IPhO 2012

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  • 1. HomeEstonia has the honour of being the host of the 43rd International Physics Olympiad (IPhO), which will be held from July 15th to 24th, 2012. The Olympiad will be held in twolocations, the capital Tallinn (team leaders), and the oldest university town in Estonia, Tartu (students). Estonian people and the organizers of the Olympiad are lookingforward to meeting young physicists and their supervisors from all over the world, and to introducing them to the innovative country, which values education, has richcultural heritage and beautiful land.International Physics Olympiads are aiming at propagating natural and exact sciences amongst school students, stimulating young peoples interest in physics, and promotingscience education throughout the world by means of international contacts. IPhO is one of the oldest and largest International Science Olympiads: the first Olympiad tookplace in 1967 (cf. Section "HISTORY"), and the teams of 88 countries have already registered for IPhO2012 in Estonia. Estonian National Olympiads also have a long andhonourable history – today, Olympiads are organised in about 20 subjects, and National Physics Olympiads have been held since 1953.According to the IPhO statute, each national team will comprise of up to five students and two leaders. The contestants shall be students of general or technical secondaryschools; students who have graduated from their school in the year of the competition can be members of the team as long as they have not commenced their universitystudies. The governing body of the IPhO is the International Board that consists of the delegation leaders of each country attending the IPhO.Welcome to IPhO2012 in EstoniaInternational IPhO2012 in physicsEstonia has the honor of organizing the 43rd this summer International Physics International physics Olympian (IPhO). July 24th-15th Olympiad to take place about 90countries are expected to IPhO2012 best young physicists, together with instructors. March 1, the countries had already registered for the 88thTogether with the guestsand the organizers of the participants about the 1000thIPhO is achieved through the organization of the United States on a rotating basis. Estonia joined the international movement of pupils teadusolümpiaadide as an independentstate in 1992. In 1996, the then Education Minister Jaak minister to promise IPhO in 2012, lead the organization.Estonian young physicists have been very successful - from the previous Olympiads, our students brought two gold medals (2011 and Yegor Gužvin Ants Remm 1999),6 silver and 12 bronze medals (for details seewww.teaduskool.ut.ee / medalists ) . Estonias national teams are selected for the final round füüsikaolümpiaadi the best in thecompetition organized by the choice of tutoring sessions and the results of the Estonian upper secondary füüsikaolümpiaadi place team winner is assured a priori. In recent1
  • 2. years, the team announced the final composition of the Estonian-Finnish maavõistluse physical basis of the results. Estonian team in preparation for international Olympiads,UT School of Science, in preparation for participation in the contests of Education and Science.IPhO the organization is to popularize physics and stimulate the development of young talent, while also promoting the natural world and the precision of scientific andtechnical education of international contacts.According to IPhO statute, any State Party to put up a 5-member team, contestants must not have begun his studies at the university and they must not be older than20 years old at the time of the Olympics. Each state team must also be accompanied by two physicists tutor, tutors countries make up the international jury for the Olympics.Olympiad is a theoretical and experimental round, each lasting 5 hours. To prepare the country organizing the Olympics. Tasks will be translated by each respective teammentors into the earth and the solutions can take the form of the contestants in their own language. Võistlusvoorude before and during contact between instructors andstudents are excluded.Evaluation of entries will take place in two stages. First, students put their work supervisors, points, and then protect them from an impartial Commission. Tutor from aninternational jury decides to award diplomas and medals section boundaries. Medals and diplomas recognized by the number of competitors may be IPhO the statuteaccording to 67% of the total number of participants. Given that