Circular motion
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Sep 02, 2012
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About This Presentation
Circular Motion
Slide Content
Prof. MukeshN. Tekwani
Department of Physics
I. Y. College,
Mumbai
[email protected]
1© Prof. Mukesh N Tekwani, 2011
Department of Physics
I. Y. College,
Mumbai
[email protected]
1© Prof. Mukesh N Tekwani, 2011
Circular Motion
2© Prof. Mukesh N Tekwani, 2011
2© Prof. Mukesh N Tekwani, 2011
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, 2011 3
•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, 2011 3
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 NOTnecessary that
the body should complete a full circle.
•Even motion along arc of a circle is circular
motion
© Prof. Mukesh N Tekwani, 2011 4
•It is defined as the motion of a particle along a
complete circle or part of a circle.
•Fore circular motion, it is NOTnecessary that
the body should complete a full circle.
•Even motion along arc of a circle is circular
motion
© Prof. Mukesh N Tekwani, 2011 4
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
90
0
or π/2
© Prof. Mukesh N Tekwani, 2011
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
90
0
or π/2
© Prof. Mukesh N Tekwani, 2011
6
Circular Motion
is the angular position.
Angular displacement:if
Note: angles measured Clockwise (CW) are negative and
angles measured (CCW) are positive. is measured in
radians.
2radians = 360= 1 revolution
x
y
i
f
© Prof. Mukesh N Tekwani, 2011
Circular Motion
is the angular position.
Angular displacement:if
Note: angles measured Clockwise (CW) are negative and
angles measured (CCW) are positive. is measured in
radians.
2radians = 360= 1 revolution
x
y
i
f
© Prof. Mukesh N Tekwani, 2011
Angular Displacement
•Angular displacement is defined as the angle
described by the radius vector
© Prof. Mukesh N Tekwani, 2011 7
a
Initial position of particle is a
Final position of particle is b
Angular displacement in time t is Θ
•Angular displacement is defined as the angle
described by the radius vector
© Prof. Mukesh N Tekwani, 2011 7
a
Initial position of particle is a
Final position of particle is b
Angular displacement in time t is Θ
Angular Displacement
© Prof. Mukesh N Tekwani, 2011 8
S =r Θ
© Prof. Mukesh N Tekwani, 2011 8
S =r Θ
9
x
y
i
f
r
arclength= s = rr
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
© Prof. Mukesh N Tekwani, 2011
x
y
i
f
r
arclength= s = rr
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
© Prof. Mukesh N Tekwani, 2011
Right Hand Rule
© Prof. Mukesh N Tekwani, 2011 11
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.
© Prof. Mukesh N Tekwani, 2011 11
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
© Prof. Mukesh N Tekwani, 2011 12
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.
© Prof. Mukesh N Tekwani, 2011 12
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 acceleratedmotion. 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 centripetalforce.
•Therefore, Centripetal force is required for circular
motion. No centripetal force +> no circular motion.
25© Prof. Mukesh N Tekwani, 2011
•UCM is an acceleratedmotion. 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 centripetalforce.
•Therefore, Centripetal force is required for circular
motion. No centripetal force +> no circular motion.
25© Prof. Mukesh N Tekwani, 2011
Velocity and Speed in UCM
Is speedchanging?
No, speedis constant
Is velocitychanging?
Yes, velocityis changing
because velocity is a
vector –and direction is
changing at every point
26© Prof. Mukesh N Tekwani, 2011
Is speedchanging?
No, speedis constant
Is velocitychanging?
Yes, velocityis changing
because velocity is a
vector –and direction is
changing at every point
26© Prof. Mukesh N Tekwani, 2011
Velocity and Speed in UCM
Is speedchanging?
No, speedis constant
Is velocitychanging?
Yes, velocityis changing
because velocity is a
vector –and direction is
changing at every point
27© Prof. Mukesh N Tekwani, 2011
Is speedchanging?
No, speedis constant
Is velocitychanging?
Yes, velocityis changing
because velocity is a
vector –and direction is
changing at every point
27© Prof. Mukesh N Tekwani, 2011
Examples of Centripetal force
A body tied to a string
and whirled in a
horizontal circle –CPF
is provided by the
tensionin the string.
28© Prof. Mukesh N Tekwani, 2011
A body tied to a string
and whirled in a
horizontal circle –CPF
is provided by the
tensionin the string.
28© Prof. Mukesh N Tekwani, 2011
Examples of Centripetal force
•For a car travelling
around a circular
road with uniform
speed, the CPF is
provided by the
force of static
friction between
tyresof the car
and the road.
