Circular motion

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Circular MotionProf. Mukesh N. TekwaniDepartment of PhysicsI. Y. College,[email protected] Prof. Mukesh N Tekwani, 2011

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Circular Motion

2 Prof. Mukesh N Tekwani, 2011

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Why study Circular Motion?To understand Motion of planetsMotion of electrons around the nucleusMotion of giant wheelMotion of space stationsMotion of moon and satellites

Prof. Mukesh N Tekwani, 20113

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Circular MotionIt is defined as the motion of a particle along a complete circle or part of a circle.Fore circular motion, it is NOT necessary that the body should complete a full circle.Even motion along arc of a circle is circular motion Prof. Mukesh N Tekwani, 20114

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5Circular Motion

X = 0y

if

How do we locate something on a circle?Give its angular position What is the location of

900 or /2


5Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site (www.mhhe.com/grr), Instructor Resources: CPS by eInstruction, Chapter 5, Questions 1, 2, 3, 13, 14, 16, and 17.

6Circular Motion is the angular position.Angular displacement:

Note: angles measured Clockwise (CW) are negative and angles measured (CCW) are positive. is measured in radians.2 radians = 360 = 1 revolution

xy

if


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Angular DisplacementAngular displacement is defined as the angle described by the radius vector


aInitial position of particle is aFinal position of particle is bAngular displacement in time t is

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Angular Displacement Prof. Mukesh N Tekwani, 20118

S = r

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xy

if

rarclength = s = r

is a ratio of two lengths; it is a dimensionless ratio! This is a radian measure of angleIf we go all the way round s =2r and =2 Prof. Mukesh N Tekwani, 2011

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Circular Motion10 Prof. Mukesh N Tekwani, 2011

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Right Hand Rule Prof. Mukesh N Tekwani, 201111

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.

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Stop & Think Pg 2 Prof. Mukesh N Tekwani, 201112

Are the following motions same or different?The motion of the tip of second hand of a clock.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.

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Angular Velocity

13 Prof. Mukesh N Tekwani, 2011Consider aparticleperforming U.C.M. in an anticlockwise direction.

In a very small time interval dt, theparticlemoves from the point P1to the point P2.

Distance travelled along the arc is ds.

In the same time interval,the radiusvector rotates through an angle d.

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Angular Velocity

14 Prof. Mukesh N Tekwani, 2011Definition:

For a particle performing circular motion, the rate of change of angular displacement with respect to time, is called as angular velocity.If angular displacement is in a short time interval t , then,Angular velocity = lim 0

Therefore, =

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Angular Velocity15 Prof. Mukesh N Tekwani, 2011Important points about Angular Velocity:

Angular velocity is a vector quantity.Its direction is in the direction of angular displacementIts direction is given by the right hand rule.Units : rad / sec

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Angular Speed16 Prof. Mukesh N Tekwani, 2011For a particle performing circular motion, the angle swept out by the radius vector per unit time is called the angular speed.

Angular speed is a scalar quantity.It has no directionUnits : rad / sec

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Angular Acceleration17 Prof. Mukesh N Tekwani, 2011For a particle performing circular motion, the rate of change of angular velocity with respect to time is called the angular acceleration.

Angular acceleration () is a vector quantity.Its direction is the same as the direction of angular velocity As the particle is speeding up, its ang. accln is in direction of . Units : rad / sec2

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Relation Between Linear Velocity and Angular Velocity18 Prof. Mukesh N Tekwani, 2011

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Relation Between Linear Velocity and Angular Velocity19 Prof. Mukesh N Tekwani, 2011

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Tangential Acceleration20 Prof. Mukesh N Tekwani, 2011

Definition:

For a particle performing circular motion, the linear acceleration tangential to the path that producers a change in the linear speed of the particle is called the tangential acceleration.Tangential acceleration when linear speed is (a) increasing, (b) decreasing

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Tangential Acceleration21 Prof. Mukesh N Tekwani, 2011Explanation:

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Acceleration of a Particle Performing Circular Motion22 Prof. Mukesh N Tekwani, 2011

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


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Velocity and Speed in UCM

Is speed changing?No, speed is constantIs velocity changing?

