MODULE 4 –
KINEMATICS
RECTILINEAR MOTION
Kinematics of a Particle
We have seen in the previous chapters that any force system can be reduced to a
resultant force-couple system. In statical equilibrium equations both the resultant force and the
resultant couple are zero. In dynamics the resultant force-couple system is not zero and causes a
change in the state of motion of the body on which they act.
When a body does not change its position with respect to other bodies then the body is
said to be in the state of rest. If the body changes its position with respect to other bodies, it is
said to be in the state of motion. Thus, motion is the change of position of an object with respect
to other bodies. For example, you move from your house to your site of work. Moon moves
around the earth while both together move around the sun.
Types of Motion
There are two types of motion.
(a) Linear motion
(b) Rotary motion
(a) Linear motion
When the body moves along a straight line, then motion is called linear motion. For
example, a car moving on a straight level road or a stone dropped from the top of a building are
in motion in a straight line. There are two types of linear motion. One is “rectilinear motion” and
the other is “curvilinear motion”. In rectilinear motion, body moves in a straight line and in
curvilinear motion, body moves along a curve. Rectilinear motion is of three types. One, the
motion in which a body moves with constant velocity. Second, the motion in which body moves
with non-uniform velocity but with constant acceleration. Third, the motion in which body
moves with non-uniform velocity and non-uniform acceleration.
(b) Rotary motion
When a body rotates about a centre or axis, then the motion is called as rotary motion.
Motion of electric fan is rotary motion.
Speed
The rate of change of a body is known as speed or it is the distance travelled by an
object in unit time interval.
Speed =
Distance
Time
It is scalar quantity, because it shows magnitude but does not show the direction. The
unit of speed is m/sec, km/hr, etc.
“Uniform” or “Constant speed”: When a body covers equal distance in equal interval to
time, then the speed is called as uniform speed.
Average Speed =
Total distance travelled
Time taken to cover total distance
Displacement
If the initial position of a body is P and after time t, it occupies final position Q, then the
shortest distance between P & Q in time t is called as displacement PQ in time t. Therefore, the
displacement is the shortest distance covered by a moving body. The distance covered may be
“linear” or “angular”. Linear displacement is the distance travelled by a moving body in a
straight line.
Angular displacement occurs in rotary motion.
Velocity
The velocity of a moving object is the rate of change of its displacement, in a definite
direction.
Velocity =
Displacement
Time

Mathematically,
S
t

ds
dt
It is a vector quantity because it possess both magnitude and direction.
Its unit is m/sec.
Acceleration
It is rate of change of velocity of a moving body.
Acceleration =
f 
=
Change of velocity
Time
Final velocity - Initial velocity
Time
ν-u
t
Mathematically,
f=
dv
dt
It is vector quantity, its unit is m/sec2.
Speed Time Graph
If a graph is plotted between the speed of a body and time taken y it, then such a graph
is called as speed-time graph. It is used to find the distance travelled by a moving body. The area
of graph is equal to the distance travelled in a time t.
In the graph, time is plotted on X-axis and speed is plotted on Y-axis as shown in
Figure. Figure (a) shows a speed-time graph of a body moving with uniform speed.
Figure (b) shows a speed time graph of a body moving with non-uniform speed. From
the Figure (a), we get
Distance travelled = Area of graph
= u×t
From Figure (b),
Distance travelled = Area of graph
Area of rectangle + Area of triangle
1
S  ut  (v - u ).t
2
S  ut 



