Download Introduction to Biomechanics and Vector Resolution

Linear Motion: Kinematics and
Kinetics
Applied Kinesiology
420:151
Agenda




Introduction to motion
Linear kinematics
Linear kinetics
Analysis of linear motion
Introduction to Motion

Linear motion (translatory)





Rectilinear
Curvilinear
Circular
Angular motion (rotary)
General motion
Shoulder = angular
or linear?
Figures 11.1, 2, 3, 4, Hamilton
Agenda




Introduction to motion
Linear kinematics
Linear kinetics
Analysis of linear motion
Linear Kinematics


Linear: A point moving along a line
Kinematics: The study of motion (of a
point moving along a line) in respect to
displacement, velocity and acceleration
Displacement




Displacement: Change in position
Vector quantity  Magnitude and
direction
Displacement vs. distance
SI unit = m
Velocity

Velocity: Rate of displacement




V = displacement/time
Vector quantity
Velocity vs. speed
SI unit = m/s
Average vs. Instantaneous
Velocity

Average velocity = displacement/time


Entire displacement  start to finish
Instantaneous: Velocity at any
particular instant within the entire
displacement

Still average velocity however time periods
much smaller therefore “essentially”
instantaneous
Instantaneous?
s (m)
tjohnson (s)
tlewis (s)
Vjohnson (m/s)
Vlewis (m/s)
10
1.86
1.88
5.38
5.32
20
2.87
2.96
6.97
6.76
30
3.8
3.88
7.89
7.73
40
4.66
4.77
8.58
8.39
50
5.55
5.61
9.01
8.91
60
6.38
6.45
9.40
9.30
70
7.21
7.29
9.71
9.60
80
8.11
8.12
9.86
9.85
90
8.98
8.99
10.02
10.01
100
9.83
9.86
10.17
10.14
Velocity Figure - Johnson vs. Lewis (1988 Summer Olympics, Seoul Korea)
11.00
Velocity (m/s)
10.00
9.00
Johnson
8.00
Lewis
7.00
6.00
5.00
10
20
30
40
50
60
Meters (m)
70
80
90
100
(m)
Splits BJ (s)
Splits CL (s)
Vinst. BJ
Vinst. CL
10
1.86
1.88
5.38
5.32
10 20
1.01
1.08
9.90
9.26
20 30
0.93
0.92
10.75
10.87
30 40
0.86
0.89
11.63
11.24
40 50
0.89
0.84
11.24
11.90
50 60
0.83
0.84
12.05
11.90
60 70
0.83
0.84
12.05
11.90
70 80
0.90
0.83
11.11
12.05
80 90
0.87
0.87
11.49
11.49
90 100
0.85
0.87
11.76
11.49
0
Instantaneous Velocity Figure - Johnson vs. Lewis (1988 Summer Olympics,
Seoul Korea)
13.00
12.00
Velocity (m/s)
11.00
10.00
Johnson
9.00
Lew is
8.00
7.00
6.00
5.00
0 10
10 20
20 30
30 40
40 50
50 60
Meters (m)
60 70
70 80
80 90
90 100
Acceleration

Acceleration: Rate of change of velocity




A = vf – vi
Vector quantity
SI unit = m/s/s or m/s2
Uniform acceleration


Very rare
Projectiles (more later)
Average vs. Instantaneous
Acceleration


Average acceleration = Rate of change in
velocity  assumes uniform acceleration
Instantaneous: Acceleration between
smaller time periods