Yes, velocity is changing because velocity is a vector and direction is changing at every point 26 Prof. Mukesh N Tekwani, 2011

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Velocity and Speed in UCMIs speed changing?No, speed is constantIs velocity changing?

Yes, velocity is changing because velocity is a vector and direction is changing at every point 27 Prof. Mukesh N Tekwani, 2011

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Examples of Centripetal forceA body tied to a string and whirled in a horizontal circle CPF is provided by the tension in the string.28 Prof. Mukesh N Tekwani, 2011

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Examples of Centripetal forceFor a car travelling around a circular road with uniform speed, the CPF is provided by the force of static friction between tyres of the car and the road.29 Prof. Mukesh N Tekwani, 2011

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Examples of Centripetal forceIn case of electrons revolving around the nucleus, the centripetal force is provided by the electrostatic force of attraction between the nucleus and the electrons

In case of the motion of moon around the earth, the CPF is provided by the ______ force between Earth and Moon30 Prof. Mukesh N Tekwani, 2011

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Centripetal ForceCentripetal forceIt is the force acting on a particle performing UCM and this force is along the radius of the circle and directed towards the centre of the circle.

REMEMBER!Centripetal force- acting on a particle performing UCM- along the radius- acting towards the centre of the circle.


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Properties of Centripetal ForceCentripetal force is a real force CPF is necessary for maintaining UCM.CPF acts along the radius of the circleCPF is directed towards center of the circle.CPF does not do any workF = mv2/ r


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Radial Acceleration

P(x, y)vOXYMNr

xyLet P be the position of the particle performing UCM

r is the radius vector

= t . This is the angular displacement of the particle in time t secs

V is the tangential velocity of the particle at point P.

Draw PM OX

The angular displacement of the particle in time t secs is

LMOP = = t 33 Prof. Mukesh N Tekwani, 2011

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Radial Acceleration

P(x, y)vOXYMNr

xyThe position vector of the particle at any time is given by:

r = ix + jy

From POM

sin = PM/OP

sin = y / r

y = r sin

But = t

y = r sin t 34 Prof. Mukesh N Tekwani, 2011

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Radial Acceleration

P(x, y)vOXYMNr

xySimilarly,

From POM

cos = OM/OP

cos = x / r

x = r cos

But = t

x = r cos t 35 Prof. Mukesh N Tekwani, 2011

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Radial Acceleration36 Prof. Mukesh N Tekwani, 2011The velocity of particle at any instant (any time) is called its instantaneous velocity.

The instantaneous velocity is given by

v = dr / dt

v = d/dt [ ir cos wt + jr sin wt]

v = - i r w sin wt + j r w cos wt

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Radial Acceleration37 Prof. Mukesh N Tekwani, 2011The linear acceleration of the particle at any instant (any time) is called its instantaneous linear acceleration.

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Radial Acceleration38 Prof. Mukesh N Tekwani, 2011Therefore, the instantaneous linear acceleration is given by

a = - w2 r

Importance of the negative sign: The negative sign in the above equation indicates that the linear acceleration of the particle and the radius vector are in opposite directions.

ar

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Relation Between Angular Acceleration and Linear Acceleration39 Prof. Mukesh N Tekwani, 2011The acceleration of a particle is given by

. (1)

But v = r w

a = .... (2)

r is a constant radius,

But

is the angular acceleration

a = r (3)

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Relation Between Angular Acceleration and Linear Acceleration40 Prof. Mukesh N Tekwani, 2011

v = w x r

Differentiating w.r.t. time t,

But

and

linear acceleration

a = aT + aR

aT is called the tangential component of linear acceleration

aR is called the radial component of linear acceleration

For UCM, w = constant, so

a = aR in UCM, linear accln is centripetal accln

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Centrifugal Force41 Prof. Mukesh N Tekwani, 2011

Centrifugal force is an imaginary force (pseudo force) experienced only in non-inertial frames of reference.

This force is necessary in order to explain Newtons laws of motion in an accelerated frame of reference.

Centrifugal force is acts along the radius but is directed away from the centre of the circle.

Direction of centrifugal force is always opposite to that of the centripetal force.

Centrifugal force

Centrifugal force is always present in rotating bodies

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Examples of Centrifugal Force42 Prof. Mukesh N Tekwani, 2011

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.