29© Prof. Mukesh N Tekwani, 2011
•For a car travelling
around a circular
road with uniform
speed, the CPF is
provided by the
force of static
friction between
tyresof the car
and the road.
29© Prof. Mukesh N Tekwani, 2011
Examples of Centripetal force
•In case of electronsrevolving around the
nucleus, the centripetal force is provided by
the electrostaticforceof 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© Prof. Mukesh N Tekwani, 2011
•In case of electronsrevolving around the
nucleus, the centripetal force is provided by
the electrostaticforceof 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© Prof. Mukesh N Tekwani, 2011
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 towardsthe centreof the circle.
31© Prof. Mukesh N Tekwani, 2011
•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 towardsthe centreof the circle.
31© Prof. Mukesh N Tekwani, 2011
Properties of Centripetal Force
1.Centripetal force is a realforce
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 = mv
2
/ r
32© Prof. Mukesh N Tekwani, 2011
1.Centripetal force is a realforce
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 = mv
2
/ r
32© Prof. Mukesh N Tekwani, 2011
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 secsis
LMOP = Θ= ωt
33© Prof. Mukesh N Tekwani, 2011
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 secsis
LMOP = Θ= ωt
33© Prof. Mukesh N Tekwani, 2011
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
From ∆POM
sin θ= PM/OP
∴sin θ= y / r
∴y = r sin θ
But θ= ωt
∴y = r sin ωt
34© Prof. Mukesh N Tekwani, 2011
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
From ∆POM
sin θ= PM/OP
∴sin θ= y / r
∴y = r sin θ
But θ= ωt
∴y = r sin ωt
34© Prof. Mukesh N Tekwani, 2011
Radial Acceleration
P(x, y)
v
O
X
Y
M
N
r
θ
x
y
Similarly,
From ∆POM
cosθ= OM/OP
∴cosθ= x / r
∴x = r cosθ
But θ= ωt
∴x = r cosωt
35© Prof. Mukesh N Tekwani, 2011
P(x, y)
v
O
X
Y
M
N
r
θ
x
y
Similarly,
From ∆POM
cosθ= OM/OP
∴cosθ= x / r
∴x = r cosθ
But θ= ωt
∴x = r cosωt
35© Prof. Mukesh N Tekwani, 2011
Radial Acceleration
36© Prof. Mukesh N Tekwani, 2011
The velocity of particle at any instant (any time) is called its
instantaneousvelocity.
The instantaneous velocity is given by
v = dr/ dt
∴v= d/dt[ ircoswt + jrsin wt]
∴v= -ir w sin wt + jr wcoswt
36© Prof. Mukesh N Tekwani, 2011
The velocity of particle at any instant (any time) is called its
instantaneousvelocity.
The instantaneous velocity is given by
v = dr/ dt
∴v= d/dt[ ircoswt + jrsin wt]
∴v= -ir w sin wt + jr wcoswt
Radial Acceleration
37© Prof. Mukesh N Tekwani, 2011
The linear acceleration of the particle at any instant (any time) is
called its instantaneouslinearacceleration.
37© Prof. Mukesh N Tekwani, 2011
The linear acceleration of the particle at any instant (any time) is
called its instantaneouslinearacceleration.
Radial Acceleration
38© Prof. Mukesh N Tekwani, 2011
Therefore, the instantaneous linear
acceleration is given by
∴a= -w
2
r
Importance of the negative sign:
The negativesign in the above
equation indicates that the linear
accelerationof the particle and the
radius vector are in opposite
directions.
a
r
38© Prof. Mukesh N Tekwani, 2011
Therefore, the instantaneous linear
acceleration is given by
∴a= -w
2
r
Importance of the negative sign:
The negativesign in the above
equation indicates that the linear
accelerationof the particle and the
radius vector are in opposite
directions.
