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  • Ask what happens when go to fast? (slide up and over top of curve)
    Ask what happens when go to slow? (slide down curve)
  • Circular

    1. 1. Circular Motion The conical pendulum and beyond
    2. 2. Circular motion And Centripetal Force What you need to know before Circular motion. 1.Force produces acceleration. F = ma(Newton 2) 2.How do I know that an object is accelerating. What is acceleration? 3.Acceleration causes changing velocity. 4.Distance(d) around a circle is: •the circumference of a circle •d = 2πr ( r = radius of a circle)
    3. 3. 5.A rotation (or revolution) means once around a circle. 6.The time to go once around a circle is called the period (T) of rotation. 7. The number of rotations in one second is the frequency (f) 8.Frequency is measured in Hertz (hz) or (s -1) 9. We will only be working with CONSTANT CIRCULAR motion ( v = d/t) 10. What is a tangent???
    4. 4. Circular Motion Definitions •An object moving at constant speed in a circular path has the direction of its velocity changing  is accelerating. •The direction of this acceleration is towards the centre of the circle. •The acceleration is perpendicular to the velocity •This acceleration is called centripetal acceleration. •The force which produces this acceleration, and therefore produces the circular motion, is called centripetal force.
    5. 5. Circular Motion • Different forces can provide a centripetal force, e.g. friction pushing inwards or a rope pulling inwards. • If the centripetal force is removed then the object will travel at a tangent to the circle at a constant speed. Ta v nge nti al ve loc it y ,
    6. 6. Circular Motion Centripetal acceleration : ac (m s-2) = v2 r Centripetal force Substitute in F = ma Fc (N) = mv2 r v : Tangential velocity (m s-1) r : Radius of the circular path (m) m : Mass of the moving object (kg) radius,r ac Fc
    7. 7. Circular motion • When an object undergoes circular motion it must experience a centripetal force • This produces an acceleration towards the centre of the circle 7
    8. 8. Angular Speed Centripetal Force 8
    9. 9. 9
    10. 10. 10
    11. 11. 11
    12. 12. Tangential Velocity • v = d ÷t • v = circumference ÷ period • v = 2πr ÷ T 12
    13. 13. Acceleration • The acceleration towards the centre of the circle is • a = v2/r Centripetal Force Equation • • • • The general force equation is F = ma so the centripetal force equation is F = mv2/r 13
    14. 14. Centripetal Force and acceleration • Circular motion is accelerated motion as the direction is changing • Thus the has to be an unbalanced/net force • This is the centripetal force • This centripetal force can come from tension in a string or friction on a corner etc • or any other force that pulls inward
    15. 15. Friction prevents a 1.5kg mass from falling off a rotating turntable, which rotates at 30rpm. If the mass is 10cm from the centre of rotation, what is the friction preventing the mass moving outwards, and falling off? Solution: 30 revolutions per minute means 1 revolution takes 2 seconds T = 2s r = 0.1m m = 1.5 kg v = 2π r ÷ T = 2π x 0.1 ÷ 2 = 0.63/2 = 0.32 F = v2 m r = 0.322 X 1.5 0.1 = 1.5 N Course BOOK page 48 no 1 - 4
    16. 16. Conical Pendulum` • An object swung slowly on the end of a string moves in a circle But the string is not horizontal. The string moves round in the shape of a cone
    17. 17. Centripetal Force • Since the object is moving round in a circle we know that there must be a force towards the centre. This is called the centripetal force
    18. 18. Force Diagram We can see that there are two forces acting Gravity downwards Tension along the string
    19. 19. Resultant Force • With no force acting towards the centre we might look at the resultant force. The resultant force is towards the centre There is no overall vertical force component
    20. 20. Calculations • Unfortunately there is no obvious force acting to the centre • So how do the forces relate to each other, and to the speed of circular motion? • The same basic formulae apply as before • a=v2/r v=2πr/T θ r v
    21. 21. Vertical Forces • When we consider the vertical components of force • The total is zero T cos θ T θ So vertical component of tension = gravity •T cos θ = m g •Giving T = m g/ cos θ W=mg
    22. 22. Horizontal Forces • Horizontally the only real force is the component of Tension F = T sin θ This is equal to the centripetal force T sin θ = m v2 / r T sin θ
