Mass Moment of Inertia.
5) Explain Area Radius of Gyration.
Unit 2
Kinetic Energy
Contents
2.1 Introduction
2.2 Objectives
2.3 What is Kinetic Energy?
2.4 Kinetic Energy
2.5 Problems on Kinetic energy
2.6 Physics circles rotational kinetic energy
2.7 Summary
2.8 Keywords
2.9 Exercise
2.1 Introduction
The concept of energy is central to physics. Indeed, it could be said that, since Einstein showed
that mass and energy are equivalent through the relation
, physics is the study of
energy and little else. However, it’s a difficult quantity to define. What exactly is energy? The
standard definition is that it is “the capacity to do work”. In that sense it is broadly the same
definition as in everyday language. Someone with a lot of energy is able to do a lot of work or
undertake a lot of activity.
This definition, however, merely transfers the problem from that of defining energy to that of
defining work. In physics, `work’ has a precise definition, which is that it is the energy produced
by the action of a force over a certain distance. We can use this definition together with
Newton’s laws to get some idea of how to measure the energy in a system of objects.
Newton’s laws are encapsulated in what may be the second most famous equation in
physics:
, which states that a force
applied to a mass
results in an acceleration .
Note that this relation is an assumption; that is, it is a statement of how Newton viewed the world,
and is derived from observations, not from any prior theory.
To apply Newton’s law to determine the energy produced by a force acting on a mass, we can
look at a simple system where a constant force acts on a constant mass over some distance .
Using the definition of work above, the amount of work done by the force is therefore
. If we
assume further that this force is unopposed, the entire effect of the force is to accelerate the mass.
How can we relate the acceleration to the distance over which the mass moves?
Acceleration is the rate of change of velocity, and velocity is the rate of change of position. For
this simple case, we’ll make the further assumptions that (a) the mass starts at rest and with zero
acceleration and (b) that the motion is in one dimension (along a straight line). This means that
we can treat the acceleration, velocity and position as scalar quantities (that is, quantities that can
be specified by giving a numerical value only, rather than both a value and a direction) rather
than vectors, which simplifies things a fair bit. We’ll deal with vectors in another post.
2.2 Objectives
At the end of this chapter you will be able to:
•
Know what is Kinetic Energy?
•
Give equation to Kinetic Energy
•
Solve Problems on Kinetic energy
•
Explain Physics circles rotational kinetic energy
2.3 What is Kinetic Energy?
Answer: Many problems in physics require an application of kinetic energy. Kinetic energy is a
form of energy that represents the energy of motion. It is a scalar quantity, which means it has a
magnitude but not a direction. It is, therefore, always positive (as will be evident when we see
the equation that defines it).
Deriving Kinetic Energy
Kinetic energy is closely linked with the concept ofwork, which is the scalar product (or dot
product) of force and the displacement vector over which the force is applied.
Using some basic kinematics equations, we obtain an equation for the acceleration of an object
which changes speed. (In the following equation, the term x - x0 has been replaced by s, a term
which represents the total distance of displacement.)
v2 = v02 + 2as
therefore,
a = ( v2 - v02 ) / 2s
Applying Newton's Second Law of Motion, F = ma, we get:
F = ma = m ( v2 - v02 ) / 2s
and, multiplying by the distance s (for work) and breaking it apart, we get:
W = Fs = 0.5mv2 - 0.5mv02
The kinetic energy, K (or sometimes Ek) is, therefore, defined as:
K = 0.5mv2
It should be noted that, as mentioned before, this quantity will always be a non-zero scalar
quantity. If the object has mass and is moving, it will always be positive. It will be zero in the
case of a massless object or an object at rest (zero velocity). The kinetic energy equation,
therefore, gives us no information about the direction of the motion, only about the speed.
Work-Energy Theorem
The work-energy theorem comes from the above derivation, and indicates that the work done by
an external force on a particle is equal to the change in kinetic energy of the particle.
Mathematically, then, you get:
Wtot = K2 - K1 = delta-K
Using Kinetic Energy
In addition to obtaining the work done, the kinetic energy equation is used frequently in
conjunction with other forms of energy. Due to the law of conservation of energy, we know that
the total energy in a closed system will remain constant. Therefore, analyzing the kinetic energy
along with, say, gravitational potential energy allows us to figure out certain factors of the
motion. (For an example of this, see the Free Falling Body problem.)
2.4 Kinetic Energy
Kinetic energy is the energy of motion. An object that has motion - whether it is vertical or
horizontal motion - has kinetic energy. There are many forms of kinetic energy - vibrational (the
energy due to vibrational motion), rotational (the energy due to rotational motion), and
translational (the energy due to motion from one location to another). To keep matters simple,
we will focus upon translational kinetic energy. The amount of translational kinetic energy (from
here on, the phrase kinetic energy will refer to translational kinetic energy) that an object has
depends upon two variables: the mass (m) of the object and the speed (v) of the object. The
following equation is used to represent the kinetic energy (KE) of an object.
where m = mass of object
v = speed of object
This equation reveals that the kinetic energy of an object is directly proportional to the square of
its speed. That means that for a twofold increase in speed, the kinetic energy will increase by a
factor of four. For a threefold increase in speed, the kinetic energy will increase by a factor of
nine. And for a fourfold increase in speed, the kinetic energy will increase by a factor of sixteen.
The kinetic energy is dependent upon the square of the speed. As it is often said, an equation is
not merely a recipe for algebraic problem solving, but also a guide to thinking about the
relationship between quantities.
Kinetic
energy
is
a scalar
quantity;
it
does
not
have
a
direction.
Unlike velocity, acceleration, force, and momentum, the kinetic energy of an object is
completely described by magnitude alone. Like work and potential energy, the standard metric
unit of measurement for kinetic energy is the Joule. As might be implied by the above equation,
1 Joule is equivalent to 1 kg*(m/s)^2.
2.5 Problems on Kinetic energy
Use your understanding of kinetic energy to answer the following questions. Then click the
button to view the answers.
Example 1. Determine the kinetic energy of a 625-kg roller coaster car that is moving with a
speed of 18.3 m/s.
KE = 0.5*m*v2
KE = (0.5) * (625 kg) * (18.3 m/s)2
KE = 1.05 x105 Joules
Example 2. If the roller coaster car in the above problem were moving with twice the speed, then
what would be its new kinetic energy?
If the speed is doubled, then the KE is quadrupled. Thus, KE = 4 * (1.04653 x 105 J) = 4.19 x
105 Joules.
or
KE = 0.5*m*v2
KE = 0.5*625 kg*(36.6 m/s)2
KE = 4.19 x 105 Joules
Example 3. Missy Diwater, the former platform diver for the Ringling Brother's Circus, had a
kinetic energy of 12 000 J just prior to hitting the bucket of water. If Missy's mass is 40 kg, then
what is her speed?
KE = 0.5*m*v2
12 000 J = (0.5) * (40 kg) * v2
300 J = (0.5) * v2
600 J = v2
v = 24.5 m/s
Example 4. A 900-kg compact car moving at 60 mi/hr has approximately 320 000 Joules of
kinetic energy. Estimate its new kinetic energy if it is moving at 30 mi/hr. (HINT: use the kinetic
energy equation as a "guide to thinking.")
KE = 80 000 J
The KE is directly related to the square of the speed. If the speed is reduced by a factor of 2 (as
in from 60 mi/hr to 30 mi/hr) then the KE will be reduced by a factor of 4. Thus, the new KE is
(320 000 J)/4 or 80 000 J.
2.6 Physics circles rotational kinetic energy
The Basic Idea
In translational motion we discussed kinetic energy, the energy of motion. Things that are spinning
are moving - if you've ever been the one who had to turn the playground merry-go-round when
smaller children were on it, you know for sure that getting something spinning takes work. So
spinning objects have kinetic energy. The equation is very similar to the familiar translation
equation,
. We will just have Moment of inertia (I) Instead of mass and Angular velocity
(W) Instead of velocity.
Some objects (such as walking people or flying airplanes) have only translational kinetic energy,
some (such as CD's in use or motors) have only rotational kinetic energy. But many objects have
both, and to find their total kinetic energy, we will have to add the rotational and kinetic energies.
Most balls in sports spin while they move - bowling balls roll, footballs on long pass spin on their
axes, baseball pitches spin. The rolling motion of a bowling ball makes an easy physics problem
because the rotation is connected to the translation and the velocities match.
The Equation
Where:
E = rotational kinetic energy (J)
I = moment of inertia (kg m2)
W = angular velocity (rad/s)
How to Do the Problems:
Sometimes you will be asked to calculate the rotational kinetic energy of an object that is just
spinning. Other times you will have to calculate the total kinetic energy of an object that is both
spinning and moving through space. If the object is rolling without slipping, then these
calculations are easy because the angular velocity is connected to the linear velocity by the
equation v = wr. If the rotation of the object is not connected to its linear movement (like a
spinning baseball pitch), then the calculation is more tedious, but the problem must also give you
separate information about linear velocity and angular velocity.
Step One: Write the information.
Step Two: Convert the units To the correct units for calculation.
These are:
Radius - meters
Time - seconds
Velocity - m/s
Angular velocity - rad/s
Step Three: Write an equation for the total kinetic energy.
