UMass Amherst KIN 296 - angular-kinetics-1 (1) (8 pages)

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angular-kinetics-1 (1)



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angular-kinetics-1 (1)

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8
School:
University of Massachusetts Amherst
Course:
Kin 296 - Independent Study
Independent Study Documents
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Angular Kinetics of Human Movement Ch 11 Torque Torques Moments and Levers Center of Mass Gravity Moment of Inertia Angular Analogues of Newton s Laws Angular Momentum and Impulse Angular Work and Power Equations of Motion Segmental Analysis Inverse Dynamics Torque Force Couple Two equal but oppositely directed forces F1 F2 0 separated by a distance will form a force couple d The force couple will produce a torque T F1 d causing rotation However there will be no translation because F1 and F2 will cancel each other out in terms of translation A net centric force will cause translation A net eccentric force will cause translation rotation Calculating Torque Torque Force perp distance perpendicular distance r Copyright 2017 Brian R Umberger Ph D University of Massachusetts Amherst Force F Axis or Center of Rotation Lever Torque is equal to force times perpendicular distance T F r Torque is also referred to as moment of force The perpendicular distance r from the axis of rotation to the force is known as the moment arm 1 Calculating Torque Calculating Torque Two ways to calculate torque created by a force Force F Axis or Center of Rotation Lever Often the force does not act perpendicular to the lever to which it is applied using moment arm r2 of the resultant force using moment arm r1 of the rotary component FMUS FMUS How do you calculate the torque in such cases r2 r1 Muscle Torque Tmus Fmus r Fmus FNR What is the perpendicular distance r in this case FR T F r r1 r2 r1 sin so FR FMUS sin so T FMUS r1 sin T FMUS sin r1 Muscle Torque The torque produced by a muscle is equal to muscle force Fmus times muscle moment arm r FMUS FR FMUS FR FMUS Tmus Fmus r FNR r Center of Joint Rotation The muscle moment arm is the perpendicular distance from the line of action of the muscle force to the center of rotation of the joint If muscle attaches to bone at an angle only the component of muscle force perpendicular to the bone FR rotary component contributes to the muscle torque 2 Example Tendon Rupture Example Tendon Rupture Zernicke et al were analyzing weightlifting performance during a competition when a world class athlete ruptured his patellar tendon during a maximal effort lift The joint torque moment was divided by the estimated patellar tendon moment arm to compute force in the tendon at rupture The lift was a clean and jerk of 175 kg 2 1 his body mass Given a peak knee extension torque of 560 N m and the approximate moment arm of 0 038 m the computed tendon force at rupture was 14 700 N Rupture occurred as he jerked the weight overhead He had already won the competition on his prior lift knee joint torque during the lift C joint center of rotation moment arm distance M joint torque moment P patellar tendon force Muscle Moment Arm 14 700 N 3303 lb 1 7 tons Muscle Moment Arm Muscle moment arms are generally not constant but rather change with joint angle Largest moment arm occurs at an angle of pull 90 May not correspond to a joint angle of 90 For some muscles e g deltoid angle of pull is far less than 90 r r r FMUS biceps moment arm at its largest moment arm is smaller with elbow flexed or extended 3 Net Joint Torque Net Joint Torque Net joint torque is the summation of the torques produce by all of the muscles acting simultaneously at a joint Under normal circumstances many muscles will be active at a joint simultaneously Net torque balances other torques due to weight of body segments m g motion of body segments m a I external forces GRF wind resistance When net joint torque is same direction as the joint motion concentric opposite direction of joint motion eccentric Net Joint Torque Example hamstrings torque hamstrings force 200 N mom arm 0 06 m quadriceps force 1000 N mom arm 0 05 m e g hamstrings active during knee extension Agonist and antagonist muscle groups agonist muscles control direction of movement antagonist muscle activity may help with fine control and or enhance movement stability Levers Knee extension produced by quadriceps agonist activity with simultaneous hamstring antagonist activity quadriceps torque Muscles that act as antagonists to the desired movement may also be active quadriceps torque TQ 1000 N 0 05 m 50 N m hamstring torque TH 200 N 0 06 m 12 N m net knee joint torque TN T TQ TH TN 50 N m 12 N m Lever a simple machine that consist of a rigid bar like body that is free to rotate about an axis Fulcrum the point about which the lever can rotate the axis Three classes of levers First Second Third TN 38 N m 4 First Class Lever Second Class Lever Resistance Force Applied Force Applied Force Resistance Force Fulcrum Fulcrum Fulcrum is between resistance applied force Resistance is between fulcrum applied force Examples Examples Third Class Lever Lever Systems Applied Force Resistance Force Fulcrum Applied force is between resistance fulcrum Examples mechanical advantage moment arm force moment arm resistance When force ma resistance ma mechanical advantage 1 0 low force needed to overcome the resistance resistance is moved through a limited ROM When resistance ma force ma mechanical advantage 1 0 high force needed to overcome resistance resistance is moved through a greater ROM 5 2 5 m Anatomical Levers Static Equilibrium In the human body most lever systems are third class mechanical advantage 1 0 This arrangement promotes third class lever system Large joint range of motion High joint angular speeds However muscle forces must be greatly in excess of the resistance forces Muscle F moment arms are small while resistance R moments arms tend to be large Center of Mass The point about which an object s mass is evenly balanced The CM of a symmetrical object of uniform density is exactly at the center of the object When density is not uniform CM shifts in the direction of greater density Importance of knowing CM location Point at which weight vector force of gravity acts While airborne CM of jumper follows a parabolic path just as a thrown ball would A system is in a state of rotational static equilibrium when all of the torques sum to zero T 0 d 700 N 1 5 m 650 N 825 N How far d from the pivot point does the girl on the left have to sit for the system to be in static equilibrium T 0 700 N d 650 N 1 5 m 825 N 2 5 m 0 700 N d 3037 5 N m d 4 34 m Center of Mass For the human body in the anatomical position CM is at the intersection of the 3 cardinal planes sagittal frontal and transverse Moving body segments away from anatomical position causes CM location to change 6


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