Circular Motion Resisted Motion

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Title: Circular Motion Resisted Motion

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Title: Circular Motion Resisted Motion

1
No there is no hand-out. You get the powerpoint
presentation for free. Enough trees have died
already. Details at the end of the talk.
2
Mechanics

Morris Needleman

Part 1 Circular Motion
3
What do you have to do

Watch out for Buffy.

4
What do you have to do

Watch out for Buffy.
When the music starts you should be thinking!

5
What do you have to do

Watch out for Buffy.
When the music starts you should be thinking!

6
Do you understand
7

please turn off your mobile phone

8
Circular Motion in a horizontal plane

P moves around a circle of radius r.

9
Circular Motion in a horizontal plane

P moves around a circle of radius r.
As P moves both the arc length PT change and the
angle q changes

10
Circular Motion in a horizontal plane

P moves around a circle of radius r.
As P moves both the arc length PT change and the
angle q changes
The angular velocity of P is given by

11
Circular Motion in a horizontal plane

P moves around a circle of radius r.
As P moves both the arc length PT change and the
angle q changes
The angular velocity of P is given by
Force mass acceleration

12
Circular Motion in a horizontal plane

P moves around a circle of radius r.
As P moves both the arc length PT change and the
angle q changes
The angular velocity of P is given by
Force mass acceleration

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This equation is important since it links angular
and linear velocity

This equation is important since it links angular
and linear velocity

To discuss acceleration we should consider the
motion in terms of horizontal and vertical
components.

To discuss acceleration we should consider the
motion in terms of horizontal and vertical
components.

P( xy)
r
q
T
21

To discuss acceleration we should consider the
motion in terms of horizontal and vertical
components.

P( xy)
r
q
T
22

To discuss acceleration we should consider the
motion in terms of horizontal and vertical
components.

P( xy)
r
q
T
23

To discuss acceleration we should consider the
motion in terms of horizontal and vertical
components.

P( xy)
r
q
T
24

To discuss acceleration we should consider the
motion in terms of horizontal and vertical
components.

P( xy)
r
q
T
25
P( xy)

To simplify life we are going to consider that
the angular velocity remains constant throughout
the motion.

r
q
T
26
P( xy)
r

To simplify life we are going to consider that
the angular velocity remains constant throughout
the motion.
This will be the case in any problem you do but
you should be able to prove these results for
variable angular velocity.

q
T
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It is important to note that the vectors
demonstrate that the force is directed along the
radius towards the centre of the circle.
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Summary
(angular velocity)
(links angular velocity and linear velocity)
force is directed along the radius towards the
centre of the circle.
(horizontal and vertical components of
acceleration)
(gives the size of the force towards the centre)
40
Summary
(angular velocity)
(links angular velocity and linear velocity)
force is directed along the radius towards the
centre of the circle.
(horizontal and vertical components of
acceleration)
(gives the size of the force towards the centre)
41

A body of mass 2kg is revolving at the end of a
light string 3m long on a smooth horizontal
table with uniform angular speed of 1 revolution
per second.

A body of mass 2kg is revolving at the end of a
light string 3m long on a smooth horizontal
table with uniform angular speed of 1 revolution
per second.

Draw a neat diagram to represent the forces.
43

A body of mass 2kg is revolving at the end of a
light string 3m long on a smooth horizontal
table with uniform angular speed of 1 revolution
per second.

(a) Find the tension in the string.
44

A body of mass 2kg is revolving at the end of a
light string 3m long on a smooth horizontal
table with uniform angular speed of 1 revolution
per second.

(a) Find the tension in the string.
N
T
P
mg
45

A body of mass 2kg is revolving at the end of a
light string 3m long on a smooth horizontal
table with uniform angular speed of 1 revolution
per second.

(a) Find the tension in the string.
N
T
P
mg
The tension in the string is the resultant of the
forces acting on the body.
46

A body of mass 2kg is revolving at the end of a
light string 3m long on a smooth horizontal
table with uniform angular speed of 1 revolution
per second.

