1. Principles of Technology/Physics in Context (PT/PIC) Unit 5 Circular Motion and Gravitation 1

2. Key Objectives At the conclusion of this chapter you’ll be able to: Identify the direction of an object’s velocity when it is undergoing uniform circular motion. Define the terms centripetal acceleration and centripetal force, and identify the directions of these quantities when an object undergoes uniform circular motion.

3. Key Objectives At the conclusion of this chapter you’ll be able to: State the equations for calculating centripetal force and centripetal acceleration. Solve problems involving uniform circular motion. Define the term period of revolution and relate it to the equations of uniform circular motion. State Kepler’s three laws of planetary motion.

4. 5.1 INTRODUCTION In this chapter we study another aspect of motion in a plane, namely, motion in a circle. The subject of circular motion leads, in turn, to a study of gravitation because both natural and artificial satellites travel in nearly circular paths.

5. 5.2 CIRCULAR MOTION An object moves in a circular path at constant speed as indicated in the diagram below.

6. 5.2 CIRCULAR MOTION At various points in the path, the direction of the velocity, v, of the object is tangent to the circle, as shown in the diagram.

7. 5.2 CIRCULAR MOTION Since the direction of the object’s motion is changing, the object must be subjected to an unbalanced force and is, therefore, accelerating.

8. Assessment Question 1 All of the following are TRUE statements concerning the diagram EXCEPT A.The object moves in a circular path at a constant speed. B.The direction of the velocity, v, of the object is tangent to the circle at various points. C.The direction of the object’s motion is always changing. D.The forces acting on the object are balanced. E.The object is accelerating.

9. 5.2 CIRCULAR MOTION Here is an example of an object that accelerates even though its speed does not change.

10. 5.2 CIRCULAR MOTION This is due to the fact that acceleration is a change in the velocity of an object, and this change can be in the magnitude and/or direction of the velocity.

11. Assessment Question 2 All of the following are TRUE statements concerning the diagram EXCEPT A.The object that accelerates even though its speed does not change. B.Acceleration is a change in velocity magnitude. C.Acceleration is a change in velocity direction. D.The magnitude of the velocity of the object is always changing. E.The direction of the velocity of the object is always changing.

12. 5.2 CIRCULAR MOTION The centripetal acceleration of an object is calculated by means of this equation: where v is the linear speed of the object and r is the radius of the circular path.

13. 5.2 CIRCULAR MOTION The unbalanced force associated with the centripetal acceleration is called the centripetal force (F c ). It also points toward the center of the circular path and is given by this relationship:

14. 5.2 CIRCULAR MOTION We can think of the centripetal force as the force needed to keep the object in its circular path.

15. 5.2 CIRCULAR MOTION If the mass of the object were increased, or if the speed of the object were increased, more force would be needed to keep the object in its circular path. If, however, the radius of the path were increased, less force would be required for this purpose.

16. 5.2 CIRCULAR MOTION PROBLEM A 5.0-kilogram object travels clockwise in a horizontal circle with a speed of 20. meters per second, as shown in the diagram. The radius of the circular path is 25 meters.

17. 5.2 CIRCULAR MOTION PROBLEM (a) Calculate the centripetal acceleration on the object.

18. 5.2 CIRCULAR MOTION SOLUTION (a) Calculate the centripetal acceleration on the object.

19. ASSESSMENT QUESTION 3 Calculate the centripetal acceleration on the object. a c = v 2 /r a c = (250 m/s) 2 / 5200 m A.0.48 m/s 2 B.12 m/s 2 C.550 m/s 2 D.13000 m/s 2 E.325000 m/s 2

20. 5.2 CIRCULAR MOTION PROBLEM (b) Calculate the centripetal force on the object.

21. 5.2 CIRCULAR MOTION SOLUTION (b) Calculate the centripetal force on the object.

22. ASSESSMENT QUESTION 4 Calculate the centripetal force on the object. F c = ma c F c = 50 kg / 0.48 m/s 2 A.0.09 N B.24 N C.55 N D.100 N E.548 N

23. 5.2 CIRCULAR MOTION The period of revolution is the time an object takes to complete one revolution in a circular path. In one revolution, the distance the object travels equals the circumference of the circle (2πr).

24. 5.2 CIRCULAR MOTION PROBLEM (c) In the position shown, indicate the direction of the velocity and of the centripetal force of the object.

25. 5.2 CIRCULAR MOTION Measuring the speed of an object in circular motion is not always easy, but we can measure the speed of the object indirectly by making use of a quantity known as the period (T).

