Transcript 355 Angular Kinetics

Torque
• The perpendicular distance from the line of action to
the pivot point is called the moment arm (r).
r
r
F
F
• Torque is the force multiplied by the moment arm.
T = F*r, Units: N-m
• If the line of action of a force vector does not go
through the pivot point of an object, it will tend to
rotate the object.
Torque
• centric force
– applied through axis (center) of rotation
– creates no torque so causes no rotation
• eccentric force
– applied some distance away from axis of
rotation
– creates torque so causes rotation
Will the green
person go up or
down?
Which door is easier to open?
What causes a limb to rotate force or torque?
• angular motion occurs at
a joint, so torque causes
the limb to rotate
• torque is developed
because a force acts at a
distance from the axis of
rotation
muscle force (Fm)
muscle torque
(Tm=Fm*dm)
Perpendicular distance between
pt of application and joint axis
(dm)
Calculation of Muscle Torque
400 N
NOTE: The torque created by the muscle depends on
1) the size of the muscle force
2) the angle at which the muscle pulls
3) the perpendicular distance from the muscle to the joint axis
Example
Fm= ?
0.45 m
0.02 m
0.15 m
Wfa=13.35 N
Wbb=44.5 N
Give me a lever long enough
and a fulcrum strong enough,
and, single-handedly, I can
move the world.
--Archimedes (287 - 212 B.C.)
Levers
• A lever consists of two forces (motive and
resistance forces) acting around a pivot point
(axis or fulcrum).
• The perpendicular distance from the line of
action of the effort force to the fulcrum is called
the motive arm.
• The perpendicular distance from the line of
action of the resistance force to the fulcrum is
called the resistance arm.
Elements of a Lever
resistance force
motive force
(effort force)
axis
(fulcrum)
motive
arm
resistance
arm
Mechanical Advantage of a Lever
motive arm
resistance arm
The ratio of the motive arm to the resistance arm is
called the mechanical advantage (MA).
If MA is approximately 1 the lever simply acts to
redirect the applied force.
If MA is > 1 the lever acts to amplify the force.
If MA is < 1 the lever acts to amplify the speed and
range of motion.
• MA =
•
•
•
•
MA = 1
motive force
direction of the force
vector is redirected
motive arm = resistance arm
MA > 1
motive force
motive arm > resistance arm
force is amplified
MA < 1
motive force
motive arm < resistance arm
ROM / speed is amplified
Muscles have MA <1
0.45 m
(resistance arm)
0.02 m
(motive arm)
0.15 m
(resistance arm)
motive arm < resistance arms
Mechanical Advantage
• Muscles typically have a MA in ROM and
speed.
Classes of Levers
• Classified according to the relative positions of
the axis, motive force and resistive force.
How to remember the class of lever
ARM
1st class Axis is between resistance and motive
force.
2nd class Resistance force is in between the axis
and the motive force.
3rd class Motive force is in between the axis and
the resistance force.
1st Class Lever
• axis in the middle
• e.g. see-saw
• most versatile lever because it can be used
for any type of mechanical advantage
• e.g. in body
– pushing down gas pedal
– elbow/triceps extensions overhead
2nd Class Lever
• resistance in middle
• force advantage usually exists for motive
force
• e.g. push-up
– body is lever, feet are axis, resistance is weight
of body and motive is arms
3rd Class Lever
• motive in middle
• most musculoskeletal arrangements are 3rd
class levers
• muscle is motive force
• advantage in ROM and speed but
disadvantage in F
Gears
• A gear is similar in
function to a lever.
• The torque on the wheel
and the gear is the same.
• The moment arms are
different and therefore
the forces are different.
Equilibrium – Linear
Components
V
W
W
Fr
Fp
Ry
if Ry = W
then resultant force = 0
if v = 0 and SF = 0
STATIC EQUILIBRIUM
Ry
Fr = resistive force
Fp = propulsive force
if v = 0 and SF = 0
DYNAMIC EQUILIBRIUM
Equilibrium – Angular
Components
T1
T2
If the object is rotating
with a non-zero velocity
and…
if T1 = T2
then resultant torque = 0
if w = 0 and ST = 0
STATIC EQUILIBRIUM
T1=T2
DYNAMIC EQUILIBRIUM
b/c w = constant and ST = 0
Stability
• Stability is the resistance to linear and
angular acceleration.
• There are 3 major factors that influence the
stability of an object.
1) The line of gravity with respect to the base of
support.
unstable
stable
2) The height of the center of mass.
10 kg
100kg
unstable
stable
3) The mass.