Provides more information
Johnson vs. Lewis
v BJ (m/s) v CL (m/s)
0
0
5.38
5.32
6.97
6.76
7.89
7.73
8.58
8.39
9.01
8.91
9.40
9.30
9.71
9.60
9.86
9.85
10.02
10.01
10.17
10.14
Average acceleration for Ben Johnson?
A = (vf – vi) / t
A = (10.17 m/s – 0 m/s) / 9.83 s
A = (10.17 m/s) / 9.83 s
A = 1.03 m/s2
Enough information?
Average acceleration for Carl Lewis?
A = (vf – vi) / t
A = (10.14 m/s – 0 m/s) / 9.86 s
A = (10.14 m/s) / 9.86 s
A = 1.03 m/s2
Instantaneous?
s (m)
t BJ (s)
t CL (s)
v BJ (m/s)
v CL (m/s)
a BJ (m/s2)
a CL (m/s2)
0
0
0
0
0
0
0
10
1.86
1.88
5.38
5.32
2.89
2.83
20
2.87
2.96
6.97
6.76
0.55
0.49
30
3.8
3.88
7.89
7.73
0.24
0.25
40
4.66
4.77
8.58
8.39
0.15
0.14
50
5.55
5.61
9.01
8.91
0.08
0.09
60
6.38
6.45
9.40
9.30
0.06
0.06
70
7.21
7.29
9.71
9.60
0.04
0.04
80
8.11
8.12
9.86
9.85
0.02
0.03
90
8.98
8.99
10.02
10.01
0.02
0.02
100
9.83
9.86
10.17
10.14
0.02
0.01
Acceleration Figure - Johnson vs. Lewis (1988 Summer Olympics, Seoul Korea)
Acceleration (m/s/s)
3
2.5
2
Johnson
1.5
Lewis
1
0.5
0
0
10
20
30
40
50
60
Distance (m)
70
80
90
100
(m)
Splits BJ (s)
Splits CL (s)
Vinst. BJ
Vinst. CL
10
1.86
1.88
5.38
5.32
2.89
2.83
10 20
1.01
1.08
9.90
9.26
4.48
3.65
20 30
0.93
0.92
10.75
10.87
0.92
1.75
30 40
0.86
0.89
11.63
11.24
1.02
0.41
40 50
0.89
0.84
11.24
11.90
-0.44
0.80
50 60
0.83
0.84
12.05
11.90
0.98
0.00
60 70
0.83
0.84
12.05
11.90
0.00
0.00
70 80
0.90
0.83
11.11
12.05
-1.04
0.17
80 90
0.87
0.87
11.49
11.49
0.44
-0.64
0
a BJ (m/s2)
a CL (m/s2)
Instantaneous Acceleration Figure - Johnson vs. Lewis (1988 Summer
Olympics, Seoul Korea)
5.00
Velocity (m/s/s)
4.00
3.00
2.00
Johnson
Lew is
1.00
0.00
0 10
10 20
20 30
30 40
40 50
50 60
-1.00
-2.00
Meters (m)
60 70
70 80
80 90
90 100
Positive and Negative
Acceleration

Motion to the right or up is considered
in the positive direction


Velocities in these directions also positive
Motion to the left or down is considered
in the negative direction

Velocities in these directions also negative
Positive and Negative
Acceleration




Increasing speed in the positive
direction is + acceleration
Decreasing speed in the positive
direction is – acceleration
Increasing speed in the negative
direction is – acceleration
Decreasing speed in the negative
direction is + acceleration
Use positive/negative acceleration as opposed to
acceleration and deceleration
Uniform Acceleration

Most common form of uniform acceleration is
projectile motion


9.8 m/s2
Laws of uniform acceleration


Acceleration is constant
Allows you to solve for:



Velocity: Initial or final
Distance traveled by implement
Time in the air
Uniform Acceleration



Solve for final velocity when initial
velocity, time and acceleration are
known
vf = vi + at
Manipulate the formula



vi = vf – at
t = vf – vi /a
What if initial or final velocity = 0?
What is the final velocity
at 4 seconds if you know
the velocity at 2 seconds?
How long was it in the air if
you know the initial and
final velocities?
vf = vi + at
Vf = -19.6 m/s + (-9.8 m/s2)(2s)
Vf = -39.2 m/s
t = -49 m/s – 0 m/s /-9.8 m/s2
t = -49 m/s / -9.8 m/s2
t=5s
What is the initial velocity
if you know the velocity at
5 seconds?
vi = vf – at
Vi = -49 m/s – (-9.8 m/s2)(5s)
Vi = 0 m/s
Uniform Acceleration



