2. T. K. Ng, HKUST

3. Lecture IV. Mechanics of rigid bodies

4. Mechanics of Rigid Bodies 1) Mathematics description of rigid bodies 2)Rigid Body in static equilibrium 3)Dynamics of rigid bodies

5. 1). Mathematical Description of Rigid Bodies Before discussing the laws of motion for rigid bodies, first we have to understand how many ways can a rigid object move?

6. Example: how many ways a dumbbell can move (assuming the solids at the two ends of the dumbbell can be regarded as point particles) and how can we describe its motion?

7. Dumbbell can (i) move as a whole, and (ii) rotate upon its center To describe motion as a whole, need (i) x (t) (x = position of center for example), and (ii) angles  (t) and  (t), describing the angular orientation of the dumbbell with respect to a chosen coordinate system (rotation).

8.   0

9. Exercise!! How can we describe mathematically the motion of the following objects: (i) bowling ball rolling on floor, (ii) rigid triangle moving in space, (iii) A piece of rock (iv) A piece of rubber

10. Question: how many ways can a general rigid body move? (or how many mathematical variables do we need to describe it’s motion)

11. Answer: Translation (3-degrees of freedom) + Rotation (3-degrees of freedom) (why?)

12. Therefore, to understand motion of rigid bodies, we just need to know how Newton’s Law governs Translation & Rotation!

13. To describe translation and rotation, we introduce the notion of center of mass

14. Recall: center of mass The center of mass is a special point in a rigid body with position defined by This point stays at rest or with uniform motion when there is no net force acting on the body

15. Center of mass (CM) For rigid bodies the rest of the body may be rotating upon this point which acts as the center of rotation. i.e., the motion of a rigid body can be described as motion of CM (translation) + rotation of the rest of the body upon CM.

16. Proof of the above statements Assumption: We imagine a rigid body as composed of many small point masses. mimi riri

17. Proof of the above statements Therefore, A rigid body is characterized by the set {m i,r i }. mimi riri

18. Newton’s Law for translation of rigid body We consider motion of center of mass Newton’s Law for CM motion

19. Newton’s Law for rotation (1)Angular momentum & rotation The angular momentum of a rigid body defined by a set {m i,r i } is

20. Newton’s Law for rotation (1)Angular momentum & rotation It is convenient to separate the CM motion by writing r i = R + r i ’ (where R = CM coordinate) since

21. Newton’s Law for rotation (1)Angular momentum & rotation It is convenient to separate the CM motion by writing r i = R + r i ’ (where R = CM coordinate) CM motion Rotation about CM

22. Newton’s Law for rotation (1)Angular momentum & rotation CM motion Rotation about CM

23. Newton’s Law for rotation (1)Angular momentum & torque The rate of change of angular momentum is given by CM motion Rotation about CM

24. Newton’s Law for rotation (1)Angular momentum & torque Force directed along the line joining the particles (central force)

25. Newton’s Law for rotation (1)Angular momentum & torque i.e., the rate of change of angular momentum is govern by external force only. Notice: internal force is needed to maintain rigidity of the body

26. Necessary conditions for a rigid body in static equilibrium (a)net force acting on the body=0 (b)net torque acting on the body=0 Q. Are these conditions sufficient?

27. Necessary conditions for a rigid body in static equilibrium Notice: A rigid body can translate with a uniform velocity + rotate with a uniform speed about CM even if external force = external torque = 0

28. Rigid body in static equilibrium: stability problem Question: Which configuration shown below is stable? And why? a b

29. Rigid body in static equilibrium: stability problem a is unstable because the orange block will fall if displaced a little bit away from equilibrium. a b

30. Rigid body in static equilibrium: stability problem Question: A piece of wood with uniform density is put on the table in two ways (a) and (b). Which way is more stable? And why? a b

31. Rigid body in static equilibrium: stability problem Ans: (b) is more stable because its center of mass is lower, and is more difficult to be turned over by a force. (or its potential energy is lower) a b

32. Sufficient conditions for a rigid body in static equilibrium : When the rigid body is displaced a little bit away from its equilibrium position, the force it felt pushes the body back to its equilibrium position

33. Sufficient conditions for a rigid body in static equilibrium : A rigid body is more stable if it has a configuration with lower potential energy.

