2. Chapter 6.2. Uniform Circular Motion

3. Centripetal forces keep these children moving in a circular path.

4. Uniform Circular Motion Uniform circular motion Uniform circular motion is motion along a circular path in which there is no change in speed, only a change in direction. Constant force toward center. Constant velocity tangent to path. v FcFc Question: Is there an outward force on the ball?

5. Uniform Circular Motion (Cont.) The question of an outward force can be resolved by asking what happens when the string breaks! When central force is removed, ball continues in straight line. v Ball moves tangent to path, NOT outward as might be expected. Centripetal force is needed to change direction.

6. Examples of Centripetal Force Car going around a curve.Car going around a curve. You are sitting on the seat next to the outside door. What is the direction of the resultant force on you as you turn? Is it away from center or toward center of the turn? Force ON you is toward the center. FcFcFcFc

7. Car Example Continued There is an outward force, but it does not act ON you. It is the reaction force exerted BY you ON the door. It affects only the door. The centripetal force is exerted BY the door ON you. (Centrally) FcFc F’Reaction

8. 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.

9. Deriving Central Acceleration Consider initial velocity at A and final velocity at B: R vovo vfvf vfvf -v o A B R vovo vv s

10. Deriving Acceleration (Cont.) vfvf -v o R vovo vv s a c = vtvt Definition: = vvvv sRsR Similar Triangles = vtvt vsRtvsRt a c = = v vvRvvR mass m Centripetal acceleration:

11. Circular Motion is often described in terms of frequency and period. The frequency f – the number of revolutions per second.The frequency f – the number of revolutions per second. The period T – time required for one complete revolution.The period T – time required for one complete revolution. T=1/f

12. Example 1: A 3-kg rock swings in a circle of radius 5 m. If its constant speed is 8 m/s, what is the centripetal acceleration? R v m R = 5 m; v = 8 m/s m = 3 kg F = (3 kg)(12.8 m/s 2 ) F c = 38.4 N

13. Car Negotiating a Flat Turn R v What is the direction of the force the car? What is the direction of the force ON the car? Ans. Toward Center FcFcFcFc This central force is exerted BY the road ON the car.

14. Car Negotiating a Flat Turn R v Is there also an outward force acting ON the car? Ans. No, but the car does exert a outward reaction force ON the road. FcFcFcFc

15. Car Negotiating a Flat Turn The centripetal force F c is that of static friction f s : The central force F C and the friction force f s are not two different forces that are equal. There is just one force on the car. The nature of this central force is static friction. F c = f s R v m FcFc n mg fsfs R

16. Finding the maximum speed for negotiating a turn without slipping. F c = f s f s =  s mg F c = mv 2 R The car is on the verge of slipping when F C is equal to the maximum force of static friction f s. R v m FcFc F c = f s n mg fsfs R

17. Maximum speed without slipping (Cont.) F c = f s mv 2 R =  s mgv =  s gR Velocity v is maximum speed for no slipping. n mg fsfs R R v m FcFc

18. Example 4: A car negotiates a turn of radius 70 m when the coefficient of static friction is 0.7. What is the maximum speed to avoid slipping? v = 21.9 m/s R v m FcFc  s = 0.7 f s =  s mg F c = mv 2 R From which: v =  s gR g = 9.8 m/s 2 ; R = 70 m

19.  slow speed  Optimum Banking Angle By banking a curve at the optimum angle, the normal force n can provide the necessary centripetal force without the need for a friction force. fast speed optimum n f s = 0 ww n fsfs w n fsfs R v m FcFc

20. Free-body Diagram n mg   Acceleration a is toward the center. Set x axis along the direction of a c, i. e., horizontal (left to right). nmg  n sin  n cos  + ac+ ac+ ac+ ac  n mg x

21. Optimum Banking Angle (Cont.) n mg  n sin  n cos   F x = m a c  F y = 0 n cos  = mg mv 2 R n sin  Apply Newton’s 2nd Law to x and y axes.  n mg

22. Optimum Banking Angle (Cont.)  n cos  = mg mv 2 R  n sin   n mg n   n sin   n cos 

23. Optimum Banking Angle (Cont.) n mg   n sin   n cos  n mg Optimum Banking Angle 

24. Example 5: A car negotiates a turn of radius 80 m. What is the optimum banking angle for this curve if the speed is to be equal to 12 m/s? n mg   n sin   n cos   tan  = = v 2 gR (12 m/s) 2 (9.8 m/s 2 )(80 m)  tan  = 0.184  n mg How might you find the centripetal force on the car, knowing its mass?  = 10.40

25. The Conical Pendulum A conical pendulum consists of a mass m revolving in a horizontal circle of radius R at the end of a cord of length L.  h T L R mg T  T sin   T cos  Note: The inward component of tension T sin  gives the needed central force.