a
r
Relation Between Angular
Acceleration and Linear Acceleration
39© Prof. Mukesh N Tekwani, 2011
The acceleration of a particle
is given by
………………. (1)
But v = r w
∴
∴a = .... (2)
∵r is a constant radius,
∴
But
αis the angular acceleration
∴a = r α ………………………(3)
Acceleration and Linear Acceleration
39© Prof. Mukesh N Tekwani, 2011
The acceleration of a particle
is given by
………………. (1)
But v = r w
∴
∴a = .... (2)
∵r is a constant radius,
∴
But
αis the angular acceleration
∴a = r α ………………………(3)
Relation Between Angular
Acceleration and Linear Acceleration
40© Prof. Mukesh N Tekwani, 2011
v = w x r
Differentiating w.r.t. time t,
But
and
∴
∴linear acceleration
a = a
T+ a
R
a
T
is called the tangentialcomponent of
linear acceleration
a
R
is called the radialcomponent of
linear acceleration
For UCM, w = constant, so
∴a = a
R
∴in UCM, linear acclnis centripetal accln
Acceleration and Linear Acceleration
40© Prof. Mukesh N Tekwani, 2011
v = w x r
Differentiating w.r.t. time t,
But
and
∴
∴linear acceleration
a = a
T+ a
R
a
T
is called the tangentialcomponent of
linear acceleration
a
R
is called the radialcomponent of
linear acceleration
For UCM, w = constant, so
∴a = a
R
∴in UCM, linear acclnis centripetal accln
Centrifugal Force
41© Prof. Mukesh N Tekwani, 2011
1.Centrifugal force is an imaginaryforce(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
41© Prof. Mukesh N Tekwani, 2011
1.Centrifugal force is an imaginaryforce(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© Prof. Mukesh N Tekwani, 2011
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.
CentripetalandCentrifugalforcesDONOTconstituteanaction-
reactionpair.Centrifugalforceisnotarealforce.Foraction-
reactionpair,bothforcesmustbereal.
42© Prof. Mukesh N Tekwani, 2011
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.
CentripetalandCentrifugalforcesDONOTconstituteanaction-
reactionpair.Centrifugalforceisnotarealforce.Foraction-
reactionpair,bothforcesmustbereal.
Banking of Roads
43© Prof. Mukesh N Tekwani, 2011
1.Whenacarismovingalongacurvedroad,itisperforming
circularmotion.Forcircularmotionitisnotnecessarythatthe
carshouldcompleteafullcircle;anarcofacircleisalso
treatedasacircularpath.
2.Weknowthatcentripetalforce(CPF)isnecessaryforcircular
motion.IfCPFisnotpresent,thecarcannottravelalonga
circularpathandwillinsteadtravelalongatangentialpath.
43© Prof. Mukesh N Tekwani, 2011
1.Whenacarismovingalongacurvedroad,itisperforming
circularmotion.Forcircularmotionitisnotnecessarythatthe
carshouldcompleteafullcircle;anarcofacircleisalso
treatedasacircularpath.
2.Weknowthatcentripetalforce(CPF)isnecessaryforcircular
motion.IfCPFisnotpresent,thecarcannottravelalonga
circularpathandwillinsteadtravelalongatangentialpath.
Banking of Roads
44© Prof. Mukesh N Tekwani, 2011
3.Thecentripetalforceforcircularmotionofthecarcanbe
providedintwoways:
•Frictionalforcebetweenthetyresofthecarandtheroad.
•BankingofRoads
44© Prof. Mukesh N Tekwani, 2011
3.Thecentripetalforceforcircularmotionofthecarcanbe
providedintwoways:
•Frictionalforcebetweenthetyresofthecarandtheroad.
•BankingofRoads
Friction between Tyresand Road
45© Prof. Mukesh N Tekwani, 2011
Thecentripetalforceforcircularmotionofthecarisprovidedby
thefrictionalforcebetweenthetyresofthecarandtheroad.
Letm=massofthecar
V=speedofthecar,and
R=radiusofthecurvedroad.
Sincecentripetalforceisprovidedbythefrictionalforce,
CPF=frictionalforce(“provideby”means“equalto”)
(µiscoefficientoffrictionbetweentyres&road)
So and
45© Prof. Mukesh N Tekwani, 2011
Thecentripetalforceforcircularmotionofthecarisprovidedby
thefrictionalforcebetweenthetyresofthecarandtheroad.
Letm=massofthecar
V=speedofthecar,and
R=radiusofthecurvedroad.
Sincecentripetalforceisprovidedbythefrictionalforce,
CPF=frictionalforce(“provideby”means“equalto”)
(µiscoefficientoffrictionbetweentyres&road)
So and
Friction between Tyresand Road
46© Prof. Mukesh N Tekwani, 2011
Thus,themaximumvelocitywithwhichacarcansafelytravel
alongacurvedroadisgivenby
Ifthespeedofthecarincreasesbeyondthisvalue,thecarwillbe
thrownoff(skid).
Ifthecarhastomoveatahigherspeed,thefrictionalforce
shouldbeincreased.Butthiscausewearandtearoftyres.
Thefrictionalforceisnotreliableasitcandecreaseonwetroads
Sowecannotrelyonfrictionalforcetoprovidethecentripetal
forceforcircularmotion.
46© Prof. Mukesh N Tekwani, 2011
Thus,themaximumvelocitywithwhichacarcansafelytravel
alongacurvedroadisgivenby
Ifthespeedofthecarincreasesbeyondthisvalue,thecarwillbe
thrownoff(skid).
