1. Predicting the path of a loosened ball Centripetal force Centripetal force and inertia Check-point 2 Daily examples of uniform circular motion Check-point 3 Check-point 4 7.2Uniform circular motion and centripetal force Book 2 Section 7.2 Uniform circular motion and centripetal force

2. P.2 Book 2 Section 7.2 Uniform circular motion and centripetal force Predicting the path of a loosened ball A hammer is swung in a circular path in a horizontal plane. If the wire suddenly breaks during the swing, which path will the ball follow? It will follow path C. Why?

3. P.3 Book 2 Section 7.2 Uniform circular motion and centripetal force 1 Centripetal force Centripetal force: the net force causing centripetal acceleration. centripetal force = mass  centripetal acceleration F = mv 2 r or F = mr  2 pointing towards the centre of the circle  the movement of the object (F  s)  does NO work on the object

4. P.4 Book 2 Section 7.2 Uniform circular motion and centripetal force 1 Centripetal force Centripetal force is not a new kind of force. Hammer throw: Conical pendulum: provided by the tension of the wire provided by the tension of the string

5. P.5 Book 2 Section 7.2 Uniform circular motion and centripetal force 1 Centripetal force Resolving T into two components. Horizontal: provides the centripetal force T sin  = mr  2  T = m  2 L Vertical: balances the weight T cos  = mg Simulation 7.1 Horizontal circular motion

6. P.6 Book 2 Section 7.2 Uniform circular motion and centripetal force 1 Centripetal force Expt 7a Verifying the equation for centripetal force

7. P.7 Book 2 Section 7.2 Uniform circular motion and centripetal force Experiment 7a Verifying the equation for centripetal force 1 Construct the apparatus. 2 Measure: mass of a rubber bung and some screw nuts Weight of screw nuts  Tension T in the string

8. P.8 Book 2 Section 7.2 Uniform circular motion and centripetal force 3 Measure the length L of nylon string from the rubber bung to the glass tube. Experiment 7a Verifying the equation for centripetal force Mark L with paper marker. 4 Whirl rubber bung around. Keep the paper marker just below the glass tube. T = m  2 L holds for all  ? Video 7.1 Expt 7a - Verifying the equation for the centripetal force

9. P.9 Book 2 Section 7.2 Uniform circular motion and centripetal force 2 Centripetal force and inertia When the wire breaks suddenly, the centripetal force due to T disappears. By Newton’s first law, the ball will continue with the same speed along a straight line  Fly off tangentially to the circle (line C )

10. P.10 Book 2 Section 7.2 Uniform circular motion and centripetal force 2 Centripetal force and inertia When the ball is in circular path, When it leaves the rail, N disappears. The ball move on straight track due to inertia. ball circular metal rail plastic beads Video 7.2 Centripetal force and inertia normal reaction N from the rail provides centripetal force. centripetal force

11. P.11 Book 2 Section 7.2 Uniform circular motion and centripetal force Example 3 Centripetal force in hammer throw 2 Centripetal force and inertia

12. P.12 Book 2 Section 7.2 Uniform circular motion and centripetal force Example 3 Centripetal force in hammer throw Alan releases a 7-kg hammer at 90 km h –1. Length of wire connecting the ball = 1.2 m Alan stretches his arm from the centre of his body by 1 m. Assume: The hammer moves in horizontal circular motion. Ignore the mass of the wire and the handle of the hammer

13. P.13 Book 2 Section 7.2 Uniform circular motion and centripetal force Example 3 Centripetal force in hammer throw (a) Tension in the wire just before release = ? T = mv 2 r 7  90 3.6 2 1.2 + 1 = = 1.99  10 3 N (b) Work done by the tension on the metal ball in the last cycle of rotation = ? ∵ Tension  movement of ball  Work done on the ball is zero.

14. P.14 Book 2 Section 7.2 Uniform circular motion and centripetal force Check-point 2 – Q1 A ball enters a channel at P. What would the path look like when the ball exits at Q ?

