1. The Conical Pendulum

2. The Conical Pendulum
A

P

3. The Conical Pendulum
A

h

O r P

4. The Conical Pendulum
A Force Diagram

h

O r P

5. The Conical Pendulum
A Force Diagram

h

O r P

6. The Conical Pendulum
A Force Diagram
 T= tension in the string
T (always away from object)
h

O r P

7. The Conical Pendulum
A Force Diagram
 T= tension in the string
T (always away from object)
h
 mg
O r P

8. The Conical Pendulum
A Force Diagram
 T= tension in the string
T (always away from object)
h    string makes with vertical
 mg
O r P

9. The Conical Pendulum
A Force Diagram
 T= tension in the string
T (always away from object)
h    string makes with vertical
 mg   angular velocity of pendulum
O r P

10. The Conical Pendulum
A Force Diagram
 T= tension in the string
T (always away from object)
h    string makes with vertical
 mg   angular velocity of pendulum
O r P
Resultant Forces

11. The Conical Pendulum
A Force Diagram
 T= tension in the string
T (always away from object)
h    string makes with vertical
 mg   angular velocity of pendulum
O r P
Resultant Forces
mv 2
m 
x
r

12. The Conical Pendulum
A Force Diagram
 T= tension in the string
T (always away from object)
h    string makes with vertical
 mg   angular velocity of pendulum
O r P
Resultant Forces
mv 2
m 
x m  0
y
r

13. The Conical Pendulum
A Force Diagram
 T= tension in the string
T (always away from object)
h    string makes with vertical
 mg   angular velocity of pendulum
O r P
Resultant Forces
mv 2
m 
x m  0
y
r
mv 2
horizontal forces 
r

14. The Conical Pendulum
A Force Diagram
 T= tension in the string
T (always away from object)
h    string makes with vertical
 mg   angular velocity of pendulum
O r P
Resultant Forces
mv 2
m 
x m  0
y
r
mv 2
horizontal forces  vertical forces  0
r

15. The Conical Pendulum
A Force Diagram
 T= tension in the string
 T (always away from object)
h    string makes with vertical
 mg   angular velocity of pendulum
O r P
Resultant Forces
mv 2
m 
x m  0
y
r
mv 2
horizontal forces  vertical forces  0
r

16. The Conical Pendulum
A Force Diagram
 T= tension in the string
 T (always away from object)
h    string makes with vertical
 mg   angular velocity of pendulum
O r P
Resultant Forces
mv 2
m 
x m  0
y
r
mv 2
horizontal forces  vertical forces  0
r

17. The Conical Pendulum
A Force Diagram
 T= tension in the string
 T (always away from object)
h    string makes with vertical
 mg   angular velocity of pendulum
O r P
Resultant Forces
mv 2
m 
x m  0
y
r
mv 2
horizontal forces  vertical forces  0
r
T sin 

18. The Conical Pendulum
A Force Diagram
 T= tension in the string
 T (always away from object)
h    string makes with vertical
 mg   angular velocity of pendulum
O r P
Resultant Forces
mv 2
m 
x m  0
y
r
mv 2
horizontal forces  vertical forces  0
r
T sin  2
mv
T sin  
r

19. The Conical Pendulum
A Force Diagram
 T= tension in the string
 T (always away from object)
h    string makes with vertical
 mg   angular velocity of pendulum
O r P
Resultant Forces
mv 2
m 
x m  0
y
r
mv 2
horizontal forces  vertical forces  0
r
T sin  2
T sin  
mv  mr 2
r

20. The Conical Pendulum
A Force Diagram
 T= tension in the string
 T (always away from object)
h    string makes with vertical
 mg   angular velocity of pendulum
O r P
Resultant Forces
mv 2
m 
x m  0
y
r
mv 2
horizontal forces  vertical forces  0
r
T sin  2
T sin  
mv  mr 2
r

