Transcript presentation source

Dynamics
Dynamics deals with forces, accelerations
and motions produced on objects by these
forces.
Newton’s Laws

First Law of Motion: Every body continues in
its state of rest or of uniform speed in a straight
line unless it is compelled to change that state
by forces acting on it.


Second Law of Motion: The acceleration of an
object is directly proportional to the net force
acting on it and is inversely proportional to its
mass. The direction of the acceleration is in the
direction of the applied net force.
Third Law of Motion: Whenever one object
exerts a force on a second object, the second
object exerts an equal and opposite force on the
first.
Law of Universal Gravitation: Every
particle in the universe attracts every other
particle with a force that is proportional to
the product of their masses and inversely to
the square of the distance between them.
This force acts along the line joining them.
m1 m2
FG 2
r
A force is a vector, having both magnitude
and direction, expressed as a symbol with
an arrow above it.
F
Acceleration is also a vector.
Mass is a scalar, having only magnitude.
Equations of Motion
Since, F  m  a
and a  M
t
then,Fnet
M
Fnet M
m
or

t
m
t
where,
M  wind velocity and direction
Writing this in the U, V components of the
wind gives:Fx net U
Fy net V


m
t
m
t

Uat t + t   Ut 
U
Since we can write, 
t
t
then substituting into the above
Fx net
equations
gives,
U(t + t )  Ut 
 t
m
Fy net
V (t+ t )  Vt  
 t
m

So, if we know the initial wind speed
components and the forces acting on air parcels,
we should be able to determine the wind speed
at some time in the future, (t + t).
Forces acting on air parcels
Pressure Gradient Force
Advection
Centrigugal / Centripital Force
Coriolis Force
Turbulent Drag Force
Advection
Advection is the process of transport of an
atmospheric property solely by the mass
motion of the atmosphere.
The motion of the air (wind) moves air of
different velocities from one location to
another.
Advection terms are written as:
Fx AD
U
U
 U
V
m
x
y
Fy A D
V
V
 U
V
m
x
y
where, U , U , V , V are gradient terms.
x
y
x
y
Pressure Gradient Force
Always acts from High Pressure toward
Low Pressure, perpendicular to isobars.
It starts air parcels moving. Without the
pressure gradient force, the air wouldn’t
move.
Fy P G
Fx PG
1 P
1 P


m
 y
m
 x
Centripital Acceleration
Consider a ball of mass, m, attached to a
string and whirled through a circle of radius
r at a constant velocity w.
From the point of view of an observer in fixed
space, the speed of the ball is constant, but its
direction of travel is continuously changing, so its
velocity is not constant; i.e., it
has an acceleration.
To compute the acceleration,
we consider the change in
velocity, v, which occurs
for a time increment, t, during which the ball
rotates through an angle Q. Q is also the angle
between the vectors v and v + v.
v  v Q
If we divide by t and note that in the limit as
t  0 , v is directed toward the axis of
dv
dQ
rotation, (it is a -v), then:
 v
dt
dt
dQ
v  w r and
w
dt
Since, tangential velocity
dv
Then,   w 2 r
dt
dM
2
In our convention we can write:  w r
dt
Therefore, when viewed from fixed
coordinates, the motion is one of uniform
acceleration directed toward the axis of
rotation and equal to the square of the
angular velocity times the distance from the
axis of rotation. This acceleration is the
Centripital Acceleration.
The resulting force on the ball to produce
this acceleration toward the axis of rotation
is the Centripital Force.
Now, suppose the motion is observed in a
coordinate system rotating with the ball.
Now, the ball appears stationary, but the centripital
force is still acting on the ball, namely the pull of
the string.
In order to apply Newton’s second law to describe
the motion relative to this rotating coordinate
system, we must include an additional (apparent)
force, the Centrifugal Force, which just balances
the force of the string on the ball. It is acting
outward, away from the axis of rotation.
Thus, the Centrifugal Force is equivalent to the
inertial reaction of the ball on the string and is equal
but opposite to the centripital acceleration.
In a fixed system, the rotating ball undergoes
constant Centripital acceleration in response to the
force exerted by the string.
In a moving system, the ball is stationary and the
force exerted by the string is balanced by a
dMcentrifugal
Centrifugal force.
 w 2r
dt
The Centrifugal force can be considered the force
necessary to balance all the other forces acting on
the ball.
Since, for wind at rest on Earth’s surface,
2
M
Fcentrif ugal
dM
M
M = wr, then w 
and


