R = 5 m ; v = 8 m/s

1. 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=

2. 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.

3. 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.

4. 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 FC and the friction force fsare not two different forces that are equal. There is just one force on the car. The nature of this central force is static friction.

5. 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

6. 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.

7. 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=

8. 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

9. 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

10. 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

11. 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

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

13. n mg v2 gR (12 m/s)2 (9.8 m/s2)(80 m) tan q = = q= 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?

14. 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.

15. 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

16. 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 =

17. 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 =

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

19. 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 =

20. 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.

21. + 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. At top: 10 N + T = 40 N T = Bottom: T – 10 N = 40 N T = __?___

22. 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:

23. 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:

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

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

26. 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

27. CONCLUSION: Chapter 10Uniform Circular Motion