ΣF = r r v. Question 213. Checkpoints Chapter 6 CIRCULAR MOTION

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1 Unit 3 Physics Cicula Motion Page 1 of 9 Checkpoints Chapte 6 CIRCULAR MOTION Question 13 Question 8 In unifom cicula motion, thee is a net foce acting adially inwads. This net foce causes the elocity to change (in diection). Since the speed is constant, the KE (½ ) is also constant. B C D Question 9 Fo all cicula motion the net foce is always adially inwads. The foce is always towads Giselda. Question ΣF N Question 11 x Nx m m/s As soon as the foce ceases to be applied, the hamme will continue in the diection it was taelling. The hamme will leae tangentially to the cicle it was moing in. ΣF + N N N Question 14 ΣF m/s Question 15 Question 1 The ball will lose contact and fall off, at the point C, if the acceleation at the point C, due to its motion, is less than ms -. This will be because when this occus the weight foce will cause the ball to fall. Σa ms -, which is less than ms -. Yes the ball will fall off Because the caiage is moing, it is undegoing cicula motion at this point, so the foce fom the seat must be geate than the weight foce, the net foce is acting up.

2 Unit 3 Physics Cicula Motion Page of 9 Question 16 ΣF R - R + The eaction foce, R +. R >, so Jim 'feels' heaie. C Question 17 Σa 5 ms - Question 18 The eaction foce, R R N Jim will feel times as heay 65 Jim will feel 5% heaie C Question 19 At the top of a hill the foce diagam looks like this: Question Σa Question 1 a ms - ΣF - R R - R m - m R This means that since the oad is not pushing up on the tuck, then the tuck is not pushing down on the oad. Theefoe thee isn't a contact foce between the tuck and the oad, so the tuck is not in contact with the oad. Question The loss in PE must equal the gain in KE. Δ h Δ KE ½ m/s 1 m/s Question 3 KE D KE c - h J ½ m/s Question 4 ΣF - R R - R is less than, so Jim 'feels' lighte.

3 Unit 3 Physics Cicula Motion Page 3 of 9 ΣF + N N N Question 5 With no othe foces, the combined weight of the ca and passenges is W 5. The nomal eaction at the top of the loop is 5% of this alue, hence the passenges will feel a 5% eduction in thei weight. Question 6 Speed distance taelled time taken π T π 3.14 m/s So at the top, because N -, you actually feel lighte and at the bottom N +, makes you feel heaie. Question 9 Question 7 ΣF N Question 3 Question 8 At the top of the wheel ΣF + N N - At the bottom of the wheel ΣF N - N + Since the ca is moing in a hoizontal cicle, the net foce must be acting adially inwads (hoizontally). Fo this you can't esole the weight foce, because it is pependicula to the net foce. You hae to esole the nomal eaction.

4 Unit 3 Physics Cicula Motion Page 4 of 9 lage enough foce acting adially inwads. The oad is going to need to supply this exta foce. Line B epesents the foce fom the oad. The component of this foce acting hoizontal and adially inwads will assist the ca to complete the cone. ΣF Nsin15 D Question 31 Question 3 Ncos15 + Ncos15 C ΣF Nsin15 ΣF N sin15 Ncos15 + N cos15 sin15 cos15 gtan15 g tan tan15 1 m/s Question 33 Line C is acting etically downwads fom the cente of mass, it is the gaity foce on the weight. Line A is acting pependicula to the oad suface, if it was shown acting fom the oad, not the cente of mass, then it would be the nomal eaction foce of oad on ca. As the ca is taelling faste than the ecommended speed it is going to need some exta assistance (fom the oad) in getting a Question 34 Line C is acting etically downwads fom the cente of mass, it is the gaity foce on the weight. Line A is acting pependicula to the oad suface, if it was shown acting fom the oad, not the cente of mass, then it would be the nomal eaction foce of oad on ca. As the ca is taelling slowe than the ecommended speed it is going to exet a foce on the oad that is outwad as in line D Question 35 The net foce is calculated fom Fm Be caeful with you substitutions; the mass of g is. kg and the adius of 8 cm is.8 m 5 F..8 The magnitude of the net foce is F 3.1 N Question 36 Question 37 Weight Tension T pole The net foce acting adially inwads is 3.15N. the hoizontal component of the tension Tsinθ gies: Tsinθ 3.15 and the etical component of the tension Tcosθ gies:

5 Unit 3 Physics Cicula Motion Page 5 of 9 Tcosθ Tcosθ. 1 Squae both equations T sin θ + T cos θ T (sin θ + cos θ).7656 T.7656 T 3.8 N Question 38 Use Tcosθ 1 3.8(11)cosθ 1 1 cosθ θ 7 Question 39 At the top of the ide, Σ Fm ΣF N Question 4 The actual foces applied to the cage ae the tensions fom each cable and the weight Tcos T + 5 T (315 5) 1.73 T 36.8 T is ey close to 36N Question m The total enegy of the system will emain constant. At the top Total Enegy KE top + GPE top ½mu + h top whee u is the speed at the top At the bottom, Total Enegy KE bottom + GPE bottom ½ + h bottom whee is the speed at the bottom ½mu + h top ½ + h bottom h top - h bottom ½ - ½mu (h top - h bottom ) ½ 5 ½ 5 5 (16) (whee the distance between the top and the bottom is the diamete of 16m) m/s Question 4 At the bottom of the ide, ΣFm ΣF The actual foces applied to the cage ae the tensions fom each cable and the foce of gaity Tcos3 (Assuming the positie diection is up) T 5 T ( ) 1.73 T N Theefoe if the speed at the bottom of the ide is.5 m/s, then the tension in the cable at the bottom of the ide is geate than the tension in the cable at the top of the ide (36N fom befoe). Question 43 At the bottom of the ide, ΣFm ΣF The actual foces applied to the cage ae the tensions fom each cable and the foce of gaity. Tcos3 (Assuming the positie diection is up) 1.73T 5 T ( + 5) 1.73 T 598 N

