Circular motion

Mukesh Tekwani
mukeshtekwani@outlook.com
1
Prof. Mukesh N Tekwani

Circular Motion
2

Why study Circular Motion?
• To understand
– Motion of planets
– Motion of electrons around the nucleus
– Motion of giant wheel
– Motion of space stations
– Motion of moon and satellites
Prof. Mukesh N Tekwani 3

Circular Motion
• It is defined as the motion of a particle along a complete circle
or part of a circle.
• For circular motion, it is NOT necessary that the body should
complete a full circle.
• Even motion along arc of a circle is circular motion

5
Circular Motion
X
θ = 0
y
i
f

How do we locate something on a circle?
Give its angular position θ
What is the location of
900 or π/2

6
Circular Motion
 is the angular position.
Angular displacement:
i
f 

 


Note: angles measured Clockwise (CW) are negative and
angles measured (CCW) are positive.  is measured in
radians.
2 radians = 360 = 1 revolution
x
y
i
f


Angular Displacement
• Angular displacement is defined as the angle described by the
radius vector
a
Initial position of particle is a
Final position of particle is b
Angular displacement in time t is Θ

Relation Between Angular Displacement & Linear
Displacement
S =r Θ

9
x
y
i
f

r
arclength = s = r
r
s


 is a ratio of two lengths;
it is a dimensionless ratio!
This is a radian measure of
angle
If we go all the way round s
=2πr and Δθ =2 π

Right Hand Rule
If the fingers of the
right hand are curled
in the direction of
revolution of the
particle, then the
outstretched thumb
gives the direction of
the angular
displacement vector.

Stop & Think
Are the following motions same or different?
1. The motion of the tip of second hand of a
clock.
2. The motion of the entire second hand of a
clock.
The motion of the tip of second hand of a clock
is uniform circular motion.
The motion of the entire second hand is a
rotational motion.

Centripetal Force
• UCM is an accelerated motion. Why?
• UCM is accelerated motion because the velocity of
the body changes at every instant (i.e. every
moment)
• But, according to Newton’s Second Law, there must
be a force to produce this acceleration.
• This force is called the centripetal force.
• Therefore, Centripetal force is required for circular
motion. No centripetal force means no circular
motion.
25

Linear Velocity and Speed in UCM
Is speed changing?
No, speed is constant
Is velocity changing?
Yes, velocity is changing
because velocity is a
vector – and direction is
changing at every point
26

Velocity and Speed in UCM
Is speed changing?
No, speed is constant
Is velocity changing?
Yes, velocity is changing
because velocity is a
vector – and direction is
changing at every point
27

Examples of Centripetal force
A body tied to a string
and whirled in a
horizontal circle – CPF
is provided by the
tension in the string.
28

• For a car travelling
around a circular
road with uniform
speed, the CPF is
provided by the
force of static
friction between
tyres of the car
and the road.
29

• In case of electrons revolving around the nucleus, the
centripetal force is provided by the electrostatic force of
attraction between the nucleus and the electrons
• In case of the motion of moon around the earth, the CPF is
provided by the ______ force between Earth and Moon
30

Centripetal Force
• Centripetal force
– It is the force acting on a particle performing UCM and
this force is along the radius of the circle and directed
towards the centre of the circle.
REMEMBER!
Centripetal force
- acting on a particle performing UCM
- along the radius
- acting towards the centre of the circle.
31

Properties of Centripetal Force
1. Centripetal force is a real force
2. CPF is necessary for maintaining UCM.
3. CPF acts along the radius of the circle
4. CPF is directed towards center of the circle.
5. CPF does not do any work
6. F = mv2/ r
32

Radial Acceleration
P(x, y)
v
O X
Y
M
N
r
θ
x
y
Let P be the position of the particle
performing UCM
r is the radius vector
Θ = ωt . This is the angular displacement of
the particle in time t secs
V is the tangential velocity of the particle at
point P.
Draw PM ┴ OX
The angular displacement of the particle in
time t secs is
LMOP = Θ = ωt
33

Radial Acceleration
P(x, y)
v
O X
Y
M
N
r
θ
x
y
The position vector of the particle at
any time is given by:
r = ix + jy ……by composition of vectors
From ∆POM
sin θ = PM/OP
∴ sin θ = y / r
∴y = r sin θ
But θ = ωt
∴ y = r sin ωt
34

Radial Acceleration
P(x, y)
v
O X
Y
M
N
r
θ
x
y
Similarly,
From ∆POM
cos θ = OM/OP
∴ cos θ = x / r
∴ x = r cos θ
But θ = ωt
∴ x = r cos ωt
35

Radial Acceleration
36
The velocity of particle at any instant (any time) is called its
instantaneous velocity.
The instantaneous velocity is given by
v = dr / dt
∴ v = d/dt [ ir cos wt + jr sin wt]
∴ v = - i r w sin wt + j r w cos wt

Radial Acceleration
37
The linear acceleration of the particle at any instant (any time) is
called its instantaneous linear acceleration.

