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- 1. PROJECTILE MOTION constant acceleration in a straight line; free fall under gravity, projectile motion; Relative motion, change in velocity, velocity vector components, circular motion (constant speed with one force only providing centripetal force). 1. Recall that in the absence of friction a falling object will have a constant acceleration of 10 ms-2 and that this value is referred to as “gravitational acceleration, g” 2. Use F = mg to show that Nkg-1 is an equivalent unit to ms-2. 3. Explain the effect that air friction has on the acceleration of a falling object (resulting in free fall) 4. Define the term projectile. 5. Describe projectile motion in terms of its uniform horizontal motion and it accelerated vertical motion. 6. Use equations of motion to calculate time, distance, velocity and acceleration. Monday, 24 May 2010
- 2. FREE FALL IN THE ABSENCE OF FRICTION Mass & weight The mass of an object, m is a measure of the amount of matter in that object. The weight of an object, Fw is a measure of the force due to gravity on that object. Gravitational constant, g The force exerted on an object by the earth’s gravitational field is called the gravitational constant, g. What g = 10 Nkg -1 does this mean? This means that for every kg of mass there is a force of 10N acting on it because of gravity This causes all objects to fall with an acceleration of 10 ms-2 We can also say g = 10 ms-2 which is called gravitational acceleration Fw = mg Monday, 24 May 2010
- 3. FOR AN OBJECT THROWN VERTICALLY UPWARDS THE ACCELERATION IS CONSTANT & NEGATIVE It slows down as it moves in the upwards + up (positive) direction and - down speeds up as it moves in the downwards (negative) direction. Velocity - time graph for the object v (ms-1) t (s) Constant negative slope shows a constant negative acceleration Monday, 24 May 2010
- 4. FOR AN OBJECT THROWN VERTICALLY UPWARDS THE ACCELERATION IS CONSTANT & NEGATIVE It slows down as it moves in the upwards + up (positive) direction and - down speeds up as it moves in the downwards (negative) direction. Velocity - time graph for the object v (ms-1) t (s) Constant negative slope shows a constant negative acceleration Monday, 24 May 2010
- 5. FOR AN OBJECT THROWN VERTICALLY UPWARDS THE ACCELERATION IS CONSTANT & NEGATIVE It slows down as it moves in the upwards + up (positive) direction and - down speeds up as it moves in the downwards (negative) direction. Velocity - time graph for the object v (ms-1) t (s) Constant negative slope shows a constant negative acceleration Monday, 24 May 2010
- 6. FOR AN OBJECT THROWN VERTICALLY UPWARDS THE ACCELERATION IS CONSTANT & NEGATIVE It slows down as it moves in the upwards + up (positive) direction and - down speeds up as it moves in the downwards (negative) direction. A symmetrical path Velocity - time graph for the object v (ms-1) t (s) Constant negative slope shows a constant negative acceleration Monday, 24 May 2010
- 7. FOR AN OBJECT THROWN VERTICALLY UPWARDS THE ACCELERATION IS CONSTANT & NEGATIVE It slows down as it moves in the upwards + up (positive) direction and - down speeds up as it moves in the downwards (negative) direction. A symmetrical path Velocity - time graph for the object v (ms-1) x t (s) Constant negative slope shows a constant negative acceleration Monday, 24 May 2010
- 8. FOR AN OBJECT THROWN VERTICALLY UPWARDS THE ACCELERATION IS CONSTANT & NEGATIVE It slows down as it moves in the upwards + up (positive) direction and - down speeds up as it moves in the downwards (negative) direction. A symmetrical path Velocity - time graph for the object v (ms-1) x x t (s) Constant negative slope shows a constant negative acceleration Monday, 24 May 2010
- 9. FOR AN OBJECT THROWN VERTICALLY UPWARDS THE ACCELERATION IS CONSTANT & NEGATIVE It slows down as it moves in the upwards + up (positive) direction and - down speeds up as it moves in the downwards (negative) direction. A symmetrical path Velocity - time graph for the object v (ms-1) x x t (s) Constant negative x slope shows a constant negative acceleration Monday, 24 May 2010
- 10. FOR AN OBJECT THROWN VERTICALLY UPWARDS THE ACCELERATION IS CONSTANT & NEGATIVE It slows down as it moves in the upwards + up (positive) direction and - down speeds up as it moves in the downwards (negative) direction. A symmetrical path Velocity - time graph for the object v (ms-1) x x t (s) Constant negative x slope shows a constant negative acceleration Monday, 24 May 2010
- 11. FOR AN OBJECT THROWN VERTICALLY UPWARDS THE ACCELERATION IS CONSTANT & NEGATIVE It slows down as it moves in the upwards + up (positive) direction and - down speeds up as it moves in the downwards (negative) direction. A symmetrical path Velocity - time graph for the object Decreasing speed in v (ms-1) x a positive direction x t (s) Constant negative x slope shows a constant negative acceleration Monday, 24 May 2010