(a) Find the tension in the string.
N
T
P
mg
The tension in the string is the resultant of the
forces acting on the body.
47

A body of mass 2kg is revolving at the end of a
light string 3m long on a smooth horizontal
table with uniform angular speed of 1 revolution
per second.

(a) Find the tension in the string.
N
T
P
mg
The tension in the string is the resultant of the
forces acting on the body.
48

A body of mass 2kg is revolving at the end of a
light string 3m long on a smooth horizontal
table with uniform angular speed of 1 revolution
per second.

(a) Find the tension in the string.
N
T
P
mg
The tension in the string is the resultant of the
forces acting on the body.
49

A body of mass 2kg is revolving at the end of a
light string 3m long on a smooth horizontal
table with uniform angular speed of 1 revolution
per second.

(a) Find the tension in the string.
N
T
P
mg
The tension in the string is the resultant of the
forces acting on the body.
50
(b) If the string would break under a tension of
equal to the weight of 20 kg find the greatest
possible speed of the mass.
51
(b) If the string would break under a tension of
equal to the weight of 20 kg find the greatest
possible speed of the mass.
52
(b) If the string would break under a tension of
equal to the weight of 20 kg find the greatest
possible speed of the mass.
53
(b) If the string would break under a tension of
equal to the weight of 20 kg find the greatest
possible speed of the mass.
54
(b) If the string would break under a tension of
equal to the weight of 20 kg find the greatest
possible speed of the mass.
55
(b) If the string would break under a tension of
equal to the weight of 20 kg find the greatest
possible speed of the mass.
56
An interesting problem solving method
57
Conical Pendulum

If a particle is tied by a string to a fixed
point by means of a string and moves in a
horizontal circle so that the string describes a
cone and the mass at the end of the string
describes a horizontal circle then the string
and the mass describe a conical pendulum.

58
Conical Pendulum

If a particle is tied by a string to a fixed
point by means of a string and moves in a
horizontal circle so that the string describes a
cone and the mass at the end of the string
describes a horizontal circle then the string
and the mass describe a conical pendulum.

59
Conical Pendulum
60
Conical Pendulum
Dimensions Diagram
Forces Diagram
Vertically
61
Conical Pendulum
q
Dimensions Diagram
Forces Diagram
mg
Vertically
62
Conical Pendulum
q
N
Dimensions Diagram
Forces Diagram
mg
Vertically
63
Conical Pendulum
q
T
N
q
mg
Dimensions Diagram
Forces Diagram
Vertically
N (-mg) 0 N mg but cos q
N/T so N T cos q T cos q mg
64
Conical Pendulum
q
T
N
q
mg
Dimensions Diagram
Forces Diagram
Vertically
Horizontally
T sin q mrw2 Since the only horizontal force
is directed along the radius towards the
centre
N (-mg) 0 N mg but cos q
N/T so N T cos q T cos q mg
65
Conical Pendulum
q
T
N
q
mg
Dimensions Diagram
Forces Diagram
Vertically
Horizontally
T sin q mrw2 T cos q mg
T sin q mrw2 Since the only horizontal force
is directed along the radius towards the
centre
N (-mg) 0 N mg but cos q
N/T so N T cos q T cos q mg
66
Conical Pendulum
q
T
N
q
mg
Dimensions Diagram
Forces Diagram
Vertically
Horizontally
T sin q mrw2 T cos q mg
T sin q mrw2 Since the only horizontal force
is directed along the radius towards the
centre
N (-mg) 0 N mg but cos q
N/T so N T cos q T cos q mg
67
An important result
h
T sin q mrw21 T cos q mg 2 v
rw
68
An important result
h
T sin q mrw21 T cos q mg 2 v
rw
69
An important result
h
T sin q mrw21 T cos q mg 2 v
rw
70
An important result
h
T sin q mrw21 T cos q mg 2 v
rw
71
An important result
h
T sin q mrw21 T cos q mg 2 v
rw
72
An important result
h
T sin q mrw21 T cos q mg 2 v
rw
73

Example 2
A string of length 2 m fixed at one end A
carries at the other end a particle of mass 6 kg
rotating in a horizontal circle whose centre is
1m vertically below A. Find the tension in the
string and the angular velocity of the particle.

