2013 Checkpoints Chapter 6 CIRCULAR MOTION

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1 013 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, the KE (½ ) is also constant. B C D Question 10 Fo all cicula motion the net foce is always adially inwads. The foce is always towads Giselda. Question 11 x x m m/s Question 1 Once the net foce no longe applied the hamme will continue in the diection it was taelling. The hamme will leae tangentially to the cicle it was moing in. Question 13 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 10 ms -. This will be because when this occus the weight foce will cause the ball to fall. Σa ms -, which is less than 10 ms -. Yes the ball will fall off Question 14 ΣF

2 Question 15 ΣF Question m/s Foce fom seat Weight foce (must be dawn fom the cente of mass) 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. Question 17 ΣF R + R - The eaction foce, R + C. R >, so Jim 'feels' heaie. Question Σa 0 5 ms- Question 19 The eaction foce, R R Jim will feel times as heay 650 Jim will feel 50% heaie C

3 Question 0 At the top of a hill the foce diagam looks like this: Foce fom seat - R R - R is less than, so Jim 'feels' lighte. Weight foce (must be dawn fom the cente of mass) Question 1 Σa 10 a ms - Question ΣF - R R - R 10m - 10m R 0 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 3 1 m/s The loss in PE must equal the gain in KE. h KE ½ m/s 1 m/s Question 4 KE D KE c - h J ½ m/s

4 Question 5 ΣF Question 6 With no othe foces, the combined weight of the ca and passenges is W 500. The nomal eaction at the top of the loop is 50% of this alue, hence the passenges will feel a 50% eduction in thei weight. Question 7 distance taelled Speed time taken π T π m/s Question 8 ΣF Question 9 At the top of the wheel ΣF + - At the bottom of the wheel ΣF + -

5 So at the top, because heaie. Question 30 -, you actually feel lighte and at the bottom omal (actually acts on all tyes) +, makes you feel F (can act on all tyes) Question 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. sin 15 0 cos 15 0 ΣF sin15 0 D Question 3 cos cos15 C

6 Question 33 ΣF sin15 0 ΣF sin15 0 cos cos15 sin15 cos15 0 gtan g tan tan15 1 m/s 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 faste than the ecommended speed it is going to need some exta assistance (fom the oad) in getting a 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. Question 35 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 36 The net foce is calculated fom F m Be caeful with you substitutions; the mass of 100g is 0.100kg and the adius of 80cm is 0.80m 5 F The magnitude of the net foce is F 3.1 Question 37 Tension T pole Weight

7 Question 38 The net foce acting adially inwads is the hoizontal component of the tension Tsinθ gies: Tsinθ 3.15 and the etical component of the tension Tcosθ gies: Tcosθ Tcosθ Squae both equations T sin θ + T cos θ T (sin θ + cos θ) T T 3.8 Question 39 Use Tcosθ 1 3.8(11)cosθ 1 1 cosθ θ θ 7 0 Question 40 At the top of the ide, Σ F m ΣF Weight m Tension T θ Tsinθ Tcosθ Question 41 The actual foces applied to the cage ae the tensions fom each cable and the weight Tcos T T ( ) 1.73 T T is ey close to 360 Question 4 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 50 ½ (16) (whee the distance between the top and the bottom is the diamete of 16m) m/s

8 Question 43 At the bottom of the ide, Σ F m ΣF The actual foces applied to the cage ae the tensions fom each cable and the foce of gaity Tcos30 0 (Assuming the positie diection is up) T 500 T ( ) 1.73 T Theefoe if the speed at the bottom of the ide is 0.5 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 (360 fom befoe). Question 44 At the bottom of the ide, ΣF m ΣF The actual foces applied to the cage ae the tensions fom each cable and the foce of gaity. 000 Tcos T 500 T ( ) 1.73 T 598 (Assuming the positie diection is up) Question 45 ΣF net T Question 46 The ca is moing in cicula motion, theefoe the net foce must be adially inwads. ΣF net is inwads P Question 47 Since the motocyclist is iding in a cicle the net foce acting on the ide can be calculated by the following: F net m 3 Fnet Fnet 560 Question 48 When taelling in a cicle the net foce is always towads the cente of the cicle. et

9 Question 49 To not leae the ails the tolley must hae a net foce of: Fnet m At the top of the loop the net foce is poided by the weight foce of the tolley. So: m V 8.4 m s -1 Question 50 F net m x 8.3 m/s 50 Question 51 Acceleation 50 but it is also equal to F m/s m 700 (In this case using F m is bette, because it is data that is poided in the question stem.) You could use this answe to check you answe to the peious question. Question 55 The two foces acting ae the weight, which acts etically down, and the omal eaction, which is pependicula to the suface. The sum of the two ectos should look like a hoizontal line pointing adially inwads. Question 53 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. If the nomal ecto is esoled into two components, one etical (cosθ ) and one hoizontal (sinθ m. We can substitute into the hoizontal equation, as this will eliminate the cosθ unknowns, and m. sinθ cosθ On substitution, tanθ θ m, which will gie tanθ g

Question 54 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 10 10 10 ms -1.

10 Question 54 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 55 This is unifom cicula motion so use ΣF ΣF 0 ΣF 3375 ΣF 3.38 x 10 3 This is a ey lage ide o a ey heay bike!!! Question 56 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. ΣF You could esole the omal ecto into two components, one pependicula and the othe hoizontal. sinθ θ cosθ Fom this we get that sinθ and cosθ Question 57 Use Tanθ Rg 15 tanθ 0 10 θ

11 Question 58 When the ca is just about to leae the ails, the nomal eaction is zeo. At A, ΣF + +, but 0 g g ms -1 Rounded coectly to one decimal place Question 59 Since the nomal is zeo, Melanie will be appaently weightless. This is because she is in fee fall acceleating down at 10ms -. 0 Question 60 At the bottom,

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

ΣF = r r v. Question 213. Checkpoints Chapter 6 CIRCULAR MOTION Unit 3 Physics 16 6. 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