PRECESSION-UNIT 1.pptx

PRECESSION

1. GYROSCOPES By C Naveen Raj Assistant Professor, Dept. of Mechanical Engineering, Vidya Jyothi Institute of Technology

2. Key Contents Introduction Precession Gyroscopes Applications Stability of Aeroplane while taking a turn Stability of ship Stability of two wheeler Stability of four wheeler Some example problems

3. Introduction Angular motion : A rigid body spinning at a constant angular velocity about a spin axis through the mass centre . The angular momentum of the spinning body is represented by vector whose magnitude is ‘I ω ’. The direction of angular momentum can be found from the right hand thumb rule

4. PRECESSION

5. PRECESSION CONTINUES.. Precession is the change in the orientation of the axis of the rotating body It always acts along the vertical axis. This precession causes off balance to the toy and gravity creates a moment of force that accelerates it in required direction. To balance this orientation , a device is used known as gyroscope. This torque induced precession is known as gyroscopic precession.

6. GYROSCOPE It is a device used for controlling the orientation and angular velocity of the rotating bodies.

7. GYROSCOPE CONTINUES… It is a spinning disc in which axis of rotation is free to assume any orientation by itself. Applications : inertial navigation systems in space telescope, tunnel mining, gyrocompasses in ships, aircrafts and space crafts and vehicles, to assist stability in bicycles, motorcycles and ships, MEMS gyroscopes used in smart phones.

8. GYROSCOPIC COUPLE AND ITS MAGNITUDE

9. GYROSCOPIC COUPLE CONTINUES… Consider a rotating body of mass m having radius of gyration (k) mounted on shaft supported at two bearings. Let the rotor spin about X-axis with constant angular velocity ( ω ) rad/s. The X-axis is therefore called spin axis , Y-axis is called precession axis and Z-axis is called couple or torque axis .

10. GYROSCOPOIC COUPLE CONTINUES… The angular momentum of the rotating mass is H = mk 2 ω = Iω If the shaft axis precesses through a small angle ẟ θ about Y-axis in the plane XOZ, then the angular momentum is represented by ab ab = oa x ẟ θ ẟH = H x ẟ θ = I ω ẟ θ The rate of change of angular momentum is

11. GYROSCOPIC COUPLE CONTINUES… C = = = I ω C = I ω ω p Where, C = Gyroscopic Couple ω = Angular velocity of rotating body, rad/s ω p = Angular velocity of Precession NOTE : ω p = V/R ω = V/r  

12. DIRECTION OF SPIN VECTOR,PRECESSION VECTOR AND TORQUE VECTOR

13. Continues… To determine the direction of spin vector, precession vector and torque vector right hand rule is used. The fingers represent the rotation of the rotor and the thumb shows the direction of spin, precession and torque vector.

14. Method of determining couple vector Turn the spin vector 90 in the direction of precession. The turned spin vector will then correspond to the direction of active gyroscopic couple vector. The reactive gyroscopic couple vector is taken opposite to active gyroscopic couple vector direction.

15. GYROSCOPIC EFFECT ON AEROPLANE Gyroscopic couple on aeroplane is C = I ω ω p

16. EFFECT OF REACTIVE GYROSCOPIC COUPLE ON AEROPLANE CASE 1 : ROTOR ROTATES IN CLOCKWISE DIRECTION WHEN SEEN FROM REAR END AEROPLANE TAKES LEFT TURN. ANALYSIS:

17. Continues… Case 2: ROTOR ROTATES IN CLOCKWISE DIRECTION WHEN SEEN FROM REAR END AEROPLANE TAKES RIGHT TURN ANALYSIS:

18. Continues… Case 3: ROTOR ROTATES IN ANTI-CLOCKWISE DIRECTION WHEN SEEN FROM REAR END AEROPLANE TAKES LEFT TURN ANALYSIS:

19. Continues… Case 4: ROTOR ROTATES IN ANTI-CLOCKWISE DIRECTION WHEN SEEN FROM REAR END AEROPLANE TAKES RIGHT TURN ANALYSIS:

20. PROBLEMS 1. A disc of 5 kg mass with radius of gyration 70 mm is mounted at span on a horizontal shaft spins at 720 rpm in clockwise direction when viewed from the right hand bearing. If the shaft precesses about the vertical axis at 30 rpm in clockwise direction when viewed from the top, determine the reactions at each bearing due to mass of the disc and gyroscopic effect.

