The Conical Pendulum

The Conical Pendulum
A

P

The Conical Pendulum
A

h

O r P

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

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

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

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

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

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

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

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

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

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

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

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

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

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 

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

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

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

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

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

T sin  mv 2 1
 
T cos r mg

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

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

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

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

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

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 

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

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.

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.

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.

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.

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

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

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

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

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

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

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 

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

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

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

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

1
 tan   mr 2 
mg
r 2

g

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

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

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


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


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


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


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

(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.

(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.

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

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

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

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

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

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

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

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

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

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

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

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

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 

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

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

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

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 

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

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

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

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

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

T sin   N sin   mg

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

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

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 

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

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

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

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;

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 

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 

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 

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 

Exercise 9C; all