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Theory Of Machines - R S Khurmi

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Chapter 1 : Introduction � 1
1
Introduction
1
Features
1. Definition.
2. Sub-divisions of Theory of
Machines.
3. Fundamental Units.
4. Derived Units.
5. Systems of Units.
6. C.G.S. Units.
7. F.P.S. Units.
8. M.K.S. Units.
9. International System of
Units (S.I. Units).
10. Metre.
11. Kilogram.
12. Second.
13. Presentation of Units and
their Values.
14. Rules for S.I. Units.
15. Force.
16. Resultant Force.
17. Scalars and Vectors.
18. Representation of Vector
Quantities.
19. Addition of Vectors.
20. Subtraction of Vectors.
1.1. Definition
The subject Theory of Machines may be defined as
that branch of Engineering-science, which deals with the study
of relative motion between the various parts of a machine,
and forces which act on them. The knowledge of this subject
is very essential for an engineer in designing the various parts
of a machine.
Note:A machine is a device which receives energy in some
available form and utilises it to do some particular type of work.
1.2. Sub-divisions of Theory of Machines
The Theory of Machines may be sub-divided into
the following four branches :
1. Kinematics. It is that branch of Theory of
Machines which deals with the relative motion between the
various parts of the machines.
2. Dynamics. It is that branch of Theory of Machines
which deals with the forces and their effects, while acting
upon the machine parts in motion.
3. Kinetics. It is that branch of Theory of Machines
which deals with the inertia forces which arise from the com-
bined effect of the mass and motion of the machine parts.
4. Statics. It is that branch of Theory of Machines
which deals with the forces and their effects while the ma-
chine parts are at rest. The mass of the parts is assumed to be
negligible.
2 � Theory of Machines
1.3.1.3.1.3.1.3.1.3. Fundamental UnitsFundamental UnitsFundamental UnitsFundamental UnitsFundamental Units
The measurement of
physical quantities is one of the
most important operations in
engineering. Every quantity is
measured in terms of some
arbitrary, but internationally
accepted units, called
fundamental units. All
physical quantities, met within
this subject, are expressed in
terms of the following three
fundamental quantities :
1. Length (L or l ),
2. Mass (M or m), and
3. Time (t).
1.4.1.4.1.4.1.4.1.4. Derived UnitsDerived UnitsDerived UnitsDerived UnitsDerived Units
Some units are expressed in terms of fundamental units known as derived units, e.g., the units
of area, velocity, acceleration, pressure, etc.
1.5.1.5.1.5.1.5.1.5. Systems of UnitsSystems of UnitsSystems of UnitsSystems of UnitsSystems of Units
There are only four systems of units, which are commonly used and universally recognised.
These are known as :
1. C.G.S. units, 2. F.P.S. units, 3. M.K.S. units, and 4. S.I. units.
1.6.1.6.1.6.1.6.1.6. C.G.S. UnitsC.G.S. UnitsC.G.S. UnitsC.G.S. UnitsC.G.S. Units
In this system, the fundamental units of length, mass and time are centimetre,gram and
second respectively. The C.G.S. units are known as absolute units or physicist's units.
1.7.1.7.1.7.1.7.1.7. FFFFF.P.P.P.P.P.S..S..S..S..S. Units Units Units Units Units
In this system, the fundamental units of length, mass and time are foot, pound and second
respectively.
1.8.1.8.1.8.1.8.1.8. M.K.S. UnitsM.K.S. UnitsM.K.S. UnitsM.K.S. UnitsM.K.S. Units
In this system, the fundamental units of length, mass and time are metre, kilogram and second
respectively. The M.K.S. units are known as gravitational units or engineer's units.
1.9.1.9.1.9.1.9.1.9. InterInterInterInterInternananananational System of Units (S.I.tional System of Units (S.I.tional System of Units (S.I.tional System of Units (S.I.tional System of Units (S.I. Units) Units) Units) Units) Units)
The 11th general conference* of weights and measures have recommended a unified and
systematically constituted system of fundamental and derived units for international use. This system
is now being used in many countries. In India, the standards of Weights and Measures Act, 1956 (vide
which we switched over to M.K.S. units) has been revised to recognise all the S.I. units in industry
and commerce.
* It is known as General Conference of Weights and Measures (G.C.W.M.). It is an international organisation,
of which most of the advanced and developing countries (including India) are members. The conference
has been entrusted with the task of prescribing definitions for various units of weights and measures, which
are the very basic of science and technology today.
Stopwatch Simple balance
Chapter 1 : Introduction � 3
In this system of units, the fundamental units are metre (m), kilogram (kg) and second (s)
respectively. But there is a slight variation in their derived units. The derived units, which will be
used in this book are given below :
Density (mass density) kg/m3
Force N (Newton)
Pressure Pa (Pascal) or N/m2 ( 1 Pa = 1 N/m2)
Work, energy (in Joules) 1 J = 1 N-m
Power (in watts) 1 W = 1 J/s
Absolute viscosity kg/m-s
Kinematic viscosity m2/s
Velocity m/s
Acceleration m/s2
Angular acceleration rad/s2
Frequency (in Hertz) Hz
The international metre, kilogram and second are discussed below :
1.10. Metre
The international metre may be defined as the shortest distance (at 0°C) between the two
parallel lines, engraved upon the polished surface of a platinum-iridium bar, kept at the International
Bureau of Weights and Measures at Sevres near Paris.
1.11. Kilogram
The international kilogram may be defined as the mass of the platinum-iridium cylinder,
which is also kept at the International Bureau of Weights and Measures at Sevres near Paris.
1.12. Second
The fundamental unit of time for all the three systems, is second, which is 1/24 × 60 × 60
= 1/86 400th of the mean solar day. A solar day may be defined as the interval of time, between the
A man whose mass is 60 kg weighs 588.6 N (60 × 9.81 m/s2) on earth, approximately
96 N (60 × 1.6 m/s2) on moon and zero in space. But mass remains the same everywhere.
4 � Theory of Machines
instants, at which the sun crosses a meridian on two consecutive days. This value varies slightly
throughout the year. The average of all the solar days, during one year, is called the mean solar day.
1.13. Presentation of Units and their Values
The frequent changes in the present day life are facilitated by an international body known as
International Standard Organisation (ISO) which makes recommendations regarding international
standard procedures. The implementation of ISO recommendations, in a country, is assisted by its
organisation appointed for the purpose. In India, Bureau of Indian Standards (BIS) previously known
as Indian Standards Institution (ISI) has been created for this purpose. We have already discussed that
the fundamental units in
M.K.S. and S.I. units for
length, mass and time is metre,
kilogram and second respec-
tively. But in actual practice, it
is not necessary to express all
lengths in metres, all masses in
kilograms and all times in sec-
onds. We shall, sometimes, use
the convenient units, which are
multiples or divisions of our
basic units in tens. As a typical
example, although the metre is
the unit of length, yet a smaller
length of one-thousandth of a
metre proves to be more con-
venient unit, especially in the
dimensioning of drawings. Such convenient units are formed by using a prefix in front of the basic
units to indicate the multiplier. The full list of these prefixes is given in the following table.
