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Circular motion, centripetal force, centrifugal force

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- 1. Circular Motion<br />Prof. Mukesh N. Tekwani<br />Department of Physics<br />I. Y. College,<br />Mumbai<br />mukeshtekwani@hotmail.com<br />1<br />Prof. Mukesh N Tekwani<br />
- 2. Circular Motion<br />2<br />Prof. Mukesh N Tekwani<br />
- 3. Circular Motion<br />3<br />Prof. Mukesh N Tekwani<br />
- 4. Relation Between Linear Velocity and Angular Velocity<br />4<br />Prof. Mukesh N Tekwani<br />Consider a particle performing U.C.M. in an anticlockwise direction. <br />In a very small time interval dt, the particle moves from the point P1 to the point P2.<br />Distance travelled along the arc is ds. <br />In the same time interval, the radius vector rotates through an angle dθ.<br />
- 5. Relation Between Linear Velocity and Angular Velocity<br />5<br />Prof. Mukesh N Tekwani<br />
- 6. Centripetal Force<br />UCM is an accelerated motion. Why?<br />UCM is accelerated motion because the velocity of the body changes at every instant (i.e. every moment)<br />But, according to Newton’s Second Law, there must be a force to produce this acceleration. <br />This force is called the centripetal force.<br />Therefore, Centripetal force is required for circular motion. No centripetal force +> no circular motion.<br />6<br />Prof. Mukesh N Tekwani<br />
- 7. Velocity and Speed in UCM<br />Is speed changing?<br />No, speed is constant<br />Is velocity changing?<br />Yes, velocity is changing because velocity is a vector – and direction is changing at every point <br />7<br />Prof. Mukesh N Tekwani<br />
- 8. Velocity and Speed in UCM<br />Is speed changing?<br />No, speed is constant<br />Is velocity changing?<br />Yes, velocity is changing because velocity is a vector – and direction is changing at every point <br />8<br />Prof. Mukesh N Tekwani<br />
- 9. Examples of Centripetal force<br />A body tied to a string and whirled in a horizontal circle – CPF is provided by the tension in the string.<br />9<br />Prof. Mukesh N Tekwani<br />
- 10. Examples of Centripetal force<br />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.<br />10<br />Prof. Mukesh N Tekwani<br />
- 11. Examples of Centripetal force<br />In case of electrons revolving around the nucleus, the centripetal force is provided by the electrostaticforce of attraction between the nucleus and the electrons<br />In case of the motion of moon around the earth, the CPF is provided by the ______ force between Earth and Moon<br />11<br />Prof. Mukesh N Tekwani<br />
- 12. Centripetal Force<br />Centripetal force<br />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.<br />REMEMBER!<br />Centripetal force<br /> - acting on a particle performing UCM<br /> - along the radius<br /> - acting towards the centre of the circle.<br />12<br />Prof. Mukesh N Tekwani<br />
- 13. Properties of Centripetal Force<br />Centripetal force is a real force <br />CPF is necessary for maintaining UCM.<br />CPF acts along the radius of the circle<br />CPF is directed towards center of the circle.<br />CPF does not do any work<br />F = mv2/ r<br />13<br />Prof. Mukesh N Tekwani<br />
- 14. Radial Acceleration<br />Let P be the position of the particle performing UCM<br />r is the radius vector<br />Θ = ωt . This is the angular displacement of the particle in time t secs<br />V is the tangential velocity of the particle at point P. <br />Draw PM ┴ OX <br />The angular displacement of the particle in time t secs is <br />LMOP = Θ = ωt <br />Y<br />v<br />P(x, y)<br />N<br />r<br />y<br />θ<br />O<br />M<br />x<br />X<br />14<br />Prof. Mukesh N Tekwani<br />
