Chapter 8. Dynamics II: Motion in a Plane

1. Chapter 8. Dynamics II: Motion in a Plane Chapter Goal: To learn how to solve problems about motion in a plane, especially circular motion.

2. Student Learning Objectives – Ch. 8 • To understand the dynamics of uniform circular motion. • To learn the basic ideas of orbital motion. • To answer “How does the water stay in the bucket?” and related questions.

3. Uniform Circular Motion The acceleration of uniform circular motion points to the center of the circle. Thus the acceleration vector has only a radial component ar. This acceleration is conveniently written in the rtz-coordinate system as

4. Uniform Circular Motion The acceleration of uniform circular motion points to the center of the circle. Why? – Let’s investigate… with a ball on a string, and on a track (of sorts)

5. Dynamics of Uniform Circular Motion Newton’s second law in terms of the r-, t-, and z-components, is as follows: There is no such thing as a “centripetal force”. It is a normal, tension or other kind of force that results in a centripetal acceleration.

6. Mass on a turntable What kind of force provides the centripetal acceleration? a. tension b. static friction c. kinetic friction d. normal

7. Mass on a turntable The turntable slowly speeds up. Which mass will slide off first? • m because r > r2 • m2 because r > r2 • same time, since ω is the same for both masses. m2•

8. Car at constant speed in a straight line – No net force on the car. y n x fr fs FG

9. Car turning a corner at constant speed Is there a net force on the car as it negotiates the turn? Where did it come from?

10. Car turning a “circular” corner at constant speed When the driver turns the wheel the tires turn. To continue along a straight line, the car must overcome static friction and slide. If the static friction force is less than the maximum, the tire cannot skid and so has no choice but to roll in the direction of the turn. fs-max = μs |n| skid…. blue arrow represents direction car would slide in the absence of friction

11. End of chapter problem #33 A 5 g coin is placed 15 cm from the center of a turntable. The table speeds up to 60 rpm. • Does the coin slide off? • If not, at what speed (in rpms) will it slide? Us = .80, uk = .50

12. End of chapter problem #33 A 5 g coin is placed 15 cm from the center of a turntable. The table speeds up to 60 rpm. • Does the coin slide off? - No • If not, at what speed (in rpms) will it slide? – 69 rpms

13. The banked curve (EOC #8) • A highway curve of radius 500 m is designed for traffic moving at 90 km/hour. What is the correct banking angle for the road?

14. The banked curve (EOC #8) • A highway curve of radius 500 m is designed for traffic moving at 90 km/hour. What is the correct banking angle for the road? • The radial axis is in the true horizontal direction, relative to the angle of bank.

15. The banked curve (EOC #8) • A highway curve of radius 500 m is designed for traffic moving at 90 km/hour. What is the correct banking angle for the road? • θ = tan-1 (v02/rg) • Answer is 7.3° • The speed which you can travel on a banked curve with no friction:

16. Comparison of level curve and banked curve Level curve with friction: fs = mv2/r. As speed increases, so does frictional force until fsmax = µs mg and vmax = (µs rg)1/2 . At any v < vmax one can negotiate the curve (i.e move in uniform circular motion). Banked curve: mg tan θ = mv2/r. There is only one value of speed v = (rg tan θ)1/2 where one can negotiate the curve (move in a circle) without some help from friction.

17. mg tan θ = mv02/rWhat happens if v >v0 ? • object slides up • object slides down • object moves around circle faster • object leaves the ground

18. mg tan θ = mv2/r.What happens if v >v0 ? For v >v0, mg tan θ is too small and car will slide up the bank, without friction. If there is friction, it points down and towards the circle, making a range of speeds >v0 that can negotiate the curve.

19. mg tan θ = mv02/rWhat happens if v < v0 ? n has decreased, but mg has not, so there is a component of net force in the negative z direction.

20. mg tan θ = mv2/rWhat happens if v < v0 ? For v<v0, mg tan θ is too large and car will slide down the bank without friction. If there is friction, it points up and away from the circle, making a range of speeds <v0 that can negotiate the curve.

21. Banked Turn with friction A 100-m radius curve in a road is constructed with a bank angle of 100. On a rainy day, the static coefficient of friction can be as low as 0.10. What are the highest and lowest speeds at which a car can make this turn without sliding in those conditions?

22. Comparison of level curve and banked curve Greatest speed is 16.6 m/s Slowest speed is 8.57 m/s

23. A ride at a carnival consists of chairs that are swung in a circle by 15.0-m cables attached to a vertical rotating pole, as shown in the drawing. The total mass of a chair and occupant is 179 kg. a) Draw a free body diagram for this situation. b) Determine the Tension in cable c) speed (v) of chair

24. A ride at a carnival consists of chairs that are swung in a circle by 15.0-m cables attached to a vertical rotating pole, as shown in the drawing. The total mass of a chair and occupant is 179 kg. a) Draw a free body diagram for this situation. b) Determine the Tension in cable c) speed (v) of chair 60° • 3510 N • 14.9 m/s

25. Circular Orbits

26. Circular Orbits An object moving around the earth in a circular orbit of radius r at speed vorbit will have centripetal acceleration of That is, if an object moves parallel to the surface with the speed then the free-fall acceleration provides exactly the centripetal acceleration needed for a circular orbit of radius r. An object with any other speed will not follow a circular orbit.

