2. Torque The perpendicular distance from the line of action to the pivot point is called the moment arm (r). Torque is the force multiplied by the moment arm. T = F*r, Units: N-m If the line of action of a force vector does not go through the pivot point of an object, it will tend to rotate the object. F r F r

3. Torque centric force –applied through axis (center) of rotation –creates no torque so causes no rotation eccentric force –applied some distance away from axis of rotation –creates torque so causes rotation

4. Which door is easier to open? Will the green person go up or down?

5. What causes a limb to rotate - force or torque? angular motion occurs at a joint, so torque causes the limb to rotate torque is developed because a force acts at a distance from the axis of rotation muscle force (F m ) Perpendicular distance between pt of application and joint axis (d m ) muscle torque (T m =F m *d m )

6. Calculation of Muscle Torque 400 N NOTE: The torque created by the muscle depends on 1) the size of the muscle force 2) the angle at which the muscle pulls 3) the perpendicular distance from the muscle to the joint axis

7. Example F m = ? W fa =13.35 NW bb =44.5 N 0.02 m 0.15 m 0.45 m

8. Give me a lever long enough and a fulcrum strong enough, and, single-handedly, I can move the world. --Archimedes (287 - 212 B.C.)

9. Levers A lever consists of two forces (motive and resistance forces) acting around a pivot point (axis or fulcrum). The perpendicular distance from the line of action of the effort force to the fulcrum is called the motive arm. The perpendicular distance from the line of action of the resistance force to the fulcrum is called the resistance arm.

10. Elements of a Lever motive force (effort force) axis (fulcrum) motive arm resistance arm resistance force

11. Mechanical Advantage of a Lever MA = The ratio of the motive arm to the resistance arm is called the mechanical advantage (MA). If MA is approximately 1 the lever simply acts to redirect the applied force. If MA is > 1 the lever acts to amplify the force. If MA is < 1 the lever acts to amplify the speed and range of motion.

12. MA = 1 motive force direction of the force vector is redirected motive arm = resistance arm

13. MA > 1 motive force motive arm > resistance arm force is amplified

14. MA < 1 motive arm < resistance arm ROM / speed is amplified motive force

15. Muscles have MA <1 0.02 m (motive arm) 0.15 m (resistance arm) 0.45 m (resistance arm) motive arm < resistance arms

16. Mechanical Advantage Muscles typically have a MA in ROM and speed. Classes of Levers Classified according to the relative positions of the axis, motive force and resistive force.

17. How to remember the class of lever 1st class Axis is between resistance and motive force. 2nd class Resistance force is in between the axis and the motive force. 3rd class Motive force is in between the axis and the resistance force. ARM

18. 1st Class Lever axis in the middle e.g. see-saw most versatile lever because it can be used for any type of mechanical advantage e.g. in body –pushing down gas pedal –elbow/triceps extensions overhead

19. 2nd Class Lever resistance in middle force advantage usually exists for motive force e.g. push-up –body is lever, feet are axis, resistance is weight of body and motive is arms

20. 3rd Class Lever motive in middle most musculoskeletal arrangements are 3rd class levers muscle is motive force advantage in ROM and speed but disadvantage in F

21. Gears A gear is similar in function to a lever. The torque on the wheel and the gear is the same. The moment arms are different and therefore the forces are different.

22. Equilibrium – Linear Components W RyRy if R y = W then resultant force = 0 if v = 0 and  F = 0 STATIC EQUILIBRIUM W RyRy FpFp FrFr F r = resistive force F p = propulsive force if v = 0 and  F = 0 DYNAMIC EQUILIBRIUM V

23. Equilibrium – Angular Components if T1 = T2 then resultant torque = 0 if  = 0 and  = 0 STATIC EQUILIBRIUM T1=T2 DYNAMIC EQUILIBRIUM b/c  = constant and  = 0 T1 T2 If the object is rotating with a non-zero velocity and…

24. Stability Stability is the resistance to linear and angular acceleration. There are 3 major factors that influence the stability of an object.

25. 1) The line of gravity with respect to the base of support.

26. 2) The height of the center of mass. unstable stable

27. 3) The mass. 10 kg unstable 100kg stable

28. Center of Mass The center of mass is the point about which the body's mass is evenly distributed. The line of gravity is the line that defines the center of mass in the transverse plane. The sum of the torques about an axis caused by the weights of multiple particles is equal to the distance from the axis to the center of mass multiplied by the sum of the weights.

29. Center of Mass The center of mass is the point about which the body's mass is evenly distributed. Symmetric distribution CM in the middle balance point Asymmetric distribution CM closer to larger weight

30. The sum of the torques about an axis caused by the weights of multiple particles is equal to the distance from the axis to the center of mass multiplied by the sum of the weights. Y X 2 kg (1,3) 1 kg (1,1) 3 kg (3,3) COM (2, 2.67) 1 2 3 line of gravity (if Y is vertical)

31. m 1 x 1 + m 2 x 2 + m 3 x 3 = Mx cm 2kg(1m) + 1kg(1m) + 3kg(3m)= 6kgx cm x cm = = 2 m Y X 2 kg (1, 3) 1 kg (1,1) 3 kg (3, 3) COM (2, 2.67) 1 2 3 COM Location in the x direction

32. m 1 y 1 + m 2 y 2 + m 3 y 3 =My cm 2kg(3m) + 1kg(1m) + 3kg(3m)=6y cm y cm = = 2.67 m Y X 2 kg (1, 3) 1 kg (1,1) 3 kg (3, 3) COM (2, 2.67) 1 2 3 COM Location in the y direction

33. General Formulas: where, x i is the distance from the y-axis to the i th mass y i is the distance from the x-axis to the i th mass m i is the mass of the i th element (segment)

34. An alternative approach is to use the proportion of each mass (p i ) instead of the actual masses. For the human body the proportion can be found in many text books for each body segment.

