X2 t06 04 uniform circular motion (2012)

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  • how can you prove the triangle in the first slide is similar to triangleAOP? you can't just say they are similar because sides in proportional. the condition is not enough to let you say so.
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X2 t06 04 uniform circular motion (2012)

  1. 1. Acceleration with Uniform Circular Motion
  2. 2. Acceleration with Uniform Circular MotionUniform circular motion is when a particle moves with constant angularvelocity.
  3. 3. Acceleration with Uniform Circular MotionUniform circular motion is when a particle moves with constant angularvelocity.  the magnitude of the linear velocity will also be constant 
  4. 4. Acceleration with Uniform Circular MotionUniform circular motion is when a particle moves with constant angularvelocity.  the magnitude of the linear velocity will also be constant  A r O
  5. 5. Acceleration with Uniform Circular MotionUniform circular motion is when a particle moves with constant angularvelocity.  the magnitude of the linear velocity will also be constant  A particle moves from A to P with constant A angular velocity. r O  P
  6. 6. Acceleration with Uniform Circular MotionUniform circular motion is when a particle moves with constant angularvelocity.  the magnitude of the linear velocity will also be constant  A particle moves from A to P with constant A angular velocity. r O  The acceleration of the particle is the change in velocity with respect to time. P
  7. 7. Acceleration with Uniform Circular MotionUniform circular motion is when a particle moves with constant angularvelocity.  the magnitude of the linear velocity will also be constant  A particle moves from A to P with constant A angular velocity. r v O  The acceleration of the particle is the change in velocity with respect to time. P
  8. 8. Acceleration with Uniform Circular MotionUniform circular motion is when a particle moves with constant angularvelocity.  the magnitude of the linear velocity will also be constant  A particle moves from A to P with constant A angular velocity. r v O  The acceleration of the particle is the change in velocity with respect to time. P v’
  9. 9. Acceleration with Uniform Circular MotionUniform circular motion is when a particle moves with constant angularvelocity.  the magnitude of the linear velocity will also be constant  A particle moves from A to P with constant A angular velocity. r v O  The acceleration of the particle is the change in velocity with respect to time. P v’ v
  10. 10. Acceleration with Uniform Circular MotionUniform circular motion is when a particle moves with constant angularvelocity.  the magnitude of the linear velocity will also be constant  A particle moves from A to P with constant A angular velocity. r v O  The acceleration of the particle is the change in velocity with respect to time. P v’ v v’
  11. 11. Acceleration with Uniform Circular MotionUniform circular motion is when a particle moves with constant angularvelocity.  the magnitude of the linear velocity will also be constant  A particle moves from A to P with constant A angular velocity. r v O  The acceleration of the particle is the change in velocity with respect to time. P v’ v v’ v
  12. 12. Acceleration with Uniform Circular MotionUniform circular motion is when a particle moves with constant angularvelocity.  the magnitude of the linear velocity will also be constant  A A particle moves from A to P with constant angular velocity. r v O  The acceleration of the particle is the change in velocity with respect to time. P v’ v’  v v
  13. 13. Acceleration with Uniform Circular MotionUniform circular motion is when a particle moves with constant angularvelocity.  the magnitude of the linear velocity will also be constant  A A particle moves from A to P with constant angular velocity. r v O  The acceleration of the particle is the change in velocity with respect to time. P v’  v v’ This triangle of vectors is similar to OAP v