26. Angle  and velocity v:  h T L R mg T   T sin   T cos   T cos  = mg mv 2 R  T sin   Solve two equations to find angle  tan  = v 2 gR

27. Example 6: A 2-kg mass swings in a horizontal circle at the end of a cord of length 10 m. What is the constant speed of the mass if the rope makes an angle of 30 0 with the vertical? R = L sin 30 0 = (10 m)(0.5)R = 5 m 1. Draw & label sketch. 2. Recall formula for pendulum. Find: v = ? 3. To use this formula, we need to find R = ?  h T L R  

28. Example 6(Cont.): Find v for  = 30 0 R = 5 m v = 5.32 m/s g = 9.8 m/s 2 Solve for v = ? 4. Use given info to find the velocity at 30 0.  h T L R   R = 5 m

29. Example 7: Now find the tension T in the cord if m = 2 kg,  = 30 0, and L = 10 m.  h T L R mg T   T sin   T cos    F y = 0: T cos  - mg = 0; T cos  = mg T = = mg  cos  (2 kg)(9.8 m/s 2 ) cos 30 0 T = 22.6 N 2 kg

30. Example 8: Find the centripetal force F c for the previous example.  h T L R mg T   T sin   T cos  m = 2 kg; v = 5.32 m/s; R = 5 m; T = 22.6 N F c = 11.3 N 2 kg F c = mv 2 R or F c = T sin 30 0 FcFc  = 30 0

31. Swinging Seats at the Fair  h T L R d This problem is identical to the other examples except for finding R. R = d + b  R = L sin  + b  tan  = v 2 gR and  v = gR tan  b

32. Example 9. If b = 5 m and L = 10 m, what will be the speed if the angle  = 26 0 ? R = d + b R = 4.38 m + 5 m = 9.38 m  tan  = v 2 gR  T L R d b d = (10 m) sin 26 0 = 4.38 m v = 6.70 m/s

33. Motion in a Vertical Circle Consider the forces on a ball attached to a string as it moves in a vertical loop. Note also that the positive direction is always along acceleration, i.e., toward the center of the circle. Note changes as you click the mouse to show new positions. + T mg vBottom Maximum tension T, W opposes F c + v T mg Top Right Weight has no effect on T + T mg v Top Right Weight causes small decrease in tension T v T mg + Left Side Weight has no effect on T + T mg v Bottom v T Top of Path Tension is minimum as weight helps F c force +

34. R v v As an exercise, assume that a central force of F c = 40 N is required to maintain circular motion of a ball and W = 10 N. The tension T must adjust so that central resultant is 40 N. At top: 10 N + T = 40 N Bottom: T – 10 N = 40 N T = __?___ T = 50 N T = 30 N T = _?_ T 10 N+ + T

35. Motion in a Vertical Circle R v v Resultant force toward center F c = mv 2 R Consider TOP of circle: AT TOP: T mg T + mg + T = mv 2 R T = - mg mv 2 R

36. Vertical Circle; Mass at bottom R v v Resultant force toward center F c = mv 2 R Consider bottom of circle: AT Bottom: T mg +T - mg = mv 2 R T = + mg mv 2 R T mg

37. Visual Aid: Assume that the centripetal force required to maintain circular motion is 20 N. Further assume that the weight is 5 N. R v v F C Resultant central force F C at every point in path! F C = 20 N W Weight vector W is downward at every point. W = 5 N, down F C = 20 N F C = 20 N at top AND at bottom.

38. Visual Aid: The resultant force (20 N) is the vector sum of T and W at ANY point in path. R v v F C = 20 N F C = 20 N at top AND at bottom. Top: T + W = F C T + 5 N = 20 N T = 20 N - 5 N = 15 N T - W = F C T - 5 N = 20 N T = 20 N + 5 N = 25 N Bottom: WTWT + + TWTW

39. For Motion in Circle R v v AT TOP: T mg + AT BOTTOM: T mg + T = - mg mv 2 R T = + mg mv 2 R

40. Example 10: A 2-kg rock swings in a vertical circle of radius 8 m. The speed of the rock as it passes its highest point is 10 m/s. What is tension T in rope? R v v T mg mg + T = mv 2 R At Top: T = - mg mv 2 R T = 25 N - 19.6 N T = 5.40 N

41. Example 11: A 2-kg rock swings in a vertical circle of radius 8 m. The speed of the rock as it passes its lowest point is 10 m/s. What is tension T in rope? R v v T mg T - mg = mv 2 R At Bottom: T = + mg mv 2 R T = 25 N + 19.6 N T = 44.6 N

42. Example 12: What is the critical speed v c at the top, if the 2-kg mass is to continue in a circle of radius 8 m? R v v T mg mg + T = mv 2 R At Top: v c = 8.85 m/s v c occurs when T = 0 0 mg = mv 2 R v = gR = (9.8 m/s 2 )(8 m) v c = gR

43. The Loop-the-Loop Same as cord, n replaces T AT TOP: n mg + AT BOTTOM: n mg + n = - mg mv 2 R n = + mg mv 2 R R v v

44. The Ferris Wheel AT TOP: n mg + AT BOTTOM: n mg + mg - n = mv 2 R n = + mg mv 2 R R v v n = mg - mv 2 R

45. Example 13: What is the apparent weight of a 60-kg person as she moves through the highest point when R = 45 m and the speed at that point is 6 m/s? n mg + R v v mg - n = mv 2 R n = mg - mv 2 R Apparent weight will be the normal force at the top: n = 540 N

46. Summary Centripetal acceleration: v =  s gR tan  = v 2 gR  v = gR tan  Conical pendulum:

47. Summary: Motion in Circle v R v AT TOP: T mg + T = - mg mv 2 R AT BOTTOM: T mg + T = + mg mv 2 R

48. Summary: Ferris Wheel AT TOP: n mg + AT BOTTOM: n mg + mg - n = mv 2 R n = + mg mv 2 R R v v n = mg - mv 2 R

49. CONCLUSION: Chapter 10 Uniform Circular Motion