Determine the shape from the descriptions in the problem and choose a formula for the moment
of inertia. For rolling objects, you must use
, but you can simplify the
calculation greatly by substituting for either linear velocity or angular velocity (whichever the
problem doesn't tell you) using v = WR. It will pay off to do the algebra now, before you plug in
numbers because usually a bunch of stuff cancels out.
Step Four: Plug in numbers.
Example - 1
Find the total kinetic energy of a bowling ball of mass 5 kg and diameter 15 cm rolling at
5 m/s.
Solution:
Step 1: Write the information.
"Find the total kinetic energy" E =?
"Mass 5 kg" m = 5 kg
"Diameter 15 cm" d = 0.15 m
"Rolling at 5 m/s" v = 5 m/s
Step 2: Convert to the correct units.
Instead of diameter, we want radius r = 0.075m
Step 3: Write an equation for the total kinetic energy.
for a solid sphere,
We notice that the problem does not give the angular velocity (w) but, since v =ωR,
And
Step 4: Plug in numbers.
E = 87.5 J
Example - 2
Calculate the rotational kinetic energy of a 10 - kg disk spinning at 15 rpm if it has a
radius of 20 cm.
Solution:
Step 1: Write the information.
"Calculate the rotational kinetic energy" E =?
"10-kg disk" m = 10 kg
"Spinning at 15 rpm" ω = 15 rpm
"Radius of 20 cm" r = 0.20 m
Step 2: Convert the units.
ω = 15 rpm should be rad/s
ω = 1.57 rad/sg
Step 3: Write an equation for the total kinetic energy
for a disk
Step 4: Plug in numbers.
.
E = 0.247 J
2.7 Summary
kinetic energy, form of energy that an object or a particle has by reason of its motion. If work,
which transfers energy, is done on an object by applying a net force, the object speeds up and
thereby gains kinetic energy. Kinetic energy is a property of a moving object or particle and
depends not only on its motion but also on its mass. The kind of motion may be translation (or
motion along a path from one place to another), rotation about an axis, vibration, or any
combination of motions.
Translational kinetic energy of a body is equal to one-half the product of its mass, m, and the
square of its velocity, v, or 1/2mv2. This formula is valid only for low to relatively high speeds;
for extremely high-speed particles it yields values that are too small. When the speed of an object
approaches that of light (3 × 108 metres per second, or 186,000 miles per second), its mass
increases, and the laws of relativity must be used. Relativistickinetic energy is equal to the
increase in the mass of a particle over that which it has at rest multiplied by the square of
the speed of light.
The unit of energy in the metre-kilogram-second system is the joule. A two-kilogram mass
(something weighing 4.4 pounds on Earth) moving at a speed of one metre per second (slightly
more than two miles per hour) has a kinetic energy of one joule. In the centimetre-gram-second
system the unit of energy is the erg, 10−7 joule, equivalent to the kinetic energy of a mosquito in
flight. Other units of energy also are used, in specific contexts, such as the still smaller unit, the
electron volt, on the atomic and subatomic scale.
2.8 Keywords
•
Kinetic Energy
•
Motion
•
Relativity
•
Joule
•
Electron volt
2.9 Exercise
1) What is Kinetic Energy?
2) Give equation to Kinetic Energy
3) Explain Physics circles rotational kinetic energy
Unit 3
Angular Momentum
Contents
3.1 Introduction
3.2 Objectives
3.3 Angular momentum about a point
3.4 Angular momentum for a particle in rotation
3.5 Summary
3.6 Keywords
3.7 Exercise
3.1 Introduction
Objects executing motion around a point possess a quantity called angular momentum. This is an
important physical quantity because all experimental evidence indicates that angular momentum
is rigorously conserved in our Universe: it can be transferred, but it cannot be created or
destroyed. For the simple case of a small mass executing uniform circular motion around a much
larger mass (so that we can neglect the effect of the center of mass) the amount of angular
momentum takes a simple form. As the adjacent figure illustrates the magnitude of the angular
momentum in this case is L = mvr, where L is the angular momentum, m is the mass of the small
object, v is the magnitude of its velocity, and r is the separation between the objects.
3.2 Objectives
At the end of this chapter you will be able to:
•
Explain Angular momentum about a point
•
Explain Angular momentum for a particle in rotation
3.3 Angular momentum about a point
Angular momentum is associated with a particle in motion. The motion need not be rotational
motion, but any motion. Importantly, it is measured with respect to a fixed point.
Angular momentum of a particle about a point is defined as a vector, denoted as "ℓ".
ℓ=rxp
(1)
where "r" is the linear vector connecting the position of the particle with the "point" about which
angular momentum is measured and "p" is the linear momentum vector. In case, the point
coincides with the origin of coordinate system, the vector "r" becomes the position vector.
We should note here that small letter "ℓ" is used to denote angular momentum of a particle. The
corresponding capital letter "L" is reserved for angular momentum of a system of particle or rigid
body. This convention helps to distinguish the context and may be adhered to.
The SI unit of angular momentum is kg−m2s , which is equivalent to J-s.
Magnitude of angular momentum
Like in the case of torque, the magnitude of angular momentum can be obtained using any of the
following relations :
Angular momentum of a particle
Fig.3.1: Angular
momentum
in
terms
of
enclosed angle.
1: Angular momentum in terms of angle enclosed
ℓ=rpsinθ
(2)
2: Angular momentum in terms of force perpendicular to position vector
ℓ=rp⊥
3: Angular momentum in terms of moment arm
(3)
ℓ=r⊥p
(4)
If the particle is moving with a velocity "v", then the expression of angular momentum becomes
:
ℓ=rxp=m(rxv)
(5)
Again, we can interpret this vector product as in the case of torque. Its magnitude can be
obtained using any of the following relations :
ℓ=mrvsinθℓ=mrv⊥ℓ=mr⊥v
(6)
EXAMPLE 1
Problem : A particle of mass, "m", moves with a constant velocity "v" along a straight line
parallel to x-axis as shown in the figure. Find the angular momentum of the particle about the
origin of the coordinate system. Also discuss the nature of angular momentum in this case.
Angular momentum of a particle
Fig. 3.2: The particle is moving with a constant
velocity.
Solution : The magnitude of the angular momentum is given by :
ℓ=mrvsinθ
This expression can be rearranged as :
ℓ=mv(rsinθ)
From the ∆OAC, it is clear that :
Angular momentum of a particle
Fig.3.3: The particle is moving with a constant
velocity.
sinθ=AC
At another instant, we have :
r'sinθ'=BD
But the perpendicular distance between two parallel lines are same (AC = BD). Thus,
⇒rsinθ=a constant
Also, the quantities "m" and "v" are constants. Therefore, angular momentum of the moving
particle about origin "O" is a constant.
ℓ=mv(rsinθ)=a constant
Since angular momentum is constant, its rate of change with time is zero. But, time rate of
change of angular momentum is equal to torque (we shall develop this relation in next module).
It means that torque on the particle is zero as time derivate of a constant is zero. Indeed it should
be so as the particle is not accelerated. This result underlines the fact that the concept of angular
momentum is consistent even for the description of linear motion as set out in the beginning of
this module.
Direction of angular momentum
Angular momentum is perpendicular to the plane formed by the pair of position and linear
momentum vectors or by the pair of position and velocity vector, depending upon the formula
used. Besides, it is also perpendicular to each of operand vectors. However, the vector relation
by itself does not tell which side of the plane formed by operands is the direction of torque.
In order to decide the orientation of the angular momentum, we employ right hand vector
product rule. The procedure involved is same as that in the case of torque. See the module
titled Torque about a point.
Angular momentum in component form
Angular momentum, being a vector, can be evaluated in component form with the help of unit
vectors along the coordinate axes. The various expressions involved in the vector algebraic
analysis are as given here :
1: In terms of position and linear momentum vectors
ℓ=rxp
ℓ=|i
j
k|
|x
y
z|
| px py pz|
ℓ=(ypz−zpy)i+(zpx−xpz)j+(xpy−ypx)k
2: In terms of position and velocity vectors
ℓ=m(rxv)
ℓ=m|i
|x
y
j k|
z|
(7)
| vx vy
vz|
ℓ=m(yvz−zvy)i+(zvx−xvz)j+(xvy−yvx)k
(8)
3.4 Angular momentum for a particle in rotation
In rotation, a particle rotates about a fixed axis as shown in the figure. We consider haere a
particle, which rotates about z-axis along a circular path in a plane parallel to "xy" plane. By the
nature of the rotational motion, linear velocity,"v", and hence linear momentum, "p" are
tangential to circular path and are perpendicular to the position vector, "r".
A particle rotating about an axis
Fig.3.4: The position vector is perpendicular to
velocity vector.
The angle between position vector "r" and velocity vector "v" is always 90°. There may be some
difficulty in visualizing the angle here. In order to visualize the same in a better perspective, we
specifically consider a time instant when position vector and moment arm, r⊥ , are in "xz" plane.
At this instant, the velocity vector, "v'", is tangential to the circle and is perpendicular to the "xz"
plane. This figure clearly shows that position vector is indeed perpendicular to velocity vector.
Angular momentum
Fig.3.5: Angular momentum of a particle
rotating about an axis.
