1. Angular Kinematics Chapter 11

2. Angular Motion All parts of a body move through the same angle, in the All parts of a body move through the same angle, in the same, direction, in the same time same, direction, in the same time More prevalent component of general motion More prevalent component of general motion Measurement of angles

3. Joint or relative angles Angle formed from the long axes of two adjacent body segments

4. Segmental angles or absolute angles Orientation of a single body segment with a fixed line of reference

5. Measurement of Angular Velocity ω = (Θ F – Θ I )/∆t = ∆Θ/∆t Period Hip Knee 1-2 s ω = (170 o -95 o )/1s ω = (175 o -100 o )/1s = 75 o /s (Ext) = 75 o /s (Ext) = 75 o /s (Ext) = 75 o /s (Ext) 2-2.5 s ω = (95 o -170 o )/0.5s ω = (110 o -175 o )/0.5s = - 150 o /s (Flex) = -130 o /s (Flex) = - 150 o /s (Flex) = -130 o /s (Flex) 2.5-3 s ω = (85 o -95 o )/0.5s ω = (160 o -110 o )/0.5s = -20 o /s (Flex) = 100 o /s (Ext) = -20 o /s (Flex) = 100 o /s (Ext)

6. Arc length = r Θ circumference = 2 ∏ r 150 o /s = 150/57.3 = 2.6 rad/s

7. Relationship between linear and angular velocity V1V1 V2V2 r1 = 0.2 m = 0.4 m v (m/s) = r (m) ω (radians/s) If ω = 30 rad/s, what is the linear velocity of the bat at points 1 and 2? v = r ω v 1 = 0.2 x 30 = 6 m/s v 2 = 0.4 x 30 = 12 m/s

8. Angular Acceleration ⍺ = (ω F – ω I ) / ∆t = ∆ω / ∆t ⍺ = (ω F – ω I ) / ∆t = ∆ω / ∆t Skater spinning anticlockwise at 198.3º/s comes to a stop in 20s. What is her angular acceleration? ⍺ = (ω F – ω I ) / ∆t = (0 -198.3 º/s) / 20 s = - 9.92 º/s 2 ⍺ = (ω F – ω I ) / ∆t = (0 -198.3 º/s) / 20 s = - 9.92 º/s 2 = - 9.91 / 57.3 rad/s 2 = -0.17 rad/s 2 = - 9.91 / 57.3 rad/s 2 = -0.17 rad/s 2 If the skater’s hand is 0.85 m from the axis of rotation, what is the tangential acceleration of her hand? use a t = r ⍺ use a t = r ⍺ a t = r ⍺ = 0.85 (- 0.17) = -0.14 m/s 2 a t = r ⍺ = 0.85 (- 0.17) = -0.14 m/s 2

9. “Center-fleeing” “Center-seeking”

10. Angular Kinetics Angular Kinetics Angular analogue to mass/inertia → moment of inertia ( I) Resistance to angular acceleration. n I = Σ m d 2 i = 1 i = 1

13. Determining I n I = Σ m d 2 I = Σ m d 2 i = 1 i = 1 However, determined from Σ T = I ⍺ Where Σ T = sum of Torques I = moment of inertia I = moment of inertia ⍺ = angular acceleration ⍺ = angular acceleration I approximated from cadaver studies: acceleration of a I approximated from cadaver studies: acceleration of a rotating limb measured after applying a known torque rotating limb measured after applying a known torque Once determined, value characterized by using the formula I = mk 2 where I = moment of inertia; m = total mass; k = radius of gyration

14. I = mk 2 Radius of gyration – k Represents the objects mass distribution with respect to an axis of rotation. It is the distance from the axis of rotation to a point at which the mass can be theoretically concentrated without altering the inertial characteristics of the rotating body The length of the radius of gyration changes as the axis of rotation changes

15. k for a particular segment can be obtained from anthropometric tables k for a particular segment can be obtained from anthropometric tables The mass of a particular segment can also be obtained from such tables The mass of a particular segment can also be obtained from such tables Therefore, net joint torques can be determined for human subjects by measuring angular accelerations and applying the equation Therefore, net joint torques can be determined for human subjects by measuring angular accelerations and applying the equation Σ T = I ⍺ Σ T = I ⍺ Whole body moment of inertia Different with respect to different axes of rotation About which axis is I the smallest?

16. Anteroposterior Mediolateral Mediolateral Longitudinal Longitudinal 12-15 kg.m 2 10.5-13.0 kg.m 2 4.0-5.0 kg.m 2 1.0-1.2 kg.m 2 2.0-2.5 kg.m 2 Whole body moment of Inertia (I) values

17. Angular momentum Quantity of angular motion (vector) H = I ω = (m k 2 ) ω Conservation of angular momentum The total angular momentum of a given system (e.g., the body) remains constant in the absence of external torques

18. If gravity is the only external force, there are no external torques on the body and angular momentum is conserved ω H = I ω I ↑ ω ↓ I ↓ ω ↑

20. ω A H A = -I A ω A (negative) ω L H L = I L ω L(positive) H A = -I A ω A is equal in magnitude but opposite in direction to H L = I L ω L I L (of legs) > I A (of arms) so arm has > ω ω A > ω L

21. Conservation of angular momentum When a body is in the air (angular momentum conserved), if the angular momentum of one body part is increased, then all or part of the rest of the body must experience a decrease in angular momentum When a body is in the air (angular momentum conserved), if the angular momentum of one body part is increased, then all or part of the rest of the body must experience a decrease in angular momentum While in the air, angular momentum is conserved, but it can be transferred While in the air, angular momentum is conserved, but it can be transferred

22. Angular impulse (I ω) B + ΣT∆t = (I ω) A A diver produces angular impulse at take-off resulting in the angular momentum that he/she possess in the air

23. Newton’s Laws of Angular Motion 1 st Law A rotating body will maintain a state of constant angular motion unless acted upon by some net external torque 2 nd Law A net external torque produces angular acceleration of a body that is directly proportional to the magnitude of the net torque, in the same direction as the net torque, and inversely proportional to the body’s moment of inertia ΣT = I ⍺ 3 rd Law For every torque exerted by one body on another, there is an equal and opposite torque exerted by the second body on the first