- 12. FOR AN OBJECT THROWN VERTICALLY UPWARDS THE ACCELERATION IS CONSTANT & NEGATIVE It slows down as it moves in the upwards + up (positive) direction and - down speeds up as it moves in the downwards (negative) direction. A symmetrical path Velocity - time graph for the object Decreasing speed in v (ms-1) x a positive direction Stationary for an instant x t (s) Constant negative x slope shows a constant negative acceleration Monday, 24 May 2010
- 13. FOR AN OBJECT THROWN VERTICALLY UPWARDS THE ACCELERATION IS CONSTANT & NEGATIVE It slows down as it moves in the upwards + up (positive) direction and - down speeds up as it moves in the downwards (negative) direction. A symmetrical path Velocity - time graph for the object Decreasing speed in v (ms-1) x a positive direction Stationary for an instant x t (s) Constant negative Increasing speed in a slope shows a negative direction x constant negative acceleration Monday, 24 May 2010
- 14. Examples 1. A plane drops a Red Cross package from a height of 1200 m. If the package had no parachute (and by this you can assume negligible air friction & g = 10 ms-2) (a) How fast will the package be travelling just before it hits the ground? (b) How many seconds will the package take to fall? 2. A 1 kg object is dropped from a tower 120 m high. (a) Calculate the time it will take for the object to fall to the ground. (b) Calculate the objects final speed on reaching the ground. (c) How long does it take to reach a speed of 35 ms-1 ? Monday, 24 May 2010
- 15. 3. A shell is fired straight up with an initial speed of 96 ms-1. (a) Calculate the time it will take for the object to fall to the ground. (b) When will the shell have an upwards speed of 48 ms-1 ? (c) Calculate the time for the shell to reach its maximum height. (d) Calculate the maximum height reached by the shell. (e) What is the shells acceleration at the top of its motion? Monday, 24 May 2010
- 16. FREE FALL - IN THE PRESENCE OF FRICTION If the object is falling fast: Air friction is too large to ignore. Here we consider the object to have two forces acting on it. Air friction and gravity. A. When the object is initially falling slowly, air friction is much smaller than the force of gravity on the object. B. As the object speeds up the air friction increases. C. The object will eventually reach a speed at which the air friction balances the force of gravity on the object. The object cannot fall any faster than at this speed (unless it changes shape). This speed is called terminal velocity. This situation is illustrated by the following example: B A C Vector arithmetic with notes Monday, 24 May 2010
- 17. WHAT IS A PROJECTILE? Brainstorm Examples of Projectiles “What do all these examples have in common?” “Why do you think that a ball dropping vertically (in free fall) is not an example of projectile motion?” Monday, 24 May 2010
- 18. PROJECTILES - HORIZONTAL AND VERTICAL MOTION • A projectile is an object that has constant horizontal velocity and constant vertical acceleration due to gravity. In other words it is moving horizontally at the same time it is in free fall. The projectiles velocity is the sum of these two velocities. • The path of a projectile is parabolic and symmetrical • Air resistance is considered to be negligible • The force of gravity is the only force acting on the object. Once the projectile is launched there is no thrust force. v = the size of the velocity of the projectile To determine the velocity: vx vx = the horizontal component of the vy projectile’s velocity add the horizontal and v vy = the vertical component of the vertical components projectile's velocity Monday, 24 May 2010
- 19. PROJECTILES - HORIZONTAL AND VERTICAL MOTION • A projectile is an object that has constant horizontal velocity and constant vertical acceleration due to gravity. In other words it is moving horizontally at the same time it is in free fall. The projectiles velocity is the sum of these two velocities. • The path of a projectile is parabolic and symmetrical • Air resistance is considered to be negligible • The force of gravity is the only force acting on the object. Once the projectile is launched there is no thrust force. v = the size of the velocity of the projectile To determine the velocity: vx vx = the horizontal component of the vy projectile’s velocity add the horizontal and v vy = the vertical component of the vertical components projectile's velocity Monday, 24 May 2010
- 20. PROJECTILES - HORIZONTAL AND VERTICAL MOTION • A projectile is an object that has constant horizontal velocity and constant vertical acceleration due to gravity. In other words it is moving horizontally at the same time it is in free fall. The projectiles velocity is the sum of these two velocities. • The path of a projectile is parabolic and symmetrical • Air resistance is considered to be negligible • The force of gravity is the only force acting on the object. Once the projectile is launched there is no thrust force. v = the size of the velocity of the projectile To determine the velocity: vx vx = the horizontal component of the vy projectile’s velocity add the horizontal and v vy = the vertical component of the vertical components projectile's velocity Monday, 24 May 2010
