Chapter VI: Motion in the 2-D Plane

Transcription

1 Chapter VI: Motion in the -D Plane Now that we have developed and refined our vector calculus concepts, we can ove on to specific application of otion in the plane. In this regard, we will deal with: projectile otion (not uch tie on this) geopotential and central force otion angular oentu (iportant for hurricane, tornado and the atospheric general circulations; 6.1. Conservative Force Before we begin, let's re-introduce an iportant concept related to forces conservation. A General Definition of Conservative Force Be definition, a force is said to be conservative if the work done by it is independent of the path, or equivalently, the work done by this force around a closed path is identically zero, i.e., if F dl = 0, then F is said to be conservative. Why? Recall the Stokes' Theore: ( V) nds= Vdr (6.1) this says that, if we replace V by F, the force F is conservative if F = 0 then the right hand side the integral along a closed path is zero! because Therefore, one convenient way to test if a force is conservative is to see if its curl ( F ) is zero! In addition, if a force can be written as the gradient of a potential function, then it is autoatically a conservative force. Why? Because for any scalar φ, φ = 0 (show it yourself!), thus, if F = φ, then F = 0 autoatically. 6-1

2 Eaple: Let's verify that gravity is a conservative force. Recall that GM F = r rˆ. (6.) To show that it is conservative, we siply need to show that F = 0. Let's rewrite F as Now, let GM = Q, so that Let's now copute GM r GM F = = ( iˆ+ yj ˆ+ zkˆ ). 3 r r r GM r Q F = = ( iˆ+ yj ˆ+ zkˆ ). 3 r r r F (do it on your own): iˆ ˆj kˆ F = y z Q Qy Qz 3 r 3 r 3 r Let's take only the î coponent: ˆ Qz Qy i 3 3 y r z r Note that 3 3/ r = ( + y + z ), doing the differentiation gives ˆ zy yz i 3Q + 3Q = r r. Siilarly, you will find that all three coponents equal to zero gravity is a conservative force! 6-

3 6.. Solving for Potential Function Recall that in Chapter 4 when we dealt with 1-D probles, a conservative force F can be written in ters of the gradient of potential energy V, because by definition, the work done by a conservative force is a function of the spatial coordinate only. We call the work done by the force to bring an object to a standard location the potential energy: s V( ) = Fd= Fd (6.a) s therefore dv F =. (6.b) d Siilar is true for ulti-diensional proble. Analogously, we define potential energy r V( r) = V(, y, z) = F dr rs (6.c) and the right hand side integration should be independent of the path fro rtor s. Eq.(6.c) actually cae fro the integration of dv( r) = F dr, (6.d) which eans that the change in the potential energy = the negative of work done oving the object fro rto r + dr. Using the definition of dv and Eq.(6.d), we have V V dv( r) = d+ dy = Fd + Fdy y y ( ) (6.e) therefore or F = V V V F =, Fy = y (6.f) (6.g) Therefore, we have showed that a conservative force can always be written in ters of the gradient of a scalar function, and we call it the potential energy. 6-3

4 Given F, how do we obtain the potential V? For 1-D probles, we can obtain V by siply integrate F according (6.a). For ulti-diensional probles, the ethod is analogous though a little ore coplicated. We show this using an eaple. Eaple: Let F = A ˆ i + Byj ˆ= Fiˆ+ F ˆj. (6.3) If you take F, you will find that it is zero, and thus the force is conservative and can be written as F = V y (6.4) V V then F =, Fy = y (6.5) therefore V = A. (6.6) Integrate it with respect to, we get V = A 3 /3 + f ( y) (6.7) where f(y) is the integration "constant" which can still be a function of y. The partial derivative of f(y) with respect to goes to zero. How do you deterine f(y)? Notice we haven't used the second equation, F y V = = By y (6.8) yet. The solution V should also satisfy this equation. Substituting (6.7) into (6.8): 3 [ A /3 + f( y)] = By y f( y) = By y (6.9) By f ( y) = + C (6.10) Here, C is truly a constant because f = f(y) only, so it can't a function of either or y. And this constant can be arbitrarily chosen since for the force, it is the gradient of the 6-4

5 potential function that atters. C is usually chosen based on convention. For eaple, for the gravity, we often choose V = 0 at the sea level, which deterines C. The final solution for V is V 3 A By = + C (6.11) Projectile Motion Projectile otion itself has little to do with the atosphere we eaine it nonetheless because it provides an ecellent, siple eaple of solving coupled equations of otion. Until now we have only looked at 1-D otion. Of course, eteorologists are involved in ballistic research when it coes to clouds, frictional properties of air and aerosols, etc. The vector equation of otion governing projectile is dv ( ) = F. (6.1) For a single projectile with ass, subjecting to gravity only, the equation is dr = gkˆ. (6.13) In reality, this includes 3 equations that can be written in coponent for as follows: d = 0, d y = 0, (6.14) dz = g In this siplest case with gravity only (no friction etc.), each of the three equations can be solved independently. The solutions are: = + ut 0 0 y= y + vt 0 0 z = z + wt gt 0 0 / (6.15) 6-5

