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Circular motion
1.
Prof. Mukesh N.
Tekwani Department of Physics I. Y. College, Mumbai mukeshtekwani@outlook.com © Prof. Mukesh N Tekwani, 2011 1
2.
Circular Motion
© Prof. Mukesh N Tekwani, 2011 2
3.
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
4.
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
5.
Circular Motion How do
we locate something on a circle? Give its angular position θ y f What is the location of i X 900 or π/2 θ=0 © Prof. Mukesh N Tekwani, 2011 5
6.
Circular Motion
y f is the angular position. i Angular displacement: x f i Note: angles measured Clockwise (CW) are negative and angles measured (CCW) are positive. is measured in radians. 2 radians = 360Tekwani, 1 revolution © Prof. Mukesh N = 2011 6
7.
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
8.
Angular Displacement
S =r Θ © Prof. Mukesh N Tekwani, 2011 8
9.
arclength = s
= r y s f r r i is a ratio of two lengths; x it is a dimensionless ratio! This is a radian measure of angle If we go all the way round s =2πr and Δθ =2 π © Prof. Mukesh N Tekwani, 2011 9
10.
Right Hand Rule If
the fingers of the right hand are curled in the direction of revolution of the particle, then the outstretched thumb gives the direction of the angular displacement vector. © Prof. Mukesh N Tekwani, 2011 11
11.
Stop & Think
– Pg 2 Are the following motions same or different? 1. The motion of the tip of second hand of a clock. 2. The motion of the entire second hand of a clock. The motion of the tip of second hand of a clock is uniform circular motion. The motion of the entire second hand is a rotational motion. © Prof. Mukesh N Tekwani, 2011 12
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
13.
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
14.
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
15.
Examples of Centripetal
force A body tied to a string and whirled in a horizontal circle – CPF is provided by the tension in the string. © Prof. Mukesh N Tekwani, 2011 28
16.
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
17.
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
18.
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
19.
Properties of Centripetal
Force 1. Centripetal force is a real force 2. CPF is necessary for maintaining UCM. 3. CPF acts along the radius of the circle 4. CPF is directed towards center of the circle. 5. CPF does not do any work 6. F = mv2/ r © Prof. Mukesh N Tekwani, 2011 32
20.
Radial Acceleration
Let P be the position of the particle Y performing UCM v P(x, y) r is the radius vector N Θ = ωt . This is the angular displacement of r the particle in time t secs y θ V is the tangential velocity of the particle at O x M X point P. Draw PM ┴ OX The angular displacement of the particle in time t secs is LMOP = Θ = ωt © Prof. Mukesh N Tekwani, 2011 33
21.
Radial Acceleration Y
The position vector of the particle at v any time is given by: P(x, y) r = ix + jy N r From ∆POM y θ sin θ = PM/OP O x M X ∴ sin θ = y / r ∴y = r sin θ But θ = ωt ∴ y = r sin ωt © Prof. Mukesh N Tekwani, 2011 34
22.
Radial Acceleration Y
Similarly, v From ∆POM P(x, y) N cos θ = OM/OP r y ∴ cos θ = x / r θ O x M X ∴ x = r cos θ But θ = ωt ∴ x = r cos ωt © Prof. Mukesh N Tekwani, 2011 35
23.
Radial Acceleration The velocity
of particle at any instant (any time) is called its instantaneous velocity. The instantaneous velocity is given by v = dr / dt ∴ v = d/dt [ ir cos wt + jr sin wt] ∴ v = - i r w sin wt + j r w cos wt © Prof. Mukesh N Tekwani, 2011 36
24.
Radial Acceleration The linear
acceleration of the particle at any instant (any time) is called its instantaneous linear acceleration. © Prof. Mukesh N Tekwani, 2011 37
25.
Radial Acceleration Therefore, the
instantaneous linear acceleration is given by ∴ a = - w2 r r a Importance of the negative sign: The negative sign in the above equation indicates that the linear acceleration of the particle and the radius vector are in opposite directions. © Prof. Mukesh N Tekwani, 2011 38
26.
Relation Between Angular Acceleration
and Linear Acceleration The acceleration of a particle ∵ r is a constant radius, is given by ………………. (1) ∴ But v = r w But ∴ α is the angular acceleration ∴a = .... (2) ∴ a=rα ………………………(3) © Prof. Mukesh N Tekwani, 2011 39
27.
Relation Between Angular Acceleration
and Linear Acceleration ∴ v=wxr Differentiating w.r.t. time t, ∴ linear acceleration a = a T + aR aT is called the tangential component of linear acceleration aR is called the radial component of linear acceleration But For UCM, w = constant, so and ∴ a = aR ∴ in UCM, linear accln is centripetal accln © Prof. Mukesh N Tekwani, 2011 40
28.