Ifthecarhastomoveatahigherspeed,thefrictionalforce
shouldbeincreased.Butthiscausewearandtearoftyres.
Thefrictionalforceisnotreliableasitcandecreaseonwetroads
Sowecannotrelyonfrictionalforcetoprovidethecentripetal
forceforcircularmotion.
Friction between Tyresand Road
47© Prof. Mukesh N Tekwani, 2011
R
1and R
2are reaction forces due
to the tyres
mg is the weightof the car,
acting vertically downwards
F
1and F
2are the frictionalforces
between the tyresand the road.
These frictional forces act
towards the centre of the
circular path and providethe
necessary centripetal force.
Center of
circular
path
47© Prof. Mukesh N Tekwani, 2011
R
1and R
2are reaction forces due
to the tyres
mg is the weightof the car,
acting vertically downwards
F
1and F
2are the frictionalforces
between the tyresand the road.
These frictional forces act
towards the centre of the
circular path and providethe
necessary centripetal force.
Center of
circular
path
Friction between Tyresand Road
48© Prof. Mukesh N Tekwani, 2011
48© Prof. Mukesh N Tekwani, 2011
Friction between Tyresand Road –
Car Skidding
49© Prof. Mukesh N Tekwani, 2011
Car Skidding
49© Prof. Mukesh N Tekwani, 2011
Banked Roads
50© Prof. Mukesh N Tekwani, 2011
What is banking of roads?
The process of raisingthe
outeredgeof a road over
the inner edge through a
certain angle is known as
banking of road.
50© Prof. Mukesh N Tekwani, 2011
What is banking of roads?
The process of raisingthe
outeredgeof a road over
the inner edge through a
certain angle is known as
banking of road.
Banking of Roads
51© Prof. Mukesh N Tekwani, 2011
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 tyresdue to friction
3.To avoid skidding
4.To avoid overturning of vehicles
51© Prof. Mukesh N Tekwani, 2011
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 tyresdue to friction
3.To avoid skidding
4.To avoid overturning of vehicles
Banked Roads
52© Prof. Mukesh N Tekwani, 2011
52© Prof. Mukesh N Tekwani, 2011
Banked Roads
53© Prof. Mukesh N Tekwani, 2011
What is angle of
banking?
R
Θ
R cosθ
Θ
W = mg
R sin θ
The anglemade by the
surface of the road with
the horizontal surface is
called as angle of banking.
Horizontal
53© Prof. Mukesh N Tekwani, 2011
What is angle of
banking?
R
Θ
R cosθ
Θ
W = mg
R sin θ
The anglemade by the
surface of the road with
the horizontal surface is
called as angle of banking.
Horizontal
Banked Roads
54© Prof. Mukesh N Tekwani, 2011
Consider a car moving
along a banked road.
Let
m = mass of the car
V = speed of the car
θ isangleof banking
R
Θ
R cosθ
Θ
W = mg
R sin θ
54© Prof. Mukesh N Tekwani, 2011
Consider a car moving
along a banked road.
Let
m = mass of the car
V = speed of the car
θ isangleof banking
R
Θ
R cosθ
Θ
W = mg
R sin θ
Banked Roads
55© Prof. Mukesh N Tekwani, 2011
The forces acting on the
car are:
(i) Its weightmgacting
vertically downwards.
(ii) The normal
reactionRacting
perpendicular to the
surface of the road.
R
Θ
R cosθ
Θ
W = mg
R sin θ
55© Prof. Mukesh N Tekwani, 2011
The forces acting on the
car are:
(i) Its weightmgacting
vertically downwards.
(ii) The normal
reactionRacting
perpendicular to the
surface of the road.
R
Θ
R cosθ
Θ
W = mg
R sin θ
Banked Roads
56© Prof. Mukesh N Tekwani, 2011
The normal reaction can be
resolved (broken up) into
two components:
1.R cosθis the vertical
component
2.R sinθis the horizontal
component
R
Θ
R cosθ
Θ
W = mg
R sin θ
56© Prof. Mukesh N Tekwani, 2011
The normal reaction can be
resolved (broken up) into
two components:
1.R cosθis the vertical
component
2.R sinθis the horizontal
component
R
Θ
R cosθ
Θ
W = mg
R sin θ
Banked Roads
57© Prof. Mukesh N Tekwani, 2011
Since the vehicle has no
vertical motion, the weight
is balanced by the vertical
component
R cosθ= mg …………… (1)
(weight is balanced by
vertical component means
weight is equal to vertical
component)
R
Θ
R cosθ
Θ
W = mg
R sin θ
57© Prof. Mukesh N Tekwani, 2011
Since the vehicle has no
vertical motion, the weight
is balanced by the vertical
component
R cosθ= mg …………… (1)
(weight is balanced by
vertical component means
weight is equal to vertical
component)
R
Θ
R cosθ
Θ
W = mg
R sin θ
Banked Roads
58© Prof. Mukesh N Tekwani, 2011
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 cosθ
Θ
W = mg
R sin θ
58© Prof. Mukesh N Tekwani, 2011
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 cosθ
Θ
W = mg
R sin θ
Banked Roads
59© Prof. Mukesh N Tekwani, 2011
Dividing (2) by (1), we get
R sinθ=
mg
R cosθ
θ= tan
-1 ( )
So,
tan θ=
Therefore, the angle of banking
is independentof the mass of
the vehicle.