1 2
ft
2
v-u

 f , (v - u )  ft 
t

1
1
Also, S  ut  vt - ut
2
2
=
1
1
 u+v 
ut + vt = 
 t = Average velocity × time
2
2
 2 
Resultant Velocity
If a body is in motion by two or more velocities then the effect produced by resultant
velocity will be same as that produced by two or more velocities.
All these velocities will be called components of their resultant.
Determination of resultant velocity:
Resultant velocity can be determined by law of parallelogram of velocities, which states
“If two velocities, on a moving body, be represented in magnitude and direction by two
adjacent sides of a parallelogram, then their resultant velocity is represented in magnitude and
direction by the diagonal of the parallelogram passing through that point.”
Let u and v be the two velocities acting on a body O at an angle  and let R is their
resultant velocity. Let  is the inclination of R with OX.
In OCD
OC2 = OD2 + DC2
R2 = (OA + AD)2 + DC2
R2 = (u  v cos ) 2  (v sin ) 2
= u2  v2 cos2   2 uv cos +v2 sin 2 
= u 2  v 2 (cos 2   sin 2 )  2 uv cos 
R2 = u 2  v2  2 uv cos 
tan  
CD
v sin 

OD u  v cos 
Special cases
(a)
or
When u = v
R=
u 2  u 2  2u 2 cos   2u 2 (1  cos )
R=
2u 2 (1  cos )  2u 2  2cos2
R = 2u cos

2

2
tan  
sin 

1  cos 


cos
2
2  tan 

2
2 cos 2
2
2sin

2


(b)
When   0o , then R = u + v
(c)
when  
(d)
when    , R =
if
u>v

,R=
2
v 2  u 2 tan  
v
u
u 2  v 2  2uv  u  v
Relative velocity
Change in velocity
Let
v1 = velocity of body at any instant
v2 = velocity of body at time ‘t’
t = time taken
Assuming that body moves in a straight line, then

Change of velocity = v2 – v1
Mean acceleration =
v2  v1
t
Assuming that body moves in a curve.
Let at any instant body is at A, having velocity v1. After time ‘t’ body reaches at B,
where its velocity becomes v2.
Draw tangent at A and tangent at B. Let  be the angle between two tangents.
Let oa and ob represented v1 and v2.
oa  ab  ob (from law of triangle)

Change of velocity = ob  oa = ab
ab makes an angle  with v1
consider  oab, ab is obtained from relation.
ab2 = oa2 +ob2 – 2 oa.ob cos 
Also applying sine rule, value of  can be obtained.
ob
ab

sin(180  ) sin 
Relative velocity
The velocity of A relative to B is defined as velocity of A minus velocity of B.
Thus if
vA = Velocity of A
vB = Velocity of B
vC = Velocity of C
Then
vA,B = vA – vB
Where
vA,B = velocity of A, relative to B
= Relative velocity of A with regard to B.
Also
vB,A = vB – vA
Where
vB,A = velocity of B, relative to A
= Relative velocity of B with regard to A.
Also
vB,C = vB – (- vC)
where
vB,C = velocity of B, relative to C.
= vB + vC
(Note that vC is in opposite direction)
Evidently the relative velocity of A with regard to B is the velocity with which A
appears to B. Consider, for example in Figure two trains A and B moving in the same direction
on parallel tracks at 60 KPH and 45 KPH respectively. To an observer  on ground, A is
moving with 60 KPM while B is going at 45 KPH. But to an observer on B, A will appear to
move at (60 – 45) = 15 KPH. The relative velocity of A with regard to B is thus 15 KPH.
The motion of bodies A and B may not be on parallel paths. It may be in different
directions. But the relative velocity of A with regard to B will always be equal to the velocity of
A minus the velocity of B.
Since the velocity is a vector quantity, the relative velocity will thus be the vector
difference of the velocities of A and B.
Rectilinear Motion under uniform acceleration
Equations of Rectilinear Motion
Let us consider a body, which is moving in a straight line.
Let
u = Initial velocity of body in m/sec.
v = Final velocity of body in m/sec.
t = Time in seconds, during which velocity changes from u to v.
S = Distance travelled by the body in time t.
f = acceleration of the body.
(a) From the definition of acceleration, we get
f 

vu
t
v = u + ft
(1)
(b) From the definition of average velocity we get
Average velocity =
uv
2
Distance covered = Average velocity × time
uv
S= 
t
 2 
 u  u  ft 
=
t
2


