Tissue mechanics: mechanical properties of bone, muscle, tendon, etc.

• 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

Overall course plan

Kinematics

• 1-D linear kinematics• Angular kinematics• 2-D linear kinematics• Locomotion

Reading

• Linear: Enoka chapter 1 pp. 3-23• Loco: Enoka chapter 4 pp. 141-146• Angular: Enoka chapter1 pp. 23-29

Kinematics


• 60-6000 images/sec• Put markers on

anatomical landmarks• Computer tracks

marker position

Video Kinematics

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=

http://www.vicon.com/applications/life_sciences.html

http://www.youtube.com/watch?v=USOYUMN5nwU

http://www.youtube.com/watch?v=USOYUMN5nwU

http://www.youtube.com/watch?v=IJ4tndpwL-o&mode=related&search

http://www.youtube.com/watch?v=IJ4tndpwL-o&mode=related&search

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

• Measurement Rules• Definitions of terms used in kinematics• Equations and examples for constant

acceleration motion• 1D Kinematics example

Units

• SI system– Length : meters (m)– Mass : kilograms (kg)– Time : seconds (s)

• Other units– Angle : radians (rad)

• 2p radians = 360o

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!

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• Displacement = Change in position

= ∆r = (rf - ri)

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

Positi

on r,

(m)

Time (s)

ri

rf

ti tf

Kinematics terms: Velocity (v)• rate of change of position• vave = ∆r / ∆t

• vinstantaneous = dr / dt (slope of position vs. time)• Units = m / s

Velo

city

v, (m

/s)

Time (s)

vi

ti tf

Positi

on r,

(m)

Time (s)

ri

ti tf

rf

vf

Kinematics terms: acceleration (a)

• rate of change of velocity• aave = ∆v / ∆t

• ainstantaneous = dv / dt = slope of velocity vs. time• units = m / s2

Velo

city

v, (m

/s)

Time (s)

vi

ti tf

Positi

on r,

(m)

Time (s)

ri

ti tf

rf

vf

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

• 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

• A comparison of the kinematics at the beginning and end of the flight reveals: vi = - vf & vi

2 = vf2

ri = rf & rf - ri = 0

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 = 0vf = at

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• 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).

A) - 0.3 sB) 0.03 sC) - 0.03 sD) 0.3 s

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, thenrf - ri = ½ at2

Example using equation 2What 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?

A) 46 mB) 46 cmC) 1.34 mD) -1.34 m

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 = vi

2 + 2a (rf - ri)

• If vi = 0, thenvf

2 = 2a (rf - ri)

Example using Equation 3

A runner starts at a standstill and acceleratesuniformly 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 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?A) 5 mB) 10 mC) 50 mD) 100 m

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 sprint

100 meter sprint

vmax reached at 6 seconds, at 50 meters.

100 meter sprint

vmax = 10 m/s aave = 1.66 m/s2

(0 - 6 seconds)

Acceleration (m / s2)

Kinematics


Position• Linear

symbol = runit = meter

• Angularsymbol = unit = radians or degrees6.28 rad = 360°

Joint angles

• Ankle: foot - shank• Knee: shank - thigh• Hip: thigh - trunk• Wrist• Elbow• ShoulderKnee

Hip

Ankle

Knee

Hip

Ankle

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

Segment angles

Shank

Trunk

Trunk

Thigh

Shank

Displacement• Linear

symbol = ∆r∆r = rf - ri

unit = meter• Angular

symbol = ∆∆ = f - i

unit = radian

∆ 1

2

Angular ---> linear displacement• s = distance travelled (arc)• s = radius • ∆∆ must be in radians

∆

s

radius1

2

Joint angle change: Flexion

• Flexion: decrease in angle between two adjacent body segments

• Bird’s eye view of arm on table

Elbow

Flexion

Joint angle change: Extension

• Extension: increase in angle between two adjacent body segments

Extension

Elbow

Kneeextension

Knee extension

Knee FLEXION

Kneeflexion

knee

Ankleext

A Squat Jump

Hipext

Kneeext

EMG: Electromyography• Measures electrical activity of muscles

– indicates when a muscle is active

Ankleext

Hipext

Kneeext

Time (s)

Hipangle

Kneeangle

Ankleangle

0.15 0.300

Hip extensor EMG

Knee extensor EMG

Ankle extensor EMG

Alternative joint angle definition

Full extension: alt = 0 (degrees or rad)

alt

norm

0

1

2

0

1

Run

Walk

Flexion

Time (% of stride)

Enoka 4.5Kneeangle(rad)

Flex-Ext Flex-Ext

SmallFlex-Ext Flex-Ext

Stance Swing

stance

stance

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

∆

• v = r

Linear velocity (v) & angular velocity ()

v

r

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)

A) 1 rad/s, 0.5 m/s, 0.25rad

B) 0.5 rad/s, 0.5 rad/s, 0.25rad

C) 1 rad/s, 0.5 m/s, 0.25 m

D) 0.25 rad/s, 0.5 m/s, 0.25m

Angular acceleration• Linear

– symbol = a– a = ∆v /∆t– unit = m / s2

• Angular– symbol = – = “alpha”– = D / Dt– unit = rad / s2

Arm Motion example

Muscles act as “motors” and “brakes”But they only pull

Displacement, velocity, accelerationdr/dt, dv/dt

Time

(rad)

(rad/s)

(rad/s2)

E

F

E

F

E= extension, F = flexion

1

2

3

1 2 3E

F

(rad)

(rad/s)

(rad/s2)

Triceps brachialis (ext)

Brachioradialis (flex)

Biceps brachialis (flex)

Flex

Ext

Enoka example1.8

• 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 = r2 = v2 / r– ar is never zero during circular motion

Linear acceleration (a) in angular motion:radial and tangential acceleration

ar

vi

vf

• a = aradial + atangential (vector sum)

• atangential (m/s2): causes ∆ & ∆v

• atangential = r = ∆v / ∆tat is zero if |v| or is constant

ar

at

Linear acceleration (a) in angular motion

v

v

Total linear acceleration (a)

• a = at + ar (vector sum)

• |a| = (at2 + ar

2)0.5

• = tan-1(at / ar)

ar

aat

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?

