Circular motion
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Circular Motion

Circular Motion

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Circular motion Circular motion Presentation Transcript

  • Prof. Mukesh N. Tekwani Department of Physics I. Y. College, Mumbaimukeshtekwani@outlook.com © Prof. Mukesh N Tekwani, 2011 1
  • Circular Motion © Prof. Mukesh N Tekwani, 2011 2
  • Why study Circular Motion?• To understand – Motion of planets – Motion of electrons around the nucleus – Motion of giant wheel – Motion of space stations – Motion of moon and satellites © Prof. Mukesh N Tekwani, 2011 3 View slide
  • Circular Motion• It is defined as the motion of a particle along a complete circle or part of a circle.• Fore circular motion, it is NOT necessary that the body should complete a full circle.• Even motion along arc of a circle is circular motion © Prof. Mukesh N Tekwani, 2011 4 View slide
  • Circular MotionHow do we locate something on a circle?Give its angular position θ y f What is the location of i X 900 or π/2 θ=0 © Prof. Mukesh N Tekwani, 2011 5
  • Circular Motion y f is the angular position. i Angular displacement: x f iNote: angles measured Clockwise (CW) are negative andangles measured (CCW) are positive. is measured inradians. 2 radians = 360Tekwani, 1 revolution © Prof. Mukesh N = 2011 6
  • Angular Displacement• Angular displacement is defined as the angle described by the radius vector Initial position of particle is a Final position of particle is b Angular displacement in time t is Θ a © Prof. Mukesh N Tekwani, 2011 7
  • Angular Displacement S =r Θ © Prof. Mukesh N Tekwani, 2011 8
  • arclength = s = ry s f r r i is a ratio of two lengths; x it is a dimensionless ratio! This is a radian measure of angle If we go all the way round s =2πr and Δθ =2 π © Prof. Mukesh N Tekwani, 2011 9
  • Right Hand RuleIf the fingers of theright hand are curledin the direction ofrevolution of theparticle, then theoutstretched thumbgives the direction ofthe angulardisplacement vector. © Prof. Mukesh N Tekwani, 2011 11
  • Stop & Think – Pg 2Are 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
  • Centripetal Force• UCM is an accelerated motion. Why?• UCM is accelerated motion because the velocity of the body changes at every instant (i.e. every moment)• But, according to Newton’s Second Law, there must be a force to produce this acceleration.• This force is called the centripetal force.• Therefore, Centripetal force is required for circular motion. No centripetal force +> no circular motion. © Prof. Mukesh N Tekwani, 2011 25
  • Velocity and Speed in UCM Is speed changing? No, speed is constant Is velocity changing? Yes, velocity is changing because velocity is a vector – and direction is changing at every point © Prof. Mukesh N Tekwani, 2011 26
  • Velocity and Speed in UCM Is speed changing? No, speed is constant Is velocity changing? Yes, velocity is changing because velocity is a vector – and direction is changing at every point © Prof. Mukesh N Tekwani, 2011 27
  • Examples of Centripetal forceA body tied to a stringand whirled in ahorizontal circle – CPFis provided by thetension in the string. © Prof. Mukesh N Tekwani, 2011 28
  • 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 tyres of the car and the road. © Prof. Mukesh N Tekwani, 2011 29
  • Examples of Centripetal force• In case of electrons revolving around the nucleus, the centripetal force is provided by the electrostatic force of attraction between the nucleus and the electrons• In case of the motion of moon around the earth, the CPF is provided by the ______ force between Earth and Moon © Prof. Mukesh N Tekwani, 2011 30
  • Centripetal Force• Centripetal force – It is the force acting on a particle performing UCM and this force is along the radius of the circle and directed towards the centre of the circle. REMEMBER! Centripetal force - acting on a particle performing UCM - along the radius - acting towards the centre of the circle. © Prof. Mukesh N Tekwani, 2011 31
  • Properties of Centripetal Force1. Centripetal force is a real force2. CPF is necessary for maintaining UCM.3. CPF acts along the radius of the circle4. CPF is directed towards center of the circle.5. CPF does not do any work6. F = mv2/ r © Prof. Mukesh N Tekwani, 2011 32
