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Overall course plan. Tissue mechanics: mechanical properties of bone, muscle, tendon, etc. Kinematics: quantification of motion, with no regard for the forces that govern motion Kinetics: forces in human movement. Kinematics. 1 -D linear kinematics Angular kinematics

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Presentation Transcript
slide1

Overall course plan

  • Tissue mechanics: mechanical properties of bone, muscle, tendon, etc.
  • Kinematics: quantification of motion, with no regard for the forces that govern motion
  • Kinetics: forces in human movement
kinematics
Kinematics
  • 1-D linear kinematics
  • Angular kinematics
  • 2-D linear kinematics
  • Locomotion
reading
Reading
  • Linear: Enoka chapter 1 pp. 3-23
  • Loco: Enoka chapter 4 pp. 141-146
  • Angular: Enoka chapter1 pp. 23-29
kinematics1
Kinematics
  • 1-D linear kinematics
  • Angular kinematics
  • 2-D linear kinematics
  • Locomotion
slide5

Video Kinematics

  • 60-6000 images/sec
  • Put markers on anatomical landmarks
  • Computer tracks marker position
kinematics example video
Kinematics example video
  • http://www.vicon.com/applications/life_sciences.html
  • http://www.youtube.com/watch?v=USOYUMN5nwU
  • http://www.youtube.com/watch?v=IJ4tndpwL-o&mode=related&search=
  • http://www.youtube.com/watch?v=frNpkIH8_vo&mode=related&search=
kinematics2
Kinematics
  • 1D = movement in one direction.
    • e.g just the forward movement of a human runner during a sprint
  • 2D = movement in a plane, components in 2 directions
    • e.g. side view of hip of a person walking
  • 3D = movement anywhere in space
    • an owl swooping to catch jackrabbit, knee movements
  • Angular = movements that include rotation
    • e.g., a biceps curl, alligator biting, figure skater
1 dimensional linear kinematics
1 dimensional linear kinematics
  • Measurement Rules
  • Definitions of terms used in kinematics
  • Equations and examples for constant acceleration motion
  • 1D Kinematics example
units
Units
  • SI system
    • Length : meters (m)
    • Mass : kilograms (kg)
    • Time : seconds (s)
  • Other units
    • Angle : radians (rad)
      • 2p radians = 360o
prefixes and changing units
Prefixes and Changing Units
  • Giga (G) 1x109
  • Mega (M) 1x106
  • Kilo (k) 1x103
  • Deci (d) 1x10-1
  • Centi (c) 1x10-2
  • Milli (m) 1x10-3

When changing units, always be sure to CANCEL units appropriately!

1 dimensional linear kinematics1
1 dimensional linear kinematics
  • Measurement Rules
  • Definitions of terms used in kinematics
  • Equations and examples for constant acceleration motion
  • 1D Kinematics example
kinematics terms position
Kinematics terms: position
  • position (r) =location in space
  • Units : meters (m)
  • Must be defined with respect to a baseline value or axis
kinematics terms displacement
Kinematics terms: Displacement
  • Displacement = Change in position

= ∆r = (rf - ri)

  • rf = final position; ri = initial position
  • Units = meters (m)
  • Involves space and time

rf

Position r, (m)

ri

ti

tf

Time (s)

kinematics terms velocity v
Kinematics terms: Velocity (v)
  • rate of change of position
  • vave = ∆r / ∆t
  • vinstantaneous = dr / dt (slope of position vs. time)
  • Units = m / s

rf

Velocity v, (m/s)

Position r, (m)

vi

vf

ri

ti

tf

ti

tf

Time (s)

Time (s)

kinematics terms acceleration a
Kinematics terms: acceleration (a)
  • rate of change of velocity
  • aave = ∆v / ∆t
  • ainstantaneous = dv / dt = slope of velocity vs. time
  • units = m / s2

rf

Velocity v, (m/s)

Position r, (m)

vi

vf

ri

ti

tf

ti

tf

Time (s)

