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What is Circular motion? Give example.

Motion of an object along circumference of a circle is called circular motion.

E.g. Motion of earth round the sun is approximately circular.

Electrons in an atom revolve around nucleus in circular orbits.

In circular motion, linear velocity at any point is directed along the tangent drawn to circle at

that point.

If stone is held in a string and whirled along a circle, it performs circular motion. When

released from any point of its circular path, the stone moves along tangential path.

Explain the terms: Radius vector, Angular displacement (θ) angular velocity ( ) angular

acceleration ( )

Radius vector: A vector drawn from the center of a circle to the position of the particle is called

radius vector.

If A is position of particle , radius vector is vectorOA

. At position B radius vector is vectorOB

.

Angular displacement: “Angular displacement is defined as angle described by radius vector in

given time at center of circle.”

OR

“Angle through which radius vector rotates in given time is called angular displacement”. It is

measured in radians.

In above figure as particle move from A to B, radius vector rotates by angle Hence angular

displacement is .

If s is length of the arc described by particle then s

r

s r

In vector form s r

Angular Velocity ( ): The rate of change of angular displacement with time is known as Angular

Velocity ( ).

1. Circular Motion

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If is small angular displacements in small time t ,then the average angular velocity in the

interval is given by t

The limit of this expression as t 0 is known as instantaneous angular velocity

0limt t

SI unit of angular velocity radians/sec and dimensions are [M0L

0T

-1 ].

Angular displacement and Angular velocity are vectors. The direction of angular displacement and

angular velocity is along the axis perpendicular to plane of circle given by right hand rule.

Right hand rule: Imagine an axis of rotation to be held in right hand with fingers curled round the

axis and thumb stretched along the axis. If curled finger denotes sense of rotation, then thumb will

denote direction of the vector.

Angular acceleration ( ): The rate of change of angular velocity with time t is known as Angular

acceleration ( )

If is change in angular velocity in time t then the average angular acceleration is given by

The limit of this expression as t 0 is known as instantaneous angular acceleration at time t.

0limt t

SI unit of is rad/sec2 Dimensions are [M

0L

0T

-2]

Angular acceleration is vector quantity. If angular velocity is increasing, direction of is same as

that of angular velocity ( ). But if angular velocity is decreasing then is oppositely directed to .

Obtain the Relationship between linear velocity and angular velocity.

d

dt

d

dt

2 1

t t

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Consider a particle performing circular motion in an anticlockwise sense along circumference

of a circle of radius r with linear speed V.

Suppose that in time t particle moves from A to B. Distance covered in time t is length of

arc AB i.e. s

is angular displacement in given time.

As time interval t is very small length s will also be very small and can be considered as

straight line.

Linear Velocity =0

limt

sv

t

…………………1

but by definition of angle in radian , Angular displacement is given by s

r

s r

Putting in equation 1

0limt

rv

t

0limt

v rt

But 0

limt t

In vector form

position Vector

angular Velocity

v Linear Velocity

r

We have the relation s r

Dividing by t

sr

t t

0 0lim limt t

sr

t t

What is Uniform circular motion (UCM)?

Motion of a particle along circumference of a circle with constant linear speed is called uniform

circular motion.

In ucm direction of linear velocity changes continuously but its magnitude remains constant.

UCM is periodic motion

Any motion, which goes on repeating in equal intervals of time, is called periodic motion.

v r

v r

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Uniform circular motion is periodic motion because particle-performing ucm has constant speed; it

takes the same time to complete each revolution.

Period of UCM (T): The time taken by a particle performing uniform circular motion to complete

one revolution is called its period.

In one complete revolution the particle describes an angle 2 radians and time taken is one

period i.e. T

Therefore angular velocity = angular displacement /time

2

T

Define Frequency of UCM (n).

It is the number of revolutions completed by the particle per second.

Since particle takes time T to complete one revolution it will complete 1/T revolution in one

second.

Unit of frequency is Hertz ( rev / sec)

Obtain the expression for Acceleration in uniform circular motion.

Or

Prove that in UCM acceleration is always directed towards the center of circle along radius

vector.

When a particle performs uniform circular motion the magnitude of linear velocity remains

constant but direction of linear velocity changes continuously. Hence the particle must have

linear acceleration.

Uniform circular motion of a particle along a circle of radius r is as shown in the figure. Let

the particle move from A to B in time interval t .Corresponding angular displacement is

.

AP

And BQ

are the linear velocities of the particle at A and B respectively. Magnitude of

velocity remains constant let, | AP

| = |BQ

| = v

Let vector BR

be equal and parallel to vector AP

From the triangle law of vectors In BRQ we get

BR RQ BQ

2T

n = 1

T

2n

2 n

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RQ BQ BR

RQ

= velocity at B – velocity at A = v

Thus RQ

represents the change in velocity ( v

) of the particle in time t. The acceleration (

a

) of the particle is given by

0limt

va

t

0limt

RQa

t

The magnitude of acceleration is given by

0 0lim limt t

v RQa

t t

QBR = AOB as the angle between two tangents is equal to the angle between the

corresponding radii.

Since angular displacement is very small hence from QBR we can write (angle in

radians = arc/radius)

RQ

V

v = RQ= V

Therefore Acceleration a is given by

0limt

Va

t

0limt

a Vt

As v is constant

But 0

limt t

a V

But V r

Acceleration of particle performing UCM is in same direction as that of change in velocity v

. As 0t then point B will approach the point A and v

will be perpendicular to the

tangent i.e. v

will be along the radius and directed towards the center. Thus in UCM

acceleration is always directed towards the center along radius vector. Hence it is called as

centripetal acceleration.

Alternate Derivation (calculus method)

let

P-Position of particle performing U.C.M.

r

-position vector or radius vector.

t Angular displacement of particle in time t

(x, y) – Co ordinates of Point P

v – Tangential velocity of particle performing U.C.M. at P.

Consider a particle performing U.C. M. along a circumference of circle of radius r in

anticlockwise sense.

22V

a V rr

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Let P(x,y) be position of particle performing U.C.M. at instant t.