15. P.15 Book 2 Section 7.2 Uniform circular motion and centripetal force Check-point 2 – Q2 Tom is sitting next to the window of a revolving restaurant. His linear speed = 4.36  10 –2 m s –1 His mass = 65 kg Centripetal force experienced = ? Centripetal force = mv 2 r = 65(4.36  10 –2 ) 2 25 = 4.94  10 –3 N

16. P.16 Book 2 Section 7.2 Uniform circular motion and centripetal force 3 Daily examples of uniform circular motion a Car making a turn on a level road When a car turns around a corner, it performs a circular motion. Friction f between road and tyres = centripetal force keeping the car on the curve mv 2 r f =

17. P.17 Book 2 Section 7.2 Uniform circular motion and centripetal force depending on nature of the two contact surfaces Max. friction f max possible =  N :: coefficient of friction, N :N : Normal reaction between two contact surfaces Insufficient friction  car skids off the road Video 7.3 Overturning and skidding a Car making a turn on a level road

18. P.18 Book 2 Section 7.2 Uniform circular motion and centripetal force Max. speed for a car to make a turn without skidding off the road: f max = mv max 2 r mv max 2 r  mg = v max 2 =  gr Example 4 Limiting speed for dry and wet road  Skidding will occur if v > a Car making a turn on a level road

19. P.19 Book 2 Section 7.2 Uniform circular motion and centripetal force Example 4 Limiting speed for dry and wet road  for a normal tyre on a dry road = 0.9 (a) Find the limiting speed. = v max = = 13.4 m s –1 (48.3 km h –1 ) A car makes a turn where radius of curvature is 20 m.

20. P.20 Book 2 Section 7.2 Uniform circular motion and centripetal force Example 4 Limiting speed for dry and wet road (b) Find the new limiting speed on a wet day when  is 0.4. = = 8.94 m s –1 (32.2 km h –1 ) v max =

21. P.21 Book 2 Section 7.2 Uniform circular motion and centripetal force Example 4 Limiting speed for dry and wet road (c) If a car is turning with 50 km h –1, how should the radius of curvature be changed to make a safe turn? ∵ 50 km h –1 > limiting speed of dry road  The radius of curvature should be increased, in particular on a rainy day or when the road is wet.

22. P.22 Book 2 Section 7.2 Uniform circular motion and centripetal force 3 Daily examples of uniform circular motion b Car making a turn on a banked road Some roads are built at an angle.  Reduce the reliance of frictional force between the road and tyres. Horizontal component of N on the car contributes to centripetal force.

23. P.23 Book 2 Section 7.2 Uniform circular motion and centripetal force In ideal situation, horizontal component of N is fully responsible for the centripetal force, no friction is needed. Horizontal: N sin  = mv 2 r Vertical: N cos  = mg r..... (1) ……… (2)  Ideal banking angle: tan  = v 2 gr b Car making a turn on a banked road

24. P.24 Book 2 Section 7.2 Uniform circular motion and centripetal force Example 5 Ideal banking angle b Car making a turn on a banked road

25. P.25 Book 2 Section 7.2 Uniform circular motion and centripetal force Example 5 Ideal banking angle Radius of curvature of a highway = 300 m Average speed of car = 80 km h –1 Ideal banking angle = ? tan  = v 2grv 2gr = 10  300  = 9.35 

26. P.26 Book 2 Section 7.2 Uniform circular motion and centripetal force Example 6 Direction of frictional force b Car making a turn on a banked road

27. P.27 Book 2 Section 7.2 Uniform circular motion and centripetal force Example 6 Direction of frictional force Indicate the direction of friction required to make the car turning on the banked road if (a) speed of car < average road speed (b) speed of car > average road speed f N N mg f

28. P.28 Book 2 Section 7.2 Uniform circular motion and centripetal force At the bend of the MTR railway track, the outer rail is slightly higher. c Trains at a bend At a certain speed, no lateral thrust is exerted by outer rail on the flanges to make the turn. 3 Daily examples of uniform circular motion  The train is slightly tilted.