The maximum speed with
which the vehicle can safely
travel along the curved road is
59© Prof. Mukesh N Tekwani, 2011
Dividing (2) by (1), we get
R sinθ=
mg
R cosθ
θ= tan
-1 ( )
So,
tan θ=
Therefore, the angle of banking
is independentof the mass of
the vehicle.
The maximum speed with
which the vehicle can safely
travel along the curved road is
Banked Roads
60© Prof. Mukesh N Tekwani, 2011 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?
60© Prof. Mukesh N Tekwani, 2011 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?
Maximum Speed of a Vehicle on a
Banked Road with Friction
61© Prof. Mukesh N Tekwani, 2011
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.
Banked Road with Friction
61© Prof. Mukesh N Tekwani, 2011
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.
Maximum Speed of a Vehicle on a
Banked Road with Friction
62© Prof. Mukesh N Tekwani, 2011
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 tyresand road.
Forces N and frictional force f are now resolved into two
components
Banked Road with Friction
62© Prof. Mukesh N Tekwani, 2011
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 tyresand road.
Forces N and frictional force f are now resolved into two
components
Maximum Speed of a Vehicle on a
Banked Road with Friction
63© Prof. Mukesh N Tekwani, 2011
The normal reaction N is
resolved into 2 components:
1)N cosθis vertical
component of N
2)N sin θis horizontal
component of N
Resolving the Normal reaction
Banked Road with Friction
63© Prof. Mukesh N Tekwani, 2011
The normal reaction N is
resolved into 2 components:
1)N cosθis vertical
component of N
2)N sin θis horizontal
component of N
Resolving the Normal reaction
Maximum Speed of a Vehicle on a
Banked Road with Friction
64© Prof. Mukesh N Tekwani, 2011
The frictional force fis
resolved into 2 components:
1)f cosθis horizontal
component of f
2)fsin θis vertical
component of f
fsin θ
f
fcosθ
θ
Resolving the frictional force
Banked Road with Friction
64© Prof. Mukesh N Tekwani, 2011
The frictional force fis
resolved into 2 components:
1)f cosθis horizontal
component of f
2)fsin θis vertical
component of f
fsin θ
f
fcosθ
θ
Resolving the frictional force
Maximum Speed of a Vehicle on a
Banked Road with Friction
65© Prof. Mukesh N Tekwani, 2011
All forces acting on vehicle
fsin θf
fcosθ
θ
Ncosθ
N
θ
Nsin θ
mg
The vertical component N cosθis
balanced by the weightof the
vehicle and the component f sin θ
∴N cosθ= mg+ f sin θ
∴mg = N cosθ -f sin θ.…. (1)
Banked Road with Friction
65© Prof. Mukesh N Tekwani, 2011
All forces acting on vehicle
fsin θf
fcosθ
θ
Ncosθ
N
θ
Nsin θ
mg
The vertical component N cosθis
balanced by the weightof the
vehicle and the component f sin θ
∴N cosθ= mg+ f sin θ
∴mg = N cosθ -f sin θ.…. (1)
Maximum Speed of a Vehicle on a
Banked Road with Friction
66© Prof. Mukesh N Tekwani, 2011
All forces acting on vehicle
fsin θf
fcosθ
θ
Ncosθ
N
θ
Nsin θ
mg
The horizontal component N sin θ
and f cosθprovide the centripetal
force for circular motion
∴N sin θ+ f cosθ=
∴ = N sin θ+ f cosθ… (2)r
mv
2 r
mv
2
Banked Road with Friction
66© Prof. Mukesh N Tekwani, 2011
All forces acting on vehicle
fsin θf
fcosθ
θ
Ncosθ
N
θ
Nsin θ
mg
The horizontal component N sin θ
and f cosθprovide the centripetal
force for circular motion
∴N sin θ+ f cosθ=
∴ = N sin θ+ f cosθ… (2)r
mv
2 r
mv
2
Maximum Speed of a Vehicle on a
Banked Road with Friction
67© Prof. Mukesh N Tekwani, 2011
All forces acting on vehicle
fsin θf
fcosθ
θ
Ncosθ
N
θ
Nsin θ
mg
Dividing (2) by (1), we getr
mv
2
mg
=
N sin θ+ f cosθ
N cosθ -f sin θ
∴rg
v
2 =
N sin θ+ f cosθ
N cosθ -f sin θ
Banked Road with Friction
67© Prof. Mukesh N Tekwani, 2011
All forces acting on vehicle
fsin θf
fcosθ
θ
Ncosθ
N
θ
Nsin θ
mg
Dividing (2) by (1), we getr
mv
2
mg
=
N sin θ+ f cosθ
N cosθ -f sin θ
∴rg
v
2 =
N sin θ+ f cosθ
N cosθ -f sin θ
Maximum Speed of a Vehicle on a
Banked Road with Friction
68© Prof. Mukesh N Tekwani, 2011
fsin θf
fcosθ
θ
Ncosθ
N
θ
Nsin θ
mg
Let V
maxbe the maximum speed of
the vehicle.