S = ut 
1 2
ft
2
(c) From equation,
t
v u
(from definition of acceleration)
f
Substituting the value of ‘t’ in equation
 v u  1  v u 
S = u
 f 

 f  2  f 
 u 1  v  u 
= (v  u )   

 f 2  f 
 2u  v  u 
= (v  u ) 

 2f

 v u 
=
 v  u 
 2f 
2
(2)
 v2  u 2 
= 

 2f 
v 2  u 2  2 fs

(3)
In equations 1, 2, 3, f is the linear acceleration. If instead of acceleration, the retardation
is given, then the value of ‘f’ in these equations should be taken negative. Retardation is negative
acceleration.
Derivation of Equation of Rectilinear Motion from Integration
dv d  ds  d 2 s 
ds 
f 
   2  v 
dt dt  dt  dt 
dt 
We know

d 2s
 f
dt 2
or
d  ds 
  f
dt  dt 
or
 ds 
d    f .dt
 dt 
Integrating the above equation.
ds
= ft + C1 where C1 = constant of integration
dt
When

t = 0, then velocity is u
C1 = u
ds
 ft  u
dt
Hence
Integrating
s
ft 2
 ut  C2 ,
2
where C2 = constant of integration
when

t = 0, then distance travelled s = 0
C2 = 0
1 2
ft
2
Hence
s  ut 
Again
ds
 ft  u
dt
or
v = u + ft
...(i)
...(ii)
dv dv ds dv
 .  ,v
dt ds dt ds
Also,
f 

f  v.
or
v.dv  f .ds
Integrating,
v2
 fs  C3 where C3 = constant of integration
2
when
s = 0, then velocity = u  C3 
dv
ds
u2
2
Substituting the value of C3,
v2
u2
 fs 
2
2
v 2  u 2  2 fs
...(iii)
Distance tavelled in the nth second
Let
u = Initial velocity of a body
f = Acceleration
Sn = Distance covered in n seconds
Sn–1 = Distance covered in (n–1) seconds
The distance travelled in the nth second
= Distance travelled in n seconds – Distance travelled in (n – 1) seconds
= Sn – Sn–1
Distance travelled in n seconds is obtained by substituting t=n in equation
1
S  ut ft 2
2
1

Sn  u.n fn 2
2
1
Similarly,
Sn1  u (n  1) f (n  1)2
2
1
[By putting t=(n–1) in equation S  ut ft 2 ]
2
 Distance travelled in the nth second = Sn  Sn1
1
1
= un  fn 2  [u (n  1)  f (n  1) 2 ]
2
2
1
1
= un  fn 2  [un  u  f (n 2  2n  1)]
2
2
1 2
1 2
f
= un  fn  [un  u  fn  fn  )]
2
2
2
=
u
f
(2n  1)
2
Rectilinear Motion under Gravity
Acceleration Due to Gravity
When a body falls freely, then its velocity increases as it approaches the earth. The
increase in the velocity of a falling body is due to “acceleration due to gravity”, i.g. ‘g’. This fact
was first established by Sir Issac Newton. The value of g changes from place to place. The value
of ‘g’ is taken as 9.81 m/sec2 in S.I. units, 981 cm/sec2 in CGS units and 32.2 ft/sec2 in FPS units,
unless otherwise specified.
Equation of Motion of Bodies Falling Vertically Downwards
In previous articles, we have developed “equations of motion” for rectilinear motion
and used ‘f’ as acceleration. Now when we consider motion under gravity then this ‘f’ will be
replaced by ‘g’ in the equation of motion. Hence desired equations of motion under gravity will
be
v=u+gt
h = ut + ½ g t2
v2 = u2 + 2g h
Distance travelled in nth second = u 
g
(2n  1)
2
Special case:When a body falls freely from a height, then u = 0 and above equations will be then,
v=gt
h = ½ g t2
v2 = 2g h
1
Distance travelled in nth second = g (2n  1)
2
Equation of Motion of Bodies Thrown Vertically Upwards
When the bodies are thrown in vertically upward direction, then ‘g’ becomes a retarding
quantity and due to this the velocity of the body decreases gradually till it becomes zero. After
attaining a certain height it again starts falling down with ‘+g’. Hence equations of motion of
bodies when thrown in vertically upward direction are:
v=u–gt
h = ut – ½ g t2
v2 = u2 – 2g h
1
Distance travelled in nth second = u  g (2n  1)
2
Important Results: Some of the important results are obtained from the above mentioned
equations, which are as follows:(a) Greatest height attained:
Suppose that maximum height = H, where v = 0, then from equ. 22
0 = u2 – 2gH
u2 = 2gH
u2
H=
2g
(b) Time taken in attaining maximum height:
Suppose time T is required in attaining maximum height then at that time v=0.
From eq. 12.20, we get
0 = u – gT
u