Table 1.1. Prefixes used in basic units
Factor by which the unit Standard form Prefix Abbreviation
is multiplied
1 000 000 000 000 1012 tera T
1 000 000 000 109 giga G
1 000 000 106 mega M
1 000 103 kilo k
100 102 hecto* h
10 101 deca* da
0.1 10–1 deci* d
0.01 10–2 centi* c
0.001 10–3 milli m
0. 000 001 10–6 micro µ
0. 000 000 001 10–9 nano n
0. 000 000 000 001 10–12 pico p
 With rapid development of Information Technology, computers are
playing a major role in analysis, synthesis and design of machines.
* These prefixes are generally becoming obsolete probably due to possible confusion. Moreover, it is becoming
a conventional practice to use only those powers of ten which conform to 103x, where x is a positive or
negative whole number.
Chapter 1 : Introduction � 5
1.14. Rules for S.I. Units
The eleventh General Conference of Weights and Measures recommended only the funda-
mental and derived units of S.I. units. But it did not elaborate the rules for the usage of the units. Later
on many scientists and engineers held a number of meetings for the style and usage of S.I. units. Some
of the decisions of the meetings are as follows :
1. For numbers having five or more digits, the digits should be placed in groups of three sepa-
rated by spaces* (instead of commas) counting both to the left and right to the decimal point.
2. In a four digit number,** the space is not required unless the four digit number is used in a
column of numbers with five or more digits.
3. A dash is to be used to separate units that are multiplied together. For example, newton
metre is written as N-m. It should not be confused with mN, which stands for millinewton.
4. Plurals are never used with symbols. For example, metre or metres are written as m.
5. All symbols are written in small letters except the symbols derived from the proper names.
For example, N for newton and W for watt.
6. The units with names of scientists should not start with capital letter when written in full. For
example, 90 newton and not 90 Newton.
At the time of writing this book, the authors sought the advice of various international
authorities, regarding the use of units and their values. Keeping in view the international reputation of
the authors, as well as international popularity of their books, it was decided to present units*** and
their values as per recommendations of ISO and BIS. It was decided to use :
4500 not 4 500 or 4,500
75 890 000 not 75890000 or 7,58,90,000
0.012 55 not 0.01255 or .01255
30 × 106 not 3,00,00,000 or 3 × 107
The above mentioned figures are meant for numerical values only. Now let us discuss about
the units. We know that the fundamental units in S.I. system of units for length, mass and time are
metre, kilogram and second respectively. While expressing these quantities we find it time consum-
ing to write the units such as metres, kilograms and seconds, in full, every time we use them. As a
result of this, we find it quite convenient to use some standard abbreviations.
We shall use :
m for metre or metres
km for kilometre or kilometres
kg for kilogram or kilograms
t for tonne or tonnes
s for second or seconds
min for minute or minutes
N-m for newton × metres (e.g. work done )
kN-m for kilonewton × metres
rev for revolution or revolutions
rad for radian or radians* In certain countries, comma is still used as the decimal mark.
** In certain countries, a space is used even in a four digit number.
*** In some of the question papers of the universities and other examining bodies, standard values are not used.
The authors have tried to avoid such questions in the text of the book. However, at certain places, the
questions with sub-standard values have to be included, keeping in view the merits of the question from the
reader’s angle.
6 � Theory of Machines
1.15. Force
It is an important factor in the field of Engineering science, which may be defined as an agent,
which produces or tends to produce, destroy or tends to destroy motion.
1.16. Resultant Force
If a number of forces P,Q,R etc. are acting simultaneously on a particle, then a single force,
which will produce the same effect as that of all the given forces, is known as a resultant force. The
forces P,Q,R etc. are called component forces. The process of finding out the resultant force of the
given component forces, is known as composition of forces.
A resultant force may be found out analytically, graphically or by the following three laws:
1. Parallelogram law of forces. It states, “If two forces acting simultaneously on a particle
be represented in magnitude and direction by the two adjacent sides of a parallelogram taken in order,
their resultant may be represented in magnitude and direction by the diagonal of the parallelogram
passing through the point.”
2. Triangle law of forces. It states, “If two forces acting simultaneously on a particle be
represented in magnitude and direction by the two sides of a triangle taken in order, their resultant
may be represented in magnitude and direction by the third side of the triangle taken in opposite
order.”
3. Polygon law of forces. It states, “If a number of forces acting simultaneously on a particle
be represented in magnitude and direction by the sides of a polygon taken in order, their resultant may
be represented in magnitude and direction by the closing side of the polygon taken in opposite order.”
1.17. Scalars and Vectors
1. Scalar quantities are those quantities, which have magnitude only, e.g. mass, time, volume,
density etc.
2. Vector quantities are those quantities which have magnitude as well as direction e.g. velocity,
acceleration, force etc.
3. Since the vector quantities have both magnitude and direction, therefore, while adding or
subtracting vector quantities, their directions are also taken into account.
1.18. Representation of Vector Quantities
The vector quantities are represented by vectors. A vector is a straight line of a certain length
Chapter 1 : Introduction � 7
possessing a starting point and a terminal point at which it carries an arrow head. This vector is cut off
along the vector quantity or drawn parallel to the line of action of the vector quantity, so that the
length of the vector represents the magnitude to some scale. The arrow head of the vector represents
the direction of the vector quantity.
1.19. Addition of Vectors
(a) (b)
Fig. 1.1. Addition of vectors.
Consider two vector quantities P and Q, which are required to be added, as shown in Fig.1.1(a).
Take a point A and draw a line AB parallel and equal in magnitude to the vector P. Through B,
draw BC parallel and equal in magnitude to the vector Q. Join AC, which will give the required sum
of the two vectors P and Q, as shown in Fig. 1.1 (b).
1.20. Subtraction of Vector Quantities
Consider two vector quantities P and Q whose difference is required to be found out as
shown in Fig. 1.2 (a).
 (a) (b)
Fig. 1.2. Subtraction of vectors.
Take a point A and draw a line AB parallel and equal in magnitude to the vector P. Through B,
draw BC parallel and equal in magnitude to the vector Q, but in opposite direction. Join AC, which
gives the required difference of the vectors P and Q, as shown in Fig. 1.2 (b).
8 � Theory of Machines
2.1. Introduction
We have discussed in the previous Chapter, that the
subject of Theory of Machines deals with the motion and
forces acting on the parts (or links) of a machine. In this chap-
ter, we shall first discuss the kinematics of motion i.e. the
relative motion of bodies without consideration of the forces
causing the motion. In other words, kinematics deal with the
geometry of motion and concepts like displacement, velocity
and acceleration considered as functions of time.
2.2. Plane Motion
When the motion of a body is confined to only one
plane, the motion is said to be plane motion. The plane mo-
tion may be either rectilinear or curvilinear.
2.3. Rectilinear Motion
It is the simplest type of motion and is along a straight
line path. Such a motion is also known as translatory motion.
2.4. Curvilinear Motion
It is the motion along a curved path. Such a motion,
when confined to one plane, is called plane curvilinear
motion.