- 15. Radial Acceleration<br />The position vector of the particle at any time is given by:<br />r = ix + jy<br />From ∆POM<br />sin θ = PM/OP<br />∴ sin θ = y / r<br />∴y = r sin θ<br />But θ = ωt<br />∴ y = r sin ωt<br />Y<br />v<br />P(x, y)<br />N<br />r<br />y<br />θ<br />O<br />M<br />x<br />X<br />15<br />Prof. Mukesh N Tekwani<br />
- 16. Radial Acceleration<br />Similarly, <br />From ∆POM<br />cosθ = OM/OP<br />∴ cosθ = x / r<br />∴ x = r cosθ<br />But θ = ωt<br />∴ x = r cosωt<br />Y<br />v<br />P(x, y)<br />N<br />r<br />y<br />θ<br />O<br />M<br />x<br />X<br />16<br />Prof. Mukesh N Tekwani<br />
- 17. Radial Acceleration<br />17<br />Prof. Mukesh N Tekwani<br />The velocity of particle at any instant (any time) is called its instantaneousvelocity.<br />The instantaneous velocity is given by<br />v = dr/ dt<br />∴ v = d/dt [ ircos wt + jr sin wt]<br />∴ v = - i r w sin wt + j r wcos wt<br />
- 18. 18<br />Prof. Mukesh N Tekwani<br />Radial Acceleration<br />The linear acceleration of the particle at any instant (any time) is called its instantaneouslinearacceleration.<br />
- 19. Radial Acceleration<br />19<br />Prof. Mukesh N Tekwani<br />Therefore, the instantaneous linear acceleration is given by<br />∴ a = - w2 r<br />Importance of the negative sign: <br />The negative sign in the above equation indicates that the linear acceleration of the particle and the radius vector are in opposite directions.<br />r<br />a<br />
- 20. 20<br />Prof. Mukesh N Tekwani<br />Relation Between Angular Acceleration and Linear Acceleration<br />The acceleration of a particle is given by<br /> ………………. (1)<br />But v = r w<br />∴ <br />∴ a = .... (2)<br />∵ r is a constant radius, <br />∴ <br />But <br />α is the angular acceleration<br />∴ a = r α ………………………(3)<br />
- 21. 21<br />Prof. Mukesh N Tekwani<br />Relation Between Angular Acceleration and Linear Acceleration<br />∴<br />∴ linear acceleration <br />a = aT+ aR<br />aTis called the tangential component of linear acceleration <br />aRis called the radial component of linear acceleration <br />For UCM, w = constant, so <br />∴ a = aR<br />∴ in UCM, linear accln is centripetal accln<br />v = w x r <br />Differentiating w.r.t. time t, <br />But <br />and <br />
- 22. 22<br />Prof. Mukesh N Tekwani<br />Centrifugal Force<br />Centrifugal force is an imaginaryforce (pseudo force) experienced only in non-inertial frames of reference.<br />This force is necessary in order to explain Newton’s laws of motion in an accelerated frame of reference.<br />Centrifugal force is acts along the radius but is directed away from the centre of the circle.<br />Direction of centrifugal force is always opposite to that of the centripetal force.<br />Centrifugal force <br />Centrifugal force is always present in rotating bodies<br />
- 23. 23<br />Prof. Mukesh N Tekwani<br />Examples of Centrifugal Force<br />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.<br />The children sitting in a merry-go-round experience an outward force as the merry-go-round rotates about the vertical axis.<br />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. <br />
- 24. 24<br />Prof. Mukesh N Tekwani<br />Banking of Roads<br />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.<br />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.<br />
- 25. 25<br />Prof. Mukesh N Tekwani<br />Banking of Roads<br />The centripetal force for circular motion of the car can be provided in two ways:<br /><ul><li>Frictional force between the tyres of the car and the road.
