Final Comprehensive Exam Physical Mechanics Friday December 15, Total 100 Points Time to complete the test: 120 minutes

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1 Final Comprehenive Exam Phyical Mechanic Friday December 15, 000 Total 100 Point Time to complete the tet: 10 minute Pleae Read the Quetion Carefully and Be Sure to Anwer All Part! In cae that you have difficulty getting to the final anwer, try at leat to put down the tep of olution with the right equation. Try to ue ketche to help you undertand the quetion whenever poible. Some quetion require only imple anwer Good Luck! Anwer 1. (10%) State in word the Newton' Second Law of motion and the reference frame in which it i valid. The Newton' Second Law of motion ay that in an inertial reference frame, the change in the momentum (or the ma time acceleration) of an object i equal to the net force applied to thi object. It' valid only in an inertial (non-accelerating) reference frame but can be ued if the acceleration of the reference frame i taken into account, in the form apparent force.. (10%) Aume the atmophere ha a contant lape rate ( γ = an air parcel undergoing adiabatic vertical motion ha an adiabatic lape dtparcel ( Γ d = ) of 9.76K/km. dz T atmophere ) of 5K/km, (a) (3%) For an air parcel diplaced vertically from it initial height level z=0, what kind of force will it be ubject to? Since the air parcel' temperature decreae with height fater than it' urrounding atmophere, it become cooler (warmer) than it urrounding when diplaced upward (downward), therefore it i ubjecting to a retoring force. (b) (3%) What' the tability (table, neutral or untable) of the atmophere? Since the diplaced parcel i ubjecting to a retoring force, the atmophere i table. 1

2 (c) (4%) Sketch the vertical coordinate, z(t), of thi diplaced air parcel a a function of time, with and without friction. Without friction, the parcel undergoe imple harmonic ocillation with contant amplitude. z t With friction, the parcel undergoe damped harmonic ocillation with the amplitude exponentially decreaing in time. z t 3. (10%) Give the phyical definition of the term in equation F df = V F. t (5%) F i the local or Eulerian rate of change in F t df i the total or ubtantial or material rate of change in F V F i the patial advection of F due to motion in the direction of gradient in F. If an eat-wet oriented cold front i moving outh and air parcel move together with the front adiabatically (no heating or cooling to the parcel), determine the ign of each of thoe term when the above equation i apply to temperature.

3 (5%) Becaue the motion i adiabatic, dt =0. T T V T = v < 0 becaue v < 0 and < 0 y y becaue T t = T v. y 4. (0%) An air parcel of 1 kg in ma in a tornado vortex i initially circulating at 50 m/ around the center of tornado in a circle of 1 km radiu. Under the action of a central force (preure gradient force in thi cae), it i brought into a circular trajectory of 500 m in radiu. The frictional force and Corioli force due to earth rotation can be neglected. (a) (4%) What will be the new peed of thi air parcel? Since the force i central, the angular momentum i conerved: Vr = Vr 11 V = Vr / r = 50 m/ 1000m / 500m = 100 m/. 11 (b) (4%) I the kinetic energy of the air parcel conerved? 1 K1 = mv1 = 0.5 1kg 50 m / = 150J 1 K = mv = 0.5 1kg 100 m / = 5000J They are not equal therefore not conerved! (c) (4%) If not, what force, viewed in a reference frame rotating with the tornado, i cauing thi change in the kinetic energy? (Hint: Think of the earth rotating coordinate analogy). Thi quetion i better aked like thi: If not, what force i cauing thi change in the kinetic energy? Anwer: A the parcel move from the trajectory of 100m radiu to one with 50m radiu, it ha to overcome centrifugal force. It' the preure gradient force (a central force) that pull the parcel from the bigger circle to the maller one, and in the proce it doe work. The PGF ha to be equal to the centrifugal force. A the parcel move toward the center, the Corioli turn it to the right (auming initially the circulation i counter-clockwie), converting the radial velocity created by PGF into a tangential velocity. The Corioli force doe not however create or detroy kinetic energy, however, becaue it i alway perpendicular to 3

4 the velocity vector therefore the trajectory it doe not caue any diplacement in the direction of the force therefore it doe no work! The Corioli force i an apparent force, no apparent (not real) force can do any work! Energy cannot be created or detroyed without work done by a real force. However, becaue the original quetion might appear mileading, you will get free credit if you anwered centrifugal force or Corioli force. (d) (4%) I thi force a true or apparent force? The PGF i a real force. If you anwered apparent force that i conitent with the anwer for (d), you will get free credit. (f) (4%) How much work doe thi force need to do to caue thi much, if any, change in kinetic energy? Thi work done equal to the change in kinetic energy (work-energy theorem), therefore W = K - K 1 = 5000J 150J = 3750J. Since we aid the work i done by PGF, let' ee if we get the ame anwer from the definition of work. The PGF i equal to the centrifugal force in magnitude but oppoite in direction. V V F = ( rˆ) = rˆ r r r rv rv W = F dr = rˆ dr = dr r 1 r1 r r 1 r r (50 100) r r r r V r = dr = V r = = 3750 J r 1 The ame a the anwer from work-energy theorem. In the above, we ued Vr = Vr (50%) The vector equation of motion for an air parcel of unit volume can be written a dv 1 + Ω V = p+ g ρ or dv 1 = p+ gnet Ω V ρ net (1) () 4