Centrifugal Force 1. Centrifugal
force is an imaginary force (pseudo force) experienced only in non-inertial frames of reference. 2. This force is necessary in order to explain Newton’s laws of motion in an accelerated frame of reference. 3. Centrifugal force is acts along the radius but is directed away from the centre of the circle. 4. Direction of centrifugal force is always opposite to that of the centripetal force. 5. Centrifugal force 6. Centrifugal force is always present in rotating bodies © Prof. Mukesh N Tekwani, 2011 41
29.
Examples of Centrifugal
Force 1. When a car in motion takes a sudden turn towards left, passengers in the car experience an outward push to the right. This is due to the centrifugal force acting on the passengers. 2. The children sitting in a merry-go-round experience an outward force as the merry-go-round rotates about the vertical axis. Centripetal and Centrifugal forces DONOT constitute an action- reaction pair. Centrifugal force is not a real force. For action- reaction pair, both forces must be real. © Prof. Mukesh N Tekwani, 2011 42
30.
Banking of Roads 1.
When a car is moving along a curved road, it is performing circular motion. For circular motion it is not necessary that the car should complete a full circle; an arc of a circle is also treated as a circular path. 2. We know that centripetal force (CPF) is necessary for circular motion. If CPF is not present, the car cannot travel along a circular path and will instead travel along a tangential path. © Prof. Mukesh N Tekwani, 2011 43
31.
Banking of Roads 3.
The centripetal force for circular motion of the car can be provided in two ways: • Frictional force between the tyres of the car and the road. • Banking of Roads © Prof. Mukesh N Tekwani, 2011 44
32.
Friction between Tyres
and Road The centripetal force for circular motion of the car is provided by the frictional force between the tyres of the car and the road. Let m = mass of the car V = speed of the car, and R = radius of the curved road. Since centripetal force is provided by the frictional force, CPF = frictional force (“provide by” means “equal to” ) (µ is coefficient of friction between tyres & road) So and © Prof. Mukesh N Tekwani, 2011 45
33.
Friction between Tyres
and Road 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. © Prof. Mukesh N Tekwani, 2011 46
34.
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
35.
Friction between Tyres
and Road © Prof. Mukesh N Tekwani, 2011 48
36.
Friction between Tyres
and Road – Car Skidding © Prof. Mukesh N Tekwani, 2011 49
37.
Banked Roads What is
banking of roads? The process of raising the outer edge of a road over the inner edge through a certain angle is known as banking of road. © Prof. Mukesh N Tekwani, 2011 50
38.
Banking of Roads Purpose
of Banking of Roads: Banking of roads is done: 1. To provide the necessary centripetal force for circular motion 2. To reduce wear and tear of tyres due to friction 3. To avoid skidding 4. To avoid overturning of vehicles © Prof. Mukesh N Tekwani, 2011 51
39.
Banked Roads
© Prof. Mukesh N Tekwani, 2011 52
40.
Banked Roads What is
angle of R cos θ R banking? Θ The angle made by the R sin θ surface of the road with the horizontal surface is called as angle of banking. Θ Horizontal W = mg © Prof. Mukesh N Tekwani, 2011 53
41.
Banked Roads Consider a
car moving R cos θ R along a banked road. Θ R sin θ Let m = mass of the car V = speed of the car Θ θ is angle of banking W = mg © Prof. Mukesh N Tekwani, 2011 54
42.
Banked Roads The forces
acting on the R cos θ R car are: Θ R sin θ (i) Its weight mg acting vertically downwards. Θ (ii) The normal reaction R acting perpendicular to the W = mg surface of the road. © Prof. Mukesh N Tekwani, 2011 55
43.
Banked Roads The normal
reaction can be R cos θ resolved (broken up) into R Θ two components: R sin θ 1. R cosθ is the vertical component Θ 2. R sinθ is the horizontal component W = mg © Prof. Mukesh N Tekwani, 2011 56
44.
Banked Roads Since the
vehicle has no R cos θ vertical motion, the weight R Θ is balanced by the vertical component R sin θ R cosθ = mg …………… (1) Θ (weight is balanced by vertical component means weight is equal to vertical component) W = mg © Prof. Mukesh N Tekwani, 2011 57
45.
Banked Roads The horizontal
component R cos θ is the unbalanced R Θ component . This horizontal component acts towards R sin θ the centre of the circular path. Θ This component provides the centripetal force for circular motion W = mg R sinθ = …………… (2) © Prof. Mukesh N Tekwani, 2011 58
46.
Banked Roads Dividing (2)
by (1), we get θ = tan-1 ( ) R sinθ = mg Therefore, the angle of banking R cos θ is independent of the mass of the vehicle. So, The maximum speed with which the vehicle can safely tan θ = travel along the curved road is © Prof. Mukesh N Tekwani, 2011 59
47.
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
48.
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
49.