Draw PM perpendicular to OX.

The angular displacement of particle in time t seconds

POM t

The position vector r

of particle at instant t is given by,

r i x jy

cos sinr i r t j r t

Instantaneous linear velocity of particle at instant t is given by,

( cos sin )dr d

v i r t j r tdt dt

sin cosv i r t j r t

Instantaneous linear acceleration of particle at instant t is given by,

dva

dt

( sin cos )d

i r t j r tdt

2 2cos . sina i r t j r t

2( cos sin )a i r t j r t

2a r

Negative sign indicates that acceleration of particle performing U.C.M. and radius vector are

oppositely directed.

Obtain Relation between Linear and Angular Acceleration

1. Consider a particle moving along a circle of radius r with uniform angular acceleration ( ).

2. The particle has linear acceleration (a). Linear acceleration is the rate of change of its linear

velocity with time.

a = 0

limt

v

t

a = dv

dt but v r

a = ( )d r

dt

rd

adt

(As r is constant)

But By definition, is the rate of change of angular velocity with time.

d

dt

Putting in above equation

a r

Therefore, Linear acceleration = radius x angular acceleration

Radial and tangential Acceleration in circular motion

We know that v r

Differentiating

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dv d

rdt dt

= d d r

rdt dt

but d

dt

=

, d r

dt

= v

dvr v

dt

a r v

t ra a a

Where ta

= r

= tangential component of linear acceleration

ra

= v

= radial component of linear acceleration.

Resultant acceleration in magnitude is 2 2

t ra a a

Remember in UCM d

dt

=

= 0, hence ta

=0. Therefore acceleration in ucm is radial.

Remember

Acceleration in accelerated circular motion

For Particle performing accelerated circular motion linear acceleration can be resolved into

two components centripetal acceleration (ac) ant tangential acceleration. (at)

Centripetal acceleration Tangential acceleration

It changes direction of velocity of particle

performing circular motion at each instant

It changes magnitude of velocity of

particle performing circular motion.

Direction: along radius vector toward the center

of circle.

Direction: tangential

This acceleration is necessary for circular

motion. If centripetal acceleration is not present

particle will move tangentially.

If tangential acceleration is absent linear

velocity has constant magnitude and

motion becomes UCM

22

c

va v r

r

2 1.ta r rt

Two components, centripetal acceleration (ac) ant tangential acceleration.(at) are

perpendicular to each other and resultant acceleration is 2 2

c ta a a

What do you mean by centripetal force? (Or Define Centripetal force)

Uniform circular motion is an accelerated motion in which acceleration is always directed

along the radius towards the center of the circle.

According to Newton‟s second law this acceleration must be produced due to force acting in

same direction. It is called as centripetal force.

Definition: Particle performing uniform circular motion is acted upon by a force directed

along the radius towards the center of the circle. This force is called as centripetal force.

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Centripetal force = mass of particle centripetal acceleration

=

2Vm

r

In vector form C.F.

2

2

Vf m r

r

or 2f m r

It is real force.

Uniform circular motion is produced and maintained by centripetal force.

Centripetal force is responsible for changing the direction of velocity at every instant.

If the centripetal force is absent circular motion is not possible.

Centripetal force does no work.

Examples of centripetal force

1. In the case of a stone whirled round at the end of the string whose other end is held in the hand,

the centripetal force is supplied by the tension of the string.

2. In case of a planet revolving round the sun, centripetal force is provided by the gravitational

attraction between them.

3. In an atom electron revolves round the nucleus in circular orbit. In this case centripetal force is

provided by electrostatic force of attraction between positively charged nucleus and negatively

charged electron.

4. When a vehicle is taking a turn on curved road, force of friction between tyres and road surface

acts as centripetal force.

What is Centrifugal Force?

Particle performing Uniform circular motion is having a constant centripetal acceleration.

Hence it is in accelerated frame of reference. (Or non inertial frame of reference)

In order to make Newton‟s laws applicable to Non inertial frame, we have to consider the

existence of Pseudo force. Therefore we have to consider existence of pseudo force acting on

particle performing UCM. This force is called as centrifugal force.

Definition: Centrifugal force is pseudo force in UCM which acts along the radius and

directed away from center of circle.

Magnitude of centrifugal force is same as that of centripetal force, but it is appositely directed

i.e. directed along the radius vector away from the center.

Centrifugal force = mass X acceleration of frame of reference

=

2Vm

r

In vector form C.F.

2

2

Vf m r

r

or 2f m r

22

cp

Vf mV m m r

r

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Give examples of centrifugal force.

1. When vehicle moves along a curved path, a passenger in the vehicle experiences a push in the

outward direction due to centrifugal force.

2. A bucket full of water is rotated in a vertical circle at particular speed water will not fall down.

This is because weight of water is balanced by centripetal force.

3. If a gramophone disc with coin kept on it is set into rotation, as the speed of rotation increases,

coin will slide away from center of disc due to centrifugal force.

4. In dryer of washing machine, wet clothes are given spinning motion at high speed. Water particles

are separated because of centrifugal force.

5. The bulging of earth at equator and flattening at the poles is due to centrifugal force acting on it.

6. Suspended particles from liquid can be separated using centrifuge. It consists of two steel tubes

suspended from ends of horizontal bars, which can be rotated at high speed in horizontal plane.

Liquid is taken in tube and bar is rotated. At high speed suspended particles are collected at the

end due to centrifugal force.

Why centrifugal force is called as Pseudo force?

1. Particle performing Uniform circular motion is in accelerated frame of reference.

2. In order to make Newton‟s laws applicable to Non inertial frame, we have to consider the

existence of centrifugal force.

3. All real forces are classified as gravitational, electromagnetic or nuclear force interaction.

Centrifugal force doesn‟t belong to any of these interactions. It appears due to acceleration of

frame of reference. Hence it is called as pseudo force.

Centripetal force Centrifugal Force

1. It is a real force 1. It is imaginary or pseudo force

2. Directed along radius vector towards the center

of the circle.

2. Directed along radius vector away from the

center of the circle.

