1. Dynamics  Dynamics deals with forces, accelerations and motions produced on objects by these forces.  Newton’s Laws l 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.

2. l 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. l 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.

3.  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.

4.  A force is a vector, having both magnitude and direction, expressed as a symbol with an arrow above it.  Acceleration is also a vector.  Mass is a scalar, having only magnitude.

5. Equations of Motion  Since, and then, where,

6.  Writing this in the U, V components of the wind gives: Since we can write, then substituting into the above equations gives,

7.  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).

8. Forces acting on air parcels  Pressure Gradient Force  Advection  Centrigugal / Centripital Force  Coriolis Force  Turbulent Drag Force

9. 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.

10.  Advection terms are written as: where, are gradient terms.

11. 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.

12. Centripital Acceleration  Consider a ball of mass, m, attached to a string and whirled through a circle of radius r at a constant velocity .

13.  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 .  is also the angle between the vectors v and v +  v.

14.  If we divide by  t and note that in the limit as,  v is directed toward the axis of rotation, (it is a -  v), then: Since, tangential velocity Then, In our convention we can write:

15.  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.

16.  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.

17.  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 Centrifugal force.  The Centrifugal force can be considered the force necessary to balance all the other forces acting on the ball.

18.  Since, for wind at rest on Earth’s surface, M =  r, then and To get the components in the x and y direction, remember that and to get the sign right, (Centrifugal force going in the proper direction), consider the following low pressure center.

19.  By writing the equations as: and using the sign below, the direction of the force is proper.

20. 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,  2 R, where  is the angular speed of rotation of the earth and R is the distance of the particle from the axis of rotation.

21.  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.

22. 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,  Since it is moving in a circular path, its Centripital acceleration is toward the axis of rotation of magnitude:

23.  Expanding gives:  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 just:

24.  The discrepancy between the magnitude of the two is:  The Centrifugal Force (acceleration) we know is:  The Coriolis Force (acceleration) is:also directed outward, like the Centrifugal Force.

25.  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).

26.  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 is called the Coriolis parameter. It is positive in NH and negative in SH.

27. Objects moving north or south.  Consider an object, of unit mass, initially at rest on the surface of the Earth at latitude . It starts moving southward and moves . As the object moves equatorward, it will conserve its angular momentum in the absence of torques in the east- west direction.

28.  Its initial angular momentum, L, is:, where  is initially equal to zero. For other than unit mass it is:  Since the distance to the axis of rotation, R, increases to R +  R for the object moving equatorward, a relative westward velocity, must develop, (  becomes smaller - or negative), if the object is to conserve its angular momentum.

29.  We can write the following showing conservation of angular momentum. Where, is the angular momentum of the unit mass object at rest on the surface of the earth at latitude .  is the change in west-east velocity which must occur because R is increasing to R+  R.  Remember that tangential velocity, U, is related to angular velocity by:, so angular velocity for the object is:

30.  If we expand the right side we get: If we neglect second-order, and higher differentials, we get: and,

31.  We can see that  Arc length =  and  So,  Dividing by a unit time increment gives the horizontal acceleration in the west-east direction.  Note: “a” is the radius of the Earth.

32.  But, is just the northward velocity component, V, so:  The magnitude of the Coriolis acceleration, or Coriolis Force per unit mass, is given by 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.)

33. Turbulent Drag Force  Also called Friction and Frictional Drag.  In the atmosphere, friction includes: l (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.

34. l Stress: The amount of frictional force per contact area where the force is parallel to the area rather than vertical, as is pressure.

35. l (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.

37.  The Turbulent Drag Force per mass (acceleration) for horizontal motion is: where, w T = the Turbulent Transport Velocity, z i = height above ground.

38.  For windy conditions of near statically neutral conditions, turbulence is generated primarily by shear, then the Turbulent Transport Velocity is given by: where, C D = Drag Coefficient M = wind speed.

39.  For statically unstable conditions of light winds and strong surface heating, the Turbulent Transport Velocity is given by: where, w B = buoyancy velocity scale(the effectiveness of thermals in producing vertical transport) b D = 1.83 x 10 -3.

40. Equations of Motion  The forces per unit mass (acceleration) is then:  Note: Other forces may occur in particular areas: Molecular friction near Earth’s surface, Mountain-wave drag, Covective mixing above the boundary layer.

41.  Centrifugal force is not included because it is an imbalance of the forces in the full equations of motion.

42. 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 Pressure- gradient force and Coriolis force.

43. 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.

44.  If we let G represent the Geostrophic Wind and M r be the Gradient wind, then 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.

45.  Solving the quadratic equation for M r gives for low pressure centers: and for high pressure centers: R = the radius of curvature of the airflow.  The term is the Rossby number. Small values indicate flow that is nearly geostrophic, little curvature.

46. 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.

47.  These can be written in terms of the Geostrophic Wind as

48.  Boundary layer winds around curved isobars require the Centrifugal force term.

49. Cyclostrophic Winds  In intense vortices, tornadoes, waterspouts, hurricanes, Pressure gradient force and Centrifugal force dominate. Others are small in magnitude.

50.  Problems: N1 (b, h), N2 (b, e), N3, N4 (a, d, e) (assume z = 1000m), N5 (a, d, k) N8, (a, d, g) (assume z i = 1000 m), N9 (b, d), N10 (a, c), N11 (b, e), N13, N16 (a, g)

51. End