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Mechanics Morris Needleman Part 1 – Circular Motion

What do you have to do ? Watch out for Buffy.

What do you have to do ? Watch out for Buffy. When the music starts – you should be writing!

What do you have to do ? Watch out for Buffy. When the music starts – you should be writing!

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Circular Motion in a horizontal plane P moves around a circle of radius r. O q P r T

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 O q P r T

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 O q P r T

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 O q P r T

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 O q P r T

O q P r T

O q P r T

O q P r T

O q P r T

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. O q P r T

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

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

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

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

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

P( x,y) To simplify life we are going to consider that the angular velocity remains constant throughout the motion. O r q T

P( x,y) O 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

Forces Diagram

It is important to note that the vectors demonstrate that the force is directed along the radius towards the centre of the circle.

Summary Forces Diagram (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)

Summary Forces Diagram (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)

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.

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.

(a) Find the tension in the string. 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

(a) Find the tension in the string. 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.

(a) Find the tension in the string. 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.

(a) Find the tension in the string. 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.

(a) Find the tension in the string. 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.

(a) Find the tension in the string. 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.

(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.

(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.

(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.

(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.

(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.

(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.

An interesting problem solving method… 9

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.

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.

Conical Pendulum m R q L

Conical Pendulum m R q L m R q L Forces Diagram Dimensions Diagram Vertically

Conical Pendulum m R q L q Forces Diagram Dimensions Diagram mg Vertically

Conical Pendulum m R q L q N Forces Diagram Dimensions Diagram mg Vertically

Conical Pendulum m R q L q T N q mg Forces Diagram Dimensions Diagram Vertically N + (-mg) = 0 N = mg but cos q = N/T so N = T cos q  T cos q = mg

Conical Pendulum m R q L q T N q mg Forces Diagram Dimensions Diagram Vertically Horizontally N + (-mg) = 0 N = mg but cos q = N/T so N = T cos q  T cos q = mg T sin q = mrw2 Since the only horizontal force is directed along the radius towards the centre

Conical Pendulum m R q L q T N q mg Forces Diagram Dimensions Diagram Vertically Horizontally T sin q = mrw2 T cos q = mg N + (-mg) = 0 N = mg but cos q = N/T so N = T cos q  T cos q = mg T sin q = mrw2 Since the only horizontal force is directed along the radius towards the centre

Conical Pendulum m R q L q T N q mg Forces Diagram Dimensions Diagram Vertically Horizontally T sin q = mrw2 T cos q = mg N + (-mg) = 0 N = mg but cos q = N/T so N = T cos q  T cos q = mg T sin q = mrw2 Since the only horizontal force is directed along the radius towards the centre

An important result N mg q Forces Diagram T m R q L h T sin q = mrw2……1 T cos q = mg … … 2 v = rw

An important result N mg q Forces Diagram T m R q L h T sin q = mrw2……1 T cos q = mg … … 2 v = rw

An important result N mg q Forces Diagram T m R q L h T sin q = mrw2……1 T cos q = mg … … 2 v = rw

An important result N mg q Forces Diagram T m R q L h T sin q = mrw2……1 T cos q = mg … … 2 v = rw

An important result N mg q Forces Diagram T m R q L h T sin q = mrw2……1 T cos q = mg … … 2 v = rw

An important result N mg q Forces Diagram T m R q L h T sin q = mrw2……1 T cos q = mg … … 2 v = rw

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 Forces Diagram Dimensions Diagram

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 Forces Diagram Dimensions Diagram

Example 2 Find the tension in the string and the angular velocity of the particle. r q L = 2 Dimensions Diagram mg Forces Diagram T N h = 1 Vertically Horizontally

Example 2 Find the tension in the string and the angular velocity of the particle. r q L = 2 Dimensions Diagram mg Forces Diagram T N h = 1 Vertically Horizontally

Example 2 Find the tension in the string and the angular velocity of the particle. r q L = 2 Dimensions Diagram mg Forces Diagram T N h = 1 Vertically Horizontally

Motion on a Banked Track Dimensions diagram h d q P

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

Motion on a Banked Track Dimensions diagram Forces diagram N h d q F q P P centre of the circle mg Vertical Forces Horizontal Forces

Motion on a Banked Track Dimensions diagram Forces diagram N h d q 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 …

Motion on a Banked Track Dimensions diagram Forces diagram N h d q 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 …

If there is no tendency to slip at v = v0 then F = 0 and the equations are …

If there is no tendency to slip at v = v0 then F = 0 and the equations are …

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.