3. It is responsible to maintain UCM 3. It is not responsible for maintaining UCM

4. It acts on the body performing UCM due to

some known interaction like gravitational or

electromagnetic.

4. It appears only due to acceleration of frame of

reference in non inertial frame.

Obtain the expression for maximum velocity of Vehicle moving along curved road,

1. While taking turn vehicle perform circular motion. For performing circular motion centripetal

force must act on the vehicle.

2. For unbanked road the required centripetal force is provided by frictional force between road

surface and tyre

3. If m = mass of vehicle moving along curved road, v = velocity of vehicle, r = radius of curvature

of road and, = Coefficient of friction between road and curved surface.

Maximum frictional force = mg

Centripetal force =

2mv

r

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Maximum velocity of vehicle on curved road:

Centripetal force = Frictional force

2mv

r= mg

V2 = rg

This is maximum velocity with which vehicle can move along curved road of radius r. If velocity is

more than this limit vehicle will skid.

Explain the necessity of Banking of Road .What is banking of road? Define the angle of

banking. Obtain the expression for maximum velocity of vehicle on banked road.

Necessity of Banking of Road:

The car moving along the curve at a high speed is performing circular motion and centripetal

force is required for this. In absence of necessary centripetal force vehicle cannot take turn

and skid.

Normally centripetal force is provided by force of friction.

Let m = mass of vehicle moving along curved road, v = velocity of vehicle, r = radius of

curvature of road and = Coefficient of friction between road and curved surface.

Maximum velocity of vehicle on curved road:

Centripetal force = Frictional force

2mv

r= mg

V2 = rg

V= rg

Above formula gives maximum velocity of vehicle on curved road. In order to increase

maximum safety speed force of friction should be increased. This can be done by making

road surface rougher. However this increases the wear and tear of the tyres, also force of

friction is not always reliable because its magnitude decreases when the road becomes wet

due to rain or oil from moving vehicles falls on the road.

Banking of road is reliable method to ensure safety of vehicle on curved road.

Banking of Road: The process of raising outer edge of road over its inner edge through certain

angle is known as banking of road. The angle made by the surface of road with horizontal is

called angle of banking.

Expression for maximum velocity of vehicle on banked road:

V rg

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= Weight of vehicle

= normal reaction

= frictional force

mg

N

F

2va

r = centripetal aceleration

cos horizonatl component of frictional forcef

sin vertical component of frictional forcef

cos vertical component of normal reactionN

sin horizontal component of normal reactionN

Consider a vehicle having mass of m moving with velocity v along a curved road of radius r

banked at an angle . Let f be frictional force between tyres and road surface,

The other forces acting on the vehicle are

Weight mg acting vertically downwards

Normal reaction N acting at right angles to the road surface.

Frictional force can be resolved into two components

cos horizonatl component of frictional forcef

sin verticaly downward component of frictional forcef

Normal reaction can be resolved into two components

cos vertical component of normal reactionN

sin horizontal component of normal reactionN

Component cosN of normal reaction is balanced by weight mg and component sinf of

frictional force

cos sin

cos sin

N mg f

mg N f

………………………………….(1)

Horizontal component of normal reaction sinN along with component cosf of frictional

force provides necessary centripetal force 2

sin cosmv

N fR

…………………………………(2)

Dividing equation (2) by (1)

2 sin cos

cos sin

v N f

Rg N f

Magnitude of frictional force depends on speed of vehicle for given road surface and tyres

Let maxv denotes maximum speed of vehicle , the frictional force produced at this speed is

max sf N

2

max sin cos

cos sin

v N f

Rg N f

2

max sin cos

cos sin

s

s

v N N

Rg N N

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2

max sin cos

cos sin

s

s

v

Rg

2

max tan

1 tan

s

s

v

Rg

For horizontal road max sv Rg

It is clear that maximum safe speed on banked road is greater than that on horizontal road.

If s = 0

tan 0

1 0 tanov Rg

At this speed frictional force is not needed to provide necessary centripetal force . Their wear

and tear of tyres if vehicle is driven at this speed on banked road. ov is called optimum speed .

2

tan ov

Rg

Above formula gives angle of banking required for optimum speed ov . Angle of banking is

independent of mass of vehicle.

Conical Pendulum:

Conical pendulum is a simple pendulum, which is given such a motion that bob describes a

horizontal circle with constant speed and the string describes a cone in space.

Let, m be the mass of the bob and r be the radius of the circle along which the bob moves. Let

l be the length of the pendulum, i. e. the distance between the point of suspension O and the

centre of gravity of the bob. During the motion of the bob, the string is inclined to the vertical

at an angle .

max

tan

1 tan

s

s

v Rg

tanov Rg

21tan ov

Rg

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In any position of the bob, there are two forces acting upon it. These forces are (i) its weight

mg acting vertically downwards and (ii) the tension T in the string.

The tension can be resolved into a vertical component cosT and a horizontal compound

sinT

The component cosT balances the weight of the bob.

cosT mg -------------------(1)

The component sinT acts as the centripetal force necessary for the uniform circular

motion of the bob.

2

sinmv

Tr

-----------------(2)

Where v is the magnitude of the velocity of the bob. Dividing equation (2) by (1) equation

2

tanv

rg -----------------------(3)

tanv rg ----------------(4)

This expression gives the magnitude of the linear velocity of the bob.

Also V r

tanr rg

2 2 tanr rg But tanr

h

2 =g

h

The period of revolution (t) of the bob is given by

2 rT

v

2

tan

rT

rg

2tan

rT

g

sin2

tan

lT

g

Also cos , cosh

h ll

sin , sinr

r ll

cos2

lT

g

2h

Tg

=g

h

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Thus period of conical pendulum is same as that of simple pendulum of height h , where h is

axial height of cone.

For small angles made with vertical cos 1

2l

Tg

. Thus for small Angle of

inclination with vertical , period of conical pendulum is independent of .

Factors affecting period of conical pendulum

Length

Inclination with vertical if is large

Acceleration due to gravity.

Period of conical pendulum is independent of mass of bob.