- 21. PROJECTILES - HORIZONTAL AND VERTICAL MOTION • A projectile is an object that has constant horizontal velocity and constant vertical acceleration due to gravity. In other words it is moving horizontally at the same time it is in free fall. The projectiles velocity is the sum of these two velocities. • The path of a projectile is parabolic and symmetrical • Air resistance is considered to be negligible • The force of gravity is the only force acting on the object. Once the projectile is launched there is no thrust force. v = the size of the velocity of the projectile To determine the velocity: vx vx = the horizontal component of the vy projectile’s velocity add the horizontal and v vy = the vertical component of the vertical components projectile's velocity Monday, 24 May 2010
- 22. PROJECTILES - HORIZONTAL AND VERTICAL MOTION • A projectile is an object that has constant horizontal velocity and constant vertical acceleration due to gravity. In other words it is moving horizontally at the same time it is in free fall. The projectiles velocity is the sum of these two velocities. • The path of a projectile is parabolic and symmetrical • Air resistance is considered to be negligible • The force of gravity is the only force acting on the object. Once the projectile is launched there is no thrust force. v = the size of the velocity of the projectile To determine the velocity: vx vx = the horizontal component of the vy projectile’s velocity add the horizontal and v vy = the vertical component of the vertical components projectile's velocity Monday, 24 May 2010
- 23. PROJECTILES - HORIZONTAL AND VERTICAL MOTION • A projectile is an object that has constant horizontal velocity and constant vertical acceleration due to gravity. In other words it is moving horizontally at the same time it is in free fall. The projectiles velocity is the sum of these two velocities. • The path of a projectile is parabolic and symmetrical • Air resistance is considered to be negligible • The force of gravity is the only force acting on the object. Once the projectile is launched there is no thrust force. v = the size of the velocity of the projectile To determine the velocity: vx vx = the horizontal component of the vy projectile’s velocity add the horizontal and v vy = the vertical component of the vertical components projectile's velocity Monday, 24 May 2010
- 24. PROJECTILES - HORIZONTAL AND VERTICAL MOTION • A projectile is an object that has constant horizontal velocity and constant vertical acceleration due to gravity. In other words it is moving horizontally at the same time it is in free fall. The projectiles velocity is the sum of these two velocities. • The path of a projectile is parabolic and symmetrical • Air resistance is considered to be negligible • The force of gravity is the only force acting on the object. Once the projectile is launched there is no thrust force. vx v = the size of the velocity of the projectile To determine the velocity: vx vx = the horizontal component of the vy projectile’s velocity add the horizontal and v vy = the vertical component of the vertical components projectile's velocity Monday, 24 May 2010
- 25. PROJECTILES - HORIZONTAL AND VERTICAL MOTION • A projectile is an object that has constant horizontal velocity and constant vertical acceleration due to gravity. In other words it is moving horizontally at the same time it is in free fall. The projectiles velocity is the sum of these two velocities. • The path of a projectile is parabolic and symmetrical • Air resistance is considered to be negligible • The force of gravity is the only force acting on the object. Once the projectile is launched there is no thrust force. vx vx v = the size of the velocity of the projectile To determine the velocity: vx vx = the horizontal component of the vy projectile’s velocity add the horizontal and v vy = the vertical component of the vertical components projectile's velocity Monday, 24 May 2010
- 26. PROJECTILES - HORIZONTAL AND VERTICAL MOTION • A projectile is an object that has constant horizontal velocity and constant vertical acceleration due to gravity. In other words it is moving horizontally at the same time it is in free fall. The projectiles velocity is the sum of these two velocities. • The path of a projectile is parabolic and symmetrical • Air resistance is considered to be negligible • The force of gravity is the only force acting on the object. Once the projectile is launched there is no thrust force. vx vx vx v = the size of the velocity of the projectile To determine the velocity: vx vx = the horizontal component of the vy projectile’s velocity add the horizontal and v vy = the vertical component of the vertical components projectile's velocity Monday, 24 May 2010