Example 2
A string of length 2 m fixed at one end A
carries at the other end a particle of mass 6 kg
rotating in a horizontal circle whose centre is
1m vertically below A. Find the tension in the
string and the angular velocity of the particle.

Example 2
A string of length 2 m fixed at one end A
carries at the other end a particle of mass 6 kg
rotating in a horizontal circle whose centre is
1m vertically below A. Find the tension in the
string and the angular velocity of the particle.

q
q
L 2
T
N
h 1
q
r
mg
Dimensions Diagram
Forces Diagram
76

Example 2
A string of length 2 m fixed at one end A
carries at the other end a particle of mass 6 kg
rotating in a horizontal circle whose centre is
1m vertically below A. Find the tension in the
string and the angular velocity of the particle.

q
q
L 2
T
T cos q
h 1
q
r
mg
Dimensions Diagram
Forces Diagram
77

Example 2
Find the tension in the string and the angular
velocity of the particle.

Vertically
Horizontally
78

Example 2
Find the tension in the string and the angular
velocity of the particle.

Vertically
Horizontally
79

Example 2
Find the tension in the string and the angular
velocity of the particle.

Vertically
Horizontally
80

Motion on a Banked Track

Dimensions diagram
P
81

Motion on a Banked Track

Dimensions diagram
Forces diagram
N
F
q
P
P
centre of the circle
mg
82

Motion on a Banked Track

Dimensions diagram
Forces diagram
N
F
q
P
P
centre of the circle
mg
Vertical Forces
Horizontal Forces
83

Motion on a Banked Track

Dimensions diagram
Forces diagram
N
F
q
P
P
centre of the circle
mg
Vertical Forces
Horizontal Forces
If there is no tendency to slip then F 0 and
the equations are
84

Motion on a Banked Track

Dimensions diagram
Forces diagram
N
F
q
P
P
centre of the circle
mg
Vertical Forces
Horizontal Forces
If there is no tendency to slip then F 0 and
the equations are
85
If there is no tendency to slip at v v0 then F
0 and the equations are
86
If there is no tendency to slip at v v0 then F
0 and the equations are
87
If there is no tendency to slip at v v0 then F
0 and the equations are
This is the method used by engineers to measure
the camber of a road.
88
2004 HSC question 6(c)
A smooth sphere with centre O and radius R is
rotating about the vertical diameter at a uniform
angular velocity w radians per second. A marble
is free to roll around the inside of the sphere.
Assume that the can be considered as a point P
which is acted upon by gravity and the normal
reaction force N from the sphere. The marble
describes a horizontal circle of radius r with
the same uniform angular velocity w radians per
second. Let the angle between OP and the
vertical diameter be q.
89
2004 HSC question 6(c)
A smooth sphere with centre O and radius R is
about the vertical diameter at a uniform angular
velocity w radians per second. A marble is free
to roll around the inside of the sphere.
Assume that the marble can be considered as a
point P which is acted upon by gravity and the
normal reaction force N from the sphere. The
marble describes a horizontal circle of radius r
with the same uniform angular velocity w radians
per second. Let the angle between OP and the
vertical diameter be q.
90
2004 HSC question 6(c)
Assume that the marble can be considered as a
point P which is acted upon by gravity and the
normal reaction force N from the sphere. The
marble describes a horizontal circle of radius r
with the same uniform angular velocity w radians
per second. Let the angle between OP and the
vertical diameter be q. (i) Explain why

q
q
N
q
P
91
2004 HSC question 6(c)
Assume that the marble can be considered as a
point P which is acted upon by gravity and the
normal reaction force N from the sphere. The
marble describes a horizontal circle of radius r
with the same uniform angular velocity w radians
per second. Let the angle between OP and the
vertical diameter be q. (i) Explain why

q
Net vertical force is 0
q
N
q
P
92
2004 HSC question 6(c)
Assume that the marble can be considered as a
point P which is acted upon by gravity and the
normal reaction force N from the sphere. The
marble describes a horizontal circle of radius r
with the same uniform angular velocity w radians
per second. Let the angle between OP and the
vertical diameter be q. (i) Explain why