21. Solution for problem 1

22. CONTINUES… 2. An aeroplane flying at a speed of 300 kmph takes right turn with a radius of 50 m. The mass of engine and propeller is 500 kg and radius of gyration is 400 mm. If the engine runs at 1800 rpm in clockwise direction when viewed from tail end, determine the gyroscopic couple and state its effect on the aeroplane . What will be the effect if the aeroplane turns to its left instead of right?

23. Solution for problem 2

24. GYROSCOPIC EFFECT ON SHIP SHIP TERMINOLOGY(SIDE VIEW & TOP VIEW)

25. CONTINUES… STEERING PITCHING ROLLING

26. CONTINUES… STEERING : Turning of a ship in a curved path while moving forward PITCHING : The upward and downward movement of the ship from horizontal position ROLLING : The sideway motion of the ship

27. CONTINUES… GYROSCOPIC COUPLE ON SHIP DURING STEERING : C = I ω ω p GYROSCOPIC COUPLE ON SHIP DURING PITCHING :

28. CONTINUES…

29. CONTINUES… GYROSCOPIC COUPLE ON SHIP DURING ROLLING : The axis of the ship is mounted along the longitudinal axis of the ship, therefore there will be no precession of this axis . hence, there is no effect of gyroscopic couple on the ship.

30. PROBLEMS 1. A turbine rotor of a ship has a mass of 3500 kg and rotates at a speed of 2000 rpm. The rotor has a radius of gyration of 0.5 m and rotates in clockwise direction when viewed from the stern (rear) end. Determine the magnitude of gyroscopic couple and its direction for the following conditions ( i ) When the ship runs at a speed of 12 knots and steers to the left in a curve of 70 m radius (ii) When the ship pitches 6° above and 6° below the horizontal position and the bow (Front) end is lowered. The pitching motion is simple harmonic with periodic time 30 sec. ( iii)When the ship rolls and at a certain instant, it has an angular velocity of 0.05 rad/s clockwise when viewed from the stern Also find the maximum angular acceleration during pitching.

31. Solution for problem 1

32. Continues…

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35. problems 2. A ship is propelled by a rotor of mass of 2000 kg rotates at a speed of 2400 rpm. The radius of gyration of rotor is 0.4 m and spins clockwise direction when viewed from bow (front) end . Find the gyroscopic couple and its effect when ; ( i ) the ship takes left turn at a radius of 350 m with a speed of 35 kmph (ii) the ship pitches with the bow rising at an angular velocity of 1 rad/s (iii)the ship rolls at an angular velocity of 0.15 rad/s

36. Solution for problem 2

37. Solution for problem 2

38. Solution for problem 2

39. Solution for problem 2

40. GYROSCOPIC EFFECT ON TWO WHEELER

41. Magnitude of gyroscopic couple Let m = Mass of the vehicle and its rider in kg, W = Weight of the vehicle and its rider in newtons = m.g , h = Height of the centre of gravity of the vehicle and rider , r w = Radius of the wheels , R = Radius of track or curvature , I w = Mass moment of inertia of each wheel , I E = Mass moment of inertia of the rotating parts of the engine , ω W = Angular velocity of the wheels , ω E = Angular velocity of the engine rotating parts , G = Gear ratio = ω E / ω W , v = Linear velocity of the vehicle = ω W × r W , θ = Angle of heel. It is inclination of the vehicle to the vertical for equilibrium.