34. Examples: Questions (2) & (3) in assignment I. Questions (6) & (7) in last year assignment II + ….

35. Dynamics of rigid bodies a) Rotation about a single axis passing through CM. b) Rotation ….. + translational motion of CM.

36. Rotation about a single axis Example: rotation of an object about a given axis (z) passing through CM.

37. Rotation about a single axis Let the angular speed of rotation be . Therefore

38. Newton’s Law for rotation Show that Notice:

39. Newton’s Law for rotation With external torque i.e. torque  change in rate of rotation

40. Newton’s Law for rotation With external torque Notice, I Z acts as “mass” (inertia) in Newton’s Law of translation

41. Examples What is the moment of inertia for the following objects? (a) a sphere with uniform density? (b)a cylinder with uniform density?

42. Examples What is the moment of inertia for the following objects? (a) a sphere with uniform density

43. Examples What is the moment of inertia for the following objects? (b)a cylinder with uniform density About z-axis z-axis length = l radius=R

44. Examples (b)a cylinder with uniform density About z-axis

45. Question: For a cylinder with uniform density under a constant torque. Will it’s angular acceleration larger if the torque is acting upon x- or z- directions?

46. Rotational kinetic energy

47. Example (Physical pendulum): What determines the oscillation frequency of a rigid body suspended and free to swing under it’s own weight about a fixed axis of rotation? I CM

48. Example (Physical pendulum): Moment of inertial about fixed point I CM R CM I

49. Example (Physical pendulum): Newton’s Law for rotation CM  I

50. Example (Physical pendulum): Small angle of oscillation: CM  I Oscillation frequency

51. Example 1: A person moving with velocity v and mass m jumps on a merry-go-around (MGA) which is initially at rest. The mass of the MGA is M and its radius is R. The person lands on the MGA at a point x with distance r from the origin. The velocity v is perpendicular to the line joining the origin and x. What is the final angular velocity of the MGA.

52. Example 1: A person moving with velocity v and mass m jumps on a merry-go-around (MGA) which is initially at rest. The mass of the MGA is M and its radius is R. The person lands on the MGA at a point x with distance r from the origin. The velocity v is perpendicular to the line joining the origin and x. What is the final angular velocity of the MGA. How long will it take for the MGA to stop if the friction coefficient between ground and MGA is  ?

53. Solution: What are the conservation laws that hold in example (1)? Energy? Momentum?Angular Momentum? (N) (N) (Y)

54. Solution (1): Conservation of angular momentum 

55. Solution (1): Let’s check energy change

56. Solution (1): the torque due to friction is

57. Newton’s Law implies Therefore, time taken for MGA to slow down is

58. Example 2: A person moving with velocity v and mass m jumps on a merry-go-around (MGA) which is initially at rest. The mass of the MGA is M and its radius is R. The person lands on the MGA at a point x with distance r from the origin. The velocity v is perpendicular to the line joining the origin and x. What is the final angular velocity of the MGA. What happens if the MGA is replaced by a disc of same mass and is resting on a frictionless ground?

59. This is an example of translational motion of CM + rotation about CM (one axis). Solution: We decompose the motion into (i)Translational motion of CM (ii)Rotation of the body above CM.

60. This is an example of translational motion of CM + rotation about CM (one axis). Conservation Laws: Translational motion of CM – conservation of momentum Rotation about CM – conservation of angular momentum

61. CM position:

62. Solution (2): conservation of momentum (translational motion of CM) Final velocity

63. Solution (2): conservation of angular momentum (rotation about CM) Final frequency of rotation about CM

64. Solution (2): conservation of angular momentum (rotation about CM)

65. Notice that the system is rotating about CM, NOT about center of MGA! CM Center of MGA

66. Example: A dumbbell is formed by 2 equal masses m joint by a light rod of length l. Discuss the system’s motion if a force F acts on the first mass for a very short time  t. The angle between the force and the rod joining the masses is . 

67. End