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Torques, Moments of Force, & Angular Impulse Course Reader: p. 61 - 85.

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Presentation on theme: "Torques, Moments of Force, & Angular Impulse Course Reader: p. 61 - 85."— Presentation transcript:

1 Torques, Moments of Force, & Angular Impulse Course Reader: p

2 Causes of Motion Linear Translation F = m*a What happens when you move the point of force application?

3 Causes of Motion M = F * d  MOMENT (N*m): cause of angular rotation Force (N) applied a perpendicular distance (m) from the axis of rotation. dddd M F

4 Axis of Rotation

5 Moment Arm d  (m) Perpendicular distance from the point of force application to the axis of rotation dddd dddd dddd

6 MOMENT M = F * d  F dddd M Known: F = 100 N d  = 0.25 m Unknown:M_____________________ M = 100 N * 0.25 m M = 25 Nm

7 MOMENT (Nm) is a vector; magnitude & direction F dddd M M M + - CCW CW M = F * d  “Right-hand Rule”

8 Right-hand Rule Positive Torques Up Out of the page Negative Torques Down Into the page Thumb Orientation: M +CCW

9 Moments at the Joint Level Static Equilibrium  M = 0 M +CCW Known: Ws = 71 N W A&H = 4 N d  S = 0.4 m d  W = 0.2 m d  FM = 0.01 m  M = 0 Unknown: F m FmFmFmFm W A&H WSWSWSWS

10 Axis of Rotation: Center of Mass Center of Mass (CM, CoG, TBCM) The balance point of an object Object of uniform density; CM is located at the Geometric Center

11 Axis of Rotation: Center of Mass Center of Mass (CM, CoG, TBCM) The balance point of an object Object of non-uniform density; CM is dependent upon mass distribution & segment orientation / shape.

12 Moments are taken about the total body center of mass. Axis of Rotation: TBCM CM location is dependent upon mass distribution & segment orientation CM

13 Moments about the total body center of mass (TBCM) Moments about the total body center of mass (TBCM) Long jump take-off CM FvFh dddd dddd Known: Fv = 7500N Fh = 5000N d  = 0.4m d  = 0.7m MhMhMhMh MvMvMvMv

14 Moments about the TBCM Moments about the TBCM Long jump take-off CM Fv dddd Known: Fv = 7500N d  = 0.4m Unknown: M v ___________________________ M v = Fv * d  M v = 7500 N * 0.4 m M v = 3000 Nm (+) MvMvMvMv

15 Moments about the TBCM Moments about the TBCM Long jump take-off CM Fh dddd MhMhMhMh Known: Fh = 5000 N d  = 0.7 m Unknown: M h ___________________________ M h = Fh * d  M h = 5000 N * 0.7 m M h = 3500 Nm (-)

16 Moments about the TBCM Moments about the TBCM Long jump take-off CM FvFh dddd dddd M Net Net Rotational Effect M Net = M v + M h M Net = 3000 Nm + (-3500 Nm) M Net = -500 Nm Angular Impulse Moment applied over a period of time  M cm  t = I cm 

17 Angular Impulse taken about an object’s CM = the object’s change in angular momentum Angular Momentum - the quantity of angular motion  M cm = I cm   M cm = I cm  /  t  M cm  t = I cm  where I cm = moment of inertia, resistance to rotation about the CM Note: The total angular momentum about the TBCM remains constant. An athlete can control their rate of rotation (angular velocity) by adjusting the radius of gyration, distribution (distance) of segments relative to TBCM.

18 CM FvFh dddd dddd Known: Fv = 1000 N Fh = 700 N d  = 0.3 m d  = 0.4 m Moments about the TBCM Moments about the TBCM sprint startMv Mh

19 CM Fv dddd Moments about the TBCM Moments about the TBCM sprint start Known: Fv = 1000 N d  = 0.3 m Unknown: M v ___________________________ M v = Fv * d  M v = 1000 N * 0.3 m M v = 300 Nm (-) Mv

20 CM Fh dddd Moments about the TBCM Moments about the TBCM sprint start Known: Fh = 700 N d  = 0.4 m Unknown: M h ___________________________ M h = Fh * d  M h = 700 N * 0.4 m M h = 280 Nm (+) Mh

21 CM FvFh dddd dddd Moments about the TBCM Moments about the TBCM sprint start Net Rotational Effect M Net = M v + M h M Net = (-300 Nm) + (280 Nm) M Net = -20 Nm M Net Angular Impulse Moment applied over a period of time  M cm  t = I cm 

22 VRF BACK Somersault time prior to take-offtake-off FVFVFVFV FHFHFHFH FVFVFVFV dddd dddd Reposition your CM relative to Reaction Force Creating Rotation Reposition your CM relative to Reaction Force

23 FrontBackInwardReverse Rotational Demands of a Diver FVFVFVFV FHFHFHFH Force primarily responsible for Net rotation: FVFVFVFV FHFHFHFH FVFVFVFV FHFHFHFH

24 Take-home Messages M (Nm) = F (N) * d  (m) Right-hand Rule: used to determine moment direction Static Equilibrium:  M = 0 Center of Mass (CM, TBCM) –balance point of an object –Position dependent upon mass distribution & segment orientation At the total-body level, moment created by the GRF’s taken about TBCM. Where moment arm length = perpendicular distance from CP location to TBCM location (dx & dy) Moments are generated to satisfy the mechanical demands of a given task (total body, joint level, etc)


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