42. CONTINUES… We know V = ω w x r w ω E = G x ω w = G x (V/ r w ) Angular momentum due to wheels = 2 x I w ω w Angular momentum due to engine = I E x ω E Total angular momentum(I ω ) = 2I w ω w ± I E ω E = 2I w (V/ r w ) ± I E G x (V/ r w ) = V/ r w [2I w ± I E G]

43. Continues…

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45. Continues…

46. Continues…

47. problems 1) A motorcycle and its rider together weighs 2000 N and their combined centre of gravity is 550 mm above the road when motorcycle is upright. Each wheel is of 580 mm diameter and has a moment of inertia of 1.0 kgm2 . The moment of inertia of rotating parts of engine is 0.15 kg m2 . The engine rotates at 5 times the speed of the vehicle and the same sense. Determine the angle of heel necessary when motorcycle is taking a turn over a track of 35 m radius at a speed of 60 kmph .

48. Solution

49. Continues…

50. GYROSCOPIC EFFECT ON FOUR WHEELER

51. Continues… Let m = Mass of the vehicle (kg) W = Weight of the vehicle (N) = m.g , h = Height of the centre of gravity of the vehicle (m ) r w = Radius of the wheels (m ) R = Radius of track or curvature (m ) I w = Mass moment of inertia of each wheel (kg-m 2 ) IE = Mass moment of inertia of the rotating parts of the engine (kg-m 2 ) ω w = Angular velocity of the wheels (rad/s ) ω E = Angular velocity of the engine (rad/s ) G = Gear ratio = ω E / ω W , v = Linear velocity of the vehicle (m/s)= ω w × r w , x = Wheel track (m ) b = Wheel base (m)

52. Continues… Reaction due to weight of the Vehicle : Assuming the weight of the vehicle is equally distributed over all four wheels Therefore, force on each wheel will be acting downwards of magnitude mg/4 and reaction by the road acts upwards. Therefore, Rw = mg/4 Rw = mg/4

53. Continues… Effect of gyroscopic couple due to wheel : Gyroscopic couple due to four wheels, Effect of gyroscopic couple due to engine : Gyroscopic couple due to rotating parts of engine , Therefore total gyroscopic couple, Cg = Cw ± Ce Cw = 4I ωω p Ce = I ω E ω P = IG ω W ω P

54. Continues… Analysis: let car takes left turn

55. Continues… The effect of reactive gyroscopic couple is it tends to press the outer wheels and lift the inner wheels.

56. Continues… Due to reactive gyroscopic couple, reactions will be acting vertically upwards at outer wheels and vertically downwards at inner wheels.

57. Continues… Let the magnitude of this reaction at two outer and inner wheels be P. then, P x X = Cg Road reaction on each outer/inner wheel, P/2 = Cg/2X

58. Continues… Effect of centrifugal couple: When a vehicle moves on a curved path, a centrifugal force acts on the vehicle in outward direction through the centre of gravity of the vehicle.

59. Continues… Centrifugal force, F = mV 2 / R Centrifugal couple, Cc = (mV 2 / R) x h This centrifugal couple tends to press the outer wheels and lift the inner wheels .

60. Continues… Due to centrifugal couple, reactions will acts vertically upwards at the outer wheels and vertically downwards at the inner wheels.

61. Continues… Let the magnitude of this reaction at two outer and inner wheels be F, then Road reaction on each outer/inner wheel, F/2 = Cc/2x The reactions on the outer/inner wheel are Total vertical reaction at each outer wheels Ro = W/4 + P/2 + F/2

62. Continues… Total vertical reaction at each inner wheels Ri = W/4 – P/2 – F/2

63. Problems An automobile car is travelling along a track of 100 m mean radius. The moment of inertia of 500 mm diameter wheel is 1.8 kg m2 . The engine axis is parallel to the rear axle and crank shaft rotates in the same sense as the wheel. The moment of inertia of rotating parts of the engine is 1 kg m2 . The gear ratio is 4 and the mass of the vehicle is 1500 kg . If the centre of gravity of the vehicle is 450 mm above the road level and width of the track of the vehicle is 1.4 m , determine the limiting speed of the vehicle for condition that all four wheels maintain contact with the road surface.

64. solution

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66. Continues… The end