29. P.29 Book 2 Section 7.2 Uniform circular motion and centripetal force c Trains at a bend If the railway track is not banked,  easily damaged the outer rails and the flanges of the wheels of the train will be subjected to large stress.

30. P.30 Book 2 Section 7.2 Uniform circular motion and centripetal force 3 Daily examples of uniform circular motion d Cyclist around a circular track  turning moment on bicycle  bicycle overturns When a cyclist turns a corner, To avoid overturning, cyclist has to lean inwards. friction  centripetal force

31. P.31 Book 2 Section 7.2 Uniform circular motion and centripetal force Vertical direction: N = mg Horizontal direction: f =f = mv 2 r Take moment about the c.g. of the cyclist and the bike. fh = Nd = Nh tan  tan  fNfN v 2 gr = = d Cyclist around a circular track

32. P.32 Book 2 Section 7.2 Uniform circular motion and centripetal force Check-point 3 – Q1 (a) A bus turns a corner at 50 km h –1. radius of the corner = 50 m  s between tyres and road surface = 0.4 Will the bus stay on the road? = = 14.1 m s –1 Speed of the bus = = 13.9 m s –1 < v max  The bus will stay on the road. v max = 50 3.6

33. P.33 Book 2 Section 7.2 Uniform circular motion and centripetal force Check-point 3 – Q1 (b) Mass of bus = 2000 kg Find the max possible friction acting on the bus. f max =  N =  mg = 0.4  2000  10 = 8000 N

34. P.34 Book 2 Section 7.2 Uniform circular motion and centripetal force Check-point 3 – Q2 A car turns a corner at 75 km h –1. Radius of the circular path = 32 m Find the ideal banking angle . tan  = v 2grv 2gr  = 53.6  tan  = 10  32 75 3.6 2

35. P.35 Book 2 Section 7.2 Uniform circular motion and centripetal force 3 Daily examples of uniform circular motion e Overturning of a car on level road A car turning a corner to the left: Horizontal direction: f 1 + f 2 = mv 2 r Vertical direction: N 1 + N 2 = mg..... (3)..... (4)

36. P.36 Book 2 Section 7.2 Uniform circular motion and centripetal force At equilibrium, no net moment about c.g. Clockwise moment = anticlockwise moment f 1 h + f 2 h + N 1 d = N 2 d......... (5) By (3), mv 2 r h = (N 2 – N 1 )d Solving (4) and (6), N 1 = m 1212 N 2 = m 1212, e Overturning of a car on level road..... (6)

37. P.37 Book 2 Section 7.2 Uniform circular motion and centripetal force 1 N 2 always > 0  right wheel always touching the road 2 When v 2 =, N 1 = 0. drg h 3 For safety turning, N 1 > 0, i.e. v 2 < drg h 4 For  gr <, drg h the car skids rather than overturn. dhdh  about to overturn about the right wheel e Overturning of a car on level road i.e.  <

38. P.38 Book 2 Section 7.2 Uniform circular motion and centripetal force e Overturning of a car on level road Example 7 Formula One racing car Simulation 7.2 Making a turn

39. P.39 Book 2 Section 7.2 Uniform circular motion and centripetal force Example 7 Formula One racing car The c.g. of a racing car is 0.3 m from the ground. The width between the wheels is 1.5 m. (a) What features of an F1 car prevent it from overturning when turning a corner? Low c.g., wide separation of the wheels

40. P.40 Book 2 Section 7.2 Uniform circular motion and centripetal force Example 7 Formula One racing car (b) Radius of curvature of the corner = 100 m Just before overturning, N 1 = 0. Vertical: N 2 = mg Horizontal: f 1 + f 2 = mv 2 r... (1)... (2) v max (in km h –1 ) before overturning = ?