21. The Conical Pendulum
A Force Diagram
 T= tension in the string
 T (always away from object)
h    string makes with vertical
 mg   angular velocity of pendulum
O r P
Resultant Forces
mv 2
m 
x m  0
y
r
mv 2
horizontal forces  vertical forces  0
r
T cos
T sin  2
T sin  
mv  mr 2 mg
r

22. The Conical Pendulum
A Force Diagram
 T= tension in the string
 T (always away from object)
h    string makes with vertical
 mg   angular velocity of pendulum
O r P
Resultant Forces
mv 2
m 
x m  0
y
r
mv 2
horizontal forces  vertical forces  0
r
T cos T cos  mg  0
T sin  2
T sin  
mv  mr 2 mg T cos  mg
r

23. T sin  mv 2 1
 
T cos r mg

24. T sin  mv 2 1
 
T cos r mg
v2
tan  
rg

25. T sin  mv 2 1
 
T cos r mg
v2  r 2 
tan    
rg  g 

26. T sin  mv 2 1
 
T cos r mg
v2  r 2 
tan    
rg  g 
r
But in AOP tan  
h

27. T sin  mv 2 1
 
T cos r mg
v2  r 2 
tan    
rg  g 
r
But in AOP tan  
h
v2 r
 
rg h

28. T sin  mv 2 1
 
T cos r mg
v2  r 2 
tan    
rg  g 
r
But in AOP tan  
h
v2 r
 
rg h
r2g
h 2
v

29. T sin  mv 2 1
 
T cos r mg
v2  r 2 
tan    
rg  g 
r
But in AOP tan  
h
v2 r
 
rg h
r2g  g 
h 2  
v  2 

30. T sin  mv 2 1
 
T cos r mg
v2  r 2 
tan    
rg  g 
r
But in AOP tan  
h
v2 r
 
rg h
r2g  g 
h 2  
v  2 
Implications

31. T sin  mv 2 1
 
T cos r mg
v2  r 2 
tan    
rg  g 
r
But in AOP tan  
h
v2 r
 
rg h
r2g  g 
h 2  
v  2 
Implications
•depth of the pendulum below A is independent of the length of the
string.

32. T sin  mv 2 1
 
T cos r mg
v2  r 2 
tan    
rg  g 
r
But in AOP tan  
h
v2 r
 
rg h
r2g  g 
h 2  
v  2 
Implications
•depth of the pendulum below A is independent of the length of the
string.
•as the speed increases, the particle (bob) rises.

33. e.g. The number of revolutions per minute of a conical pendulum
increases from 60 to 90.
Find the rise in the level of the bob.

34. e.g. The number of revolutions per minute of a conical pendulum
increases from 60 to 90.
Find the rise in the level of the bob.

35. e.g. The number of revolutions per minute of a conical pendulum
increases from 60 to 90.
Find the rise in the level of the bob.
T

36. e.g. The number of revolutions per minute of a conical pendulum
increases from 60 to 90.
Find the rise in the level of the bob.
T
mg

37. e.g. The number of revolutions per minute of a conical pendulum
increases from 60 to 90.
Find the rise in the level of the bob.
 T
h
r
mg

38. e.g. The number of revolutions per minute of a conical pendulum
increases from 60 to 90.
Find the rise in the level of the bob.
 T
h
r
mg
mv 2
horizontal forces 
r

39. e.g. The number of revolutions per minute of a conical pendulum
increases from 60 to 90.
Find the rise in the level of the bob.
 T
h
r
mg
mv 2
horizontal forces  vertical forces  0
r

40. e.g. The number of revolutions per minute of a conical pendulum
increases from 60 to 90.
Find the rise in the level of the bob.
 T
h
r
mg
mv 2
horizontal forces  vertical forces  0
r

41. e.g. The number of revolutions per minute of a conical pendulum
increases from 60 to 90.
Find the rise in the level of the bob.
 T
h
r
mg
mv 2
horizontal forces  vertical forces  0
r
T sin 

