Motion 2007 mansw

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Published on January 10, 2008

Author: 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:  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 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 :  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) 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:  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. Slide47:  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. Slide48:  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. Slide49:  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. 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 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:  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 T sin q = mrw2 T cos q = mg 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 T sin q = mrw2 T cos q = mg 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:  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. Slide75:  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. r q L = 2 Dimensions Diagram mg q Forces Diagram T N q h = 1 Slide76:  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. r q L = 2 Dimensions Diagram mg q Forces Diagram T T cos q q h = 1 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:  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 Net vertical force is 0 q q P N q Slide92:  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 Net vertical force is 0 q q P Net horizontal force force N q Slide93:  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 Net vertical force is 0 q q P Net horizontal force force N q 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 More Problem Solving…:  More Problem Solving… 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 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…….:  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 :  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 :  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…….:  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…….:  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 - 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...:  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...:  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:  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 Slide148:  The equation of motion is given by…... Slide149:  The equation of motion is given by…... Slide150:  The equation of motion is given by…... Slide151:  The equation of motion is given by…... Slide152:  The equation of motion is given by…... Slide153:  The equation of motion is given by…... Slide154:  The equation of motion is given by…... Slide155:  The equation of motion is given by…... Motion downwards - what the syllabus says…….:  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…….:  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…….:  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...:  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...:  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:  Forces diagram 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. t = 0, v = 0 Forces diagram:  Forces diagram 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. t = 0, v = 0 Slide176:  Now we can find the terminal velocity two ways: 1. Consider what happens to v as t ® ¥. Slide177:  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 Slide178:  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….:  What to do when you are stuck…. Draw a picture! What to do when you are stuck….:  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….:  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….:  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):  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):  HSC Bloopers 1988 (4 unit) In trying to answer this question I have looked into the depths of the abyss……there is nothing there. HSC Bloopers 1992 (2 unit):  HSC Bloopers 1992 (2 unit) “The lines are parallel because eternal angels are equal.” HSC Bloopers 1994 (2 unit):  HSC Bloopers 1994 (2 unit) y = ln (5x + 1). Find the derivative. HSC Bloopers 1994 (2 unit):  HSC Bloopers 1994 (2 unit) y = ln (5x + 1). Find the derivative. HSC Bloopers 1994 (2 unit):  HSC Bloopers 1994 (2 unit) y = ln (5x + 1). Find the derivative. HSC Bloopers 1994 (2 unit):  HSC Bloopers 1994 (2 unit) y = ln (5x + 1). Find the derivative. HSC Bloopers 1994 (2 unit):  HSC Bloopers 1994 (2 unit) y = ln (5x + 1). Find the derivative. HSC Bloopers 1994 (2 unit):  HSC Bloopers 1994 (2 unit) y = ln (5x + 1). Find the derivative. HSC Bloopers 1994 (2 unit):  HSC Bloopers 1994 (2 unit) y = ln (5x + 1). Find the derivative. Slide193:  www.mansw.edu.au Student Services

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