    23. 23. Combined Formulae • We can relate the centripetal force to mass. • Combining equations T sin θ = m v2 / r and T = m g/ cos θ Gives the relationship m g tan θ = m v2 / r tan θ = sin θ / cos θ
    24. 24. Example 1 Henry is swinging a 2.50 kg hammer round on the end of a 1.80 m long rope. The angle of the rope to the vertical is 45o. Calculate: a) The weight of the hammer (g = 9.8 m s-2) b) The tension force in the rope c) The resultant force towards the centre d) The radius of the circular motion e) The velocity of the hammer
    25. 25. Worked Answers 1 a) Weight = m g = 2.5 x 9.8 = 24.5 N b) From vertical motion T = m g / cos θ T = 24.5 / cos 45 = 34.6 N c) Horizontally the resultant force F = T sin θ F = 34.6 sin 45 = 24.5 N d) Radius is found by trigonometry r = l cos θ = 1.80 cos 45 = 1.27 m e) the velocity of the hammer is found by F = m v2 / r 24.5 = 2.5 v2 / 1.27 v2 = 12.5 v = 3.53 m s-1
    26. 26. Other Circular Motions • This approach can be used whenever the circular motion is caused by two or more forces whose resultant is towards the centre Cars or bikes on Banked Curves Aeroplanes Banking
    27. 27. Banked Curves • We will assume a Smooth (icy, slippery, oily, wet) surface. • i.e. no friction • Here the two forces Reaction Force acting are: Gravity downwards Normal Reaction at θ 90o to the surface Weight Force
    28. 28. Banked Curve Formulae • We can relate the forces in the same way as with the conical pendulum The resultant is towards the centre (horizontal) Reaction Force T = m g/ cos θ (Vertically) T sin θ = m v2 / r (Horizontally) θ Weight Force Resultant Force
    29. 29. Derivation of Banked Curves • Divide the x by the y mv FN sin θ = r FN cosθ = mg • Gives 2 • Notice mass is not involved v2 = rg tan ϴ v tan θ = rg 2
    30. 30. Example 2 A motor bike is travelling round a smooth curve banked at 20o. The radius of the curve is 55 m. If the mass of the bike and rider is 275 kg find the maximum safe speed round the bend. How would the maximum safe speed change if there was friction between the tyres and the road? Explain your answer.
    31. 31. Worked Answers 2 g tan θ = v2 / r m = 275 kg, g = 9.8 ms-2, r = 55 m, q = 20o 9.8 tan 20 = v2 / 55 v2 = 196 v = 14 m s-1 With friction the bike can go faster. This is because the friction force will be down the slope, stopping the bike sliding off the outside of the track. It will have a component towards the centre. This will increase the centripetal force, and hence the maximum speed.
    32. 32. Why do motorcyclists lean when cornering?
    33. 33. Consider the forces Straight line
    34. 34. Consider the forces Cornering upright Centripetal force Which way is he turning? What will happen? (consider Torque about center of mass)
    35. 35. Consider the forces Cornering leaning Centripetal force What will happen? (consider Torque about center of mass) clockwise torque = anticlockwise torque
    36. 36. Planes Banking • With planes it is the lift force created by the wings that keeps the plane flying In order to turn corners planes must be tilted, or banked A bigger angle of tilt means a sharper turn Note banking the plane causes it to turn
    37. 37. Forces on Planes • The lift force is perpendicular to the wings • The weight force is downwards When banking the lift must balance the weight AND provide the centripetal force θ Lift Force Resultant Force Weight Force L = m g/ cos θ (Vert.), L sin θ = m v2 / r (Hor.), g tan θ = v2 / r
    38. 38. Example 3 The picture shows a plane banking. The plane is travelling at 400 m s-1. By examining the picture: a Estimate the angle of banking. b Estimate the mass of the plane. For your estimates work out: c The radius of the circular path d The size of the lift force required by the wings
    39. 39. Worked Answers 3 a Angle is about 25o (20-30) b Mass is about 5 tonnes = 5000 kg c g tan θ = v2 / r g = 9.8 m s-2, θ = 25o, v = 400 m s-1 9.8 tan25 = 4002 / r r = 35000 m = 35 km L = m g/ cos θ d L = 5000 x 9.8 / cos 25 = 54000 N
    40. 40. Circular Motion Summary under the action of two forces •Real forces may not act towards the centre •The resultant force always acts towards the centre •The same basic formulae apply as before a=v2/r v=2πr/T •The two forces acting are related by the formulae T = m g/ cos θ (Vertically) T sin θ = m v2 / r (Horizontally) θ is the angle of tilt to the vertical •Conical Pendulum, Smooth Banked Curves and Planes Banking are examples of circular motion caused by two combined forces.
