Overall course plan Tissue mechanics: mechanical properties of bone, muscle, tendon, etc. Kinematics: quantification of motion, with no regard for the.

Name: Overall course plan Tissue mechanics: mechanical properties of bone, muscle, tendon, etc. Kinematics: quantification of motion, with no regard for the.
Uploaded: 2017-08-29T02:09:40+00:00
Duration: PTM40S19
Channel: Matthew Marshall
Description: Overall course plan Tissue mechanics: mechanical properties of bone, muscle, tendon, etc. Kinematics: quantification of motion, with no regard for the.

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
Locomotion

Reading Linear: Enoka chapter 1 pp. 3-23
Loco: Enoka chapter 4 pp Angular: Enoka chapter1 pp

Locomotion

Video Kinematics images/sec Put markers on anatomical landmarks Computer tracks marker position

Kinematics example video

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 Other units 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 rf Position r, (m) ri ti tf Time (s)

Kinematics terms: Velocity (v)
rate of change of position vave = ∆r / ∆t vinstantaneous = dr / dt (slope of position vs. time) Units = m / s rf Position r, (m) Velocity v, (m/s) vi vf ri ti tf ti tf Time (s) Time (s)

Kinematics terms: acceleration (a)
rate of change of velocity aave = ∆v / ∆t ainstantaneous = dv / dt = slope of velocity vs. time units = m / s2 rf Position r, (m) Velocity v, (m/s) vi vf ri ti tf ti tf Time (s) Time (s)

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 & vi2 = 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 = 0 vf = 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). - 0.3 s 0.03 s s 0.3 s

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
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

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
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 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-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) vmax = 10 m/s aave = 1.66 m/s2 (0 - 6 seconds)

Locomotion

Position Linear Angular symbol = r unit = meter symbol = 
unit = radians or degrees 6.28 rad = 360° 

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

Segment angles Measured relative to a fixed axis (e.g., horizontal)
Knee Hip Ankle Segment angles Measured relative to a fixed axis (e.g., horizontal) Trunk Trunk Thigh Shank Shank

Displacement Linear Angular symbol = ∆r ∆r = rf - ri unit = meter
∆ = f - i unit = radian ∆ 1 2

Angular ---> linear displacement
s = distance travelled (arc) s = radius • ∆ ∆ must be in radians 2 s ∆ 1 radius

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

Knee extension Knee extension

Knee FLEXION knee Knee flexion

A Squat Jump Hip ext Knee ext Ankle ext

EMG: Electromyography
Measures electrical activity of muscles indicates when a muscle is active Hip ext Knee ext Ankle ext

Hip angle Knee angle Ankle angle 0.15 0.30 Hip extensor EMG
Knee extensor EMG Ankle angle Ankle extensor EMG 0.15 0.30 Time (s)

Alternative joint angle definition
Full extension: alt = 0 (degrees or rad) norm alt

Knee angle (rad) 2 Run 1 Walk stance stance Enoka 4.5 Flexion Stance
Swing 1 2 Run Walk Flex-Ext Small stance stance Time (% of stride)

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  ∆

Linear velocity (v) & angular velocity ()
v =  r v r 

A person sits in a chair and does a knee extension
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 Linear symbol = a a = ∆v /∆t unit = m / s2
= “alpha”  = D / Dt unit = rad / s2 

Arm Motion example Muscles act as “motors” and “brakes”
But they only pull Displacement, velocity, acceleration dr/dt, dv/dt

1 2 3  (rad)  (rad/s)  (rad/s2) Time   
F  2 E  (rad/s) F  3 E  (rad/s2) F Time E= extension, F = flexion

Triceps brachialis (ext)
 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
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 ar vi

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  ar at v

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

at = 981 m/s2; ar = 2 m/s2 at = 0 m/s2; ar = 4 m/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 = 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

at = r= (2 m)• (0 rad/s2) = 0 m/s2 ar = r2  = 1 rad/sec
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
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?

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.

Locomotion

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. v v

Hip moves vertically & horizontally.

Horizontal & vertical velocity components

Y Y X X X Running direction

Y Y X X X Z Z Z

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

Vectors in kinematics VECTOR: magnitude and direction. Small vertical
Length of vector indicates magnitude. Direction of vector indicates direction of motion. Small vertical velocity Large horizontal velocity

Which Vectors Are Equal?
A&B A,B&E C&E 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 vx = horizontal component
vy   Right triangle ---> use trigonometry. |v| = magnitude of vector  = angle of vector with the horizontal vx = horizontal component vy = vertical component

Determining vector components
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 

Example of determining components
|v| = 2m/sec vy 60° 60° vx 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

How to find resultant vector
vy vy  vx vx Finding  tan  = vy / vx thus,  = tan-1(vy / vx) Finding |v| |v| = (vx 2 + vy2)0.5

Example: Find resultant vector
vx = 2m/s vy = 3m/s  |v| = ?  = ?  = tan-1(vy /vx)   = tan-1(3 / 2) = 56.3° |v| = (vx2 + vy2)0.5 |v| = ( )0.5 = (13)0.5 = 3.6m/s

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 & vi2 = vf2 ri = rf & rf - ri = 0

Example: A long jumper leaves the ground with |v| = 7
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.
7.6 m/s vi vyi 23° vxi vyi = vi * sin  vyi = 7.6 * sin 23° = 3 m/s vxi = vi * cos  vxi = 7.6 * cos 23° = 7 m/s

Step 2: From vyi, calculate how long the jumper stayed in the air.
7.6 m/s 7 m/s 3 m/s  = 23° 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.

vx is constant throughout jump. ∆rx = vx * t
Step 3: From the time in the air and vx, calculate the horizontal distance travelled while in the air. 7.6 m/s 3 m/s  = 23° 7 m/s vx is constant throughout jump. ∆rx = vx * t ∆rx = 7 * 0.6 = 4.2 meters

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

Locomotion

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: 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)
Stride Length (m) 1.03 – 1.35 1.51 – 3.0 4.60 – 4.50 Cadence (steps/min) Stride Rate (Hz) 0.65 – 0.98 1.10 – 1.38 1.75 – 2.0 Cycle Time (s) 1.55 – 1.02 Stance (% of gait cycle) 66 -60 25-20 Swing (% of gait cycle) 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.
True False 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)
Stride Length (m) 1.03 – 1.35 1.51 – 3.0 4.60 – 4.50 Cadence (steps/min) Stride Rate (Hz) 0.65 – 0.98 1.10 – 1.38 1.75 – 2.0 Cycle Time (s) 1.55 – 1.02 Stance (% of gait cycle) 66 -60 25-20 Swing (% of gait cycle) 75- 80

Walk Run • Hip lowest at mid-stance. • Leg more bent. • Hip highest at mid-stance. • Leg is straighter.

Run 0.25 0.50 0.75 Time (s)

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.

Locomotion

Overall course plan Tissue mechanics: mechanical properties of bone, muscle, tendon, etc. Kinematics: quantification of motion, with no regard for the.

Similar presentations

Presentation on theme: "Overall course plan Tissue mechanics: mechanical properties of bone, muscle, tendon, etc. Kinematics: quantification of motion, with no regard for the."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Overall course plan Tissue mechanics: mechanical properties of bone, muscle, tendon, etc. Kinematics: quantification of motion, with no regard for the.

Similar presentations

Presentation on theme: "Overall course plan Tissue mechanics: mechanical properties of bone, muscle, tendon, etc. Kinematics: quantification of motion, with no regard for the."— Presentation transcript:

Similar presentations

About project

Feedback