- 27. PROJECTILES - HORIZONTAL AND VERTICAL MOTION • A projectile is an object that has constant horizontal velocity and constant vertical acceleration due to gravity. In other words it is moving horizontally at the same time it is in free fall. The projectiles velocity is the sum of these two velocities. • The path of a projectile is parabolic and symmetrical • Air resistance is considered to be negligible • The force of gravity is the only force acting on the object. Once the projectile is launched there is no thrust force. vx vx vx vx v = the size of the velocity of the projectile To determine the velocity: vx vx = the horizontal component of the vy projectile’s velocity add the horizontal and v vy = the vertical component of the vertical components projectile's velocity Monday, 24 May 2010
- 28. PROJECTILES - HORIZONTAL AND VERTICAL MOTION • A projectile is an object that has constant horizontal velocity and constant vertical acceleration due to gravity. In other words it is moving horizontally at the same time it is in free fall. The projectiles velocity is the sum of these two velocities. • The path of a projectile is parabolic and symmetrical • Air resistance is considered to be negligible • The force of gravity is the only force acting on the object. Once the projectile is launched there is no thrust force. vx vx vy vx vx v = the size of the velocity of the projectile To determine the velocity: vx vx = the horizontal component of the vy projectile’s velocity add the horizontal and v vy = the vertical component of the vertical components projectile's velocity Monday, 24 May 2010
- 29. PROJECTILES - HORIZONTAL AND VERTICAL MOTION • A projectile is an object that has constant horizontal velocity and constant vertical acceleration due to gravity. In other words it is moving horizontally at the same time it is in free fall. The projectiles velocity is the sum of these two velocities. • The path of a projectile is parabolic and symmetrical • Air resistance is considered to be negligible • The force of gravity is the only force acting on the object. Once the projectile is launched there is no thrust force. vy vx vx vy vx vx v = the size of the velocity of the projectile To determine the velocity: vx vx = the horizontal component of the vy projectile’s velocity add the horizontal and v vy = the vertical component of the vertical components projectile's velocity Monday, 24 May 2010
- 30. PROJECTILES - HORIZONTAL AND VERTICAL MOTION • A projectile is an object that has constant horizontal velocity and constant vertical acceleration due to gravity. In other words it is moving horizontally at the same time it is in free fall. The projectiles velocity is the sum of these two velocities. • The path of a projectile is parabolic and symmetrical • Air resistance is considered to be negligible • The force of gravity is the only force acting on the object. Once the projectile is launched there is no thrust force. vy vx vx vy = 0 vy vx vx v = the size of the velocity of the projectile To determine the velocity: vx vx = the horizontal component of the vy projectile’s velocity add the horizontal and v vy = the vertical component of the vertical components projectile's velocity Monday, 24 May 2010
- 31. PROJECTILES - HORIZONTAL AND VERTICAL MOTION • A projectile is an object that has constant horizontal velocity and constant vertical acceleration due to gravity. In other words it is moving horizontally at the same time it is in free fall. The projectiles velocity is the sum of these two velocities. • The path of a projectile is parabolic and symmetrical • Air resistance is considered to be negligible • The force of gravity is the only force acting on the object. Once the projectile is launched there is no thrust force. vy vx vx vy = 0 vy vx vy vx v = the size of the velocity of the projectile To determine the velocity: vx vx = the horizontal component of the vy projectile’s velocity add the horizontal and v vy = the vertical component of the vertical components projectile's velocity Monday, 24 May 2010
- 32. PROJECTILES - HORIZONTAL AND VERTICAL MOTION • A projectile is an object that has constant horizontal velocity and constant vertical acceleration due to gravity. In other words it is moving horizontally at the same time it is in free fall. The projectiles velocity is the sum of these two velocities. • The path of a projectile is parabolic and symmetrical • Air resistance is considered to be negligible • The force of gravity is the only force acting on the object. Once the projectile is launched there is no thrust force. vy vx vx vy = 0 vy v vx vy vx v = the size of the velocity of the projectile To determine the velocity: vx vx = the horizontal component of the vy projectile’s velocity add the horizontal and v vy = the vertical component of the vertical components projectile's velocity Monday, 24 May 2010
