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
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
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
2rad/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
2rad/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
2rad/s 4 2
r
But tan
h when 90rev/min
r 2 r
180
g h rad/s
60
h 2
g 3rad/s
52. 1
tan mr 2 when 60rev/min h
g
mg
120
2 2
r 2 rad/s g
60 m
g
2rad/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 3rad/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
2rad/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 3rad/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
79. T sin N sin mg
T N sin mg
mg
TN
sin
80. T sin N sin mg T cos N cos mr 2
T N sin mg
mg
TN
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
TN 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
TN 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
TN 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
TN 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
TN 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
TN 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
TN 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
TN 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
TN 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
