Circular motion, centripetal force, centrifugal force

1. Circular Motion Prof. Mukesh N. Tekwani Department of Physics I. Y. College, Mumbai mukeshtekwani@hotmail.com 1 Prof. Mukesh N Tekwani

2. Circular Motion 2 Prof. Mukesh N Tekwani

3. Circular Motion 3 Prof. Mukesh N Tekwani

4. Relation Between Linear Velocity and Angular Velocity 4 Prof. Mukesh N Tekwani Consider a particle performing U.C.M. in an anticlockwise direction. In a very small time interval dt, the particle moves from the point P1 to the point P2. Distance travelled along the arc is ds. In the same time interval, the radius vector rotates through an angle dθ.

5. Relation Between Linear Velocity and Angular Velocity 5 Prof. Mukesh N Tekwani

6. 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. 6 Prof. Mukesh N Tekwani

7. 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 7 Prof. Mukesh N Tekwani

8. 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 8 Prof. Mukesh N Tekwani

9. 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. 9 Prof. Mukesh N Tekwani

10. 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. 10 Prof. Mukesh N Tekwani

11. Examples of Centripetal force In case of electrons revolving around the nucleus, the centripetal force is provided by the electrostaticforce 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 11 Prof. Mukesh N Tekwani

12. 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. 12 Prof. Mukesh N Tekwani

13. Properties of Centripetal Force Centripetal force is a real force CPF is necessary for maintaining UCM. CPF acts along the radius of the circle CPF is directed towards center of the circle. CPF does not do any work F = mv2/ r 13 Prof. Mukesh N Tekwani

14. Radial Acceleration Let P be the position of the particle performing UCM r is the radius vector Θ = ωt . This is the angular displacement of the particle in time t secs V is the tangential velocity of the particle at point P. Draw PM ┴ OX The angular displacement of the particle in time t secs is LMOP = Θ = ωt Y v P(x, y) N r y θ O M x X 14 Prof. Mukesh N Tekwani

15. Radial Acceleration The position vector of the particle at any time is given by: r = ix + jy From ∆POM sin θ = PM/OP ∴ sin θ = y / r ∴y = r sin θ But θ = ωt ∴ y = r sin ωt Y v P(x, y) N r y θ O M x X 15 Prof. Mukesh N Tekwani

16. Radial Acceleration Similarly, From ∆POM cosθ = OM/OP ∴ cosθ = x / r ∴ x = r cosθ But θ = ωt ∴ x = r cosωt Y v P(x, y) N r y θ O M x X 16 Prof. Mukesh N Tekwani

17. Radial Acceleration 17 Prof. Mukesh N Tekwani The velocity of particle at any instant (any time) is called its instantaneousvelocity. The instantaneous velocity is given by v = dr/ dt ∴ v = d/dt [ ircos wt + jr sin wt] ∴ v = - i r w sin wt + j r wcos wt

18. 18 Prof. Mukesh N Tekwani Radial Acceleration The linear acceleration of the particle at any instant (any time) is called its instantaneouslinearacceleration.

19. Radial Acceleration 19 Prof. Mukesh N Tekwani Therefore, 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. r a

20. 20 Prof. Mukesh N Tekwani Relation Between Angular Acceleration and Linear Acceleration The acceleration of a particle is given by ………………. (1) But v = r w ∴ ∴ a = .... (2) ∵ r is a constant radius, ∴ But α is the angular acceleration ∴ a = r α ………………………(3)

21. 21 Prof. Mukesh N Tekwani Relation Between Angular Acceleration and Linear Acceleration ∴ ∴ linear acceleration a = aT+ aR aTis called the tangential component of linear acceleration aRis called the radial component of linear acceleration For UCM, w = constant, so ∴ a = aR ∴ in UCM, linear accln is centripetal accln v = w x r Differentiating w.r.t. time t, But and

22. 22 Prof. Mukesh N Tekwani Centrifugal Force Centrifugal force is an imaginaryforce (pseudo force) experienced only in non-inertial frames of reference. This force is necessary in order to explain Newton’s 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

23. 23 Prof. Mukesh N Tekwani Examples of Centrifugal Force 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.