5 where V i the earth-relative velocity. (a) (5%) What i the firt principle / fundamental law that we ue to obtain thee equation? Newton' econd law of motion, dv m = F (b) (5%) Give a phyical definition for each of the term in the equation. dv the acceleration relative to the earth Ω V Corioli acceleration due to earth rotation and earth-relative motion 1 p Preure gradient force ρ g apparent (net) gravity including centrifugal force due to earth rotation net Ω V Corioli force, all for a unit ma. (c) (5%) In the local Carteian coordinate fixed to the earth, velocity V = uiˆ+ vj ˆ+ wkˆ and the earth' angular velocity Ω=Ω co( φ) ˆj+Ωin( φ) kˆ. Expre the lat term on the right hand ide of Eq.() in it component form, i.e., in the form of Ω V = iˆ() + ˆj( ) + kˆ (). iˆ ˆj kˆ Ω V = Ωx Ωy Ωz u v w = iˆ(ωvinφ Ω wco φ) + ˆj( Ω uin φ) + kˆ (Ω uco φ) (d) (5%) Making ue of the reult of (c) and Eq.(), write down the equation of motion for the three Carteian component of velocity u, v and w. You can ue f = Ω in( φ) and f = Ωco( φ) to implify the notation. du 1 p = + fv fw ρ x dv 1 p = fu ρ y dw 1 p = g+ fu ρ real. 5

6 (e) (5%) For large-cale upper-level atmopheric flow with mall or zero curvature, the flow acceleration i typically much maller than the other term in the equation of motion and the flow i quai-two-dimenional (i.e., w i negligibly mall). If we alo neglect friction, What approximate/implified equation can you obtain for thi ituation (write down the equation)? 1 p 0 = + ρ x 1 p 0 = ρ y fv fu Which i the wind velocity governed by uch implified equation called? Geotrophic wind velocity. (f) (5%) In a weather chart plotted at a contant height (10km) level, the eat-wet preure contour are traight and the preure decreae northward by 1 mb over 100 km. Auming parameter f = and the air denity there i 0. kg m -3, determine the wind peed and direction from the implified equation obtained in (e). 1 p v = = 0 f ρ x 1 p 1 100Pacal u = = = 50 m/ f ρ y kg/ m 10 m Therefore, the wind i eatward and the peed i 50m/. (g) (5%) If the preure contour are not traight, but exhibit a trough pattern to the wet and a ridge pattern to the eat, and the radiu of curvature i 500 km for both (pay attention to it ign), determine the gradient wind peed at the bottom of trough and the top of ridge. Note that you are dealing with normal/regular low and normal/regular high cae. There i a problem with the number given for thi quetion with R = 500km, you get negative number inide the radical for the ridge cae (okay for trough cae) not phyical. It' a cae where the centrifugal force plu the preure gradient force i too large for it to be balanced by the Corioli force. R = 5000km i a more realitic number (thi alo how why preure contour are rarely very tight in high-preure ytem we aume that PGF i the ame for both ridge and trough region in thi problem). If you howed that you knew how to olve the 6

7 equation, and plugged in the right number, your anwer i conidered correct. We will ue R = 5000 km in our olution. Ue the given equation in natural coordinate to olve for the gradient wind peed. V R = 1 p ρ n fv V R p + + frv = 0 ρ n V fr f R R p = ± 4 ρ n 1/. For the trough, R > 0, need the poitive ign otherwie V < 0, no allowed. fr f R R p V = + 4 ρ n 1/ / 4 6 ( ) 5 10 ( 100) = = 45.8 m/ For the ridge, R < 0. fr f R R p V = ± 4 ρ n 1/ / 4 6 ( ) ( 5 10 )( 100) = ( 5 10 ) ± 5 = 50± m/ The poitive ign give an unrealitic peed of m/ and correponding to the abnormal high cae. The negative ign give u a peed = 56.4 m/, the normal high cae we want. (h) (5%) I the wind peed at the ridge larger or maller than that at the trough? Larger. Thi i expected, noting the qualitative dicuion in the Note on upergeotrophic flow that ridge and ubgeotrophic flow at the trough. Do you have flow divergence or convergence between the trough and ridge? Divergence. Becaue the flow peed increae along the trajectory. 7

8 Auming that the tropopaue located above 10 km level act like a rigid lid (i.e., w = 0 there), do you expect acending or decending motion between the tough and ridge below the 10km level? Becaue of the divergence, auming the atmophere i quai-incompreible, there need to be acending motion to compenate the divergence at the upperlevel. No decending motion can come from the above becaue of the rigid lid. (i) (5%) Aume the bottom of trough and the top of ridge are 1000 km apart, calculate the horizontal velocity divergence between the trough and ridge. I your anwer conitent with your expectation? Uing the natural coordinate, the horizontal divergence i u uridge utrough h Vh = = = = km Since the divergence i poitive, it preent divergent flow, conitent with our previou dicuion. (j) (5%) Aume the divergence remain contant between the 5 km and 10 km height level, ue the kinematic method to determine the vertical velocity at the 5 km level. u w Uing V = + = 0 w u = = km 1000km 10km w u uridge utrough dz = d = = = 1000km w(5 km) = =.1 10 m/ 5000 Since w i poitive, it repreent upward/acending motion, again conitent with the expectation. Since the anwer to (h), (i) and (j) depend on the reult of (g), I will be looking for clue that you know the method and equation to get to the right olution and you are clear about the phyical concept. 8

9 Equation that might be ueful to you: D equation of motion in natural coordinate: dv = 1 p ρ and V R = 1 p fv. ρ n u v w w 3D divergence i V = + +. When V = 0, = h Vh. x y u w In natural coordinate, V = + auming no velocity gradient in the trajectorynormal direction. Angular momentum for D circular motion = ωr. 9