When all the particles of a body travel in concentric
circular paths of constant radii (about the axis of rotation
perpendicular to the plane of motion) such as a pulley rotating
8
Kinematics of
Motion
2
Features
1. 1ntroduction.
2. Plane Motion.
3. Rectilinear Motion.
4. Curvilinear Motion.
5. Linear Displacement.
6. Linear Velocity.
7. Linear Acceleration.
8. Equations of Linear Motion.
9. Graphical Representation of
Displacement with respect to
Time.
10. Graphical Representation of
Velocity with respect to Time.
11. Graphical Representation of
Acceleration with respect to
Time.
12. Angular Displacement.
13. Representation of Angular
Displacement by a Vector.
14. Angular Velocity.
15. Angular Acceleration.
16. Equations of Angular Motion.
17. Relation Between Linear
Motion and Angular Motion.
18. Relation Between Linear and
Angular Quantities of
Motion.
19. Acceleration of a Particle
along a Circular Path.
Chapter 2 : Kinematics of Motion � 9
about a fixed shaft or a shaft rotating about its
own axis, then the motion is said to be a plane
rotational motion.
Note: The motion of a body, confined to one plane,
may not be either completely rectilinear nor completely
rotational. Such a type of motion is called combined
rectilinear and rotational motion. This motion is dis-
cussed in Chapter 6, Art. 6.1.
2.5. Linear Displacement
It may be defined as the distance moved
by a body with respect to a certain fixed point.
The displacement may be along a straight or a
curved path. In a reciprocating steam engine, all
the particles on the piston, piston rod and cross-
head trace a straight path, whereas all particles
on the crank and crank pin trace circular paths,
whose centre lies on the axis of the crank shaft. It will be interesting to know, that all the particles on
the connecting rod neither trace a straight path nor a circular one; but trace an oval path, whose radius
of curvature changes from time to time.
The displacement of a body is a vector quantity, as it has both magnitude and direction.
Linear displacement may, therefore, be represented graphically by a straight line.
2.6. Linear Velocity
It may be defined as the rate of
change of linear displacement of a body with
respect to the time. Since velocity is always
expressed in a particular direction, therefore
it is a vector quantity. Mathematically, lin-
ear velocity,
v = ds/dt
Notes: 1. If the displacement is along a circular
path, then the direction of linear velocity at any
instant is along the tangent at that point.
2. The speed is the rate of change of linear displacement of a body with respect to the time. Since the
speed is irrespective of its direction, therefore, it is a scalar quantity.
2.7. Linear Acceleration
It may be defined as the rate of change of linear velocity of a body with respect to the time. It
is also a vector quantity. Mathematically, linear acceleration,2
2
dv d ds d s
a
dt dt dt dt
 
= = =   ...
ds
v
dt
 
=  �
Notes: 1. The linear acceleration may also be expressed as follows:
dv ds dv dv
a v
dt dt ds ds
= = × = ×
2. The negative acceleration is also known as deceleration or retardation.
Spindle
(axis of rotation)
Axis of rotation
Reference
line
θ
∆θ
θο
r
10 � Theory of Machines
2.8.2.8.2.8.2.8.2.8. Equations of Linear MotionEquations of Linear MotionEquations of Linear MotionEquations of Linear MotionEquations of Linear Motion
The following equations of linear motion are
important from the subject point of view:
1. v = u + a.t 2. s = u.t + 
1
2
 a.t2
3. v2 = u2 + 2a.s
4.
( )
2 av
u v
s t v t
+
= × = ×
where u = Initial velocity of the body,
v = Final velocity of the body,
a = Acceleration of the body,
s = Displacement of the body in time t seconds, and
vav = Average velocity of the body during the motion.
Notes: 1. The above equations apply for uniform
acceleration. If, however, the acceleration is variable,
then it must be expressed as a function of either t, s
or v and then integrated.
2. In case of vertical motion, the body is
subjected to gravity. Thus g (acceleration due to grav-
ity) should be substituted for ‘a’ in the above equa-
tions.
3. The value of g is taken as + 9.81 m/s2 for
downward motion, and – 9.81 m/s2 for upward mo-
tion of a body.
4. When a body falls freely from a height h,
then its velocity v, with which it will hit the ground is
given by
2 .v g h=
2.9.2.9.2.9.2.9.2.9. GraGraGraGraGraphical Reprphical Reprphical Reprphical Reprphical Representaesentaesentaesentaesentation oftion oftion oftion oftion of
Displacement with RespectDisplacement with RespectDisplacement with RespectDisplacement with RespectDisplacement with Respect
to to to to to TimeTimeTimeTimeTime
The displacement of a moving body in a given time may be found by means of a graph. Such
a graph is drawn by plotting the displacement as ordinate and the corresponding time as abscissa. We
shall discuss the following two cases :
1. When the body moves with uniform velocity. When the body moves with uniform velocity,
equal distances are covered in equal intervals of time. By plotting the distances on Y-axis and time on
X-axis, a displacement-time curve (i.e. s-t curve) is drawn which is a straight line, as shown in Fig. 2.1
(a). The motion of the body is governed by the equation s = u.t, such that
Velocity at instant 1 = s1 / t1
Velocity at instant 2 = s2 / t2
Since the velocity is uniform, therefore
31 2
1 2 3
tan
ss s
t t t
= = = θ
where tan θ is called the slope of s-t curve. In other words, the slope of the s-t curve at any instant
gives the velocity.
t = 0 s
v = 0 m/s
t = 2 s
v = 19.62 m/s
t = time
v = velocity (downward)
g = 9.81 m/s2 = acceleration
due to gravity
t = 1 s
v = 9.81 m/s
Chapter 2 : Kinematics of Motion � 11
2. When the body moves with variable velocity. When the body moves with variable velocity,
unequal distances are covered in equal intervals of time or equal distances are covered in unequal intervals
of time. Thus the displacement-time graph, for such a case, will be a curve, as shown in Fig. 2.1 (b).
(a) Uniform velocity. (b) Variable velocity.
Fig. 2.1. Graphical representation of displacement with respect to time.
Consider a point P on the s-t curve and let this point travels to Q by a small distance δs in a
small interval of time δt. Let the chord joining the points P and Q makes an angle θ with the horizontal.
The average velocity of the moving point during the interval PQ is given by
tan θ = δs / δt . . . (From triangle PQR )
In the limit, when δt approaches to zero, the point Q will tend to approach P and the chord PQ
becomes tangent to the curve at point P. Thus the velocity at P,
vp = tan θ = ds /dt
where tan θ is the slope of the tangent at P. Thus the slope of the tangent at any instant on the s-t curve
gives the velocity at that instant.
2.10. Graphical Representation of Velocity with Respect to Time
We shall consider the following two cases :
1. When the body moves with uniform velocity. When the body moves with zero acceleration,
then the body is said to move with a uniform
velocity and the velocity-time curve (v-t
curve) is represented by a straight line as
shown by A B in Fig. 2.2 (a).
