Upcoming SlideShare
×

# Circular motion slideshare

2,780 views

Published on

Basic theory of uniform circular motion, content first created by the late James Taylor and further edited by Morris Needleman.

• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

### Circular motion slideshare

1. 1. MechanicsPart 1 – Circular Motion
2. 2. Circular Motion in a horizontal plane P moves around a circle of radius r. P r θ O T
3. 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. 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. 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. 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. 7. dθ &ω= =θ dt PLet PT = x r Oθ T
8. 8. dθ &ω= =θ dt PLet PT = x r x = rθ Oθ T
9. 9. dθ &ω= =θ dt PLet PT = x r x = rθ Oθ T dx dθ =r dt dt
10. 10. dθ &ω= =θ dt PLet PT = x r x = rθ Oθ T dx dθ =r dt dt v = rω
11. 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. 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. 13. v = rωTo discuss acceleration weshould consider the motion interms of horizontal and vertical Pcomponents. r θ O T
14. 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. 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. 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. 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. 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. 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. 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 TThis will be the case in anyproblem you do , but you shouldbe able to prove these results forvariable angular velocity.
21. 21. v = rωx = − rω sin θ , y = rω cosθ& &
22. 22. v = rω x = −rω sin θ , y = rω cosθ & & d dx = −rω sin θ , && = rω cosθ&& y dt dt
23. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 52. Conical Pendulum θ L R m
53. 53. Conical Pendulum θ θ L L R R m mForces Diagram Dimensions DiagramVertically
54. 54. Conical Pendulum θ θ L R mForces Diagram Dimensions Diagram mgVertically
55. 55. Conical Pendulum θ θ L N R mForces Diagram Dimensions DiagramVertically mg
56. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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