The angular momentum of the particle, therefore, is :
ℓ=mrvsinθ=mrvsin900=mrv
The direction of angular momentum is perpendicular to the plane formed by position and
velocity vectors. For the specific situation as shown in the figure above, the direction of angular
momentum is obtained by first shifting the velocity vector to the origin and then applying right
hand rule. Importantly, the angular momentum vector makes an angle with extended x-axis in
opposite direction as shown here.
Angular momentum
Fig.3.6: Direction of angular momentum.
We observe that particle is restrained to move along a circular path perpendicular to axis of
rotation i.e. z -axis. Thus, only the component of angular momentum in the direction of axis is
relevant in the case of rotation. The component of angular momentum in z -direction is :
ℓz=mrvsinα
From geometry, we see that :
rsinα=r⊥
Hence,
ℓz=mr⊥v=mr⊥xωr⊥=mr⊥2ω
But, we know that mr⊥2=I . Hence,
ℓz=Iω
This has been the expected relation corresponding to p=mv for translational motion. The product
of moment of inertia and angular velocity about a common axis is equal to component of angular
momentum about the rotation axis.
If we define angular momentum of the particle for rotation as the product of linear momentum
and moment arm about the axis of rotation (not position vector with respect to point "O" as in the
general case), then we can say that product of moment of inertia and angular velocity about a
common axis is equal to the angular momentum about the rotation axis. Dropping the suffix
referring the axis of rotation, we have :
ℓ=Iω
(9)
We must ensure, however, that all quantities in the equation above refer to the same axis of
rotation. Also, we should also keep in mind that the definition of angular momentum for rotation
about an axis has been equal to the component of linear momentum about the axis of rotation and
is different to the one about a point as in the general case. In the nutshell, we find that the angular
momentum in rotation is a subset of angular momentum about a point in general motion.
It should be amply clear that the expression of angular momentum in terms of moment of inertia
and angular velocity is valid only for rotational motion.
3.5 Summary
1: Angular momentum of a particle in general motion is given as :
ℓ=rxp=m(rxv)
The interpretation of above relation differs for the reference with respect to which linear
distance is measured. In the case of point reference, the vector “r” denotes position vector
from the point, whereas it denotes radius vector from the center of circle in rotation.
2: The magnitude of angular momentum is evaluated, using any of the following six
relations :
ℓ=rpsinθℓ=rp⊥ℓ=r⊥pℓ=mrvsinθℓ=mrv⊥ℓ=mr⊥v
3: The direction of the angular momentum, being a vector product, is evaluated in the
same manner as that in the case of torque.
3: In the component form, the angular momentum is expressed as :
ℓ=|i
j
k|
|x
y
z|
| px py
pz|
and
ℓ=m|i
|x
j k|
y
| vx vy
z|
vz|
4: For rotation of a particle, angular momentum has additional expression in terms of
moment of inertia and angular velocity as :
ℓ=Iω=mr2ω
where “r “ is the radius of the circle from the center lying on the axis of rotation.
3.6 Keywords
•
Angular momentum
•
Torque about a point
3.7 Exercise
1. Explain Angular momentum about a point
2. Explain Angular momentum for a particle in rotation
Unit 4
Kepler’s Law
Content
4.1 Introduction
4.2 Objectives
4.3 Kepler’s First Law
4.4 Kepler’s Second Law
4.5 Kepler’s Third Law
4.6 Summary
4.7 Keywords
4.8 Exercise
4.1 Introduction
Kepler took the data that Brahe had spent his life collecting and used it (especially the
information on Mars) to create three laws that apply to any object that is orbiting something else.
•
Although Kepler’s math was essentially wrong, the three laws he came up with were
correct!
•
It would be like you writing a test, and even though you did all the work on a question
wrong, you somehow get the correct final answer.
•
Kepler’s Three Laws of Planetary Motion are still the basis for work done in the field of
astronomy to this day.
4.2 Objectives
At the end of this chapter you will be able to:
•
Kepler’s First Law
•
Solve the problem of Kepler’s law
4.3 Kepler’s First Law
Kepler’s First Law went against the major assumption that scientists of the time had about
orbits… in fact it is probably against the image of orbits that you have!
•
If I asked you to describe or draw a sketch of the Earth’s orbit around the sun, how would
you draw it. Think about in your head.
•
You’d probably stick the sun in the centre and draw a circle around it to show the path
the Earth takes.
•
In fact, this is totally wrong, as Kepler’s First Law states…
“The
path
of
any
object
in
an
orbit
follows
the
shape
of
an
ellipse,
with the orbited body at one of the foci.”
So what does all that mean?
•
An ellipse is shaped like a circle that someone has sat on. It’s squished in the middle and
bulges out on the ends.
•
Foci (the plural form of the word focus) are two points inside the ellipse.
•
If you were to push stick pins into the foci and put a loop of string around them, you
could draw an ellipse.
•
This means we have a shape that looks like this…
Fig. 4.1
Remember that the object being orbited sits at one focus, and the other object follows the path of
the ellipse.
•
The sketch of the Earth orbiting the Sun should look like this…
Fig.4. 2
•
This means that sometimes the Earth is closer to the Sun, and sometimes further away.
•
This is not the reason for summer and winter!
•
The seasons on Earth are created by Earth’s tilt on its axis.
The diagrams I have drawn here are exaggerated quite a bit to show the elliptical shape and focus
clearly.
•
The true orbit of planets isn’t this much flattened out.
•
This elliptical shape does not apply to just planets orbiting the sun. It works for any
object orbiting any other object.
•
If you measured the orbit of the Moon around the Earth, it would have an elliptical shape,
and so would any satellite in orbit around the Earth.
•
Since orbits around the Earth are quite small, their shapes are almost a circle.
Newton was able to show that his laws of gravitation gave the same results as Kepler's. In fact,
Newton was able to take things farther with his strong math background to show that the shape
of the orbits were conic sections (for those of you that have seen that stuff in math). We'll be
looking more at Newton's contributions in a lesson coming up.
4.4 Kepler’s Second Law
Kepler’s Second Law is based on the speed of the object as it orbits.
•
In the Earth-Sun example shown in Figure 4. 2, the Earth will travel faster and faster as it
gets closer to the sun.
•
As the Earth moves away from the sun, it will move slower and slower.
•
It’s almost like the Earth is being “slingshot” around the sun very quickly as it passes
near it.
Kepler didn’t talk about speed when he wrote out his second law. Instead, he looked at a
mathematical detail that pops out because we are talking about ellipses.
“An imaginary line from the sun to the planet
sweeps out equal areas in equal times.”
If we were to look at the area the Earth sweeps out in a 15 day period, first when close to the sun
(Figure 3) and then when far away (Figure 4 ), we would get diagrams that look like this.
Fig.4. 3
Fig. 4. 4
Notice how in Figure 4.3 we have a stubby, fat, (basically) triangular area that was swept out by
the line, but in Figure 4.4 we have a tall, thin, (basically) triangular area swept out.
•
If we calculate the area that I have (more or less) shaded in as triangles, you would find
that they are equal.
•
This just shows that the planet is moving a lot faster when it is closer to the sun, since
you can see that it traveled a greater distance along its orbit during that time.
4.5 Kepler’s Third Law
The big mathematical accomplishment for Kepler is in his Third Law, where he relates the
radius of an orbit to it’s period of orbit (the time it takes to complete one orbit).
•
Now, here's the weird part. Students in Alberta need to know about this law, be able to
basically explain it, but are not required to do actual calculations with it.
•
After the definition is explained, and the formula has been shown so you can recognize it,
you can stop. Just stop where I put "the sign" if you want to. The rest of the notes are
purely for the interest of people that want to know how this formula works.
“The square of the period of orbit,
divided by the cube of the radius of the orbit,
is equal to a constant (Kepler’s Constant) for that one object being orbited.”
The formula looks like this...
T = period (in any unit)
r = radius (in any unit)
K = Kepler's Constant
There's a few weird things about this formula compared to many other physics formulas:
•
You can measure the period and radius in any units you want, as long as you keep them
consistent for the whole question.
•
The radius is the average radius of the orbit. When Kepler did his original calculations he
assumed a circular orbit in these calculations, even though this went against his
own First Law.
•
Kepler's Constant is only a constant if the object being orbited stays the same. So,
anything orbiting the sun has the same Kepler's Constant, just like anything orbiting the
Earth has the same Kepler's Constant. The Sun and Earth Kepler's Constants will be
different from each other.
Figure 5
•
You can see for these values for a three different planets all orbiting the Sun I get (within
experimental error of course) the same value for Kepler's Constant.
Example 1: Using the values for Kepler’s Constant that were calculated above for Earth, Mars,
and Jupiter orbiting the sun…
a) determine the average Kepler’s Constant for anything orbiting our sun.
b) Neptune has an average orbit of 4.5e12 m from the sun. Determine how long it takes to
complete one orbit.
a) The average of the three values is 3.99e-29.
b) T2 = Kr3 = (3.99e-29) (4.5e12m)3 = 6.0e4 days = 165 years!
It takes Neptune about 165 years to go once around the sun!!!
You can also write out this formula a couple of other different ways. They are just ways of
rearranging things.
•
We know that Kepler’s constant will be the same for any objects orbiting the same thing.
•
It would be like doing one set of calculations for Earth and another for Mars like we did
in the table above. The K values are the same, so we can just stick their formulas
together.
e -> values for Earth
m -> values for Mars
•
This formula is usually re-written to look like this…
•
You don’t have to have “e” and “m” in the formula. I just used those because we were
talking about Earth and Mars in the last example.
