Centripetal forces keep these children moving in a circular path.

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

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. v When central force is removed, ball continues in straight line. Centripetal force is needed to change direction.

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? Fc Car going around a curve. Force ON you is toward the center.

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

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

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

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

Example 1: A 3-kg rock swings in a circle of radius 5 m 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 m = 3 kg R = 5 m; v = 8 m/s F = (3 kg)(12.8 m/s2) Fc = 38.4 N

Example 2: A skater moves with 15 m/s in a circle of radius 30 m 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 450 N 30 m v = 15 m/s R Fc m=? Speed skater m = 60.0 kg

Newton’s 2nd law for circular motion: 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? Draw and label sketch r = ? m = 80 kg; v = 4 m/s2 Newton’s 2nd law for circular motion: Fc = 600 N r = 2.13 m

Car Negotiating a Flat Turn v 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.

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

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

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

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

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

Optimum Banking Angle n n n m 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. R v m Fc n fs = 0 n fs q slow speed q n fs q w w w fast speed optimum

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

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

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

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

Example 5: A car negotiates a turn of radius 80 m 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? q n mg tan q = = v2 gR (12 m/s)2 (9.8 m/s2)(80 m) q = 10.40 n mg q n sin q n cos q tan q = 0.184

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

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

2. Recall formula for pendulum. 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 q h T L R 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

Example 6(Cont.): Find v for q = 300 h T L R q = 300 R = 5 m 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

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

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

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

Example 9. If b = 5 m and L = 10 m, what will be the speed if the angle q = 260? tan q = v2 gR q T L R d b 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

Motion in a Vertical Circle + T mg v Top Right Weight causes small decrease in tension T Consider the forces on a ball attached to a string as it moves in a vertical loop. v T mg Top of Path Tension is minimum as weight helps Fc force + + v T mg Top Right Weight has no effect on T v T mg + Left Side Weight has no effect on T + T mg v Bottom Maximum tension T, W opposes Fc + T mg v Bottom 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.

The tension T must adjust so that central resultant is 40 N. 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. R v T 10 N + + T 10 N The tension T must adjust so that central resultant is 40 N. At top: 10 N + T = 40 N T = _?_ T = 30 N Bottom: T – 10 N = 40 N T = 50 N T = __?___

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

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

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 Resultant central force FC at every point in path! FC = 20 N Weight vector W is downward at every point. W = 5 N, down FC = 20 N at top AND at bottom.

FC = 20 N 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. R v FC = 20 N at top AND at bottom. Top: T + W = FC WT + T + 5 N = 20 N T = 20 N - 5 N = 15 N + T W Bottom: T - W = FC T - 5 N = 20 N T = 20 N + 5 N = 25 N

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

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

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

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? mg + T = mv2 R At Top: R v T mg vc occurs when T = 0 mg = mv2 R vc = gR vc = 8.85 m/s v = gR = (9.8 m/s2)(8 m)

The Loop-the-Loop Same as cord, n replaces T v AT TOP: n = - mg mv2 R n mg + AT BOTTOM: n mg + n= + mg mv2 R

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

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

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

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

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

CONCLUSION: Chapter 10 Uniform Circular Motion