T
g
(c) Time taken in attaining a given height:
1
From eq. 12.21, h  ut  gt 2
2
2
gt  2ut  2h  0
or
u  u 2  2 gh

t
g
When u2 > 2gh, then there will be two real roots of t. Smaller value of t represents the
time taken in ascending to a given height and larger value of t represents the time taken for
attaining maximum height & coming back to given height.
If a body is thrown from pt. O, maximum height is at A, given height is at pt.P, then
Smaller value of t = time taken in covering distance O to P
Larger value of t = time taken in covering distance O to A +
time taken to reach from A to P.
(d) Velocity at a given height:
From (c),
v2 = u2 – 2gh
v  u 2  2 gh
EXAMPLES
Ex.1. A body is moving with a velocity of 2m/sec. After 4 seconds the velocity of the body
becomes 6 m/sec. Find the acceleration of the body.
Sol.
Given
Initial velocity, u
Final velocity, v
Time, t
Let f
Using expression
v
or
6
or
f
=
=
=
=
=
=
=
2 m/sec
6 m/sec
4 sec
acceleration,
u + ft
2+f×4
1 m/sec2
Ans.
Ex.2. A boat crosses a river with a velocity of 3 km/hour. If the velocity of current is 4
km/hour, find the resultant velocity of the boat.
Sol.
Given
Velocity of current OA
Velocity of Boat OC
Let resultant velocity of Boat
Hence, from fig.
OB2
v2
v
= 4 Km/hr
= 3 Km/hr
= v Km/hr
= OC2 + BC2
= 32 + 42
= 5 Km/hr.
Ex.3. A car is moving with a velocity of 15 m/sec. The car is brought to rest by applying breaks
in 5 seconds. Determine (a) the retardation, (b) Distance travelled by car after applying
brakes.
Sol.
(a)

(b)
Given
Initial velocity, u
Final velocity, v
Time, t
Let f
Using equation v
or
0

f
Let S
=
=
=
=
=
=
=
=
Using equation S
=
=
=
15 m/sec
0 (car is brought to rest)
5 sec
acceleration of car
u + ft
15 + f × 5
–3 m/sec2 (– sign is for retardation)
Ans.
Distance travelled by car after applying brakes.
1
ut  ft 2
2
15 × 5 + ½(–3) × 52
37.5 m
Ans.
Ex.4. A body is subjected to four velocities of magnitude 4, 3, 2 and 1 m/sec. If the angle
between first and second velocity is 30o, between second and third is 90o and between third and
fourth is 120o, find the resultant velocity.
Sol.
Let, V
=
Resultant velocity,

=
angle of inclination of V with OX
Resolving all velocities in horizontal and vertical direction.
V cos 
=
4 + 3 cos 30o + 2 cos 120 o + 1 cos 240 o
=
4 + 2.598 – 1 – 0.5
=
5.098
...(i)
o
o
o
V sin 
=
3 sin 30 + 2 sin 120 + 1 sin 240
=
1.5 + 1.732 – 0.866
=
2.365
...(ii)
Squaring and adding equation (i) & (ii)
v2
=
5.597 + 25.989 = 31.586
V
=
5.62 m/sec
tan 
=
2.365
 0.4639
5.098