The children sitting in a merry-go-round experience an outward force as the merry-go-round rotates about the vertical axis.

Centripetal and Centrifugal forces DONOT constitute an action-reaction pair. Centrifugal force is not a real force. For action-reaction pair, both forces must be real.

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Banking of Roads43 Prof. Mukesh N Tekwani, 2011

When a car is moving along a curved road, it is performing circular motion. For circular motion it is not necessary that the car should complete a full circle; an arc of a circle is also treated as a circular path.

We know that centripetal force (CPF) is necessary for circular motion. If CPF is not present, the car cannot travel along a circular path and will instead travel along a tangential path.

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

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Friction between Tyres and Road45 Prof. Mukesh N Tekwani, 2011

The centripetal force for circular motion of the car is provided by the frictional force between the tyres of the car and the road.

Let m = mass of the carV = speed of the car, andR = radius of the curved road.

Since centripetal force is provided by the frictional force, CPF = frictional force (provide by means equal to )

( is coefficient of friction between tyres & road)

So and

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Thus, the maximum velocity with which a car can safely travel along a curved road is given by

If the speed of the car increases beyond this value, the car will be thrown off (skid).

If the car has to move at a higher speed, the frictional force should be increased. But this cause wear and tear of tyres.

The frictional force is not reliable as it can decrease on wet roads

So we cannot rely on frictional force to provide the centripetal force for circular motion.

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R1 and R2 are reaction forces due to the tyres

mg is the weight of the car, acting vertically downwards

F1 and F2 are the frictional forces between the tyres and the road.

These frictional forces act towards the centre of the circular path and provide the necessary centripetal force.

Center of circular path

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Friction between Tyres and Road Car Skidding49 Prof. Mukesh N Tekwani, 2011

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Banked Roads50 Prof. Mukesh N Tekwani, 2011

What is banking of roads?

The process of raising the outer edge of a road over the inner edge through a certain angle is known as banking of road.

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Purpose of Banking of Roads:Banking of roads is done:To provide the necessary centripetal force for circular motionTo reduce wear and tear of tyres due to frictionTo avoid skiddingTo avoid overturning of vehicles

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What is angle of banking?

R

R cos W = mgR sin The angle made by the surface of the road with the horizontal surface is called as angle of banking.Horizontal Surface of road

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Consider a car moving along a banked road.

Letm = mass of the carV = speed of the car isangleof banking

R

R cos W = mgR sin

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The forces acting on the car are:

(i) Its weightmgacting vertically downwards.

(ii) The normal reactionRacting perpendicular to the surface of the road.

R

R cos W = mgR sin

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The normal reaction can be resolved (broken up) into two components:

R cos is the vertical component

R sin is the horizontal component

R

R cos W = mgR sin

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Since the vehicle has no vertical motion, the weight is balanced by the vertical component

R cos = mg (1)

(weight is balanced by vertical component means weight is equal to vertical component)

R

R cos W = mgR sin

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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 sin = (2)

R

R cos W = mgR sin

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Dividing (2) by (1), we get

R sin =

mgR cos = tan-1 ( )

So,

tan =

Therefore, the angle of banking is independent of the mass of the vehicle.

The maximum speed with which the vehicle can safely travel along the curved road is

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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?

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Maximum Speed of a Vehicle on a Banked Road with Friction61 Prof. Mukesh N Tekwani, 2011

Consider a vehicle moving along a curved banked road.

Let m = mass of vehicler = radius of curvature of road = angle of bankingF = frictional force between tyres and road.

The forces acting on the vehicle are shown in the diagram.

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The forces acting on the vehicle are:Weight of the vehicle mg, acting vertically downwardsNormal reaction N acting on vehicle, perpendicular to the surface of the road.Friction force between tyres and road.

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

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The normal reaction N is resolved into 2 components:

N cos is vertical component of NN sin is horizontal component of N

Resolving the Normal reaction

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The frictional force f is resolved into 2 components:

f cos is horizontal component of f

f sin is vertical component of f f sin f f cos

Resolving the frictional force

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All forces acting on vehiclef sin f f cos

N cos N

N sin mgThe vertical component N cos is balanced by the weight of the vehicle and the component f sin

N cos = mg + f sin

mg = N cos - f sin .. (1)

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N cos N

N sin mgThe horizontal component N sin and f cos provide the centripetal force for circular motion

N sin + f cos =

= N sin + f cos (2)

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N cos N

N sin mgDividing (2) by (1), we get

mg = N sin + f cos N cos - f sin

= N sin + f cos N cos - f sin

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f sin f f cos

N cos N

N sin mgLet Vmax be the maximum speed of the vehicle.