24. 24 Prof. Mukesh N Tekwani Banking of Roads 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 centripetalforce (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.

27. 27 Prof. Mukesh N Tekwani 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.

28. 28 Prof. Mukesh N Tekwani Friction between Tyres and Road R1and R2 are reaction forces due to the tyres mg is the weight of the car, acting vertically downwards F1and 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

29. 29 Prof. Mukesh N Tekwani Friction between Tyres and Road

30. 30 Prof. Mukesh N Tekwani Friction between Tyres and Road – Car Skidding

31. 31 Prof. Mukesh N Tekwani Banked Roads What is banking of roads? The process of raising the outeredge of a road over the inner edge through a certain angle is known as banking of road.

32. 32 Prof. Mukesh N Tekwani Banking of Roads Purpose of Banking of Roads: Banking of roads is done: To provide the necessary centripetal force for circular motion To reduce wear and tear of tyres due to friction To avoid skidding To avoid overturning of vehicles

33. 33 Prof. Mukesh N Tekwani Banked Roads

34. 34 Prof. Mukesh N Tekwani Banked Roads R cosθ What is angle of banking? R Θ Surface of road R sin θ The angle made by the surface of the road with the horizontal surface is called as angle of banking. Θ Horizontal W = mg

35. 35 Prof. Mukesh N Tekwani Banked Roads R cosθ Consider a car moving along a banked road. Let m = mass of the car V = speed of the car θ is angle of banking R Θ R sin θ Θ W = mg

36. 36 Prof. Mukesh N Tekwani Banked Roads R cosθ The forces acting on the car are: (i) Its weight mg acting vertically downwards. (ii) The normal reaction R acting perpendicular to the surface of the road. R Θ R sin θ Θ W = mg

37. 37 Prof. Mukesh N Tekwani Banked Roads The normal reaction can be resolved (broken up) into two components: R cosθ is the vertical component R sinθis the horizontal component R cosθ R Θ R sin θ Θ W = mg

38. 38 Prof. Mukesh N Tekwani Banked Roads Since the vehicle has no vertical motion, the weight is balanced by the vertical component R cosθ = mg …………… (1) (weight is balanced by vertical component means weight is equal to vertical component) R cosθ R Θ R sin θ Θ W = mg

39. 39 Prof. Mukesh N Tekwani Banked Roads The horizontal component is the unbalanced component . This horizontal component acts towards the centre of the circular path. This component provides the centripetal force for circular motion R sinθ = …………… (2) R cosθ R Θ R sin θ Θ W = mg

40. 40 Prof. Mukesh N Tekwani Banked Roads Dividing (2) by (1), we get R sinθ= θ = tan-1 ( ) mg 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 R cosθ So, tan θ =

41. 41 Prof. Mukesh N Tekwani Banked Roads A car travels at a constant speed around two curves. Where is the car most likely to skid? Why? Smaller radius: larger centripetal force is required to keep it in uniform circular motion.

42. 42 Prof. Mukesh N Tekwani 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.

43. 43 Prof. Mukesh N Tekwani 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.

44. 44 Prof. Mukesh N Tekwani Conical Pendulum Consider a bob of mass m revolving in a horizontal circle of radius r. Let v = linear velocity of the bob h = height T = tension in the string Θ = semi vertical angle of the cone g = acceleration due to gravity l = length of the string T cos θ θ T sin θ

45. 45 Prof. Mukesh N Tekwani Conical Pendulum 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 θ θ T sin θ

46. 46 Prof. Mukesh N Tekwani Conical Pendulum The tension T in the string can be resolved (broken up) into 2 components as follows: Tcosθ acting vertically upwards. This force is balanced by the weight of the bob T cos θ = mg ……………………..(1) T cos θ θ T sin θ