•
You could just as easily use “a” and “b”, just as long as you put both the “a” on top, and
both the “b” on the bottom.
•
You can also use this version of the formula to do tricks if you are missing some
information, as the next example shows.
Example 2: If the orbit of Mars is 1.52 times greater than the orbit of Earth,determine how
much time it takes Mars to complete one orbit.
If I was able to look up the orbit of Earth in a book, this would be a fast question. All I would do
is multiply that number by 1.52 to get the orbit of Mars, and then I would have both the radii I
need. But what if I was doing this question on an exam and I didn’t know the value for Earth’s
orbit.
Note: Do not use the “Radius of Earth” number on your data sheet. That is the distance from the
centre of the Earth to the surface of the Earth, not the size of Earth’s orbit.
Here’s what I do know…
rm = 1.52 re
Te = 365 days
Let’s see if we can substitute that into the formula…
Tm = 684 days
4.6 Summary
Johannes Kepler, working with data painstakingly collected by Tycho Brahe without the aid of a
telescope, developed three laws which described the motion of the planets across the sky.
1. The Law of Orbits: All planets move in elliptical orbits, with the sun at one focus.
2. The Law of Areas: A line that connects a planet to the sun sweeps out equal areas in equal
times.
3. The Law of Periods: The square of the period of any planet is proportional to the cube of the
semi-major axis of its orbit.
Kepler's laws were derived for orbits around the sun, but they apply to satellite orbits as well.
4.7 Keywords
•
Keple
•
Orbit
•
Radius
4.8 Exercise
1) Derive Kepler’s First Law
2)
Suppose a small planet is discovered that is 14 times as far from the sun as the Earth's
distance is from the sun (1.5 x 1011 m). Use Kepler's law of harmonies to predict the
orbital period of such a planet. GIVEN: T2/R3 = 2.97 x 10-19 s2/m3
3) The average orbital distance of Mars is 1.52 times the average orbital distance of the
Earth. Knowing that the Earth orbits the sun in approximately 365 days, use Kepler's law
of harmonies to predict the time for Mars to orbit the sun.
Unit 1
Active and Passive remote sensing
Contents
1.1 Introduction
1.2 Objectives
1.3 Active & Passive Remote Sensing
1.4 Types of Remote Sensing
1.5 Passive vs. Active Sensing
1.6 Summary
1.7 Keywords
1.8 Exercise
1.1 Introduction
Remote sensing is the science and art of obtaining and interpreting information about an object,
area, or phenomenon through the analysis of data acquired by a sensor that is not in contact with
the object, area, or phenomenon being observed. There are four major characteristics of a remote
sensing system, namely, an electromagnetic energy source, transmission path, target, and sensor.
The Sun is a common source of electromagnetic energy. It radiates solar energy in all directions.
Earth reflects the energy from the Sun and emits some energy in the form of heat.
Based on the energy source, remote sensing systems can be grouped into two types, passive and
active systems. Passive remote sensing systems detect radiation that is reflected and/or emitted
from the surface features of Earth. Examples are the Landsat and European SPOT satellite
systems. Active remote sensing systems provide their own energy source. For example, the
Radarsat-1 synthetic aperture radar (SAR) system has an antenna that beams pulses of
electromagnetic energy towards the target.
The transmission path is the space between the electro-magnetic energy source and the target,
and back to the sensor. In the case of Earth observation, the transmission path is usually the
atmosphere of Earth. While passing through Earth's atmosphere, the electromagnetic energy can
be scattered by minute particles or absorbed by gases such that its strength and spectral
characteristics are modified before being detected by the sensor.
The target could be a particular object, an area, or phenomenon. For example, it could be a ship,
city, forest cover, mineralized zone, and water body contaminated by oil slick, a forest fire, or a
combination thereof.
Electromagnetic energy that hits a target, called incident radiation, interacts with matter or the
target in several ways. The energy could be reflected, absorbed, or transmitted. When incident
radiation hits a smooth surface, it is reflected or bounced in the opposite direction like light
bouncing off a mirror. If it hits a relatively rough surface, it could be scattered in all directions in
a diffuse manner. When incident radiation is absorbed, it loses its energy largely to heating the
matter. A portion of the energy may be emitted by the heated substance, usually at longer
wavelengths. When incident radiation is transmitted, it passes through the substance such as
from air into water.
The sensor is a device that detects reflected and/or emitted energy. Passive remote sensing
systems carry optical sensors that detect energy in the visible, infrared, and thermal infrared
regions of the electromagnetic spectrum. Common sensors used are cameras and charge-coupled
detectors (CCD) mounted on either airborne or space-borne platforms. In active remote sensing
systems, the same antenna that sends out energy pulses detects the return pulse.
Present applications of remote sensing are numerous and varied. They include land cover
mapping and analysis, land use mapping, agricultural plant health monitoring and harvest
forecast,
water
resources,
wildlife
ecology,
archeological
investigations,
snow
and ice monitoring, disaster management, geologic and soil mapping, mineral exploration,
coastal resource management, military surveillance, and many more.
1.2 Objectives
At the end of this chapter you will be able to:
•
Explain Active & Passive Remote Sensing
•
List the Types of Remote Sensing
•
Know Passive vs. Active Sensing
1.3 Active & Passive Remote Sensing
Various references have been made to the sun as a source of energy or radiation in remote
sensing. The sun provides a very convenient source of energy for remote sensing. The sun's
energy is either reflected, as it is for visible wavelengths, or absorbed and then re-emitted, as it is
for thermal infrared wavelengths.
Remote sensing systems which measure energy that is naturally available are called Passive
Sensors. Passive sensors can only be used to detect energy when the naturally occurring energy
is available. For all reflected energy, this can only take place during the time when the sun is
illuminating the Earth.
There is no reflected energy availabl
availablee from the sun at night. Energy that is naturally emitted
(such as thermal infrared) can be detected day or night, as long as the amount of energy is large
enough to be recorded.
Active sensors,, on the other hand, transmit short bursts or 'pulses' of electromagnetic energy in
the direction of interest and record the origin and strength of the backscatter received from
objects within the system's field of view. Passive systems sense low level microwave radiation
given off by all objects in the natural eenvironment.
Advantages for active sensors include the ability to obtain measurements anytime, regardless of
the time of day or season. Active sensors can be used for examining wavelengths that are not
sufficiently provided by the sun, such as microwaves, oorr to better control the way a target is
illuminated.
However, active systems require the generation of a fairly large amount of energy to adequately
illuminate targets. Some examples of active sensors are a laser fluorosensor and a synthetic
aperture radar (SAR)
Fig.1.1
Radar imaging systems such as ERS (European Remote Sensing Satellite), JERS (Japan Earth
Resources Satellite), and RADARSAT
RADARSAT-1 are active systems. They both transmit and receive
energy.
Microwave
scanning
radiometers
only
receive
microwave
energy.
The
Japanese MOS (Marine Observation Satellite) and JERS satellite systems employ microwave
scanning units. Since the source is a very low level of electromagnetic energy, this type of data is
prone to noise.
One advantage of active radar sensing systems is that, since they provide their own source of
energy, they can collect data at any time of the day or night. Passive sensors - optical, thermal,
and microwave - rely on receiving the naturally emitted or the sun's reflected energy from tthe
Earth's surface and can therefore image only during the daytime.
1.4 Types of Remote Sensing
Fig.1.2 Satellite sensors record radiation reflected, emitted and scattered from the Earth's surface
Depending on the source of the predominant source of electromagnetic energy in the remote sensing
system,
remote
sensing
can
be
classified
into
passive
and
active
remote
sensing.
Passive remote sensing depends on a natural source to provide energy. The sun is the most powerful
and commonly used source
rce of energy for passive remote sensing. The satellite sensor in this case records
primarily the radiation that is reflected from the target. Remote sensing in the visible part of the
electromagnetic
spectrum
is
an
example
of
passive
(reflected)
remote
se
sensing.
A portion of the sun’s radiation that is not reflected back to the sensor is absorbed by the target, raising
the temperature of target material. The absorbed radiation is later emitted by the material at a different
wavelength. Passive remote sens
sensing
ing can also be carried out in the absence of the sun. In this latter case,
the source of energy is the target material itself and the sensor records primarily emitted radiation.
Remote sensing in the thermal infrared portion of the electromagnetic spectrum is an example of passive
(emitted) remote sensing.
Active remote sensing uses an artificial source for energy. For example the satellite itself can send a
pulse of energy which can interact with the target. In active remote sensing, humans can control the
nature (wavelength, power, duration) of the source energy. Remote sensing in the microwave region of
the electromagnetic spectrum (radar remote sensing) is an example of active remote sensing. Active
remote sensing can be carried out during day and night and in all weather conditions.
1.5 Passive vs. Active Sensing
So far, throughout this chapter, we have made various references to the sun as a source of energy
or radiation. The sun provides a very convenient source of energy for remote sensing. The sun's
energy is either reflected, as it is for visible wavelengths, or absorbed and then re-emitted, as it
is for thermal infrared wavelengths. Remote sensing systems which measure energy that is
naturally available are called
passive sensors. Passive sensors can only be used to detect
energy when the naturally occurring energy is available. For all reflected energy, this can only
take place during the time when the sun is illuminating the Earth. There is no reflected energy
available from the sun at night. Energy that is naturally emitted (such as thermal infrared) can be
detected day or night, as long as the amount of energy is large enough to be recorded.