  14. 14. Av’  v r O  v r
  15. 15. Av’  v r O  v r Pv v AP r
  16. 16. A v’  v r O  v r Pv v AP r v  APv  r
  17. 17. A v’  v r O  v r Pv v AP r v  APv  r v a t
  18. 18. A v’  v r O  v r Pv v AP r v  APv  r v a t v  AP a r  t
  19. 19. A v’  v r O  v r P v v  AP r v  AP v  r v a t v  AP a r  tBut, as Δt  0, AP  arcAP
  20. 20. A v’  v r O  v r P v v  AP r v  AP v  r v a t v  AP a r  tBut, as Δt  0, AP  arcAP v  arcAP  a  lim t 0 r  t
  21. 21. A v’  v r distance  speed  time O  arcAP  v  t v r P v v  AP r v  AP v  r v a t v  AP a r  tBut, as Δt  0, AP  arcAP v  arcAP  a  lim t 0 r  t
  22. 22. A v’  v r distance  speed  time O  arcAP  v  t v r P v v  v  v  t AP r  a  lim t 0 r  t v  AP v  r v2  lim v t 0 r a t v2  v  AP r a r  tBut, as Δt  0, AP  arcAP v  arcAP  a  lim t 0 r  t
  23. 23. A v’  v r distance  speed  time O  arcAP  v  t v r P v v  v  v  t AP r  a  lim t 0 r  t v  AP v  r v2  lim v t 0 r a t v2  v  AP r a r  t  the acceleration in uniform circularBut, as Δt  0, AP  arcAP v2 v  arcAP motion has magnitude and is  a  lim r t 0 r  t directed towards the centre
  24. 24. Acceleration Involved inUniform Circular Motion v2 a r
  25. 25. Acceleration Involved inUniform Circular Motion v2 a r OR a  r 2
  26. 26. Acceleration Involved in Forces Involved inUniform Circular Motion Uniform Circular Motion v2 a r OR a  r 2
  27. 27. Acceleration Involved in Forces Involved inUniform Circular Motion Uniform Circular Motion v2 mv 2 a F r r OR a  r 2
  28. 28. Acceleration Involved in Forces Involved inUniform Circular Motion Uniform Circular Motion v2 mv 2 a F r r OR OR a  r 2 F  mr 2
  29. 29. Acceleration Involved in Forces Involved in Uniform Circular Motion Uniform Circular Motion v2 mv 2 a F r r OR OR a  r 2 F  mr 2e.g. (i) (2003) A particle P of mass m moves with constant angular velocity  on a circle of radius r. Its position at time t is given by; x  r cos y  r sin  , where   t
  30. 30. Acceleration Involved in Forces Involved in Uniform Circular Motion Uniform Circular Motion v2 mv 2 a F r r OR OR a  r 2 F  mr 2 e.g. (i) (2003) A particle P of mass m moves with constant angular velocity  on a circle of radius r. Its position at time t is given by; x  r cos y  r sin  , where   ta) Show that there is an inward radial force of magnitude mr 2 acting on P.
  31. 31. y P r y  r sin   x  r cos x
  32. 32. x  r cos y P r y  r sin   x  r cos x
  33. 33. x  r cos y y  r sin  P r y  r sin   x  r cos x
  34. 34. x  r cos y y  r sin  dx   r sin   P dt r y  r sin    r sin   x  r cos x
  35. 35. x  r cos y y  r sin  d dx   r sin   P y  r cos   dt dt r y  r sin    r sin   r cos  x  r cos x
  36. 36. x  r cos y y  r sin  d d x   r sin    P y  r cos   dt dt r y  r sin    r sin   r cos  d x  r cos x   r cos x dt   r 2 cos   2 x
  37. 37. x  r cos y y  r sin  d d x   r sin    P y  r cos   dt dt r y  r sin    r sin   r cos    r cos x d x  r cos x    r sin   d y dt dt   r 2 cos   r 2 sin    x2   2 y
  38. 38. x  r cos y y  r sin  d d x   r sin    P y  r cos   dt dt r y  r sin    r sin   r cos    r cos x d x  r cos x    r sin   d y dt dt   r 2 cos   r 2 sin    x 2   2 y y  x 
  39. 39. x  r cos y y  r sin  d d x   r sin    P y  r cos   dt dt r y  r sin    r sin   r cos    r cos x d x  r cos x    r sin   d y dt dt   r 2 cos   r 2 sin    x 2   2 y a y   x 
  40. 40. x  r cos y y  r sin  d d x   r sin    P y  r cos   dt dt r y  r sin    r sin   r cos    r cos x d x  r cos x    r sin   d y dt dt   r 2 cos   r 2 sin    x 2 a      2 x 2 y 2   2 y a y   x 
  41. 41. x  r cos y y  r sin  d d x   r sin    P y  r cos   dt dt r y  r sin    r sin   r cos    r cos x d x  r cos x    r sin   d y dt dt   r 2 cos   r 2 sin    x 2 a      2 x 2 y 2   2 y   4 x2   4 y2   4 x 2  y 2  a y    4r 2  x 
  42. 42. x  r cos y y  r sin  d d x   r sin    P y  r cos   dt dt r y  r sin    r sin   r cos    r cos x d x  r cos x    r sin   d y dt dt   r 2 cos   r 2 sin    x 2 a      2 x y 2 2   2 y   4 x2   4 y2   4 x 2  y 2  a y    4r 2 a  r 2  x 
