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- 1. MechanicsPart 1 – Circular Motion
- 2. Circular Motion in a horizontal plane P moves around a circle of radius r. P r θ O T
- 3. Circular Motion in a horizontal plane P moves around a circle of radius r. As P moves both the arc length PT change and the P angle θ changes r θ O T
- 4. Circular Motion in a horizontal plane P moves around a circle of radius r. As P moves both the arc length PT change and the P angle θ changes r The angular velocity of P is θ given by O T
- 5. Circular Motion in a horizontal plane P moves around a circle of radius r. As P moves both the arc length PT change and the angle θ changes P The angular velocity of P is r given by dθ & θ ω= =θ O T dt Force = mass × acceleration
- 6. Circular Motion in a horizontal plane P moves around a circle of radius r. As P moves both the arc length PT change and the angle θ changes P The angular velocity of P is r given by dθ & θ ω= =θ O T dt Force = mass × acceleration
- 7. dθ &ω= =θ dt PLet PT = x r Oθ T
- 8. dθ &ω= =θ dt PLet PT = x r x = rθ Oθ T
- 9. dθ &ω= =θ dt PLet PT = x r x = rθ Oθ T dx dθ =r dt dt
- 10. dθ &ω= =θ dt PLet PT = x r x = rθ Oθ T dx dθ =r dt dt v = rω
- 11. dθ & ω= =θ dt P Let PT = x r x = rθ Oθ T dx dθ =r dt dt v = rωThis equation is important since it links angular and linear velocity
- 12. dθ & ω= =θ dt P Let PT = x r x = rθ Oθ T dx dθ =r dt dt v = rωThis equation is important since it links angular and linear velocity
- 13. v = rωTo discuss acceleration weshould consider the motion interms of horizontal and vertical Pcomponents. r θ O T
- 14. v = rωTo discuss acceleration weshould consider the motion interms of horizontal and vertical P( x,y)components. rx = r cosθ , y = r sin θ O θ T
- 15. v = rωTo discuss acceleration weshould consider the motion interms of horizontal and vertical P( x,y)components. rx = r cosθ , y = r sin θ θ O Tdx d dy d = r cosθ , = r sin θdt dt dt dt
- 16. v = rωTo discuss acceleration weshould consider the motion interms of horizontal and vertical P( x,y)components. rx = r cosθ , y = r sin θ O θ Tdx d dy d = r cosθ , = r sin θdt dt dt dt d dθ d dθx = r cosθ& , y = r sin θ & dθ dt dθ dt
- 17. v = rωTo discuss acceleration weshould consider the motion interms of horizontal and vertical P( x,y)components. rx = r cosθ , y = r sin θ θ O Tdx d dy d = r cosθ , = r sin θdt dt dt dt d dθ d dθx = r cosθ& , y = r sin θ & dθ dt dθ dtx = − rω sin θ , y = rω cosθ & &
- 18. v = rωTo discuss acceleration weshould consider the motion interms of horizontal and vertical P( x,y)components. rx = r cosθ , y = r sin θ θ O Tdx d dy d = r cosθ , = r sin θdt dt dt dt d dθ d dθx = r cosθ , y = r sin θ& & dθ dt dθ dt x = −rω sin θ , y = rω cosθ & &
- 19. v = rω x = − rω sin θ , y = rω cosθ & & P( x,y)To simplify life we are going rto consider that the angularvelocity remains constant θ O Tthroughout the motion.
- 20. v = rω x = − rω sin θ , y = rω cosθ & &To simplify life we are going P( x,y)to consider that the angular rvelocity remains constant θthroughout the motion. O TThis will be the case in anyproblem you do , but you shouldbe able to prove these results forvariable angular velocity.