Active sensors, on the other hand, provide their own energy source for illumination. The
sensor emits radiation which is directed toward the target to be investigated. The radiation
reflected from that target is detected and measured by the sensor. Advantages for active sensors
include the ability to obtain measurements anytime, regardless of the time of day or season.
Active sensors can be used for examining wavelengths that are not sufficiently provided by the
sun, such as microwaves, or to better control the way a target is illuminated. However, active
systems require the generation of a fairly large amount of energy to adequately illuminate
targets. Some examples of active sensors are a laser fluorosensor and a synthetic aperture radar
(SAR).
1.6 Summary
Remote sensing is the acquisition of information about an object or phenomenon, without
making physical contact with the object. In modern usage, the term generally refers to the use of
aerial sensor technologies to detect and classify objects on Earth (both on the surface, and in
the atmosphere and oceans) by means of propagated signals.
1.7 Keywords
•
Passive Sensors
•
Active sensors
•
SAR
•
ERS
•
MOS
•
Reflected
•
Re-emitted
1.8 Exercise
1) What is meant by Remote Sensing? State some of its fields of application.
2) What are the main differences between active and passive remote sensing?
3) Briefly describe the main elements of Remote Sensing.
Unit 2
Microwave Remote Sensing
Contents
2.1 Introduction
2.2 Objectives
2.3 Microwave Remote Sensing
2.4 Interaction between Microwaves and Earth's Surface
2.5 All-Weather Imaging
2.6 Summary
2.7 Keywords
2.8 Exercise
2.1 Introduction
Electromagnetic radiation at long wavelengths (0.1 to 30 centimeters) falls into a segment of the
spectrum commonly called the microwave region. At still longer wavelengths (centimeters to
meters) the radiation is known as radio waves (these can be generated by manmade transmitters
or occur naturally [e.g., beamed from energetic stars]). Remote sensing has utilized passive
microwaves, emanating from thermally activated bodies. But, in much more common use (since
World War II) is another manmade device, radar, an active (transmitter-produced) microwave
system that sends out radiation, some of which is reflected back to a receiver. The varying signal,
which changes with the positions and shapes of target bodies, and is influenced by their
properties, can be used to form kinds of images that superficially resemble those recorded by
Landsat-like sensors. This first page introduces certain basic principles, describes the common
radar bands in use, and shows a typical radar image.
2.2 Objectives
At the end of this chapter you will be able to:
•
Explain Microwave Remote Sensing
•
Explain Interaction between Microwaves and Earth's Surface
2.3 Microwave Remote Sensing
Fig.2.1
Electromagnetic radiation in the microwave wavelength region is used in remote sensing to
provide useful information about the Earth's atmosphere, land and ocean. A microwave
radiometer is a passive device which records the natural microwave emission from the earth. It
can be used to measure the total water content of the atmosphere within its field of view.
A radar altimeter sends out pulses of microwave signals and record the signal scattered back
from the earth surface. The height of the surface can be measured from the time delay of the
return signals. A wind scatterometer can be used to measure wind speed and direction over the
ocean surface. it sends out pulses of microwaves along several directions and records the
magnitude of the signals backscattered from the ocean surface. The magnitude of the
backscattered signal is related to the ocean surface roughness, which in turns is dependent on the
sea surface wind condition, and hence the wind speed and direction can be derived. orne
platforms to generate high resolution images of the earth surface using microwave energy.
Synthetic Aperture Radar (SAR)
In synthetic aperture radar (SAR) imaging, microwave pulses are transmitted by an antenna
towards the earth surface. The microwave energy scattered back to the spacecraft is measured.
The SAR makes use of the radar principle to form an image by utilising the time delay of the
backscattered signals.
Fig.2.2
Fig.2.3
A radar pulse is transmitted from the antenna to the ground
The radar pulse is scattered by the ground targets back to the antenna.
In real aperture radar imaging, the ground resolution is limited by the size of the microwave
beam sent out from the antenna. Finer details on the ground can be resolved by using a narrower
beam. The beam width is inversely proportional to the size of the antenna, i.e. the longer the
antenna, the narrower the beam.
Fig.2.4
The microwave beam sent out by the antenna illuminates an area on the ground (known as the
antenna's "footprint"). In radar imaging, the recorded signal strength depends on the microwave
energy backscattered from the ground targets inside this footprint. Increasing the length of the
antenna will decrease the width of the footprint.
It is not feasible for a spacecraft to carry a very long antenna which is required for high
resolution imaging of the earth surface. To overcome this limitation, SAR capitalises on the
motion of the space craft to emulate a large antenna (about 4 km for the ERS SAR) from the
small antenna (10 m on the ERS satellite) it actually carries on board.
Fig.2.5
Imaging geometry for a typical strip-mapping synthetic aperture radar imaging system. The
antenna's footprint sweeps out a strip parallel to the direction of the satellite's ground track.
2.4 Interaction between Microwaves and Earth's Surface
When microwaves strike a surface, the proportion of energy scattered back to the sensor depends
on many factors:
•
Physical factors such as the dielectric constant of the surface materials which also
depends strongly on the moisture content;
•
Geometric factors such as surface roughness, slopes, orientation of the objects relative to
the radar beam direction;
•
The types of landcover (soil, vegetation or man-made objects).
•
Microwave frequency, polarisation and incident angle.
2.5 All-Weather Imaging
Due to the cloud penetrating property of microwave, SAR is able to acquire "cloud-free" images
in all weather. This is especially useful in the tropical regions which are frequently under cloud
covers throughout the year. Being an active remote sensing device, it is also capable of nighttime operation.
2.6 Summary
RADAR stands for "RAdio Detection And Ranging". By virtue of sending out pulses of
microwave electromagnetic radiation this type of instrument can be classified as an "active
sensor" - it measures the time between pulses and their reflected components to determine
distance. Different pulse intervals, different wavelengths, different geometry and polarizations
can be combined to roughness characteristics of the earth surface. Radar wavelengths range
between less than 1 millimeter to 1 meter.
2.7 Keywords
•
Microwave radiometer
•
SAR
•
Incident angle
2.8 Exercise
1) Explain Microwave Remote Sensing
2) Explain Interaction between Microwaves and Earth's Surface
Unit 3
Types of Sound Waves
Contents
3.1 Introduction
3.2 Objectives
3.3 What is sound and how does it travel?
3.4 What do waves consist of?
3.5 What do sound waves look like?
3.6 Different types of waves
3.7 What are the characteristics of sound waves?
3.8 Summary
3.9 Keywords
3.10 Exercise
3.1 Introduction
Mechanical Waves are waves which propagate through a material medium (solid, liquid, or gas)
at a wave speed which depends on the elastic and inertial properties of that medium. There are
two
basic
types
of
wave
motion
for
mechanical
waves: longitudinal waves
and transverse waves. The section below demonstrate both types of wave and illustrate the
difference between the motion of the wave and the motion of the particles in the medium through
which the wave is travelling.
3.2 Objectives
At the end of this chapter you will be able to:
•
Know What is sound and how does it travel?
•
Tell What do waves consist of?
•
Explain What do sound waves look like?
•
List Different types of waves
•
List the characteristics of sound waves?
3.3 What is sound and how does it travel?
All of the sounds you heard on the previous page occurred because mechanical energy produced
by your computer speaker was transferred to your ear through the movement of atomic particles.
Sound is a pressure disturbance that moves through a medium in the form of mechanical waves.
When a force is exerted on an atom, it moves from its rest or equilibrium position and exerts a
force on the adjacent particles. These adjacent particles are moved from their rest position and
this continues throughout the medium. This transfer of energy from one particle to the next is
how sound travels through a medium. The words "mechanical wave" are used to describe the
distribution of energy through a medium by the transfer of energy from one particle to the next.
Waves of sound energy move outward in all directions from the source. Your vocal chords and
the strings on a guitar are both sources which vibrate to produce sound waves. Without energy,
there would be no sound. Let's take a closer look at sound waves.
3.4 What do waves consist of?
Sound or pressure waves are made up of compressions and rarefactions. Compression happens
when particles are forced, or pressed, together. Rarefaction is just the opposite, it occurs when
particles are given extra space and allowed to expand. Remember that sound is a type of kinetic
energy. As the particles are moved from their rest position, they exert a force of the adjacent
particles and pass the kinetic energy. Thus sound energy travels outward from the source.
Sound travels through air, water, or a block of steel; thus, all are mediums for sound. Without a
medium there are no particles to carry the sound waves. The word "particle" suggests a tiny
concentration of matter capable of transmitting energy. A particle could be an atom or molecule.
In places like space, where there is no atmosphere, there are too few atomic particles to transfer
the sound energy.
Let's look at the example of a stereo speaker. To produce sound, a thin surfaced cone, called
a diaphragm, is caused to vibrate using electromagnetic energy. When the diaphragm moves to
the right, its energy pushes the air molecules on the right together, opening up space for the
molecules on the left to move into. We call the molecules on the right compressed and the
molecules on the left rarefied. When the diaphragm moves to the left, the opposite happens.
Now, the molecules to the left become compressed and the molecules to the right are rarefied.