- 26. Banking of Roads</li></li></ul><li>26<br />Prof. Mukesh N Tekwani<br />Friction between Tyres and Road<br />The centripetal force for circular motion of the car is provided by the frictional force between the tyres of the car and the road.<br />Let m = mass of the car<br />V = speed of the car, and<br />R = radius of the curved road.<br />Since centripetal force is providedby the frictional force, <br />CPF = frictional force (“provide by” means “equal to” )<br /> (µ is coefficient of friction between tyres & road)<br />So and <br />
- 27. 27<br />Prof. Mukesh N Tekwani<br />Friction between Tyres and Road<br />Thus, the maximum velocity with which a car can safely travel along a curved road is given by <br />If the speed of the car increases beyond this value, the car will be thrown off (skid).<br />If the car has to move at a higher speed, the frictional force should be increased. But this cause wear and tear of tyres. <br />The frictional force is not reliable as it can decrease on wet roads<br />So we cannot rely on frictional force to provide the centripetal force for circular motion.<br />
- 28. 28<br />Prof. Mukesh N Tekwani<br />Friction between Tyres and Road<br />R1and R2 are reaction forces due to the tyres<br />mg is the weight of the car, acting vertically downwards<br />F1and F2 are the frictional forces between the tyres and the road. <br />These frictional forces act towards the centre of the circular path and provide the necessary centripetal force.<br />Center of circular path<br />
- 29. 29<br />Prof. Mukesh N Tekwani<br />Friction between Tyres and Road<br />
- 30. 30<br />Prof. Mukesh N Tekwani<br />Friction between Tyres and Road – Car Skidding<br />
- 31. 31<br />Prof. Mukesh N Tekwani<br />Banked Roads<br />What is banking of roads?<br />The process of raising the outeredge of a road over the inner edge through a certain angle is known as banking of road.<br />
- 32. 32<br />Prof. Mukesh N Tekwani<br />Banking of Roads<br />Purpose of Banking of Roads:<br />Banking of roads is done:<br />To provide the necessary centripetal force for circular motion<br />To reduce wear and tear of tyres due to friction<br />To avoid skidding<br />To avoid overturning of vehicles<br />
- 33. 33<br />Prof. Mukesh N Tekwani<br />Banked Roads<br />
- 34. 34<br />Prof. Mukesh N Tekwani<br />Banked Roads<br />R cosθ<br />What is angle of banking?<br />R<br />Θ <br />Surface of road<br />R sin θ<br />The angle made by the surface of the road with the horizontal surface is called as angle of banking.<br />Θ <br />Horizontal <br />W = mg<br />
- 35. 35<br />Prof. Mukesh N Tekwani<br />Banked Roads<br />R cosθ<br />Consider a car moving along a banked road.<br />Let<br />m = mass of the car<br />V = speed of the car<br />θ is angle of banking<br />R<br />Θ <br />R sin θ<br />Θ <br />W = mg<br />
- 36. 36<br />Prof. Mukesh N Tekwani<br />Banked Roads<br />R cosθ<br />The forces acting on the car are:<br />(i) Its weight mg acting vertically downwards.<br />(ii) The normal reaction R acting perpendicular to the surface of the road.<br />R<br />Θ <br />R sin θ<br />Θ <br />W = mg<br />
- 37. 37<br />Prof. Mukesh N Tekwani<br />Banked Roads<br />The normal reaction can be resolved (broken up) into two components:<br />R cosθ is the vertical component<br />R sinθis the horizontal component<br />R cosθ<br />R<br />Θ <br />R sin θ<br />Θ <br />W = mg<br />
- 38. 38<br />Prof. Mukesh N Tekwani<br />Banked Roads<br />Since the vehicle has no vertical motion, the weight is balanced by the vertical component<br />R cosθ = mg …………… (1)<br />(weight is balanced by vertical component means weight is equal to vertical component)<br />R cosθ<br />R<br />Θ <br />R sin θ<br />Θ <br />W = mg<br />