- 21. v = rωx = − rω sin θ , y = rω cosθ& &
- 22. v = rω x = −rω sin θ , y = rω cosθ & & d dx = −rω sin θ , && = rω cosθ&& y dt dt
- 23. v = rω x = −rω sin θ , y = rω cosθ & & d dx = −rω sin θ , && = rω cosθ&& y dt dt d dθ d dθx = −rω sin θ , && = rω cosθ&& y dθ dt dθ dt
- 24. v = rω x = −rω sin θ , y = rω cosθ & & d dx = −rω sin θ , && = rω cosθ&& y dt dt d dθ d dθx = −rω sin θ , && = rω cosθ&& y dθ dt dθ dtx = −rω cosθω , && = rω ( − sin θ ) ω&& y
- 25. v = rωx = −rω sin θ , y = rω cosθ& & d dx = −rω sin θ , y = rω cosθ&& & dt dt d dθ d dθx = −rω sin θ , && = rω cosθ&& y dθ dt dθ dtx = −rω cosθω , && = rω ( − sin θ ) ω&& yx = −rω cosθ , && = −rω sin θ&& 2 y 2
- 26. v = rωx = −rω sin θ , y = rω cosθ& & d dx = −rω sin θ , y = rω cosθ&& & dt dt d dθ d dθx = −rω sin θ , && = rω cosθ&& y dθ dt dθ dtx = −rω cosθω , && = rω ( − sin θ ) ω&& yx = −rω cosθ , && = −rω sin θ&& 2 y 2
- 27. v = rω x = −rω sin θ , y = rω cosθ & & x = −rω 2 cosθ , && = −rω 2 sin θ && yForces Diagram −rω 2 sin θ −rω 2 cosθ
- 28. x = −rω 2 cosθ , && = −rω 2 sin θ&& y a −rω 2 sin θ −rω 2 cosθIt is important to note that the vectors demonstratethat the force is directed along the radius towardsthe centre of the circle.
- 29. x = −rω 2 cosθ , && = −rω 2 sin θ&& y a −rω 2 sin θ −rω 2 cosθ a = ( − rω cosθ ) + ( −rω sin θ ) 2 2 2 2 2
- 30. x = −rω 2 cosθ , && = −rω 2 sin θ&& y a −rω 2 sin θ −rω 2 cosθ a = ( − rω cosθ ) + ( −rω sin θ ) 2 2 2 2 2 a = r 2ω 4 cos 2 θ + r 2ω 4 sin 2 θ a = r 2ω 4 ( cos 2 θ + sin 2 θ )
- 31. x = −rω 2 cosθ , && = −rω 2 sin θ&& y a −rω 2 sin θ −rω 2 cosθ a = ( − rω cosθ ) + ( −rω sin θ ) 2 2 2 2 2 a = r 2ω 4 cos 2 θ + r 2ω 4 sin 2 θ a = r 2ω 4 ( cos 2 θ + sin 2 θ ) a = r 2ω 4
- 32. x = −rω 2 cosθ , && = −rω 2 sin θ && y a −rω 2 sin θ −rω 2 cosθ va = ( − rω cosθ ) + ( − rω sin θ ) a = rω but v = rω → ω = 2 2 2 2 2 2 r 2a = r ω cos θ + r ω sin θ 2 4 2 2 4 2 v a = r ÷a = r 2ω 4 ( cos 2 θ + sin 2 θ ) r v2a = r 2ω 4 a= ra = rω 2
- 33. dθ &ω= dt =θ Summary Forces Diagram(angular velocity)v = rω −rω 2 sin θ(links angular velocity andlinear velocity) −rω 2 cosθx = −rω sin θ , y = rω cosθ& & force is directed along the(horizontal and vertical radius towards the centrecomponents of acceleration) of the circle.a = rω 2(gives the size of the forcetowards the centre)
- 34. dθ &ω= dt =θ Summary Forces Diagram(angular velocity)v = rω −rω 2 sin θ(links angular velocity andlinear velocity) −rω 2 cosθx = −rω sin θ , y = rω cosθ& & force is directed along the(horizontal and vertical radius towards the centrecomponents of acceleration) of the circle. 2 va = rω 2 a= r(gives the size of the forcetowards the centre)
- 35. 1. A body of mass 2kg is revolving at the end of a light string 3m long, on a smooth horizontal table with uniform angular speed of 1 revolution per second.
- 36. 1. A body of mass 2kg is revolving at the end of a light string 3m long, on a smooth horizontal table with uniform angular speed of 1 revolution per second. Draw a neat diagram to represent the forces.
- 37. 1. A body of mass 2kg is revolving at the end of a light string 3m long, on a smooth horizontal table with uniform angular speed of 1 revolution per second.(a) Find the tension in the string.