    41. 41. Vertical Circular Motion
    42. 42. • Why doesn’t he fall at the highest point?
    43. 43. What happens to the penguin if the lift is stationary? Why? acceleration = 0 Gravity pulls the penguin down, the elevator produces an equal and opposite force up
    44. 44. What happens to the penguin if the lift is accelerates up at 10ms-2? Why? acceleration Gravity pulls it down, the elevator produces a force up that is twice its weight.
    45. 45. What happens to the penguin if the lift is accelerates down at 10ms-2? Why? acceleration The penguin is in freefall. There is no upward force on the penguin from the elevator. When we feel “weight”, it is the upward force from the floor. The penguin thinks she is weightless
    46. 46. What happens to the penguin if the lift is accelerates down at 15ms-2? Why? acceleration The penguin is stuck to the ceiling of the elevator. She is accelerating at more than 10 ms-2, so an extra force is needed. (this is from the elevator ceiling)
    47. 47. Roller Coaster Why does the car not fall? It is falling!! But it’s also moving sideways. If the centripetal acceleration is greater than 10 ms-2 the car will stay on the track This is like being in an elevator accelerating down at › 10ms-2. You will be on the ceiling.
    48. 48. Gravity and Satellites • How are you able to watch the All Blacks on TV when they are playing in France? • How are you able to have a cellphone conversation with your friend in the USA?
    49. 49. International Space Station
    50. 50. Other Uses of Satellites • Tracking endangered animals
    51. 51. Satellite Weather Forecasting
    52. 52. Satellites Satellites are above the atmosphere (no air resistance), so the only force on them is…………. Gravity
    53. 53. Why don’t satellites fall? • link to Newton's mountain • Satellites are falling, but they are moving sideways, • They fall at the same rate that the earth curves away under them
    54. 54. Weightlessness. • You don’t feel the gravity force on your body, you feel the support force pushing up. • You can feel weightless for two reasons. 1:You are weightless in outer space. There is no gravity force so there is no reaction force
    55. 55. You feel weightless in orbit • You feel “weight” due to the support force on you. • The astronaut and their satellite are in free fall. • There is no support force acting on them
    56. 56. Newton’s Law of Gravitation m1m2 F =G 2 r
    57. 57. Can a satellite have any speed? • At a certain height, a satellite must have one particular speed. • Satellite Speed: ms mE ms v 2 F =G 2 = r r 2 mE v G 2 = r r mE 2 G =v r mE v = G r
    58. 58. Period of a satellite The period of a satellite orbiting around the earth only depends on the altitude.
    59. 59. mE v =G r 2 d 2π r v= = t T mE 4π r =G 2 T r 2 2 4π r 2 =T GmE 2 3
    60. 60. 4π r 2 =T GmE 2 3 • Notice the mass of the satellite doesn’t appear in the equation. • All masses have the same acceleration due to gravity: F a= m weight g= mass
    61. 61. • A geostationary satellite has a period of 24 hours • It must have the earth’s centre as its orbital centre • It must be above the equator
    62. 62. Geostationary Satellites AU S NZ
    63. 63. AU S NZ
    64. 64. AU S NZ
    65. 65. AU S NZ
    66. 66. S AU NZ
    67. 67. AU S NZ
    68. 68. AU S NZ