  43. 43. x  r cos y y  r sin  d d x   r sin    P y  r cos   dt dt r y  r sin    r sin   r cos    r cos x d x  r cos x    r sin   d y dt dt   r 2 cos   r 2 sin    x 2 a      2 x y 2 2   2 y   4 x2   4 y2   4 x 2  y 2  a y    4r 2 a  r 2  x  F  ma  F  mr 2
  44. 44. x  r cos y y  r sin  d d x   r sin    P y  r cos   dt dt r y  r sin    r sin   r cos    r cos x d x  r cos x    r sin   d y dt dt   r 2 cos   r 2 sin  y    x 2 a      2 x y 2 2   tan 1   2 y x    4 x2   4 y2   4 x 2  y 2  a y    4r 2 a  r 2  x  F  ma  F  mr 2
  45. 45. x  r cos y y  r sin  d d x   r sin    P y  r cos   dt dt r y  r sin    r sin   r cos    r cos x d x  r cos x    r sin   d y dt dt   r 2 cos   r 2 sin  y    x 2 a      2 x y 2 2   tan 1   2 y x    4 x2   4 y2   2 y   tan 1     4 x 2  y 2    x  2 a y    4r 2 y  tan 1 a  r 2 x  F  ma  x   F  mr 2
  46. 46. x  r cos y y  r sin  d d x   r sin    P y  r cos   dt dt r y  r sin    r sin   r cos    r cos x d x  r cos x    r sin   d y dt dt   r 2 cos   r 2 sin  y    x 2 a      2 x y 2 2   tan 1   2 y x    4 x2   4 y2   2 y   tan 1     4 x 2  y 2    x  2 a y    4r 2 y  tan 1 a  r 2 x  F  ma  x   F  mr 2  There is a force , F  mr 2 , acting towards the centre
  47. 47. b) A telecommunications satellite, of mass m, orbits Earth with constant angular velocity  at a distance r from the centre of the Earth. The gravitational force exerted by Earth on the satellite is Am where r2 A is a constant. By considering all other forces on the satellite to be negligible, show that; A r 3  2
  48. 48. b) A telecommunications satellite, of mass m, orbits Earth with constant angular velocity  at a distance r from the centre of the Earth. The gravitational force exerted by Earth on the satellite is Am where r2 A is a constant. By considering all other forces on the satellite to be negligible, show that; A r 3  2
  49. 49. b) A telecommunications satellite, of mass m, orbits Earth with constant angular velocity  at a distance r from the centre of the Earth. The gravitational force exerted by Earth on the satellite is Am where r2 A is a constant. By considering all other forces on the satellite to be negligible, show that; A r 3  2 m  mr 2 x
  50. 50. b) A telecommunications satellite, of mass m, orbits Earth with constant angular velocity  at a distance r from the centre of the Earth. The gravitational force exerted by Earth on the satellite is Am where r2 A is a constant. By considering all other forces on the satellite to be negligible, show that; A r 3  2 Am m  mr 2 x r2
  51. 51. b) A telecommunications satellite, of mass m, orbits Earth with constant angular velocity  at a distance r from the centre of the Earth. The gravitational force exerted by Earth on the satellite is Am where r2 A is a constant. By considering all other forces on the satellite to be negligible, show that; A r 3  2 Am mr 2  Am r2 m  mr 2 x r2
  52. 52. b) A telecommunications satellite, of mass m, orbits Earth with constant angular velocity  at a distance r from the centre of the Earth. The gravitational force exerted by Earth on the satellite is Am where r2 A is a constant. By considering all other forces on the satellite to be negligible, show that; A r 3  2 Am mr 2  Am r2 m  mr 2 x r2 Am r3  m 2 A  2 
  53. 53. b) A telecommunications satellite, of mass m, orbits Earth with constant angular velocity  at a distance r from the centre of the Earth. The gravitational force exerted by Earth on the satellite is Am where r2 A is a constant. By considering all other forces on the satellite to be negligible, show that; A r 3  2 Am mr 2  Am r2 m  mr 2 x r2 Am r3  m 2 A  2  A r3 2
  54. 54. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it. A mass of 2kg is attached to one end of the string and it is revolved in a circle. Find the greatest angular velocity which may be imparted without breaking the string.