We know that distance covered by a
body in time t second
= Area under the v-t curve A B
= Area of rectangle OABC
Thus, the distance covered by a
body at any interval of time is given by the
area under the v-t curve.
2. When the body moves with
variable velocity. When the body moves with
constant acceleration, the body is said to move with variable velocity. In such a case, there is equal
variation of velocity in equal intervals of time and the velocity-time curve will be a straight
line AB inclined at an angle θ, as shown in Fig. 2.2 (b). The equations of motion i.e. v = u + a.t, and
s = u.t + 12 a.t
2
 may be verified from this v-t curve.
12 � Theory of Machines
Let u = Initial velocity of a moving body, and
v = Final velocity of a moving body after time t.
Then,
Changein velocity
tan Acceleration ( )
Time
BC v u
a
AC t
−θ = = = =
(a) Uniform velocity. (b) Variable velocity.
Fig. 2.2. Graphical representation of velocity with respect to time.
Thus, the slope of the v-t curve represents the acceleration of a moving body.
Now tan
−
= θ = =BC v ua
AC t
or v = u + a.t
Since the distance moved by a body is given by the area under the v-t curve, therefore
distance moved in time (t),
s = Area OABD = Area OACD + Area ABC
21 1
2 2
. ( ) . .u t v u t u t a t= + − = + ... (� v – u = a.t)
2.11. Graphical Representation of Acceleration with Respect to Time
(a) Uniform velocity. (b) Variable velocity.
Fig. 2.3. Graphical representation of acceleration with respect to time.
We shall consider the following two cases :
1. When the body moves with uniform acceleration. When the body moves with uniform
acceleration, the acceleration-time curve (a-t curve) is a straight line, as shown in Fig. 2.3(a). Since
the change in velocity is the product of the acceleration and the time, therefore the area under the
a-t curve (i.e. OABC) represents the change in velocity.
2. When the body moves with variable acceleration. When the body moves with variable
acceleration, the a-t curve may have any shape depending upon the values of acceleration at various
instances, as shown in Fig. 2.3(b). Let at any instant of time t, the acceleration of moving body is a.
Mathematically, a = dv / dt or dv = a.dt
Chapter 2 : Kinematics of Motion � 13
Integrating both sides,
2 2
1 1
.
v t
v t
dv a dt=∫ ∫ or 212 1 .− =∫ ttv v a dt
where v1 and v2 are the velocities of the moving body at time intervals t1 and t2 respectively.
The right hand side of the above expression represents the area (PQQ1P1) under the a-t curve
between the time intervals t1 and t2 . Thus the area under the a-t curve between any two ordinates
represents the change in velocity of the moving body. If the initial and final velocities of the body are
u and v, then the above expression may be written as
0
.
t
v u a d t− = =∫ Area under a-t curve A B = Area OABC
Example 2.1. A car starts from rest and
accelerates uniformly to a speed of 72 km. p.h. over
a distance of 500 m. Calculate the acceleration and
the time taken to attain the speed.
If a further acceleration raises the speed to
90 km. p.h. in 10 seconds, find this acceleration and
the further distance moved. The brakes are now
applied to bring the car to rest under uniform
retardation in 5 seconds. Find the distance travelled
during braking.
Solution. Given : u = 0 ; v = 72 km. p.h. = 20 m/s ; s = 500 m
First of all, letus consider the motion of the car from rest.
Acceleration of the car
Let a = Acceleration of the car.
We know that v2 = u2 + 2 a.s
∴ (20)2 = 0 + 2a × 500 = 1000 a or a = (20)2/ 1000 = 0.4 m/s2 Ans.
Time taken by the car to attain the speed
Let t = Time taken by the car to attain the speed.
We know that v = u + a.t
∴ 20 = 0 + 0.4 × t or t = 20/0.4 = 50 s Ans.
Now consider the motion of the car from 72 km.p.h. to 90 km.p.h. in 10 seconds.
Given : * u = 72 km.p.h. = 20 m/s ; v = 96 km.p.h. = 25 m/s ; t = 10 s
Acceleration of the car
Let a = Acceleration of the car.
We know that v = u + a.t
25 = 20 + a × 10 or a = (25 – 20)/10 = 0.5 m/s2 Ans.
Distance moved by the car
We know that distance moved by the car,
2 21 1
2 2
. . 20 10 0.5(10) 225ms u t a t= + = × + × =
 Ans.
* It is the final velocity in the first case.
14 � Theory of Machines
Now consider the motion of the car during the application of brakes for brining it to rest in
5 seconds.
Given : *u = 25 m/s ; v = 0 ; t = 5 s
We know that the distance travelled by the car during braking,
25 0 5 62.5 m
2 2
u v
s t
+ +
= × = × =
 Ans.
Example 2.2. The motion of a particle is given by a = t3 – 3t2 + 5, where a is the acceleration
in m/s2 and t is the time in seconds. The velocity of the particle at t = 1 second is 6.25 m/s, and the
displacement is 8.30 metres. Calculate the displacement and the velocity at t = 2 seconds.
Solution. Given : a = t3 – 3t2 + 5
We know that the acceleration, a = dv/dt. Therefore the above equation may be written as
3 23 5= − +dv t t
dt
or 3 2( 3 5)= − +dv t t dt
Integrating both sides
4 3 4
3
1 1
3 5 5
4 3 4
= − + + = − + +
t t t
v t C t t C ...(i)
where C1 is the first constant of integration. We know that when t = 1 s, v = 6.25 m/s. Therefore
substituting these values of t and v in equation (i),
6.25 = 0.25 – 1 + 5 + C1 = 4.25 + C1 or C1 = 2
Now substituting the value of C1 in equation (i),
34 5 2
4
= − + +
t
v t t ...(ii)
Velocity at t = 2 seconds
Substituting the value of t = 2 s in the above equation,
4
32 2 5 2 2 8 m/s
4
v = − + × + =
 Ans.
Displacement at t = 2 seconds
We know that the velocity, v = ds/dt, therefore equation (ii) may be written as
4 4
3 35 2 or 5 2
4 4
ds t t
t t ds t t dt
dt
 
= − + + = − + +   
Integrating both sides,
5 4 2
2
5 2
20 4 2
t t t
s t C= − + + +
...(iii)
where C2 is the second constant of integration. We know that when t = 1 s, s = 8.30 m. Therefore
substituting these values of t and s in equation (iii),
2 2
1 1 58.30 2 4.3
20 4 2
C C= − + + + = + or C2 = 4
* It is the final velocity in the second case.
Chapter 2 : Kinematics of Motion � 15
Substituting the value of C2 in equation (iii),
5 4 25 2 4
20 4 2
= − + + +
t t t
s t
Substituting the value of t = 2 s, in this equation,
5 4 22 2 5 2 2 2 4 15.6 m
20 4 2
×
= − + + × + =s Ans.
Example 2.3. The velocity of a
train travelling at 100 km/h decreases by
10 per cent in the first 40 s after applica-
tion of the brakes. Calculate the velocity
at the end of a further 80 s assuming that,
during the whole period of 120 s, the re-
tardation is proportional to the velocity.
