Circular Motion Resisted Motion

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MANSW 2009 MOTION Morris Needleman Reddam House
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The Past
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Gifted and Talented? Yes - They Do Extension 2
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Of course there is gifted talented and then there
is TERRY TAO
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How do we attract students to Extension 2?
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How do we attract students to Extension 2?
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Mechanics

• Morris Needleman

Part 1 Circular Motion
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What do you have to do ?

• Watch out for Buffy.

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What do you have to do ?

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

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What do you have to do ?

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

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Do you understand?
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• please turn off your mobile phone

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Circular Motion in a horizontal plane

• P moves around a circle of radius r.

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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

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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

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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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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

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• This equation is important since it links angular
  and linear velocity

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

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

P( x,y)
r
q
T
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• To discuss acceleration we should consider the
  motion in terms of horizontal and vertical
  components.

P( x,y)
r
q
T
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• To discuss acceleration we should consider the
  motion in terms of horizontal and vertical
  components.

P( x,y)
r
q
T
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• To discuss acceleration we should consider the
  motion in terms of horizontal and vertical
  components.

P( x,y)
r
q
T
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• To discuss acceleration we should consider the
  motion in terms of horizontal and vertical
  components.

P( x,y)
r
q
T
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P( x,y)

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

r
q
T
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P( x,y)
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)
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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)
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• 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.

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• 1. 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.
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• 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.
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• 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
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• 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.
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• 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.
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• 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.
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• 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.
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• 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.
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(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.
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(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.
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(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.
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(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.
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(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.
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(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.
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An interesting problem solving method
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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.

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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.

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Conical Pendulum
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Conical Pendulum
Dimensions Diagram
Forces Diagram
Vertically
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Conical Pendulum
q
Dimensions Diagram
Forces Diagram
mg
Vertically
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Conical Pendulum
q
N
Dimensions Diagram
Forces Diagram
mg
Vertically
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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
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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
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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
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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
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An important result
h
T sin q mrw21 T cos q mg 2 v
rw
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An important result
h
T sin q mrw21 T cos q mg 2 v
rw
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An important result
h
T sin q mrw21 T cos q mg 2 v
rw
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An important result
h
T sin q mrw21 T cos q mg 2 v
rw
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An important result
h
T sin q mrw21 T cos q mg 2 v
rw
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An important result
h
T sin q mrw21 T cos q mg 2 v
rw
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• 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.

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• 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.

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

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

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

Vertically
Horizontally
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• Motion on a Banked Track

Dimensions diagram
P
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• Motion on a Banked Track

Dimensions diagram
Forces diagram
N
F
q
P
P
centre of the circle
mg
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• Motion on a Banked Track

Dimensions diagram
Forces diagram
N
F
q
P
P
centre of the circle
mg
Vertical Forces
Horizontal Forces
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• 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
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• 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
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If there is no tendency to slip at v v0 then F
0 and the equations are
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If there is no tendency to slip at v v0 then F
0 and the equations are
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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.
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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.
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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.
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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
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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
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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
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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
q
P
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2004 HSC question 6(c)
(ii) Show that either q 0 or
O
q
R
r
P
q
N
q
P
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2004 HSC question 6(c)
(ii) Show that either q 0 or
O
q
R
r
P
q
N
q
P
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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
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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
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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
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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
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The present
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Resisted Motion
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More Problem Solving
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More Problem Solving
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From 3 Unit An important proof
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From 3 Unit An important proof
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From 3 Unit An important proof
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From 3 Unit An important proof
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From 3 Unit An important proof
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From 3 Unit An important proof
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There are 2 important results for resisted motion
here
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There are 2 important results for resisted motion
here
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Three types of resisted motion

• 1. Along a straight line

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Three types of resisted motion

• 1. Along a straight line
• 2. Going up

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Three types of resisted motion

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

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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.
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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.
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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.
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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.
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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.
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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.
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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.
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How do you approach a problem?

• Draw a forces diagram

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How do you approach a problem?

• Draw a forces diagram
• Understand that force mass acceleration

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How do you approach the problems ?

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

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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

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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

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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

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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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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

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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

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

Derive expressions for velocity as functions of
time and displacement where possible Solve
problems by using expressions derived for acc,
vel and displacement.
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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.

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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.

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Forces diagram

t 0, v U
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Forces diagram

• gravity

-g
When you are going up gravity acts against you -
so make it negative.
t 0, v U
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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
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The equation of motion is given by...
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The equation of motion is given by...
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The equation of motion is given by...
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The equation of motion is given by...
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The equation of motion is given by...
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The equation of motion is given by...
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The equation of motion is given by...
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The equation of motion is given by...
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Motion downwards
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Motion downwards
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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

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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
• find terminal velocity

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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
• 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.
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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.

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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.

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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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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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terminal velocity
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Now we can find the terminal velocity two
ways 1. Consider what happens to v as t
.
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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
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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
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The future?
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What to do when you are stuck.

• Draw a picture!

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What to do when you are stuck.

• Draw a picture!

HERE LIES THE BODY OF A FAILED MATHEMATICIAN
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What to do when you are stuck.

• Draw a picture!

HERE LIES THE BODY OF A FAILED MATHEMATICIAN
NEVER DREW A PICTURE
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What to do when you are stuck.

• Draw a picture!

HERE LIES THE BODY OF A FAILED MATHEMATICIAN
NEVER DREW A PICTURE
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HSC Bloopers 1988 (4 unit)

• In trying to answer this question I have looked
  into the depths of the abyss

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HSC Bloopers 1988 (4 unit)

• In trying to answer this question I have looked
  into the depths of the abyssthere is nothing
  there.

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HSC Bloopers 1992 (2 unit)

• The lines are parallel because eternal angels
  are equal.

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HSC Bloopers 1994 (2 unit)

• y ln (5x 1). Find the derivative.

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HSC Bloopers 1994 (2 unit)

• y ln (5x 1). Find the derivative.

201
HSC Bloopers 1994 (2 unit)

• y ln (5x 1). Find the derivative.

202
HSC Bloopers 1994 (2 unit)

• y ln (5x 1). Find the derivative.

203
HSC Bloopers 1994 (2 unit)

• y ln (5x 1). Find the derivative.

204
HSC Bloopers 1994 (2 unit)

• y ln (5x 1). Find the derivative.

205
HSC Bloopers 1994 (2 unit)

• y ln (5x 1). Find the derivative.

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