q
Net vertical force is 0
q
N
Net horizontal force force
q
P
93
2004 HSC question 6(c)
Assume that the marble can be considered as a
point P which is acted upon by gravity and the
normal reaction force N from the sphere. The
marble describes a horizontal circle of radius r
with the same uniform angular velocity w radians
per second. Let the angle between OP and the
vertical diameter be q. (i) Explain why

q
Net vertical force is 0
q
N
Net horizontal force force
q
P
94
2004 HSC question 6(c)
(ii) Show that either q 0 or
O
q
R
r
P
q
N
q
P
95
2004 HSC question 6(c)
(ii) Show that either q 0 or
O
q
R
r
P
q
N
q
P
96
2004 HSC question 6(c)
(ii) Show that either q 0 or
O
q
R
Now either r 0 and the marble is stationary or
r 0 and.
r
P
q
N
q
P
97
2004 HSC question 6(c)
(ii) Show that either q 0 or
O
q
R
Now either r 0 and the marble is stationary or
r 0 and.
r
P
q
N
q
P
98
2004 HSC question 6(c)
(ii) Show that either q 0 or
O
q
R
Now either r 0 and the marble is stationary or
r 0 and.
r
P
q
N
q
P
99
2004 HSC question 6(c)
(ii) Show that either q 0 or
O
q
R
Now either r 0 and the marble is stationary or
r 0 and.
r
P
q
N
q
P
100
Resisted Motion
101
More Problem Solving
102
From 3 Unit An important proof
103
From 3 Unit An important proof
104
From 3 Unit An important proof
105
From 3 Unit An important proof
106
From 3 Unit An important proof
107
From 3 Unit An important proof
108
There are 2 important results for resisted motion
here
109
There are 2 important results for resisted motion
here
110
Three types of resisted motion

1. Along a straight line

111
Three types of resisted motion

1. Along a straight line
2. Going up

112
Three types of resisted motion

1. Along a straight line
2. Going up
3. Coming down

113
Type 1 - along a horizontal line

resistance

-mkv (say)
When you move on a surface you need not include
gravity in your equation . Resistance always acts
against you so make it negative.
114
Type 1 - along a horizontal line

resistance

-mkv (say)
When you move on a surface you need not include
gravity in your equation . Resistance always acts
against you so make it negative.
115
Type 1 - along a horizontal line

resistance

-mkv (say)
When you move on a surface you need not include
gravity in your equation . Resistance always acts
against you so make it negative.
116
Type 2 - going up

gravity resistance

-mg -mkv (say)
When you are going up gravity acts against you -
so make it negative. Resistance always acts
against you so make it negative as well.
117
Type 2 - going up

gravity resistance

-mg -mkv (say)
When you are going up gravity acts against you -
so make it negative. Resistance always acts
against you so make it negative as well.
118
Type 2 - going up

gravity resistance

-mg -mkv (say)
When you are going up gravity acts against you -
so make it negative. Resistance always acts
against you so make it negative as well.
119
Type 3 - going down

gravity resistance

mg -mkv (say)
When you are going down gravity acts with you -
so make it positive. Resistance always acts
against you so make it negative as well.
120
How do you approach a problem

Draw a forces diagram

121
How do you approach a problem

Draw a forces diagram
Understand that force mass acceleration

122
How do you approach the problems

Draw a forces diagram
Understand that force mass acceleration
Write down an initial equation

123
Along a horizontal line - what the syllabus
says.

Derive from Newtons Laws of motion the equation
of motion of a particle moving in a single
direction under a resistance proportional to a
power of the speed

124
Along a horizontal line - what the syllabus
says.

Derive from Newtons Laws of motion the equation
of motion of a particle moving in a single
direction under a resistance proportional to a
power of the speed

Derive expressions for velocity as functions of
time and position where possible

125
Along a horizontal line - what the syllabus
says.

Derive from Newtons Laws of motion the equation
of motion of a particle moving in a single
direction under a resistance proportional to a
power of the speed

Derive expressions for velocity as functions of
time and displacement where possible
Derive an expression for displacement as a
function of time

126
HSC 1987 - straight line

A particle unit mass moves in a straight line
against a resistance numerically equal to v v3
where v is the velocity. Initially the particle
is at the origin and is travelling with velocity
Q where Q gt 0.
(a) Show that v is related to the displacement x
by the formula

127
HSC 1987 - straight line

A particle unit mass moves in a straight line
against a resistance numerically equal to v v3
where v is the velocity. Initially the particle
is at the origin and is travelling with velocity
Q where Q gt 0.
(a) Show that v is related to the displacement x
by the formula

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Motion upwards - what the syllabus says.