These alternating compressions and rarefactions produce a wave. One compression and one
rarefaction is called a wavelength. Different sounds have different wavelengths.
Fig.3.1
3.5 What do sound waves look like?
We cannot see the energy in sound waves, but a sound wave can be modeled in two ways. One
way is to create a graph of the diaphragm's position at different times. Think of a number line.
We call the diaphragm's rest position zero. As it travels to the right, it moves to an increasing
increasingly
positive position along the number line. As is travels to the left, its position becomes more and
more negative. The graph of the diaphragm’s position as it vibrates looks like the sine graph,
with its highest point when the diaphragm is the farthest ri
right
ght and its lowest point when it is
farthest left.
Another graph can be made using the amount of force on the molecules versus time. The force is
greatest when the diaphragm is moving through its original position. This is similar to the way
we feel the greatest
eatest force on a swing as we move through the center, where we started. As the
diaphragm moves to the right, there is less and less force. At its rightmost position, it is exerting
no force (due to pressure) and begins its trip the opposite way. Similarly
Similarly,, the diaphragm is
exerting no force at its leftmost position. For our graph, we say the force is least when the
diaphragm moves through its starting position heading the opposite way. When the force is
exerting a pulling force, we assign negative values ttoo it. A graph of the force versus time also
resembles the sine graph.
More about compression and rarefaction
Compression
and
rarefaction
are
terms
defining
the
molecules
near
the
diaphragm. Compression is the point when the most force is being applied to a molecule
and rarefaction is the point when the least force is applied. It is important to note that when a
molecule to the right of the diaphragm is experiencing compression, a molecule to the
diaphragm's left is experiencing rarefaction. For right side molecules, compression occurs when
the diaphragm is in its original position, moving towards the right. This is where the molecule
experiences the most force. Rarefaction happens when the diaphragm is once again in the center,
this time moving towards the left. At this point, the molecule is experiencing the least force. Of
course, this is the opposite for molecules to the diaphragm's left.
Longitude
Fig.3.2
Shear
Fig.3.3
Surface
Fig.3.4
Plate(Symmetric)
Fig.3.5
Plate(Asymmetric)
Fig.3.6
3.6 Different types of waves
As the diaphragm vibrates back and forth, the sound waves produced move the same direction
(left and right). Waves that travel in the same direction as the particle movement are
called longitudinal waves. Longitudinal sound waves are the easiest to produce and have the
highest speed. However, it is possible to produce other types. Waves which move perpendicular
to the direction particle movement are called shear waves or transverse waves. Shear waves
travel at slower speeds than longitudinal waves, and can only be made in solids. Think of a
stretched out slinky, you can create a longitudinal wave by quickly pushing and pulling one end
of the slinky. This causes longitudinal waves for form and propagate to the other end. A shear
wave can be created by taking one end of the slinky and moving it up and down. This generates a
wave that moves up and down as it travels the length of the slinky.
Another type of wave is the surface wave. Surface waves travel at the surface of a material with
the particles move in elliptical orbits. They are slightly slower than shear waves and fairly
difficult to make. A final type of sound wave is the plate wave.
Transverse Waves
For transverse waves the displacement of the medium is perpendicular to the direction of
propagation of the wave. A ripple on a pond and a wave on a string are easily visualized
transverse waves.
Fig.3.7
Transverse waves cannot propagate in a gas or a liquid because there is no mechanism for
driving motion perpendicular to the propagation of the wave.
Longitudinal Waves
In longitudinal waves the displacement of the medium is parallel to the propagation of the wave.
A wave in a "slinky" is a good visualization. Sound waves in air are longitudinal waves.
Fig.3.8
3.7 What are the characteristics of sound waves?
Sound waves are often characterized by four basic qualities, though many more are related:
Frequency, Amplitude, Wave shape and Phase*
Some sound waves are periodic, in that the change from equilibrium (average atmospheric
pressure) to maximum compression to maximum rarefaction back to equilibrium is repetitive.
The 'round trip' back to the starting point just described is called a cycle. Periodic motion is
classically demonstrated by the up and down motion of a dropped weight (mass) attached to a
spring or by observing the motion of a pendulum. The amount of time a single cycle takes is
called a period.
Fig.3.9
It is possible to measure frequency in seconds per cycle or periods, but it is far more common for
sound measurements to use cycles per second.
Periodic motion depends on two prime factors; 1) elasticity, in that medium being distorted
return to its original state (equilibrium), and 2) a source of energy to initiate and sustain motion.
In the case of sound waves, the atmospheric pressure will return to the ambient pressure without
an energy source to disturb it, and any vibrating surface will constitute an energy or excitation
source.
Simple harmonic motion, the motion described by mass/spring example above, is represented
in sound as a sine wave, which traces the mathematical shape of it namesake. A sinusoidal wave
(which also includes a cosine wave) is the only wave shape that produces a singles frequency, as
we will see in the waveform chapter. With any minute deviations in the sine shape, additional
frequencies will be generated.
Noise is characterized as being aperiodic or having a non-repetitive pattern. There are many
different types of noise, depending primarily on the random distribution of frequencies. For
example, some types of noise may sound brighter than others.
Fig.3.10
Some periodic waveforms can be complex enough to be perceived as noise if our ears cannot
detect perceptible pitches. Many real-world sounds, such as the "chiffy" attack of a flute note
contain some combination of periodic and aperiodic components.
3.8 Summary
In air, sound travels by the compression and rarefaction of air molecules in the direction of
travel. However, in solids, molecules can support vibrations in other directions, hence, a number
of different types of sound waves are possible. Waves can be characterized in space by
oscillatory patterns that are capable of maintaining their shape and propagating in a stable
manner. The propagation of waves is often described in terms of what are called “wave modes.”
3.9 Keywords
•
Compression
•
Rarefaction
•
Diaphragm
•
Wavelength
•
Longitudinal waves
•
Transverse waves
3.10 Exercise
1)
Discuss the relationship between the speed of sound and speed of light.
2)
Describe what the sound barrier is.
3)
What is sound and how does it travel?
4)
List Different types of waves
5)
List the characteristics of sound waves?
Unit 4
Acoustics
Contents
4.1 Introduction
4.2 Objectives
4.3 Acoustics
4.4 Measuring the speed of sound
4.5 Modern advances
4.6 Amplifying, recording, and reproducing
4.7 Architectural acoustics
4.8 Summary
4.9 Keywords
4.10 Exercise
4.1 Introduction
In general, the experimental and theoretical science of sound and its transmission; in particular,
that branch of the science that has to do with the phenomena of sound in a particular space such
as a room or theatre. In architecture, the sound-reflecting character of an internal space.
Acoustic engineering is concerned with the technical control of sound, and involves architecture
and construction, studying control of vibration, soundproofing, and the elimination of noise. It
also includes all forms of sound recording and reinforcement, the hearing and perception of
sounds, and hearing aids.
4.2 Objectives
At the end of this chapter you will be able to:
•
Explain Acoustics
•
Know Measuring the speed of sound
•
Define Amplifying, recording, and reproducing
•
Explain Architectural acoustics
4.3 Acoustics
Acoustics, the science concerned with the production, control, transmission, reception, and
effects of sound. The term is derived from the Greek akoustos, meaning “hearing.”
Beginning with its origins in the study of mechanical vibrations and the radiation of these
vibrations through mechanical waves, acousticshas had important applications in almost every
area of life. It has been fundamental to many developments in the arts—some of which,
especially in the area of musical scales and instruments, took place after long experimentation by
artists and were only much later explained as theory by scientists. For example, much of what is
now known about architectural acoustics was actually learned by trial and error over centuries of
experience and was only recently formalized into a science.
Other applications of acoustic technology are in the study of geologic, atmospheric, and
underwater phenomena. Psychoacoustics, the study of the physical effects of sound on
biological systems, has been of interest since Pythagoras first heard the sounds of vibrating
strings and of hammers hitting anvils in the 6th century BC, but the application of modern
ultrasonic technology has only recently provided some of the most exciting developments in
medicine. Even today, research continues into many aspects of the fundamental physical
processes involved in waves and sound and into possible applications of these processes in
modern life.
Sound waves follow physical principles that can be applied to the study of all waves; these
principles are discussed thoroughly in the article mechanics of solids. The article ear explains in
detail the physiological process of hearing—that is, receiving certain wave vibrations and
interpreting them as sound.
Early experimentation
The
origin
of
the
science
of acoustics is
generally
attributed
to
the
Greek
philosopher Pythagoras (6th century BC), whose experiments on the properties of vibrating
strings that produce pleasing musical intervals were of such merit that they led to a tuning system
that bears his name. Aristotle (4th century BC) correctly suggested that a sound wave propagates
in air through motion of the air—a hypothesis based more on philosophy than on experimental
physics; however, he also incorrectly suggested that high frequencies propagate faster than low
frequencies—an error that persisted for many centuries. Vitruvius, a Roman architectural
engineer of the 1st century BC, determined the correct mechanism for the transmission of sound
waves, and he contributed substantially to the acoustic design of theatres. In the 6th century AD,
the Roman philosopher Boethiusdocumented several ideas relating science to music, including
a suggestion that the human perception of pitch is related to the physical property of frequency.