6 Therefore, the objection oves in the horizontal directions at constant velocities u andv, and in the vertical, it behaves eactly like a 1-D upward thrown object. 0 0 Let's assue that the projectile starts at (, y, z) = (0, 0, 0) with otion only in the -z plane v 0 = 0. Then, = ut 0 y = 0 z = w gt 0 / (6.16) That's it! If we solve the first equation above for t and use it in the 3rd equation, we have: z w 1 g 0 = (6.17) u0 u0 which tells you the height of the projectile at any location instead of at any tie. Equation (6.17) is a parabolic equation, which describes the projectile trajectory, as shown blow. z We can find that aiu height of the trajectory by coputing dz/d and setting it to zero. Doing this: dz w g 0 = = d u u a wu g 0 0 = - the position of the peak of the trajectory. w0 The peak is reached at ta = =, u g 0 6-6

7 and the peak height is z w0 wu g wu 0 0 w0 a = = u0 g u0 g g. The peak height is actually independent of the horizontal coponent of the initial velocity! The sae peak height will be reached if the object is projected vertically at initial speed of w 0 (with zero initial horizontal velocity)! Often the projectile proble will be given in ters of the angle at which the projectile is fired. You need to copute u 0 and w 0 fro the initial speed and angle. What has this analysis neglected? - air resistance (friction) - air otion - the earth's rotation a big thing for long-range ballistic issiles These additional effects ake the three coponents equations coupled to each other, but we will not go into details here. The last topic will be covered in net chapter. In the following, we will look at another special type of otion before getting into angular oentu, the otion under the influence of central force Central Force Motion The otion of a central force, which we will define oentarily, arose out of Newton's law of gravitation and his successful use in eplaining Kepler's elliptic orbits for the planets. A central force is one in which the force has the for: F = Frr ()ˆ, (6.18) i.e., the force acts along a line connecting one or ore bodies and is soehow related to the distance between the. Eaples include gravitational attraction, the force that causes an electron to orbit the nucleus of an ato, etc. In ost physically iportant cases, the force follows an "inverse square law", i.e., F ~ 1/r. Two interacting bodies is a siple atter, but ore than two is etreely cople referenced to as the "n-body proble" where n >. To keep life siple, consider bodies, in isolation, that are under the influence of a central force. Let the asses be 1 and and their position vectors be r 1 and r : 6-7

8 r R r = r 1 r 1 r 1 Let r = r 1 r and r = r. The equations of otion for the two bodies are dr1 1 = frr ()ˆ (6.19) dr = frr ()ˆ The force is attractive for f(r) < 0 and repulsive for f(r)>0. In reality, both objects are attracted to one another, but this is a atheatically convenient way to deal with the force think of Newton's law of gravitation for two bodies. Note that these two equations of otion are coupled via r. It turns out that life is sipler if we replace r 1 and r by r and by the vector to the center of ass of the bodies: r 11+ r R = + 1 (Note, for an N body syste, the center of ass is defined as R = i i r i i ) i (6.19a) The center of ass is special the syste behaves as if all the ass is concentrated at this single point. For the above -body syste, since there is no eternal force (all forces are internal interacting forces), therefore the equation of otion for ( 1+ ) at R is: 6-8

9 d 1 ( + ) = 0 (6.19b) R = R + Vt, 0 which is the position vector of the center of ass. What about the positions of the individual bodies? To find the, we will see that it's easier to first fine r. Recall our original equations of otion in (6.19) for r 1 and r. If we divide the by 1 and, respectively, and subtract the two resultant equations, we have dr1 dr 1 1 = + frr ()ˆ 1 1 dr1 dr frr () ˆ = 1+ 1 If µ + 1 = reduced ass (definition), and with µ = dr frr () ˆ dr1 dr dr =, we have. (6.0) Equations (6.19b) and (6.0) fully describe the syste. Unlike the coupled equations in (6.19), these two are uncoupled and are analogous to the equation for a 1-particle syste! Note that one cannot do this for 3 or ore particles, and to this day no solution eists for those syste! Goal: We now want to solve this differential equation for rt (), and fro that we can easily get r 1 and r :, r1 = R+ r r = R r 1+ where is the final solution to the two body proble! SHOW THIS FOR YOURSELVES! 6-9

10 6.5. Angular Moentu and Its Conservation We can now use the general equation µ = dr frr () ˆ (6.1) to look at all sorts of otion involving one or any asses. (If we have one ass, then µ = ). As an eaple, consider otion in a plane as preface to the topic of angular oentu. Definition: Torque (a vector) = r F. Torque causes oentu to change, while force causes regular oentu to change. In a -D plane, the torque (also called oent of force) about an origin = radius r the coponent of the force perpendicular to the radius vector, consistent with the above vector definition. For siplicity, consider a single particle of ass subjecting to a central force: z F r y Because this is central force otion, F lies in the plane along r - and thus F can ecert no torgue ( = r F ) on ass ( F and r are parallel). 6-10