Maximum Speed of
a Vehicle on a Banked Road with Friction The forces acting on the vehicle are: 1) Weight of the vehicle mg, acting vertically downwards 2) Normal reaction N acting on vehicle, perpendicular to the surface of the road. 3) Friction force between tyres and road. Forces N and frictional force f are now resolved into two components © Prof. Mukesh N Tekwani, 2011 62
50.
Maximum Speed of
a Vehicle on a Banked Road with Friction Resolving the Normal reaction The normal reaction N is resolved into 2 components: 1) N cos θ is vertical component of N 2) N sin θ is horizontal component of N © Prof. Mukesh N Tekwani, 2011 63
51.
Maximum Speed of
a Vehicle on a Banked Road with Friction Resolving the frictional force The frictional force f is resolved into 2 components: f cos θ 1) f cos θ is horizontal component of f f θ f sin θ 2) f sin θ is vertical component of f © Prof. Mukesh N Tekwani, 2011 64
52.
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 θ f cos θ f θ f sin θ ∴ N cos θ = mg + f sin θ mg ∴ mg = N cos θ - f sin θ .…. (1) © Prof. Mukesh N Tekwani, 2011 65
53.
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 θ f cos θ mv 2 f θ f sin θ ∴ N sin θ + f cos θ = 2 r mg mv ∴ = N sin θ + f cos θ … (2) r © Prof. Mukesh N Tekwani, 2011 66
54.
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 θ f cos θ θ = f f sin θ N cos θ - f sin θ mg mg v2 N sin θ + f cos θ ∴ = rg N cos θ - f sin θ © Prof. Mukesh N Tekwani, 2011 67
55.
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 N f cos θ θ f f sin θ 2 N sin θ + fm cos θ mg v max ∴ = rg N cos θ - fm sin θ © Prof. Mukesh N Tekwani, 2011 68
56.
Maximum Speed of
a Vehicle on a Banked Road with Friction But fm = µs N N sin θ µs N cos θ + 2 N cos θ N cos θ v 2 N sin θ + µs N cos θ v ∴ max = ∴ max = rg N cos θ - µs N sin θ rg N cos θ µs N sin θ - N cos θ N cos θ Dividing numerator and 2 tan θ + µs v max = denominator of RHS by ∴ cos θ, we get rg 1 - µs tan θ © Prof. Mukesh N Tekwani, 2011 69
57.
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 tan This is the maximum velocity with which a vehicle can travel on a banked road with friction. © Prof. Mukesh N Tekwani, 2011 70
58.
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
59.
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
60.
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
61.
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
62.
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
63.
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
64.
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
65.
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
66.
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
67.
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
68.
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
69.
Vertical Circular Motion
Due to Earth’s Gravitation A Consider an object v1 v3 of mass m tied to mg the end of an T1 r inextensible string O C and whirled in a T2 vertical circle of radius r. B v2 © Prof. Mukesh N Tekwani, 2011 82
70.
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
71.
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
72.
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
73.
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
74.
Vertical Circular Motion
Due to Earth’s Gravitation Linear velocity of object at highest A v1 point A: v3 The object must have a certain mg minimum velocity at point A so as to T1 r continue in circular path. O C This velocity is called the critical T2 velocity. Below the critical velocity, the string becomes slack and the tension T1 disappears (T1 = 0) B v2 © Prof. Mukesh N Tekwani, 2011 87
75.
Vertical Circular Motion
Due to Earth’s Gravitation Linear velocity of object at highest A v1 point A: v3 mg T1 r O C T2 B v2 © Prof. Mukesh N Tekwani, 2011 88
76.
Vertical Circular Motion
Due to Earth’s Gravitation Linear velocity of object at highest A v1 point A: v3 mg This is the minimum velocity that T1 the object must have at the r O C highest point A so that the string does not become slack. T2 If the velocity at the highest point is less than this, the object can not B v2 continue in circular orbit and the string will become slack. © Prof. Mukesh N Tekwani, 2011 89
77.
Vertical Circular Motion
Due to Earth’s Gravitation Linear velocity of object at lowest A v1 point B: v3 When the object moves from the lowest position to the highest position, the mg increase in potential energy is mg x 2r T1 r O C By the law of conservation of energy, KEA + PEA = KEB + PEB T2 B v2 © Prof. Mukesh N Tekwani, 2011 90
78.
Vertical Circular Motion
Due to Earth’s Gravitation Linear velocity of object at lowest A v1 point B: v3 At the highest point A, the minimum mg velocity must be T1 r O C Using this in T2 we get, B v2 © Prof. Mukesh N Tekwani, 2011 91
79.
Vertical Circular Motion
Due to Earth’s Gravitation Linear velocity of object at lowest A v1 point B: v3 mg Therefore, the velocity of the particle is T1 highest at the lowest point. r If the velocity of the particle is less than O C this it will not complete the circular path. T2 B v2 © Prof. Mukesh N Tekwani, 2011 92
80.
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
81.
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
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