42. e.g. The number of revolutions per minute of a conical pendulum
increases from 60 to 90.
Find the rise in the level of the bob.
 T
h
r
mg
mv 2
horizontal forces  vertical forces  0
r
T sin 
T sin   mr 2

43. e.g. The number of revolutions per minute of a conical pendulum
increases from 60 to 90.
Find the rise in the level of the bob.
 T
h
r
mg
mv 2
horizontal forces  vertical forces  0
r
T sin 
T sin   mr 2

44. e.g. The number of revolutions per minute of a conical pendulum
increases from 60 to 90.
Find the rise in the level of the bob.
 T
h
r
mg
mv 2
horizontal forces  vertical forces  0
r
T sin  T cos
T sin   mr 2 mg

45. e.g. The number of revolutions per minute of a conical pendulum
increases from 60 to 90.
Find the rise in the level of the bob.
 T
h
r
mg
mv 2
horizontal forces  vertical forces  0
r
T sin  T cos T cos  mg  0
T sin   mr 2 mg T cos  mg

46. 1
 tan   mr 2 
mg
r 2

g

47. 1
 tan   mr 2 
mg
r 2

g
r
But tan  
h

48. 1
 tan   mr 2 
mg
r 2

g
r
But tan  
h
r 2 r
 
g h
g
h 2


49. 1
 tan   mr 2  when   60rev/min
mg
120
r 2  rad/s
 60
g
 2rad/s
r
But tan  
h
r 2 r
 
g h
g
h 2


50. 1
 tan   mr 2  when   60rev/min h
g
mg
120
2 2
r 2  rad/s g
 60  m
g
 2rad/s 4 2
r
But tan  
h
r 2 r
 
g h
g
h 2


51. 1
 tan   mr 2  when   60rev/min h
g
mg
120
2 2
r 2  rad/s g
 60  m
g
 2rad/s 4 2
r
But tan  
h when   90rev/min
r 2 r
  180
g h  rad/s
60
h 2
g  3rad/s


52. 1
 tan   mr 2  when   60rev/min h
g
mg
120
2 2
r 2  rad/s g
 60  m
g
 2rad/s 4 2
r
But tan  
h when   90rev/min g
h

r 2 r
 180
3 2
g h  rad/s g
60  m
g  3rad/s 9 2
h 2


53. 1
 tan   mr 2  when   60rev/min h
g
mg
120
2 2
r 2  rad/s g
 60  m
g
 2rad/s 4 2
r
But tan  
h when   90rev/min g
h

r 2 r
 180
3 2
g h  rad/s g
60  m
g  3rad/s 9 2
h 2

 g  g m
 rise in height   2
 4 9 2 

 0.14m

54. (ii) (2002)
A particle of mass m is suspended by a string of length l from a point
directly above the vertex of a smooth cone, which has a vertical axis.
The particle remains in contact with the cone and rotates as a conical
pendulum with angular velocity .
The angle of the cone at its vertex is 2

where   , and the string makes an
4
angle of  with the horizontal as shown in
the diagram. The forces acting on the
particle are the tension in the string T, the
normal reaction N and the gravitational
force mg.

55. (ii) (2002)
A particle of mass m is suspended by a string of length l from a point
directly above the vertex of a smooth cone, which has a vertical axis.
The particle remains in contact with the cone and rotates as a conical
pendulum with angular velocity .
The angle of the cone at its vertex is 2

where   , and the string makes an
4
angle of  with the horizontal as shown in
the diagram. The forces acting on the
particle are the tension in the string T, the
normal reaction N and the gravitational
force mg.
Note: whenever a particle makes contact with a surface there will be
a normal force perpendicular to the surface.

