Motion 2007 mansw

Carmina

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Slide1:

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.

Mechanics:

Mechanics Morris Needleman Part 1 – Circular Motion

What do you have to do ?:

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

What do you have to do ?:

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

What do you have to do ?:

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

Do you understand?:

Do you understand?

Slide7:

please turn off your mobile phone

Circular Motion in a horizontal plane:

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

Circular Motion in a horizontal plane:

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

Circular Motion in a horizontal plane:

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

Circular Motion in a horizontal plane:

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

Circular Motion in a horizontal plane:

Slide17:

This equation is important since it links angular and linear velocity

Slide18:

This equation is important since it links angular and linear velocity

Slide19:

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

Slide20:

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

Slide21:

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

Slide22:

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

Slide23:

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

Slide24:

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

Slide25:

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

Slide26:

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 P( x,y) r T

Slide34:

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

Summary :

Summary force is directed along the radius towards the centre of the circle. (angular velocity) (links angular velocity and linear velocity) (horizontal and vertical components of acceleration) (gives the size of the force towards the centre)

Summary :

Slide41:

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.

Slide42:

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.

Slide43:

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.

Slide44:

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 mg P T

Slide45:

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 mg P T The tension in the string is the resultant of the forces acting on the body.

Slide46:

Slide47:

Slide48:

Slide49:

Slide50:

(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.

Slide51:

(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.

Slide52:

(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.

Slide53:

(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.

Slide54:

(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.

Slide55:

(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.

Slide56:

An interesting problem solving method…

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:

Conical Pendulum

Conical Pendulum:

Conical Pendulum Dimensions Diagram Forces Diagram Vertically

Conical Pendulum:

Conical Pendulum Dimensions Diagram mg q Forces Diagram Vertically

Conical Pendulum:

Conical Pendulum Dimensions Diagram mg q Forces Diagram Vertically N

Conical Pendulum:

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

Conical Pendulum:

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

Conical Pendulum:

An important result:

An important result T sin q = mrw2……1 T cos q = mg … … 2 v = rw h

An important result:

An important result T sin q = mrw2……1 T cos q = mg … … 2 v = rw h

An important result:

An important result T sin q = mrw2……1 T cos q = mg … … 2 v = rw h

An important result:

An important result T sin q = mrw2……1 T cos q = mg … … 2 v = rw h

An important result:

An important result T sin q = mrw2……1 T cos q = mg … … 2 v = rw h

An important result:

An important result T sin q = mrw2……1 T cos q = mg … … 2 v = rw h

Slide73:

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.

Slide74:

Slide75:

Slide76:

Slide77:

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

Slide78:

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

Slide79:

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

Slide80:

Motion on a Banked Track Dimensions diagram P

Slide81:

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

Slide82:

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

Slide83:

Motion on a Banked Track Dimensions diagram N P q Forces diagram P mg F centre of the circle Vertical Forces Horizontal Forces If there is no tendency to slip then F = 0 and the equations are …

Slide84:

Motion on a Banked Track Dimensions diagram N P q Forces diagram P mg F centre of the circle Vertical Forces Horizontal Forces If there is no tendency to slip then F = 0 and the equations are …

Slide85:

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

Slide86:

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

Slide87:

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.

Slide88:

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.

Slide89:

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.

Slide90:

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 P N q

Slide91:

Slide92:

Slide93:

Slide94:

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

Slide95:

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

Slide96:

2004 HSC question 6(c) (ii) Show that either q = 0 or q q P N P q R r O Now either r = 0 and the marble is stationary, or r  0 and….

Slide97:

2004 HSC question 6(c) (ii) Show that either q = 0 or q q P N P q R r O Now either r = 0 and the marble is stationary, or r  0 and….

Slide98:

2004 HSC question 6(c) (ii) Show that either q = 0 or q q P N P q R r O Now either r = 0 and the marble is stationary, or r  0 and….

Slide99:

2004 HSC question 6(c) (ii) Show that either q = 0 or q q P N P q R r O Now either r = 0 and the marble is stationary, or r  0 and….

Resisted Motion:

Resisted Motion

From 3 Unit: An important proof:

From 3 Unit: An important proof

From 3 Unit: An important proof:

From 3 Unit: An important proof

From 3 Unit: An important proof:

From 3 Unit: An important proof

From 3 Unit: An important proof:

From 3 Unit: An important proof

From 3 Unit: An important proof:

From 3 Unit: An important proof

From 3 Unit: An important proof:

From 3 Unit: An important proof

There are 2 important results for resisted motion here:

There are 2 important results for resisted motion here

There are 2 important results for resisted motion here:

There are 2 important results for resisted motion here

Three types of resisted motion:

Three types of resisted motion 1. Along a straight line

Three types of resisted motion:

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

Three types of resisted motion:

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

Type 1 - along a horizontal line:

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:

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:

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:

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:

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:

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:

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?:

How do you approach a problem? Draw a forces diagram

How do you approach a problem?:

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

How do you approach the problems ?:

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…….:

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…….:

HSC 1987 - straight line :

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 :

Motion upwards - what the syllabus says…….:

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…….:

Motion upwards - a problem...:

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...:

Forces diagram:

Forces diagram t = 0, v = U

Forces diagram:

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

Forces diagram:

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