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Centripetal forces keep these children moving in a circular path.

Centripetal forces keep these children moving in a circular path. Objectives: After completing this module, you should be able to:. Apply your knowledge of centripetal acceleration and centripetal force to the solution of problems in circular motion.

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Centripetal forces keep these children moving in a circular path.

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

  2. Objectives: After completing this module, you should be able to: Apply your knowledge of centripetal acceleration and centripetal force to the solution of problems in circular motion. Define and apply concepts of frequency and period, and relate them to linear speed. Solve problems involving banking angles, the conical pendulum, and the vertical circle.

  3. v Constant velocity tangent to path. Constant force toward center. 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. Fc Question: Is there an outwardforce on the ball?

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

  5. Fc Examples of Centripetal Force 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. Car going around a curve.

  6. Reaction The centripetal force is exerted BY the door ON you. (Centrally) Fc F’ 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.

  7. R Disappearing platform at fair. Fc Another Example What exerts the centripetal force in this example and on what does it act? The centripetal force is exerted BY the wall ON the man. A reaction force is exerted by the man on the wall, but that does not determine the motion of the man.

  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. Fc n W v R Centripetal Acceleration Consider ball moving at constant speed v in a horizontal circle of radius Rat end of string tied to peg on center of table. (Assume zero friction.) Force Fcand acceleration ac toward center. W = n

  10. vf vf -vo vo vo R R Deriving Central Acceleration Consider initial velocity at A and final velocity at B: B Dv s A

  11. Dv t Definition: ac = vf Similar Triangles -vo Dv s vo R mass m vs Rt Dv t vv R ac = = = Dv v s R = Deriving Acceleration (Cont.) Centripetal acceleration:

  12. v m R 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? m = 3 kg R = 5 m; v = 8 m/s F = (3 kg)(12.8 m/s2) Fc = 38.4 N

  13. v = 15 m/s Fc R 450 N 30 m m=? Speed skater Example 2:A skater moves with 15 m/s in a circle of radius 30 m. The ice exerts a central force of 450 N. What is the mass of the skater? Draw and label sketch m = 60.0 kg

  14. Draw and label sketch Newton’s 2nd law for circular motion: m = 80 kg; v = 4 m/s2 Fc = 600 N r = ? Example 3.The wall exerts a 600 N force on an 80-kg person moving at 4 m/s on a circular platform. What is the radius of the circular path? r = 2.13 m

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

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

  17. n Fc R fs m R v mg Car Negotiating a Flat Turn The centripetal force Fc is that of static friction fs: Fc = fs The central force FCand the friction force fs 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.

  18. n Fc Fc = fs R fs m R v mv2 R Fc = mg Finding the maximum speed for negotiating a turn without slipping. The car is on the verge of slipping when FC is equal to the maximum force of static friction fs. fs = msmg Fc = fs

  19. n fs R Fc R m mg v mv2 R = msmg v = msgR Maximum speed without slipping (Cont.) Fc = fs Velocity v is maximum speed for no slipping.

  20. m Fc R mv2 R Fc = ms = 0.7 v v = msgR 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? fs = msmg From which: g = 9.8 m/s2; R = 70 m v = 21.9 m/s

  21. Fc R m v fs = 0 fs n n n fs w w w q q q slow speed Optimum Banking Angle By banking a curve at theoptimum angle, the normal force ncan provide the necessary centripetal force without the need for a friction force. fast speed optimum

  22. n mg n q +ac mg q Free-body Diagram Acceleration ais toward the center. Set x axis along the direction of ac , i. e., horizontal (left to right). x n cos q n q n sin q q mg

  23. n mg n cos q n q mv2 R Apply Newton’s 2nd Law to x and y axes. n sin q = n sin q mg q Optimum Banking Angle (Cont.) SFx = mac n cos q = mg SFy = 0

  24. n cos q n q n n sin q mg mg mv2 R n sin q = q Optimum Banking Angle (Cont.) n cos q= mg

  25. n cos q n n q n sin q mg mg Optimum Banking Angle q q Optimum Banking Angle (Cont.)

  26. n mg v2 gR (12 m/s)2 (9.8 m/s2)(80 m) tan q = = q= 10.40 n cos q n q n sin q q mg 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? tan q = 0.184 How might you find the centripetal force on the car, knowing its mass?

  27. T L q h T mg R 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. T cos q q T sin q Note: The inward component of tension T sin q gives the needed central force.

