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INTRODUCTION TO BIOMECHANICS

INTRODUCTION TO BIOMECHANICS. SECTION 4.2: CENTRE OF MASS. CENTER OF MASS (CENTER OF GRAVITY). We model the body as a system of linked segments, where each segment is a mass.

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INTRODUCTION TO BIOMECHANICS

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  1. INTRODUCTION TO BIOMECHANICS SECTION 4.2: CENTRE OF MASS

  2. CENTER OF MASS (CENTER OF GRAVITY) • We model the body as a system of linked segments, where each segment is a mass. • We don't generally attempt to define mass, and instead appreciate it intuitively as one of the fundamental properties (along with time, length, and electric charge) of our physical world.

  3. CENTRE OF MASS • Definition: (geometric) point around which every particle of a body's mass is equally distributed. A body behaves as if its entire mass acts or is acted upon at its center of gravity.

  4. WHOLE BODY • COG located at sacral promontory, anterior to S2 (PSIS), at 55% of body height

  5. The COG of the entire lower extremity, the body segment that lies distal to the hip (pelvi-femoral) joint, is located just proximal to the knee (Smith, Weiss, & Lehmkuhl, 1996, p.55). • This information comes from an anthropometric table that specifies typical segmental masses and centers of gravity.

  6. Human Anthropometric Data • Remember that "proximal" means closest to the centre of the trunk [or closest to the heart in some cases] • Data are from Winter DA (1979) Biomechanics of human movement, p. 151 - Now published as Biomechanics and Motor Control of Human Movement

  7. WHETHER A MASS IS STABLE OR MOBILE ... • depends on its size, the location of its COG, the size of the mass' base of support (BOS), and the location of the COG's vertical projection into that base of support

  8. CENTRE OF MASSSTABILITY VS MOBILITY

  9. CENTRE OF MASSREVISITED • The c/m is the point in the body where all of the torques causing rotation in one direction are balanced by all the torques causing rotation in the other direction. • -Analogy to weights on a balance

  10. CENTRE OF MASS • IMPORTANT TO UNDERSTAND AND LOCATE THE CENTRE OF MASS- greatly contributes to our understanding of motion X

  11. 1. GAIT ANALYSISIn walking the path of the c/m forms a sinusoidal curve • The > the displacement of C/M the > the amount of energy expended. The smoother the curve the less energy expended

  12. The normal amount of rise and fall in an adult male = approx 5 cm • If 1000 steps/day – raise and lower the body 5000cm • Work = (force)(distance) The > the work done the > the energy expended • By translating the c/m through a smooth undulating pathway of low amplitude the body conserves energy

  13. C/M AND POSTURE • The bones of the skeletal system make up a series of links connected by joints and held upright by muscle and ligaments • If these are stacked so that the line of gravity goes directly through the supporting links the lever arms (FORCE ARMS) = 0 (or close to it) • If 0 moment arms then 0 torques established • -> only a resulting downward force

  14. No torque

  15. As soon as the c/m shifts a moment is established. To remain in equilibrium forces must be exerted to create an equal and opposite moment • Muscular effort increases -> strain • When one segment moves forward another must move back to compensate

  16. x http://www.phy.ntnu.edu.tw/java/block/block.html

  17. SEGMENTAL METHOD • Process for calculating the location of the total body centre of mass from projected images of the body. • The location of the total body c/m is a function of the location of the respective segmental C/M’s • The calculation finds the location in 2 or 3 D where the sum of the torques = 0

  18. Find the position of the c/m • Sum of the individual torques = resultant torque • T1+T2+T3=resultant torque Resultant Torque = (FR) (Fa) .45m .35m .2m 7N 5N 15N

  19. .45m .35m = Force arm .2m X 7N 5N 15N FR

  20. Force arm X FR (15)(.2)+(7)(.35)+(5)(.45)=27(Fa) Fa = .29m The Force arm gives the location of the centre of mass relative to the axis

  21. EXAMPLE • Find the position of the centre of mass of the following system F3 F1 2.8 5 2.3 F2 1.9 10 8 x 1.7 4.6 6.3

  22. y F3 F1 2.8 5 2.3 F2 1.9 10 8 • F Fax Tx Fay Ty • 2.8 28 1.7 17 x 1.7 4.6 6.3 23Fax = 54.7 Fax = 2.38 23Fay = 85.3 Fay=3.71 8 1.9 15.2 4.6 36.8 5 2.3 11.5 6.3 31.5 SUM Tx = 54.7 SUM Ty = 85.3

  23. (3.71, 2.38) 3.71 y C/m 2.38 x

  24. Finding C/M in 2D when absolute masses unknown • - therefore weight unknown • - since actual segmental weights are unknown must use proportional weights • The proportional weights are expressed as a percentage of the total body weight

  25. The segmental method employs each segmental weight as a separate force acting at some distance from an arbitrary axis

  26. MASS F Fax Tx Fay Ty • Prop prop • .45 .45 7 3.15 3 1.35 • .43 .43 4 1.72 5 2.15 • .12 .12 5 .6 7 .84 • Sum F = 1 Sum Tx =5.47 Sum Ty = 4.34 • About the x axis • sum Tx = (Fresultant) ( Fax) • 5.47 = (1)(Fax) • Fax = 5.47 • About the y axis • sum Ty =(Fresultant) ( Fay) • 4.34 = (1)(Fay) • Fay = 4.34

  27. 4.34,5.47) X

  28. Center of massThe segmental method • We need to know: • - Location of segment c/m expressed • as % distance from prox end • Segment mass expressed as a % • of total body mass • - Position of segments ->clear image

  29. http://flash.lakeheadu.ca/~health/menu.html

  30. Fay Fax

  31. Total body Centre of mass

  32. TEXT BOOK • PAGE 436-441

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