Time (s)

1 dimensional linear kinematics2
1 dimensional linear kinematics
  • Measurement Rules
  • Definitions of terms used in kinematics
  • Equations and examples for constant acceleration motion
  • 1D Kinematics example
special case constant acceleration
Special case: Constant acceleration
  • Example: Projectile motion when air resistance is negligible
    • Gravitational acceleration = 9.81 m / s2
      • in the downward direction!!!
    • g = 9.81 m/ s2
  • Useful for analyzing a jump (frogs, athletes), the aerial phase of running, falling
projectile motion
Projectile Motion
  • The only significant force that the object experiences while in the air will be that due to gravity
  • Flight time depends on vertical velocity at release and the height of release above landing surface
symmetry in projectile motion
Symmetry in projectile motion
  • A comparison of the kinematics at the beginning and end of the flight reveals:
    • vi = - vf & vi2 = vf2
    • ri = rf & rf - ri = 0
equation 1 for constant acceleration
Equation 1 for constant acceleration
  • Expression for final velocity (vf) based on initial velocity (vi), acceleration (a) and time (t)

vf = vi + at

  • If vi = 0

vf = at

problem solving strategy
Problem solving strategy
  • Define your coordinate system!
  • List all known variables (with UNITS!)
  • Write down which variable you need to know
    • (with UNITS!)
  • Figure out which equations allow you to solve for the unknown variable from the known variables.
example using equation 1
Example using equation 1
  • An animal jumps vertically. The animal’s takeoff or initial velocity is 3 m/s. How long does it take to reach the highest point of the jump? (Air resistance negligible).
  • - 0.3 s
  • 0.03 s
  • - 0.03 s
  • 0.3 s
equation 2 for constant acceleration
Equation 2 for constant acceleration
  • Expression for change in position (rf - ri) in terms of initial velocity (vi), final velocity (vf), acceleration (a) and time (t).

rf - ri = vit + ½ at2

  • If vi = 0, then

rf - ri = ½ at2

example using equation 2
Example using equation 2

What was the net jump height of the animal in example #1. In other words, how high did it jump once its feet left the ground?

  • 46 m
  • 46 cm
  • 1.34 m
  • -1.34 m
equation 3 for constant acceleration
Equation 3 for constant acceleration
  • An expression for the final velocity (vf) in terms of the inital velocity (vi), acceleration (a) and the distance travelled (rf - ri).

vf2 = vi2 + 2a (rf - ri)

  • If vi = 0, then

vf2 = 2a (rf - ri)

example using equation 3
Example using Equation 3

A runner starts at a standstill and accelerates

uniformly at 1 m/s2. How far has she travelled at the point when she reaches her maximum speed of 10 m/s?

example using equation 31
Example using Equation 3

A runner starts at a standstill and accelerates uniformly at 1 m/s2. How far has she travelled at the point when she reaches her maximum speed of 10 m/s?

  • 5 m
  • 10 m
  • 50 m
  • 100 m
1 dimensional linear kinematics3
1 dimensional linear kinematics
  • Measurement Rules
  • Definitions of terms used in kinematics
  • Equations and examples for constant acceleration motion
  • 1D Kinematics example
1 d kinematic analysis of a 100 meter race
1-D kinematic analysis of a 100 meter race
  • Question to answer: How long does it take to reach top speed?
  • Videotape the sprint
  • Measure the position of a hip marker in each frame of video
100 meter sprint1
100 meter sprint

vmax reached at 6 seconds, at 50 meters.