Tension in the string of conical pendulum

We know

cosT mg --------------(1) and ,

2

sinmv

Tr

-----------------(2)

Squaring and adding

2

222 mv

T mgr

But

2

tanv

Rg =

r

h

22 r g

vh

2

22 mrgT mg

h

2

221

rT mg

h

Velocities of object at various points in circular motion in vertical plane

2

1r

T mgh

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Consider object of mass m tied at end of inextensible string and whirled in vertical circle of

radius r. A is topmost and B is lowest position of object Let v1 and v2 be the corresponding

velocities of the body at the points A and B.

Forces acting on object at highest point A are

a) Tension T1 acting in downward direction

b) Weight mg acting in downward direction.

Centripetal force acting on object at A is provided partly by tension and partly by self weight

T1 + mg =

2

1mv

r

T1 =

2

1mvmg

r

Similarly at lowest point B, forces on particle are a) Tension T2 acting in upward direction

b) weight mg acting in downward direction.

T2 - mg =

2

2mv

r

T2 =

2

2mvmg

r

Linear velocity at highest point A

There is certain velocity called as critical velocity or minimum velocity V1 of object at A

below which string becomes slack i.e. tension T1 vanishes. (T1=0)

T1 =

2

1mvmg

r

0 =

2

1mvmg

r

2

1mvmg

r

Above equation gives required minimum velocity at highest point so that string doesn‟t slack.

If the velocity at the highest point is less than gr the string will become slack and the body

will not be able to continue its circular motion. Object will perform vertical circular motion

only if its velocity at highest point is more than rg .

Linear velocity lowest point B

Minimum velocity v2 at the lowest point B should be such that, when the body reaches the

highest point A, its velocity reduces to gr and the body is able to move along its

circular path. The minimum velocity at the lowest point A can be found by applying the

law of conservation of energy.

Let us assume that the potential energy at the horizontal level passing through the point A is

zero. As the body moves from the lowest point A to the highest point B, it is raised

through a vertical distance equal to AOB = 2r. Hence, according to the law of

conservation of energy,

K. E. at B = K. E. at A + P.E. at A

2 2

2 1

1 1(2 )

2 2mv mv mg r

2 2

2 1 4v v gr

v1= rg

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Substituting the value of V1 from equation we get,

2

2 4 5v gr gr gr

This expression gives the minimum velocity which the body must possess at the lowest point,

so that it can safely travel along the vertical circle of radius r.

Linear velocity at midway (horizontal position)

Let us now find the velocity (v3) of the body at the point C. In the position C, the body is

raised above the point A through the vertical distance r. Hence the body possesses both K. E.

and P.E. at the point C.

K. E. at 2

3

1

2C mv

P.E. at C = mgr

According to the law of conservation of energy,

K.E at A = K.E. at C + P. E. at C

2 2

1 3

1 1

2 2mv mv mgr

2 2

1 3 2v v gr

But 1 5v gr

2

35 2gr v gr

2

3 3v gr

This expression gives the velocity of the body at midway between top and bottom position.

Energy of particle performing circular motion at various points.

(Conservation of energy in vertical circular motion)

Consider object of mass m tied at end of inextensible string and whirled in vertical circle of

radius r. A is topmost and B is lowest position of object Let v1 and v2 be the corresponding

velocities of the body at the points A and B. C is horizontal position where velocity is 3V .

Total energy at lowest point B

Total energy = kinetic energy + potential energy

E= Ek + Ep

= ½ mv22 + 0 (let Ep=0 at lowest position)

But minimum velocity required at lowest point 2 5v gr

5

2E mgr

V2= 5rg

V3= 3rg

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The total energy (E) at highest point A

E = K.E. at B + P.E. at B

As body move from A to B, potential energy increases by mg (2r)

E2

1

1(2 )

2mv mg r

But minimum velocity required at highest point 1v gr

1 5

22 2

E mgr mgr mgr

The total energy (E) of the body at the point C

E = K.E. at C. + P. E. at C.

2

3

1

2mv mgr

Substituting 3 3v gr in the above expression, we get

3 5

2 2E mgr mgr mgr

At any other point of the vertical circle, the body possesses both K. E. and P.E. However the

minimum total energy at every point is the same .

Difference in tensions at lowest and highest points

At lowest point B,

2

22

mvT mg

r

2

22

mvT mg

r …………….(1)

At highest point A ,

2

11

mvT mg

r

2

11

mvT mg

r ……………(2)

Now

2 22 1

2 1

mv mvT T mg mg

r r

5

2E mgr

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2 2

2 12 1 2

mv mvT T mg

r r

2 22 1 2 1

2m

T T mgv vr

………..(3)

By law of conservation of energy

. . . . . . . .at B at AP E K E P E K E

2 2

2 1

1 10 2

2 2mv mg r mv

2 2

2 1

12

2m mgrv v

2 2

2 14grv v

Put in equation (3)

2 1 24m

T T mggrr

Kinematical equations for circular motion in analogy with linear motion

Angular displacement is analogous to linear displacement, angular velocity is analogous to

linear velocity, and angular acceleration is analogous to linear acceleration.

If a particle moving along a straight line with a linear velocity u is subjected to a uniform linear

acceleration a, it‟s linear velocity goes on changing with time. Let v be its linear velocity after

time t and s be its linear displacement in time t. Then,

v = u+ at

21

2s ut at

2 2 2v u as

In analogy with linear motion, we can write three kinematical equations for uniformly

accelerated circular motion, in the following manner.

Consider a particle moving along a circle with an angular velocity 0 . Suppose that the particle

is subjected to a uniform angular acceleration , so that its angular velocity goes on

changing with time. If is its angular velocity after time t and is its angular displacement

in time t, then the three kinematical equations for uniformly accelerated circular motion can

be stated as follows.

Additional point

If frequency of particle performing circular motion changes from n1 to n2 in time then

no of revolutions in time t= 1 2

2

n navg frequency time t

0 t

2

0

1

2t at

2 2

0 2

2 1 6T T mg

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Vertical Circular motion under gravity.