41. P.41 Book 2 Section 7.2 Uniform circular motion and centripetal force Example 7 Formula One racing car Taking moment about the c.g.: h (f 1 + f 2 ) =N 2 d Sub (1) and (2) into (3), h = mgd mv 2rmv 2r v 2 = drghdrgh = 1.5 2  100  10 0.3 v = 50 m s –1 = 180 km h –1...... (3)

42. P.42 Book 2 Section 7.2 Uniform circular motion and centripetal force 3 Daily examples of uniform circular motion f Aeroplane making a turn Video 7.4 A tilted aeroplane Larger tilting angle A tilted aeroplane can make a turn in the air.  sharper turn

43. P.43 Book 2 Section 7.2 Uniform circular motion and centripetal force Vertical: L cos  = mg L sin  = mv 2rmv 2r Horizontal: Combining (7) and (8): tan  = v 2 gr......... (7).... (8) When a plane flies in the air, a normal lifting force L acts on it. In order to turn in air, a plane has to incline. f Aeroplane making a turn

44. P.44 Book 2 Section 7.2 Uniform circular motion and centripetal force f Aeroplane making a turn Example 8 The sharp turns of an aeroplane

45. P.45 Book 2 Section 7.2 Uniform circular motion and centripetal force Example 8 The sharp turns of an aeroplane An aeroplane makes a turn. Tilted angle = 30 , speed = 300 km h –1 Radius of curvature r of the sharp turn = ? By tan  =, v 2grv 2gr = 1203 m r =r = v 2 g tan  = 10 tan 30  300 3.6 2

46. P.46 Book 2 Section 7.2 Uniform circular motion and centripetal force 3 Daily examples of uniform circular motion g The ‘rotor’ in amusement park Rotor rotates at high speeds and retracts its floor after reaching full speed. Horizontal: Normal reaction N acting by the wall provides centripetal force N =N = mv 2 r

47. P.47 Book 2 Section 7.2 Uniform circular motion and centripetal force Example 9 ‘Rotor’ ride Vertical: friction balances the weight of the person f = mg g Rotor in amusement park  N =  mv 2 r      gr v 2 Min. coefficient of friction  min required to prevent the person from falling:   min = gr v 2

48. P.48 Book 2 Section 7.2 Uniform circular motion and centripetal force Example 9 ‘Rotor’ ride A man rides a rotor of diameter 4 m. Take  = 0.556. Rotor rotates at min. angular velocity  min. 4 m

49. P.49 Book 2 Section 7.2 Uniform circular motion and centripetal force Example 9 ‘Rotor’ ride (a) Min. angular velocity  min = ? Radius r of the rotor = 2 m = g2rg2r  min = = = 3.00 rad s –1 gr v 2    gr (  r ) 2 = 

50. P.50 Book 2 Section 7.2 Uniform circular motion and centripetal force Example 9 ‘Rotor’ ride (b) Linear speed of the man = ? v = r  = 2  3.00 = 6.00 m s –1 (c) Express centripetal acceleration in terms of g. By a = r  2 and  min =, a = r = gg = g 0.556 = 1.80g

51. P.51 Book 2 Section 7.2 Uniform circular motion and centripetal force Check-point 4 – Q1 A racing car turns around a corner with radius of curvature r on a banked road. List the equations required. (a) Along the vertical direction, net force is zero. (N 1 + N 2 ) cos  = mg

52. P.52 Book 2 Section 7.2 Uniform circular motion and centripetal force Check-point 4 – Q1 (b) Along the horizontal direction, net force is equal to the centripetal force. (N 1 +N 2 ) sin  +(f 1 +f 2 ) cos  = mv 2rmv 2r

53. P.53 Book 2 Section 7.2 Uniform circular motion and centripetal force Check-point 4 – Q2 Speed of an aeroplane = 250 m s –1 Radius of circular path = 1.5 km  between its wings and the horizontal = ? By tan  =, v 2 gr tan  = 250 2 10  1.5  10 3  = 76.5 

54. P.54 Book 2 Section 7.2 Uniform circular motion and centripetal force The End