1. The Conical Pendulum

2. The Conical PendulumA
P


3. The Conical PendulumA

h
rO P


4. The Conical PendulumA Force Diagram

h
rO P


5. The Conical PendulumA Force Diagram

h
rO P


6. The Conical PendulumA Force Diagram

h
rO P

T
T= tension in the string
(always away from object)

7. The Conical PendulumA
mg
Force Diagram

h
rO P

T
T= tension in the string
(always away from object)

8. The Conical PendulumA
mg
Force Diagram

h
rO P

T
T= tension in the string
(always away from object)
icalwith vertmakesstring

9. The Conical PendulumA
mg
Force Diagram

h
rO P

T
T= tension in the string
(always away from object)
icalwith vertmakesstring
pendulumoflocityangular ve

10. The Conical PendulumA
mg
Force Diagram

h
rO P

T
T= tension in the string
(always away from object)
icalwith vertmakesstring
pendulumoflocityangular ve
Resultant Forces

11. The Conical PendulumA
r
mv
xm
2

mg
Force Diagram

h
rO P

T
T= tension in the string
(always away from object)
icalwith vertmakesstring
pendulumoflocityangular ve
Resultant Forces

12. The Conical PendulumA
r
mv
xm
2
 0ym 
mg
Force Diagram

h
rO P

T
T= tension in the string
(always away from object)
icalwith vertmakesstring
pendulumoflocityangular ve
Resultant Forces

13. The Conical PendulumA
r
mv
xm
2
 0ym 
mg
Force Diagram

h
rO P

T
T= tension in the string
(always away from object)
icalwith vertmakesstring
pendulumoflocityangular ve
Resultant Forces
r
mv2
forceshorizontal 

14. The Conical PendulumA
r
mv
xm
2
 0ym 
mg
Force Diagram

h
rO P

T
T= tension in the string
(always away from object)
icalwith vertmakesstring
pendulumoflocityangular ve
Resultant Forces
r
mv2
forceshorizontal  0forcesvertical 

15. The Conical PendulumA
r
mv
xm
2
 0ym 
mg
Force Diagram

h
rO P

T
T= tension in the string
(always away from object)
icalwith vertmakesstring
pendulumoflocityangular ve
Resultant Forces
r
mv2
forceshorizontal 

0forcesvertical 

16. The Conical PendulumA
r
mv
xm
2
 0ym 
mg
Force Diagram

h
rO P

T
T= tension in the string
(always away from object)
icalwith vertmakesstring
pendulumoflocityangular ve
Resultant Forces
r
mv2
forceshorizontal 

0forcesvertical 

17. The Conical PendulumA
r
mv
xm
2
 0ym 
mg
Force Diagram

h
rO P

T
T= tension in the string
(always away from object)
icalwith vertmakesstring
pendulumoflocityangular ve
Resultant Forces
r
mv2
forceshorizontal 

sinT
0forcesvertical 

18. The Conical PendulumA
r
mv
xm
2
 0ym 
mg
Force Diagram

h
rO P

T
T= tension in the string
(always away from object)
icalwith vertmakesstring
pendulumoflocityangular ve
Resultant Forces
r
mv2
forceshorizontal 

sinT
r
mv
T
2
sin 
0forcesvertical 

19. The Conical PendulumA
r
mv
xm
2
 0ym 
mg
Force Diagram

h
rO P

T
T= tension in the string
(always away from object)
icalwith vertmakesstring
pendulumoflocityangular ve
Resultant Forces
r
mv2
forceshorizontal 

sinT
r
mv
T
2
sin   2
mr
0forcesvertical 

20. The Conical PendulumA
r
mv
xm
2
 0ym 
mg
Force Diagram

h
rO P

T
T= tension in the string
(always away from object)
icalwith vertmakesstring
pendulumoflocityangular ve
Resultant Forces
r
mv2
forceshorizontal 

sinT
r
mv
T
2
sin   2
mr
0forcesvertical 

21. The Conical PendulumA
r
mv
xm
2
 0ym 
mg
Force Diagram

h
rO P

T
T= tension in the string
(always away from object)
icalwith vertmakesstring
pendulumoflocityangular ve
Resultant Forces
r
mv2
forceshorizontal 

sinT
r
mv
T
2
sin   2
mr
0forcesvertical 
mg
cosT

22. The Conical PendulumA
r
mv
xm
2
 0ym 
mg
Force Diagram

h
rO P

T
T= tension in the string
(always away from object)
icalwith vertmakesstring
pendulumoflocityangular ve
Resultant Forces
r
mv2
forceshorizontal 

sinT
r
mv
T
2
sin   2
mr
0forcesvertical 
mg
cosT
mgT
mgT




cos
0cos

23. mgr
mv
T
T 1
cos
sin 2




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

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

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

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

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

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

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

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

32. mgr
mv
T
T 1
cos
sin 2



rg
v2
tan  






g
r 2

h
r
AOP  taninBut
h
r
rg
v

2
2
2
v
gr
h  




 2

g
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.
mg
T

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.
mg
T
h
r

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.
mg
T
h
r
r
mv2
forceshorizontal 