Position Tension in the

string

Minimum velocity

Required

Potential

energy

KE

bottom 2

lmvT mg

r

5rg 0 5

2mgr

Horizontal 2mvT

r

3rg mgr 3

2mgr

Top 2

lmvT mg

r

rg 2mgr 1

2mgr

Numerical Problems

1. The minute hand of a clock is 8 cm long. Calculate the linear speed of an ant sitting at its

tip. (M'2005 , 2-Marks) ( 1.395 X 10-4 m/s)

2. A body of mass 2 kg is tied to the free end of a string of length 1.5 m and revolved along a

horizontal circle with the other end fixed. The body makes 300 rpm, calculate the linear

velocity, the centripetal acceleration and the force acting on it. (M'93)

(Ans: 47.1 m/s, 1.48 X 103 m/s

2, 2.96 X 10

3 N)

3. The frequency of a particle performing uniform circular motion change from 60 r.p.m to

180 r.p.m in 20 seconds. Calculate the angular acceleration. (O-98 2-Marks)

(0.628 rad/s2)

4. A 0.5 kg mass is rotated in a horizontal circle of radius 20 cm. Calculate the centripetal

force acting on it if its angular speed of revolution is 0.6 rad/s. (M-88 , 2-Mark)

(Ans:23.6 10 N )

5. Find the maximum speed of a car which can be safely driven along a curve of radius 100

m; coefficient of friction between the tyres and the road is 0.2 (0'88 , 2-Marks)

(Ans: 14 m/s)

6. To simulate the acceleration of large rockets, astronauts are spun at the end of a long

rotating beam of length 9.8 m. What angular velocity is required to generate a centripetal

acceleration 8 times the acceleration of gravity? ( Ans : 2.82 rad/s) (M'90, 2-Marks)

7. Find the angle of banking of the railway track of radius of curvature of 1600 m if the

optimum velocity of the train is 20 m/s. Also find the elevation of the outer track above

the inner track if the distance between the two tracks is 1.8 m (M'90 , 3-Marks)

(4.6 cm)

8. A train of mass105 kg rounds a curve of radius 150 m at a speed of 20 m/s. find the

horizontal thrust on the outer rail if the track is not banked. At what angle must the track

be banked in order that there is no thrust on the rail? (0'91 , 3-Marks) ( 15013’)

9. A certain string breaks under a tension of 45 kg-wt. A mass of 100 g is attached to this

string of length 500 cm and whirled in a horizontal circle. Find the maximum number of

revolutions per second without breaking the string. (M'92 , 3-Mark) ( 4.73 Hz)

10. One end of a string 1 m long is fixed and a body of mass 500 gm is tied to the other end. If

breaking tension is 98 N; find the maximum angular velocity of the body that the string

can withstand when rotated in horizontal circle. (14 rad/sec)( 0'94)

11. Calculate the maximum speed with which a car can be driven safely along a curved road

of radius 30 m and banked at 30° with the horizontal. (M'96 , 2-Marks) (13.03 m/s)

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12. An object of mass 400 g is whirled in a horizontal circle of radius 2 m. If it performs 60

rpm, calculate the centripetal force acting on it. (0'96, 2-Marks) (31.55 N)

13. Find the angle which the bicycle and its rider will make with the vertical when going

round a curve at 27 km/hr on a horizontal curved road of radius 10 m.

(M'98, 2-Marks) (29051’)

14. Find the angle of banking of curved railway track of radius 600 m, if the maximum safety

speed limit is 54 km/hr. If the distance between tile rails is 1.6 m, find the elevation of the

outer track above the inner track. (0'98,3-Marks) ( Ans : 2011’ , 0.304 m)

15. A motor cyclist rides in a vertical circle in a hollow sphere of radius 3 m. Find the

minimum speed required so that he does not lose contact with the sphere at the highest

point. (M-87, 2-Marks) ( Ans : 5.42 m/s)

16. The vertical section of a road over a bridge in the direction of its length is in the form of

an arc of a circle of radius 4.4 m. Find the greatest velocity at which a vehicle can cross

the bridge without losing contact with the road at the highest point, if the center of gravity

of the vehicle is 0.5 m from the ground (0'2001 , 2-Marks) (Ans :6.93 m/s)

17. Find the period of conical pendulum having length of string 1m, revolving at an angle 600

with vertical .( g=9.8 m/s2) ( Ans : 1.418 sec)

18. A stone of mass 1 kg is whirled in a horizontal circle at the end of 1m long string. If the

string makes an angle of 300 with vertical, calculate i) period ii) centripetal force (g= 9.8

m/s2). ( Ans : 1.866 sec , 6.53 N)

19. Find the minimum velocity required to be given to simple pendulum of length 5m so that

it can perform circular motion in vertical plane. ( Ans : 15.65 m/s)

20. A wheel rotating at 300 RPM slows down uniformly to 240 RPM while making 20

revolutions. Find angular acceleration and time taken for revolution.

21. A pulley 1m in diameter rotating at 600 RPM is brought to rest in 80 sec by constant force

of friction on its shaft. How many revolutions does it make before coming to rest?

22. A sphere of mass 0.2 kg is attached to inextensible string of length 1.3 m whose upper end

is fixed to the ceiling. The sphere is made to describe a horizontal circle of radius 50 cm.

Calculate time period of revolution.

23. A 1 kg stone tied at the end of 1m long string and is whirled in a vertical circle with

constant speed of 4m/s. If tension in the string at certain instant is 6.2N then find position

of stone.

24. Wheel initially at rest is subjected to uniform angular acceleration about its axis. In first

two seconds it rotates through the angle 1 and in next two seconds it rotates through angle

2 . Find ratio of 1

2

( Ans : 1:3)

SAPTARS HI


State Newton’s Law of Gravitation. Give its mathematical form.

“Every Particle of matter attracts every other particle of matter with a force, which is directly

proportional to product of their masses and inversely proportional to the square of the

distance between them.”

If two particles are having mass m1, m2 separated by distance r, according to Newton‟s law

these particles attract each other with a force given by:

1 2

2

m mF

r

1 2

2

Gm mF

r Where G is Gravitational Constant.