  • Radial Acceleration Let P be the position of the particleY performing UCM v P(x, y) r is the radius vectorN Θ = ωt . This is the angular displacement of r the particle in time t secs y θ V is the tangential velocity of the particle at O x M X point P. Draw PM ┴ OX The angular displacement of the particle in time t secs is LMOP = Θ = ωt © Prof. Mukesh N Tekwani, 2011 33
  • Radial AccelerationY The position vector of the particle at v any time is given by: P(x, y) r = ix + jyN r From ∆POM y θ sin θ = PM/OP O x M X ∴ sin θ = y / r ∴y = r sin θ But θ = ωt ∴ y = r sin ωt © Prof. Mukesh N Tekwani, 2011 34
  • Radial AccelerationY Similarly, v From ∆POM P(x, y)N cos θ = OM/OP r y ∴ cos θ = x / r θ O x M X ∴ x = r cos θ But θ = ωt ∴ x = r cos ωt © Prof. Mukesh N Tekwani, 2011 35
  • Radial AccelerationThe velocity of particle at any instant (any time) is called itsinstantaneous velocity.The instantaneous velocity is given byv = dr / dt∴ v = d/dt [ ir cos wt + jr sin wt]∴ v = - i r w sin wt + j r w cos wt © Prof. Mukesh N Tekwani, 2011 36
  • Radial AccelerationThe linear acceleration of the particle at any instant (any time) iscalled its instantaneous linear acceleration. © Prof. Mukesh N Tekwani, 2011 37
  • Radial AccelerationTherefore, the instantaneous linearacceleration is given by∴ a = - w2 r r aImportance of the negative sign:The negative sign in the aboveequation indicates that the linearacceleration of the particle and theradius vector are in oppositedirections. © Prof. Mukesh N Tekwani, 2011 38
  • Relation Between AngularAcceleration and Linear AccelerationThe acceleration of a particle ∵ r is a constant radius,is given by ………………. (1) ∴But v = r w But∴ α is the angular acceleration∴a = .... (2) ∴ a=rα ………………………(3) © Prof. Mukesh N Tekwani, 2011 39
  • Relation Between AngularAcceleration and Linear Acceleration ∴v=wxrDifferentiating w.r.t. time t, ∴ linear acceleration a = a T + aR aT is called the tangential component of linear acceleration aR is called the radial component of linear accelerationBut For UCM, w = constant, soand ∴ a = aR ∴ in UCM, linear accln is centripetal accln © Prof. Mukesh N Tekwani, 2011 40
  • Centrifugal Force1. Centrifugal force is an imaginary force (pseudo force) experienced only in non-inertial frames of reference.2. This force is necessary in order to explain Newton’s laws of motion in an accelerated frame of reference.3. Centrifugal force is acts along the radius but is directed away from the centre of the circle.4. Direction of centrifugal force is always opposite to that of the centripetal force.5. Centrifugal force6. Centrifugal force is always present in rotating bodies © Prof. Mukesh N Tekwani, 2011 41
  • Examples of Centrifugal Force1. When a car in motion takes a sudden turn towards left, passengers in the car experience an outward push to the right. This is due to the centrifugal force acting on the passengers.2. The children sitting in a merry-go-round experience an outward force as the merry-go-round rotates about the vertical axis.Centripetal and Centrifugal forces DONOT constitute an action-reaction pair. Centrifugal force is not a real force. For action-reaction pair, both forces must be real. © Prof. Mukesh N Tekwani, 2011 42
  • Banking of Roads1. When a car is moving along a curved road, it is performing circular motion. For circular motion it is not necessary that the car should complete a full circle; an arc of a circle is also treated as a circular path.2. We know that centripetal force (CPF) is necessary for circular motion. If CPF is not present, the car cannot travel along a circular path and will instead travel along a tangential path. © Prof. Mukesh N Tekwani, 2011 43
  • Banking of Roads3. 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 © Prof. Mukesh N Tekwani, 2011 44
  • Friction between Tyres and RoadThe centripetal force for circular motion of the car is provided bythe 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” ) (µ is coefficient of friction between tyres & road)So and © Prof. Mukesh N Tekwani, 2011 45