- 33. PROJECTILES - HORIZONTAL AND VERTICAL MOTION • A projectile is an object that has constant horizontal velocity and constant vertical acceleration due to gravity. In other words it is moving horizontally at the same time it is in free fall. The projectiles velocity is the sum of these two velocities. • The path of a projectile is parabolic and symmetrical • Air resistance is considered to be negligible • The force of gravity is the only force acting on the object. Once the projectile is launched there is no thrust force. vy v vx vx vy = 0 vy v vx vy vx v = the size of the velocity of the projectile To determine the velocity: vx vx = the horizontal component of the vy projectile’s velocity add the horizontal and v vy = the vertical component of the vertical components projectile's velocity Monday, 24 May 2010
- 34. PROJECTILES - HORIZONTAL AND VERTICAL MOTION • A projectile is an object that has constant horizontal velocity and constant vertical acceleration due to gravity. In other words it is moving horizontally at the same time it is in free fall. The projectiles velocity is the sum of these two velocities. • The path of a projectile is parabolic and symmetrical • Air resistance is considered to be negligible • The force of gravity is the only force acting on the object. Once the projectile is launched there is no thrust force. vy v v vx vx vy = 0 vy v vx vy vx v = the size of the velocity of the projectile To determine the velocity: vx vx = the horizontal component of the vy projectile’s velocity add the horizontal and v vy = the vertical component of the vertical components projectile's velocity Monday, 24 May 2010
- 35. PROJECTILES - HORIZONTAL AND VERTICAL MOTION • A projectile is an object that has constant horizontal velocity and constant vertical acceleration due to gravity. In other words it is moving horizontally at the same time it is in free fall. The projectiles velocity is the sum of these two velocities. • The path of a projectile is parabolic and symmetrical • Air resistance is considered to be negligible • The force of gravity is the only force acting on the object. Once the projectile is launched there is no thrust force. vy v v vx vx vy = 0 vy v vx vy v vx v = the size of the velocity of the projectile To determine the velocity: vx vx = the horizontal component of the vy projectile’s velocity add the horizontal and v vy = the vertical component of the vertical components projectile's velocity Monday, 24 May 2010
- 36. SYMMETRICAL FLIGHT PATH • The path of a projectile is parabolic and symmetrical Symmetrical Path It is often convenient to work with half the flight path of a projectile. Remember When the projectile is at the maximum height the vertical velocity is zero. At t : vy = 0 2 Monday, 24 May 2010
- 37. SYMMETRICAL FLIGHT PATH • The path of a projectile is parabolic and symmetrical Symmetrical Path It is often convenient to work with half the flight path of a projectile. d Remember When the projectile is at the maximum height the vertical velocity is zero. At t : vy = 0 2 Monday, 24 May 2010
- 38. SYMMETRICAL FLIGHT PATH • The path of a projectile is parabolic and symmetrical Symmetrical Path It is often convenient to work with half the flight path of a projectile. t 2 d Remember When the projectile is at the maximum height the vertical velocity is zero. At t : vy = 0 2 Monday, 24 May 2010
- 39. SYMMETRICAL FLIGHT PATH • The path of a projectile is parabolic and symmetrical Symmetrical Path It is often convenient to work with half the flight path of a projectile. t 2 t d Remember When the projectile is at the maximum height the vertical velocity is zero. At t : vy = 0 2 Monday, 24 May 2010
- 40. SYMMETRICAL FLIGHT PATH • The path of a projectile is parabolic and symmetrical Symmetrical Path It is often convenient to work with half the flight path of a projectile. t 2 t d d= the total distance travelled by a projectile t = the time taken to travel that distance t/2 = the time taken to reach maximum height Remember When the projectile is at the maximum height the vertical velocity is zero. At t : vy = 0 2 Monday, 24 May 2010
- 41. Forces The force due to gravity is the only force acting on the projectile. Not a projectile: An object in horizontal flight is not a projectile Fgrav because Fgrav is balanced by the lift force. An object that has not been launched but maintains a high horizontal speed will have air friction acting on it and therefore must have a thrust force to balance air friction. This will not be a projectile. Vertical motion calculations • Since the force of gravity is constant (10 N per kg of any object), the acceleration is also constant. (10 ms-2 for any object) • Kinematic equations must be used for analysing the vertical motion Horizontal motion calculations • A projectile progresses horizontally with constant speed. • Horizontal speed can be calculated in the usual way: Speed = distance travelled v=d time taken t Units of v: ms-1 Monday, 24 May 2010