Newton’s Law Of gravitation in vector form

If 12r

is vector drawn from m1 to m2 . And r

is unit vector along it. Then Force acting on

m1 is

1 2

2

Gm mF r

r

or 1 2

123

Gm mF r

r

Force acting on m2

1 2

2

Gm mF r

r

Or 1 2

123

Gm mF r

r

Gravitational force acting on both masses has same magnitude but opposite direction.

Unit and dimension of G

1 2

2

Gm mF

r

2

1 2

FrG

m m Hence SI unit of G is Nm

2/Kg

2

Dimensions

2

1 2

F rG

m m

1 1 2 2

2

M LT L

M

1 3 2M L T

Newton‟s law of gravitation holds good for all material objects irrespective of their sizes or

distances between them. Therefore it is called universal law

Value of Gravitational constant (G=6.673x 10-11

Nm2/kg

2) is same irrespective of mass or

medium between two bodies. Hence it is called universal constant.

2. Gravitation

SAPTARS HI


Gravitational force is independent of medium between two masses.

Concept of gravitational field

To explain how two masses attract each other even though there is no physical contact

between them, concept of gravitational field was introduced. According to this concept, there

exist gravitational field in the space surrounding any Mass.

When another mass is brought in this space, it is acted upon gravitational force of attraction.

What is satellite?

An object revolving round a planet is called satellite. (e.g. Moon is natural satellite of earth)

Many manmade satellites are recently launched called as artificial satellite.

Artificial satellites are used for many purposes such as telecommunication, television

broadcasting, meteorology and weather forecasting etc.

In order to launch a satellite multistage rocket is used. Why?

(Or minimum two stage rocket is required to launch a satellite)

For launching a satellite firstly it has to be raised to sufficient height above the surface of

earth and then specific horizontal velocity is required to be given to it. These two things are

not possible by single stage rocket.

In two stage rocket, satellite is placed at tip of rocket, and fuel in first stage of rocket is

ignited so that rocket rises to desired height.

Then by remote control empty first stage of rocket is detached, rocket is rotated through

900.The fuel in second stage is ignited so that it gets push in horizontal direction.

When fuel in second stage is completely burned, empty second stage also gets detached. If

velocity attained by satellite is equal to critical velocity, satellite revolves in circular orbit

round the earth.

What are the possible paths of satellite depending upon horizontal velocity?

In order to launch a satellite it is raised above the surface of earth to a suitable height. And

then horizontal velocity is given to it. Motion of satellite depends on magnitude of horizontal

velocity given to it.

If horizontal velocity is less than critical velocity satellite moves along elliptical path with

point of projection as apogee and the satellite comes closer to the earth with its perigee point

lying at 1800 . If it enters the atmosphere while coming towards perigee it will lose energy and

spiral down before falling down to earth. If it does not enters the atmosphere it will continue

to move in elliptical orbit.

If horizontal velocity is equal to critical velocity satellite moves along circular orbit round the

earth.

SAPTARS HI


If horizontal velocity is greater than critical velocity and less than escape velocity satellite

moves in an elliptical orbit round the earth. Its perigee is point of projection and its apogee

will be at greater distance than projection height.

If horizontal velocity is greater than escape velocity then the orbit will be hyperbolic then

satellite leaves earth‟s gravitational field and escape in to space

Obtain the Relationship between Constant of gravitation (G ) and gravitational acceleration (g)

Let M and R be mass and radius of earth respectively and g is value of acceleration due to

gravity on the earth‟s surface,

Consider a mass m situated on surface of earth.

Weight mg of object is equal to gravitational force exerted by earth on it.

2

GMmmg

R

If is density of earth then 34

3M R

3

2

4

3G R

gR

Define Critical velocity vc . Obtain the expression for it

Definition: Horizontal velocity of projection for which satellite revolves in circular orbit

round the earth is called critical velocity or orbital velocity of satellite.

Expression of critical velocity:

Let M be the mass and R be radius of earth. Consider a satellite of mass m is raised to height

h and projected in horizontal direction with critical velocity vc. It will revolve in circular

orbit of radius r = R+h.

The centripetal force required for circular motion of satellite is provided by gravitational

force of attraction exerted on satellite by earth.

2

GMg

R

4

3g GR

SAPTARS HI


2

2

c

centrepital force Gravitational force

mv GMm

R h R h

From above formula Critical velocity is independent of mass of satellite.

Also gravitational acceleration at height h is given by 2( )

h

GMg

R h

( )h

GMg R h

R h

Putting in above expression

For Satellite moving very close to earth h << R hence h is neglected.

2

2

2

c

c

GMV

R

GMbut g

R

GM gR

gRV

R

Define Period of a satellite. Show that square of period of satellite is directly proportional to

cube of its radius.

Definition: Time taken by satellite to complete one revolution round the earth is called as

period of satellite (T).

Let M and R are mass and radius of earth respectively.

Consider satellite revolving in circular orbit of radius r with critical velocity. Distance

covered by it in one revolution is equal to circumference of its circular orbit.

circumferance

critical speedPeriod

2

c

rV

T

2

c

rT

V

But expression for critical velocity is c

GMV

r

2 rT

GMr

T =

3/ 22 r

GM

c

GMV

R h

( )C hV g R h

cV gR

2 32 4 r

TGM

SAPTARS HI


Period of satellite does not depend on mass of satellite. As other quantities are constant .

As 2h

GMg

r

2

hGM g r

3

22

h

rT

g r

Kepler’s laws of planetary motion

1) Statement of Kepler‟s first law (law of orbit)

Every planet revolves in an elliptical orbit round the sun, with the sun situated at one focus of

the ellipse.

2) Statement of Kepler‟s second law (law of equal areas)

The radius vector drawn from the sun to any planet sweeps out equal areas in equal intervals

of time. This law is called the law of areas i.e. the areal velocity of the radius vector is

constant.

According to Kepler‟s second law, when a planet is close to the sun its speed is maximum and

when it is farthest from the sun its speed is minimum.

3) Statement of Kepler‟s third law (law of period)

The square of the period of revolution of the planet round the sun is directly proportional to

the cube of the semi-major axis of the elliptical orbit. This law is called the harmonic law

(law of periods.)