A) at = 981 m/s2; ar = 2 m/s2

B) at = 0 m/s2; ar = 4 m/s2

C) at = 0 m/s2; ar = 2 m/s2

D) at = 0 m/s2; ar = 2 rad/s2

E) at = 981 m/s2; ar = 4 rad/s2

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 = r2

= 1 rad/secar = (2 meters) • (1 rad/s)2 = 2 m/s2

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

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?

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.

Angular Kinematics Example # 2

Kinematics


2-D linear kinematics• 1-D vs. 2-D kinematics• conventions• vectors• constant acceleration motion• applications

1-D analysis: 100 meter sprint example

• We only considered the horizontal motion.

vv

Hip moves vertically & horizontally.

• Horizontal & vertical velocity components

vv

Running direction

Y

XX

Y

X

Y

X

Z

X

Z

X

Z

Y

Sign conventions• Y axis (vertical): Positive is upward• X axis (horizontal): Positive is in direction of

movement.

Vectors in kinematics• VECTOR: magnitude and direction.

– Length of vector indicates magnitude.– Direction of vector indicates direction of motion.

Large horizontalvelocity

Small verticalvelocity

Which Vectors Are Equal?

A) A&BB) A,B&EC) C&ED) None of them are equal

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

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

= angle of vector with the horizontal

• vx = horizontal component

vy = vertical component

|v| v

vx

vy

• 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

Determining vector components

v

vx

vy

|v|

Example of determining components

• 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

60°

|v| = 2m/sec

vx

vy

60°

How to find resultant vector

• Finding tan = vy / vx thus, = tan-1(vy / vx)

• Finding |v||v| = (vx 2 + vy

2)0.5

v = ?

vx

vy

=vx

vy

Example: Find resultant vector

= tan-1(vy /vx) = tan-1(3 / 2) = 56.3° |v| = (vx

2 + vy2)0.5

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

|v|

vx = 2m/s

vy = 3m/s

|v| = ? = ?

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.

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 motion

• A comparison of the kinematics at the beginning and end of the flight reveals: vi = - vf & vi

2 = vf2

ri = rf & rf - ri = 0

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?1. vf = vi + at2. rf - ri = vit + ½ at2

3. vf2 = vi

2 + 2a (rf - ri)

A) 1,2B) 2,3C) 1,2,3D) 2

Step 1: Break the vector into components.

• vyi = vi * sin

• vyi = 7.6 * sin 23° = 3 m/s

• vxi = vi * cos

• vxi = 7.6 * cos 23° = 7 m/s

23°

7.6 m/s vi

vxi

vyi

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.

7.6 m/s

7 m/s

3 m/s = 23°

• vx is constant throughout jump.

• ∆rx = vx * t

• ∆rx = 7 * 0.6 = 4.2 meters

Step 3: From the time in the air and vx, calculate the horizontal distance travelled while in the air.

7.6 m/s

7 m/s

3 m/s = 23°

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 ExampleA 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

vf = vi + atrf - ri = vit + ½ at2

vf2 = vi

2 + 2a (rf - ri)

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

What defines a walk vs. a run?

• Kinematic• Kinetic

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


• What defines gaits?• Components of a stride• Vertical displacement of the body• Horizontal velocity of the body

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 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

• 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

Walking vs Running

Variable Walking Running Sprint

Speed (m/s) 0.67-1.32 1.65-4.0 8.0-9.0

Stride Length (m) 1.03 – 1.35 1.51 – 3.0 4.60 – 4.50

Cadence (steps/min) 79 - 118 132-200

Stride Rate (Hz) 0.65 – 0.98 1.10 – 1.38 1.75 – 2.0

Cycle Time (s) 1.55 – 1.02 0.91- 0.73 0.57-0.5

Stance (% of gait cycle) 66 -60 59 - 30 25-20

Swing (% of gait cycle) 34 - 40 41 - 70 75- 80

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

To increase my walking speed my steps must get longer.

A)TrueB) FalseC) It depends

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 * 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?

Walking vs Running

Variable Walking Running Sprint

Speed (m/s) 0.67-1.32 1.65-4.0 8.0-9.0

Stride Length (m) 1.03 – 1.35 1.51 – 3.0 4.60 – 4.50

Cadence (steps/min) 79 - 118 132-200

Stride Rate (Hz) 0.65 – 0.98 1.10 – 1.38 1.75 – 2.0

Cycle Time (s) 1.55 – 1.02 0.91- 0.73 0.57-0.5

Stance (% of gait cycle) 66 -60 59 - 30 25-20

Swing (% of gait cycle) 34 - 40 41 - 70 75- 80

Walk Run

• Hip highest atmid-stance.• Leg is straighter.

• Hip lowest atmid-stance.• Leg more bent.

Run

Time (s)

0.25 0.750.500

Make a graph of the horizontal velocity vs. time for a walking stride

Walk and run• Forward velocity fluctuates, reaching its slowest

value at the middle of the stance phase.• Vx never reaches zero

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.

Kinematics


Tissue mechanics: mechanical properties of bone, muscle, tendon, etc.

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