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Lesson 2

Honors/AP Physics: An Electronic Textbook Chapter 5 Moving in a Circle: Circular Motion, Gravity and Torque. Lesson 2. Centripetal Forces in 1-Dimension. Introduction. Why do amusement park rides feel the way they do?. Introduction. Why do they employ so many circles or parts of

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Lesson 2

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  1. Honors/AP Physics: An Electronic TextbookChapter 5Moving in a Circle: Circular Motion, Gravity and Torque

  2. Lesson 2 Centripetal Forces in 1-Dimension

  3. Introduction Why do amusement park rides feel the way they do?

  4. Introduction Why do they employ so many circles or parts of circles?

  5. Introduction In this lesson we will apply Newton’s 2nd Law to circular motion situations to explain why you feel the way you do at different parts of various rides.

  6. Introduction In Lesson 5-1, we learned that when an object moves in a circle, its velocity is tangent to its path. v v v Velocity is Tangent to ThePath v v v R v v v v

  7. Introduction The period of revolution (T) is defined as the length of time to complete one revolution. The frequency (f) is defined as the number of revolutions in a set period of time. Frequency and period are related by the equation f = 1/T. v v v f = 1/T v v v R v v v v

  8. Introduction The speed of an object moving in a circle is determined by the equation vc = 2pR/T, where R is the radius of the circle and T is the period. v v v f = 1/T v v v R v v v v

  9. v v v v a a a a v v a a v v a a a a v v a a v v a a a a v v v v Introduction Even though the speed of an object moving in a circle may remain constant, its direction is constantly changing. Because acceleration is defined as the rate of change of velocity, an object that is changing direction is accelerating. Centri petal Acceleration Center toward

  10. v v v v a a a a v v a a v v a a a a v v a a v v a a a a v v v v Introduction The direction of the acceleration of an object moving in a circle is toward the center of the circle. We call this kind of acceleration “centripetal”. Centri petal Acceleration Center toward

  11. v v v v a a a a v v a a v v a a a a v v a a v v a a a a v v v v Introduction The equations for centripetal acceleration are • a = v2/R • a = 4p2R/T2 • a = 2pv/T a is the acceleration, v is the speed, R is the radius and T is the period. Centri petal Acceleration Center toward

  12. v v v Fg v v FT v v v v Introduction Centripetal acceleration is caused by the same forces discussed in previous chapters. Tension, Gravity, Friction and Normal forces may each cause centripetal acceleration if they happen to act toward the center. FN FFr

  13. v v v Fg v v FT v v v v Introduction In each situation we analyze, we will use Fnet = ma to solve. The only difference between circular and linear situations is that in the circular situations the acceleration points toward the center and equals v2/R. FN FFr

  14. C D B A Example Problem #1 Example 1: A 100 kg man rides a fast-moving Ferris Wheel. The wheel has a radius of 50 meters and the man moves at a constant speed of 10 m/s (note: this speed is significantly faster than most real Ferris Wheels.) How large is the Normal force acting on the man at position A? Question 1a: Draw all forces acting on the man and his acceleration at A.

  15. C D B FN a A FW Example Problem #1 Answer 1a: The forces and acceleration are shown on the diagram. The Weight point down. The Normal force from the seat points up. The acceleration points up because it points toward the center of the circle. Question 1b: Use Newton’s 2nd Law to determine the size of the Normal force acting on the man.

  16. + FN a A FW - Example Problem #1 Answer 1b: 1200 n. Setting the direction of the acceleration (up) to be positive: FNet = FN – FW = ma FN – mg = mv2/R FN = mg + mv2/R = m(g + v2/R) FN = (100 kg)(10 m/s2 + (10 m/s)2/50 m) FN = 1200 n

  17. + FN = 1200 n a A FW = 1000 n - Example Problem #1 Answer 1b: FN = 1200 n. The Normal force is a measure of how heavy the man feels in the ride. If he were sitting on a bathroom scale, it would read his Normal force, 1200 n, not his Weight of 1000 n. The man feels heavier than his Weight at point A. Note that this is the same feeling he would have if he were in an elevator accelerating upward at 2 m/s2.

  18. Example Problem #1 Answer 1b: FN = 1200 n. You probably have noticed that when you are at the bottom of a curve in a roller coaster you feel pressed down into your seat – i.e. you feel heavier than you actually are.