- 39. 39<br />Prof. Mukesh N Tekwani<br />Banked Roads<br />The horizontal component is the unbalanced component . This horizontal component acts towards the centre of the circular path. <br />This component provides the centripetal force for circular motion<br />R sinθ = …………… (2)<br />R cosθ<br />R<br />Θ <br />R sin θ<br />Θ <br />W = mg<br />
- 40. 40<br />Prof. Mukesh N Tekwani<br />Banked Roads<br />Dividing (2) by (1), we get<br />R sinθ=<br />θ = tan-1 ( )<br />mg<br />Therefore, the angle of banking is independent of the mass of the vehicle.<br />The maximum speed with which the vehicle can safely travel along the curved road is <br />R cosθ<br />So, <br />tan θ = <br />
- 41. 41<br />Prof. Mukesh N Tekwani<br />Banked Roads<br />A car travels at a constant speed around two curves. Where is the car most likely to skid? Why? <br />Smaller radius: larger centripetal force is required to keep it in uniform circular motion.<br />
- 42. 42<br />Prof. Mukesh N Tekwani<br />Conical Pendulum<br />Definition:<br />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.<br />
- 43. 43<br />Prof. Mukesh N Tekwani<br />Conical Pendulum<br />Definition:<br />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.<br />
- 44. 44<br />Prof. Mukesh N Tekwani<br />Conical Pendulum<br />Consider a bob of mass m revolving in a horizontal circle of radius r. <br />Let<br />v = linear velocity of the bob<br />h = height<br />T = tension in the string<br />Θ = semi vertical angle of the cone<br />g = acceleration due to gravity<br />l = length of the string<br />T cos θ<br />θ<br />T sin θ<br />
- 45. 45<br />Prof. Mukesh N Tekwani<br />Conical Pendulum<br />The forces acting on the bob at position A are:<br />Weight of the bob acting vertically downward<br />Tension T acting along the string.<br />T cos θ<br />θ<br />T sin θ<br />
- 46. 46<br />Prof. Mukesh N Tekwani<br />Conical Pendulum<br />The tension T in the string can be resolved (broken up) into 2 components as follows:<br />Tcosθ acting vertically upwards. This force is balanced by the weight of the bob<br />T cos θ = mg ……………………..(1)<br />T cos θ<br />θ<br />T sin θ<br />
- 47. 47<br />Prof. Mukesh N Tekwani<br />Conical Pendulum<br />(ii) T sinθacting along the radius of the circle and directed towards the centre of the circle<br />T sinθ provides the necessary centripetal force for circular motion.<br />∴ T sinθ = ……….(2)<br />Dividing (2) by (1) we get,<br /> ………………….(3)<br />T cos θ<br />θ<br />T sin θ<br />
- 48. 48<br />Prof. Mukesh N Tekwani<br />Conical Pendulum<br />This equation gives the speed of the bob.<br />But v = rw<br />∴ rw =<br />Squaring both sides, we get<br />T cos θ<br />θ<br />T sin θ<br />
- 49. 49<br />Prof. Mukesh N Tekwani<br />Conical Pendulum<br />From diagram, tan θ = r / h<br />∴ r 2w2 = rg<br />T cos θ<br />θ<br />T sin θ<br />
- 50. 50<br />Prof. Mukesh N Tekwani<br />Conical Pendulum<br />Periodic Time of Conical Pendulum<br />But <br />Solving this & substituting sin θ = r/l <br />we get,<br />T cos θ<br />θ<br />T sin θ<br />
- 51. 51<br />Prof. Mukesh N Tekwani<br />Conical Pendulum<br />Factors affecting time period of conical pendulum:<br />The period of the conical pendulum depends on the following factors:<br />Length of the pendulum<br />Angle of inclination to the vertical<br />Acceleration due to gravity at the given place<br />Time period is independent of the mass of the bob<br />T cos θ<br />θ<br />T sin θ<br />
- 52. 52<br />Prof. Mukesh N Tekwani<br />Vertical Circular Motion Due to Earth’s Gravitation<br />Consider an object of mass m tied to the end of an inextensible string and whirled in a vertical circle of radius r.<br />A<br />v1<br />v3<br />mg<br />T1<br />r<br />O<br />C<br />T2<br />B<br />v2<br />
