Circular motion

Prof. Mukesh N. Tekwani
Department of Physics
I. Y. College,
Mumbai
mukeshtekwani@outlook.com
© Prof. Mukesh N Tekwani, 2011 1

Circular Motion


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


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


Circular Motion
How do we locate something on a circle?
Give its angular position θ

y

f
What is the location of
i
X 900 or π/2
θ=0


Circular Motion
y

f
is the angular position.

i Angular displacement:
x f i

Note: angles measured Clockwise (CW) are negative and
angles measured (CCW) are positive. is measured in
radians.
2 radians = 360Tekwani, 1 revolution
© Prof. Mukesh N
= 2011 6

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


Angular Displacement

S =r Θ


arclength = s = r
y

s
f
r r
i
is a ratio of two lengths;
x 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 – Pg 2
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 +> no circular motion.

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

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.


• 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.


• 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


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.

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


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

Draw PM ┴ OX

The angular displacement of the particle in
time t secs is

LMOP = Θ = ωt

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

∴y = r sin θ

But θ = ωt

∴ y = r sin ωt

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

But θ = ωt

∴ x = r cos ωt


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


Radial Acceleration
Therefore, the instantaneous linear
acceleration is given by

∴ a = - w2 r r

a
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.


Relation Between Angular
Acceleration and Linear Acceleration
The acceleration of a particle ∵ r is a constant radius,
is given by

………………. (1)
∴
But v = r w
But
∴
α is the angular acceleration

∴a = .... (2) ∴ a=rα ………………………(3)


Relation Between Angular
Acceleration and Linear Acceleration
∴
v=wxr
Differentiating w.r.t. time t, ∴ linear acceleration
a = a T + aR
aT is called the tangential component of
linear acceleration

aR is called the radial component of
linear acceleration
But For UCM, w = constant, so

and ∴ a = aR
∴ in UCM, linear accln is centripetal accln

Centrifugal Force
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
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
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
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
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 (“provide by” means “equal to” )

(µ is coefficient of friction between tyres & road)

So and

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 cause 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.

R1 and R2 are reaction forces due
to the tyres

mg is the weight of the car,
Center of acting vertically downwards
circular
path
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.

Friction between Tyres and Road –
Car Skidding


Banked Roads
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
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


Banked Roads

What is angle of R cos θ
R
banking? Θ

The angle made by the R sin θ
surface of the road with
the horizontal surface is
called as angle of banking. Θ
Horizontal

W = mg

Banked Roads

Consider a car moving R cos θ
R
along a banked road. Θ

R sin θ
Let
m = mass of the car
V = speed of the car Θ

θ is angle of banking

W = mg

Banked Roads

The forces acting on the R cos θ
R
car are: Θ

R sin θ
(i) Its weight mg acting
vertically downwards.
Θ
(ii) The normal
reaction R acting
perpendicular to the W = mg
surface of the road.

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

Banked Roads
Since the vehicle has no R cos θ
vertical motion, the weight R
Θ
is balanced by the vertical
component R sin θ
R cosθ = mg …………… (1)
Θ
(weight is balanced by
vertical component means
weight is equal to vertical
component) W = mg

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

Banked Roads
Dividing (2) by (1), we get
θ = tan-1 ( )
R sinθ = mg
Therefore, the angle of banking
R cos θ
is independent of the mass of
the vehicle.

So, The maximum speed with
which the vehicle can safely
tan θ = travel along the curved road is


Banked Roads
A car travels at a constant speed
around two curves. Where is the
car most likely to skid? Why?

2
mv
F = ma
r
Smaller radius: larger centripetal
force is required to keep it in
uniform circular motion.

Maximum Speed of a Vehicle on a
Banked Road with Friction
Consider a vehicle moving along a
curved banked road.

Let
m = mass of vehicle
r = radius of curvature of road
θ = angle of banking
F = frictional force between tyres
and road.
The forces acting on the vehicle
are shown in the diagram.

The forces acting on the vehicle are:
1) Weight of the vehicle mg, acting vertically downwards
2) Normal reaction N acting on vehicle, perpendicular to
the surface of the road.
3) Friction force between tyres and road.