100 meter sprint2

Acceleration (m / s2)

100 meter sprint

vmax = 10 m/s

aave = 1.66 m/s2

(0 - 6 seconds)

kinematics3
Kinematics
  • 1-D linear kinematics
  • Angular kinematics
  • 2-D linear kinematics
  • Locomotion
position

Position
  • Linear

symbol = r

unit = meter

  • Angular

symbol = 

unit = radians or degrees

6.28 rad = 360°

joint angles

Hip

Knee

Ankle

Joint angles
  • Ankle: foot - shank
  • Knee: shank - thigh
  • Hip: thigh - trunk
  • Wrist
  • Elbow
  • Shoulder
segment angles

Hip

Knee

Ankle

Segment angles
  • Measured relative to a fixed axis (e.g., horizontal)

Trunk

Trunk

Thigh

Shank

Shank

displacement

2

∆

1

Displacement
  • Linear

symbol = ∆r

∆r = rf - ri

unit = meter

  • Angular

symbol = ∆

∆ = f - i

unit = radian

angular linear displacement
Angular ---> linear displacement
  • s = distance travelled (arc)
  • s = radius • ∆

∆ must be in radians

2

s

∆

1

radius

joint angle change flexion

Elbow

Joint angle change: Flexion
  • Flexion: decrease in angle between two adjacent body segments
  • Bird’s eye view of arm on table

Flexion

joint angle change extension
Joint angle change: Extension
  • Extension: increase in angle between two adjacent body segments

Extension

Elbow

knee extension
Knee extension

Knee

extension

a squat jump
A Squat Jump

Hip

ext

Knee

ext

Ankle

ext

emg electromyography
EMG: Electromyography
  • Measures electrical activity of muscles
    • indicates when a muscle is active

Hip

ext

Knee

ext

Ankle

ext

slide45

Hip

angle

Hip extensor EMG

Knee

angle

Knee extensor EMG

Ankle

angle

Ankle extensor EMG

0

0.30

0.15

Time (s)

alternative joint angle definition
Alternative joint angle definition

Full extension: alt = 0 (degrees or rad)

norm

alt

slide47

Knee

angle

(rad)

2

Run

Enoka 4.5

Flex-Ext

Flex-Ext

1

0

1

Small

Flex-Ext

Walk

Flex-Ext

0

Flexion

Swing

Stance

stance

stance

Time (% of stride)

velocity
Velocity
  • Linear
    • symbol = v
    • v = ∆r / ∆t
    • unit = meters per second
  • Angular
    • symbol =  = “omega””
    •  = ∆ / ∆t
    • unit = radians per second
    • human body: we define extension as positive

∆

slide50

A person sits in a chair and does a knee extension. The knee angle changes by 0.5 radians in 0.5 seconds. What is the magnitude of the angular velocity of the knee, the linear velocity of the foot and the linear displacement of the foot? (Shank length is 0.5 meters)

  • 1 rad/s, 0.5 m/s, 0.25rad
  • 0.5 rad/s, 0.5 rad/s, 0.25rad
  • 1 rad/s, 0.5 m/s, 0.25 m
  • 0.25 rad/s, 0.5 m/s, 0.25m
angular acceleration
Angular acceleration
  • Linear
    • symbol = a
    • a = ∆v /∆t
    • unit = m / s2
  • Angular
    • symbol = 
    • = “alpha”
    •  = D / Dt
    • unit = rad / s2

arm motion example
Arm Motion example

Muscles act as “motors” and “brakes”

But they only pull

Displacement, velocity, acceleration

dr/dt, dv/dt

slide53

2

1

3

1

E

 (rad)

F

2

E

 (rad/s)

F

3

E

 (rad/s2)

F

Time

E= extension, F = flexion

slide54

Flex

Ext

 (rad)

 (rad/s)

 (rad/s2)

Triceps brachialis (ext)

Brachioradialis (flex)

Biceps brachialis (flex)

Enoka example1.8

linear acceleration a in angular motion radial and tangential acceleration

ar

Linear acceleration (a) in angular motion:radial and tangential acceleration
  • Even if vi =vf, there is still a radial acceleration
  • Even if i =f, there is still a radial acceleration
  • aradial (m/s2): ∆ direction of linear velocity vector.
  • aradial = r2 = v2 / r
    • ar is never zero during circular motion

vf

vi

linear acceleration a in angular motion

at

ar

Linear acceleration (a) in angular motion
  • a = aradial + atangential (vector sum)
  • atangential (m/s2): causes ∆ & ∆v
  • atangential = r= ∆v / ∆t

at is zero if |v| or  is constant

v

v

total linear acceleration a

at

a

ar

Total linear acceleration (a)
  • a = at + ar (vector sum)
  • |a| = (at2 + ar2)0.5
  • = tan-1(at / ar)
slide58

