# UNIFORM CIRCULAR MOTION - PowerPoint PPT Presentation Download Presentation UNIFORM CIRCULAR MOTION

UNIFORM CIRCULAR MOTION Download Presentation ## UNIFORM CIRCULAR MOTION

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript

1. UNIFORM CIRCULAR MOTION Uniform circular motion is motion in which there is no change in speed, only a change in direction.

2. v Constant velocity tangent to path. Constant force toward thecenter. Uniform circular motion is motion along a circular path in which there is no change in speed, only a change in direction. Fc

3. CENTRIPETAL ACCELERATION An object experiencing uniform circular motion is continually accelerating. The position and velocity of a particle moving in a circular path of radius r are shown at two instants in the figure. When the particle is at point A, its velocity is represented by vector v1. After a time interval t,its velocity is represented by the vectorv2.

4. The acceleration is given by: as Δtbecomes smaller and smaller, the chord length becomes equal to the arc lengths = v Δt

5. This acceleration is called the centripetal, and it points toward the center of the circle. Units: m/s2

6. The term centripetalmeans that the acceleration a is always directed toward the center. The velocity and the acceleration are not in the same direction; v points in the direction of motion which is tangential to the circle. vand aare perpendicular at every point.

7. vand aare perpendicular at every point.

8. The period T is the time for one complete revolution. So the linear speed can be found by dividing the period into the circumference: Units: m/s Another useful parameter in engineeringproblems is the rotational speed, expressed in revolutions per minute (rpm), revolutions per second (rev/s) or Hz (s-1). This quantity is called the frequency f of rotation and is given by the reciprocal of the period. Units: s-1 = Hz

9. 4.11 A 2 kg body is tied to the end of a cord and whirled in a horizontal circle of radius 2 m. If the body makes three complete revolutions every second, determine its linear speed and its centripetal acceleration. UCM m = 2 kg r = 2 m f = 3 rev/s = 0.33 s = 38.07 m/s = 725 m/s2

10. 4.12 A ball is whirled at the end of a string in a horizontal circle 60 cm in radius at the rate of 1 revolution every 2 s. Find the ball's centripetal acceleration. UCM r = 0.6 m T = 2 s = 5.92 m/s2

11. CENTRIPETAL FORCE The inward force necessary to maintain uniform circular motion is defined as centripetal force. From Newton's Second Law, the centripetal force is given by:

12. The centripetal force is not a 'special' kind of force. The centripetal force is provided by the force that keeps the object in a circle, this is called the centripetal force requirement. 4.13 a. A car makes a turn, what force is required to keep it in circular motion? As a car makes a turn, the force of friction acting upon the tires of the car provide the centripetal force required for circular motion.

13. b. A bucket of water is tied to a string and spun in a circle, what force is required to keep it in circular motion? As a bucket of water is tied to a string and spun in a circle, the force of tension acting upon the bucket provides the centripetal force required for circular motion. c. The moon orbits the Earth, what force is required to keep it in circular motion? As the moon orbits the Earth, the force of gravity acting upon the moon provides the centripetal force required for circular motion.

14. The Forbidden F-Word When the subject of circular motion is discussed, it is not uncommon to hear mention of the word "centrifugal." Centrifugal, not to be confused with centripetal, means away from the center or outward. The use of this word combined with the common sensation of an outward force when experiencing circular motion, often creates or reinforces a deadly student misconception.

15. The Forbidden F-Word The deadly misconception, is the notion that objects in circular motion are experiencing an outward force. "After all," a well-meaning student may think, "I can recall vividly the sensation of being thrown outward away from the center of the circle on that roller coaster ride. Therefore, circular motion must be characterized by an outward force."

16. Centrifugal Force is often used to describe why mud gets spun off a spinning tire, or water gets pushed out of the clothes during the spin dry cycle of your washer.  It is also used to describe why we tend to slide to the outer side of a car going around a curve.  Let’s imagine that you are riding in Granny’s car going around a curve.  Sitting on your dashboard is a cassette tape.  As you go around the curve, the tape moves to outside edge of the car.  Because you don't want to blame it on ghosts, you say “centrifugal force pushed the tape across the dashboard.” Wwrroonngg!!

17. The animation shows both views at the same time.  The top window shows you the passenger's view of the car and the tape, while the shows you the bird's eye view.

18. The tape on the slippery dashboard does not have enough friction to act as a centripetal force, so in the absence of a centripetal force the tape follows straight line motion. If the car you are riding in has the windows rolled down, then the tape will leave the car (or does the car leave the tape?) as it follows its straight line path.  If the windows are rolled up, then the window will deliver a centripetal force to the tape and keep it in a circular path.

19. Without a centripetal force, an object in motion continues along a straight-line path. With a centripetal force, an object in motion will be accelerated and change its direction.

20. We can see that the force must be inward by thinking about a ball on a string:

21. There is no centrifugal force pointing outward; what happens is that the natural tendency of the object to move in a straight line must be overcome. If the centripetal force vanishes, the object flies off tangent to the circle.