Radial Acceleration
38
Therefore, the instantaneous linear
acceleration is given by
∴ a = - w2 r
Importance of the negative sign:
The negative sign in the above
equation indicates that the linear
acceleration of the particle and the
radius vector are in opposite
directions.
a
r

Relation Between Angular
Acceleration and Linear Acceleration
39
The acceleration of a particle
is given by
………………. (1)
But v = r w
∴
∴ a = .... (2)
∵ r is a constant radius,
∴
But
α is the angular acceleration
∴ a = r α ………………………(3)

Relation Between Angular
Acceleration and Linear Acceleration
40
v = w x r
Differentiating w.r.t. time t,
But
and
∴
∴linear acceleration
a = aT + aR
aT is called the tangential component of
linear acceleration
aR is called the radial component of
linear acceleration
For UCM, w = constant, so
∴ a = aR
∴ in UCM, linear accln is centripetal accln

Centrifugal Force
41
1. Centrifugal force is an imaginary force (pseudo force)
experienced only in non-inertial frames of reference.
2. This force is necessary in order to explain Newton’s laws of
motion in an accelerated frame of reference.
3. Centrifugal force is acts along the radius but is directed away
from the centre of the circle.
4. Direction of centrifugal force is always opposite to that of the
centripetal force.
5. Centrifugal force
6. Centrifugal force is always present in rotating bodies

Examples of Centrifugal Force
42
1. When a car in motion takes a sudden turn towards left,
passengers in the car experience an outward push to the right.
This is due to the centrifugal force acting on the passengers.
2. The children sitting in a merry-go-round experience an
outward force as the merry-go-round rotates about the
vertical axis.
Centripetal and Centrifugal forces DONOT constitute an action-
reaction pair. Centrifugal force is not a real force. For action-
reaction pair, both forces must be real.

Banking of Roads
43
1. When a car is moving along a curved road, it is performing
circular motion. For circular motion it is not necessary that the
car should complete a full circle; an arc of a circle is also
treated as a circular path.
2. We know that centripetal force (CPF) is necessary for circular
motion. If CPF is not present, the car cannot travel along a
circular path and will instead travel along a tangential path.

Banking of Roads
44
3. The centripetal force for circular motion of the car can be
provided in two ways:
• Frictional force between the tyres of the car and the road.
• Banking of Roads

Friction between Tyres and Road
45
The centripetal force for circular motion of the car is provided by
the frictional force between the tyres of the car and the road.
Let m = mass of the car
V = speed of the car, and
r = radius of the curved road.
Since centripetal force is provided by the frictional force,
CPF = frictional force (“provided by” means “equal to” )
(µ is coefficient of friction between tyres & road)
So and

46
Thus, the maximum velocity with which a car can safely travel
along a curved road is given by
If the speed of the car increases beyond this value, the car will be
thrown off (skid).
If the car has to move at a higher speed, the frictional force
should be increased. But this causes wear and tear of tyres.
The frictional force is not reliable as it can decrease on wet roads
So we cannot rely on frictional force to provide the centripetal
force for circular motion.

47
R1 and R2 are reaction forces due
to the tyres
mg is the weight of the car,
acting vertically downwards
F1 and F2 are the frictional forces
between the tyres and the road.
These frictional forces act
towards the centre of the
circular path and provide the
necessary centripetal force.
Center of
circular
path

48

Friction between Tyres and Road –
Car Skidding
49

Banked Roads
50
What is banking of roads?
The process of raising the
outer edge of a road over
the inner edge through a
certain angle is known as
banking of road.

Banking of Roads
51
Purpose of Banking of Roads:
Banking of roads is done:
1. To provide the necessary centripetal force for
circular motion
2. To reduce wear and tear of tyres due to friction
3. To avoid skidding
4. To avoid overturning of vehicles

Banked Roads
52

Banked Roads
53
What is angle of
banking?
R
Θ
R cos θ
Θ
W = mg
R sin θ
The angle made by the
surface of the road with
the horizontal surface is
called as angle of banking.
Horizontal

Banked Roads
54
Consider a car moving
along a banked road.
Let
m = mass of the car
V = speed of the car
θ is angle of banking
R
Θ
R cos θ
Θ
W = mg
R sin θ

Banked Roads
55
The forces acting on the
car are:
(i) Its weight mg acting
vertically downwards.
(ii) The normal
reaction R acting
perpendicular to the
surface of the road.
R
Θ
R cos θ
Θ
W = mg
R sin θ

Banked Roads
56
The normal reaction can be
resolved (broken up) into
two components:
1. R cosθ is the vertical
component
2. R sinθ is the horizontal
component
R
Θ
R cos θ
Θ
W = mg
R sin θ

Banked Roads
57
Since the vehicle has no
vertical motion, the weight
is balanced by the vertical
component
R cosθ = mg …………… (1)
R cosθ and mg form a
action-reaction pair.
(weight is balanced by
vertical component means
weight is equal to vertical
component)
R
Θ
R cos θ
Θ
W = mg
R sin θ

Banked Roads
58
The horizontal component
is the unbalanced
component . This horizontal
component acts towards
the centre of the circular
path.
This component provides
the centripetal force for
circular motion
R sinθ = …………… (2)
R
Θ
R cos θ
Θ
W = mg
R sin θ

Banked Roads
59
Dividing (2) by (1), we get
R sinθ = mg
R cos θ
θ = tan-1 ( )
So,
tan θ =
Therefore, the angle of banking
is independent of the mass of
the vehicle.
The maximum speed with
which the vehicle can safely
travel along the curved road is

Banked Roads
60
F = ma
mv
r

2
Smaller radius: larger centripetal
force is required to keep it in
uniform circular motion.
A car travels at a constant speed
around two curves. Where is the
car most likely to skid? Why?