- 42. PROCESS FOR PROBLEM-SOLVING [PIA] 1. Read the question and underline the relevant information 2. Draw a diagram of the situation 3. List the information that relates to the vertical motion and the horizontal motion. (Keeping it separate. Use a table with two columns “V” & “H”, remembering that vertical and horizontal motion are independent of each other) 4. Show up as a positive direction and down as negative. The signs that you use for the vertical information should reflect this. 5. Select the appropriate kinematic equation when calculating a vertical quantity. 6. Use v = d/t for calculation of a horizontal quantity. Example 1 Imagine a car is driven off a 250m cliff at 30 ms-1. How far from the base of the cliff will the car be when it lands? Diagram Monday, 24 May 2010
- 43. Example 2 A hockey ball is flicked with an initial velocity of 14 ms-1, 45o to the horizontal, as shown. A spectator 1.6 m tall is standing directly in the path of the ball, 20 m away. Will the ball hit the spectator? Diagram Monday, 24 May 2010
- 44. 12 PHYSICS PROJECTILES ASSIGNMENT Name ______________________ 1. A rock is dropped from the top of a bridge. It takes 4 seconds to reach the water below. (b) Explain why this is not an example of projectile motion. ________________________________________________________________ ________________________________________________________________ (c) What would need to be done to the rock to turn it into a projectile? ________________________________________________________________ (d) How far has the rock fallen? ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ (e) A second rock is now thrown horizontally at 6 ms-1 from the same bridge. ________________________________________________________________ (f) How long will it take for the rock to reach the water? ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ (g) Calculate the range of this second rock ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ Monday, 24 May 2010
- 45. 2. A 2 kg bowling ball is launched from ground level and follows the path shown below: At each of the labelled positions A, B, C and D, state the size and direction of: (a) the acceleration of the projectile (b) the net force on the projectile 3. When an aircraft was travelling in level flight at 200 ms-1, a nut fell off part of the landing gear. Assume air friction is negligible. (a) Sketch the paths of the aircraft and the nut. Sketch The nut covered a total horizontal distance of 3.0 km. (b) Find the total time it took to fall to the ground. ________________________________________________________________ ________________________________________________________________ Monday, 24 May 2010
- 46. (c) What was the altitude of the plane when the nut fell off? ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ 4. A ball is kicked with an initial velocity as shown below. The angle of inclination is 40o. 21 ms-1 40o (a) Calculate the horizontal component of the initial velocity. __________________ (b) Calculate the vertical component of the initial velocity. ____________________ (c) What is the instantaneous velocity of the ball at the top of its path? ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ (d) Find the height of the ball at the top of its path. ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ Monday, 24 May 2010
- 47. (e) Find the time taken for the ball to reach its maximum vertical displacement. ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ (f) Calculate the range of the ball. ________________________________________________________________ ________________________________________________________________ If the ball was kicked at an angle of 60o to the horizontal, would it have travelled as far along the field? ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ 5. A potato was launched from the muzzle of a spud gun with an initial velocity of 30 ms-1 at an angle of 60o to the horizontal. Will it clear a 25 m tree that is 60 m away on the flight path? ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 6. A cricket ball was hit with the following trajectory. Find the initial velocity of the ball. 25 m 100 m Monday, 24 May 2010
- 48. 7. Johnny is competing in the javelin event of his school athletics competition. The javelin behaves like an ideal projectile. (a) Describe the shape of the path of the javelin. ________________________________________________________________ (b) Ignoring air resistance, draw arrow(s) on the drawing of the javelin below to show the force(s) acting on it when it is in the position shown. Name the forces. Joe now throws the javelin into the air at an angle of 40o above the horizontal at an initial velocity of 30 ms-1. Joe now throws the javelin into the air at an angle of 40° above the horizontal at an initial velocity of 30 m s–1 Monday, 24 May 2010
- 49. (c) Show that the horizontal component of the initial velocity of the javelin is 23 ms-1. ________________________________________________________________ ________________________________________________________________ (d) Calculate the range of the javelin under these conditions. ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ [this is a 2004 NCEA exam question] Monday, 24 May 2010