A 100 kg person is riding a bike around a corner (radius = 2 m) with a constant angular velocity of 1 rad /sec. What were the magnitudes of the tangential & radial components of the acceleration?

  • at = 981 m/s2; ar= 2 m/s2
  • at = 0 m/s2; ar= 4 m/s2
  • at = 0 m/s2; ar= 2 m/s2
  • at = 0 m/s2; ar= 2 rad/s2
  • E) at = 981 m/s2; ar= 4 rad/s2
slide59

A 100 kg person is riding a bike around a corner (radius = 2 m) with a constant angular velocity of 1 rad /sec. What were the magnitudes of the tangential & radial components of the acceleration?

at = r

 = 0; r = 2 meters

at = r= (2 m)• (0 rad/s2) = 0 m/s2

ar = r2

 = 1 rad/sec

ar = (2 meters) • (1 rad/s)2 = 2 m/s2

linear angular motion equations
Linear & Angular Motion Equations
  • On equation sheet
  • analogous

Only for constant acceleration

make a list of examples in human movement where radial and tangential accelerations are important
Make a list of examples in human movement where radial and tangential accelerations are important
angular kinematics example 1
Angular Kinematics Example # 1

A hammer thrower releases the hammer after reaching an angular velocity of 14.9 rad/s. If the hammer is 1.6 m from the shoulder joint, what is the linear velocity of release?

slide63

Angular Kinematics Example # 2

A 60 kg sprinter is running a race on an unbanked circular track with a circumference of 200 m. At the start, she increases speed at a constant rate of 5 m/s2 for 2 seconds. After two seconds, she maintains a constant speed for the duration of the race.

a. How long does it take for her to run the 200 m?  

b. What is her speed when she crosses the finish line?

c. What is her angular velocity when she finishes?

d. What was her angular acceleration during the first two seconds?

e. What is her angular acceleration when she finishes?

f. Determine her radial and tangential accelerations 1 second after the start of the race.

kinematics4
Kinematics
  • 1-D linear kinematics
  • Angular kinematics
  • 2-D linear kinematics
  • Locomotion
2 d linear kinematics
2-D linear kinematics
  • 1-D vs. 2-D kinematics
  • conventions
  • vectors
  • constant acceleration motion
  • applications
1 d analysis 100 meter sprint example

v

v

1-D analysis: 100 meter sprint example
  • We only considered the horizontal motion.
2 d linear kinematics1
2-D linear kinematics
  • 1-D vs. 2-D kinematics
  • conventions
  • vectors
  • constant acceleration motion
  • applications
slide70

Y

Y

X

X

X

Running direction

slide71

Y

Y

X

X

X

Z

Z

Z

sign conventions
Sign conventions
  • Y axis (vertical): Positive is upward
  • X axis (horizontal): Positive is in direction of movement.
2 d linear kinematics2
2-D linear kinematics
  • 1-D vs. 2-D kinematics
  • conventions
  • vectors
  • constant acceleration motion
  • applications
vectors in kinematics
Vectors in kinematics
  • VECTOR: magnitude and direction.
    • Length of vector indicates magnitude.
    • Direction of vector indicates direction of motion.