Center of Mass
• The center of mass is the point about which
the body's mass is evenly distributed.
• The line of gravity is the line that defines
the center of mass in the transverse plane.
• The sum of the torques about an axis caused
by the weights of multiple particles is equal
to the distance from the axis to the center of
mass multiplied by the sum of the weights.
Center of Mass
• The center of mass is the point about which
the body's mass is evenly distributed.
balance point
Symmetric
distribution
Asymmetric
distribution
CM in the middle
CM closer to larger weight
Y
2 kg
(1,3)
3
1
3 kg
(3,3)
COM
(2, 2.67)
1 kg
(1,1)
2
line of gravity
(if Y is vertical)
X
The sum of the torques about an axis caused by the weights
of multiple particles is equal to the distance from the axis
to the center of mass multiplied by the sum of the weights.
Y
2 kg
(1, 3) 1
3 kg
3 (3, 3)
COM Location
in the x direction
COM
(2, 2.67)
1 kg
(1,1)
2
X
m1x1 + m2x2 + m3x3 =
2kg(1m) + 1kg(1m) + 3kg(3m) =
12kg  m
xcm =
= 2m
6kg
Mxcm
6kgxcm
Y
2 kg
(1, 3) 1
COM Location
in the y direction
3 kg
3 (3, 3)
COM
(2, 2.67)
1 kg
(1,1)
2
X
m1y1 + m2y2 + m3y3 =
2kg(3m) + 1kg(1m) + 3kg(3m) =
16kg  m
ycm =
= 2.67 m
6kg
Mycm
6ycm
General Formulas:
n
m x
i
x cm =
i=1
n
m
i=1
n
m y
i
i
y cm =
i
i
i=1
n
m
i
i=1
where,
xi is the distance from the y-axis to the ith mass
yi is the distance from the x-axis to the ith mass
mi is the mass of the ith element (segment)
n
x cm =
p x
i
i
i=1
n
y cm =
p y
i
i
i=1
An alternative approach is to use the proportion
of each mass (pi) instead of the actual masses. For
the human body the proportion can be found in
many text books for each body segment.
Y
X
Segment
Relative Mass
CM Location (% Length)
(% Total Body Mass)
Head
7.3
46.4% from vertex
Trunk
50.7
43.8% from suprasternale
L Upper Arm
2.6
49.1% from shoulder
L Forearm
1.6
41.8% from elbow
L Hand
0.7
82.0% from wrist
R Upper Arm
2.6
49.1% from shoulder
R Forearm
1.6
41.8% from elbow
R Hand
0.7
82.0% from wrist
L Thigh
10.3
40.0% from hip
L Shank
4.3
41.8% from knee
L Foot
1.5
44.9% from heel
R Thigh
10.3
40.0% from hip
R Shank
4.3
41.8% from knee
R Foot
1.5
44.9% from heel
Body Segment
Parameters
Derived from
direct cadaver
measurements
Elderly, male,
Caucasian
cadavers
From these data it is apparent that to determine the center of mass
of a segment it is necessary to locate the segment endpoints
Y
X
Y
X
Y
To locate the segment CM
1st measure the length of the
segment
6.8 cm
X
Y
CM of the trunk is 43.8% of the
length of the trunk away from the
suprasternale
43.8% of 6.8 cm = 3.0 cm
3.0 cm
X
Y
to find the whole body CM you need
to express the segmental CM
locations with respect to a common
reference point - we’ll use the origin
x = p x
n
cm
i
i
i =1
n
y cm =
p y
i
i
i=1
xi
yi
X
Y
Do this for every segment
Use these distances and the
x = p x
segment masses to compute
y = p y
the whole body CM location
n
cm
i
i
i
i
i =1
n
cm
i=1
xi
yi
X
Y
Plot the final coordinates of the CM
X-distance = 120 mm
Y-distance = 92 mm
X
Straddle Jump
r
Fosbury Flop
r
A basketball player can appear to
remain at a constant height for brief
periods of time by manipulating the
body segments about their center of
mass. The COM will always follow
the path of a parabola while the body
is in the air. (Michael Jordan is very
good at this.)
Moment of Inertia
Resistance to angular
motion (like linear
motion) is dependent
on mass.
The more closely mass is
distributed to the axis of
rotation, the easier it
is to rotate.
therefore: resistance to
angular motion is dependent
on the distribution of mass
This resistance is called the
Moment of Inertia.