=
24.88o

Ans.
Ans.
CIRCULAR MOTION
Introduction
When a body moves along a circular path, then such a motion is known as circular
motion. In such motion centre of rotation remains fixed. The common examples of circular
motion are, shafts, flywheels, pulleys etc. rotating about their axis of rotation.
Angular Velocity
In circular motion, the displacement of body is known as angular displacement. Angular
displacement is always measured in terms of angle covered by the body from its initial position.
Hence the rate of change of angular displacement is known as angular velocity of a body, and is
represented by w.
Let a body move along a circular path.
Let the initial position of body be at A, and after time ‘t’, the body becomes at B,
covering an angle AOB=
Then,  = angular displacement
Angular displacement in radians
Time in seconds


=
rad/second
t
d
Mathematically,  =
dt

V
=
5.62 m/sec
Ans.
Relation between Linear Velocity & Angular Velocity
From Fig. 1,


=
t
If
V
=
Linear velocity, then
Linear displacement
V
=
Time
Here, linear displacement =
Arc AB
=
r.
( Arc = radius × angle)
r

V
=
t

V
=
r
t
V
=
r × 

(   = angular velocity)
t
Hence,
Linear velocity =
radius × angular velocity

Angular velocity
=
...(1)
...(2)
...(3)
Angular Acceleration
The rate of change of angular velocity is known as angular acceleration. Its unit is
radian/sec2, i.e. rad/sec2. It is represented by .


=
rate of change of angular velocity
d d  d  
d
=
( 
)
= 

dt
dt dt dt 
=
d2 
d t2
...(4)
Also,

d
dt
=

=
d d d
×


d dt d
d

d
...(5)
...(6)
Relation between Linear Acceleration and Angular Acceleration

V
Differentiating w.r.t. ‘t’
dV

dt
dV
But,
dt
d
and
dt
Hence,
f
=
r
=
r.
=
linear acceleration = f
=
angular acceleration = 
=
r.
d
dt
...(7)
Equations of Motion along a Circular Path
Consider that a body moves in circular path.
Let
0
=
initial angular velocity in rad/sec.

=
final angular velocity in rad/sec.
t
=
time (in seconds), taken in traversing angle 

=
angular displacement.

=
angular acceleration in rad/sec2
We know that

=
rate of change of angular velocity
Change in angular velocity
=
Time taken
Final angular velocity – Initial angular velocity
=
Time
  0
=
t
or
t
=
 – 0


=
0 +  . t
...(8)
We know also that
Angular displacement, 
=
Average angular velocity × time
 0   
=
 2 t


0  (0   t )


=
(   0  t )
t
2
 20   t 
=

t
2


1
 t2
2
0  t
From equation (8),

=
w  w0
we get
t
=

Substituting this value in equation (9),

=
0t 
...(9)
2
   0  1    0 
  0 
  

   2   
1    0 
   0  
=
0  . 


2    
  
  0 
   0  
=
0 


2 
  
   0   20    0 
=


2
  

2
2
   0    0     0
=



2
   2 
2  20  2
...(10)
Hence,
Hence equations are:In linear Motion
v = u + ft
1
S  ut  ft 2
2
v2 = u2 + 2 fs
Linear displacement in nth second
 2n  1 
= u 
f
 2 
In angular motion
  0  t
1
  0t  t 2
2
2
2
  0  2
Angular displacement in nth second
 2n  1 
= 0  

 2 
...(11)
Rolling of wheel without sliding
In the figure a wheel rolls on a horizontal surface. Consider any point P on the periphery
of wheel. The point P will be under the action of two motion, one is linear in horizontal direction
and other is tangential.
Let
v = linear velocity of the centre of the wheel, and hence
linear velocity of point P.
 = angular velocity of the wheel.
r = radius of wheel.
Considering motion of P.
(a) Point P will have velocity v in horizontal direction.
(b) Point P will have tangential velocity v such that
v = r
The resultant velocity vr is
vr  v2  v2  2v2 cos 
=
2v 2  2v 2 cos 
=
2v2 (1  cos )
=
2v 2  2cos2