- 38. 1. A body of mass 2kg is revolving at the end of a light string 3m long, on a smooth horizontal table with uniform angular speed of 1 revolution per second.(a) Find the tension in the string. N T P mg
- 39. 1. A body of mass 2kg is revolving at the end of a light string 3m long, on a smooth horizontal table with uniform angular speed of 1 revolution per second.(a) Find the tension in the string. N T P mg The tension in the string is the resultant of the forces acting on the body.
- 40. 1. A body of mass 2kg is revolving at the end of a light string 3m long, on a smooth horizontal table with uniform angular speed of 1 revolution per second.(a) Find the tension in the string. T = F = ma N T P mg The tension in the string is the resultant of the forces acting on the body.
- 41. 1. A body of mass 2kg is revolving at the end of a light string 3m long, on a smooth horizontal table with uniform angular speed of 1 revolution per second.(a) Find the tension in the string. T = F = ma N T = mrω 2 T P mg The tension in the string is the resultant of the forces acting on the body.
- 42. 1. A body of mass 2kg is revolving at the end of a light string 3m long, on a smooth horizontal table with uniform angular speed of 1 revolution per second.(a) Find the tension in the string. N T = F = ma T P T = mrω 2 = 2 × 3 × ( 2π ) 2 mg = 24π 2 N The tension in the string is the resultant of the forces acting on the body.
- 43. 1. A body of mass 2kg is revolving at the end of a light string 3m long, on a smooth horizontal table with uniform angular speed of 1 revolution per second.(a) Find the tension in the string. N T = F = ma T P T = mrω 2 = 2 × 3 × ( 2π ) 2 mg = 24π 2 N The tension in the string is the resultant of the forces acting on the body.
- 44. (b) If the string would break under a tension of equal to the weight of 20 kg, find the greatest possible speed of the mass.
- 45. (b) If the string would break under a tension of equal to the weight of 20 kg, find the greatest possible speed of the mass.
- 46. (b) If the string would break under a tension of equal to the weight of 20 kg, find the greatest possible speed of the mass. T ≤ 20 g mv 2 T= r
- 47. (b) If the string would break under a tension of equal to the weight of 20 kg, find the greatest possible speed of the mass. T ≤ 20 g mv 2 T= r mv 2 ≤ 20 g r r v ≤ 20 g × 2 m
- 48. (b) If the string would break under a tension of equal to the weight of 20 kg, find the greatest possible speed of the mass. T ≤ 20 g 3 v ≤ 20 g × 2 mv 2 2 T= r v ≤ 30 g m / s mv 2 ≤ 20 g r r v ≤ 20 g × 2 m
- 49. (b) If the string would break under a tension of equal to the weight of 20 kg, find the greatest possible speed of the mass. T ≤ 20 g 3 v ≤ 20 g × 2 mv 2 2 T= r v ≤ 30 g m / s mv 2 ≤ 20 g r r v ≤ 20 g × 2 m
- 50. Conical PendulumIf a particle is tied by a string to a fixed pointby means of a string and moves in a horizontalcircle so that the string describes a cone, andthe mass at the end of the string describes ahorizontal circle, then the string and the massdescribe a conical pendulum.
- 51. Conical PendulumIf a particle is tied by a string to a fixed pointby means of a string and moves in a horizontalcircle so that the string describes a cone, andthe mass at the end of the string describes ahorizontal circle, then the string and the massdescribe a conical pendulum.
- 52. Conical Pendulum θ L R m
- 53. Conical Pendulum θ θ L L R R m mForces Diagram Dimensions DiagramVertically
- 54. Conical Pendulum θ θ L R mForces Diagram Dimensions Diagram mgVertically
- 55. Conical Pendulum θ θ L N R mForces Diagram Dimensions DiagramVertically mg
- 56. Conical Pendulum θ θ L T N θ R m mgForces Diagram Dimensions DiagramVerticallyN + (-mg) = 0 N = mgbut cos θ = N/Tso N = T cos θ==> T cos θ = mg
- 57. Conical Pendulum θ θ L T N θ R m mgForces Diagram Dimensions DiagramVertically HorizontallyN + (-mg) = 0 T sin θ = mrω2 N = mg Since the onlybut cos θ = N/T horizontal forceso N = T cos θ is directed along==> T cos θ = mg the radius towards the centre
- 58. Conical Pendulum θ θ L T N θ R m mgForces Diagram Dimensions DiagramVertically Horizontally T sin θ = mrω2N + (-mg) = 0 T sin θ = mrω2 T cos θ = mg N = mg Since the onlybut cos θ = N/T horizontal forceso N = T cos θ is directed along==> T cos θ = mg the radius towards the centre
- 59. Conical Pendulum θ θ L T N θ R m mgForces Diagram Dimensions DiagramVertically Horizontally T sin θ = mrω2N + (-mg) = 0 T sin θ = mrω2 T cos θ = mg N = mg Since the onlybut cos θ = N/T horizontal forceso N = T cos θ is directed along==> T cos θ = mg the radius towards the centre
- 60. An important result θ θ T N L θ h R m mg Forces Diagram T sin θ = mrω2……1 T cos θ = mg … … 2 v = rωNow dividing equation 1 by equation 2..