The modern study of waves and acoustics is said to have originated with Galileo Galilei (1564–
1642), who elevated to the level of science the study of vibrations and the correlation between
pitch and frequency of the sound source. His interest in sound was inspired in part by his father,
who was a mathematician, musician, and composer of some repute. Following Galileo’s
foundation
work,
progress
in acoustics came
relatively
rapidly.
The
French
mathematician Marin Mersenne studied the vibration of stretched strings; the results of these
studies were summarized in the three Mersenne’s laws. Mersenne’s Harmonicorum Libri (1636)
provided the basis for modern musical acoustics. Later in the century Robert Hooke, an English
physicist, first produced a sound wave of known frequency, using a rotating cog wheel as a
measuring device. Further developed in the 19th century by the French physicist Félix Savart,
and now commonly called Savart’s disk, this device is often used today for demonstrations
during physics lectures. In the late 17th and early 18th centuries, detailed studies of the
relationship between frequency and pitch and of waves in stretched strings were carried out by
the French physicist Joseph Sauveur, who provided a legacy of acoustic terms used to this day
and first suggested the name acoustics for the study of sound.
One of the most interesting controversies in the history of acoustics involves the famous and
often misinterpreted “bell-in-vacuum” experiment, which has become a staple of contemporary
physics lecture demonstrations. In this experiment the air is pumped out of a jar in which a
ringing bell is located; as air is pumped out, the sound of the bell diminishes until it becomes
inaudible. As late as the 17th century many philosophers and scientists believed that
soundpropagated via invisible particles originating at the source of the sound and moving
through space to affect the ear of the observer. The concept of sound as a wave directly
challenged this view, but it was not established experimentally until the first bell-in-vacuum
experiment was performed by Athanasius Kircher, a German scholar, who described it in his
book Musurgia Universalis (1650). Even after pumping the air out of the jar, Kircher could still
hear the bell, so he concluded incorrectly that air was not required to transmit sound. In fact,
Kircher’s jar was not entirely free of air, probably because of inadequacy in his vacuum pump.
By 1660 the Anglo-Irish scientist Robert Boyle had improved vacuum technology to the point
where he could observe sound intensity decreasing virtually to zero as air was pumped out.
Boyle then came to the correct conclusion that a medium such as air is required for transmission
of sound waves. Although this conclusion is correct, as an explanation for the results of the bellin-vacuum experiment it is misleading. Even with the mechanical pumps of today, the amount of
air remaining in a vacuum jar is more than sufficient to transmit a sound wave. The real reason
for a decrease in sound level upon pumping air out of the jar is that the bell is unable to transmit
the sound vibrations efficiently to the less dense air remaining, and that air is likewise unable to
transmit the sound efficiently to the glass jar. Thus, the real problem is one of
an impedance mismatch between the air and the denser solid materials—and not the lack of a
medium such as air, as is generally presented in textbooks. Nevertheless, despite the confusion
regarding this experiment, it did aid in establishing sound as a wave rather than as particles.
4.4 Measuring the speed of sound
Once it was recognized that sound is in fact a wave, measurement of the speed of sound became
a serious goal. In the 17th century, the French scientist and philosopher Pierre Gassendi made
the earliest known attempt at measuring the speed of sound in air. Assuming correctly that
the speed of light is effectively infinite compared with the speed of sound, Gassendi measured
the time difference between spotting the flash of a gun and hearing its report over a long distance
on a still day. Although the value he obtained was too high—about 478.4 metres per second
(1,569.6 feet per second)—he correctly concluded that the speed of sound is independent of
frequency.
In
the
1650s,
Italian
physicists Giovanni
Alfonso
Borelli and Vincenzo
Viviani obtained the much better value of 350 metres per second using the same technique. Their
compatriot G.L. Bianconi demonstrated in 1740 that the speed of sound in air increases
with temperature. The earliest precise experimental value for the speed of sound, obtained at
the Academy of Sciences in Paris in 1738, was 332 metres per second—incredibly close to the
presently accepted value, considering the rudimentary nature of the measuring tools of the day. A
more recent value for the speed of sound, 331.45 metres per second (1,087.4 feet per second),
was obtained in 1942; it was amended in 1986 to 331.29 metres per second at 0° C (1,086.9 feet
per second at 32° F).
The speed of sound in water was first measured by Daniel Colladon, a Swiss physicist, in 1826.
Strangely enough, his primary interest was not in measuring the speed of sound in water but in
calculating water’s compressibility—a theoretical relationship between the speed of sound in a
material and the material’s compressibility having been established previously. Colladon came
up with a speed of 1,435 metres per second at 8° C; the presently accepted value interpolated at
that temperature is about 1,439 metres per second.
Two approaches were employed to determine the velocity of sound in solids. In 1808 JeanBaptiste Biot, a French physicist, conducted direct measurements of the speed of sound in 1,000
metres of iron pipe by comparing it with the speed of sound in air. A better measurement had
earlier been carried out by a German,Ernst Florenz Friedrich Chladni, using analysis of the
nodal pattern in standing-wave vibrations in long rods.
4.5 Modern advances
Simultaneous with these early studies in acoustics, theoreticians were developing the
mathematical theory of waves required for the development of modern physics,
including acoustics.
In
the
early
18th
century,
the
English
mathematician Brook
Taylor developed a mathematical theory of vibrating strings that agreed with previous
experimental observations, but he was not able to deal with vibrating systems in general without
the proper mathematical base. This was provided by Isaac Newton of England and Gottfried
Wilhelm Leibniz of Germany, who, in pursuing other interests, independently developed the
theory of calculus, which in turn allowed the derivation of the general wave equation by the
French mathematician and scientist Jean Le Rond d’Alembert in the 1740s. The Swiss
mathematicians Daniel Bernoulli and Leonhard Euler, as well as the Italian-French
mathematician Joseph-Louis Lagrange, further applied the new equations of calculus to waves
in strings and in the air. In the 19th century, Siméon-Denis Poisson of France extended these
developments to stretched membranes, and the German mathematician Rudolf Friedrich Alfred
Clebsch completed Poisson’s earlier studies. A German experimental physicist, August Kundt,
developed a number of important techniques for investigating properties of sound waves. These
included the Kundt’s tube, discussed below.
One of the most important developments in the 19th century involved the theory of vibrating
plates. In addition to his work on the speed of sound in metals, Chladni had earlier introduced a
technique of observing standing-wave patterns on vibrating plates by sprinkling sand onto the
plates—a demonstration commonly used today. Perhaps the most significant step in the
theoretical explanation of these vibrations was provided in 1816 by the French
mathematician Sophie Germain, whose explanation was of such elegance and sophistication
that errors in her treatment of the problem were not recognized until some 35 years later, by the
German physicist Gustav Robert Kirchhoff.
The analysis of a complex periodic wave into its spectral components was theoretically
established early in the 19th century by Jean-Baptiste-Joseph Fourier of France and is now
commonly referred to as the Fourier theorem. The German physicist Georg Simon Ohm first
suggested that the ear is sensitive to these spectral components; his idea that the ear is sensitive
to the amplitudes but not the phases of the harmonics of a complex tone is known as Ohm’s law
of hearing (distinguishing it from the more famous Ohm’s law of electrical resistance).
Hermann von Helmholtz made substantial contributions to understanding the mechanisms of
hearing and to the psychophysics of sound and music. His bookOn the Sensations of Tone As a
Physiological Basis for the Theory of Music (1863) is one of the classics of acoustics. In
addition, he constructed a set ofresonators, covering much of the audio spectrum, which were
used in the spectral analysis of musical tones. The Prussian physicist Karl Rudolph Koenig, an
extremely clever and creative experimenter, designed many of the instruments used for research
in hearing and music, including a frequency standard and the manometric flame. The flame-tube
device, used to render standing sound waves “visible,” is still one of the most fascinating of
physics classroom demonstrations. The English physical scientist John William Strutt, 3rd
Baron Rayleigh, carried out an enormous variety of acoustic research; much of it was included
in his two-volume treatise, The Theory of Sound, publication of which in 1877–78 is now
thought to mark the beginning of modern acoustics. Much of Rayleigh’s work is still directly
quoted in contemporary physics textbooks.
The study of ultrasonics was initiated by the American scientist John LeConte, who in the 1850s
developed a technique for observing the existence of ultrasonic waves with a gas flame. This
technique was later used by the British physicist John Tyndall for the detailed study of the
properties of sound waves. The piezoelectric effect, a primary means of producing and sensing
ultrasonic waves, was discovered by the French physical chemist Pierre Curie and his
brother Jacques in 1880. Applications of ultrasonics, however, were not possible until the
development in the early 20th century of the electronic oscillator and amplifier, which were used
to drive the piezoelectric element.
Among 20th-century innovators were the American physicist Wallace Sabine, considered to be
the originator of modern architectural acoustics, and the Hungarian-born American
physicist Georg von Békésy, who carried out experimentation on the ear and hearing and
validated the commonly accepted place theory of hearing first suggested by Helmholtz. Békésy’s
book Experiments in Hearing, published in 1960, is the magnum opus of the modern theory of
the ear.
4.6 Amplifying, recording, and reproducing
The earliest known attempt to amplify a sound wave was made by Athanasius Kircher, of “bellin-vacuum” fame; Kircher designed a parabolic horn that could be used either as a hearing
aid or as a voice amplifier. The amplification of body sounds became an important goal, and the
first stethoscope was invented by a French physician, René Laënnec, in the early 19th century.
