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Chapter 10. Uniform Circular Motion AA PowerPoint PowerPoint Presentation Presentation by by Paul Paul E. E. Tippens, Tippens, Professor Professor of of Physics Physics Southern Southern Polytechnic Polytechnic State State University University © 2007 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. • 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 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 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 You are are sitting sitting on on the the seat seat next next to to the the outside outside door. door. What What isis the the direction direction of of the the resultant resultant force force on on you you as as you you turn? turn? Is Is itit away away from from center center or or toward toward center center of of the the turn? 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 ON you. (Centrally) There There isis an an outward outward force, force, but but itit does does not not act act ON ON you. you. It It isis the the reaction reaction force force exerted exerted BY BY you you ON ON the the door. door. It It affects affects only only the the door. door. Another Example Disappearing platform at fair. R Fc What exerts the centripetal force in this example and on what does it act? The The centripetal centripetal force force isis exerted exerted BY BY the the wall wall ON ON the the man. man. AA reaction reaction force force isis exerted exerted by by the the man man on on the the wall, wall, but but that that does does not not determine determine the the motion motion of of the the man. man. 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. NO. Actually, Actually, itit isis the the LACK LACK of of aa force force that that allows allows the the water water to to leave leave the the clothes clothes through through holes holes in in the the circular circular wall wall of of the the rotating rotating washer. washer. 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.) 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 vf B R vo A -vo v R s v o Deriving Acceleration (Cont.) v Definition: ac = Similar Triangles ac = v t = Centripetal Centripetal acceleration: acceleration: v v vs Rt vf -vo v t = = R s R s v o mass m vv R 2 v ac ; R mv Fc mac R 2 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? 2 v v m m = 3 kg ac R R R = 5 m; v = 8 m/s 2 (8 m/s) 2 ac 12.8 m/s 5m mv Fc mac R 2 F = (3 kg)(12.8 m/s2) FFcc == 38.4 38.4 NN 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 R v Fc R 450 N 30 m m=? Speed skater (450 N)(30 m) m 2 (15 m/s) m m == 60.0 60.0 kg kg 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 Newton’s 2nd law for circular motion: m = 80 kg; v = 4 m/s2 Fc = 600 N 2 mv mv F ; r r F r=? (80 kg)(4 m/s) r 600 N 2 rr = = 2.13 2.13 m m 2 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 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: m Fc R n fs Fc = fs R v mg The The central central force force FFCC and and the the friction friction force force ffss are are not not two two different different forces forces that that are are equal. equal. There There isis just just one one force force on on the the car. car. The The nature nature of of this this central central force force isis static static friction. friction. Finding Finding the the maximum maximum speed speed for for negotiating negotiating aa turn turn without without slipping. 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 = mv2 R fs = smg Maximum Maximum speed speed without without slipping slipping (Cont.) (Cont.) n Fc = fs fs R mv2 R mg v= m v Fc R = smg sgR Velocity Velocity vv isis maximum maximum speed speed for for no no slipping. 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? m v Fc Fc = R s = 0.7 mv2 R fs = smg From which: v = sgR g = 9.8 m/s2; R = 70 m v s gR (0.7)(9.8)(70 m) vv == 21.9 21.9 m/s m/s Optimum Banking Angle Fc m R v fs w 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. n slow speed n w fs fast speed fs = 0 w n optimum Free-body Diagram Acceleration a is toward the center. Set x axis along the direction of ac , i. e., horizontal (left to right). n x mg n n cos n n sin mg mg + ac Optimum Banking Angle (Cont.) n cos n n mg n sin Apply Newton’s 2nd Law to x and y axes. mg mv2 Fx = mac n sin Fy = 0 n cos = mg R Optimum Banking Angle (Cont.) n cos n n n sin tan n cos n sin mg mg n sin mv2 R n cos = mg 2 mv 2 v R tan mg gR 1 Optimum Banking Angle (Cont.) n mg Optimum Banking Angle n cos n n sin mg 2 v tan gR 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 n n sin mg v2 gR = (12 m/s)2 (9.8 m/s2)(80 m) tan = 0.184 = 10.40 How might you 2find the mv centripetal FC force on the car, knowing R its mass? 