Solve for velocity when initial velocity,
distance and acceleration are known
vf2 = vi2 + 2as
Manipulate the formula



vi2 = vf2 - 2as
s = vf2 - vi2 /2a
What if initial or final velocity = 0?
Page 288 of text book:
Assuming that a ball is thrown upward so that it reaches a height of 5 meters
before starting to fall, what is its initial velocity? What is its final velocity?
Initial velocity:
Final velocity:
vi2 = vf2 – 2as
vf2 = vi2 + 2as
vi2 = (0 m/s)2 - 2(-9.8 m/s2)(5m)
vf2 = (0 m/s)2 + 2(-9.8 m/s2)(5m)
vi2 = 0 m2/s2 + 98 m2/s2
vf2 = 0 m2/s2 - 98 m2/s2
vi2 = 98 m2/s2 (sq. root both sides)
vf2 = -(98 m2/s2) (sq. root both sides)
vi = 9.9 m/s
vf = -9.9 m/s
Check work by solving for distance
s = vf2 - vi2 / 2a
s = (-9.9)2 – (0)2 / 2(-9.8m/s2)
s = 98 m2/s2 / 19.6 m/s2
s=5m
Uniform Acceleration



Solve for distance when initial velocity,
acceleration and time are known
s = vit + ½ at2
Manipulate the formula
How far will an object drop if you
let if fall for 3 seconds?
s = vit + ½ at2
s = 0 + ½ (-9.8 m/s2)(3 s)2
s = 0 + -4.9 m/s2 (9 s2)
s = -44.1 m
How high will an object go in the air in 2
seconds if thrown with an initial velocity of
29.4 m/s?
s = vit+ ½ at2
s = 29.4 m/s (2 s) + ½(-9.8 m/s2)(2 s)2
s = 58.8 m – 19.6 m
s = 39.2 m
Vi = 0
1s
V = -4.9 m/s
S = -4.9 m
Vf = -9.8 m/s
2s
V = -14.7 m/s
s = -14.7 m
Total distance traveled:
Vf = -19.6
m/s
3s
V = -24.5 m/s
Vf = -29.4
(-4.9)+(-14.7)+(-24.5) = -44.1 m
s = -24.5 m
Total distance traveled:
14.7+24.5 =
39.2 m
Vf = 9.8 m/s
2s
V = 14.7 m/s
S = 14.7 m
Vf = 19.6 m/s
V = 24.5 m/s
1s
Vi = 29.4 m/s
S = 24.5 m
Vacuum  no air
resistance
Real world  air
resistance
Unlimited constant
acceleration
Terminal velocity
Continued acceleration
until air resistance =
gravity
Terminal velocity of
skydiver = 120 mph
Agenda



Introduction to motion
Linear kinematics
Linear kinetics




Newton’s laws
Types of force
Work, power, energy
Analysis of linear motion
Linear Kinetics


What causes motion?  Force
Newton’s Three Laws of Motion:


Cannot be proved
Universally accepted
1st Law: Law of Inertia




Law: A body continues in its state of
rest/motion unless an unbalanced force
acts upon it
Inertia: The resistance to motion
proportionate to mass
Unbalanced forces during start?
Unbalanced forces during stop?
2nd Law: Law of Acceleration



Law: The acceleration of an object is
proportional to the force causing it and
inversely proportional to its mass
F = ma  a = F/m
Impulse: The product of force and time


F = ma = m(vf – vi)/t
Ft = m(vf – vi)
Impulse

Max acceleration of an object = max force
and max time



Hammer throw
Baseball swing (quickness vs. velocity)
Manipulation of impulse




Two basketball players of 100 kg
Player A = 2500 N over 0.25 s
Player B = 3000 N over 0.21 s
Advantage?
Impulse and Momentum




Momentum = mv
Ft = m(vf – vi) = mvf – mvi
Any change in momentum is equal to
the impulse that produced it
Applications?