Frictional force at this speed will be fm = s N

= N sin + fm cos N cos - fm sin

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But fm = s N

= N sin + s N cos N cos - s N sin Dividing numerator and denominator of RHS by cos , we get=

N sin s N cos N cos N cos +N cos s N sin N cos N cos -

= tan + s1 - s tan

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=

tan + s1 - s tan

r g =

tan + s1 - s tan

r gThis is the maximum velocity with which a vehicle can travel on a banked road with friction.For a frictionless road, s = 0 =

tan + 01 - 0

r g

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Conical Pendulum71 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 circle and the string describes a cone.

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Conical Pendulum72 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.

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Conical Pendulum Time Period73 Prof. Mukesh N Tekwani, 2011

Consider a bob of mass m revolving in a horizontal circle of radius r.

Letv = linear velocity of the bobh = heightT = tension in the string = semi vertical angle of the coneg = acceleration due to gravityl = length of the stringT cos

T sin

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The forces acting on the bob at position A are:

Weight of the bob acting vertically downward

Tension T acting along the string.

T cos

T sin

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

Tcos acting vertically upwards. This force is balanced by the weight of the bobT cos = mg ..(1)

T cos

T sin

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(ii) T sin acting along the radius of the circle and directed towards the centre of the circle

T sin provides the necessary centripetal force for circular motion.

T sin = .(2)

Dividing (2) by (1) we get,

.(3)T cos

T sin

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T cos

T sin

This equation gives the speed of the bob.

But v = rw

rw =

Squaring both sides, we get

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T cos

T sin

From diagram, tan = r / h

r 2w2 = rg

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T cos

T sin

Periodic Time of Conical Pendulum

But

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

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T cos

T sin

Periodic Time of Conical PendulumBut cos = h/l

So the eqn becomes,

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T cos

T sin

Factors affecting time period of conical pendulum: The period of the conical pendulum depends on the following factors:

Length of the pendulumAngle of inclination to the verticalAcceleration due to gravity at the given place

Time period is independent of the mass of the bob

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Vertical Circular Motion Due to Earths Gravitation82 Prof. Mukesh N Tekwani, 2011

Consider an object of mass m tied to the end of an inextensible string and whirled in a vertical circle of radius r.

mgT1Orv1AT2Bv2v3C

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mgT1Orv1AT2Bv2v3CHighest Point A:Let the velocity be v1

The forces acting on the object at A (highest point) are:Tension T1 acting in downward directionWeight mg acting in downward direction

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mgT1Orv1AT2Bv2v3CAt 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)

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mgT1Orv1AT2Bv2v3CLowest Point B:Let the velocity be v2

The forces acting on the object at B (lowest point) are:Tension T2 acting in upward directionWeight mg acting in downward direction

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mgT1Orv1AT2Bv2v3CAt the lowest point B:

(2)

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mgT1Orv1AT2Bv2v3CLinear velocity of object at highest point A:

The object must have a certain minimum velocity at point A so as to continue in circular path.

This velocity is called the critical velocity. Below the critical velocity, the string becomes slack and the tension T1 disappears (T1 = 0)

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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.

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mgT1Orv1AT2Bv2v3CLinear velocity of object at lowest point B:

When the object moves from the lowest position to the highest position, the increase in potential energy is mg x 2r

By the law of conservation of energy,KEA + PEA = KEB + PEB

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At the highest point A, the minimum velocity must be

Using this in

we get,

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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.

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Linear Velocity at a point midway between top and bottom positions in a vertical circle93 Prof. Mukesh N Tekwani, 2011

Orv1ABv2v3C

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

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Kuch Self-study bhi Karo Na!94 Prof. Mukesh N Tekwani, 2011

Derive an expression for the tension in the string of a conical pendulum.Write kinematical equations for circular motion in analogy with linear motion.Derive the expression for the tension in a string in vertical circular motion at any position.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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Education

Circular motion