35. X Y

36. Body Segment Parameters Derived from direct cadaver measurements Elderly, male, Caucasian cadavers From these data it is apparent that to determine the center of mass of a segment it is necessary to locate the segment endpoints

37. X Y

38. X Y

39. X Y 6.8 cm To locate the segment CM 1 st measure the length of the segment

40. X Y 3.0 cm CM of the trunk is 43.8% of the length of the trunk away from the suprasternale 43.8% of 6.8 cm = 3.0 cm

41. X Y to find the whole body CM you need to express the segmental CM locations with respect to a common reference point - we’ll use the origin yiyi xixi

42. X Y Do this for every segment yiyi xixi Use these distances and the segment masses to compute the whole body CM location

43. X Y Y-distance = 92 mm X-distance = 120 mm Plot the final coordinates of the CM

44. Straddle Jump 

45. Fosbury Flop 

46. A basketball player can appear to remain at a constant height for brief periods of time by manipulating the body segments about their center of mass. The COM will always follow the path of a parabola while the body is in the air. (Michael Jordan is very good at this.)

47. Resistance to angular motion (like linear motion) is dependent on mass. The more closely mass is distributed to the axis of rotation, the easier it is to rotate. therefore: resistance to angular motion is dependent on the distribution of mass This resistance is called the Moment of Inertia. Moment of Inertia

48. ANGULAR FORM OF INERTIA –resistance to changes in the state of angular motion I = mr 2 –for a single particle I is proportional to the distance squared SI unit = kg-m 2

49. Each block is.5 m by 1.5 m and has a mass of 2 kg. The mass in each block is uniformly distributed. What is the moment of inertia about the x axis? I x =  m i r i 2 = [2kg*(.25m) 2 ] + [2kg*.75m) 2 ] + [2*(.25m) 2 ] + [2kg*(.75m) 2 ] = 2.5 kg-m 2 x

50. If the mass of the above object were concentrated at a single point (the center of mass) how far from the axis would it have to be located to have the same moment of inertia? I x = 2.5kg-m 2 = mk 2 = 8kg*k 2 k = =.559 m x

51. This value is called the radius of gyration: distance from axis of rotation to a point where the body’s mass could be concentrated without altering its rotational characteristics for a system of particles I = mk 2 where k = ‘radius of gyration’ It is often expressed as a proportion of the segment length in biomechanics. Thus, I = m(  l) 2 where I is the moment of inertia  is the radius of gyration as a proportion of the segment length (l)

52. Different Axes recognize that rotation can occur about different axes –each axis has its own moment of inertia associated with it

53. Whole Body Moment of Inertia consider human movement to occur about 3 principal axes each principal axis has a principal moment of inertia associated with it when mass is distributed closer to the axis, the moment of inertia is lower

54. Angular Analog Newton’s Laws 1) A rotating body will continue to turn about its axis of rotation with constant angular momentum, unless an external couple or eccentric force is exerted upon it. linear momentum p = m*v angular momentum H = I* 

56. Angular Momentum linear momentum p = m*v angular momentum H = I*  In the linear case mass does not change but the moment of inertia can be manipulated by reorienting body segments. - ice skaters - divers

57. Angular Analog Newton’s Laws 2) The rate of change of angular momentum of a body is proportional to the torque causing it and the change takes place in the direction in which the torque acts. T = I  f -  i t T = I  or

58. Angular Analog Newton’s Laws 3) For every torque that is exerted by one body on another there is an equal and opposite torque exerted by the second body on the first.

59. TRANSFER OF ANGULAR MOMENTUM enter pike - H legs because legs slow down H trunk+arms to maintain a constant H total the opposite occurs at entry - H trunk + arms to give a clean entry H legs to maintain H total

60. Angular Momentum in the Long Jump H total = H trunk+head + H arms + H legs = constant CW To prevent trunk+head from rotating forward (CW) rotate arms and legs CW to account for H total

61. I arms and I legs are smaller than I total so  arms and  legs must be larger to produce H’s large enough to accommodate H total

62. Initiation of Rotation in Air Newton’s Laws specifically state that you can NOT initiate rotation (e.g. in the air) without an external torque being applied to you So -- can you initiate rotation while airborne? A cat does! (seemingly)

63. explanation - consider relationship between I’s of body parts that interact when rotation is initiated 1) As the cat begins to fall it bends in middle, brings its front legs in close to its head and rotates the upper body through 180 degrees.

64. 1a) In reaction to the upper body the lower body will rotate in the opposite direction. However -- since the body is piked I lower body is very large compared to upper body so the corresponding rotation is small (about 5 degrees).

65. 2) To complete the 180 degree rotation the cat brings its hind legs and tail into line with its lower trunk such that its longitudinal axis runs through its hindquarters. 2a) The reaction of upper body is again small since I upper body (about this axis of rotation) is large, so there is little rotation of upper body

66. 3) Minor adjustments are made by rotating tail in direction opposite to the desired rotation.

68. Centripetal vs. Centrifugal Force Centripetal force (center seeking force) = mass x centripetal acceleration Centrifugal force (center fleeing force) -- reaction to the centripetal force; applied to the other body

69. Consider Newton’s second law of motion:  F = ma Now substitute centripetal acceleration. In centripetal motion the centripetal acceleration is linked to a centripetal force. You can think of this force as being responsible for holding the object in a circular path.

70. Example you make a right turn in your car you feel the driver door push on you to the right (toward the center of the curvature of your curved path) the door applies a centripetal force to you, you apply a centrifugal force which is equal and opposite to the centripetal force door you F cp F cf Forces occurring along a curved path