Small vertical

velocity

Large horizontal

velocity

which vectors are equal
Which Vectors Are Equal?
  • A&B
  • A,B&E
  • C&E
  • None of them are equal
2 d kinematic analysis
2-D kinematic analysis
  • Vector analysis: divide movement into components & analyze separately
  • Reason: different forces &accelerations act in each direction
    • vertical: gravitational acceleration
    • horizontal & lateral: no grav. accel.
vector components

v

vy

vx

Vector components

|v|

  • Right triangle ---> use trigonometry.
  • |v| = magnitude of vector

 = angle of vector with the horizontal

  • vx = horizontal component

vy= vertical component

slide78

v

vy

vx

Determining vector components

|v|

  • What are the horizontal (vx) and vertical (vy) components of the vector (v)?
  • cos  = vx / v thus, vx = v cos 
  • sin  = vy / v thus, vy = v sin 
example of determining components
Example of determining components

|v| = 2m/sec

  • sin 60° = vy/ v thus, vy = v * sin 60°
  • vy = 2 * 0.87 = 1.74 m/sec
  • cos 60° = vx / v thus, vx = v * cos 60°
  • vx = 2 * 0.50 = 1 m/sec

vy

60°

60°

vx

how to find resultant vector
How to find resultant vector

v = ?

  • Finding 

tan  = vy / vx thus,  = tan-1(vy / vx)

  • Finding |v|

|v| = (vx2 + vy2)0.5

vy

vy



vx

vx

example find resultant vector

|v|

vy = 3m/s

vx = 2m/s

Example: Find resultant vector

 = tan-1(vy/vx)

  = tan-1(3 / 2) = 56.3°

|v| = (vx2 + vy2)0.5

|v| = (22 + 32)0.5 = (13)0.5 = 3.6m/s

|v| = ?

 = ?

2 d kinematic analysis1
2-D kinematic analysis
  • Vector analysis: divide movement into components & analyze separately
  • Reason: different forces/accelerations act in each direction
    • vertical: gravitational acceleration
    • horizontal & lateral: no grav. accel.
2 d linear kinematics3
2-D linear kinematics
  • 1-D vs. 2-D kinematics
  • conventions
  • vectors
  • constant acceleration motion
  • applications
projectile motion1
Projectile Motion
  • The only significant force that the object experiences while in the air will be that due to gravity
  • Flight time depends on vertical velocity at release and the height of release above landing surface
  • gravity causes the trajectory to deviate from a straight line into a parabola
  • Because there is no force in the horizontal direction, the horizontal acceleration is zero
symmetry in projectile motion1
Symmetry in projectile motion
  • A comparison of the kinematics at the beginning and end of the flight reveals:
    • vi = - vf & vi2 = vf2
    • ri = rf & rf - ri = 0
slide86

Example: A long jumper leaves the ground with |v| = 7.6 m/s at an angle of 23°. Which equation(s) do you need to calculate how far she jumps?

  • vf = vi + at
  • rf - ri = vit + ½ at2
  • vf2 = vi2 + 2a (rf - ri)

A) 1,2

B) 2,3

C) 1,2,3

D) 2

step 1 break the vector into components
Step 1: Break the vector into components.

7.6 m/s

vi

  • vyi = vi * sin 
  • vyi = 7.6 * sin 23° = 3 m/s
  • vxi = vi * cos 
  • vxi = 7.6 * cos 23° = 7 m/s

vyi

23°

vxi

step 2 from v yi calculate how long the jumper stayed in the air

7.6 m/s

3 m/s

 = 23°

7 m/s

Step 2: From vyi, calculate how long the jumper stayed in the air.
  • When air resistance is negligible,

vyf = -vyi

  • vyf = vyi + at (Eq. 1)
  • -vyi = vyi-9.81t ---> 2vyi = 9.81t ---> t = vyi / 4.9
  • t = 3/4.9 = 0.6 seconds.
slide89
Step 3: From the time in the air and vx, calculate the horizontal distance travelled while in the air.
  • vx is constant throughout jump.
  • ∆rx = vx * t
  • ∆rx = 7 * 0.6 = 4.2 meters

7.6 m/s

3 m/s

 = 23°

7 m/s

angular 2 d example 1
Angular/2-D Example # 1

A hammer thrower releases the hammer after reaching an angular velocity of 14.9 rad/s. If the hammer is 1.6 m from the shoulder joint, what is the linear velocity of release?