Frictional force at this speed will be
f
m= µ
sN
∴rg
v
2
m ax =
N sin θ+ f
mcosθ
N cosθ -f
msin θ
Banked Road with Friction
68© Prof. Mukesh N Tekwani, 2011
fsin θf
fcosθ
θ
Ncosθ
N
θ
Nsin θ
mg
Let V
maxbe the maximum speed of
the vehicle.
Frictional force at this speed will be
f
m= µ
sN
∴rg
v
2
m ax =
N sin θ+ f
mcosθ
N cosθ -f
msin θ
Maximum Speed of a Vehicle on a
Banked Road with Friction
69© Prof. Mukesh N Tekwani, 2011
But f
m= µ
sN
∴rg
v
2
m ax =
N sin θ+ µ
sN cosθ
N cosθ -µ
sN sin θ
Dividing numerator and
denominator of RHS by
cosθ, we get
=∴rg
v
2
m ax
N sin θµ
sN cosθ
N cosθ
N cosθ
+
N cosθ µ
sN sin θ
N cosθ N cosθ
-
∴rg
v
2
m ax
=
tan θ+ µ
s
1-µ
stan θ
Banked Road with Friction
69© Prof. Mukesh N Tekwani, 2011
But f
m= µ
sN
∴rg
v
2
m ax =
N sin θ+ µ
sN cosθ
N cosθ -µ
sN sin θ
Dividing numerator and
denominator of RHS by
cosθ, we get
=∴rg
v
2
m ax
N sin θµ
sN cosθ
N cosθ
N cosθ
+
N cosθ µ
sN sin θ
N cosθ N cosθ
-
∴rg
v
2
m ax
=
tan θ+ µ
s
1-µ
stan θ
Maximum Speed of a Vehicle on a
Banked Road with Friction
70© Prof. Mukesh N Tekwani, 2011
∴ =2
m axv
tan θ+ µ
s
1-µ
stan θ
r g
∴ =m axv
tan θ+ µ
s
1-µ
stan θ
r g
This is the maximum velocity
with which a vehicle can travel
on a banked road with friction.
For a frictionless road, µ
s= 0
∴ =m axv
tan θ+ 0
1-0
r gtan
maxrgv
Banked Road with Friction
70© Prof. Mukesh N Tekwani, 2011
∴ =2
m axv
tan θ+ µ
s
1-µ
stan θ
r g
∴ =m axv
tan θ+ µ
s
1-µ
stan θ
r g
This is the maximum velocity
with which a vehicle can travel
on a banked road with friction.
For a frictionless road, µ
s= 0
∴ =m axv
tan θ+ 0
1-0
r gtan
maxrgv
Conical Pendulum
71© Prof. Mukesh N Tekwani, 2011
Definition:
A conical pendulum is a
simple pendulum which is
given a motion so that the
bob describes a horizontal
circleand the string
describes a cone.
71© Prof. Mukesh N Tekwani, 2011
Definition:
A conical pendulum is a
simple pendulum which is
given a motion so that the
bob describes a horizontal
circleand the string
describes a cone.
Conical Pendulum
72© Prof. Mukesh N Tekwani, 2011
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.