the contestants is a rahvusolümpiaadide the best in the participating countries, the competition is verytight and every IPhO medal place great credit worthy.IPhO2012 organizer of the first contests in the history of Estonia offers the opportunity to prepare for everybody - IPhO2012 website has opened a web-based contestof Solving Physics IPhO2012 Cup, with more than 200 students from approximately 50 countries.IPhO2012 athletes program held in Tartu, Tallinn, running an international jury, including the opening and closing ceremonies will take place at the Nokia Concert Hall.Estonian University of Life Sports Võistlusvoorud are building - the theoretical virtues thereof On 17 June, 19.juulil resolved experimental tasks. Olympics held in oneof the goals is to introduce Estonia as well as an innovative world of small countries, which also marks the emphasis on innovative IT solutions - for example, takes placethe functions and brokerage solutions through electronic channels, it is also programmed an automated record keeping system and the jury vote.IPhO2012 will certainly be the culmination of one of the 20.July the opening of Tartu city and capital of the world of physics can be seen in the current future Nobel Prizewinners. To this day, all participants IPhO2012 Tartu. On the same day beginning in Tartu Hanseatic Days, which takes place simultaneously with the opening of the Cityof Tartu, physical delivery of the ceremony. After giving the city the capital of symbolic regalia of Physics IPhO IPhO2012 president and main guest, Nobel laureateSir Harold cro. Day programs in public science events, Toome Hill opened the Hanseatic Days Science City, is a career fair in the worlds top universities, research in the2
  • 3. participating students exhibited IPhO2012 Poster inventions or the like. The evening, Sir cro Olympiad participants of the academic lecture, followed by a reception,Mayor of Tartu.IPhO2012 program is in addition to the contest and plenty of leisure activities, thereby providing opportunities for exchanges between professionals as well as examine theEstonian land and culture.Informal avamisüritus in Rocca al Mare Open Air Museum. Trips are both Tartu and Tallinn, plus several trips to take place: the contestants willspend the day at the castle of Rakvere, coaches and guests from the island of Saaremaa. IPhOdel been seen as a traditional football tournament and a greatekstreemspordipäev Tartu Song Festival Grounds.Termination of the Olympiad and the award ceremony will be held on Monday, 23.juulil Nokia Concert Hall in Tallinn. This was followed by a closing party at TallinnSong Festival Grounds.See a video call to the world for young physicists http://www.ipho2012.ee/movie_short/Internasional IPhO2012 dalam fisikaEstonia memiliki kehormatan penyelenggaraan ke-43 musim panas ini Internasional Olimpiade Fisika Internasional fisika (IPhO). 24-15 Juli Olimpiade berlangsung sekitar 90negara diperkirakan akan IPhO2012 fisikawan muda terbaik, bersama dengan instruktur. 1 Maret negara sudah mendaftar untuk ke-88 Bersama dengan para tamu danpenyelenggara peserta tentang 1000IPhO dicapai melalui organisasi Amerika Serikat secara berputar. Estonia bergabung dengan gerakan internasional teadusolümpiaadide murid sebagai negara merdeka padatahun 1992. Pada tahun 1996, kemudian Menteri Pendidikan Jaak menteri berjanji IPhO pada tahun 2012, memimpin organisasi.Fisikawan muda Estonia telah sangat berhasil - dari Olimpiade sebelumnya, siswa kami membawa dua medali emas (2011 dan Yegor Gužvin Semut Remm 1999), 6 perakdan 12 medali perunggu (selengkapnya lihat www.teaduskool.ut.ee / peraih medali ) . Tim nasional Estonia dipilih untuk babak final füüsikaolümpiaadi yang terbaik dalamkompetisi yang diselenggarakan oleh pilihan les sesi dan hasil tempat pemenang atas Estonia tim füüsikaolümpiaadi sekunder terjamin apriori. Dalam beberapa tahunterakhir, tim mengumumkan komposisi akhir dari dasar Estonia-Finlandia fisik maavõistluse hasil. Estonia tim dalam persiapan untuk olimpiade internasional, UT SekolahIlmu, dalam persiapan untuk berpartisipasi dalam kontes Pendidikan dan Ilmu Pengetahuan.3