Attempts to record and reproduce sound waves originated with the invention in 1857 of a
mechanical sound-recording device called the phonautograph byÉdouard-Léon Scott de
Martinville. The first device that could actually record and play back sounds was developed by
the American inventor Thomas Alva Edisonin 1877. Edison’s phonograph employed grooves
of varying depth in a cylindrical sheet of foil, but a spiral groove on a flat rotating disk was
introduced a decade later by the German-born American inventor Emil Berliner in an invention
he called the gramophone. Much significant progress in recording and reproduction techniques
was made during the first half of the 20th century, with the development of highquality electromechanical transducers and linear electronic circuits. The most important
improvement on the standard phonograph record in the second half of the century was
thecompact disc, which employed digital techniques developed in mid-century that substantially
reduced noise and increased the fidelity and durability of the recording.
4.7 Architectural acoustics
Reverberation time
Although architectural acoustics has been an integral part of the design of structures for at least
2,000 years, the subject was only placed on a firm scientific basis at the beginning of the 20th
century by Wallace Sabine. Sabine pointed out that the most important quantity in determining
the acoustic suitability of a room for a particular use is its reverberation time, and he provided a
scientific basis by which the reverberation time can be determined or predicted.
When a source creates a sound wave in a room or auditorium, observers hear not only the sound
wave propagating directly from the source but also the myriad reflections from the walls, floor,
and ceiling. These latter form the reflected wave, or reverberant sound. After the source ceases,
the reverberant sound can be heard for some time as it grows softer. The time required, after the
sound source ceases, for the absolute intensity to drop by a factor of 106—or, equivalently, the
time for the intensity level to drop by 60 decibels—is defined as the reverberation time (RT,
sometimes referred to as RT60). Sabine recognized that the reverberation time of an auditorium is
related to the volume of the auditorium and to the ability of the walls, ceiling, floor, and contents
of the room to absorb sound. Using these assumptions, he set forth the empirical relationship
through which the reverberation time could be determined: RT = 0.05V/A, where RT is the
reverberation time in seconds, V is the volume of the room in cubic feet, and A is the total sound
absorption of the room, measured by the unit sabin. The sabin is the absorption equivalent to
one square foot of perfectly absorbing surface—for example, a one-square-foot hole in a wall or
five square feet of surface that absorbs 20 percent of the sound striking it.
Both the design and the analysis of room acoustics begin with this equation. Using the equation
and the absorption coefficients of the materials from which the walls are to be constructed, an
approximation can be obtained for the way in which the room will function acoustically.
Absorbers and reflectors, or some combination of the two, can then be used to modify the
reverberation time and its frequency dependence, thereby achieving the most desirable
characteristics for specific uses. Representative absorption coefficients—showing the fraction of
the wave, as a function of frequency, that is absorbed when a sound hits various materials—are
given in the Table. The absorption from all the surfaces in the room are added together to obtain
the total absorption (A).
Absorption coefficients of common materials at several frequencies
frequency (hertz)
material
125
250
500
1,000
2,000
4,000
concrete
0.01
0.01
0.02
0.02
0.02
0.03
plasterboard
0.20
0.15
0.10
0.08
0.04
0.02
acoustic board
0.25
0.45
0.80
0.90
0.90
0.90
curtains
0.05
0.12
0.25
0.35
0.40
0.45
While there is no exact value of reverberation time that can be called ideal, there is a range of
values deemed to be appropriate for each application. These vary with the size of the room, but
the averages can be calculated and indicated by lines on a graph. The need for clarity in
understanding speech dictates that rooms used for talking must have a reasonably short
reverberation time. On the other hand, the full sound desirable in the performance of music of
the Romantic era, such as Wagner operas or Mahler symphonies, requires a long reverberation
time. Obtaining a clarity suitable for the light, rapid passages of Bach or Mozart requires an
intermediate value of reverberation time. For playing back recordings on an audio system, the
reverberation time should be short, so as not to create confusion with the reverberation time of
the music in the hall where it was recorded.
Acoustic criteria
Many of the acoustic characteristics of rooms and auditoriums can be directly attributed to
specific physically measurable properties. Because the music critic or performing artist uses a
different vocabulary to describe these characteristics than does the physicist, it is helpful to
survey some of the more important features of acoustics and correlate the two sets of
descriptions.
“Liveness” refers directly to reverberation time. A live room has a long reverberation time and a
dead room a short reverberation time. “Intimacy” refers to the feeling that listeners have of being
physically close to the performing group. A room is generally judged intimate when the first
reverberant sound reaches the listener within about 20 milliseconds of the direct sound. This
condition is met easily in a small room, but it can also be achieved in large halls by the use of
orchestral shells that partially enclose the performers. Another example is a canopy placed above
a speaker in a large room such as a cathedral: this leads to both a strong and a quick first
reverberation and thus to a sense of intimacy with the person speaking.
The amplitude of the reverberant sound relative to the direct sound is referred to
as fullness. Clarity, the opposite of fullness, is achieved by reducing the amplitude of the
reverberant sound. Fullness generally implies a long reverberation time, while clarity implies a
shorter reverberation time. A fuller sound is generally required of Romantic music or
performances by larger groups, while more clarity would be desirable in the performance of
rapid passages from Bach or Mozart or in speech.
“Warmth” and “brilliance” refer to the reverberation time at low frequencies relative to that at
higher frequencies. Above about 500 hertz, the reverberation time should be the same for all
frequencies. But at low frequencies an increase in the reverberation time creates a warm sound,
while, if the reverberation time increased less at low frequencies, the room would be
characterized as more brilliant.
“Texture” refers to the time interval between the arrival of the direct sound and the arrival of the
first few reverberations. To obtain good texture, it is necessary that the first five reflections
arrive at the observer within about 60 milliseconds of the direct sound. An important corollary to
this requirement is that the intensity of the reverberations should decrease monotonically; there
should be no unusually large late reflections.
“Blend” refers to the mixing of sounds from all the performers and their uniform distribution to
the listeners. To achieve proper blend it is often necessary to place a collection of reflectors on
the stage that distribute the sound randomly to all points in the audience.
Although the above features of auditorium acoustics apply to listeners, the idea of ensemble
applies primarily to performers. In order to perform coherently, members of the ensemble must
be able to hear one another. Reverberant sound cannot be heard by the members of an orchestra,
for example, if the stage is too wide, has too high a ceiling, or has too much sound absorption on
its sides.
Acoustic problems
Certain acoustic problems often result from improper design or from construction limitations. If
large echoes are to be avoided, focusing of the sound wave must be avoided. Smooth, curved
reflecting surfaces such as domes and curved walls act as focusing elements, creating large
echoes and leading to bad texture. Improper blend results if sound from one part of the ensemble
is focused to one section of the audience. In addition, parallel walls in an auditorium reflect
sound back and forth, creating a rapid, repetitive pulsing of sound known as flutter echo and
even leading to destructive interference of the sound wave. Resonances at certain frequencies
should also be avoided by use of oblique walls.
Acoustic shadows, regions in which some frequency regions of sound are attenuated, can be
caused by diffraction effects as the sound wave passes around large pillars and corners or
underneath a low balcony. Large reflectors called clouds, suspended over the performers, can be
of such a size as to reflect certain frequency regions while allowing others to pass, thus affecting
the mixture of the sound.
External noise can be a serious problem for halls in urban areas or near airports or highways.
One technique often used for avoiding external noise is to construct the auditorium as a smaller
room within a larger room. Noise from air blowers or other mechanical vibrations can be reduced
using techniques involving impedance and by isolating air handlers.
Good acoustic design must take account of all these possible problems while emphasizing the
desired acoustic features. One of the problems in a large auditorium involves simply delivering
an adequate amount of sound to the rear of the hall. The intensity of a spherical sound wave
decreases in intensity at a rate of six decibels for each factor of two increase in distance from the
source, as shown above. If the auditorium is flat, a hemispherical wave will result. Absorption of
the diffracted wave by the floor or audience near the bottom of the hemisphere will result in even
greater absorption, so that the resulting intensity level will fall off at twice the theoretical rate, at
about 12 decibels for each factor of two in distance. Because of this absorption, the floors of an
auditorium are generally sloped upward toward the rear.
4.8 Summary
Acoustics is the interdisciplinary science that deals with the study of all mechanical waves in
gases, liquids, and solids including vibration, sound, ultrasound and infrasound. A scientist who
works in the field of acoustics is an acoustician while someone working in the field of acoustics
technology may be called anacoustical engineer. The application of acoustics can be seen in
almost all aspects of modern society with the most obvious being the audio and noise control
industries.
Hearing is one of the most crucial means of survival in the animal world, and speech is one of
the most distinctive characteristics of human development and culture. So it is no surprise that
the science of acoustics spreads across so many facets of our society—music, medicine,
architecture, industrial production, warfare and more. Art, craft, science and technology have
provoked
one
another
to
advance
the
whole,
as
in
many
other
fields
of
knowledge. Lindsay's 'Wheel of Acoustics' is a well accepted overview of the various fields in
acoustics.
4.9 Keywords
•
Acoustics
•
Amplifying
•
Recording
•
Reproducing
•
Architectural acoustics
•
Noise
4.10 Exercise
1) Explain Acoustics
2) Define Amplifying, recording, and reproducing
3) Explain Architectural acoustics