Solution. Given : Velocity in the
beginning (i.e. when t = 0), v0 = 100 km/h
Since the velocity decreases by 10
per cent in the first 40 seconds after the
application of brakes, therefore velocity at the end of 40 s,
v40 = 100 × 0.9 = 90 km/h
Let v120 = Velocity at the end of 120 s (or further 80s).
Since the retardation is proportional to the velocity, therefore,
.= − =
dv
a k v
dt or
.= −
dv k dt
v
where k is a constant of proportionality, whose value may be determined from the given conditions.
Integrating the above expression,
log
e
 v = – k.t + C ... (i)
where C is the constant of integration. We know that when t = 0, v = 100 km/h. Substituting these
values in equation (i),
log
e
100 = C or C = 2.3 log 100 = 2.3 × 2 = 4.6
We also know that when t = 40 s, v = 90 km/h. Substituting these values in equation (i),
log
e 
90 = – k × 40 + 4.6 ...( � C = 4.6 )
2.3 log 90 = – 40k + 4.6
or
4.6 2.3log 90 4.6 2.3 1.9542 0.0026
40 40
− − ×
= = =k
Substituting the values of k and C in equation (i),
log
e
 v = – 0.0026 × t + 4.6
or 2.3 log v = – 0.0026 × t + 4.6 ... (ii)
Now substituting the value of t equal to 120 s, in the above equation,
2.3 log v120 = – 0.0026 × 120 + 4.6 = 4.288
or log v120 = 4.288 / 2.3 = 1.864
∴ v120 = 73.1 km/h Ans. ... (Taking antilog of 1.864)
16 � Theory of Machines
Example 2.4. The acceleration (a) of a slider block and its displacement (s) are related by
the expression, =a k s , where k is a constant. The velocity v is in the direction of the displacement
and the velocity and displacement are both zero when time t is zero. Calculate the displacement,
velocity and acceleration as functions of time.
Solution. Given : =a k s
We know that acceleration,
dv
a v
ds
= ×
 or 
dvk s v
ds
= × ...
dv ds dv dv
v
dt dt ds ds
 
= × = ×  �
∴ v × dv = k.s1/2 ds
Integrating both sides,
1/ 2
0
. =∫ ∫v v dv k s ds or 2 3 / 2 1.2 3/ 2= +v k s C ... (i)
where C1 is the first constant of integration whose value is to be determined from the given conditions
of motion. We know that s = 0, when v = 0. Therefore, substituting the values of s and v in equation (i),
we get C1 = 0.
∴
2
3 / 22
.
2 3
=
v k s or 3/ 4
4
3
= ×
k
v s
... (ii)
Displacement, velocity and acceleration as functions of time
We know that 3 / 4
4
3
= = ×
ds k
v s
dt ... [From equation (ii)]
∴ 3/ 4
4
3
ds k dt
s
= or
3/ 4 4
3
−
=
k
s ds dt
Integrating both sides,
3 / 4
0 0
4
3
−
=∫ ∫s tks ds dt
1/ 4
2
4
1/ 4 3
= × +
s k
t C ...(iii)
where C2 is the second constant of integration. We know that displacement, s = 0 when t = 0. There-
fore, substituting the values of s and t in equation (iii), we get C2 = 0.
∴
1/ 4 4
1/ 4 3
= ×
s k
t
 or 
2 4
.
144
=
k t
s
 Ans.
We know that velocity,
2 2 3
3 .4
144 36
= = × =
ds k k t
v t
dt
 Ans.
2 4
.
... Differentiating
144
k t   
and acceleration,
2 2 2
2 .3
36 12
= = × =
dv k k t
a t
dt
 Ans.
2 3
.
... Differentiating
36
k t   
Chapter 2 : Kinematics of Motion � 17
Example 2.5. The cutting stroke of a planing
machine is 500 mm and it is completed in 1 second.
The planing table accelerates uniformly during the first
125 mm of the stroke, the speed remains constant during
the next 250 mm of the stroke and retards uniformly during
the last 125 mm of the stroke. Find the maximum cutting
speed.
Solution. Given : s = 500 mm ; t = 1 s ;
s1 = 125 mm ; s2 = 250 mm ; s3 = 125 mm
Fig. 2.4 shows the acceleration-time and veloc-
ity-time graph for the planing table of a planing machine.
Let
v = Maximum cutting speed in mm/s.
Average velocity of the table during acceleration
and retardation,
(0 ) / 2 / 2= + =avv v v
Time of uniform acceleration 11
125 250
s
/ 2av
s
t
v v v
= = =
Time of constant speed, 22
250
s
s
t
v v
= =
and time of uniform retardation,
3
3
125 250
s
/ 2av
s
t
v v v
= = =
Fig. 2.4
Since the time taken to complete the stroke is 1 s, therefore
1 2 3+ + =t t t t
250 250 250 1+ + =
v v v
 or v = 750 mm/s Ans.
2.12. Angular Displacement
It may be defined as the angle described by a particle from one point to another, with respect
to the time. For example, let a line OB has its inclination θ radians to the fixed line OA, as shown in
Planing Machine.
18 � Theory of Machines
Fig. 2.5. If this line moves from OB to OC, through an angle δθ during
a short interval of time δt, thenδθ is known as the angular
displacement of the line OB.
Since the angular displacement has both magnitude and
direction, therefore it is also a vector quantity.
2.13. Representation of Angular Displacement by
a Vector
In order to completely represent an angular displacement, by a vector, it must fix the follow-
ing three conditions :
1. Direction of the axis of rotation. It is fixed by drawing a line perpendicular to the plane
of rotation, in which the angular displacement takes place. In other words, it is fixed along the axis
of rotation.
2. Magnitude of angular displacement. It is fixed by the length of the vector drawn along
the axis of rotation, to some suitable scale.
3. Sense of the angular displacement. It is fixed by a right hand screw rule. This rule
states that if a screw rotates in a fixed nut in a clockwise direction, i.e. if the angular displacement
is clockwise and an observer is looking along the axis of rotation, then the arrow head will point
away from the observer. Similarly, if the angular displacement is anti-clockwise, then the arrow
head will point towards the observer.
2.14. Angular Velocity
It may be defined as the rate of change of angular displacement with respect to time. It is
usually expressed by a Greek letter ω (omega). Mathematically, angular velocity,
/ω = θd dt
Since it has magnitude and direction, therefore, it is a vector quantity. It may be represented
by a vector following the same rule as described in the previous article.
Note : If the direction of the angular displacement is constant, then the rate of change of magnitude of the
angular displacement with respect to time is termed as angular speed.
2.15. Angular Acceleration
It may be defined as the rate of change of angular velocity with respect to time. It is usually
expressed by a Greek letter α (alpha). Mathematically, angular acceleration,
2
2
ω θ θ 
α = = =  
d d d d
dt dt dt dt
...
d
dt
θ 
ω =  �
It is also a vector quantity, but its direction may not be same as that of angular displacement
and angular velocity.