Moment of Inertia
• ANGULAR FORM OF INERTIA
– resistance to changes in the state of angular
motion
• I = mr2
– for a single particle I is proportional to the
distance squared
• SI unit = kg-m2
x
Each block is .5 m by 1.5 m and has a mass of 2 kg. The
mass in each block is uniformly distributed. What is the
moment of inertia about the x axis?
Ix = Smiri2 = [2kg*(.25m)2] + [2kg*.75m)2] + [2*(.25m)2]
+ [2kg*(.75m)2]
= 2.5 kg-m2
x
If the mass of the above object were concentrated at a
single point (the center of mass) how far from the axis
would it have to be located to have the same moment of
inertia?
Ix = 2.5kg-m2 = mk2 = 8kg*k2
k=
2.5kg  m /8kg
2
= .559 m
This value is called the radius of gyration:
distance from axis of rotation to a point where the body’s mass
could be concentrated without altering its rotational characteristics
for a system of particles
I = mk2
where k = ‘radius of gyration’
It is often expressed as a proportion of the segment length in
biomechanics. Thus,
I = m(rl)2
where I is the moment of inertia
r is the radius of gyration as a proportion of the segment
length (l)
Different Axes
• recognize that
rotation can occur
about different axes
– each axis has its
own moment of
inertia associated
with it
Whole Body Moment of Inertia
• consider human movement to occur about 3 principal axes
• each principal axis has a principal moment of inertia
associated with it
• when mass is distributed closer to the axis, the moment of
inertia is lower
Angular Analog Newton’s Laws
1) A rotating body will continue to turn about
its axis of rotation with constant angular
momentum, unless an external couple or
eccentric force is exerted upon it.
•linear momentum
p = m*v
•angular momentum
H = I*w
Angular Momentum
•linear momentum
•In the linear case mass
does not change but the
p = m*v
•angular momentum moment of inertia can
be manipulated by
H = I*w
reorienting body
segments.
- ice skaters
- divers
Angular Analog Newton’s Laws
2) The rate of change of angular momentum
of a body is proportional to the torque
causing it and the change takes place in the
direction in which the torque acts.
wf - wi
T=I t
or
T = Ia
Angular Analog Newton’s Laws
3) For every torque that is exerted by one
body on another there is an equal and
opposite torque exerted by the second body
on the first.
TRANSFER OF
ANGULAR
MOMENTUM
enter pike - Hlegs
because legs slow down
Htrunk+arms to maintain
a constant Htotal
the opposite occurs at
entry - Htrunk + arms
to give a clean entry
Hlegs
to maintain Htotal
Angular Momentum in the Long Jump
Htotal = Htrunk+head + Harms + Hlegs = constant CW
To prevent trunk+head from rotating forward (CW)
rotate arms and legs CW to account for Htotal
Iarms and Ilegs are smaller than Itotal so
warms and wlegs must be larger to produce
H’s large enough to accommodate Htotal
Initiation of Rotation in Air
Newton’s Laws specifically
state that you can NOT
initiate rotation (e.g. in the
air) without an external torque
being applied to you
So -- can you initiate rotation
while airborne?
A cat does! (seemingly)
explanation consider relationship
between I’s of body parts
that interact when rotation
is initiated
1) As the cat begins to fall it
bends in middle, brings its
front legs in close to its
head and rotates the upper
body through 180 degrees.
1a) In reaction to the upper
body the lower body will
rotate in the opposite
direction.
However -- since the body is
piked Ilower body is very large
compared to upper body so
the corresponding rotation is
small (about 5 degrees).
2) To complete the 180 degree
rotation the cat brings its hind
legs and tail into line with its
lower trunk such that its
longitudinal axis runs through
its hindquarters.
2a) The reaction of upper body
is again small since Iupper body
(about this axis of rotation) is
large, so there is little rotation
of upper body
3) Minor adjustments are
made by rotating tail in
direction opposite to the
desired rotation.
Centripetal vs. Centrifugal Force
• Centripetal force (center
seeking force) = mass x
centripetal acceleration
Fc  mac
2
vt
m
r
 mw 2r
• Centrifugal force (center fleeing force)
-- reaction to the centripetal force;
applied to the other body
Consider Newton’s second law of motion:
SF = ma
Now substitute centripetal acceleration. In centripetal motion the
centripetal acceleration is linked to a centripetal force. You can
think of this force as being responsible for holding the object in a
circular path.
Example
you make a right turn in your car
you feel the driver door push on you to the right
(toward the center of the curvature of your
curved path)
the door applies a centripetal force to you,
you apply a centrifugal force which is
equal and opposite to the centripetal force
door
you
Fcf
Fcp
Forces occurring
along a curved path