6 Unit 3 Physics Cicula Motion Page 6 of 9 Question 44 ( Q1, m, 85%) F net m m/s 5 Question 45 ( Q, m, 87%) Acceleation 5 but it is also equal to F 11 m 7 16 m/s (In this case using F is bette, because it is m data that is poided in the question stem.) You could use this answe to check you answe to the peious question. Question 46 ( Q5, m, 47%) The two foces acting ae the weight, which acts etically down, and the Nomal eaction, which is pependicula to the suface. The sum of the two ectos should look like a hoizontal line pointing adially inwads. one hoizontal (Nsinθ substitute m. We can N cosθ into the hoizontal equation, as this will eliminate the unknowns, N and m. sinθ N cosθ m, which will gie tanθ g On substitution, tanθ.1 θ 5.7 Question 48 ( Q7, m, 85%) The ides will feel as if they hae no weight if the nomal eaction is zeo. This means that at the point C, the net foce acting on the ides must be. m g g ms -1. Question 47 ( Q6, 3m, 5%) If thee is to be no sideways fictional foce between the tyes and the tack, then the two ectos need to combine to gie a hoizontal ecto, adially inwads. Question 49 (11 Q4, 1m, 84%) This is unifom cicula motion so use ΣF 3 15 ΣF ΣF 3375 ΣF N This is a ey lage ide o a ey heay bike!!! If the nomal ecto is esoled into two components, one etical (Ncosθ ) and

7 Unit 3 Physics Cicula Motion Page 7 of 9 Question 5 (11 Q5, m, 5%) The only two foces acting ae the weight foce, acting down and the nomal eaction fom the suface, acting pependicula to the suface. They must add togethe to gie the esultant foce ΣF, which is adially inwads. N ΣF You could esole the Nomal ecto into two components, one pependicula and the othe hoizontal. Question 53 (11 Q, m, 5%) Since the nomal is zeo, Melanie will be appaently weightless. This is because she is in fee fall, acceleating down at ms -. N Question 54 (11 Q11, 3m, 54%) At the bottom, N N N 71 N Nsinθ Fom this we get that Nsinθ and Ncosθ Question 51 (11 Q6, m, 65%) N θ Ncosθ Question 55 (1 Q7a, m, 8%) Use ΣF ma T m/s Use Tanθ Rg 15 tanθ θ 48.4 Question 56 (1 Q7b, 1m, 75%) The sphee will fly off tangentially. Question 5 (11 Q9, m, 74%) When the ca is just about to leae the ails, the nomal eaction is zeo. At A, ΣF N + N +, but N g g ms -1 Rounded coectly to one decimal place

8 Unit 3 Physics Cicula Motion Page 8 of 9 Question 57 (1 Q7c, 3m, 33%) Conside the motion to be etical, at constant speed. Question 58 (13 Q4a, m, 6%) N F net T At the top of the path, thee ae two foces acting on the sphee, its weight,, and the tension fom the sting T. ΣF ma Becomes T + T -. At the bottom of the cicle, the situation looks like. T The two foces acting ae the weight foce (fom the cente of mass) acting etically downwads, and the Nomal eaction, which is a foce pependicula to the banked suface. The net foce is a ecto acting adially inwads (it had to be hoizontal). The question didn t ask you to label, and N. Good physics means that you do. Also it s bette to be safe than soy. Question 59 (13 Q4b, m, 54%) This question einfoces the need to hae the fomula on the cheat sheet, een though it is not on the couse. Use tanθ g 5 tanθ θ 7.1 ΣF ma Becomes T - T +. Since is constant, the tension at the bottom is lage than the foce at the top. the tension in the sting is geate at the bottom of the cicula path. Question 6 (13 Q5a, 1m, 68%) Since the mass is taelling at a constant speed, the net foce is adially inwads. D Question 61 (13 Q5b, 3m, 43%) At the point S thee ae two foces acting on the mass. The weight is acting down, and the tension foce is up. Since the net foce is up, ΣF ma ma T a 7 1

9 Unit 3 Physics Cicula Motion Page 9 of 9 49 m/s. 49 T. T 118 N Question 6 (14 Q4a, m, 7%) Fo May and Bob to feel weightless, the Nomal eaction must be zeo. / g / / m/s Question 63 (14 Q4b, 3m, 4%) At this speed, May and Bob ae in fee fall. The nomal eaction is zeo, so the only foce acting on them is the foce due to gaity. This is the conditions fo appaent weightlessness. Question 64 (14 Q4c, 3m, 67%) N F The weight is acting down fom the cente of mass. The nomal is etically upwads, fom the suface. The nomal ecto is lage than the weight ecto. The esultant foce is acting upwads

2013 Checkpoints Chapter 6 CIRCULAR MOTION

2013 Checkpoints Chapter 6 CIRCULAR MOTION 013 Checkpoints Chapte 6 CIRCULAR MOTIO Question 09 In unifom cicula motion, thee is a net foce acting adially inwads. This net foce causes the elocity to change (in diection). Since the speed is constant,