72© Prof. Mukesh N Tekwani, 2011
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
73© Prof. Mukesh N Tekwani, 2011
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 cosθ
T sinθ
θ
73© Prof. Mukesh N Tekwani, 2011
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 cosθ
T sinθ
θ
Conical Pendulum –Time Period
74© Prof. Mukesh N Tekwani, 2011
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 cosθ
T sinθ
θ
74© Prof. Mukesh N Tekwani, 2011
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 cosθ
T sinθ
θ
Conical Pendulum –Time Period
75© Prof. Mukesh N Tekwani, 2011
The tension T in the string can be
resolved (broken up) into 2
components as follows:
i)Tcosθacting vertically
upwards. This force is balanced
by the weight of the bob
T cosθ= mg ……………………..(1)
T cosθ
T sinθ
θ
75© Prof. Mukesh N Tekwani, 2011
The tension T in the string can be
resolved (broken up) into 2
components as follows:
i)Tcosθacting vertically
upwards. This force is balanced
by the weight of the bob
T cosθ= mg ……………………..(1)
T cosθ
T sinθ
θ
Conical Pendulum –Time Period
76© Prof. Mukesh N Tekwani, 2011
(ii) T sinθacting along the
radius of the circle and directed
towards the centre of the circle
T sinθprovides the necessary
centripetal force for circular
motion.
∴T sinθ= ……….(2)
Dividing (2) by (1) we get,
………………….(3)
T cosθ
T sinθ
θ
76© Prof. Mukesh N Tekwani, 2011
(ii) T sinθacting along the
radius of the circle and directed
towards the centre of the circle
T sinθprovides the necessary
centripetal force for circular
motion.
∴T sinθ= ……….(2)
Dividing (2) by (1) we get,
………………….(3)
T cosθ
T sinθ
θ
Conical Pendulum –Time Period
77© Prof. Mukesh N Tekwani, 2011
T cosθ
T sinθ
θ
This equation gives the speed
of the bob.
But v = rw
∴rw=
Squaring both sides, we get
77© Prof. Mukesh N Tekwani, 2011
T cosθ
T sinθ
θ
This equation gives the speed
of the bob.
But v = rw
∴rw=
Squaring both sides, we get
Conical Pendulum –Time Period
78© Prof. Mukesh N Tekwani, 2011
T cosθ
T sinθ
θ
From diagram, tanθ= r / h
∴r
2
w
2
= rg
78© Prof. Mukesh N Tekwani, 2011
T cosθ
T sinθ
θ
From diagram, tanθ= r / h
∴r
2
w
2
= rg
Conical Pendulum –Time Period
79© Prof. Mukesh N Tekwani, 2011
T cosθ
T sinθ
θ
Periodic Time of Conical Pendulum
But
Solving this & substituting sin θ= r/l
we get,
79© Prof. Mukesh N Tekwani, 2011
T cosθ
T sinθ
θ
Periodic Time of Conical Pendulum
But
Solving this & substituting sin θ= r/l
we get,
Conical Pendulum –Time Period
80© Prof. Mukesh N Tekwani, 2011
T cosθ
T sinθ
θ
Periodic Time of Conical Pendulum
But cosθ= h/l
So the eqn becomes, g
h
l
xl
T2 g
h
T2
80© Prof. Mukesh N Tekwani, 2011
T cosθ
T sinθ
θ
Periodic Time of Conical Pendulum
But cosθ= h/l
So the eqn becomes, g
h
l
xl
T2 g
h
T2
Conical Pendulum –Time Period
81© Prof. Mukesh N Tekwani, 2011
T cosθ
T sinθ
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
81© Prof. Mukesh N Tekwani, 2011
T cosθ
T sinθ
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
82© Prof. Mukesh N Tekwani, 2011
Consider an object
of mass m tied to
the end of an
inextensible string
and whirled in a
verticalcircle of
radius r.
mg
T1
O
r
v1
A
T2
B v2
v3
C
Earth’s Gravitation
82© Prof. Mukesh N Tekwani, 2011
Consider an object
of mass m tied to
the end of an
inextensible string
and whirled in a
verticalcircle of
radius r.