56. a) Show, with the aid of a diagram, that the vertical component of
N is N sin 

57. a) Show, with the aid of a diagram, that the vertical component of
N is N sin 

58. a) Show, with the aid of a diagram, that the vertical component of
N is N sin 
T

59. a) Show, with the aid of a diagram, that the vertical component of
N is N sin 
T


60. a) Show, with the aid of a diagram, that the vertical component of
N is N sin 
T
N


61. a) Show, with the aid of a diagram, that the vertical component of
N is N sin 
T
N

mg

62. a) Show, with the aid of a diagram, that the vertical component of
N is N sin 
T
N
 

63. a) Show, with the aid of a diagram, that the vertical component of
N is N sin 
T
N
  

64. a) Show, with the aid of a diagram, that the vertical component of
N is N sin 

T 
N
  

65. a) Show, with the aid of a diagram, that the vertical component of
N is N sin 

T 
N
  
mg

66. a) Show, with the aid of a diagram, that the vertical component of
N is N sin 

T c 
N
  
mg

67. a) Show, with the aid of a diagram, that the vertical component of
N is N sin 
 c
T c  sin 
N 
N
c  N sin 
  
mg

68. a) Show, with the aid of a diagram, that the vertical component of
N is N sin 
 c
T c  sin 
N 
N
c  N sin 
 
 the vertical component 
mg
of N isN sin 

69. a) Show, with the aid of a diagram, that the vertical component of
N is N sin 
 c
T c  sin 
N 
N
c  N sin 
 
 the vertical component 
mg
of N isN sin 
mg
b) Show that T  N  , and find an expression for T  N in
sin 
terms of m, l and

70. a) Show, with the aid of a diagram, that the vertical component of
N is N sin 
 c
T c  sin 
N 
N
c  N sin 
 
 the vertical component 
mg
of N isN sin 
mg
b) Show that T  N  , and find an expression for T  N in
sin 
terms of m, l and
horizontal forces  mr 2

71. a) Show, with the aid of a diagram, that the vertical component of
N is N sin 
 c
T c  sin 
N 
N
c  N sin 
 
 the vertical component 
mg
of N isN sin 
mg
b) Show that T  N  , and find an expression for T  N in
sin 
terms of m, l and
horizontal forces  mr 2

72. a) Show, with the aid of a diagram, that the vertical component of
N is N sin 
 c
T c  sin 
N 
N
c  N sin 
 
 the vertical component 
mg
of N isN sin 
mg
b) Show that T  N  , and find an expression for T  N in
sin 
terms of m, l and
horizontal forces  mr 2
T cos  N cos 

73. a) Show, with the aid of a diagram, that the vertical component of
N is N sin 
 c
T c  sin 
N 
N
c  N sin 
 
 the vertical component 
mg
of N isN sin 
mg
b) Show that T  N  , and find an expression for T  N in
sin 
terms of m, l and
horizontal forces  mr 2
T cos  N cos 
T cos   N cos   mr 2

74. a) Show, with the aid of a diagram, that the vertical component of
N is N sin 
 c
T c  sin 
N 
N
c  N sin 
 
 the vertical component 
mg
of N isN sin 
mg
b) Show that T  N  , and find an expression for T  N in
sin 
terms of m, l and
horizontal forces  mr 2 vertical forces  0
T cos  N cos 
T cos   N cos   mr 2

75. a) Show, with the aid of a diagram, that the vertical component of
N is N sin 
 c
T c  sin 
N 
N
c  N sin 
 
 the vertical component 
mg
of N isN sin 
mg
b) Show that T  N  , and find an expression for T  N in
sin 
terms of m, l and
horizontal forces  mr 2 vertical forces  0
T cos  N cos 
T cos   N cos   mr 2

76. a) Show, with the aid of a diagram, that the vertical component of
N is N sin 
 c
T c  sin 
N 
N
c  N sin 
 
 the vertical component 
mg
of N isN sin 
mg
b) Show that T  N  , and find an expression for T  N in
sin 
terms of m, l and
horizontal forces  mr 2 vertical forces  0
T cos  N cos  T sin  N sin 
T cos   N cos   mr 2 mg