47. 47 Prof. Mukesh N Tekwani Conical Pendulum (ii) T sinθacting along the radius of the circle and directed towards the centre of the circle T sinθ provides the necessary centripetal force for circular motion. ∴ T sinθ = ……….(2) Dividing (2) by (1) we get, ………………….(3) T cos θ θ T sin θ

48. 48 Prof. Mukesh N Tekwani Conical Pendulum This equation gives the speed of the bob. But v = rw ∴ rw = Squaring both sides, we get T cos θ θ T sin θ

49. 49 Prof. Mukesh N Tekwani Conical Pendulum From diagram, tan θ = r / h ∴ r 2w2 = rg T cos θ θ T sin θ

50. 50 Prof. Mukesh N Tekwani Conical Pendulum Periodic Time of Conical Pendulum But Solving this & substituting sin θ = r/l we get, T cos θ θ T sin θ

51. 51 Prof. Mukesh N Tekwani Conical Pendulum Factors affecting time period of conical pendulum: The period of the conical pendulum depends on the following factors: Length of the pendulum Angle of inclination to the vertical Acceleration due to gravity at the given place Time period is independent of the mass of the bob T cos θ θ T sin θ

52. 52 Prof. Mukesh N Tekwani Vertical Circular Motion Due to Earth’s Gravitation Consider an object of mass m tied to the end of an inextensible string and whirled in a vertical circle of radius r. A v1 v3 mg T1 r O C T2 B v2

53. 53 Prof. Mukesh N Tekwani Vertical Circular Motion Due to Earth’s Gravitation Highest Point A: Let the velocity be v1 The forces acting on the object at A (highest point) are: Tension T1 acting in downward direction Weight mg acting in downward direction A v1 v3 mg T1 r O C T2 B v2

54. 54 Prof. Mukesh N Tekwani Vertical Circular Motion Due to Earth’s Gravitation At the highest point A: The centripetal force acting on the object at A is provided partly by weight and partly by tension in the string: A v1 v3 mg T1 r O C T2 …… (1) B v2

55. 55 Prof. Mukesh N Tekwani Vertical Circular Motion Due to Earth’s Gravitation Lowest Point B: Let the velocity be v2 The forces acting on the object at B (lowest point) are: Tension T2 acting in upward direction Weight mg acting in downward direction A v1 v3 mg T1 r O C T2 B v2

56. 56 Prof. Mukesh N Tekwani Vertical Circular Motion Due to Earth’s Gravitation At the lowest point B: A v1 v3 mg …… (2) T1 r O C T2 B v2

57. 57 Prof. Mukesh N Tekwani Vertical Circular Motion Due to Earth’s Gravitation Linear velocity of object at highest point A: The object must have a certain minimum velocity at point A so as to continue in circular path. This velocity is called the criticalvelocity. Below the critical velocity, the string becomes slack and the tension T1 disappears (T1 = 0) A v1 v3 mg T1 r O C T2 B v2

58. 58 Prof. Mukesh N Tekwani Vertical Circular Motion Due to Earth’s Gravitation Linear velocity of object at highest point A: A v1 v3 mg T1 r O C T2 B v2

59. 59 Prof. Mukesh N Tekwani Vertical Circular Motion Due to Earth’s Gravitation Linear velocity of object at highest point A: A v1 v3 mg 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. T1 r O C T2 B v2

60. 60 Prof. Mukesh N Tekwani Vertical Circular Motion Due to Earth’s Gravitation Linear velocity of object at lowest point B: A v1 v3 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 mg T1 r O C T2 B v2

61. 61 Prof. Mukesh N Tekwani Vertical Circular Motion Due to Earth’s Gravitation Linear velocity of object at lowest point B: A v1 v3 At the highest point A, the minimum velocity must be mg T1 r O C Using this in T2 we get, B v2

62. 62 Prof. Mukesh N Tekwani Vertical Circular Motion Due to Earth’s Gravitation Linear velocity of object at lowest point B: A v1 v3 mg 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. T1 r O C T2 B v2