- 53. 53<br />Prof. Mukesh N Tekwani<br />Vertical Circular Motion Due to Earth’s Gravitation<br />Highest Point A:<br />Let the velocity be v1<br />The forces acting on the object at A (highest point) are:<br />Tension T1 acting in downward direction<br />Weight mg acting in downward direction<br />A<br />v1<br />v3<br />mg<br />T1<br />r<br />O<br />C<br />T2<br />B<br />v2<br />
- 54. 54<br />Prof. Mukesh N Tekwani<br />Vertical Circular Motion Due to Earth’s Gravitation<br />At the highest point A:<br />The centripetal force acting on the object at A is provided partly by weight and partly by tension in the string:<br />A<br />v1<br />v3<br />mg<br />T1<br />r<br />O<br />C<br />T2<br />…… (1)<br />B<br />v2<br />
- 55. 55<br />Prof. Mukesh N Tekwani<br />Vertical Circular Motion Due to Earth’s Gravitation<br />Lowest Point B:<br />Let the velocity be v2<br />The forces acting on the object at B (lowest point) are:<br />Tension T2 acting in upward direction<br />Weight mg acting in downward direction<br />A<br />v1<br />v3<br />mg<br />T1<br />r<br />O<br />C<br />T2<br />B<br />v2<br />
- 56. 56<br />Prof. Mukesh N Tekwani<br />Vertical Circular Motion Due to Earth’s Gravitation<br />At the lowest point B:<br />A<br />v1<br />v3<br />mg<br />…… (2)<br />T1<br />r<br />O<br />C<br />T2<br />B<br />v2<br />
- 57. 57<br />Prof. Mukesh N Tekwani<br />Vertical Circular Motion Due to Earth’s Gravitation<br />Linear velocity of object at highest point A:<br />The object must have a certain minimum velocity at point A so as to continue in circular path. <br />This velocity is called the criticalvelocity. Below the critical velocity, the string becomes slack and the tension T1 disappears (T1 = 0)<br />A<br />v1<br />v3<br />mg<br />T1<br />r<br />O<br />C<br />T2<br />B<br />v2<br />
- 58. 58<br />Prof. Mukesh N Tekwani<br />Vertical Circular Motion Due to Earth’s Gravitation<br />Linear velocity of object at highest point A:<br />A<br />v1<br />v3<br />mg<br />T1<br />r<br />O<br />C<br />T2<br />B<br />v2<br />
- 59. 59<br />Prof. Mukesh N Tekwani<br />Vertical Circular Motion Due to Earth’s Gravitation<br />Linear velocity of object at highest point A:<br />A<br />v1<br />v3<br />mg<br />This is the minimum velocity that the object must have at the highest point A so that the string does not become slack. <br />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.<br />T1<br />r<br />O<br />C<br />T2<br />B<br />v2<br />
- 60. 60<br />Prof. Mukesh N Tekwani<br />Vertical Circular Motion Due to Earth’s Gravitation<br />Linear velocity of object at lowest point B:<br />A<br />v1<br />v3<br />When the object moves from the lowest position to the highest position, the increase in potential energy is mg x 2r<br />By the law of conservation of energy,<br />KEA + PEA = KEB + PEB<br />mg<br />T1<br />r<br />O<br />C<br />T2<br />B<br />v2<br />
- 61. 61<br />Prof. Mukesh N Tekwani<br />Vertical Circular Motion Due to Earth’s Gravitation<br />Linear velocity of object at lowest point B:<br />A<br />v1<br />v3<br />At the highest point A, the minimum velocity must be <br />mg<br />T1<br />r<br />O<br />C<br />Using this in <br />T2<br />we get,<br />B<br />v2<br />
- 62. 62<br />Prof. Mukesh N Tekwani<br />Vertical Circular Motion Due to Earth’s Gravitation<br />Linear velocity of object at lowest point B:<br />A<br />v1<br />v3<br />mg<br />Therefore, the velocity of the particle is highest at the lowest point.<br />If the velocity of the particle is less than this it will not complete the circular path.<br />T1<br />r<br />O<br />C<br />T2<br />B<br />v2<br />

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