  • Friction between Tyres and RoadThus, the maximum velocity with which a car can safely travelalong a curved road is given byIf the speed of the car increases beyond this value, the car will bethrown off (skid).If the car has to move at a higher speed, the frictional forceshould be increased. But this cause wear and tear of tyres.The frictional force is not reliable as it can decrease on wet roadsSo we cannot rely on frictional force to provide the centripetalforce for circular motion. © Prof. Mukesh N Tekwani, 2011 46
  • Friction between Tyres and Road R1 and R2 are reaction forces due to the tyres mg is the weight of the car,Center of acting vertically downwards circular path F1 and F2 are the frictional forces between the tyres and the road. These frictional forces act towards the centre of the circular path and provide the necessary centripetal force. © Prof. Mukesh N Tekwani, 2011 47
  • Friction between Tyres and Road © Prof. Mukesh N Tekwani, 2011 48
  • Friction between Tyres and Road – Car Skidding © Prof. Mukesh N Tekwani, 2011 49
  • Banked RoadsWhat is banking of roads?The process of raising theouter edge of a road overthe inner edge through acertain angle is known asbanking of road. © Prof. Mukesh N Tekwani, 2011 50
  • Banking of RoadsPurpose of Banking of Roads:Banking of roads is done:1. To provide the necessary centripetal force for circular motion2. To reduce wear and tear of tyres due to friction3. To avoid skidding4. To avoid overturning of vehicles © Prof. Mukesh N Tekwani, 2011 51
  • Banked Roads © Prof. Mukesh N Tekwani, 2011 52
  • Banked RoadsWhat is angle of R cos θ Rbanking? ΘThe angle made by the R sin θsurface of the road withthe horizontal surface iscalled as angle of banking. Θ Horizontal W = mg © Prof. Mukesh N Tekwani, 2011 53
  • Banked RoadsConsider a car moving R cos θ Ralong a banked road. Θ R sin θLetm = mass of the carV = speed of the car Θθ is angle of banking W = mg © Prof. Mukesh N Tekwani, 2011 54
  • Banked RoadsThe forces acting on the R cos θ Rcar are: Θ R sin θ(i) Its weight mg actingvertically downwards. Θ(ii) The normalreaction R actingperpendicular to the W = mgsurface of the road. © Prof. Mukesh N Tekwani, 2011 55
  • Banked RoadsThe normal reaction can be R cos θresolved (broken up) into R Θtwo components: R sin θ1. R cosθ is the vertical component Θ2. R sinθ is the horizontal component W = mg © Prof. Mukesh N Tekwani, 2011 56
  • Banked RoadsSince the vehicle has no R cos θvertical motion, the weight R Θis balanced by the verticalcomponent R sin θR cosθ = mg …………… (1) Θ(weight is balanced byvertical component meansweight is equal to verticalcomponent) W = mg © Prof. Mukesh N Tekwani, 2011 57
  • Banked RoadsThe horizontal component R cos θis the unbalanced R Θcomponent . This horizontalcomponent acts towards R sin θthe centre of the circularpath. ΘThis component providesthe centripetal force forcircular motion W = mgR sinθ = …………… (2) © Prof. Mukesh N Tekwani, 2011 58
  • Banked RoadsDividing (2) by (1), we get θ = tan-1 ( )R sinθ = mg Therefore, the angle of bankingR cos θ is independent of the mass of the vehicle.So, The maximum speed with which the vehicle can safelytan θ = travel along the curved road is © Prof. Mukesh N Tekwani, 2011 59
  • Banked Roads A car travels at a constant speed around two curves. Where is the car most likely to skid? Why? 2 mv F = ma r Smaller radius: larger centripetal force is required to keep it in uniform circular motion. © Prof. Mukesh N Tekwani, 2011 60
  • Maximum Speed of a Vehicle on a Banked Road with Friction Consider a vehicle moving along a curved banked road. Let m = mass of vehicle r = radius of curvature of road θ = angle of banking F = frictional force between tyres and road. The forces acting on the vehicle are shown in the diagram. © Prof. Mukesh N Tekwani, 2011 61
  • Maximum Speed of a Vehicle on a Banked Road with FrictionThe forces acting on the vehicle are:1) Weight of the vehicle mg, acting vertically downwards2) Normal reaction N acting on vehicle, perpendicular to the surface of the road.3) Friction force between tyres and road.Forces N and frictional force f are now resolved into twocomponents © Prof. Mukesh N Tekwani, 2011 62
  • Maximum Speed of a Vehicle on a Banked Road with FrictionResolving the Normal reaction The normal reaction N is resolved into 2 components: 1) N cos θ is vertical component of N 2) N sin θ is horizontal component of N © Prof. Mukesh N Tekwani, 2011 63