Derive from Newtons Laws of motion the equation
of motion of a particle moving vertically upwards
in a medium with resistance proportional to the
first or second power of the speed

141
Motion upwards - what the syllabus says.

Derive from Newtons Laws of motion the equation
of motion of a particle moving vertically upwards
in a medium with resistance proportional to the
first or second power of the speed

Derive expressions for velocity as functions of
time and displacement where possible
142
Motion upwards - what the syllabus says.

Derive from Newtons Laws of motion the equation
of motion of a particle moving vertically upwards
in a medium with resistance proportional to the
first or second power of the speed

Derive expressions for velocity as functions of
time and displacement where possible Solve
problems by using expressions derived for acc
vel and displacement.
143
Motion upwards - a problem...

A particle of unit mass is thrown vertically
upwards with velocity of U into the air and
encounters a resistance of kv2. Find the greatest
height H achieved by the particle and the
corresponding time.

144
Motion upwards - a problem...

A particle of unit mass is thrown vertically
upwards with velocity of U into the air and
encounters a resistance of kv2. Find the greatest
height H achieved by the particle and the
corresponding time.

145
Forces diagram

t 0 v U
146
Forces diagram

gravity

-g
When you are going up gravity acts against you -
so make it negative.
t 0 v U
147
Forces diagram

gravity resistance

-g -kv2
When you are going up gravity acts against you -
so make it negative. Resistance always acts
against you so make it negative as well.
t 0 v U
148
The equation of motion is given by...
149
The equation of motion is given by...
150
The equation of motion is given by...
151
The equation of motion is given by...
152
The equation of motion is given by...
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The equation of motion is given by...
154
The equation of motion is given by...
155
The equation of motion is given by...
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Motion downwards - what the syllabus says.

Derive from Newtons Laws of motion the equation
of motion of a particle falling in a medium with
resistance proportional to the first or second
power of the speed

162
Motion downwards - what the syllabus says.

Derive from Newtons Laws of motion the equation
of motion of a particle falling in a medium with
resistance proportional to the first or second
power of the speed
find terminal velocity

163
Motion downwards - what the syllabus says.

Derive from Newtons Laws of motion the equation
of motion of a particle falling in a medium with
resistance proportional to the first or second
power of the speed
find terminal velocity

Derive expressions for velocity as functions of
time and displacement where possible Solve
problems by using expressions derived for acc
vel and displacement.
164
Motion downwards - a problem...

A particle of unit mass falls vertically from
rest in a medium and encounters a resistance of
kv. Find the velocity in terms of time and use
two different methods to find the terminal
velocity.

165
Motion downwards - a problem...

A particle of unit mass falls vertically from
rest in a medium and encounters a resistance of
kv. Find the velocity in terms of time and use
two different methods to find the terminal
velocity.

166
Forces diagram
t 0 v 0

gravity resistance

g -kv
When you are going down gravity acts with you -
so make it positive. Resistance always acts
against you so make it negative.
167
Forces diagram
t 0 v 0

gravity resistance

g -kv
When you are going down gravity acts with you -
so make it positive. Resistance always acts
against you so make it negative.
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Now we can find the terminal velocity two
ways 1. Consider what happens to v as t
.
177
Now we can find the terminal velocity two
ways 1. Consider what happens to v as t
. 2. Or alternatively we can just let the
acceleration equal zero
178
Now we can find the terminal velocity two
ways 1. Consider what happens to v as t
. 2. Or alternatively we can just let the
acceleration equal zero
179
What to do when you are stuck.