Shot put: Effect of force, arm length?
Catching a fastball
Why does a 5 story fall
typically result in death?
40 m/s or 90 mph
If you want to stop this pitch
with your bare hands (change
mv from 5.6 kg*m/s to 0), this
change in momentum will
equal the impulse that creates
the change in momentum!
Weight = 5 oz.
Mass = 0.14 kg
Momentum = mv
Momentum = 40 m/s (0.14 kg)
Momentum = 5.6 kg*m/s
Ft = mvf - mvi
If you try to stop the pitch in 0.05
seconds, it will require 112 N of
force (5.6/0.05 = 112). This is
approximately 25 lbs of force!
112(0.05) = 5.6 - 0
Why wasn’t he able to hold on?
(15 July 1999, Alabama) A 25-year-old soldier died
of injuries sustained from a 3-story fall,
precipitated by his attempt to spit farther than his
buddy. His plan was to hurl himself towards a
metal guardrail while expectorating, in order to
add momentum to his saliva. In a tragic
miscalculation, his momentum carried him right
over the railing, which he caught hold of for a few
moments before his grip slipped, sending him
plummeting 24 feet to the cement below. The
military specialist had a blood alcohol content of
0.14%, impairing his judgment and paving the
way for his opportunity to win a Darwin Award.
3rd Law: Law of Reaction



Law: For every action there is an equal and
opposite reaction
Objects at rest are in equilibrium  the weight of
the object is balanced by the force of the surface
pushing back against it
Ground reaction forces


Cement vs. sand
Examples:

Objects colliding? Conservation of momentum


Momentum lost by one body = momentum gained by other
body
Unequal masses = unequal accelerations
Agenda



Introduction to motion
Linear kinematics
Linear kinetics




Newton’s laws
Types of force
Work, power, energy
Analysis of linear motion
Types of Forces





Gravity
Ground reaction forces
Friction
Rebound
Air resistance (drag)
Gravity


“The force that causes objects to fall to earth,
the moon to orbit the earth and the planets
to orbit the sun”
Weight: The attractive force of the earth 
9.8 m/s/s



Weight = mg
1 N = 0.2248 lbs
Gravity is a force vector


Magnitude = weight
Direction = straight down
Figure 12.18, Hamilton
Ground Reaction Force


GRF: The reaction force from
a surface upon which one is
moving
GRFs are force vectors



Magnitude: Equal to amount of
force expressed into ground
Direction: Opposite direction
from which force was expressed
into ground
Energy can be transformed (ie
sand)
Friction




Friction: the force that opposes efforts
to slide or roll one body across/over
another
Can work for us or against us
More friction can be good
Less friction can be good
Figure 12.14, Hamilton
Friction

Starting vs. sliding friction
Friction depends on:


Nature of surfaces (coefficient of friction)
Forces involved
Ff
Coefficient of friction
Push
Weight of book
Rebound


Rebound: The force that causes objects to
rebound from each other after contact
Amount of rebound depends on several
factors




Elasticity
Angle of rebound
Spin
Momentum
Rebound - Elasticity




Objects that receive a stress will strain
(become distorted/deformed)
Elasticity: Ability to return to original
shape once stress is removed
Coefficient of restitution = stress/strain
Greater elasticity = greater rebound
Rebound – Angle of Rebound


Assume COR = 1.0  angle of incidence =
angle of reflection
As COR decreases, so does angle of reflection
relative to angle of incidence
Figure 12.23, Hamilton
Rebound - Spin



Topspin: Increase Hv, decrease Vv
Backspin: Increase Vv, decrease Hv
Sidespin: Rebound in direction of spin
Rebound - Momentum


Momentum = mv
As mass and/or velocity increase, so
does rebound
Air Resistance (Drag)

Air resistance: The force that occurs as a result of fluid
pressure at the leading edge of an object and the
backward pull created by turbulence on the trailing
edge
Figure 12.25, Hamilton
Agenda



Introduction to motion
Linear kinematics
Linear kinetics




Newton’s laws
Types of force
Work, power, energy
Analysis of linear motion
Work, Power, Energy