If the release angle is 30 degrees, and release height is 1.5 m, how far away did the hammer land?

2 d example
2-D Example

A baseball is thrown with a velocity of 31 m/s at an angle of 40 degrees from a height of 1.8m. Calculate the:

  • vertical and horizontal velocity components
  • time to peak trajectory
  • height of trajectory from point of release
  • total height of parabola
  • time from peak to the ground
  • total flight time
  • range of the throw
equations for constant acceleration
Equations for Constant Acceleration

vf = vi + at

rf - ri = vit + ½ at2

vf2 = vi2 + 2a (rf - ri)

kinematics5
Kinematics
  • 1-D linear kinematics
  • Angular kinematics
  • 2-D linear kinematics
  • Locomotion
locomotion kinematics
Locomotion kinematics
  • What defines gaits?
  • Components of a stride
  • Vertical displacement of the body
  • Horizontal velocity of the body
  • How SF and SL change with speed
walking and running speeds
Walking and running speeds
  • Walking speeds: 0.5 - 2 m/s
  • Running speeds: 2.5 m/s - max
    • Peak for elite male athletes = 12 m/s
locomotion kinematics1
Locomotion kinematics
  • What defines gaits?
  • Components of a stride
  • Vertical displacement of the body
  • Horizontal velocity of the body
kinematic terms for gait analysis
Kinematic terms for gait analysis
  • Stride: One complete cycle from an event (e.g., right foot touch-down) to the next time that event occurs.
  • Stance phase: Time when a limb is in contact with the ground.
  • Swing phase: Time when a limb is NOT in contact with the ground.
kinematic terms for gait analysis1
Kinematic terms for gait analysis
  • Double support: Time when 2 limbs are in contact with the ground.
  • Aerial phase: Time when no feet are on the ground.
stance vs swing phase
Stance vs Swing phase
  • In walking: 60%/40% : Stance/Swing
  • As speed increases:
    • absolute time spent in stance decreases
    • relative time spent in stance decreases
    • true for both walking and running
comparison of walk vs run
Comparison of walk vs. run
  • stance time: run < walk
  • swing time: run > walk
  • stride time: run << walk
  • aerial time: run >> walk (0)
  • double support: run (0) << walk
speed sf sl
Speed = SF * SL
  • Speed (m/s) = SF (strides/s) * SL (m/strides)
    • stride frequency: number of strides per second
    • stride length: distance covered by one stride
  • Can increase speed by increasing SF, SL or both
  • But step length can only be increased so much
speed sf sl1
Speed = SF * SL
  • Speed (m/s) = SF (strides/s) * SL (m/strides)
  • Question: Do people increase speed by increasing SF and/or SL during walking and running?
locomotion kinematics2
Locomotion kinematics
  • What defines gaits?
  • Components of a stride
  • Vertical displacement of the body
  • Horizontal velocity of the body
slide115

Walk

Run

• Hip lowest at

mid-stance.

• Leg more bent.

• Hip highest at

mid-stance.

• Leg is straighter.

slide117

Run

0

0.25

0.50

0.75

Time (s)

locomotion kinematics3
Locomotion kinematics
  • What defines gaits?
  • Components of a stride
  • Vertical displacement of the body
  • Horizontal velocity of the body
walk and run
Walk and run
  • Forward velocity fluctuates, reaching its slowest value at the middle of the stance phase.
  • Vx never reaches zero
slide123

Walking vs. running

  • Double stance?
  • Aerial phase?
  • Position of hip:
    • highest at mid-stance in running.
    • lowest at mid-stance in walking.
  • Leg posture during stance phase.
    • straighter in walking than running
  • Both walk & run: forward velocity reaches minimum at the middle of the stance phase.
kinematics6
Kinematics
  • 1-D linear kinematics
  • Angular kinematics
  • 2-D linear kinematics
  • Locomotion