11 Angular Moentu For ass oving about a point (origin) O, the vector angular oentu is defined as L= r V. (6.) where r is the position vector fro O to the particle with ass and velocity V. Velocity and Acceleration in Plane Polar Coordinates For a rotational otion in a -D plane, it's easier to work in plane polar coordinates. y ˆ θ ˆr r θ In plane polar coordinates, the unit vectors are ˆr and ˆ θ. Let's now epress our equation of otion in these coordinates: dr = frr () ˆ (6.3) Fro the diagra, r = rrˆ( θ ) where the unit vector ˆr is a function of θ. dr dr drˆ = rˆ + r = V (6.3a) 6-11

12 ˆ θ rˆ+ drˆ dr ˆ ˆr The change in ˆr as a result of change in θ by the aount of dθ in the in direction of ˆ θ, according to the above figure, and is given by therefore drˆ= rˆ dθθˆ= dθθˆ drˆ θˆ dθ = thus (6.3a) becoes dr dr dθ drˆ = rˆ+ r Vrˆ ˆ = r + Vθ θ dθ therefore V = Vrˆ + V ˆ θ θ r (6.4) where V r the radial coponent, and V θ is the tangential or aziuthal coponent. One can differentiate the velocity as well to obtain the acceleration, which, of course, is iportant if we want to apply F = a. dv d r dr d d dr d a θ θ θ = = = r rˆ r ˆ + + θ 6-1

13 Now let's use this in our central force equation of otion: dr = frr () ˆ. Equating "like" coponents on each side of this vector equation, and using the convention that a 'dot' denotes d( )/, we have r ( rθ ) = f() r (6.5a) r ( θ + r θ ) = 0 (6.5b) They are basically equations of otion (based on Newton's second law) applied in the radial and aziuthal directions, respectively. They are coupled second-order differential equations in tie, are not very easy to solve. Angular Moentu Conservation Recall that the direction of the angular oentu is a constant, i.e., the otion itself is in the -y plane. There are other iportant quantities in this syste: the agnitude of the angular oentu; the total energy. Let's look only at the forer recall that conservation properties are very iportant in eteorology, and that conservation eans that d/ = 0 (Lagrangian derivative following the otion). Let's first copute the agnitude of the angular oentu: L = ˆr V = rˆ V sin(90 ) = rvθ = r( rθ ) L = r θ (6.6) Is the property conserved following the otion? 6-13

14 What we are about to do is soething you will see a lot in eteorology taking the basic equations of otion (here, in plane polar coordinates with a central force) and anipulating then to obtain a conservation equation. If we want to show that L = r θ is conserved, we want to show that d. ( r θ ) = 0 Epanding gives ( rr θ + r θ ) = 0. (6.7) Going back to the equations of otion, ultiplying Eq.(6.5b) gives r ( θ + rr θ) = 0 d which is eactly the sae (6.7). Therefore L = 0. So, we have shown that ANGULAR MOMENTUM IS CONSERVED FOLLOWING THE MOTION. But, when is this true? Only when the torque is zero, or in other words, when the applied force acts along (parallel to) the position vector of the particle (see previously drawing). This is true when every force acting on the particle is a central force. This property is very iportant in rotating atosphere flows and is the "ice skater" analogy that you have heard so uch about. Let's look closer: If angular oentu is conserved, then d ( r θ ) = 0 d or ( rv θ ) = 0 ( rvθ) point1 ( rvθ) point =, 6-14

15 where, via interpretation of the Lagrangian derivative, points 1 and are separated in space by soe interval of tie. Eaple: Suppose an air parcel is rotating around a esocyclone of radius k with a tangential speed of 30 /s. How fast will the parcel be going if it spirals into a radius of 0.5 k? ( r V θ ) 1 = ( r V θ ) ( 000) (30 /s) = (500 ) ( V θ ) ( V θ ) = 10 /s! The point is that, as a parcel decreases its distance to the ais of rotation, its tangential speed ust increases proportionally to keep A.M. constant. Angular Moentu Theore Let's now consider pure rotation about the vertical ais --- recall vertical vorticity and its iportance in eteorology. With the vertical ais of rotation fied, we can look at the vertical coponent of A.M. Lz = r θ = I ω where I r = oent of inertia. It depends on the distribution of ass and the distance of the ass fro the ais of rotation. ω θ. Let's now look at the general 3-D case: Now, τ = r F = torgue and L= r V. 6-15

16 dl d = ( r V) dr dv = V + r = V V + [ r a] = r F = τ Thus, dl =τ Therefore the tie rate of change in angular oentu equals the torque torque causes angular oentu to change, analogous to the fact that force causes oentu to change. The above is the Angular Moentu Theore! dl As we saw earlier, the A.M. is conserved ( 0 = ) if the torque is zero. The above equation quantifies this fact! Again, note the Lagrangian derivative. Let's return briefly to the z coponent of A.M. L z = I ω. By the above, we know that 0 dlz d di dω τz = = ( Iω) = ω + I = Iα τz = Iα where α = angular acceleration = d θ/. This equation looks very uch like F = a, though is applicable to bodies undergoing rotary otion. We can define the kinetic energy here as 1/ Iω (I is analogous to ass). We have the following correspondences: τ F I α a 6-16