 2v cos
2
2
( cos A  2 cos 2
A
1 )
2

Since vr  2v cos , therefore at point B,   0 and hence vr  2v , at peak point B on
2
the wheel.
Also at point A,   180o , therefore,
vr  2v cos90o  0
Centripetal Force
When a body moves along a circle, then tendency of body is to move in tangential
direction due to inertial force. The inertial force is counter acted by a force known as centripetal
force to keep the body in circular motion. The centripetal force always acts radially towards the
centre of the circle. Hence centripetal force is that force which acts along the radius of the circle
at every point and is always directed towards the centre of the circle.
Centrifugal Force
As per Newton’s third law, to every action there is always equal and opposite reaction.
Therefore, the force which acts radially outward and equal in magnitude of centripetal force is
known as centrifugal force. Hence, centrifugal force is equal and opposite to centripetal force,
and the tendency of the centrifugal force is to throw the body away from the centre of circle.
Normal Acceleration
When a body moves uniformly along a circular path,
is subjected to normal acceleration. Let at any instant body is
A and its tangential velocity is v along AM. Let the body come
B in time ‘t’ traversing an angle  at ‘O’ the centre of the
circular path. At pt. B, tangential velocity is again v because
moves with uniform velocity. Resolving the tangential velocity
direction along AM and AO, we get v cos  and v sin 
components respectively. Now change in velocity in the body
the direction AM.
= v cos  – v (since  is very small,  sin  =  and cos  = 1)
= v–v
= 0
Hence, rate of change of velocity in direction AM
i.e. acceleration along AM is zero.
Also change in velocity in the body along the direction AO
= v sin  – 0 (since at pt A velocity along AO is zero.)
= v
Hence, rate of change of velocity in direction AO i.e. acceleration along AO
=
v
t
= v.




 
t
= v.2



v



v
 
r
v
r
=
v.
=
v2
r
t 

r 
then it
at pt.
at pt.
body
v
in
along
= angular acceleration = 

=
v2
r
= r . 2
= centripetal acceleration
(14)
Centrifugal Force Acting on a Body Moving along a Circular Path
When a body moves along the path of a circle then it is always acted upon by a
centrifugal force in radially outward direction. Thus the tendency of this force is to throw the
body outward. Its magnitude can be determined from Newton’s second law.
Let
m
=
mass of the body
r
=
radius of circle
v
=
constant velocity of the body

=
angular acceleration

=
v2
from art 14.11
r
Centrifugal force
=
mass × centrifugal acceleration
P
=
m.
P
=
m r 2
Then,
or
v2
r
(15)
(16)
Application of Centrifugal Force and Centripetal Force
(1) Tension in the string attached to a body rotating in a horizontal
plane.
Let a body of mass m is attached to one end of a string of
‘r’ and other end is attached to a fixed point. If the body now
in a circular path in a horizontal plane about the fixed point as
of circle then radius of circle will be equal to the length of the
length
rotates
centre
string.
Let T be the tension in the string, due to the centrifugal
force
in string.
T
=
m r 2 = centrifugal tension
T
=
m.
v2
r
(17)
where v = velocity,  = angular velocity
Thus, the tension in the string remains the same.
(2) Tension in the string attached to a body rotating in a vertical
plane.
Let one end of a string be attached to a fixed point O, and the other end be attached to a
body of mass ‘m’ rotating in a vertical plane. When the body comes at lowest point A, then body
is acted upon by centrifugal force and the weight (mg). Hence tension (T1) in the string is
T1
v2
 mg  mr 2  mg
r
=
m
=
m(r 2  g )
(18)
When the body comes to point B, i.e., the highest point of circle, then body is acted
mv 2
upon by centrifugal force
and weight (mg).
r
Hence tension (T2) in the string, when body is at B,
is
T2
=
mv 2
 mg
r
=
mr 2  mg
=
m(r 2  g )
(19)
T2 is therefore centripetal tension in the string.
Hence: (a) When body is at A, there will be maximum tension in the string.
(b) When body is at B, there will be minimum tension in the string.
Rotation and Translation
In the figure, AB is a link. At any instant it
occupies a new position A1 B1.
It means AB has translatory motion in
occupying new position A1B which is parallel to AB
then rotary motion with centre as A to come to
position A1B1. Thus, the motion is combined
translation and rotation. Similarly, we can also
consider the motion as first rotary as AB and the translatory as
The combined motion of translation and rotation can be
assumed to be a motion of entirely rotation about a certain point.
point is known as instantaneous centre or rotation. The position
instantaneous centre of rotation can be calculated by considering
next figure. Here the link AB after certain interval of time changes
position to A1B1. Join AA1 and BB1. Draw perpendicular bisectors
and M2 respectively. Now extend these perpendicular bisectors to
a point O. This point O is called as instantaneous centre of rotation
link AB. This means that link AB as a whole has rotated about
O.
Let
vA
=
Linear velocity of point A
and
A1B1.
This
of this
the
its
at M1
cut at
of the
point
vB
=
Linear velocity of point B

