SOLUTIONS FOR THEORETICAL COMPETITION

Transcription

1 VI Intrnational Zhautykov Olympiad Thortical Comptition/Solutions Pag /5 SOLUTIONS FOR THEORETICAL COMPETITION Thortical Qustion A Potntial nrgy of th rigid rod U=mgl/sinα transforms to th kintic nrgy of its rotation E=J /, whr J=m is its inrtia momnt with rspct to th vrtical support, ω is th angular vlocity Making balanc of ths nrgis, on can find angular vlocity of th cntr of mass around th axis of th support, and aftr that on can gt instantanous angular vlocity of rotation of th cntr of mass around th axis of th support, and with th hlp of th obtaind xprssion w find normal acclration W find th tangnt acclration from th dynamics quation M=βJ, whr M=mgl/cosα is momntum of th gravity forc with rspct to th axis of rotation, β is angular acclration of th cntr of mass rlatd to tangnt acclration : g cosα Cntr of mass acclration а is found from th quation P+N=ma whr Р is gravity forc, N is raction forc of th support Dcomposing this quation into vrtical and horizontal componnts, and accounting for w find,,,,, Marking schm Itms No Writing down balanc quation for th kintic and potntial nrgis and dtrmination of th angular vlocity of rotation of th cntr of mass Points 05

2 VI Intrnational Zhautykov Olympiad Thortical Comptition/Solutions Pag /5 Dtrmination of th normal acclration 05 3 Dtrmination of th tangnt acclration from th quation of 075 dynamics of rotation 4 Dtrmination of th componnt of th raction forc of th support from th nd Nwton law 5 В First of all lt us find dpndncis of th liquid dnsity and prssur upon th hight masurd from th vssl bottom: ( ) h жидкости h = H ; () H h H h () ( ) ( ) p h = x gdx = gh On can s that prssur gos to zro at th top lvl of th vssl It mans that in this rgion volum of th gas will xcd th volum of th tst-tub, so th gas bulbs will com out Until th gas dosn t com out from th tst-tub, its isothrmal xpansion taks plac, so p ( h ) V ( h ) = ghv (3) From () and (3) on can obtain: V V ( h) = (4) h H V h < V, i for Formula (4) is valid for ( ) 0 V h < H (5) V 0 On can writ down th avrag dnsity of th gas in th tst-tub taking into account th mass of its walls from (4)-(5): h V, h < H ; M H V 0 газа ( h ) = = (6) V ( h) V 0, h H, V 0 whr 0, = MV0, Various kinds of dpndncis of upon hh ar plottd on Fig-3 Curv corrsponds to th liquid, whil curv corrsponds to th avrag dnsity of th gas in th tst-tub taking into account th mass of its walls Fig Fig

3 VI Intrnational Zhautykov Olympiad Thortical Comptition/Solutions Pag 3/5 Fig3 Fig4 Curvs and don t intrsct (Fig), if th condition газа > жидкости is fulfilld for V h = H, i V 0 > V V 0 0 (7) In this cas th tst-tub will always sink If th conditions 0 V <, >, (8) V 0 ar fulfilld in th sam tim, curvs intrsct in two points A and B (Fig): (point А); (9) (point B) (0) Point A is unstabl bcaus th tst-tub will sink du to th shift down, and it will ris du to th shift up Analysis of stability of th point B will b prsntd blow If conditions 0 V <, <, () V 0 ar fulfilld, only th intrsction point B xists (Fig3) But on should tak into account that at th horizontal part of curv motion to th right along this curv rsults to th gas flow out of th tsttub Consquntly, motion to th lft along th curv will b quit diffrnt It will tak plac along th parabola corrsponding to (6) (uppr lin of th formula), but with th nw (largr) valu of corrsponding to th amount of gas rmaind in th tst-tub Grading schm Prssur dpndncy upon th hight 05 Dpndncy of th avrag gas dnsity in th tst-tub upon th hight: a) taking into account th gas flow from th tst-tub (complt answr) 0 b) without taking into account th gas flow from th tst-tub (incomplt answr) 05 3 Comparison of dpndncis of liquid and gas dnsitis upon th hight: a) for thr cass (complt answr) 5 b) for two cass (incomplt answr) 075 c) for on cas (incomplt answr) 05 4 Dtrmination of th hight corrsponding to th intrsction points: a) two points (complt answr) 05 b) on point (incomplt answr) 05 5 Study of stability for point A 05

4 VI Intrnational Zhautykov Olympiad Thortical Comptition/Solutions Pag 4/5 6 Study of stability for point B 05 Totally (imum) 40 C Th form of shap dscribd in th problm is xplaind by th apparanc of th full shad (dark rctangl) and th smi-shadow (lightr outr rctangl) Fig illustrats th cours of outr rays forming a shadow and th smi-shadow ( and ) in cross sction prpndicular to on sids of th sourc Dnot th full width of th shadow -,, th width of th smishadow - Ths valus can b xprssd through th gomtric dimnsions of th sourc and th plat From th similarity of th triangls and it follows () From th similarity of th triangls and it follows () From th drawing of shadows, w dfin th rquird sizs Using formula () w find Hnc, w find From formula () w find on of th transvrs sourc siz Similar calculations for th prpndicular cross sction givs th following rsults: Thn, it follows that Th rsults indicat that th long sid of th sourc is placd horizontally (if you us fig from th conditions of th problm) Marking schm No Solution itm numbr Points Plotting of th ray tracing to xplain th mrgnc of th 0,5 shadow and smi-shadow Gomtric rlationships btwn siz of th sourc, location х0,5 and siz of th plat and siz of th shadow and th smishadow ()-() 3 Calculation of hight of th plat abov th floor (with a 0,5 numrical valu) 4 Calculation of th dimnsions of th sourc (with numrical х0,5