mg
T1
O
r
v1
A
T2
B v2
v3
C
Vertical Circular Motion Due to
Earth’s Gravitation
83© Prof. Mukesh N Tekwani, 2011
mg
T1
O
r
v1
A
T2
B v2
v3
C
Highest Point A:
Let the velocity be v
1
The forces acting on the
object at A (highest point)
are:
1.Tension T
1acting in
downward direction
2.Weight mg acting in
downward direction
Earth’s Gravitation
83© Prof. Mukesh N Tekwani, 2011
mg
T1
O
r
v1
A
T2
B v2
v3
C
Highest Point A:
Let the velocity be v
1
The forces acting on the
object at A (highest point)
are:
1.Tension T
1acting in
downward direction
2.Weight mg acting in
downward direction
Vertical Circular Motion Due to
Earth’s Gravitation
84© Prof. Mukesh N Tekwani, 2011
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)
Earth’s Gravitation
84© Prof. Mukesh N Tekwani, 2011
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)
Vertical Circular Motion Due to
Earth’s Gravitation
85© Prof. Mukesh N Tekwani, 2011
mg
T1
O
r
v1
A
T2
B v2
v3
C
Lowest Point B:
Let the velocity be v
2
The forces acting on the
object at B (lowest point) are:
1.Tension T
2acting in
upward direction
2.Weight mg acting in
downward direction
Earth’s Gravitation
85© Prof. Mukesh N Tekwani, 2011
mg
T1
O
r
v1
A
T2
B v2
v3
C
Lowest Point B:
Let the velocity be v
2
The forces acting on the
object at B (lowest point) are:
1.Tension T
2acting in
upward direction
2.Weight mg acting in
downward direction
Vertical Circular Motion Due to
Earth’s Gravitation
86© Prof. Mukesh N Tekwani, 2011
mg
T1
O
r
v1
A
T2
B v2
v3
C
At the lowest point B:
…… (2)
Earth’s Gravitation
86© Prof. Mukesh N Tekwani, 2011
mg
T1
O
r
v1
A
T2
B v2
v3
C
At the lowest point B:
…… (2)
Vertical Circular Motion Due to
Earth’s Gravitation
87© Prof. Mukesh N Tekwani, 2011
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 T
1disappears (T
1= 0)
Earth’s Gravitation
87© Prof. Mukesh N Tekwani, 2011
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 T
1disappears (T
1= 0)
Vertical Circular Motion Due to
Earth’s Gravitation
88© Prof. Mukesh N Tekwani, 2011
mg
T1
O
r
v1
A
T2
B v2
v3
C
Linear velocity of object at highest
point A:
Earth’s Gravitation
88© Prof. Mukesh N Tekwani, 2011
mg
T1
O
r
v1
A
T2
B v2
v3
C
Linear velocity of object at highest
point A:
Vertical Circular Motion Due to
Earth’s Gravitation
89© Prof. Mukesh N Tekwani, 2011
mg
T1
O
r
v1
A
T2
B v2
v3
C
Linear velocity of object at highest
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.
Earth’s Gravitation
89© Prof. Mukesh N Tekwani, 2011
mg
T1
O
r
v1
A
T2
B v2
v3
C
Linear velocity of object at highest
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.
Vertical Circular Motion Due to
Earth’s Gravitation
90© Prof. Mukesh N Tekwani, 2011
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,
KE
A+ PE
A= KE
B+ PE
B
Earth’s Gravitation
90© Prof. Mukesh N Tekwani, 2011
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,
KE
A+ PE
A= KE
B+ PE
B
Vertical Circular Motion Due to
Earth’s Gravitation
91© Prof. Mukesh N Tekwani, 2011
mg
T1
O
r
v1
A
T2
B v2
v3
C
Linear velocity of object at lowest
point B:
At the highest point A, the minimum
velocity must be
Using this in
we get,
Earth’s Gravitation
91© Prof. Mukesh N Tekwani, 2011
mg
T1
O
r
v1
A
T2
B v2
v3
C
Linear velocity of object at lowest
point B:
At the highest point A, the minimum
velocity must be
Using this in
we get,
Vertical Circular Motion Due to
Earth’s Gravitation
92© Prof. Mukesh N Tekwani, 2011
mg
T1
O
r
v1
A
T2
B v2
v3
C
Linear velocity of object at lowest
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.
Earth’s Gravitation
92© Prof. Mukesh N Tekwani, 2011
mg
T1
O
r
v1
A
T2
B v2
v3
C
Linear velocity of object at lowest
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.
Linear Velocity at a point midway between top
and bottom positions in a vertical circle
93© Prof. Mukesh N Tekwani, 2011
O
r
v1
A
B v2
v3
C
The total energy of a body performing
circular motion is constant at all points on
the path.
By law of conservation of energy,
Total energy at B = Total energy at C
and bottom positions in a vertical circle
93© Prof. Mukesh N Tekwani, 2011
O
r
v1
A
B v2
v3
C
The total energy of a body performing
circular motion is constant at all points on
the path.
By law of conservation of energy,
Total energy at B = Total energy at C
KuchSelf-study bhiKaroNa!
94© Prof. Mukesh N Tekwani, 2011
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.
94© Prof. Mukesh N Tekwani, 2011
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.
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