  • Maximum Speed of a Vehicle on a Banked Road with FrictionResolving the frictional force The frictional force f is resolved into 2 components: f cos θ 1) f cos θ is horizontal component of f f θ f sin θ 2) f sin θ is vertical component of f © Prof. Mukesh N Tekwani, 2011 64
  • Maximum Speed of a Vehicle on a Banked Road with Friction All forces acting on vehicle N The vertical component N cos θ is N cos θ balanced by the weight of the θ vehicle and the component f sin θ N sin θf cos θ f θ f sin θ ∴ N cos θ = mg + f sin θ mg ∴ mg = N cos θ - f sin θ .…. (1) © Prof. Mukesh N Tekwani, 2011 65
  • Maximum Speed of a Vehicle on a Banked Road with Friction All forces acting on vehicle N The horizontal component N sin θ N cos θ and f cos θ provide the centripetal θ force for circular motion N sin θf cos θ mv 2 f θ f sin θ ∴ N sin θ + f cos θ = 2 r mg mv ∴ = N sin θ + f cos θ … (2) r © Prof. Mukesh N Tekwani, 2011 66
  • Maximum Speed of a Vehicle on a Banked Road with Friction All forces acting on vehicle N Dividing (2) by (1), we get N cos θ 2 mv θ N sin θ r N sin θ + f cos θf cos θ θ = f f sin θ N cos θ - f sin θ mg mg v2 N sin θ + f cos θ ∴ = rg N cos θ - f sin θ © Prof. Mukesh N Tekwani, 2011 67
  • Maximum Speed of a Vehicle on a Banked Road with Friction Let Vmax be the maximum speed of N the vehicle. N cos θ θ Frictional force at this speed will be N sin θ fm = µs Nf cos θ θ f f sin θ 2 N sin θ + fm cos θ mg v max ∴ = rg N cos θ - fm sin θ © Prof. Mukesh N Tekwani, 2011 68
  • Maximum Speed of a Vehicle on a Banked Road with FrictionBut fm = µs N N sin θ µs N cos θ + 2 N cos θ N cos θ v 2 N sin θ + µs N cos θ v∴ max = ∴ max = rg N cos θ - µs N sin θ rg N cos θ µs N sin θ - N cos θ N cos θDividing numerator and 2 tan θ + µs v max =denominator of RHS by ∴cos θ, we get rg 1 - µs tan θ © Prof. Mukesh N Tekwani, 2011 69
  • Maximum Speed of a Vehicle on a Banked Road with Friction tan θ + µs For a frictionless road, µs = 0 2∴ v m ax = rg 1 - µs tan θ tan θ + 0 ∴ v m ax = r g 1 - 0 tan θ + µs∴ v m ax = rg 1 - µs tan θ vmax rg tanThis is the maximum velocitywith which a vehicle can travelon a banked road with friction. © Prof. Mukesh N Tekwani, 2011 70
  • Conical Pendulum Definition: A conical pendulum is a simple pendulum which is given a motion so that the bob describes a horizontal circle and the string describes a cone. © Prof. Mukesh N Tekwani, 2011 71
  • Conical Pendulum Definition: A conical pendulum is a simple pendulum which is given such a motion that the bob describes a horizontal circle and the string describes a cone. © Prof. Mukesh N Tekwani, 2011 72
  • Conical Pendulum – Time Period Consider a bob of mass m revolving in a horizontal circle of radius r. T cos θ Let v = linear velocity of the bob h = height θ T = tension in the string Θ = semi vertical angle of the cone g = acceleration due to gravity T sin θ l = length of the string © Prof. Mukesh N Tekwani, 2011 73
  • Conical Pendulum – Time Period The forces acting on the bob at position A are: T cos θ 1) Weight of the bob acting vertically downward θ 2) Tension T acting along the string. T sin θ © Prof. Mukesh N Tekwani, 2011 74
  • Conical Pendulum – Time Period The tension T in the string can be resolved (broken up) into 2 components as follows: T cos θ i) Tcosθ acting vertically upwards. This force is balanced θ by the weight of the bob T cos θ = mg ……………………..(1) T sin θ © Prof. Mukesh N Tekwani, 2011 75
  • Conical Pendulum – Time Period (ii) T sinθ acting along the radius of the circle and directed towards the centre of the circle T cos θ T sinθ provides the necessary centripetal force for circular θ motion. ∴ T sinθ = ……….(2) T sin θ Dividing (2) by (1) we get, ………………….(3) © Prof. Mukesh N Tekwani, 2011 76
  • Conical Pendulum – Time Period This equation gives the speed T cos θ of the bob. But v = rw θ ∴ rw = T sin θ Squaring both sides, we get © Prof. Mukesh N Tekwani, 2011 77
  • Conical Pendulum – Time Period From diagram, tan θ = r / h ∴ r 2w2 = rg T cos θ θ T sin θ © Prof. Mukesh N Tekwani, 2011 78
  • Conical Pendulum – Time Period Periodic Time of Conical Pendulum T cos θ But θ T sin θ Solving this & substituting sin θ = r/l we get, © Prof. Mukesh N Tekwani, 2011 79
  • Conical Pendulum – Time Period Periodic Time of Conical Pendulum But cos θ = h/l T cos θ So the eqn becomes, l θ lx T 2 h g T sin θ h T 2 g © Prof. Mukesh N Tekwani, 2011 80
