2 Centripetal forces keep these children moving in a circular path.
3 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.
4 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.vFcConstant force toward center.Question: Is there an outward force on the ball?
5 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.vWhen central force is removed, ball continues in straight line.Centripetal force is needed to change direction.
6 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?FcCar going around a curve.Force ON you is toward the center.
7 The centripetal force is exerted BY the door ON you. (Centrally) Car Example ContinuedThe centripetal force is exerted BY the door ON you. (Centrally)F’ReactionFcThere 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.
8 Another Example R Disappearing platform at fair. Fc 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.
9 Spin Cycle on a WasherHow 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.
10 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.)nFcRvWForce Fc and acceleration ac toward center. W = n
11 Deriving Central Acceleration Consider initial velocity at A and final velocity at B:vfvfBRvo-voDvRvosA
13 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?Rvmm = 3 kgR = 5 m; v = 8 m/sF = (3 kg)(12.8 m/s2)Fc = 38.4 N
14 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 sketch450 N30 mv = 15 m/sRFcm=?Speed skaterm = 60.0 kg
15 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 sketchr = ?m = 80 kg; v = 4 m/s2Newton’s 2nd law for circular motion:Fc = 600 Nr = 2.13 m
16 Car Negotiating a Flat Turn vFcWhat is the direction of the force ON the car?Ans. Toward CenterThis central force is exerted BY the road ON the car.
17 Car Negotiating a Flat Turn vFcIs there also an outward force acting ON the car?Ans. No, but the car does exert a outward reaction force ON the road.
18 Car Negotiating a Flat Turn The centripetal force Fc is that of static friction fs:RvmFcnFc = fsfsRmgThe 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.
19 Finding the maximum speed for negotiating a turn without slipping. Fc = fsnmgfsRRvmFcThe car is on the verge of slipping when FC is equal to the maximum force of static friction fs.Fc =mv2Rfs = msmgFc = fs
20 Maximum speed without slipping (Cont.) Fc = fsnmgfsRvmFcmv2R= msmgv = msgRVelocity v is maximum speed for no slipping.
21 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?RvmFcms = 0.7Fc =mv2Rfs = msmgFrom which:v = msgRg = 9.8 m/s2; R = 70 mv = 21.9 m/s
22 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.RvmFcnfs = 0nfsqslow speedqnfsqwwwfast speedoptimum
23 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).nmgqxn cos qnmgqnq+ acn sin qqmg
24 Optimum Banking Angle (Cont.) mgqn sin qn cos qqnmgmv2Rn sin q =Apply Newton’s 2nd Law to x and y axes.SFx = macn cos q = mgSFy = 0
25 Optimum Banking Angle (Cont.) mgqn sin qn cos qqnmgmv2Rn sin q =n cos q = mg
26 Optimum Banking Angle (Cont.) mgqn sin qn cos qOptimum Banking Angle q
27 How might you find the centripetal force on the car, knowing its mass? 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?qnmgtan q = =v2gR(12 m/s)2(9.8 m/s2)(80 m)q = 10.40nmgqn sin qn cos qtan q = 0.184How might you find the centripetal force on the car, knowing its mass?
28 The Conical PendulumA 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 qqhTLRTqmgT sin qNote: The inward component of tension T sin q gives the needed central force.
29 Solve two equations to find angle q Angle q and velocity v:qhTLRmgT sin qT cos qSolve two equations to find angle qmv2RT sin q =tan q =v2gRT cos q = mg
30 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 = 300qhTLR2. 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
31 Example 6(Cont.): Find v for q = 300 hTLRq = 300R = 5 m4. Use given info to find the velocity at 300.R = 5 mg = 9.8 m/s2Solve for v = ?v = 5.32 m/s
32 Example 7: Now find the tension T in the cord if m = 2 kg, q = 300, and L = 10 m. mgT sin qT cos q2 kgSFy = 0: T cos q - mg = 0; T cos q = mgT = =mgcos q(2 kg)(9.8 m/s2)cos 300T = 22.6 N
33 Example 8: Find the centripetal force Fc for the previous example. qhTLRmgT sin qT cos qq = 300Fc2 kgm = 2 kg; v = m/s; R = 5 m; T = 22.6 NFc =mv2Ror Fc = T sin 300Fc = 11.3 N
34 Swinging Seats at the Fair This problem is identical to the other examples except for finding R.qhTLRbdR = d + bR = L sin q + btan q =v2gRandv = gR tan q
35 Example 9. If b = 5 m and L = 10 m, what will be the speed if the angle q = 260? tan q =v2gRqTLRdbR = d + bd = (10 m) sin 260 = 4.38 mR = 4.38 m + 5 m = 9.38 mv = 6.70 m/s
36 Motion in a Vertical Circle +TmgvTop RightWeight causes small decrease in tension TConsider the forces on a ball attached to a string as it moves in a vertical loop.vTmgTop of PathTension is minimum as weight helps Fc force++vTmgTop RightWeight has no effect on TvTmg+Left SideWeight has no effect on T+TmgvBottomMaximum tension T, W opposes Fc+TmgvBottomNote 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.
37 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.RvT10 N++T10 NThe tension T must adjust so that central resultant is 40 N.At top: 10 N + T = 40 NT = _?_T = 30 NBottom: T – 10 N = 40 NT = 50 NT = __?___
38 Motion in a Vertical Circle Fc =mv2RResultant force toward centerTmgConsider TOP of circle:mg + T =mv2RAT TOP:+TmgT = mgmv2R
39 Vertical Circle; Mass at bottom Fc =mv2RResultant force toward centerTmgConsider bottom of circle:T - mg =mv2RAT Bottom:Tmg+T = mgmv2R
40 Visual Aid: Assume that the centripetal force required to maintain circular motion is 20 N. Further assume that the weight is 5 N.RvResultant central force FC at every point in path!FC = 20 NWeight vector W is downward at every point.W = 5 N, downFC = 20 N at top AND at bottom.
41 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.RvFC = 20 N at top AND at bottom.Top: T + W = FCWT+T + 5 N = 20 NT = 20 N - 5 N = 15 N+TWBottom:T - W = FCT - 5 N = 20 NT = 20 N + 5 N = 25 N
42 For Motion in Circle v AT TOP: mv2 R + T = - mg R mg T AT BOTTOM: mv2
43 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 =mv2RAt Top:RvTmgT = mgmv2RT = 5.40 NT = 25 N N
44 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 =mv2RAt Bottom:RvTmgT = mgmv2RT = 44.6 NT = 25 N N
45 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 =mv2RAt Top:RvTmgvc occurs when T = 0mg =mv2Rvc = gRvc = 8.85 m/sv = gR = (9.8 m/s2)(8 m)
46 The Loop-the-Loop Same as cord, n replaces T vAT TOP:n = mgmv2Rnmg+AT BOTTOM:nmg+n= mgmv2R
47 The Ferris Wheel n n = mg - n n = + mg R v mg - n = mv2 R AT TOP: mg + AT BOTTOM:nmg+n = mgmv2R
48 Apparent weight will be the normal force at the top: mg+RvExample 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 =mv2Rn = mg -mv2Rn = 540 N
49 Centripetal acceleration: SummaryCentripetal acceleration:tan q =v2gRv = msgRv = gR tan qConical pendulum:
50 Summary: Motion in Circle vRAT TOP:Tmg+T = mgmv2AT BOTTOM:T = mg
51 Summary: Ferris Wheel n n = mg - n = + mg AT TOP: mg + AT BOTTOM: mv2Rn = mgvn = mg -