  28. T cos q q T T sin q L q h mv2 R Solve two equations to find angle q T sin q = T v2 gR mg R tan q = Angle qand velocity v: T cos q = mg

  29. L q h T R 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 300 with the vertical? 1. Draw & label sketch. q = 300 2. Recall formula for pendulum. Find: v = ? 3. To use this formula, we need to find R = ? R = L sin 300 = (10 m)(0.5) R = 5 m

  30. R = 5 m q = 300 L q h T R Example 6(Cont.): Find vfor q = 300 4. Use given info to find the velocity at 300. R = 5 m g = 9.8 m/s2 Solve for v = ? v = 5.32 m/s

  31. T cos q q T T sin q L q h T mg R mg cos q (2 kg)(9.8 m/s2) cos 300 T = = Example 7: Now find the tension T in the cord if m= 2 kg, q= 300, and L= 10 m. 2 kg SFy = 0: T cos q - mg = 0; T cos q= mg T = 22.6 N

  32. T cos q q T T sin q mv2 R Fc = L q h T mg R Example 8: Find the centripetal force Fc for the previous example. q = 300 Fc 2 kg m= 2kg; v = 5.32m/s;R = 5 m; T = 22.6 N Fc = 11.3 N or Fc = T sin 300

  33. L q h T R v2 gR tan q= v = gR tanq Swinging Seats at the Fair b This problem is identical to the other examples except for finding R. d R = d + b R = L sin q + b and

  34. v2 gR tan q= b L q T d R Example 9. If b = 5 m and L = 10 m, what will be the speed if the angle q= 260? R = d + b d = (10 m) sin 260 = 4.38 m R = 4.38 m + 5 m = 9.38 m v = 6.70 m/s

  35. v v v + v Bottom mg T T T mg T + T T v v Top of Path Left Side + Top Right Top Right Tension is minimum as weight helps Fc force Weight has no effect on T Maximum tension T, W opposes Fc Weight causes small decrease in tension T Weight has no effect on T Bottom + + + mg mg mg mg 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.

  36. + 10 N v T T R + 10 N v As an exercise, assume that a central force of Fc = 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. T = _?_ At top: 10 N + T = 40 N T = 30 N Bottom: T – 10 N = 40 N T = 50 N T = __?___

  37. v T R mv2 R Fc = mv2 R v mg + T = mv2 R T = - mg + mg mg T Motion in a Vertical Circle Resultant force toward center Consider TOP of circle: AT TOP:

  38. v T R mv2 R Fc = mv2 R mg v T - mg = mv2 R T T = + mg + mg Vertical Circle; Mass at bottom Resultant force toward center Consider bottom of circle: AT Bottom:

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

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

  41. v mv2 R T = - mg R v AT BOTTOM: mv2 R T T = + mg + + mg mg T For Motion in Circle AT TOP:

  42. v R mg T mv2 R v mg + T = mv2 R T = - mg 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 Tin rope? At Top: T = 5.40N T = 25N - 19.6 N

  43. v mv2 R T - mg = R v T mg mv2 R T = + mg 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? At Bottom: T = 44.6N T = 25N + 19.6 N

  44. v 0 At Top: R mg T mv2 R mv2 R v mg + T = mg = vc = gR v = gR = (9.8 m/s2)(8 m) Example 12:What is the critical speed vc at the top, if the 2-kg mass is to continue in a circle of radius 8 m? vc occurs when T = 0 vc = 8.85 m/s

  45. v mv2 R n = - mg R v AT BOTTOM: mv2 R n= + mg n + + mg mg n The Loop-the-LoopSame as cord, nreplaces T AT TOP:

  46. v mv2 R mg - n = AT TOP: n R mv2 R n = mg - v AT BOTTOM: mv2 R n n = + mg + + mg mg The Ferris Wheel

  47. n v mv2 R mg - n = R v mv2 R + n = mg - mg 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? Apparent weight will be the normal force at the top: n= 540 N

  48. v2 gR tan q= Conical pendulum: v = gR tanq v = msgR Summary Centripetal acceleration:

  49. v AT TOP: mv2 R + T = - mg mg R T v AT BOTTOM: mv2 R T T = + mg + mg Summary: Motion in Circle

  50. v mv2 R mg - n = AT TOP: n R mv2 R n = mg - v AT BOTTOM: mv2 R n n = + mg + + mg mg Summary: Ferris Wheel

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