  • Conical Pendulum – Time Period Factors affecting time period of conical pendulum: T cos θ The period of the conical pendulum depends on the following factors: i) Length of the pendulum θ ii) Angle of inclination to the vertical iii) Acceleration due to gravity at the given place T sin θ Time period is independent of the mass of the bob © Prof. Mukesh N Tekwani, 2011 81
  • Vertical Circular Motion Due to Earth’s Gravitation A Consider an objectv1 v3 of mass m tied to mg the end of an T1 r inextensible string O C and whirled in a T2 vertical circle of radius r. B v2 © Prof. Mukesh N Tekwani, 2011 82
  • Vertical Circular Motion Due to Earth’s Gravitation A Highest Point A:v1 v3 Let the velocity be v1 mg The forces acting on the T1 r object at A (highest point) O C are: 1. Tension T1 acting in T2 downward direction 2. Weight mg acting in B v2 downward direction © Prof. Mukesh N Tekwani, 2011 83
  • Vertical Circular Motion Due to Earth’s Gravitation A At the highest point A:v1 v3 The centripetal force acting on mg the object at A is provided T1 r partly by weight and partly by O C tension in the string: T2 B v2 …… (1) © Prof. Mukesh N Tekwani, 2011 84
  • Vertical Circular Motion Due to Earth’s Gravitation A Lowest Point B:v1 v3 Let the velocity be v2 mg The forces acting on the T1 r object at B (lowest point) are: O C 1. Tension T2 acting in upward direction T2 2. Weight mg acting in downward direction B v2 © Prof. Mukesh N Tekwani, 2011 85
  • Vertical Circular Motion Due to Earth’s Gravitation A At the lowest point B:v1 v3 mg …… (2) T1 r O C T2 B v2 © Prof. Mukesh N Tekwani, 2011 86
  • Vertical Circular Motion Due to Earth’s Gravitation Linear velocity of object at highest Av1 point A: v3 The object must have a certain mg minimum velocity at point A so as to T1 r continue in circular path. O C This velocity is called the critical T2 velocity. Below the critical velocity, the string becomes slack and the tension T1 disappears (T1 = 0) B v2 © Prof. Mukesh N Tekwani, 2011 87
  • Vertical Circular Motion Due to Earth’s Gravitation Linear velocity of object at highest Av1 point A: v3 mg T1 r O C T2 B v2 © Prof. Mukesh N Tekwani, 2011 88
  • Vertical Circular Motion Due to Earth’s Gravitation Linear velocity of object at highest Av1 point A: v3 mg This is the minimum velocity that T1 the object must have at the r O C highest point A so that the string does not become slack. T2 If the velocity at the highest point is less than this, the object can not B v2 continue in circular orbit and the string will become slack. © Prof. Mukesh N Tekwani, 2011 89
  • Vertical Circular Motion Due to Earth’s Gravitation Linear velocity of object at lowest Av1 point B: v3 When the object moves from the lowest position to the highest position, the mg increase in potential energy is mg x 2r T1 r O C By the law of conservation of energy, KEA + PEA = KEB + PEB T2 B v2 © Prof. Mukesh N Tekwani, 2011 90
  • Vertical Circular Motion Due to Earth’s Gravitation Linear velocity of object at lowest Av1 point B: v3 At the highest point A, the minimum mg velocity must be T1 r O C Using this in T2 we get, B v2 © Prof. Mukesh N Tekwani, 2011 91
  • Vertical Circular Motion Due to Earth’s Gravitation Linear velocity of object at lowest Av1 point B: v3 mg Therefore, the velocity of the particle is T1 highest at the lowest point. r If the velocity of the particle is less than O C this it will not complete the circular path. T2 B v2 © Prof. Mukesh N Tekwani, 2011 92
  • Linear Velocity at a point midway between top and bottom positions in a vertical circle The total energy of a body performing A circular motion is constant at all points on v1 v3 the path. By law of conservation of energy, Total energy at B = Total energy at C r O C B v2 © Prof. Mukesh N Tekwani, 2011 93
  • Kuch Self-study bhi Karo Na!1. Derive an expression for the tension in the string of a conical pendulum.2. Write kinematical equations for circular motion in analogy with linear motion.3. Derive the expression for the tension in a string in vertical circular motion at any position.4. Derive the expression for the linear velocity at a point midway between the top position and the bottom position in vertical circular motion, without the string slackening at the top. © Prof. Mukesh N Tekwani, 2011 94