Forces N and frictional force f are now resolved into two
components


Resolving the Normal reaction The normal reaction N is
resolved into 2 components:

1) N cos θ is vertical
component of N
2) N sin θ is horizontal
component of N


Resolving the frictional force
The frictional force f is
resolved into 2 components:

f cos θ
1) f cos θ is horizontal
component of f
f
θ f sin θ
2) f sin θ is vertical
component of f


All forces acting on vehicle

N The vertical component N cos θ is
N cos θ
balanced by the weight of the
θ vehicle and the component f sin θ
N sin θ
f cos θ
f
θ
f sin θ ∴ N cos θ = mg + f sin θ
mg
∴ mg = N cos θ - f sin θ .…. (1)



N The horizontal component N sin θ
N cos θ
and f cos θ provide the centripetal
θ force for circular motion
N sin θ
f cos θ mv 2
f
θ
f sin θ ∴ N sin θ + f cos θ =
2
r
mg mv
∴ = N sin θ + f cos θ … (2)
r



N Dividing (2) by (1), we get
N cos θ
2
mv
θ
N sin θ r N sin θ + f cos θ
f cos θ θ =
f f sin θ N cos θ - f sin θ
mg
mg

v2 N sin θ + f cos θ
∴ =
rg N cos θ - f sin θ


Let Vmax be the maximum speed of
N
the vehicle.
N cos θ

θ Frictional force at this speed will be
N sin θ fm = µs N
f cos θ θ
f f sin θ
2 N sin θ + fm cos θ
mg v max
∴ =
rg N cos θ - fm sin θ


But fm = µs N N sin θ µs N cos θ
+
2 N cos θ N cos θ
v
2 N sin θ + µs N cos θ v
∴ max
= ∴ max
=
rg N cos θ - µs N sin θ
rg N cos θ µs N sin θ
-
N cos θ N cos θ

Dividing numerator and 2 tan θ + µs
v max =
denominator of RHS by ∴
cos θ, we get rg 1 - µs tan θ


tan θ + µs For a frictionless road, µs = 0
2
∴ v m ax = rg
1 - µs tan θ tan θ + 0
∴ v m ax = r g
1 - 0

tan θ + µs
∴ v m ax = rg
1 - µs tan θ vmax rg tan
This is the maximum velocity
with which a vehicle can travel
on a banked road with friction.

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


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


The tension T in the string can be
resolved (broken up) into 2
components as follows:
T cos θ

i) Tcosθ acting vertically
upwards. This force is balanced
θ
by the weight of the bob

T cos θ = mg ……………………..(1)
T sin θ


(ii) T sinθ acting along the
radius of the circle and directed
towards the centre of the circle
T cos θ
T sinθ provides the necessary
centripetal force for circular
θ motion.

∴ T sinθ = ……….(2)
T sin θ
Dividing (2) by (1) we get,

………………….(3)


This equation gives the speed
T cos θ of the bob.
But v = rw
θ ∴ rw =

T sin θ Squaring both sides, we get


From diagram, tan θ = r / h
∴ r 2w2 = rg
T cos θ

θ

T sin θ


Periodic Time of Conical Pendulum

T cos θ

But

θ

T sin θ
Solving this & substituting sin θ = r/l
we get,


Periodic Time of Conical Pendulum
But cos θ = h/l

T cos θ So the eqn becomes,

l
θ lx
T 2 h
g
T sin θ
h
T 2
g

Factors affecting time period of
conical pendulum:

T cos θ The period of the conical pendulum
depends on the following factors:

i) Length of the pendulum
θ ii) Angle of inclination to the
vertical
iii) Acceleration due to gravity at
the given place
T sin θ
Time period is independent of the
mass of the bob


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


A
Highest Point A:
v1
v3
Let the velocity be v1

mg The forces acting on the
T1
r object at A (highest point)
O C are:
1. Tension T1 acting in
T2
downward direction
2. Weight mg acting in
B v2 downward direction


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

B v2 …… (1)


A
Lowest Point B:
v1
v3
Let the velocity be v2

mg The forces acting on the
T1
r object at B (lowest point) are:
O C 1. Tension T2 acting in
upward direction
T2
2. Weight mg acting in
downward direction
B v2


A
At the lowest point B:
v1
v3

mg
…… (2)
T1
r
O C

T2

B v2


Linear velocity of object at highest
A
v1 point A:
v3

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


A
v1 point A:
v3

mg
T1
r
O C

T2

B v2


A
v1 point A:
v3

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

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

B v2


A
v1 point B:
v3
At the highest point A, the minimum
mg velocity must be

T1
r
O C
Using this in

T2
we get,

B v2


A
v1 point B:
v3

mg
Therefore, the velocity of the particle is
T1 highest at the lowest point.
r
If the velocity of the particle is less than
O C
this it will not complete the circular path.

T2

B v2


Linear Velocity at a point midway between top
and bottom positions in a vertical circle
The total energy of a body performing
A circular motion is constant at all points on
v1
v3 the path.

By law of conservation of energy,
Total energy at B = Total energy at C
r
O C

B v2


Kuch Self-study bhi Karo Na!
1. Derive an expression for the tension in the string of
a conical pendulum.
2. Write kinematical equations for circular motion in
analogy with linear motion.
3. Derive the expression for the tension in a string in
vertical circular motion at any position.
4. Derive the expression for the linear velocity at a
point midway between the top position and the
bottom position in vertical circular motion, without
the string slackening at the top.


Circular motion

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