Kepler‟s third law is consistent with the Newton‟s law of gravitation.

Define Binding energy and obtain the expression of it for satellites revolving round the earth.

The minimum energy required to remove an object from a point in earth‟s gravitational field

to infinity is called as the binding energy of an object

Binding energy of an orbiting satellite:

A satellite revolving in circular orbit round the earth possesses potential energy as well as

kinetic energy

If h is the height above the earth‟s surface the radius of its orbit is r = R + h where R is the

radius of the earth

Potential energy of a unit mass situated at distance r from the center of the earth i.e.

gravitational potential is equal to –GM/r where M is the mass of the earth.

Therefore potential energy of satellite = gravitational potential x mass

. .GM

P E mr

2 3T r

2h

rT

g

P.E. =GMm

r

SAPTARS HI


Satellite is revolving with critical velocity Vc hence it‟s Kinetic energy is given by,

21. .

2cK E mV

2

1. .

2

GMK E m

r

Total Energy of satellite

= PE + KE

= GMm

r

+

2

GMm

r

The negative value of the total energy indicates that the satellite is bound to earth. And if the

amount of energy 2

GMm

r is supplied to satellite it will be able to free itself from the

influence of earth‟s gravitational field.

Thus the binding energy of the orbiting satellite is the negative of the total energy

Remember

For rotating satellite

BE= KE

PE= -2 KE

TE= -KE

Obtain the expression for Binding energy of a body at rest on the earth’s surface.

Consider a body of mass m situated on earth surface. M= mass of earth R = radius of earth.

The gravitational potential at the surface of the earth is GM

R

Hence the gravitational

potential energy of a body of mass m, situated on the earth surface is given by

PE = gravitational potential X mass of body = GM

mR

=

GMm

R

As body is at rest it will not have any Kinetic Energy.

Total energy = P.E. = GMm

R

Binding energy = - total energy

K.E. =2

GMm

r

T.E. =2

GMm

r

B.E. =2

GMm

r

B.E. =GMm

R

SAPTARS HI


Define Escape velocity (Ve). Obtain expression for it .

Escape velocity is defined as minimum velocity with which body should be projected so as to

escape from earth‟s gravitation field.

Consider a body of mass m situated on earth surface. M= mass of earth R = radius of earth.

The gravitational potential at the surface of the earth is GM

R

Hence the gravitational

potential energy of a body of mass m, situated on the earth surface is given by

PE = gravitational potential X mass of body = GM

mR

=

GMm

R

As body is at rest it will not have any Kinetic Energy.

Total energy = P.E. = GMm

R

Binding energy = - total energy = GMm

R

If body is projected with escape velocity ( Ve ) its KE equal to binding energy and it will be

free from earth‟s gravitational field.

. . .K E B E

21

2e

GMmmV

R

Also g = 2

2

GMGM gR

R

Escape velocity from the earth‟s surface depends only upon the mass and radius of the earth it

does not depend upon the mass of the body. Hence its value is the same for any object at rest

on the earth‟s surface.

For body moving very close to earth its critical velocity is given by cV gR

Hence

Imp Note

Also M =34

3R

342

3e

G R

VR

2e

GMV

R

2eV gR

2e CV V

22

3eV R G

SAPTARS HI


Astronaut in orbiting satellite has a feeling of Weightlessness. Explain

When person stands on earth, he exerts force equal to his weight on the surface of earth. And

surface of earth exerts equal and opposite normal reaction on person. Because of this force of

reaction person feels weight.

Astronaut in orbiting satellite as well as satellite are attracted towards centre of earth with

same acceleration. They both are falling towards centre of earth with acceelration gh.

Therefore Astronaut does not exert any force on the floor of satellite and hence floor of

satellite do not exert any normal reaction on astronaut.

If „a‟ is centripetal acceleration of the satellite then force exerted by the floor on austronaut

= N= mgh - ma . But a = gh . Hence N = 0. Therefore astronaut has a feeling of weightlessness.

Weight of astronaut is not zero in orbiting satellite. But he feels weightless because of

absence of reactive force.

Effect of altitude on value of g

If object of mass m is at height h above the earth‟s surface where gravitational acceleration is

gh

Distance of mass from center of earth is R+h

Weight = centripetal force

2( )h

GMmmg

R h

2( )h

GMmg

R h

Comparison between g on surface of earth and g at height h

2

2( )

hg R

g R h

2

2

2 1

hg R

g hR

R

2

1hg h

g R

If h << R (higher powers of h

R can be neglected)

21hg h

g R

(Neglecting higher powers of

h

R)

Variation of g with depth:

21h

hg g

R

SAPTARS HI


Suppose that the body of mass m is placed inside the earth at a depth d below the earth‟s

surface. As seen from figure, the body is now (i) on surface of a smaller sphere of radius r = R

– d and (II) inside an outer spherical shell of thickness d.

It can be proved that the outer spherical shell does not exert any gravitational force on the

body. Hence the only gravitational force acting on the body is the force exerted by the inner

sphere of radius R-d .

Let M be mass of earth and R be radius of the earth. Acceleration due to gravity on surface of

earth is given by ,

2

GMg

R

Where, G is gravitational constant .

Let dg be gravitational acceleration at depth d from surface of earth. Gravitational

acceleration at depth d is because of inner sphere of radius R-d.

2d

GMg

R d

2

1

2( )

dg M R

g M R d

Let be the density of the earth. Then,

mass of earth volume density

34

3M R

The mass (M1) of the inner sphere is

3

1

4( )

3M R d

3

1

3

( )M R d

M R

Substitution this value in Eq. we get

3 2

3 2

( )1

( )

dg R d R R d d

g R R d R R

This relation shows that the value of g decreases with the increase in the depth below the

earth‟s surface.

At the centre of the earth, d= R, hence from Eq, gd = 0 at the centre of the earth. Thus, if the

body of mass m is taken to the centre of the earth, its weight will be equal to zero.