Conical Pendulum
61
Definition:
A conical pendulum is a
simple pendulum which is
given a motion so that the
bob describes a horizontal
circle and the string
describes a cone.

Conical Pendulum
62
Definition:
A conical pendulum is a
simple pendulum which is
given such a motion that
the bob describes a
horizontal circle and the
string describes a cone.

Conical Pendulum
63
Consider a bob of mass m revolving
in a horizontal circle of radius r.
Let
v = linear velocity of the bob
h = height
T = tension in the string
Θ = semi vertical angle of the cone
g = acceleration due to gravity
l = length of the string
T cos θ
T sin θ
θ

Conical Pendulum
64
The forces acting on the bob at
position A are:
1) Weight of the bob acting
vertically downward
2) Tension T acting along the
string.
T cos θ
T sin θ
θ

Conical Pendulum
65
The tension T in the string can be
resolved (broken up) into 2
components as follows:
i) Tcosθ acting vertically
upwards. This force is balanced
by the weight of the bob
T cos θ = mg ……………………..(1)
T cos θ
T sin θ
θ

Conical Pendulum
66
(ii) T sinθ acting along the
radius of the circle and directed
towards the centre of the circle
T sinθ provides the necessary
centripetal force for circular
motion.
∴ T sinθ = ……….(2)
Dividing (2) by (1) we get,
………………….(3)
T cos θ
T sin θ
θ

Conical Pendulum
67
T cos θ
T sin θ
θ
This equation gives the speed
of the bob.
But v = rw
∴ rw =
Squaring both sides, we get

Conical Pendulum
68
T cos θ
T sin θ
θ
From diagram, tan θ = r / h
∴ r 2w2 = rg

Conical Pendulum
69
T cos θ
T sin θ
θ
Periodic Time of Conical Pendulum
But
Solving this & substituting sin θ = r/l
we get,

Conical Pendulum
70
T cos θ
T sin θ
θ
Factors affecting time period of
conical pendulum:
The period of the conical pendulum
depends on the following factors:
i) Length of the pendulum
ii) Angle of inclination to the
vertical
iii) Acceleration due to gravity at
the given place
Time period is independent of the
mass of the bob

Vertical Circular Motion Due to
Earth’s Gravitation
71
Consider an object
of mass m tied to
the end of an
inextensible string
and whirled in a
vertical circle of
radius r.
mg
T1
O
r
v1
A
T2
B v2
v3
C

72
mg
T1
O
r
v1
A
T2
B v2
v3
C
Highest Point A:
Let the velocity be v1
object at A (highest point)
are:
1. Tension T1 acting in
downward direction
2. Weight mg acting in
downward direction

73
mg
T1
O
r
v1
A
T2
B v2
v3
C
At the highest point A:
The centripetal force acting on
the object at A is provided
partly by weight and partly by
tension in the string:
…… (1)

74
mg
T1
O
r
v1
A
T2
B v2
v3
C
Lowest Point B:
Let the velocity be v2
object at B (lowest point) are:
1. Tension T2 acting in
upward direction
2. Weight mg acting in
downward direction
mg

75
mg
T1
O
r
v1
A
T2
B v2
v3
C
At the lowest point B:
…… (2)

76
mg
T1
O
r
v1
A
T2
B v2
v3
C
Linear velocity of object at highest
point A:
The object must have a certain
minimum velocity at point A so as to
continue in circular path.
This velocity is called the critical
velocity. Below the critical velocity,
the string becomes slack and the
tension T1 disappears (T1 = 0)

77
mg
T1
O
r
v1
A
T2
B v2
v3
C
point A:

78
mg
T1
O
r
v1
A
T2
B v2
v3
C
point A:
This is the minimum velocity that
the object must have at the
highest point A so that the string
does not become slack.
If the velocity at the highest point
is less than this, the object can not
continue in circular orbit and the
string will become slack.

79
mg
T1
O
r
v1
A
T2
B v2
v3
C
Linear velocity of object at lowest
point B:
When the object moves from the lowest
position to the highest position, the
increase in potential energy is mg x 2r
By the law of conservation of energy,
KEA + PEA = KEB + PEB

80
mg
T1
O
r
v1
A
T2
B v2
v3
C
point B:
At the highest point A, the minimum
velocity must be
Using this in
we get,

81
mg
T1
O
r
v1
A
T2
B v2
v3
C
point B:
Therefore, the velocity of the particle is
highest at the lowest point.
If the velocity of the particle is less than
this it will not complete the circular path.

Circular motion

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