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 R T h mg T sin Note: The inward component of tension T sin gives the needed central force. Angle and velocity v: T cos L T h mg R Solve two equations to find angle T T sin T sin mv2 R T cos = mg tan = v2 gR 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? T L R h 1. Draw & label sketch. 2. Recall formula for pendulum. 2 v tan gR Find: v = ? 3. To use this formula, we need to find R = ? R = L sin 300 = (10 m)(0.5) R=5m Example 6(Cont.): Find v for = 300 4. Use given info to find the velocity at 300. R=5m Solve for v=? g = 9.8 m/s2 T v gR tan v gR tan v (9.8 m/s )(5 m) tan 30 2 h R v2 tan gR 2 L R=5m 0 vv == 5.32 5.32 m/s m/s 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 Fy = 0: mg cos T cos - mg = 0; = T (2 kg)(9.8 m/s2) cos 300 T sin T cos = mg TT == 22.6 22.6 NN 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 = mv2 R or Fc = T sin 300 FFcc == 11.3 11.3 NN 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 v2 gR and v= gR tan Example 9. If b = 5 m and L = 10 m, what will be the speed if the angle = 260? v2 tan = R=d+b gR L b d = (10 m) sin 260 = 4.38 m T d R = 4.38 m + 5 m = 9.38 m R v gR tan 2 v gR tan v (9.8 m/s 2 )(9.38 m) tan 260 vv == 6.70 6.70 m/s m/s Motion in a Vertical Circle v Consider the forces on a ball attached to a string as it moves in a vertical loop. + v Note also that the positive direction is always along acceleration, i.e., toward the center of the circle. v + v Bottom Tmg + T T T mg + +mg mg T v Top ofSide Path Left mg Top Right Right Top Tension is Weight hasasno Maximum minimum Weight causes Weight has no effect on T tension T, W Bottom weight helps small decrease effect on T opposes F force Ftension in Tc c Note changes as you click the mouse to show new positions. + 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 The The tension tension TT must must adjust adjust so so that that central central resultant resultant isis 40 40 NN.. At top: 10 N + T = 40 N TT = ?_N = _30 Bottom: T – 10 N = 40 N TT==__?___ 50 N Motion in a Vertical Circle v mg T mv2 Resultant force Fc = toward center R R v mg + T = AT TOP: + mg T Consider TOP of circle: T= mv2 R mv2 R - mg Vertical Circle; Mass at bottom v T Resultant force toward center R v mg T - mg = mg + R Consider bottom of circle: AT Bottom: T Fc = mv2 T= mv2 R mv2 R + mg 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 W at ANY point in path. W T + T W + v Top: T + W = FC 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 mv2 R - mg T v AT BOTTOM: T mg + T= mv2 R + 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 T in rope? 2 At Top: v mg T T= R v mg + T = mv2 mv - mg R R 2 (2 kg)(10 m/s) 2 T 2 kg(9.8 m/s ) 8m T = 25 N - 19.6 N TT == 5.40 5.40 NN 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 = mv2 mv + mg R R 2 (2 kg)(10 m/s) 2 T 2 kg(9.8 m/s ) 8m T = 25 N + 19.6 N TT == 44.6 44.6 NN 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 mv v At Top: mg + T = mg R vc occurs when T = 0 R T mv2 mg = vc = gR v R v= gR = (9.8 m/s2)(8 m) vvcc== 8.85 8.85 m/s m/s The Loop-the-Loop Same as cord, n replaces T v AT TOP: mg R v AT BOTTOM: n mg + + n= mv2 R n n= mv2 R + mg - mg The Ferris Wheel v AT TOP: n R v + mg AT BOTTOM: n mg + n= mg - n= n = mg mv2 R + mg mv2 R mv2 R Example 13: What is the n apparent weight of a 60-kg person as she moves v through the highest point mg when R = 45 m and the speed at that point is 6 m/s? Apparent weight will be the normal force at the top: mv2 mg - n = n = mg R 2 (60 kg)(6 m/s) 2 n 60 kg(9.8 m/s ) 45 m - + R v mv2 R nn == 540 540 NN Summary Centripetal Centripetal acceleration: acceleration: v= sgR Conical pendulum: 2 v ac ; R mv Fc mac R v2 tan = gR v= gR tan 2 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: n R v + mg AT BOTTOM: n mg + n= mg - n= n = mg mv2 R + mg mv2 R mv2 R CONCLUSION: Chapter 10 Uniform Circular Motion