2.16. Equations of Angular Motion
The following equations of angular motion corresponding to linear motion are important
from the subject point of view :
1. 0 .ω = ω +α t 2. 
2
0
1
. .
2
θ = ω + αt t
3. ( )22 0 2 .ω = ω + αθ 4. ( )0 2
ω +ω
θ =
t
where ω0 = Initial angular velocity in rad/s,
ω
 
= Final angular velocity in rad/s,
Fig. 2.5. Angular
displacement.
Chapter 2 : Kinematics of Motion � 19
Fig. 2.6. Motion of a body
along a circular path.
t = Time in seconds,
θ = Angular displacement in time t seconds, and
α = Angular acceleration in rad / s2.
Note : If a body is rotating at the rate of N r.p.m. (revolutions per minute), then its angular velocity,
ω
 
= 2πΝ / 60 rad/s
2.17. Relation between Linear Motion and Angular Motion
Following are the relations between the linear motion and the angular motion :
Particulars Linear motion Angular motion
Initial velocity u ω0
Final velocity v ω
Constant acceleration a α
Total distance traversed s θ
Formula for final velocity v = u + a.t ω = ω0 + α.t
Formula for distance traversed s = u.t + 12 a.t
2 θ = ω0. t + 
1
2 α.t
2
Formula for final velocity v2 = u2 + 2 a.s ω = (ω0) 2 + 2 α.θ
2.18. Relation between Linear and Angular Quantities of Motion
Consider a body moving along a circular path from A to B as shown in Fig. 2.6.
Let r = Radius of the circular path,
θ = Angular displacement in radians,
s = Linear displacement,
v = Linear velocity,
ω = Angular velocity,
a = Linear acceleration, and
α = Angular acceleration.
From the geometry of the figure, we know that
 s = r . θ
We also know that the linear velocity,
( . )
.
θ θ
= = = × = ω
ds d r d
v r r
dt dt dt
... (i)
and linear acceleration,
( . )
.
ω ω
= = = × = α
dv d r d
a r r
dt dt dt ... (ii)
Example 2.6. A wheel accelerates uniformly from rest to 2000 r.p.m. in 20 seconds. What is its
angular acceleration? How many revolutions does the wheel make in attaining the speed of 2000 r.p.m.?
Solution. Given : N0 = 0 or ω = 0 ; N = 2000 r.p.m. or ω = 2π × 2000/60 = 209.5 rad/s ; t = 20s
Angular acceleration
Let α = Angular acceleration in rad/s2.
We know that
ω = ω0 + α.t or 209.5 = 0 + α × 20
∴ α = 209.5 / 20 = 10.475 rad/s2 Ans.
20 � Theory of Machines
Number of revolutions made by the wheel
We know that the angular distance moved by the wheel during 2000 r.p.m. (i.e. when
ω = 209.5 rad/s),
( ) ( )0 0 209.5 20 2095
2 2
ω +ω +
θ = = =
t
rad
Since the angular distance moved by the wheel during one revolution is 2π radians, therefore
number of revolutions made by the wheel,
n = θ /2π = 2095/2π = 333.4 Ans.
2.19. Acceleration of a Particle along a Circular Path
Consider A and B, the two positions of a particle displaced through an angle δθ in time δt as
shown in Fig. 2.7 (a).
Let r = Radius of curvature of the circular path,
v = Velocity of the particle at A , and
v + dv = Velocity of the particle at B.
The change of velocity, as the particle moves from A to B may be obtained by drawing the
vector triangle oab, as shown in Fig. 2.7 (b). In this triangle, oa represents the velocity v and ob
represents the velocity v + dv. The change of velocity in time δt is represented by ab.
Fig. 2.7. Acceleration of a particle along a circular path.
Now, resolving ab into two components i.e. parallel and perpendicular to oa. Let ac and cb
be the components parallel and perpendicular to oa respectively.
∴ ac = oc – oa = ob cos δθ – oa = (v + δv) cos δθ – v
and cb = ob sin δθ = (v + δv) sin δθ
Since the change of velocity of a particle (represented by vector ab) has two mutually
perpendicular components, therefore the acceleration of a particle moving along a circular path has
the following two components of the acceleration which are perpendicular to each other.
1. Tangential component of the acceleration. The acceleration of a particle at any instant
moving along a circular path in a direction tangential to that instant, is known as tangential component
of acceleration or tangential acceleration.
∴ Tangential component of the acceleration of particle at A or tangential acceleration at A,
( )cos+ δ δθ −
= =
δ δt
ac v v v
a
t t
In the limit, when δt approaches to zero, then
/ .= = αta dv dt r ... (i)
2. Normal component of the acceleration. The acceleration of a particle at any instant mov-
ing along a circular path in a direction normal to the tangent at that instant and directed towards the
centre of the circular path (i.e. in the direction from A to O) is known as normal component of the
Chapter 2 : Kinematics of Motion � 21
acceleration or normal acceleration. It is also called radial or centripetal acceleration.
∴ Normal component of the acceleration of the particle at A or normal (or radial or centrip-
etal) acceleration at A ,
( )sin+ δ θ
= =
δ δn
v vcb
a
t t
In the limit, when δt approaches to zero, then
2
2
. .
θ
= × = ω = × = = ωn
d v v
a v v v r
dt r r
... (ii)
 [ ]... / , and /d dt v rθ = ω ω =�
Since the tangential acceleration (at) and the normal accelera-
tion (a
n
) of the particle at any instant A are perpendicular to each other,
as shown in Fig. 2.8, therefore total acceleration of the particle (a) is
equal to the resultant acceleration of at and an.
∴ Total acceleration or resultant acceleration,
( ) ( )2 2= +t na a a
and its angle of inclination with the tangential acceleration is given by
tan θ = a
n
/at or θ = tan–1 (an/at)
The total acceleration or resultant acceleration may also be obtained by the vector sum of at
and a
n
.
Notes : 1. From equations (i) and (ii) we see that the tangential acceleration (at ) is equal to the rateof change of
the magnitude of the velocity whereas the normal or radial or centripetal acceleration (a
n
) depends upon its
instantaneous velocity and the radius of curvature of its path.
2. When a particle moves along a straight path, then the radius of curvature is infinitely great. This
means that v2/r is zero. In other words, there will be no normal or radial or centripetal acceleration. Therefore,
the particle has only tangential acceleration (in the same direction as its velocity and displacement) whose value
is given by
at = dv/dt = α.r
3. When a particle moves with a uniform velocity, then dv/dt will be zero. In other words, there will be
no tangential acceleration; but the particle will have only normal or radial or centripetal acceleration, whose
value is given by
a
n
 = v2/r = v.ω = ω2 r
Example 2.7. A horizontal bar 1.5 metres long and of small cross-section rotates about
vertical axis through one end. It accelerates uniformly from 1200 r.p.m. to 1500 r.p.m. in an interval
of 5 seconds. What is the linear velocity at the beginning and end of the interval ? What are the
normal and tangential components of the acceleration of the mid-point of the bar after 5 seconds
after the acceleration begins ?
