Uniform Circular Motion Physics 40 S Semester Two

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Uniform Circular Motion Physics 40 S Semester Two 2020

Uniform Circular Motion Physics 40 S Semester Two 2020

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

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.

Uniform Circular Motion Uniform circular motion is motion along a circular path in which

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 v Constant velocity tangent to path. 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

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 v path, NOT outward as might be expected. 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 on the seat next to to

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

Car Example Continued Reaction Fc F’ The centripetal force is exerted BY the door

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

Another Example Disappearing platform at fair. R Fc What exerts the centripetal force in

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

Spin Cycle on a Washer How is the water removed from clothes during the

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, itit isis the LACK of of aa force that allows the water to to leave the clothes through holes in in the circular wall of of the rotating washer.

Centripetal Acceleration Consider a ball moving at constant speed v in a horizontal circle

Centripetal Acceleration Consider a 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. ) Fc W v R n Force Fc and acceleration ac toward center. W=n

Deriving Central Acceleration Consider initial velocity at A and final velocity at B: vf

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

Deriving Acceleration (Cont. ) Definition: ac = Similar Triangles ac = v t =

Deriving Acceleration (Cont. ) Definition: ac = Similar Triangles ac = v t = Centripetal acceleration: v v vs Rt vf v = = -vo v t R s v o mass m vv R ac v 2 R ; 2 mv Fc mac R

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? v v 2 m m = 3 kg ac R R R = 5 m; v = 8 m/s (8 m/s) 2 ac 12. 8 m/s 2 5 m Fc mac R FFcc = = 38. 4 NN

Example 2: A skater moves with 15 m/s in a circle of radius 30

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 Fc R mv 2 ; m 2 Fc v = 15 m/s v R Fc R 450 N 30 m m=? Speed skater (450 N)(30 m) m (15 m/s) 2 m m= = 60. 0 kg kg

Example 3. The wall exerts a 600 N force on an 80 -kg person

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 m = 80 kg; v = 4 m/s 2 Fc = 600 N r=? r 600 N Newton’s 2 nd law for circular motion: 2 mv mv F ; r r F rr = = 2. 13 m m 2

Car Negotiating a Flat Turn v Fc R What is the direction of the

Car Negotiating a Flat Turn v Fc R 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 R Is there also an outward force

Car Negotiating a Flat Turn v Fc R 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

Car Negotiating a Flat Turn The centripetal force Fc is that of static friction fs: fs m R v mg The central force FFCC and the friction force ffss are not two different forces that are equal. There isis just one force on on the car. The nature of of this central force isis static friction.

Finding the maximum speed for negotiating aa turn without slipping. n fs Fc =

Finding the maximum speed for negotiating aa turn without slipping. n fs Fc = fs m v R Fc R mg The car is on the verge of slipping when FC is equal to the maximum force of static friction fs. Fc = fs Fc = mv 2 R fs = smg

Maximum speed without slipping (Cont. ) n Fc = fs fs mg R mv

Maximum speed without slipping (Cont. ) n Fc = fs fs mg R mv 2 R = smg v= m v sg. R Velocity vv isis maximum speed for no no slipping.

Example 4: A car negotiates a turn of radius 70 m when the coefficient

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? m v Fc Fc = R s = 0. 7 mv 2 R fs = smg From which: v = sg. R g = 9. 8 m/s 2; R = 70 m v s g. R (0. 7)(9. 8)(70 m) vv = = 21. 9 m/s

Optimum Banking Angle By banking a curve at the optimum angle, the Fc m

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

Free-body Diagram n x mg n Acceleration a is toward the center. Set x

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

Optimum Banking Angle (Cont. ) n cos n n mg n sin Apply Newton’s

Optimum Banking Angle (Cont. ) n cos n n mg n sin Apply Newton’s 2 nd Law to x and y axes. mg mv 2 Fx = mac n sin Fy = 0 n cos = mg R

Optimum Banking Angle (Cont. ) n cos n n n sin tan n cos

Optimum Banking Angle (Cont. ) n cos n n n sin tan n cos n sin mg n sin mv 2 R n cos = mg tan mg R 1 v 2 g. R

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

Optimum Banking Angle (Cont. ) n mg Optimum Banking Angle n cos n n sin mg tan v 2 g. R

Example 5: A car negotiates a turn of radius 80 m. What is the

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 tan = mg n cos (12 m/s)2 = g. R (9. 8 m/s 2)(80 m) tan = 0. 184 n sin mg v 2 = 10. 40 How might you find the 2 centrip. F etal fom rcve on the C car, knowing its R mass?