Work: The product of force expended
and distance through which force
succeeds in overcoming resistance
W = Fd
SI unit = J or N*m
Distance measured vertically only
Force needed to overcome inertia of block = 20 N
Total vertical distance = 2 m
Work = 20 N * 2 m
Work = 40 J or 40 N*m
F
2m
Figure 12.30, Hamilton
Work performed climbing
stairs


Work = Fd
Force

Subject weight


Displacement

Height of each step


Typical 8 inches (20cm)
Work per step


From mass, ie 65 kg
636 N x 0.2 m = 127 Nm
Multiply by the number of steps
Work on a stair stepper


Work = Fd
Force

Push on the step


Displacement

Step Height


????
8 inches
Work per step

???N x .203 m = ???Nm
Work on a cycle ergometer


Work = Fd
Force

belt friction on the flywheel


Displacement

revolution of the pedals


mass ie 3 kg
Monark: 6 m per rev
Work per revolution

3kg x 6 m = 18 kgm
Work, Power, Energy


Work can be positive or negative
Work done in direction of force
application is positive work


Concentric actions
Work done in opposite direction of force
application is negative work

Eccentric actions
Work, Power, Energy



Power: The rate at which work is
performed
Power = Work/time or Fv
SI unit = W or J/s or N*m/s
Calculate & compare power



During the ascent phase of a rep of the
bench press, two lifters each exert an
average vertical force of 1000 N against
a barbell while the barbell moves 0.8 m
upward
Lifter A: 0.50 seconds
Lifter B: 0.75 seconds
Calculate & compare power
Lifter A
F = 1000 N
d = 0.8 m
t = 0.50 s
Lifter B?
Power
Power
Power
Power
=
=
=
=
Fd/t
1000 N*0.8 m /0.50 s
800 J/ 0.50 s
1600 W
Power on a cycle ergometer
Power = Fd/t
Power = Fd*rev/min
How much power output is there?
Force = 3 kg
d = 6 m / revolution
60 revolutions / min
Power = 3kg*6m*60 rev/min
Power = 1080 kg*m/min
Note: 1 Watt = ~ 6 kg*m/min
How many Watts? 180 Watts
Other Ways to Calculate Power

Treadmill sprinting



Vertical jumping (Lewis equation)



Work = weight*vj height
Power is estimated with an equation
Stair stepper



Work = weight*vertical displacement (angle of treadmill)
Time = length of sprint
Work = resistance of step*displacement of step
Time = time to move step
Stair running (Margaria-Kalaman test)


Work = weight*vertical displacement*# steps
Time = time it takes to get up the stairs
Work, Power, Energy



Energy: The capacity to do work
Amount of energy = work accomplished
Many types of energy


Heat, sound, light, electric, chemical,
atomic
Mechanical energy  Two main types


Potential energy
Kinetic energy
Mechanical Energy

Potential energy: Energy a body
possesses due to its position


PE = mgh
Kinetic energy: Energy a body
possesses due to its motion

KE = ½ mv2
PE = Maximum
KE = 0
PE = Decreasing
KE = Increasing
PE = 0
KE = Maximum
Energy can neither be created nor destroyed
Agenda



Introduction to motion
Linear kinematics
Linear kinetics




Newton’s laws
Types of force
Work, power, energy
Analysis of linear motion
General Analysis of Linear
Motion

Linear kinematics:




Linear displacement
Linear velocity
Linear acceleration
Linear kinetics:

Effect of a force in an instant in time


Effect of a force applied over a period of time


F = ma
Impulse and momentum
Effect of a force applied over a distance

Work, power, energy
General Phases of Skill Performance




Ritual Phase
 Full of idiosyncrasies, useful for mental focus
 Does the ritual phase affect performance negatively, positively?
Preparation Phase
 Wind up (force, velocity, accuracy, combination?)
 Is energy storage required?
Force Phase
 Is the force applied in the right direction?
 Is there enough force, too much?
Follow Through Phase
 Is there enough time to slow down body parts?
 Does the follow through promote correct application of force?