Download presentation

Presentation is loading. Please wait.

Published byJon Lemons Modified over 2 years ago

1
Torques, Moments of Force, & Angular Impulse Course Reader: p. 61 - 85

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. dddd M F

4
Axis of Rotation

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

6
MOMENT M = F * d F dddd 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 dddd 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 dddd dddd 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 dddd 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 dddd 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 dddd dddd 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 dddd dddd 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 dddd 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 dddd 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 dddd dddd 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 dddd dddd 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)

Similar presentations

OK

Cause-Effect ~ Total Body Level Total Body Center of Mass (TBCM) Free Body Diagram (FBD) Mass-Acceleration Diagram (MAD) General global coordinate.

Cause-Effect ~ Total Body Level Total Body Center of Mass (TBCM) Free Body Diagram (FBD) Mass-Acceleration Diagram (MAD) General global coordinate.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on addition of integers Ppt on sources of energy for class 8th Ppt on microcontroller based temperature sensor Ppt on animals our friends Ppt on index of industrial production Ppt on combination of resistances in parallel Ppt on rain water conservation Projector view ppt on ipad Ppt on metropolitan area network Ppt on characteristics of computer