Solution. Given : r = 1.5 m ; N0 = 1200 r.p.m. or ω0 = 2 π × 1200/60 = 125.7 rad/s ;
N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157 rad/s ; t = 5 s
Linear velocity at the beginning
We know that linear velocity at the beginning,
v0 = r . ω0 = 1.5 × 125.7 = 188.6 m/s Ans.
Linear velocity at the end of 5 seconds
We also know that linear velocity after 5 seconds,
v5 = r . ω = 1.5 × 157 = 235.5 m/s Ans.
Fig. 2.8. Total acceleration
 of a particle.
22 � Theory of Machines
Tangential acceleration after 5 seconds
Let α = Constant angular acceleration.
We know that ω = ω0+ α.t
 157 = 125.7 + α × 5 or α = (157 – 125.7) /5 = 6.26 rad/s2
Radius corresponding to the middle point,
r = 1.5 /2 = 0.75 m
∴ Tangential acceleration = α. r = 6.26 × 0.75 = 4.7 m/s2 Ans.
Radial acceleration after 5 seconds
 Radial acceleration = ω2 . r = (157)2 0.75 = 18 487 m/s2 Ans.
EXERCISES
1. A winding drum raises a cage through a height of 120 m. The cage has, at first, an acceleration
of 1.5 m/s2 until the velocity of 9 m/s is reached, after which the velocity is constant until the
cage nears the top, when the final retardation is 6 m/s2. Find the time taken for the cage to reach
the top. [ Ans. 17.1s ]
2. The displacement of a point is given by s = 2t3 + t2 + 6, where s is in metres and t in seconds.
Determine the displacement of the point when the velocity changes from 8.4 m/s to 18 m/s. Find also
the acceleration at the instant when the velocity of the particle is 30 m/s. [ Ans. 6.95 m ; 27 m/s2 ]
3. A rotating cam operates a follower which moves in a straight line. The stroke of the follower is 20
mm and takes place in 0.01 second from rest to rest. The motion is made up of uniform acceleration
for 1/4 of the time, uniform velocity for 12 of the time followed by uniform retardation. Find the
maximum velocity reached and the value of acceleration and retardation.
[ Ans. 2.67 m/s ; 1068 m/s2 ; 1068 m/s2 ]
4. A cage descends a mine shaft with an acceleration of 0.5 m/s2. After the cage has travelled 25 metres,
a stone is dropped from the top of the shaft. Determine : 1. the time taken by the stone to hit the cage,
and 2. distance travelled by the cage before impact. [ Ans. 2.92 s ; 41.73 m ]
5. The angular displacement of a body is a function of time and is given by equation :
θ = 10 + 3 t + 6 t2, where t is in seconds.
Determine the angular velocity, displacement and acceleration when t = 5 seconds. State whether or
not it is a case of uniform angular acceleration. [Ans. 63 rad/s ; 175 rad ; 12 rad/s2]
6. A flywheel is making 180 r.p.m. and after 20 seconds it is running at 140 r.p.m. How many revolu-
tions will it make, and what time will elapse before it stops, if the retardation is uniform ?
[ Ans. 135 rev. ; 90 s ]
7. A locomotive is running at a constant speed of 100 km / h. The diameter of driving wheels is 1.8 m. The
stroke of the piston of the steam engine cylinder of the locomotive is 600 mm. Find the centrip-
etal acceleration of the crank pin relative to the engine frame. [ Ans. 288 m/s2 ]
DO YOU KNOW ?
1. Distinguish clearly between speed and velocity. Give examples.
2. What do you understand by the term ‘acceleration’ ? Define positive acceleration and negative accel-
eration.
3. Define ‘angular velocity’ and ‘angular acceleration’. Do they have any relation between them ?
4. How would you find out the linear velocity of a rotating body ?
5. Why the centripetal acceleration is zero, when a particle moves along a straight path ?
6. A particle moving with a uniform velocity has no tangential acceleration. Explain clearly.
Chapter 2 : Kinematics of Motion � 23
OBJECTIVE TYPE QUESTIONS
1. The unit of linear acceleration is
(a) kg-m (b) m/s (c) m/s2 (d) rad/s2
2. The angular velocity (in rad/s) of a body rotating at N r.p.m. is
(a) π N/60 (b) 2 π N/60 (c) π N/120 (d) π N/180
3. The linear velocity of a body rotating at ω rad/s along a circular path of radius r is given by
(a) ω.r (b) ω/r (c) ω2.r (d) ω2/r
4. When a particle moves along a straight path, then the particle has
(a) tangential acceleration only (b) centripetal acceleration only
(c) both tangential and centripetal acceleration
5. When a particle moves with a uniform velocity along a circular path, then the particle has
(a) tangential acceleration only (b) centripetal acceleration only
(c) both tangential and centripetal acceleration
ANSWERS
1. (c) 2. (b) 3. (a) 4. (a) 5. (b)
24 ����� Theory of Machines
24
Kinetics of
Motion
3
Features
1. Introduction.
2. Newton's Laws of Motion.
3. Mass and Weight.
4. Momentum.
5. Force.
6. Absolute and Gravitational
Units of Force.
7. Moment of a Force.
8. Couple.
9. Centripetal and Centrifugal
Force.
10. Mass Moment of Inertia.
11. Angular Momentum or
Moment of Momentum.
12. Torque.
13. Work.
14. Power.
15. Energy.
16. Principle of Conservation of
Energy.
17. Impulse and Impulsive Force.
18. Principle of Conservation of
Momentum.
19. Energy Lost by Friction
Clutch During Engagement.
20. Torque Required to
Accelerate a Geared System.
21. Collision of Two Bodies.
22. Collision of Inelastic Bodies.
23. Collision of Elastic Bodies.
24. Loss of Kinetic Energy
During Elastic Impact.
3.1. Introduction
In the previous chapter we have discussed the
kinematics of motion, i.e. the motion without considering
the forces causing the motion. Here we shall discuss the
kinetics of motion, i.e. the motion which takes into
consideration the forces or other factors, e.g. mass or weight
of the bodies. The force and motion is governed by the three
laws of motion.
3.2. Newton’s Laws of Motion
Newton has formulated three laws of motion, which
are the basic postulates or assumptions on which the whole
system of kinetics is based. Like other scientific laws, these
are also justified as the results, so obtained, agree with the
actual observations. These three laws of motion are as
follows:
1. Newton’s First Law of Motion. It states, “Every
body continues in its state of rest or of uniform motion in
a straight line, unless acted upon by some external force.”
This is also known as Law of Inertia.
The inertia is that property of a matter, by virtue of
which a body cannot move of itself, nor change the motion
imparted to it.
Chapter 3 : Kinetics of Motion ����� 25
2. Newton’s Second Law of Motion. It states,
“The rate of change of momentum is directly
proportional to the impressed force and takes place in
the same direction in which the force acts.”
3. Newton’s Third Law of Motion. It states, “To
every action, there is always an equaland opposite
reaction.”
