X2 t06 05 conical pendulum (2012)

1. The Conical Pendulum

2. The Conical Pendulum
A
P

3. The Conical Pendulum
A

h
r
O
P

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

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

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

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

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

9. The Conical Pendulum
A
m
g
Force Diagram

h
r
O
P
T
T= tension in the string
(always away from object)
ical with vert makes string



p
endulu
m
oflocity angular ve 


10. The Conical Pendulum
A
m
g
Force Diagram

h
r
O
P
T
T= tension in the string
(always away from object)
ical with vert makes string



p
endulu
m
oflocity angular ve 

Resultant Forces

11. The Conical Pendulum
A
r
mv
xm
2

m
g
Force Diagram

h
r
O
P
T
T= tension in the string
(always away from object)
ical with vert makes string



p
endulu
m
oflocity angular ve 

Resultant Forces

12. The Conical Pendulum
A
r
mv
xm
2

0

ym
m
g
Force Diagram

h
r
O
P
T
T= tension in the string
(always away from object)
ical with vert makes string



p
endulu
m
oflocity angular ve 

Resultant Forces

13. The Conical Pendulum
A
r
mv
xm
2

0

ym
m
g
Force Diagram

h
r
O
P
T
T= tension in the string
(always away from object)
ical with vert makes string



p
endulu
m
oflocity angular ve 

Resultant Forces
r
mv
2
forces horizontal 

14. The Conical Pendulum
A
r
mv
xm
2

0

ym
m
g
Force Diagram

h
r
O
P
T
T= tension in the string
(always away from object)
ical with vert makes string



p
endulu
m
oflocity angular ve 

Resultant Forces
r
mv
2
forces horizontal 
0forces vertical


15. The Conical Pendulum
A
r
mv
xm
2


0

ym
m
g
Force Diagram

h
r
O
P
T
T
= tension in the string
(always away from object)
ical with ver
t
makes string



p
endulu
m
o
f
locity angular ve 

Resultant Forces
r
mv
2
forces horizontal 

0forces vertical


16. The Conical Pendulum
A
r
mv
xm
2


0

ym
m
g
Force Diagram

h
r
O
P
T
T
= tension in the string
(always away from object)
ical with ver
t
makes string



p
endulu
m
o
f
locity angular ve 

Resultant Forces
r
mv
2
forces horizontal 

0forces vertical


17. The Conical Pendulum
A
r
mv
xm
2


0

ym
m
g
Force Diagram

h
r
O
P
T
T
= tension in the string
(always away from object)
ical with ver
t
makes string



p
endulu
m
o
f
locity angular ve 

Resultant Forces
r
mv
2
forces horizontal 


sin
T
0forces vertical


18. The Conical Pendulum
A
r
mv
xm
2


0

ym
m
g
Force Diagram

h
r O
P
T
T= tension in the string
(always away from object)
ical with ver
t
makes string



p
endulu
m
o
f
locity angular ve 

Resultant Forces
r
mv
2
forces horizontal 


sin
T
r
mv
T
2
sin

0forces vertical


19. The Conical Pendulum
A
r
mv
xm
2


0

ym
m
g
Force Diagram

h
r O
P
T
T= tension in the string
(always away from object)
ical with ver
t
makes string



p
endulu
m
o
f
locity angular ve 

Resultant Forces
r
mv
2
forces horizontal 


sin
T
r
mv
T
2
sin



2

mr

0forces vertical


20. The Conical Pendulum
A
r
mv
xm
2


0

ym
m
g
Force Diagram

h
r O
P
T
T
= tension in the string
(always away from object)
ical with ver
t
makes string



p
endulu
m
o
f
locity angular ve


Resultant Forces
r
mv
2
forces horizontal



sin
T
r
mv
T
2
sin




2

mr

0forces vertical


21. The Conical Pendulum
A
r
mv
xm
2


0

ym
m
g
Force Diagram

h
r O
P
T
T
= tension in the string
(always away from object)
ical with ver
t
makes string



p
endulu
m
o
f
locity angular ve


Resultant Forces
r
mv
2
forces horizontal



sin
T
r
mv
T
2
sin




2

mr

0forces vertical

m
g

cos
T

22. The Conical Pendulum
A
r
mv
xm
2


0

ym
m
g
Force Diagram

h
r O
P
T
T
= tension in the string
(always away from object)
ical with ver
t
makes string