2004 HSC question 6(c) O N P mg r R 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 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.

2004 HSC question 6(c) O N P mg r R 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.

2004 HSC question 6(c) O N P mg r R 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

2004 HSC question 6(c) O N P mg r R 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

2004 HSC question 6(c) O N P mg r R 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

2004 HSC question 6(c) O N P mg r R 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

2004 HSC question 6(c) (ii) Show that either q = 0 or O q R r P q N q

2004 HSC question 6(c) (ii) Show that either q = 0 or O q R r P q N q

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

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

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

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

Resisted Motion

More Problem Solving…

From 3 Unit: An important proof 9

From 3 Unit: An important proof 9

From 3 Unit: An important proof 9

From 3 Unit: An important proof 9

From 3 Unit: An important proof 9

From 3 Unit: An important proof 9

There are 2 important results for resisted motion here

There are 2 important results for resisted motion here

Three types of resisted motion 1. Along a straight line

Three types of resisted motion 1. Along a straight line 2. Going up

Three types of resisted motion 1. Along a straight line 2. Going up 3. Coming down

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.

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.

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.

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.

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.

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.

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.

How do you approach a problem? Draw a forces diagram

How do you approach a problem? Draw a forces diagram Understand that force = mass ´ acceleration

How do you approach the problems ? Draw a forces diagram Understand that force = mass ´ acceleration Write down an initial equation

Along a horizontal line - what the syllabus says……. Derive from Newton’s 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

Along a horizontal line - what the syllabus says……. Derive from Newton’s 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

Along a horizontal line - what the syllabus says……. Derive from Newton’s 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

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 > 0. (a) Show that v is related to the displacement x by the formula

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 > 0. (a) Show that v is related to the displacement x by the formula

Motion upwards - what the syllabus says……. Derive from Newton’s 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

Motion upwards - what the syllabus says……. Derive from Newton’s 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

Motion upwards - what the syllabus says……. Derive from Newton’s 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.

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.

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.

Forces diagram t = 0, v = U

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

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

The equation of motion is given by…...

The equation of motion is given by…...

The equation of motion is given by…...

The equation of motion is given by…...

The equation of motion is given by…...

The equation of motion is given by…...

The equation of motion is given by…...

The equation of motion is given by…...

Motion downwards - what the syllabus says……. Derive from Newton’s 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

Motion downwards - what the syllabus says……. Derive from Newton’s 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

Motion downwards - what the syllabus says……. Derive from Newton’s 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.

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.

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.

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.

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.

Now we can find the terminal velocity two ways: 1. Consider what happens to v as t ® ¥.

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

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

What to do when you are stuck…. Draw a picture!

What to do when you are stuck…. Draw a picture! HERE LIES THE BODY OF A FAILED MATHEMATICIAN

What to do when you are stuck…. Draw a picture! HERE LIES THE BODY OF A FAILED MATHEMATICIAN NEVER DREW A PICTURE

What to do when you are stuck…. Draw a picture! HERE LIES THE BODY OF A FAILED MATHEMATICIAN NEVER DREW A PICTURE

HSC Bloopers 1988 (4 unit) In trying to answer this question I have looked into the depths of the abyss……

HSC Bloopers 1988 (4 unit) In trying to answer this question I have looked into the depths of the abyss……there is nothing there.

“The lines are parallel because eternal angels are equal.” HSC Bloopers 1992 (2 unit) “The lines are parallel because eternal angels are equal.”

y = ln (5x + 1). Find the derivative. HSC Bloopers 1994 (2 unit) y = ln (5x + 1). Find the derivative.

y = ln (5x + 1). Find the derivative. HSC Bloopers 1994 (2 unit) y = ln (5x + 1). Find the derivative.

y = ln (5x + 1). Find the derivative. HSC Bloopers 1994 (2 unit) y = ln (5x + 1). Find the derivative.

y = ln (5x + 1). Find the derivative. HSC Bloopers 1994 (2 unit) y = ln (5x + 1). Find the derivative.

y = ln (5x + 1). Find the derivative. HSC Bloopers 1994 (2 unit) y = ln (5x + 1). Find the derivative.

y = ln (5x + 1). Find the derivative. HSC Bloopers 1994 (2 unit) y = ln (5x + 1). Find the derivative.

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