5 VI Intrnational Zhautykov Olympiad Thortical Comptition/Solutions Pag 5/5 valus) 5 Position of th sourc with rspct to th shadow 0,5 Thortical Qustion Solution [ point] Th total inrtia momnt with rspct to th rotation axis is a sum of th inrtia momnt of th coil itslf and th mtallic wir J = J0 + mr () [ point] Th quation of th coil rotation as a rigid bod taks th form dω Jε = J = M, () dt whr ε is th angular acclration It follows from quation () that th coil stops at th tim momnt ω0j t0 = (3) M Finally, th dpndnc of th angular vlocity on tim t is found as M ω0j ω0 t, t < t 0 = ω() t = J M (4) 0, t t0 3 [ point] At th stoppag of th coil, lctrons kp on moving du to thir inrtia, as a rsult th galvanomtr rgistrs th lctric currnt Lt a = εr b th linar acclration of th coil rim If th coil is tightly rld up and th wir is rathr thin that linar acclration is dirctd along th wir At th stoppag procss lctrons ar subjctd to th inrtial forc ma opposit to th linar acclration of th coil This inrtial forc can b intrprtd as an ffctiv lctric fild ma Eff = (5) Thus, th ffctiv lctromotiv forc in th coil causd by th inrtia of frly moving lctrons is obtaind as m Emf = Eff = a (6) Thrfor, th Ohm s law for th lctric circuit is writtn as ma IR = Emf = (7) Taking into account solution of, on gts Mmr ω0j, t < t0 = It () = JR M (8) 0, t t0 4 [ points] Th lctric charg, rgistrd by th galvanomtr, is found from (8) as mω0r Q = It0 = (9) R Th charg-to-mass ratio of lctron is simply obtaind as ω0r = (0) m RQ

6 VI Intrnational Zhautykov Olympiad Thortical Comptition/Solutions Pag 6/5 5 [ point] In this cas quation (7) is rwrittn as follows di ma L + IR = Emf = () dt whr L = 0 n πr h is th coil inductanc It follows from quation () that th imal lctric currnt strngth is Mm r I = () JR Th qualitativ dpndnc of th lctric currnt strngth is plottd blow 6 [ point] Th imal lctromagntic nrgy stord in th coil quals LI 0πh nmm r W0 = = (3) JR 7 [3 points] In th stationary rgim th magntic fild inductanc B = 0nI (4) rmains constant in th coil and th lctric fild is absnt This is not tru for initial tim momnts whil th lctric currnt incrass from 0 to its imal valu dtrmind by formula () According to (4) th varying magntic fild gnrats th vortx lctric fild which causs th flux of th lctromagntic nrgy to appar Th strngth of th vortx lctric fild at th latral surfac of th coil is found from th lctromagntic induction law of Farady dφ d Emf = Eπr = = ( Bπr ), (5) dt dt as r db E = = 0 nr di (6) dt dt Th mutual orintation of th vctors E, B и S is shown blow Substituting xprssions (4) and (6) into th xprssion for th Pointing vctor and taking into account that th vctors E and B ar prpndicular, on obtains n r di S = 0 I (7) dt Thus, th lctromagntic nrgy, going inward th latral surfac of th coil whil th lctric currnt incrass, is givn by th summation (or intgrating) of (7) as 0 0π W = I πrh = n r h nmm r 4 (8) JR

7 VI Intrnational Zhautykov Olympiad Thortical Comptition/Solutions Pag 7/5 It is obvious that th sam amount of th lctromagntic nrgy gos outward whil th lctric currnt strngth dcrass 4 0n r 0πh nmmr W ' = I πrh = JR Marking schm (9) Contnt Points Total inrtia momnt () Equation of motion () 05 3 Stoppag tim (3) 05 4 Dpndnc (4) of th angular vlocity on tim t 05 5 Exprssions for th ffctiv lctric fild (5) or (6) 05 6 Th Ohm s law (7) 05 7 Dpndnc (8) of th lctric currnt strngth on tim t 05 8 Charg (9) rgistrd by th galvanomtr 9 Charg-to-mass ratio (0) for lctron 0 Equation () for th lctric currnt strngth 05 Maximal lctric currnt strngth () 05 Qualitativ dpndnc of th lctric currnt strngth 05 3 Maximal nrgy (3) 4 Magntic fild induction (4) 05 5 Elctromagntic induction law (5) 05 6 Vortx lctric fild strngth (6) 05 7 Pointing vctor (7) 05 8 Elctromagntic nrgy (8) 05 9 Elctromagntic nrgy (9) 05