  19. Example Problem #1 Answer 1b: FN = 1200 n. We define a Force Factor (F.F.) as the ratio of the Normal force to the Weight. F.F. = FN/FW In this case, the F.F. equals (1200 n)/(1000 n) = 1.2 Sometimes this is called 1.2 g’s + FN = 1200 n a A FW = 1000 n -

  20. Example Problem #1 F.F. = FN / FW = 1.2 g A F.F. of 1.2 or 1.2 g’s means that the rider feels as though he weighs 1.2 times as great as he really does. Note that the F.F. does not depend on the Weight of the person. Any rider in this situation will feel as though he/she weight 1.2 times his/her actual weight. + FN = 1200 n a A FW = 1000 n F.F. = 1.2 g -

  21. C D B A Example Problem #1 Question 1c: Determine the size and direction of the Normal force acting on the rider at point C, the top of the ride.

  22. FN C a FW D B A Example Problem #1 Answer 1c: FN = 800 n The Normal force still points up from the seat and the Weight still points down. The acceleration, however, now points down – but still toward the center of the circle. The upward Normal force should be less than the downward Weight.

  23. - FN a C FW + Example Problem #1 Answer 1c: FN = 800 n Setting the direction of the acceleration (down) to be positive: FNet = FW – FN = ma mg - FN = mv2/R FN = mg - mv2/R = m(g - v2/R) FN = (100 kg)(10 m/s2 - (10 m/s)2/50 m) FN = 800 n

  24. Example Problem #1 Question 1d: What is the Force Factor at point C and how does the rider experience that ride at that point? - FN = 800 n a C FW = 1000 n +

  25. Example Problem #1 Answer 1d: F.F. = FN/FW = (800 n)/(1000 n) = 0.8 g. The rider feels lighter – 80% of his normal weight. This is the same feeling as if accelerating downward in an elevator at 2 m/s2. - FN = 800 n a C FW = 1000 n +

  26. Example Problem #1 Answer 1d: F.F. = 0.8 g. You have probably experienced that when you go fast over a hill, you feel lighter – you generally get a queasy feeling in your abdomen due to the muscles not having to work as hard to suspend your internal organs.

  27. C D B A Example Problem #1 Question 1e: What is the fastest speed the rider can travel at on the ride without losing contact with his seat at point C? FN a FW

  28. C D B A Example Problem #1 Answer 1e: 22 m/s. The faster the speed, the greater the acceleration caused by a smaller and smaller Normal force. “Losing contact” means that the Normal force becomes zero. FNet = FW – FN = ma mg = mv2/R v = √(Rg) = √((50 m)(10 m/s2)) FN a FW

  29. C D B A Example Problem #1 Question 1f: Draw the forces acting on the rider at point B. What is the direction of the acceleration? What force(s) cause(s) the acceleration?

  30. C D B A Example Problem #1 Answer 1f: The forces and acceleration are drawn below. The Normal force from the seat is still upward, the Weight is still downward. The acceleration is to the right, toward the center of the circle. This is caused by the Normal force from the back of the seat. The rider is trying to move straight up, but the back of the seat pushes him/her into a circle. FN, Bottom of Seat a FN, Back of Seat FW

  31. FN, bottom of the seat a FN, back of the seat FW Example Problem #1 Question 1g: How large are each of the forces acting on the rider at B? What is the vertical force factor? What is the horizontal force factor? How does the rider feel at point B? B

  32. Example Problem #1 Answer 1g: FN, bottom = 1000 n; FW = 1000 n; FN, back = 200 n; FFvertical = 1; FFhorizontal = 0.2 y: FNet y = FN, bpttpm – FW = may = 0 FN, bottom = FW = mg = 1000 n Note that the acceleration is horizontal, therefore the vertical component of acceleration (ay) is zero. FN, bottom of the seat a B FN, back of the seat FW

  33. Example Problem #1 Answer 1g: FN, bottom = 1000 n; FW = 1000 n; FN, back = 200 n; FFvertical = 1; FFhorizontal = 0.2 FFvertical = FN, bottom / FW = (1000 n)/(1000 n) = 1 Vertically, the rider feels normal weight. FN, bottom of the seat a B FN, back of the seat FW

  34. Example Problem #1 Answer 1g: FN, bottom = 1000 n; FW = 1000 n; FN, back = 200 n; FFvertical = 1; FFhorizontal = 0.2 x: FNet x = FN, back = max = mv2/R FN, back = (100 kg)(10 m/s)2/(50 m) = 200 n FN, bottom of the seat a B FN, back of the seat FW