26. 5.2 CIRCULAR MOTION Using the equation v = d/t we substitute and obtain

27. 5.2 CIRCULAR MOTION PROBLEM An object traveling in a circular path makes 1200 revolutions in 1.0 hour. If the radius of the path is 10. meters, calculate the speed of the object.

28. 5.2 CIRCULAR MOTION SOLUTION If the object makes 1200 revolutions in 1.0 hour (3600 s), the time (1) for one revolution is

29. Assessment Question 5 An object traveling in a circular path makes 1800 revolutions in 1.0 hr (3600s). Calculate the period (T) of the object. T = time / revolutions = 3600 s / 1800 rev = A.0.5 seconds B.2.0 seconds C.13 seconds D.180 seconds E.6480 seconds

30. 5.2 CIRCULAR MOTION SOLUTION The speed of the object is

31. Assessment Question 6 An object traveling in a circular path with a radius of 170 m (r) with a period of 50 seconds (T). Calculate the velocity (v) of the object. v = 2πr / T = (2 ∙ 3.14 ∙ 170 m) / 50 s = A.0.5 m/s B.3.4 m/s C.21 m/s D.340 m/s E.1068 m/s

32. 5.2 CIRCULAR MOTION The equation for the centripetal acceleration can be rewritten in terms of the period of revolution rather than the speed of the object:

33. Assessment Question 7 An object traveling in a circular path with a radius of 170 m (r) with a period of 50 seconds (T). Calculate the object’s acceleration (a). a c = 4π 2 r/ T 2 = (4 ∙ 3.14 2 ∙ 170 m) / (50 s) 2 = A.2.7 m/s 2 B.44 m/s 2 C.130 m/s 2 D.670 m/s 2 E.2100 m/s 2

34. 5.3 KEPLER’S LAWS The motions of the planets in the heavens were of great interest in the sixteenth century.

35. 5.3 KEPLER’S LAWS The German astronomer Johannes Kepler summarized the laws of planetary motion; this work was based on data collected by the Danish astronomer Tycho Brahe. Kepler proposed three laws:

36. 5.3 KEPLER’S LAWS FIRST LAW The path of each planet about the Sun is an ellipse with the Sun at one focus.

37. 5.3 KEPLER’S LAWS SECOND LAW Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas of space in equal periods of time.

38. 5.3 KEPLER’S LAWS SECOND LAW.

39. 5.3 KEPLER’S LAWS SECOND LAW In the diagram above, the time it takes for the planet to move between points P and Q (Δt 1 ) is equal to the time it takes the planet to move between points R and S (Δt 2 ). As a result of Kepler’s second law, we can conclude that area A 1 is equal to Area A 2

40. 5.3 KEPLER’S LAWS THIRD LAW The cube of a planet’s average distance from the Sun (r 3 ) divided by the square of its period (T 2 ) is a constant for all planets.

41. 5.3 KEPLER’S LAWS THIRD LAW The table below illustrates the validity of Kepler’s third law for our solar system. In fact, Kepler’s three laws hold for any satellite orbiting a heavenly body.

42. Assessment Question 8 All of the following are Kepler’s Laws EXCEPT: A.The path of each planet about the Sun is an ellipse with the Sun at one focus. B.Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas of space in equal periods of time. C.The cube of a planet’s average distance from the Sun (r 3 ) divided by the square of its period (T 2 ) is a constant for all planets. D.The motion of the moons of planets can not be predicted by Kepler’s Laws.

43. 5.3 KEPLER’S LAWS

44. Assessment Question 9 All of the following are Kepler’s Laws EXCEPT: A B. CD.

45. Conclusion The unbalanced force responsible for this acceleration is called a centripetal force and is also directed toward the center of the circle.

46. Conclusion An object that travels in a circular path experiences an acceleration directed toward the center of the circle and known as a centripetal acceleration.

47. Conclusion The study of planetary motion was greatly advanced by Kepler, who deduced three laws that provided a mathematical basis for this motion.

48. Assessment Question 10 An asteroid traveling around the Sun at an average distance of 25 Gigameters (r) with a period of 0.0685 years (T) Is the object following Kepler’s Laws of Motion. 3.33 x 10 6 ≈ r 3 / T 2 = (25 Gm) 3 / (0.0685 year) 2 = A.Yes B.No C.All of the Laws EXCEPT the first Law D.All of the Laws EXCEPT the second Law E.All of the Laws EXCEPT the third Law