r
dt
r
m
To get the components in the x and y
2
2
2
direction, remember
that
and
M  U V
to get the sign right, (Centrifugal force going in
the proper direction), consider the following low
pressure center.
Fx CN
VV
 By writing the equations as:
 s 
m
R
and using the sign below, the
direction of the force is proper. Fy CN  s  U  U
m
Direction and sign
the Centrigugal Forces
should have.
2
F y centrifugal = +U
r
M = +V
M = -U
Fx centrifugal= -
V
r
2
L
M = -V
F y centrifugal
Fx centrifugal= +
M = +U
2
= -U
r
V2
r
R
Effective Gravity
A particle of unit mass at rest on the surface of the
earth observed in a reference frame rotating with
the earth, is subject to a centrifugal force, W2R,
where W is the angular speed of rotation of the
earth and R is the distance of the particle from the
axis of rotation.
Thus, the weight of the particle
of mass, m, at rest on the Earth’s
surface will generally be less
than the gravitational force, mg*,
because the centrifugal force
partly balances the gravitational
force.
The spheroid shape of the Earth
makes g (the Effective gravity)
directed normal to the level
surface.
g  g * W R
2
Coriolis Force
Consider an object moving easterward in the northern
hemisphere with velocity U with respect to the Earth’s
surface.
For an observer NOT on the Earth, its tangential (linear)
velocity,
v  R(objects angular velocity )  WR  U
Since it is moving in a circular path, its Centripital
acceleration is toward the axis of rotation of magnitude:

WR  U
R
2
or
WR  U 
R
2
Expanding gives:
WR  U 2
R
2
W 2 R2  2WRU  U 2
U
2

 W R  2WU 
R
R
For an observer on the Earth, the Earth’s surface
seems stationary, so there is no tangential velocity
due to the Earth’s motion, only due to the object’s
motion. The acceleration on the object would be
2
just: U 2
U

R
or magnitude
R
The discrepancy between
the magnitude
of
the
two
2
2

WR

U

U
2
is:
 W R  2WU 
R
R
U2

R
W R  2WU
2
The Centrifugal Force (acceleration) we know is:
W R
2
2WU
The Coriolis Force (acceleration) is:
directed outward, like the Centrifugal Force.
also
This outward directed
Coriolis Force
(acceleration) can be
divided into components
in the vertical and
parallel to the Earth’s
surface along the meridional directions
(North - South).
The horizontal force causes a change in the
direction of movement of the object (as
perceived by someone on a moving coordinate
system).
That change in direction for our object moving
eastward with velocity U, is toward the south, i.e.,
to the right of the direction of motion in the
northern hemisphere. (In other words, it is causing
a change in the north-south velocity of the parcel).
The vertical component is much smaller than the
gravitational acceleration (force) and its only effect
is to cause a very minor change in the apparent
weight of an object.
The term f c  2Wsin  is called the Coriolis
parameter. It is positive in NH and negative in
SH.
Objects moving north or south.
Consider an object, of unit mass,
initially at rest on the surface of
the Earth at latitude F. It starts
moving southward and moves
F. As the object moves
equatorward, it will conserve its angular
momentum in the absence of torques in the eastwest direction.
Its initial angular momentum,
L, is: W  wR2, where w is
initially equal to zero. For other
than unit mass it is:
L  I W w   mW  w R
2
Since the distance to the axis of rotation, R,
increases to R + R for the object moving
equatorward, a relative westward velocity, must
develop, (w becomes smaller - or negative), if
the object is to conserve its angular momentum.
We can write the following showing
conservation of angular momentum.

d U 
2
2
WR  W 
 R  d R

R  d R 
2
Where,
is the angular momentum of
WR
the unit mass object at rest on the surface of the
earth at latitude F.
 d U is the change in west-east velocity which
must occur because R is increasing to R+dR.
Remember that tangential velocity, U, is related
to angular velocity by: U  rw , so angular
velocity for the object is: w  U
R  dR
If we expand the right side we get:
2
2
R
d
U
2
R
d
R
d
U
d
R
dU
2
2
2
WR  WR  2WRdR  Wd R 


R  dR
R  dR
R  dR
If we neglect second-order, and higher differentials,
2
2
we get: 0  2WR dR  R d U
and,
d U  2WdR
We can see that
Arc length = ad F
and d R  ad Fsin F
So, d U  2WadFsin F
Dividing by a unit time
increment gives the
horizontal acceleration in the west-east
dF
direction. d U
 2Wa
sin F
dt
dt
Note: “a” is the radius of the Earth.
dF
But, a dt is just the northward velocity
dU
component, V, so:
 2WV sin F
dt
The magnitude of the Coriolis acceleration,
or Coriolis Force per unit mass, is given by
2WV sin F which acts to the right of the
direction of motion of the object. (Thus, it is
causing a change in the west-east motion of
the parcel.)
Turbulent Drag Force
Also called Friction and Frictional Drag.
In the atmosphere, friction includes:

(1) Molecular (viscous force) - force per unit
volume or per unit mass
arising from the action of
tangential stresses in a
moving viscous fluid.
• Viscous (viscosity) - That molecular property of a fluid
which enables it to support tangential stresses.

Stress: The amount of frictional force per
contact area where the force is parallel to the
area rather than vertical, as is pressure.