=
Angular velocity of link AB about centre O.
Therefore, the angular velocity of point A and angular velocity of point B about O is .
We know from equation No.3, that
Linear velocity
=
radium × angular velocity
v
=
r.w

vA
=
OA.
and

=
vA
OA
Also
vB
=
OB.
or

=
vB
OB
.....(i)
....(ii)
Equating equations (i) and (ii), we have
vA
OA
=
vB
OB
vA
vB
=
OA
OB
....(20)
Here vA is in the direction perpendicular to OA and vB is in the direction perpendicular
to OB.
Reciprocating Engine Mechanism
In the following figure, a mechanism of reciprocating engine is shown. AB is
connecting rod connected with crank BC and piston rod. The crank BC is rotating with angular
velocity  in clockwise direction. The point A is having a forward and backward motion in the
horizontal plane. Connecting rod AB is having a combined motion of rotation and translation.
Let
N
=
Revolution of crank in R.P.M.
Here,

=
angular velocity of crank BC
r
=
radius of BC, crank
L
=
Length of connecting rod AB.

=
2n
60
vB
=
r
....(i)
It is in the direction perpendicular to BC at point B, as shown in the above figure. The
velocity at A, vA is in the horizontal direction along AC for a given position of the crank. As the
direction of vA and vB are known, the position of instantaneous centre can be located by drawing
perpendicular to the direction of vA and vB. These perpendicular intersect at point O, which is
instantaneous centre for the connecting rod AB.
Let
and
0
=
angular velocity of connecting rod AB
about instantaneous centre O.
vA
=
AO.0
....(ii)
vB
=
OB.0
....(iii)
Equating (i) and (iii), we have

r.
=
OB.0
0
=
r
OB
....(21)
Here, the values of r and are known OB, is measured to scale.
We can get vA from equation (ii)
vA

.
vA
=
AO.0
=
AO.
=
AO
.r.
OB
r.
OB



0 
r. 

OB 
....(22)
Therefore, vA can be determined by measuring AO and OB and knowing values of r and
The velocity vA can also be determined as given below.
Consider figure below. Through C, erect a perpendicular CM. From B drop a
perpendicular BN on AC.
From the geometry of the figure.
AOB and BCM are similar.

