Chapter 5

1. Chapter 5 Circular Motion and Gravitation

2. Chapter 5Circular Motion and Gravitation

3. Introduction • Circular motion • Acceleration is not constant • Cannot be reduced to a one-dimensional problem • Examples • Car traveling around a turn • Parts of the motion of a roller coaster • Centrifuge • The Earth orbiting the Sun • Gravitation • Explore gravitational force in more detail • Look at Kepler’s Laws of Motion • Further details about g Introduction

4. Uniform Circular Motion • Path of Motion: Circle • The direction of the velocity is continually changing. • The vector is always tangent to the circle • Uniform circular motion (UCM) is circular motion at constant speed. Section 5.1

5. Circular Motion - Period Period T (s): time for a complete rotation r: radius of circle If n rotations happen in time t: T=t/n. v v Frequency f (s-1, Hertz-Hz): number of rotations per unit of time - f=n/t. r v v T=1/f, f=1/T, Tf=1

6. Tangential (Linear) Velocity C = 2πr = vT or

7. Exercises Set 1 1. A car’s tire rotates at 1200 RPM. a. What is the frequency? f = 1200/60s = 20 Hz b. What is the period? T= 1/f = 1/20 Hz = 0.05s c. What is the speed (R= 0.15 m) v = 2πRf = 2 x 3.14 x 0.15m x 20 Hz = 18.85 m/s

8. Ex 2 Question 1: An object is constrained to move in a circular orbit with radius 1 m. a) How far does the object travel after completing one orbit? b) If the orbital period (time to complete one orbit) is 0.5 s, what is its velocity? a) C = 2πR = 2 x 3.14 x 1 m = 6.28 m b) v = C/T = 6.28 m/0.5s = 12. 56 m/s

9. Ex3 • Problem: The earth is 1.50 x1011 m (93 million miles) from the sun. What is its speed in m/s (neglecting the motion of the sun through the galaxy)? sec

10. Angular Measure The position of an object can be described using polar coordinates—r and θ—rather than x and y. The figure at left gives the conversion between the two descriptions.

11. Angular Measure It is most convenient to measure the angle θ in radians:

12. Angular Speed and Velocity In analogy to the linear case, we define the angular speed: where θ is the angle swept by the radius in the time t. [θ] = rad/s r θ

13. Angular Speed and Velocity The angular velocity is (+) positive when the rotation is ccw and (-)negative if the rotation is cw.

14. Angular Speed and Velocity Relationship between tangential and angular speeds: This means that parts of a rotating object farther from the axis of rotation move faster.

15. Angular Speed, Period, and Frequency From: and we get the relation of the frequency to the angular speed:

16. Uniform Circular Motion and Centripetal Acceleration A careful look at the change in the velocity vector of an object moving in a circle at constant speed shows that the acceleration is toward the center of the circle.

17. Uniform Circular Motion and Centripetal Acceleration The change in the direction of the velocity vector is provided by the centripetal acceleration:

18. Exercises Set 4 • 2) A car is traveling along a circular path of 50 m radius at 22 m/s. a. What is the car’s acceleration? b. How much time to complete a circuit?

19. Example Set 5 • 3) Satellite, radius 1.3 x 107 m, g = 2.5 m/s2. a) What is the speed? b) Period?

20. Ex. 6 n = 0.56 x 60 = 33.44 rev/min

21. Uniform Circular Motion and Centripetal Force The centripetal force is the mass multiplied by the centripetal acceleration. This force is the net force on the object. As the force is always perpendicular to the velocity, it does no work.

22. Centripetal Force Features: • A force causing a centripetal acceleration acting toward the center of the circle. • It causes a change in the direction of the velocity vector • If the force vanishes, the object would move in a straight-line path tangent to the circle (“Centrifugal” effects which come from inertia) !This is not a new force, it is a new application of a force …

23. Centripetal force The centripetal force can be produced in any number of ways (for example): • Due to the tension in a string (pendulum) • Due to friction between tires and the road • Due to gravity

24. Conical Pendulum • A small object of mass m is suspended from a string of length L. The object revolves in a horizontal circle of radius r with constant speed v, as shown. • Find the speed of the object. • Find the tension, T, in the string.