- 50. UNIFORM CIRCULAR MOTION 1. Recognize that even though a body in circular motion may have a constant speed, its velocity is changing and hence it is accelerating. 2. Demonstrate that the acceleration is towards the centre, i.e. centripetal. 3. Deduce that all things in circular motion must have a centripetal force. 4. Show graphically using velocity vectors that the acceleration is towards the centre. 5. Remember the equations for uniform circular motion questions, and use them to solve problems: Read Chapter 12 (p139 to 145) Monday, 24 May 2010
- 51. BRAINSTORM Give me some everyday examples of objects in uniform circular motion Complete the functional definition below: An object in uniform circular object is _____________________________________ ___________________________________________________________________ Monday, 24 May 2010
- 52. CENTRIPETAL ACCELERATION For an object in uniform circular motion: • The speed is always the same • but the direction is always changing Therefore the velocity is always changing and so the object is always accelerating “remember that acceleration is the rate of change in velocity.” REMEMBER Centripetal acceleration: THIS --> • Constant in size • Direction is always changing • Direction is towards the centre of the circular path Monday, 24 May 2010
- 53. MORE EVIDENCE FOR AN ACCELERATION TOWARDS THE CENTRE For each 0.1 s time interval, determine ∆v. When you carry out the vector ~ ~ subtractions required, do so by leaving vf in the position drawn (below). ~ ~ ~ ∆t = 0.01 s vi ~ vf ~ . vi ~ vf ~ ∆t = 0.01 s Comment on the size and direction of the change in velocity vector: Monday, 24 May 2010
- 54. CENTRIPETAL FORCE “Acceleration is caused by an unbalanced force so now we can say something about the force acting on an object in circular motion” Centripetal force: • Constant in size REMEMBER • Direction is always changing THIS --> • Direction is towards the centre of the circular path “Does this sound familiar??” Note • Greater speeds will require greater forces. • Force in uniform circular motion changes direction not speed Monday, 24 May 2010
- 55. The carousel WHAT DO WE MEAN BY UNBALANCED FORCE? The force acting through the rope Tension The force acting vertically downwards Gravity “Tension on its own does not cause circular motion” UNBALANCED FORCE “Gravity on its own does not cause circular motion” • A combination of these two forces is called the unbalanced force. • It is the unbalanced force that causes the uniform circular motion Monday, 24 May 2010
- 56. EQUATIONS FOR UNIFORM CIRCULAR Consider a mass in uniform circular motion with speed v and a radius of circular path, r: v = velocity (ms-1) v ~ ~ m F = Force (N) F ~ ~ P a = acceleration (ms-2) a ~ ~ r r = radius of the circular path (m) (measured from the centre of the path to the centre of mass of the object. m = mass of the object (kg) velocity is at a tangent to At any point: the circular path Speed The speed is given by the distance travelled around the circular path divided by the time taken to travel that distance. v=d t Monday, 24 May 2010
- 57. Note that the distance travelled in a complete rotation is the circumference, C. C = 2 πr Period and frequency • The Period, T is the time the object takes to move through one complete rotation. (Unit: second, s) • The frequency, f is the number of revolutions performed per second. (Unit: Hertz, Hz or s-1)) T=1 f=1 and f T Acceleration and force Experiment shows that: ac = v2 ac = centripetal acceleration r Fc = centripetal force Fc = mv2 r since F = ma Monday, 24 May 2010
- 58. Examples 1. The Apollo 11 space capsule was placed in a parking orbit around the Earth before moving onwards to the Moon. The radius of the orbit was 6.56 x 104 m and the mass of the capsule was 4.4 x 104 kg. to the moon moon parking orbit Earth (a) If the centripetal force on the capsule was 407 kN while it was in the parking orbit, what was its acceleration? (b) What was the speed of the capsule in the parking orbit? (c) How long did it take the capsule to complete one orbit? Monday, 24 May 2010
- 59. 2. A game of swing-ball is played with a 100 g ball. The effective radius of the circular path of the ball is 1.4 m. Find the tension in the string (centripetal force) when the ball has a velocity of: (a) 7.5 ms-1 (b) 15 ms-1 3. A string has a breaking strain of 320 N. Find the maximum speed that a mass can be whirled around in uniform circular motion with a radius of 0.45 m if the mass is 0.2 kg. 4. An object is in uniform circular motion, tracing an angle of 30o every 0.010 s. Find: (a) the period of this motion. (b) the frequency of this motion If the radius of the object’s path is doubled but the period remains the same, what happens to: (c) its speed? (d) its acceleration? Monday, 24 May 2010
