torques, moments of force, & angular impulse course reader: p. 61 - 85
TRANSCRIPT
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Torques, Moments of Force, & Torques, Moments of Force, & Angular ImpulseAngular Impulse
Course Reader: p. 61 - 85
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Causes of MotionCauses of MotionLinear TranslationLinear Translation
F = m*aF = m*aWhat happens when you move the point of force
application?
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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
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Axis of RotationAxis of Rotation
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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
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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
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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”
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Right-hand RuleRight-hand Rule
Positive TorquesUpOut of the page
Negative TorquesDown
Into the pageThumb Thumb Orientation:Orientation:
MM+CCWCCW
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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
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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
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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.
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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
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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
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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
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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 (-)(-)
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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
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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.
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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
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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
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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
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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
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-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
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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
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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)