27. Stop to Think – Newtonian Analysis The roller coaster is moving. Is the normal force at the cars (and people) at the bottom of the moving roller coaster: a)greater than, b)less than c)equal to FG? What would the scale say? Use a Newtonian analysis to support your answer.

28. Stop to think more Fr-net = n – FG = mv2/r What changes (n, FG, both, neither) if the carts are going faster? a)n, b)FG, c)both, c)neither n FG

29. Stop to think more Fr-net = n – FG = mv2/r What does it mean physically if n = FG in this situation? What are the carts doing? Can n ever be less than FG for this situation? In what case(s)? n FG

30. EOC #14 • A roller coaster car with a scale built into the seat (I hate when that happens!) goes through a dip with a radius of 30 m. The scale reading increases by 50% at the bottom of the dip. What is car’s speed?

31. EOC #14 - answer The scale reading is the normal force: Fr-net = n – FG = mv2/r v = 12.1 m/s n FG

32. Over the Top A car is rolling over the top of a hill at speed v. At this instant: • n > FG. • n < FG. • n = FG. • We can’t tell about n without knowing v. • Use a newtonian analysis

33. A car is rolling over the top of a hill at speed v. At this instant, Fr-net = FG - n = mv2/r n < FG If you were sitting on a scale now, what would happen as you went faster? How fast could you go? Explain.

34. It is easy to show that the maximum speed you could attain and still stay on the road is What is the physical significance of n becoming zero for this instance?

35. Tension (Workbook #8) A ball swings around in a vertical circle on a string. Is the tension at the bottom of the swing greater than, less than or equal to FG? Use a Newtonian analysis to support your answer.

36. Tension T greater than FG , the ball is in circular motion, so there has to be a net force into the circle. Fr-net = T – FG = mv2/r If T=FG, what does that mean physically?

37. What about at the top? Is the tension at the top of the swing greater than, less than or equal to FG? Use a Newtonian analysis to support your answer.

38. What about at the top? For this situation: FG + T = mv2/r Can T be 0 for this situation? Can T be less than zero? (i.e. away from the circle)? • What happens physically? • What happens mathematically?

39. What about at the top? Derive an expression for minimum speed the ball can have and reach the top of the circle without falling in terms of T, m, r, and any constants:

40. What about at the top? The mimimum speed occurs when T = 0, since T cannot be negative • mg = mv2/r • v = • This value is known as the vc, the critical speed.

41. Why Does the Water Stay in the Bucket? • For this situation, what happens to the normal force and the gravitational force, as the speed decreases? • Is is physically possible for n< 0 (for the track to pull on the cart)? • What is the minimum speed necessary for the roller coaster to stay on the track?

42. Why Does the Water Stay in the Bucket? • mg = mv2/r • v = • This value is known as the vc, the critical speed. • If the track isn’t pressing against the cart, the cart can’t press against the track, i.e. it loses contact and falls.

43. Why Does the Water Stay in the Bucket? • The critical angular velocity ωcis that at which gravity alone is sufficient to cause circular motion at the top. • Since ω = v/r:

44. Why Does the Water Stay in the Bucket?

45. Vertical Circular Motion Problem An circular clothes dryer (radius = 0.32 m) is designed so that the clothes tumble gently as they dry. This means that when a piece of clothing reaches some angle above the horizontal, it loses contact with the cylinder wall and falls. How many revolutions per second should the dryer make so this occurs at θ = 700 ? Note: the axes shown are not necessarily the best choice for this problem. They simply serve to show the orientation of θ with true horizontal and vertical.

46. Vertical Circular Motion Problem Once an appropriate free body diagram is chosen, all that is left is to write Newton’s 2nd Law for circular motion for the radial axis: mg sin θ = mω2r ω = .85 rps r θ t FG

47. Nonuniform Circular Motion

48. Nonuniform Circular Motion • The force component (Fnet)r creates a centripetal acceleration and causes the particle to change directions. • The component (Fnet)t creates a tangential acceleration    and causes the particle to change speed. • Force and acceleration are related to each other   through Newton’s second law.

49. EOC #19 A toy train is released with an angular speed of 30 rpm on a 1.0 m-diameter track. The coefficient of rolling friction is 0.10 (This is a resistive friction). a. What is the magnitude of angular acceleration immediately after being released? b. How long does it take the train to stop?

50. EOC #19 - answer a. What is the magnitude of angular acceleration immediately after being released? • Use Newton’s 2nd Law to solve for Fnet in the t direction, then calculate angular accleration: α = Fnet/mr = 1.96 rad/s2. Note this opposes motion.