- 61. An important result θ θ T N L θ h R m mg Forces Diagram Tsinθ mrω 2 T sin θ = mrω2……1 = T cos θ = mg … … 2 T cos θ mg v = rωNow dividing equation 1 by equation 2..
- 62. An important result θ θ T N L θ h R m mg Forces Diagram Tsinθ mrω 2 = T cos θ mg T sin θ = mrω2……1 T cos θ = mg … … 2 rω 2 tan θ = v = rω gNow dividing equation 1 by equation 2..
- 63. An important result θ θ T N L θ h R m mg Forces Diagram Tsinθ mrω 2 = T cos θ mg T sin θ = mrω2……1 T cos θ= mg … … 2 rω 2 tan θ = v = rω g rNow dividing equation 1 by equation 2.. tan θ = h
- 64. An important result θ θ T N L h θ r m mg Tsinθ mrω 2 Forces Diagram = T cos θ mg T sin θ = mrω2……1 rω 2 T cos θ = mg … … 2 tan θ = g v = rω r tan θ =Now dividing equation 1 by equation 2.. h rω 2 r = g h g h= 2 ω
- 65. An important result θ θ T N L θ h r m mg Tsinθ mrω 2 Forces Diagram = T cos θ mg T sin θ = mrω2……1 rω 2 T cos θ = mg … … 2 tan θ = g v = rω r tan θ =Now dividing equation 1 by equation 2.. h rω 2 r = g h g h= 2 ω
- 66. Example 2A string of length 2 m, fixed at one end A carries at theother end a particle of mass 6 kg rotating in a horizontalcircle whose centre is 1m vertically below A. Find thetension in the string and the angular velocity of theparticle.
- 67. Example 2A string of length 2 m, fixed at one end A carries at theother end a particle of mass 6 kg rotating in a horizontalcircle whose centre is 1m vertically below A. Find thetension in the string and the angular velocity of theparticle.
- 68. Example 2A string of length 2 m, fixed at one end A carries at theother end a particle of mass 6 kg rotating in a horizontalcircle whose centre is 1m vertically below A. Find thetension in the string and the angular velocity of theparticle. θ θ L=2 T N h=1 θ r mg Forces Diagram Dimensions Diagram
- 69. Example 2A string of length 2 m, fixed at one end A carries at theother end a particle of mass 6 kg rotating in a horizontalcircle whose centre is 1m vertically below A. Find thetension in the string and the angular velocity of theparticle. θ θ L=2 T T cos θ h=1 θ r mg Forces Diagram Dimensions Diagram
- 70. Example 2Find the tension in the string and the angular velocity ofthe particle. θ θ L=2 T N h=1 θ r mg Forces Diagram Dimensions Diagram Vertically Horizontally T cos θ = mg mg T= cos θ 6g T= 1 2 T = 12 g
- 71. Example 2Find the tension in the string and the angular velocity ofthe particle. θ θ L=2 T N h=1 θ r mg Forces Diagram Dimensions Diagram Vertically Horizontally T cos θ = mg T sin θ = mrω 2 mg T sin θ = mL sin θω 2 T= cos θ T 6g ω = 2 T= mL 1 12 g 2 ω= = g 12 T = 12 g
- 72. Example 2Find the tension in the string and the angular velocity ofthe particle. θ θ L=2 T N h=1 θ r mg Forces Diagram Dimensions Diagram Vertically Horizontally T cos θ = mg T sin θ = mrω 2 mg T sin θ = mL sin θω 2 T= cos θ T 6g ω = 2 T= mL 1 12 g 2 ω= = g 12 T = 12 g

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