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.
mg
T
h
r
r
mv2
forceshorizontal  0forcesvertical 

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.
mg
T
h
r
r
mv2
forceshorizontal  0forcesvertical 

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.
mg
T
h
r
r
mv2
forceshorizontal 
sinT
0forcesvertical 

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.
mg
T
h
r
r
mv2
forceshorizontal 
sinT
2
sin  mrT 
0forcesvertical 

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.
mg
T
h
r
r
mv2
forceshorizontal 
sinT
2
sin  mrT 
0forcesvertical 

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.
mg
T
h
r
r
mv2
forceshorizontal 
sinT
2
sin  mrT 
0forcesvertical 
mg
cosT

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.
mg
T
h
r
r
mv2
forceshorizontal 
sinT
2
sin  mrT 
0forcesvertical 
mg
cosT
mgT
mgT




cos
0cos

46. g
r
mg
mr
2
2 1
tan





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

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

49. g
r
mg
mr
2
2 1
tan




h
r
tanBut
2
2


g
h
h
r
g
r


rad/s2
rad/s
60
120
60rev/minwhen







50. g
r
mg
mr
2
2 1
tan




h
r
tanBut
2
2


g
h
h
r
g
r


rad/s2
rad/s
60
120
60rev/minwhen






 
m
4
2
2
2


g
g
h



51. g
r
mg
mr
2
2 1
tan




h
r
tanBut
2
2


g
h
h
r
g
r


rad/s2
rad/s
60
120
60rev/minwhen






 
m
4
2
2
2


g
g
h


rad/s3
rad/s
60
180
rev/min09when







52. g
r
mg
mr
2
2 1
tan




h
r
tanBut
2
2


g
h
h
r
g
r


rad/s2
rad/s
60
120
60rev/minwhen






 
m
4
2
2
2


g
g
h


rad/s3
rad/s
60
180
rev/min09when






 
m
9
3
2
2


g
g
h



53. g
r
mg
mr
2
2 1
tan




h
r
tanBut
2
2


g
h
h
r
g
r


rad/s2
rad/s
60
120
60rev/minwhen






 
m
4
2
2
2


g
g
h


rad/s3
rad/s
60
180
rev/min09when






 
m
9
3
2
2


g
g
h


0.14m
m
94
heightinrise 22




 

gg

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
where , and the string makes an
2
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
where , and the string makes an
2
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 sinN

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

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

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

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

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

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

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

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

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

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

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

68. a) Show, with the aid of a diagram, that the vertical component of
N is sinN
N
T
mg


c
 



sin
sin
Nc
N
c


sinisof
componentverticalthe
NN


69. a) Show, with the aid of a diagram, that the vertical component of
N is sinN
N
T
mg


c
 



sin
sin
Nc
N
c


sinisof
componentverticalthe
NN



and,ofterms
inforexpressionanfindand,
sin
thatShowb)
lm
NT
mg
NT 

70. a) Show, with the aid of a diagram, that the vertical component of
N is sinN
N
T
mg


c
 



sin
sin
Nc
N
c


sinisof
componentverticalthe
NN



and,ofterms
inforexpressionanfindand,
sin
thatShowb)
lm
NT
mg
NT 
2
forceshorizontal mr

71. a) Show, with the aid of a diagram, that the vertical component of
N is sinN
N
T
mg


c
 



sin
sin
Nc
N
c


sinisof
componentverticalthe
NN



and,ofterms
inforexpressionanfindand,
sin
thatShowb)
lm
NT
mg
NT 
2
forceshorizontal mr

72. a) Show, with the aid of a diagram, that the vertical component of
N is sinN
N
T
mg


c
 



sin
sin
Nc
N
c


sinisof
componentverticalthe
NN



and,ofterms
inforexpressionanfindand,
sin
thatShowb)
lm
NT
mg
NT 
2
forceshorizontal mr
cosT cosN

73. a) Show, with the aid of a diagram, that the vertical component of
N is sinN
N
T
mg


c
 



sin
sin
Nc
N
c


sinisof
componentverticalthe
NN



and,ofterms
inforexpressionanfindand,
sin
thatShowb)
lm
NT
mg
NT 
2
forceshorizontal mr
cosT
2
coscos  mrNT 
cosN

74. a) Show, with the aid of a diagram, that the vertical component of
N is sinN
N
T
mg


c
 



sin
sin
Nc
N
c


sinisof
componentverticalthe
NN



and,ofterms
inforexpressionanfindand,
sin
thatShowb)
lm
NT
mg
NT 
2
forceshorizontal mr
cosT
2
coscos  mrNT 
cosN
0forcesvertical 