77. a) Show, with the aid of a diagram, that the vertical component of
N is N sin 
 c
T c  sin 
N 
N
c  N sin 
 
 the vertical component 
mg
of N isN sin 
mg
b) Show that T  N  , and find an expression for T  N in
sin 
terms of m, l and
horizontal forces  mr 2 vertical forces  0
T cos  N cos  T sin  N sin 
T sin   N sin   mg  0
T cos   N cos   mr 2 mg
T sin   N sin   mg

78. T sin   N sin   mg

79. T sin   N sin   mg
T  N  sin   mg
mg
TN 
sin 

80. T sin   N sin   mg T cos   N cos   mr 2
T  N  sin   mg
mg
TN 
sin 

81. T sin   N sin   mg T cos   N cos   mr 2
T  N  sin   mg T  N  cos   mr 2
mg mr 2
TN  T N 
sin  cos 

82. T sin   N sin   mg T cos   N cos   mr 2
T  N  sin   mg T  N  cos   mr 2
mg mr 2
TN  T N 
sin  cos 
r
But  cos 
l

83. T sin   N sin   mg T cos   N cos   mr 2
T  N  sin   mg T  N  cos   mr 2
mg mr 2
TN  T N 
sin  cos 
r
But  cos  T  N  ml 2
l

84. T sin   N sin   mg T cos   N cos   mr 2
T  N  sin   mg T  N  cos   mr 2
mg mr 2
TN  T N 
sin  cos 
r
But  cos  T  N  ml 2
l
c) The angular velocity is increased until N  0, that is, when the particle
is about to lose contact with the cone.
Find an expression for this value of  in terms of  , l and g

85. T sin   N sin   mg T cos   N cos   mr 2
T  N  sin   mg T  N  cos   mr 2
mg mr 2
TN  T N 
sin  cos 
r
But  cos  T  N  ml 2
l
c) The angular velocity is increased until N  0, that is, when the particle
is about to lose contact with the cone.
Find an expression for this value of  in terms of  , l and g
When N  0;

86. T sin   N sin   mg T cos   N cos   mr 2
T  N  sin   mg T  N  cos   mr 2
mg mr 2
TN  T N 
sin  cos 
r
But  cos  T  N  ml 2
l
c) The angular velocity is increased until N  0, that is, when the particle
is about to lose contact with the cone.
Find an expression for this value of  in terms of  , l and g
When N  0;
mg
T
sin 

87. T sin   N sin   mg T cos   N cos   mr 2
T  N  sin   mg T  N  cos   mr 2
mg mr 2
TN  T N 
sin  cos 
r
But  cos  T  N  ml 2
l
c) The angular velocity is increased until N  0, that is, when the particle
is about to lose contact with the cone.
Find an expression for this value of  in terms of  , l and g
When N  0;
mg
T and T  ml 2
sin 

88. T sin   N sin   mg T cos   N cos   mr 2
T  N  sin   mg T  N  cos   mr 2
mg mr 2
TN  T N 
sin  cos 
r
But  cos  T  N  ml 2
l
c) The angular velocity is increased until N  0, that is, when the particle
is about to lose contact with the cone.
Find an expression for this value of  in terms of  , l and g
When N  0; 
mg
 ml 2
sin 
mg
T and T  ml 2
sin 

89. T sin   N sin   mg T cos   N cos   mr 2
T  N  sin   mg T  N  cos   mr 2
mg mr 2
TN  T N 
sin  cos 
r
But  cos  T  N  ml 2
l
c) The angular velocity is increased until N  0, that is, when the particle
is about to lose contact with the cone.
Find an expression for this value of  in terms of  , l and g
When N  0; 
mg
 ml 2
sin 
mg g
T and T  ml 2
 2
sin  l sin 
g

l sin 

90. Exercise 9C; all