22. Spin Cycle on a Washer How is the water removed from clothes during the spin cycle of a washer? Think carefully before answering . . . Does the centripetal force throw water off the clothes? NO. Actually, it is the LACK of a force that allows the water to leave the clothes through holes in the circular wall of the rotating washer.

23. 4.14 A 75 g toy airplane is fastened to one end of a 44 cm string, and the other end is held fixed at the ceiling The plane whirls in a horizontal circle. Find the speed of the plane and the tension of the string if the angle to the vertical is 30˚. UCM FT 60º FG L = 0.44 m m = 0.075 kg = 0.22 m

24. L = 0.44 m m = 0.075 kg r = 0.22 m FT 60º FG = 1.1 m/s

25. = 0.85 N

26. 4.15 In a Rotor ride at the amusement park the room radius is 4.6 m and the rotation frequency is 0.5 revolutions per second when the floor drops out. What is the minimum coefficient of static friction so that the people will not slip down? UCM r = 4.6 m f = 0.5 rev/s T = 2 s

27. r = 4.6 m f = 0.5 rev/s T = 2 s = 0.22

28. MOTION IN A VERTICAL CIRCLE When a body moves in a vertical circle at the end of a string, the tension FTin the string varies with the body's position. The centripetal force Fcon the body at any point is the vector sum of FT and the component of the body's weight toward the center of the circle.

29. At the highest point in the circular turn the centripetal force is given by: At the lowest point in the circular turn the centripetal force is given by:

30. v mv2 R FN= - mg R v AT BOTTOM: mv2 R FN= + mg FN + + mg mg FN The Loop-the-Loop AT TOP:

31. v mv2 R AT TOP: mg - FN = FN R mv2 R FN = mg - v AT BOTTOM: mv2 R FN FN= + mg + + mg mg The Ferris Wheel

32. 4.16 A string 0.5 m long is used to whirl a 1-kg stone in a vertical circle at a uniform velocity of 5 m/s. a. What is the tension in the string when the stone is at the top of the circle UCM Top of Circle Fc = FT + Fg FT = Fc - Fg r = 0.5 m m = 1 kg v = 5 m/s = 50 - 9.8 = 40.2 N

33. b. What is the tension in the string when the stone is at the bottom of the circle? Bottom of Circle Fc = FT - Fg FT = Fc + Fg = 50 + 9.8 = 59.8 N

34. 4.17 A ball of mass M is attached to a string of length R and negligible mass. The ball moves clockwise in a vertical circle, as shown. When the ball is at point P, the string is horizontal. Point Q is at the bottom of the circle and point Z is at the top of the circle. Air resistance is negligible. Express all algebraic answers in terms of the given quantities and fundamental constants. UCM

35. a. On the figures below, draw and label all the forces exerted on the ball when it is at points P and Q. FT FT FG FG

36. b. Derive an expression for vmin the minimum speed the ball can have at point Z without leaving the circular path. The minimum speed will occur when there is no FT FG

37. c. The maximum tension the string can have without breaking is Tmax Derive an expression for vmax, the maximum speed the ball can have at point Q without breaking the string.

38. d. Suppose that the string breaks at the instant the ball is at point P. Describe the motion of the ball immediately after the string breaks. The ball will go straight up i.e. tangential to point P

39. Kepler’s Laws and Newton's Synthesis • Kepler’s laws describe planetary motion. • The orbit of each planet is an ellipse, with the Sun at one focus.

40. 2. An imaginary line drawn from each planet to the Sun sweeps out equal areas in equal times.

41. As the planet is closest the sun, the planet is moving fastest and as the planet is farthest from the sun,it is moving slowest. Nonetheless, the imaginary line adjoining the center of the planet to the center of the sun sweeps out the same amount of area in each equal interval of time.

42. KEPLER’S THIRD LAW "If T is the period and r is the length of the semi-major axis of a planet’s orbit, then the ratioT2/r3 is the same for all planets."

43. The ratio of the square of a planet’s orbital period is proportional to the cube of its mean distance from the Sun: T2/r3

44. Kepler’s laws can be derived from Newton’s laws. Irregularities in planetary motion led to the discovery of Neptune, and irregularities in stellar motion have led to the discovery of many planets outside our Solar System.

45. 4.18 The mean distance from the Earth to the Sun is 1.496x108 km and the period of its motion about the Sun is one year. The period of Jupiter’s motion around the Sun is 11.86 years. Determine the mean distance from the Sun to Jupiter. KL rE = 1.496x108 km TE = 1 year TJ = 11.86 years = 7.77x108 km

46. Newton’s Law of Universal Gravitation If the force of gravity is being exerted on objects on Earth, what is the origin of that force? Newton’s realization was that the force must come from the Earth. He further realized that this force must be what keeps the Moon in its orbit.

47. Newton’s Law of Universal Gravitation The gravitational force must be proportional to both masses. By observing planetary orbits, Newton concluded that the gravitational force must decrease as the inverse of the square of the distance between the masses. In its final form, the Law of Universal Gravitation reads: Where:

48. The magnitude of the gravitational constant G can be measured . This is the Cavendish experiment.