OA
OB
=
MC
BC
=
MC
r
tan 
=
MC
AC
MC
=
AC tan 
=
(AN + NC) tan 
=
(AB cos  + BC cos ) tan 
=
(L cos  + r cos ) tan 
=
L cos  tan  + r cos  tan 
=
L cos .
=
L.sin  + r cos  tan 
....(i)
In the ACM,

sin 
 r cos  tan 
cos 
Substituting the value of MC in equation (i)
we get,
OA
OB
=
Substituting this values of
vA
vA
1
( L sin   r cos  tan )
r
OA
in equation no.22, we get
OB
=
OA
.r.
OB
=
1
(L sin   r cos  tan ).r.
r
=
(L sin   r cos  tan )
....(23)
EXAMPLES
Ex. 1. A body is rotating with an angular velocity of 10 radian/sec. After 4 seconds, the
angular velocity of the body becomes 26 radians per second. Determine the angular
acceleration of the body.
Sol.
Let initial angular velocity
Final angular velocity
= 0 = 10 rad/sec
=  = 26 rad/sec
Time, t
Angular acceleration
Using relation,

26

= 4 sec
=  rad/sec2
= 0 + t
= 10 +  × 4
=
26  10 16
= 4 rad/sec2 Ans.

4
4
Ex. 2. A wheel rotating about a fixed axis at 40 r.p.m. is uniformly accelerated for 70 secs,
during which time it makes 100 revolutions. Find (a) Angular velocity at the end of this
interval, and (b) time required for the speed to reach 200 r.p.m.
Sol. Let
Initial r.p.m. of wheel, N0
Time t
No. of revolution in 70 sec
Angular displacement 
= 40
= 70 sec
= 100
= angular displacement
× no. of revolution
in
one
= 2 × 100 = 200 rad

=
2N
60
 Angular initial velocity, 0
=
2N 0
60
=
2 40 4
rad/sec

60
3
Using relation,
Let
Using relation,
or
Now from the relation

= angular velocity at the end of 70 sec

= Angular acceleration

=
1
0t  t 2
2
200
=
4
1
 70    702
3
2
200
= 93.34 + 2450 
2450
= 200 – 93.34 
106.66 
= 0.1368 rad/sec2
2450

=

= 0 + t
revolution

we have
=
4
 0.1368  70
3
=
4
 9.5760
3
= 4.1888 + 9.5760
= 13.7648 rad/sec.
Now final speed
N
= 200 r.p.m.
Final angular velocity,

=
Ans.
2N
2 200
=
60
60
= 20.944 rad/sec
Using relation


=
0  t
20.944
=
4
 0.1368  t
3
20.944
= 4.1888 + 0.1368 t
0.1368 t
= 16.7552
t
= 122.479
= 2 minutes 2.48 sec.
Ans.
Ex. 3. A wheel is rotating about its axis with a constant angular acceleration of 2 rad/sec2. If
the initial and final angular acceleration are 10.50 rad/sec and 21 rad/sec, find the total angle
turned through during the time interval this change of angular velocity took place.
Sol. Let
Constant angular acceleration 
Initial angular velocity, 0
Final angular velocity, 
Total angle turned
= 2 rad/sec2
= 10.50 rad/sec
= 21 rad/sec
= 
Using the relation,
2
212
4

= 02 + 2
= 10.502 + 2 × 2
= 212 – 10.502 = 441 – 110.25
= 82.6875 rad.
Ans.
Ex. 4. (a) A flywheel starts rotating from rest and is given an acceleration of 2 rad/sec2. Find
the angular velocity and speed in r.p.m. after 3 minutes.
(b) If the flywheel is brought to rest with a uniform angular retardation of 1 rad/sec2,
find the time taken by the flywheel in seconds to come to rest.
Sol. Let
Angular initial velocity, 0
(a)
Angular acceleration, 
Time t
Using the relation,

= 0
= 2 rad/sec2
= 3 minutes, t = 180 sec
= 0 +  t
= 0 + 2 × 180
= 360 rad/sec Ans.
(b)
Speed in r.p.m. is given by, 
=
2N
60

=
60 60  360

 3437.747
2
2
N
Let angular initial velocity 0
Final angular velocity, 
Angular retardation, 
Ans.
= 360 rad/sec
= 0
= 1.0 rad/sec2
Let the time taken by the flywheel in coming to rest = t1
Using the relation,

= 0 +  t
0
= 360 – 1 × t1
t1
= 360 sec
= 6 minutes.
(since  = –1.0 rad/sec2)
Ans.