- 60. 5. In a circular motion experiment, a mass is whirled around a horizontal circle of radius 0.5 m. A student times four revolutions to take 1.5 s. sinker tube (a) Calculate the speed of the mass around the circle (b) What is the direction of the velocity of the mass? (c) Calculate the centripetal acceleration of the mass. (d) What is the direction of this acceleration? (e) How does the value of the centripetal acceleration compare to the acceleration of gravity? Ex.12A All Q’s Monday, 24 May 2010
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- 70. 12 PHYSICS CIRCULAR MOTION ASSIGNMENT Name 1. A car is travelling around a bend in the road and for a few seconds is in uniform circular motion. (a) The centripetal force is being provided by the road. Name this force. ________________________________________________________________ The car passes over a patch of oil while it is rounding the bend. (b) Describe the path the car will take after it hits the oil patch and explain why this happens in terms of the forces acting. ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ 2. An object in uniform circular motion completes 10 revolutions in 0.4 seconds (a) Find the frequency of this motion. ________________________________________________________________ ________________________________________________________________ (b) Find the period of this motion. ________________________________________________________________ _______________________________________________________________ Monday, 24 May 2010
- 71. 3. A big wheel at a fair spins in a circular path of radius 20 m. Once the wheel has reached a steady speed, a student times each revolution at 13 seconds. (a) Calculate the circumference of the big wheel. ______________________________________ (b) Hence calculate the speed of the big wheel. ______________________________________ ______________________________________ (c) Calculate the centripetal acceleration of each passenger. ________________________________________________________________ ________________________________________________________________ 4. In a circular motion experiment, a mass is whirled around a horizontal circle which has a 0.50 m radius. A student time 4 revolutions to take 2.0 s. (a) Calculate the speed of the mass around the circle. ________________________________________ ________________________________________ (b) What is the direction of the velocity of the mass? ________________________________________ ________________________________________ (c) Calculate the centripetal acceleration of the mass. ________________________________________ ________________________________________ Monday, 24 May 2010
- 72. (d) What is the direction of this acceleration? _______________________________________________________________ (e) How does the value of the centripetal acceleration compare to the acceleration of gravity? ______________________________________________________ 5. Two students go to a fun park for a day where they pay to drive carts around a circular track. The track has a radius of 31.8 m and once the carts are at a maximum speed they complete a lap in 16 s. (a) What is the frequency of the cart’s motion when travelling at maximum speed? __________________________________________ (b) When travelling at maximum speed, calculate the speed of the cart. __________________________________________ __________________________________________ (c) Calculate the acceleration of the cart when travelling at maximum speed. _____________________________________________________________ _____________________________________________________________ _____________________________________________________________ _____________________________________________________________ Monday, 24 May 2010
- 73. The cart has a mass of 150 kg and one of the students, Chris has a mass of 75 kg. (d) Calculate the size of the force acting on Chris and his cart at maximum speed. ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ (e) Chris drives over a patch of oil and loses control of his cart whilst travelling at this maximum speed. On the diagram, draw his path after driving through the oil. 6. Jon and Ana are two ice-skaters. In a practiced skating move, Jon spins Ana around in a horizontal circle. Ana moves in a circle as shown: Jon Ana (a) Draw an arrow on the diagram to show the direction of the tension force that Jon’s arm exerts on Ana at the instant shown. (b) If the radius of the circle is 0.95 m and the tension force in Jon’s arm is 5.00 x 102 N, calculate the speed with which Ana (55 kg) is travelling around the circle. Give your answer to the correct number of significant figures. ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ Monday, 24 May 2010
- 74. (d) While Ana is still moving in a circle on the ice, Jon lets her go. (i) Describe her velocity (speed and direction) after he releases her. ______________________________________________________________ (ii) Explain why Ana travels with this velocity. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ Monday, 24 May 2010

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