25. y x Conical Pendulum • Need a free body diagram: • T is the force exerted by the string

26. y x Conical Pendulum • The body does not move in the vertical direction …therefore no acceleration here! • Rearranging gives

27. y x Conical Pendulum • In the x direction, there is centripetal acceleration! • Rearranging gives

28. y x Conical Pendulum • Consider the x and y equations: • We have two equations in two unknowns (v, T) • T can be eliminated by dividing 2 by 1 to give • and 1 2

29. Conical Pendulum • And solving for v

30. Conical Pendulum • To find tension, just go back to the y-component equation: • And solve for T

31. Problem Solving Strategy – Circular Motion Section 5.1 • Recognize the principle • If the object moves in a circle, then there is a centripetal force acting on it • Sketch the problem • Show the path the object travels • Identify the circular part of the path • Include the radius of the circle • Show the center of the circle • Selecting a coordinate system that assigns the positive direction toward the center of the circle is often convenient

32. Problem Solving Strategy, cont. • Solve for the quantities of interest • Check your answer • Consider what the answer means • Does the answer make sense Section 5.1 • Identify the principles • Find all the forces acting on the object • A free body diagram is generally useful • Find the components of the forces that are directed toward the center of the circle • Find the components of the forces perpendicular to the center • Apply Newton’s Second Law for both directions • The acceleration directed toward the center of the circle is a centripetal acceleration

33. Horizontal (Flat) Curve • A 1500 kg car moving on a flat, horizontal road negotiates a curve with a radius of 35.0 m. If the static coefficient of friction between the tires and the dry pavement is 0.523, what is the maximum speed with which the car can negotiate the turn?

34. Centripetal Acceleration Example: Car • A car rounding a curve travels in an approximate circle • The radius of this circle is called the radius of curvature • Forces in the y-direction • Gravity and the normal force • Forces in the x-direction • Friction is directed toward the center of the circle Section 5.1

35. Horizontal (Flat) Curve • What is the maximum force of friction FFriction? We can get that from the vertical force balance and the static friction coefficient: • So • And,

36. Car Example, cont. Section 5.1 Since friction is the only force acting in the x-direction, it supplies the centripetal force Solving for the maximum velocity at which the car can safely round the curve gives

37. Horizontal (Flat) Curve • Finish up, • Does it look okay? • Increase μs, vmax increases √ • Increase r, vmax increases √ • Put in numbers

38. Horizontal (Flat) Curve • Reality check • You might want a bit more of a margin of safety here …

39. Angular Acceleration The average angular acceleration is the rate at which the angular speed changes: In analogy to constant linear acceleration:

40. Angular Acceleration If the angular speed is changing, the linear speed must be changing as well. The tangential acceleration is related to the angular acceleration:

41. Angular Acceleration

42. Circular Motion Example: Vertical Circle • The speed of the rock varies with time • At the bottom of the circle: • Tension and gravity are in opposite directions • The tension supports the rock (mg) and supplies the centripetal force Section 5.2

43. Vertical Circle Example, cont. • At the top of the circle: • Tension and gravity are in the same direction • Pointing toward the center of the circle Section 5.2

44. Vertical Circle Example, Final • There is a minimum value of v needed to keep the string taut at the top • Let Ttop = 0 • If the speed is smaller than this, the string will become slack and circular motion is no longer possible

45. Circular Motion Example: Roller Coaster • The roller coaster’s path is nearly circular at the minimum or maximum points on the track • There is a maximum speed at which the coaster will not leave the top of the track: • If the speed is greater than this, N would have to be negative • This is impossible, so the coaster would leave the track Section 5.2

46. Newton’s Law of Gravitation, Equation • Law states: There is a gravitational attraction between any two objects. If the objects are point masses m1 and m2, separated by a distance r the magnitude of the force is Section 5.3

47. Law of Gravitation, cont. Section 5.3 • Note that r is the distance between the objects • G is the Universal Gravitational Constant • G = 6.67 x 10-11 N . m2/ kg2 • The gravitational force is always attractive • Every mass attracts every other mass • The gravitational force is symmetric • The magnitude of the gravitational force exerted by mass 1 on mass 2 is equal in magnitude to the force exerted by mass 2 on mass 1 • The two forces form an action-reaction pair

48. Gravitation and the Moon’s Orbit • The Moon follows an approximately circular orbit around the Earth • There is a force required for this motion • Gravity supplies the force Section 5.3

49. Notes on the Moon’s Motion Section 5.3 • We assumed the Moon orbits a “fixed” Earth • It is a good approximation • It ignores the Earth’s motion around the Sun • The Earth and Moon actually both orbit their center of mass • We can think of the Earth as orbiting the Moon • The circle of the Earth’s motion is very small compared to the Moon’s orbit

50. Gravitation and g • Assuming a spherical Earth, we can consider all the mass of the Earth to be concentrated at its center • The value of r in the Law of Gravitation is just the radius of the Earth Section 5.3