1. Angular Kinetics
Torque Kinetics
• study of the relationship between the forces
acting on a system and the motion of the system
Angular Motion (Rotation)
Objectives: • All points in an object or system move in a circle
• Define angular kinetics about a single axis of rotation. All points move
• Define and learn to compute moment through the same angle in the same time
arms, torque, and resultant torques Linear Kinetics
• Introduction to resultant joint torques, • The kinetics of particles, objects, or systems
anatomical torque descriptions, and force undergoing rotation
couples
Torque (or Moment) Line of Action
• A measure of the extent to which a force will cause • The line of action of a force is the imaginary line
an object to rotate about a specific axis that extends from the force vector in both directions
• A net force applied through the center of mass
produces translation
• A net force applied away from the center of mass line of action of F
(i.e. an eccentric force) produces both translation
and rotation
F
F F
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2. Moment Arm Computing a Moment Arm
• Shortest distance from a force’s line of action to the • Need to know:
axis of rotation – distance (d) from axis of rotation to point at which
• Moment arm is always perpendicular to the line of force is applied
action and passes through the axis of rotation – angle (θ) at which force is applied
• Use trigonometry to compute moment arm (d⊥)
line of action of F
d⊥ = d sin θ
90°
moment arm
of F F
axis of rotation F
θ
axis of rotation d
Moment Arm Examples Computing Torque
• Torque has:
axis of rotation d⊥ = d
d – a magnitude
– a direction (+ or –)
θ – a specific axis of rotation
d⊥= d sin θ F F • The magnitude of the torque (T) produced by a force
is the product of the force’s magnitude (F) times the
force’s moment arm (d⊥):
axis of T = F d⊥
T = F d⊥ rotation
d⊥ = d sin θ d⊥ = 0
θ F
• Units: F
d F d – English: foot-pounds (ft-lb) d⊥
– SI : Newton-meters (Nm)
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3. Direction of a Torque Example Problem #1
• Positive torque : acts counterclockwise about the Shown below are 4 muscles acting across a joint.
axis of rotation. Which muscles have the largest and smallest
• Negative torque : acts clockwise about the axis force? moment arm? torque magnitude?
• Determine direction using the right hand rule: 3 100 N
2 100 N
– Place right hand on force vector, fingers towards
arrow tip joint
50°
– Curl fingers around axis of rotation 20°
– Torque acts in direction that fingers are curled 60°
1 150 N 0.01 m
T>0 T<0 limb segment
0.02 m
90°
axis of F F 0.04 m
rotation 4 35 N
Torque Composition Resultant Joint Torque
• Process of determining a single resultant (or net)
torque from two or more torques. • The effects of all forces acting about a joint can be
• Performed by adding the torques together, taking the duplicated exactly by the combination of:
sign (direction) of the torque into account – A resultant joint force acting at the joint center
• Resultant torque has same effect on rotation as the – A resultant joint torque acting about the axis of
individual torques acting together rotation through the joint center
T3 T net = |T 1| – |T 2| + |T 3| • Resultant joint force is the vector composition of all
F3
forces acting across a joint.
T1 • Resultant joint torque is the composition of the
axis of torques produced about the joint axis by these forces.
rotation
T2 • Note: Forces that do not act across the joint (e.g.
F1 F2 weight) are not included in the resultant joint force or
torque.
Note: |T| = magnitude of torque T (≥ 0)
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4. Example Use of Resultant Joint Torque
• Typically, joint contact force, muscle forces, ligament
Fcontact forces, etc. cannot be determined individually
Tresultant
Fresultant
• We can compute resultant joint forces and torques
Facl
d ⊥acl based on data measured external to the body
knee joint center
• Except near the limits of the anatomical range of
d ⊥quads Fquads Fquads
Fcontact motion, the main contributors to the resultant joint
d ⊥hams
Fhams torque are the muscles
tibia • The resultant joint torque provides a simplified picture
Facl
Fhams of which muscle groups are most active about a joint
Tresultant = (Fquads d⊥quads) + (F acl d⊥acl) – (Fhams d⊥hams)
Example Problem #2 Force Couple
Shown is a forearm with 2 elbow flexors and 1 • For pure rotation about the center of mass, the center
elbow extensor. Find the resultant joint torque for of mass must remain stationary from Newton’s 1st
the 3 combinations of forces shown in the table: law, the net force on the object must equal zero
• Force couple : Two forces of equal magnitude,
Ft 0 0 32
Ft Fcontact Fbi applied in opposite directions. Produce pure rotation
Fbr
about the center of mass.
Fcontact 8 3.2 46.4
T = F (d⊥1 + d ⊥2 )
30° F
Fbi 16 10 20
0.025 m 0.05 m d⊥2
0.10 m
W=8N d⊥1
Fbr 0 2.4 4.8 ΣF=0
0.25 m F
RJT
Force Couple Net Effect
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5. Anatomical Torques
• Positive & negative torques depend on frame of
reference chosen:
y y
Fquad Fquad
knee knee
T>0 T<0
x x
• To avoid this problem, joint torques are typically
described by the joint motion that would occur if the
segment moved in the direction of the torque
(e.g. Fquad produces a knee extension torque)
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