The Conical Pendulum A conical pendulum consists of a mass m revolving in a

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

Angle and velocity v: T cos L T h mg R equations to find

Angle and velocity v: T cos L T h mg R equations to find angle T T T sin mv 2 R T cos = mg tan = v 2 g. R

Example 6: A 2 -kg mass swings in a horizontal circle at the end

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? L T R 1. Draw & label sketch. 2. Recall formula for pendulum. h tan v 2 Find: v = ? g. R 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 = 300 4. Use given info to find

Example 6(Cont. ): Find v for = 300 4. Use given info to find the velocity at 300. R=5 m Solve for v=? g = 9. 8 m/s 2 tan v g. R tan 2 L T R=5 m h R v 2 g. R v g. R tan v (9. 8 m/s 2 )(5 m) tan 300 vv = = 5. 32 m/s

Example 7: Now find the tension T in the cord if m = 2

Example 7: Now find the tension T in the cord if m = 2 kg, = 300, and L = 10 m. T cos L 2 kg T T= h mg R T cos - mg = 0; Fy = 0: cos = T cos 300 T sin T cos = mg TT = = 22. 6 NN

Example 8: Find the centripetal force Fc for the previous example. = 300 2

Example 8: Find the centripetal force Fc for the previous example. = 300 2 kg T cos L h T Fc mg R T T sin m = 2 kg; v = 5. 32 m/s; R = 5 m; T = 22. 6 N Fc = mv 2 R or Fc = T sin 300 Fc = 11. 3 N

Swinging Seats at the Fair This problem is identical to the other examples except

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

Example 9. If b = 5 m and L = 10 m, what will

Example 9. If b = 5 m and L = 10 m, what will be the speed if the angle = 260? v 2 tan = R=d+b g. R L b d = (10 m) sin 260 = 4. 38 m T d R = 4. 38 m + 5 m = 9. 38 m R v 2 g. R tan v (9. 8 m/s 2 )(9. 38 m) tan 260 vv = = 6. 70 m/s

Motion in a Vertical Circle Consider the forces on a ball attached to a

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.

v 10 N T As an exercise, assume that a central force of Fc

v 10 N T 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 T + 10 N v Thetension. TTmust adjustso sothatcentral resultantisis 40 40 NN. . At top: 10 N + T = 40 N T = 3_0? _N Bottom: T – 10 N = 40 N TT== 50? N

Motion in a Vertical Circle v Resultant force Fc = toward center m g

Motion in a Vertical Circle v Resultant force Fc = toward center m g T R v + mg T R Consider TOP of circle: mg + T = AT TOP: mv 2 T= mv 2 R - mg

Vertical Circle; Mass at bottom v T Resultant force toward center R v mg

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

Visual Aid: Assume that the centripetal force required to maintain circular motion is 20

Visual Aid: Assume that the centripetal force required to maintain circular motion is 20 N. Further assume that the weight is 5 N. v mv 2 FC 20 N R Resultant central force FC at every point in path! R v FC = 20 N at top AND at bottom. FC = 20 N Weight vector W is downward at every point. W = 5 N, down

Visual Aid: The resultant force (20 N) is the vector sum of T and

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

For Motion in Circle v AT TOP: R + T= mg mv 2 R

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

Example 10: A 2 -kg rock swings in a vertical circle of radius 8

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? 2 At Top: v mg T T= R v mg + T = mv 2 mv - mg R R (2 kg)(10 m/s) 2 T 2 kg(9. 8 m/s 2 ) 8 m T = 25 N - 19. 6 N TT = = 5. 40 NN

Example 11: A 2 -kg rock swings in a vertical circle of radius 8

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? 2 At Bottom: v R T mg T= v T - mg = mv 2 mv + mg R R (2 kg)(10 m/s) 2 T 2 kg(9. 8 m/s 2 ) 8 m T = 25 N + 19. 6 N TT = = 44. 6 NN

Example 12: What is the critical speed vc at the top, if the 2

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? 0 2 m v At Top: v mg + T = mg R vc occurs when T = 0 R T mv 2 mg = vc = g. R v= g. R = (9. 8 m/s 2)(8 m) vvcc= = 8. 85 m/s

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

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

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

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

Example 13: What is the apparent weight of a 60 -kg person as she

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 Apparent weight will be the normal force at the top: mv 2 mg - n = mg R (60 kg)(6 m/s) n 60 kg(9. 8 m/s ) 45 m 2 + v R v - mv 2 R nn == 540 NN

Summary Centripetal acceleration: v= sg. R Conical pendulum: ac v 2 R ; mv

Summary Centripetal acceleration: v= sg. R Conical pendulum: ac v 2 R ; mv Fc mac R v 2 tan = g. R v= g. R tan 2

Summary: Motion in Circle v AT TOP: + mg R v 2 mv T=

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

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

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