(2) Turbulence
• Which may be sub-divided into Mechanical (shear)
turbulence and Convective turbulence.
• Deals with flow in which the instantaneous
velocities exhibit irregular and apparently random
fluctuations so that in practice, only statistical
(average) properties can be recognized and
subjected to analysis.
• Can be at least an order of magnitude greater than
molecular / viscosity friction or drag.
• Turbulent stress is often called Reynolds stress
after Osborne Reynolds who related this stress to
turbulent gust velocities.
(a) Mechan ical (Shear) Turb ulence:
mean
z wind
turbu len ce
sh ear
momentu m
tran spo rt
M
feedback
(b) Co nvectiv e Tu rbulence:
sh ear
mean
z wind
turbu len ce
momentu m
tran spo rt
M
NO feedback
The Turbulent Drag Force per mass
(acceleration) for horizontal motion is:
Fy TD
Fx TD
U
V
 wT
 wT
m
zi
m
zi
where,
wT = the Turbulent Transport Velocity,
zi = height above ground.
For windy conditions of near statically
neutral conditions, turbulence is generated
primarily by shear, then the Turbulent
Transport Velocity is given by:
wT  CD M
where, CD = Drag Coefficient
M = wind speed.
For statically unstable conditions of light
winds and strong surface heating, the
Turbulent Transport Velocity is given by:
wT  bD wB
where,
wB = buoyancy velocity scale
(the
effectiveness of thermals in producing
vertical transport)
bD = 1.83 x 10-3.
Equations of Motion
The forces per unit mass (acceleration) is then:
U
U
U 1 P
U
 U
V

 f cV  wT
t
x
y  x
zi
V
V
V 1 P
V
 U
V

 f cU  wT
t
x
y  y
zi
tendency
advection
pressure Coriolis turbulent
drag
gradient
Note: Other forces may occur in particular areas:
Molecular friction near Earth’s surface,
Mountain-wave drag, Covective mixing above
the boundary layer.
Centrifugal force is not included because it
is an imbalance of the forces in the full
equations of motion.
Geostropic Winds
A theoretical wind that results from a balance
between the Pressure Gradient Force and the
Coriolis Force.
Winds above the Atmospheric Boundary Layer,
along straight isobars and away from the
equator are nearly Geostrophic.
A steady-state balance exists between Pressuregradient force and Coriolis force.
1 P
1 P
 f c U
0
 f c V 0  
 y
 x
Gradient Wind
A theoretical wind that results from an imbalance
between the Pressure Gradient Force and Coriolis
Force which is equal to the Centrigugal Force.
Other forces are considered negligable.
The parcel follows a curved path.
1 P
VV
0 
 f c V  s 
 x
R
1 P
UU
0 
 f c U  s 
 y
R
Pressure Coriolis centrifugal
gradient
If we let G represent the Geostrophic Wind and
Mr be the Gradient wind, then Mr  Ur2  Vr 2 
Mr 2
Mr  G 
fc r
where the negative sign is used around low
pressure centers and positive around high
pressure centers.
Thus, it is implied that the Gradient Wind is
slower than Geostrophic around low pressure
centers and higher around high pressure centers.
Solving the quadratic equation for Mr gives for
low pressure centers:

4 G 

Mr  0.5  f c  R  1  1 

f c  R 


and for high pressure centers:

4

G

Mr  0.5  f c  R  1  1 

f c  R 


R = the radius of curvature of the airflow.
G
The term Roc  f  R is the Rossby number.
c
Small values indicate flow that is nearly
geostrophic, little curvature.
Boundary-Layer Wind
In the boundary layer, turbulent drag becomes
important. The dominating forces along straight
line isobars are: Pressure gradient force, Coriolis
force and turbulent drag force.
1 P
U
0 
 f c V  wT 
 x
zi
1 P
V
0 
 f c U  wT 
 y
zi
Pressure Coriolis turbulent
drag
gradient
These can be written in terms of the Geostrophic
Wind as
wT V BL
UBL  Ug 
f c  zi
wT U BL
V BL  V g 
f c  zi
Boundary layer winds around curved
isobars require the Centrifugal force term.
1 P
U
VV
0 
 f c V  wT   s 
 x
zi
R
1 P
V
U U
0 
 f c U  wT   s 
 y
zi
R
Pressure Coriolis
gradient
turbulent
drag
centrifugal
Cyclostrophic Winds
In intense vortices, tornadoes, waterspouts,
hurricanes, Pressure gradient force and
Centrifugal force dominate. Others are
small in magnitude.
1 P
VV
0 
s
 x
R
1 P
UU
0 
s
 y
R
Pressure centrifugal
gradient
Problems: N1 (b, h), N2 (b, e), N3, N4 (a,
d, e) (assume z = 1000m), N5 (a, d, k) N8,
(a, d, g) (assume zi = 1000 m), N9 (b, d),
N10 (a, c), N11 (b, e), N13, N16 (a, g)
End