Torques, Moments of Force, & Torques, Moments of Force, & Angular ImpulseAngular Impulse

Course Reader: p. 61 - 85

Causes of MotionCauses of MotionLinear TranslationLinear Translation

F = m*aF = m*aWhat happens when you move the point of force

application?

Causes of MotionCauses of Motion

M = F * dM = F * d

MOMENT (N*m): cause of angular rotationMOMENT (N*m): cause of angular rotationForce (N) applied a perpendicular distance (m) from the axis of rotation.

dd

MM FF

Axis of RotationAxis of Rotation

Moment ArmMoment Armdd (m) (m)

Perpendicular distance from the point of force Perpendicular distance from the point of force application to the axis of rotationapplication to the axis of rotation

dd

dd dd

MOMENTMOMENTM = F * dM = F * d

FF

dd

MM

Known: Known: F = 100 N F = 100 N dd = 0.25 m= 0.25 m

Unknown:Unknown:MM__________________________________________

M = 100 N * 0.25 mM = 100 N * 0.25 mM = 25 NmM = 25 Nm

MOMENT (Nm) is a vector; MOMENT (Nm) is a vector; magnitude & directionmagnitude & direction

FF

dd

MM MM

MM+

-

CCWCCW

CWCW

M = F * dM = F * d

““Right-hand Rule”Right-hand Rule”

Right-hand RuleRight-hand Rule

Positive TorquesUpOut of the page

Negative TorquesDown

Into the pageThumb Thumb Orientation:Orientation:

MM+CCWCCW

Moments at the Joint LevelMoments at the Joint LevelStatic EquilibriumStatic Equilibrium

M = 0M = 0

MM+CCWCCW

Known: Ws = 71 NWs = 71 NW W A&HA&H = 4 N = 4 NdS = 0.4 mdW = 0.2 mdFM = 0.01 m

M = 0M = 0

Unknown: Fm

FFmm

WWA&HA&H WWSS

Axis of Rotation: Center of MassAxis of Rotation: Center of MassCenter of Mass (CM, CoG, TBCM)

• The balance point of an object

Object of uniform density;

CM is located at the Geometric Center

Axis of Rotation: Center of MassAxis of Rotation: Center of MassCenter 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.

Moments are taken about the total body center of mass.

Axis of Rotation: TBCMAxis of Rotation: TBCMCM location is dependent upon mass distribution CM location is dependent upon mass distribution

& segment orientation& segment orientation

CM CM CM CM

Moments about the total body center Moments about the total body center of mass (TBCM)of mass (TBCM)Long jump take-off

CM

FvFv

FhFhdd

dd

Known: Fv = 7500NFh = 5000N

d = 0.4m

d = 0.7m

MMhh

MMvv

Moments about the TBCMMoments about the TBCMLong jump take-off

CM

FvFvdd

Known: Fv = 7500N

d = 0.4m

Unknown:

Mv___________________________

M v = Fv * d

M v = 7500 N * 0.4 m

M v = 3000 Nm M v = 3000 Nm (+)(+)

MMvv

Moments about the TBCMMoments about the TBCMLong jump take-off

CM

FhFhdd

MMhh

Known: Fh = 5000 N

d = 0.7 m

Unknown:

Mh___________________________

M h = Fh * d

M h = 5000 N * 0.7 m

M h = 3500 Nm M h = 3500 Nm (-)(-)

Moments about the TBCMMoments about the TBCMLong jump take-off

CM

FvFv

FhFhdd

dd

M M NetNet

Net Rotational EffectNet Rotational EffectM M NetNet = = MMvv ++ MMhh

M M NetNet = = 3000 Nm3000 Nm ++

(-3500 Nm)(-3500 Nm)M M NetNet = -500 Nm = -500 Nm

Angular ImpulseMoment applied over a period of timeMMcmcm t = It = Icmcm

Angular Impulse taken about an object’s CM = the object’s change in angular

momentumAngular Momentum - the quantity of angular motion

Mcm = Icm Mcm = Icm / t

Mcm t = Icm

where Icm = 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.

CM

FvFv

FhFhdd

dd

Known: Fv = 1000 NFh = 700 N

d = 0.3 m

d = 0.4 m

Moments about the TBCMMoments about the TBCMsprint start

MvMv

MhMh

CM

FvFvdd

Moments about the TBCMMoments about the TBCMsprint start

Known: Fv = 1000 N

d = 0.3 m

Unknown:

Mv___________________________

M v = Fv * d

M v = 1000 N * 0.3 m

M v = 300 Nm M v = 300 Nm (-)(-)

MvMv

CM

FhFhdd

Moments about the TBCMMoments about the TBCMsprint start

Known: Fh = 700 N

d = 0.4 m

Unknown:

Mh___________________________

M h = Fh * d

M h = 700 N * 0.4 m

M h = 280 Nm M h = 280 Nm (+)(+)

MhMh

CM

FvFv

FhFhdd

dd

Moments about the TBCMMoments about the TBCMsprint start

Net Rotational EffectNet Rotational EffectM M NetNet = = MMvv ++ MMhh

M M NetNet = = (-300 Nm)(-300 Nm) ++

(280 Nm)(280 Nm)M M NetNet = -20 Nm = -20 Nm

M M NetNet

Angular ImpulseMoment applied over a period of timeMMcmcm t = It = Icmcm

-500

0

500

1000

1500

2000

2500

-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1

Time Prior to Take-off (s)F

orce

(N)

Horizontal RF

Vertical RF

VRF

BACK Somersault

time prior to take-off take-off

FFVV

FFHH

FFVV

dd

dd

Creating RotationReposition your CM relative to Reaction ForceReposition your CM relative to Reaction Force

Front Back InwardReverse

Rotational Demands of a DiverRotational Demands of a Diver

FFVV

FFHH

Force Force primarily primarily responsible responsible for Net for Net rotation:rotation:

FFVV FFHHFFVV FFHH

Take-home MessagesTake-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)