  55. 55. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it. A mass of 2kg is attached to one end of the string and it is revolved in a circle. Find the greatest angular velocity which may be imparted without breaking the string.
  56. 56. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it. A mass of 2kg is attached to one end of the string and it is revolved in a circle. Find the greatest angular velocity which may be imparted without breaking the string. m x
  57. 57. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it. A mass of 2kg is attached to one end of the string and it is revolved in a circle. Find the greatest angular velocity which may be imparted without breaking the string. m x mg
  58. 58. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it. A mass of 2kg is attached to one end of the string and it is revolved in a circle. Find the greatest angular velocity which may be imparted without breaking the string. T m x mg
  59. 59. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it. A mass of 2kg is attached to one end of the string and it is revolved in a circle. Find the greatest angular velocity which may be imparted without breaking the string. m  mg  T x T m x mg
  60. 60. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it. A mass of 2kg is attached to one end of the string and it is revolved in a circle. Find the greatest angular velocity which may be imparted without breaking the string. m  mg  T x T 0  mg  T m x mg
  61. 61. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it. A mass of 2kg is attached to one end of the string and it is revolved in a circle. Find the greatest angular velocity which may be imparted without breaking the string. m  mg  T x T 0  mg  T T  40 9.8  392 N m x mg
  62. 62. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it. A mass of 2kg is attached to one end of the string and it is revolved in a circle. Find the greatest angular velocity which may be imparted without breaking the string. m  mg  T x T 0  mg  T T  40 9.8  392 N m x mg
  63. 63. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it. A mass of 2kg is attached to one end of the string and it is revolved in a circle. Find the greatest angular velocity which may be imparted without breaking the string. m x m  mg  T x T 0  mg  T T  40 9.8  392 N m x mg
  64. 64. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it. A mass of 2kg is attached to one end of the string and it is revolved in a circle. Find the greatest angular velocity which may be imparted without breaking the string. m x T m  mg  T x T 0  mg  T T  40 9.8  392 N m x mg
  65. 65. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it. A mass of 2kg is attached to one end of the string and it is revolved in a circle. Find the greatest angular velocity which may be imparted without breaking the string. m x T m  mg  T x T  mr 2 T 0  mg  T T  40 9.8  392 N m x mg
  66. 66. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it. A mass of 2kg is attached to one end of the string and it is revolved in a circle. Find the greatest angular velocity which may be imparted without breaking the string. m x T m  mg  T x T  mr 2 0  mg  T T 392  2 0.5 2 T  40 9.8  392 N m x mg
  67. 67. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it. A mass of 2kg is attached to one end of the string and it is revolved in a circle. Find the greatest angular velocity which may be imparted without breaking the string. m x T m  mg  T x T  mr 2 0  mg  T T 392  2 0.5 2 T  40 9.8  2  392  392 N mg   392 m x   2 98rad/s
  68. 68. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it. A mass of 2kg is attached to one end of the string and it is revolved in a circle. Find the greatest angular velocity which may be imparted without breaking the string. m x T m  mg  T x T  mr 2 0  mg  T T 392  2 0.5 2 T  40 9.8  2  392  392 N mg   392 m x   2 98rad/s Exercise 9B; all

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