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Module 2.1 – Projectile Motion

Module 2.1 – Projectile Motion. Objects launched in the air follow parabolic trajectories. Both kinematics and dynamics principles can be used to study the motion of these objects. Suitable methods will be learned to predict their behaviour. Horizontal Projectiles. No horizontal forces!.

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Module 2.1 – Projectile Motion

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  1. Module 2.1 – Projectile Motion Objects launched in the air follow parabolic trajectories. Both kinematics and dynamics principles can be used to study the motion of these objects. Suitable methods will be learned to predict their behaviour.

  2. Horizontal Projectiles No horizontal forces!

  3. Falling Objects • Consider two objects, one dropped and one launched horizontally Equal times!

  4. Horizontal Motion No horizontal forces No horizontal acceleration

  5. Vertical Motion Only force of gravity For horizontally launched projectiles

  6. Example A movie stunt driver drives a car off of a cliff that is 70.0 m high. If the car has a speed of 90.0 km/h, how far away does the car land from the base of the cliff?

  7. Solution

  8. Check Your Learning A ball thrown horizontally at 12.0 m/s from the roof of a building lands 36.0 m from the base of the building. How high is the building?

  9. Projectiles Launched at Angles

  10. Parabolic Projectile Path if it returns to the same height from which it was launched

  11. Example 1 A football is kicked with a speed of 21.0 m/s at an angle of 37.0o to the ground. • How much later does it hit the ground? • How far away does it land?

  12. Solution

  13. Solution

  14. Quadratic Formula

  15. Example 2 A cricket ball is hit by a bat at a height of 0.85 m. The bat gives the ball an initial velocity of 112 km/h, at an angle of 15.0o above the horizontal. How far away does the ball land?

  16. Solution Quadratic formula

  17. Check Your Learning A soccer ball is kicked with an initial velocity of 22.4 m/s at an angle of 34.0o above the horizontal. It hits the top of a soccer net that is 2.70 m high. • How many seconds have passed between the time the ball was kicked and when it hit the net?

  18. Check Your Learning Since both times are positive, both are possible answers.

  19. Check Your Learning • Why are there two possible answers to a)? Remember that a projectile follows a parabolic trajectory. There are two possible answers to a) because the ball is at a height of 2.70 m twice – once on the way up and again on the way back down. The ball may have hit the goal after 0.238 s when it was on its way up or it may have hit it after 2.31 s when it was on its way back down.

  20. Module Summary In this module you learned that • An object launched horizontally will hit the ground at the same time as an object that is dropped; in other words, the vertical motion of a projectile is not affected by its horizontal motion. • The kinematics equations that were learned in Unit 2 can be applied separately to the horizontal and vertical motion of a projectile, if the initial velocity of the projectile is first broken up into components.

  21. Module 2.2 – Uniform Circular Motion • An object undergoing circular motion is another example of two-dimensional motion that can be analyzed using Newton’s Laws and kinematics. This module will identify the conditions necessary for an object to maintain uniform circular motion.

  22. Period and Frequency • Period – the amount of time to complete one revolution. Since it is a measure of time, the SI unit is seconds (s). • Frequency –the number of revolutions per second. A revolution per second is referred to as a hertz (Hz).

  23. Centripetal Acceleration

  24. Example A child on a merry-go-round is moving with a constant speed of 2.65 m/s when 1.60 m away from the centre of the merry-go-round. • In what direction is the child accelerating? • Calculate the acceleration of the child. • What is the frequency of the spinning merry-go-round?

  25. Solution • Since the merry-go-round is moving with a constant speed, there is no linear acceleration. The only acceleration is the centripetal acceleration due to moving in a circle. The child’s acceleration is therefore inward, toward the centre of the merry-go-round.

  26. Solution

  27. Check Your Learning A compact disc spins with a frequency of 8.00 Hz when data is being read at a distance of 2.50 cm from the centre. What is the centripetal acceleration of a point at this location? • As always, the centripetal acceleration is toward the centre of the disc.

  28. Centripetal Force Acceleration requires a net force Centripetal acceleration requires a centripetal force Centripetal force is not an actual force and should not be included in any free body diagram – it is a force requirement.

  29. Directions Consider a ball being swung on a string in a circle as shown below: • Centripetal force is always directed toward the centre of the circle • Centripetal acceleration is also always directed toward the centre of the circle • Velocity is perpendicular to the radius of the circle (tangential)

  30. Example 1 What is the maximum speed (in km/h) at which a 1300 kg car can safely travel around a circular track of radius 80.0 m if the coefficient of friction between the tire and the road is 0.30?

  31. Solution

  32. Example 2 In the previous example, consider a person sitting in the car who appears to feel a force pushing them toward the outside of the turn. As can be seen in the free body diagram used to solve the problem, however, there is no force toward the outside of the turn. Explain this discrepancy.

  33. Solution There is no force pushing the person out from the centre of the turn. Since the person is accelerating, they are not in an inertial reference frame. So Newton’s Laws cannot be applied. If we look at the person from an inertial reference frame (the ground), we see that their body is simply trying to continue going forward in a straight line. Since the car is turning, it applies a force on the person to make them turn as well. Because the person’s body is trying to continue going in a straight line (which is toward the outside door of the car), it feels as if there is a force pushing them in this direction. This fictitious force that the person “feels” is referred to as the centrifugal force.

  34. Check Your Learning How large must the coefficient of friction be between the tires and the road if a car is to round a level curve of radius 62 m at a speed of 55 km/h?

  35. Check Your Learning

  36. Vertical Circles More than one force contributes to the centripetal force

  37. Example 3 A 0.110 kg ball is being swung in a vertical circle on a 58.5 cm string. Calculate the tension in the string at the bottom of the path if the speed is 6.25 m/s.

  38. Solution

  39. Check Your Learning A 0.140 kg ball is being swung in a vertical circle on a 39.2 cm string. Calculate the tension in the string at the top of the path if the speed is 2.41 m/s.

  40. Check Your Learning

  41. Module Summary In this module you learned that • The centripetal acceleration of an object is always toward the centre of the circle and is given by • Newton’s Second Law can be applied to circular motion where the net force is inward and is referred to as the centripetal force

  42. Module 2.3 – Universal Gravitation Newton’s Law of Universal Gravitation can be applied to interactions between all objects and can be used in conjunction with circular motion principles to predict the behavior of planets and other celestial bodies.

  43. Force of Gravity • What determines the force of gravity? • Henry Cavendish found the constant G that was needed to form an equation where

  44. Example 1 Two people are standing 1.3 m apart (centre to centre). One person has a mass of 65.0 kg and the other person has a mass of 78.0 kg. • What is the force that the heavier person exerts on the lighter person? • What is the force that the lighter person exerts on the heavier person? • Why do the two people not move toward one another?

  45. Solution • Because of Newton’s 3rd Law, the lighter person exerts the same force on the heavier person as the heavier person does on the lighter person. The force is therefore also

  46. Solution • This force is extremely small compared to other forces acting on the two people, such as friction and the forces of gravity exerted by many other objects.

  47. Acceleration due to Gravity On any planet and

  48. Example 2 The radius of the Earth was actually calculated over 2000 years ago. Using today’s value of for the radius of the earth, calculate the mass of the earth.

  49. Planetary Constants

  50. Check Your Learning A 710 kg spacecraft is 12800 km above the earth's surface. • Calculate the force of gravity on the spacecraft.

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