p
endulu
m
o
f
locity angular ve


Resultant Forces
r
mv
2
forces horizontal



sin
T
r
mv
T
2
sin




2

mr

0forces vertical

m
g

cos
T
mg T
m
g
T




cos
0 cos

23. mgr
mv
T
T1
cos
sin
2




24. mgr
mv
T
T1
cos
sin
2



rg
v
2
tan


25. mgr
mv
T
T1
cos
sin
2



rg
v
2
tan






g
r
2


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

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

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

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

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

31. mgr
mv
T
T
1
cos
sin
2



rg
v
2
tan







g
r
2

h
r
AOP 

tan in But
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
v
2
tan







g
r
2

h
r
AOP 

tan in But
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 pe r 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 pe r 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 pe r 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 pe r minute of a conical pendulum
increases from 60 to 90.
Find the rise in the level of the bob.
m
g
T

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

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

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

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

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

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

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

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

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

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/min when





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/min when





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/min when





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/min when





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/min when





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
heightin rise
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

sin
N

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

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

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

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

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

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

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

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

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

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

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

68. a) Show, with the aid of a diagram, that the vertical component of
N is

sin
N
N
T
m
g


c





sin
sin
N
c
N
c


sinisof
component verticalthe
N
N


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

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

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

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

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

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

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

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

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

78. m
g
N
T




sinsin

79. m
g
N
T




sinsin




sin
sin
mg
NT
m
g
N
T



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

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

82. m
g
N
T




sinsin




sin
sin
mg
NT
m
g
N
T


2
cos cos



mr
N
T






cos
cos
2
2
mr
NT
mr NT



cosBut 
l
r

83. m
g
N
T




sinsin




sin
sin
mg
NT
m
g
N
T


2
cos cos



mr
N
T






cos
cos
2
2
mr
NT
mr NT



cosBut 
l
r
2

ml
N
T




84. m
g
N
T




sinsin




sin
sin
mg
NT
m
g
N
T


2
cos cos



mr
N
T






cos
cos
2
2
mr
NT
mr NT



cosBut 
l
r
2

ml
N
T



and , of in terms of value for this expression an Find
cone. th the contact wi lose about to is

p
article when theis,that 0,until increasedislocity angular ve The c)
gl
N
 


85. m
g
N
T




sinsin




sin
sin
mg
NT
m
g
N
T


2
cos cos



mr
N
T






cos
cos
2
2
mr
NT
mr NT



cosBut 
l
r
2

ml
N
T



and , of in terms of value for this expression an Find
cone. th the contact wi lose about to is

p
article when theis,that 0,until increasedislocity angular ve The c)
gl
N
 

;0When 
N

86. m
g
N
T




sinsin




sin
sin
mg
NT
mg
N
T


2
cos cos



mr
N
T






cos
cos
2
2
mr
NT
mr NT



cosBut 
l
r
2

ml
N
T



and , of in terms of value for this expression an Find
cone. th the contact wi lose about to is

p
article when theis,that 0,until increasedislocity angular ve The c)
gl
N
 

;0When 
N

sin
m
g
T

87. m
g
N
T




sinsin




sin
sin
mg
NT
mg
N
T


2
cos cos



mr
N
T






cos
cos
2
2
mr
NT
mr NT



cosBut 
l
r
2

ml
N
T



and , of in terms of value for this expression an Find
cone. th the contact wi lose about to is

p
article when theis,that 0,until increasedislocity angular ve The c)
gl
N
 

;0When 
N

sin
m
g
T
2
an
d

ml
T


88. m
g
N
T




sinsin




sin
sin
mg
NT
mg
N
T


2
cos cos



mr
N
T






cos
cos
2
2
mr
NT
mr NT



cosBut 
l
r
2

ml
N
T



and , of in terms of value for this expression an Find
cone. th the contact wi lose about to is

p
article when theis,that 0,until increasedislocity angular ve The c)
gl
N
 

;0When 
N

sin
m
g
T
2
an
d

ml
T

2
sin


ml
m
g



89. m
g
N
T




sinsin




sin
sin
mg
NT
mg
N
T


2
cos cos



mr
N
T






cos
cos
2
2
mr
NT
mr NT



cosBut 
l
r
2

ml
N
T



and , of in terms of value for this expression an Find
cone. th the contact wi lose about to is

p
article when theis,that 0,until increasedislocity angular ve The c)
gl
N
 

;0When 
N

sin
m
g
T
2
an
d

ml
T

2
sin


ml
m
g






sin
sin
2
l
g
l
g



90. Exercise 9C; all