1d

dg g

R

SAPTARS HI


Graphical representation of variation of g with altitude and depth

Variation of g with latitude (Effect of rotation of the earth):

Earth is rotating about its polar axis from west to east with uniform angular velocity . As

the earth rotates, any body situated on the surface of the earth performs circular motion

Consider a body situated at a point P on the surface of at the latitude . Due to the rotation of

the earth point P revolves along a circle of centre O‟ and radius PO‟. Let PO‟= r.

If R is the radius of the earth, '

cosPO

R

cosr R Centripetal acceleration needed for body at P to perform circular motion is

2

ra r

Component of centripetal acceleration along the radius of earth is

cosra a

2 cosa r 2 cos cosa R 2 2cosa R

If g

1 is the effective value of the acceleration due to gravity at point P ,

1g g a

Where g is acceleration due to gravity in the absence of rotation of earth.

Thus the effect of the rotation of the earth is to reduce the value of the acceleration due to

gravity. The reduction is very small. As latitude increases, cos decreases and g

increases.

Thus as we move from equator to pole acceleration due to gravity increases.

Case -I

At Equator

=0

cos 1

2

Eg g R

1 2 2cosg g R

SAPTARS HI


Where Eg is acceleration due to gravity at equator.

At equator maximum reduction in acceleration due to gravity takes place , hence g is

minimum at equator.

Case -II

At poles

2

cos 0

pg g

Where pg is acceleration due to gravity at poles.

At poles there is no reduction in acceleration due to gravity due to rotation of earth,

hence g is maximum at poles.

Communication Satellite and uses of satellites

An artificial satellite revolving in a circular orbit round the earth in the equatorial plane, in the

same sense of the rotation of the earth and having same period of revolution as the period of

rotation of the earth (i.e. 1 day = 24 hours = 86400 seconds) is called geo-stationary satellite

As relative angular velocity or the satellite with respective to the earth is zero, hence it

appears stationary from the earth‟s surface, therefore it is known as geo-stationary satellite or

geo-synchronous satellite.

It is mainly used for communication purposes so it is called communication satellite.

Geostationary orbit is at height 35870 Km from surface of earth. When satellite revolves in

this orbit it appears stationary from surface of earth, hence this orbit is called parking orbit.

Polar Satellite

They revolve round the earth in polar planes. (Perpendicular to equatorial planes.)

They are at low altitude (500 km to 800 km) satellites. Since its time period is 100 minutes. it

crosses any altitude many times a day.

It is mainly used for remote sensing.

Uses of satellite

For the transmission of Television and radio wave signals over large areas of the earth‟s

surface.

For broadcasting telecommunication

For military purposes

For weather forecasting and meteorological purposes

For astronomical observations.

For study of solar and cosmic radiations.

To relay distress signals from ships.

For geopositions system (GPS).

To transmit cyclone warnings to coastal villages.

Polar satellites are used for remote sensing.

Numerical Problem

(Take g = 9.8 m/s2 , M = 6 x 10

24 kg, G = 6.67 x 10

-11 Nm

2 kg

-2 , R = 6400 km)

1) Find the altitude at which acceleration due to gravity is 25% of that at the surface of the earth.

0' 96 , 2-marks) ( Ans h = R = 6400 km )

SAPTARS HI


2) Show that the critical velocity of a body orbiting in a circular orbit very close to surface

of a planet of radius R and mean density is 2R3

g (0' 99 , 2-marks) (M'97 , 2-

marks)

3) A body weights 3.5 kg-wt on the surface of the earth. What will be its weight on the surface

of a planet whose mass is 7

1 th mass of the earth and radius half that of the earth? (0' 2000 ,

2-marks) ( Ans 2 kgwt )

4) What would be the duration of the year if the distance between the earth and sun gets

doubled? Will this affect the escape velocity of an object from the surface the earth? (M' 79

, 3-Marks) ( Ans: 1032 days , no change in escape velocity )

5) A satellite is revolving round a planet in a circular orbit with a velocity of 8 km/s a height

where the value of acceleration due to gravity is 8 m/s2. How high is satellite from the surface

of the planet? Radius of the planet is 6000 km. (M' 88 , 3-Marks) ( h = 2000 km )

6) A communication satellite appears stationary from a place on the equator. Find height of the

satellite from the surface of the earth. G = 6.67 X 10-11

Nm2 kg

-2. of the earth = 6 x 10

24 kg.

Radius of the earth = 6400 km. (M' 84 , 0'83 , 4-Marks) ( h = 35.9 X 106 m)

7) What will be the binding energy of a satellite of mass 2000 kg moving in a circular orbit close

to the earth? (M' 87 , 3-marks) ( Ans : 6.253 X 1010

J )

8) Find the total energy and binding energy of an artificial satellite of mass 100 kg orbiting at a

height of 1600 km above the surface of the earth. (M' 89 , 3-marks) ( Ans - 2.501 X 109 J ,

2.501 X 109 J )

9) The binding energy of a satellite is 4 x 108 J. Calculate its kinetic energy and potential energy.

(M' 94) ( Ans KE = 4 x 108 J , PE = - 8 x 10

8 J )

10) The escape velocity of a body from the surface of earth is 11.2 km/s. The mass of moon

is 80

1 th of the earth's mass and the radius of the moon is

4

1 th of the radius of the earth.

Find the escape velocity from the moon's surface (M' 2006 , 2-marks) ( Ans 2.5 Km/s

)

11) Find the angular speed of rotation of the earth so that bodies on the equator would feel

no weight (Radius of earth = 6400 km, g = 9.8 m/s2) (M'2003 , 2-Marks) (Ans 1.237 X

10-3 rad/s)

12) A body Weighs 3.5 kgwt on the surface of the earth , what will be wight on the surface

of a planet whose mass is 1

7th that of earth and radius half that of earth. [ Ans 2 kg wt]

13) Body of having weight 50 Kgwt on surface of earth is taken to depth R/2 where R is

radius of earth. What will be weight of body? ( Ans 25 kgwt )

14) The mass of a body is 100kg on the surface of earth , what will be its mass and weight at

altitude 1000 km. [ R = 6400 m , g = 9.8 m/s2]

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