3.3. Mass and Weight
Sometimes much confu-sion and misunder-
standing is created, while using the various systems of units
in the measurements of force and mass. This happens
because of the lack of clear understanding of the
difference between the mass and the weight. The
following definitions of mass and weight should be
clearly understood :
1. Mass. It is the amount of matter contained in a
given body, and does not vary with the change in its
position on the earth's surface. The mass of a body is
measured by direct comparison with a standard mass by using a lever balance.
2. Weight. It is the amount of pull, which the earth exerts upon a given body. Since the pull
varies with distance of the body from the centre of the earth, therefore the weight of the body will
vary with its position on the earth’s surface (say latitude and elevation). It is thus obvious, that the
weight is a force.
The earth’s pull in metric units at sea
level and 45° latitude has been adopted as one
force unit and named as one kilogram of force.
Thus, it is a definite amount of force. But, unfor-
tunately, it has the same name as the unit of mass.
The weight of a body is measured by the use of a
spring balance which indicates the varying ten-
sion in the spring as the body is moved from place
to place.
Note: The confusion in the units of mass and weight
is eliminated, to a great extent, in S.I. units. In this system, the mass is taken in kg and force in newtons.
The relation between the mass (m) and the weight (W) of a body is
W = m.g or m = W/g
where W is in newtons, m is in kg and g is acceleration due to gravity.
3.4. Momentum
It is the total motion possessed by a body. Mathematically,
Momentum = Mass × Velocity
Let m = Mass of the body,
u = Initial velocity of the body,
v = Final velocity of the body,
a = Constant acceleration, and
t = Time required (in seconds) to change the velocity from u to v.
The above picture shows space shuttle.
All space vehicles move based on
Newton’s third laws.
26 ����� Theory of Machines
Now, initial momentum = m.u
and final momentum = m.v
∴ Change of momentum = m.v – m.u
and rate of change of momentum = 
. . ( )
.
m v m u m v u
m a
t t
− −
= = ...
v u
a
t
− 
=  ∵
3.5. Force
It is an important factor in the field of Engineering-science, which may be defined as an
agent, which produces or tends to produce, destroy or tends to destroy motion.
According to Newton’s Second Law of Motion, the applied force or impressed force is
directly proportional to the rate of change of momentum. We have discussed in Art. 3.4, that the rate
of change of momentum
= m.a
where m = Mass of the body, and
a = Acceleration of the body.
∴ Force , F ∝ m.a or F = k.m.a
where k is a constant of proportionality.
For the sake of convenience, the unit of force adopted is such that it produces a unit
acceleration to a body of unit mass.
∴ F = m.a = Mass × Acceleration
In S.I. system of units, the unit of force is called newton (briefly written as N). A newton
may be defined as the force while acting upon a mass of one kg produces an acceleration of
1 m/s2 in the direction of which it acts. Thus
 1 N = 1 kg × 1 m/s2 = 1 kg-m/s2
Note: A force equal in magnitude but opposite in direction and collinear with the impressed force producing
the acceleration, is known as inertia force. Mathematically,
Inertia force = – m.a
3.6. Absolute and Gravitational Units of Force
We have already discussed, that when a body of mass 1 kg is moving with an acceleration of
1 m/s2, the force acting on the body is one newton (briefly written as N). Therefore, when the same
body is moving with an acceleration of 9.81 m/s2, the force acting on the body is 9.81 newtons. But
we denote 1 kg mass, attracted towards the earth with an acceleration of 9.81 m/s2 as 1 kilogram-
force (briefly written as kgf) or 1 kilogram-weight (briefly written as kg-wt). It is thus obvious that
 1 kgf = 1 kg × 9.81 m/s2 = 9.81 kg-m/s2 = 9.81 N ... (∵ 1 N = 1 kg-m/s2)
The above unit of force i.e. kilogram-force (kgf ) is called gravitational or engineer's unit
applied force, F
W, weight (force)
f, friction force
N, normal force
Chapter 3 : Kinetics of Motion � 27
of force, whereas newton is the absolute or scientific or S.I. unit of force. It is thus obvious, that the
gravitational units are ‘g’ times the unit of force in the absolute or S.I. units.
It will be interesting to know that the mass of a body in absolute units is numerically equal
to the weight of the same body in gravitational units.
For example, consider a body whose mass, m = 100 kg.
∴ The force, with which it will be attracted towards the centre of the earth,
 F = m.a = m.g = 100 × 9.81 = 981 N
Now, as per definition, we know that the weight of a body is the force, by which it is attracted
towards the centre of the earth. Therefore, weight of the body,
W = 981 N = 981 / 9.81 = 100 kgf ... (∵ 1 kgf = 9.81 N)
In brief, the weight of a body of mass m kg at a place where gravitational acceleration is ‘g’
m/s2 is m.g newtons.
3.7. Moment of a Force
It is the turning effect produced by a force, on the body, on which it acts. The moment of a
force is equal to the product of the force and the perpendicular distance of the point about which the
moment is required, and the line of action of the force. Mathematically,
 Moment of a force = F × l
where F = Force acting on the body, and
l = Perpendicular distance of the
point and the line of action of
the force, as shown in Fig. 3.1.
3.8. Couple
The two equal and opposite parallel forces, whose lines of
action are different, form a couple, as shown in Fig. 3.2.
The perpendicular distance (x) between the lines of action of
two equal and opposite parallel forces (F) is known as arm of the
couple. The magnitude of the couple (i.e. moment of a couple) is
the product of one of the forces and the arm of the couple.
Mathematically,
Moment of a couple = F × x
A little consideration will show, that a couple does not produce any translatory motion (i.e.
motion in a straight line). But, a couple produces a motion of rota-
tion of the body, on which it acts.
3.9. Centripetal and Centrifugal Force
Consider a particle of mass m moving with a linear velocity
v in a circular path of radius r.
We have seen in Art. 2.19 that the centripetal acceleration,
 a
c
 = v2/r = ω2.r
and Force = Mass × Acceleration
∴ Centripetal force = Mass × Centripetal acceleration
or F
c
 = m.v2/r = m.ω2.r
Fig. 3.2. Couple.
Fig. 3.1. Moment of a force.
28 � Theory of Machines
This force acts radially inwards and is essential for circular motion.
We have discussed above that the centripetal force acts radially inwards. According to
Newton's Third Law of Motion, action and reaction are equal and opposite. Therefore, the particle
must exert a force radially outwards of equal magnitude. This force is known as centrifugal force
whose magnitude is given by
 F
c
= m.v2/r = m.ω2r
3.10. Mass Moment of Inertia
It has been established since long that a rigid body is
composed of small particles. If the mass of every particle of a
body is multiplied by the square of its perpendicular distance
from a fixed line, then the sum of these quantities(for the whole
body) is known as mass moment of inertia of the body. It is
denoted by I.
Consider a body of total mass m. Let it is composed of
small particles of masses m1, m2, m3, m4 etc. If k1, k2, k3, k4 are
the distances of these masses from a fixed line, as shown in Fig.
3.3, then the mass moment of inertia of the whole body is given
by
I = m1 (k1)2 + m2(k2)2 + m3 (k3)2 + m4 (k4)2 +....
If the total mass of body may be assumed to concentrate

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