  35. Example Problem #1 Answer 1g: FN, bottom = 1000 n; FW = 1000 n; FN, back = 200 n; FFvertical = 1; FFhorizontal = 0.2 FFhorizontal = FN, back / FW = (200 n)/(1000 n) = 0.2 Horizontally, the rider feels pushed back into his seat with a force equal to 20% of his weight.. FN, bottom of the seat a B FN, back of the seat FW

  36. C B D A Example Problem #2 In this ride, the passengers sit in a car that is rotated in a vertical circle at a constant speed. Unlike the Ferris Wheel, the riders are upside down at C and parallel to the ground at B and D. The ride has a radius of 8.0 meters and it takes 4.0 seconds to make one complete revolution. Question 2a: What is the size of the Normal force acting on a 50 kg girl at point A?

  37. Example Problem #2 Answer 2a: FN = 1500 n. The analysis is similar to at point A on the Ferris Wheel, except that we know the period (4.0 s) instead of the speed. FNet = FN – FW = ma FN – mg = m4p2R/T2 FN = mg + m4p2R/T2 FN = (50 kg)(10 m/s2 + 4p2(8.0 m)/(4.0)2) FN = 1500 n C B FN D a A FW

  38. Example Problem #2 Question 2b: What is the Force Factor at point A and how does the rider feel at that point? C B FN D a A FW

  39. Example Problem #2 Answer 2b: F.F. = FN /FW = (1500 n)/(500 n) = 3 The girl will feel heavy - 3 times her normal weight. C B FN D a A FW

  40. C B D A Example Problem #2 Question 2c: What is the size and direction of the Normal force acting on the girl at point C? Hint: it is not obvious which way the Normal force points – it depends on the speed. Just pick a direction and solve. If the value for the Normal force is negative, then you know you picked the wrong direction.

  41. C B D A Example Problem #2 Answer 2c: FN = 500 n down. If the girl feel the Normal force from the seat, then it repels her and points down. If she feels the force from the shoulder harness, then it repels her and points up. FN a FW

  42. Example Problem #2 Answer 2c: FN = 500 n down. In slower rides, such as the space shuttle shown here that is upside down at almost rest, the force is from the shoulder harness and points upward. For faster rides, like the roller coaster or the ride you are analyzing, the force is felt at the seat and points down.

  43. C B D A Example Problem #2 Answer 2c: FN = 500 n down. For this calculation, we will assume that the Normal force points downward. If we get a positive number for the answer, then we know that the direction of the force was correct. Because the acceleration points down, down will be positive. FN a FW +

  44. C B D A Example Problem #2 Answer 2c: FN = 500 n down. FNet = FN + FW = ma FN + mg = m4p2R/T2 FN = m4p2R/T2 - mg FN = (50 kg)(4p2(8.0 m)/(4.0)2 - 10 m/s2 ) FN = 500 n The answer is positive so the direction is correct. FN a FW +

  45. C B D A Example Problem #2 Question 2d: What is the force factor at point C? How does the girl feel at this point? If she closed her eyes, would she feel right-side-up or upside- down? FN a FW +

  46. C B D A Example Problem #2 Question 2d: Normal Weight and right-side-up. F.F. = FN /FW = (500 n)/(500 n) = 1 The girl will feel her normal weight. FN a FW +

  47. FN a C FW B + D A Example Problem #2 Question 2d: Normal Weight and right-side-up. If the girl were sitting in a chair on the ground, the Normal force would point toward her head, rather than toward her feet. At point C, the Normal force also points toward her head. She feels just as if she were sitting in a chair on the ground. FN points toward head FN

  48. v Top View R Side View Example Problem #3 A car makes rounds a curve with a radius of curvature ‘R’. The coefficient of static friction between the tires and the road is ‘m’. What is the maximum speed ‘v’ the car can travel without sliding off the road? (Problem continued on next slide)

  49. Example Problem #3 Radius of curvature represents the radius if the curve were to continue into an entire circle. Question 3a: Draw all the forces acting on the car and its acceleration on the side view. v Top View R Side View

  50. Example Problem #3 Answer 3a: The forces are drawn below. Normal points up, Weight points down, acceleration points right (toward the center of the circle) and it is caused by the Static Friction force pointing right. v Top View FN R a FFr, s FW Side View

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