75. a) Show, with the aid of a diagram, that the vertical component of
N is sinN
N
T
mg


c
 



sin
sin
Nc
N
c


sinisof
componentverticalthe
NN



and,ofterms
inforexpressionanfindand,
sin
thatShowb)
lm
NT
mg
NT 
2
forceshorizontal mr
cosT
2
coscos  mrNT 
cosN
0forcesvertical 

76. a) Show, with the aid of a diagram, that the vertical component of
N is sinN
N
T
mg


c
 



sin
sin
Nc
N
c


sinisof
componentverticalthe
NN



and,ofterms
inforexpressionanfindand,
sin
thatShowb)
lm
NT
mg
NT 
2
forceshorizontal mr
cosT
2
coscos  mrNT 
cosN
0forcesvertical 
mg
sinT sinN

77. a) Show, with the aid of a diagram, that the vertical component of
N is sinN
N
T
mg


c
 



sin
sin
Nc
N
c


sinisof
componentverticalthe
NN



and,ofterms
inforexpressionanfindand,
sin
thatShowb)
lm
NT
mg
NT 
2
forceshorizontal mr
cosT
2
coscos  mrNT 
cosN
0forcesvertical 
mg
sinT
mgNT
mgNT




sinsin
0sinsin
sinN

78. mgNT   sinsin

79. mgNT   sinsin
 


sin
sin
mg
NT
mgNT



80. mgNT   sinsin
 


sin
sin
mg
NT
mgNT


2
coscos  mrNT 

81. mgNT   sinsin
 


sin
sin
mg
NT
mgNT


2
coscos  mrNT 
 



cos
cos
2
2
mr
NT
mrNT



82. mgNT   sinsin
 


sin
sin
mg
NT
mgNT


2
coscos  mrNT 
 



cos
cos
2
2
mr
NT
mrNT


cosBut 
l
r

83. mgNT   sinsin
 


sin
sin
mg
NT
mgNT


2
coscos  mrNT 
 



cos
cos
2
2
mr
NT
mrNT


cosBut 
l
r 2
mlNT 

84. mgNT   sinsin
 


sin
sin
mg
NT
mgNT


2
coscos  mrNT 
 



cos
cos
2
2
mr
NT
mrNT


cosBut 
l
r 2
mlNT 
and,ofin termsofvaluefor thisexpressionanFind
cone.th thecontact wiloseabout tois
particlewhen theis,that0,untilincreasedislocityangular veThec)
gl
N



85. mgNT   sinsin
 


sin
sin
mg
NT
mgNT


2
coscos  mrNT 
 



cos
cos
2
2
mr
NT
mrNT


cosBut 
l
r 2
mlNT 
and,ofin termsofvaluefor thisexpressionanFind
cone.th thecontact wiloseabout tois
particlewhen theis,that0,untilincreasedislocityangular veThec)
gl
N


;0When N

86. mgNT   sinsin
 


sin
sin
mg
NT
mgNT


2
coscos  mrNT 
 



cos
cos
2
2
mr
NT
mrNT


cosBut 
l
r 2
mlNT 
and,ofin termsofvaluefor thisexpressionanFind
cone.th thecontact wiloseabout tois
particlewhen theis,that0,untilincreasedislocityangular veThec)
gl
N


;0When N
sin
mg
T 

87. mgNT   sinsin
 


sin
sin
mg
NT
mgNT


2
coscos  mrNT 
 



cos
cos
2
2
mr
NT
mrNT


cosBut 
l
r 2
mlNT 
and,ofin termsofvaluefor thisexpressionanFind
cone.th thecontact wiloseabout tois
particlewhen theis,that0,untilincreasedislocityangular veThec)
gl
N


;0When N
sin
mg
T  2
and mlT 

88. mgNT   sinsin
 


sin
sin
mg
NT
mgNT


2
coscos  mrNT 
 



cos
cos
2
2
mr
NT
mrNT


cosBut 
l
r 2
mlNT 
and,ofin termsofvaluefor thisexpressionanFind
cone.th thecontact wiloseabout tois
particlewhen theis,that0,untilincreasedislocityangular veThec)
gl
N


;0When N
sin
mg
T  2
and mlT 
2
sin


ml
mg


89. mgNT   sinsin
 


sin
sin
mg
NT
mgNT


2
coscos  mrNT 
 



cos
cos
2
2
mr
NT
mrNT


cosBut 
l
r 2
mlNT 
and,ofin termsofvaluefor thisexpressionanFind
cone.th thecontact wiloseabout tois
particlewhen theis,that0,untilincreasedislocityangular veThec)
gl
N


;0When N
sin
mg
T  2
and mlT 
2
sin


ml
mg





sin
sin
2
l
g
l
g



90. Exercise 9C; all