  • 4. IPhO organisasi adalah untuk mempopulerkan fisika dan merangsang perkembangan bakat muda, sementara juga mempromosikan alam dan presisi pendidikan ilmiah danteknis dari kontak internasional.Menurut undang-undang IPhO, setiap Negara Pihak untuk memasang tim 5-anggota, kontestan tidak harus mulai studinya di universitas dan mereka tidak boleh lebih tuadari 20 tahun pada saat Olimpiade.Setiap tim negara juga harus disertai oleh dua guru fisika, tutor negara membentuk juri internasional untuk Olimpiade. Olimpiade adalahputaran teoritis dan eksperimental, masing-masing 5 jam berlangsung.Untuk mempersiapkan negara menyelenggarakan Olimpiade. Tugas akan diterjemahkan oleh masing-masing mentor tim masing-masing ke dalam bumi dan solusi yang dapat mengambil bentuk satu kontestan dalam bahasa mereka sendiri. Võistlusvoorude sebelum danselama kontak antara instruktur dan siswa dikecualikan.Evaluasi entri akan berlangsung dalam dua tahap. Pertama, siswa menempatkan pekerjaan mereka supervisor, poin, dan kemudian melindungi mereka dari Komisimemihak. Tutor dari juri internasional memutuskan untuk diploma penghargaan medali dan batas-batas bagian. Medali dan diploma yang diakui dengan jumlah pesaingmungkin IPhO undang-undang sesuai dengan 67% dari jumlah peserta. Mengingat bahwa para kontestan adalah rahvusolümpiaadide yang terbaik di negara peserta,persaingan sangat ketat dan setiap medali IPhO kredit tempat yang bagus layak.IPhO2012 penyelenggara kontes pertama dalam sejarah Estonia menawarkan kesempatan untuk mempersiapkan semua orang - IPhO2012 situs telah membuka kontesberbasis web dari Solving Fisika IPhO2012 Piala, dengan lebih dari 200 siswa dari sekitar 50 negara.IPhO2012 atlet Program diselenggarakan di Tartu, Tallinn, menjalankan juri internasional, termasuk acara pembukaan dan penutupan akan berlangsung di Nokia ConcertHall.Universitas Estonia Kehidupan Olahraga Võistlusvoorud sedang membangun - kebajikan teoritis daripadanya Pada tanggal 17 Juni, 19.juulil diselesaikan tugaseksperimental. Olimpiade diadakan di salah satu tujuan adalah untuk memperkenalkan Estonia serta dunia inovatif dari negara kecil, yang juga menandai penekanan padainovatif solusi TI - misalnya, terjadi fungsi dan solusi broker melalui saluran elektronik, juga diprogram sebuah sistem rekaman otomatis menjaga dan suara juri.IPhO2012 tentu akan menjadi puncak dari salah satu 20.July pembukaan Tartu kota dan ibukota dunia fisika dapat dilihat dalam waktu Nobel pemenang Hadiah saatini. Sampai hari ini, semua peserta IPhO2012 Tartu. Pada awal hari yang sama di Hari Tartu Hansa, yang berlangsung bersamaan dengan pembukaan Kota Tartu,penyerahan fisik upacara. Setelah memberikan kota ibukota regalia simbolis Fisika IPhO IPhO2012 presiden dan tamu utama, pemenang Nobel Sir Harold cro. Hari programdalam kegiatan ilmu pengetahuan umum, Toome Bukit dibuka Kota Hari Ilmu Hanseatic, adalah adil karir di universitas top dunia, penelitian di mahasiswa pesertadipamerkan IPhO2012 penemuan Poster atau sejenisnya. Malam itu, Sir cro peserta Olimpiade dari kuliah akademik, diikuti dengan resepsi, Walikota Tartu.4
  • 5. IPhO2012 program selain kontes dan banyak kegiatan rekreasi, sehingga memberikan peluang untuk pertukaran antara profesional serta memeriksa tanah Estonia danbudaya. Informal avamisüritus di Rocca al Mare Open Air Museum. Perjalanan keduanya Tartu dan Tallinn, ditambah beberapa perjalanan ke terjadi: para kontestan akanmenghabiskan hari di istana Rakvere, pelatih dan para tamu dari pulau Saaremaa. IPhOdel dilihat sebagai sebuah turnamen sepak bola tradisional dan ekstreemspordipäevbesar Grounds Song Festival Tartu.Pemutusan Olimpiade dan upacara penghargaan akan diselenggarakan pada hari Senin, 23.juulil Nokia Concert Hall di Tallinn. Hal ini diikuti oleh pihak ditutup pada SongFestival Grounds Tallinn.Lihat panggilan video ke dunia untuk http://www.ipho2012.ee/movie_short/ fisikawan mudaTekan Kontak: +372 5189951 Viire Sepp, viire.sepp @ ut.eeHubungi Panitia: ipho2012@eitsa.ee5
  • 6. Tentative program for leaders and observers Sunday, 15 July 2012 Arrival and Registration in Radisson Blu Hotel Olümpia17.00 – 18.00 Departure from the hotel, transport to Open Air Museum18.00 – 21.00 Icebreaking Estonian Open Air Museum21.00 – 22.00 Arrival to the hotel Monday, 16 July 201207.00 – 09.00 Breakfast Radisson Blu Hotel Olümpia09.15 – 09.45 Walk to Opening Ceremony10.00 – 12.00 Opening Ceremony NOKIA Concert Hall12.00 – 13.30 Welcome Banquet NOKIA Concert Hall13.30 – 14.30 Walk to the hotel14.30 – 19.00 International Board Meeting: Discussion of theoretical problems Radisson Blu Hotel Olümpia Conference Centre19.00 – 21.00 Dinner Restaurant Senso (Radisson Blu Hotel Olümpia)20.30 ‐ …….. Translation of theoretical problems Radisson Blu Hotel Olümpia Conference Centre Tuesday, 17 July 201206.00 – 07.30 Breakfast Radisson Blu Hotel Olümpia07.30 Departure from the hotel6
  • 7. 07.30 – 10.00 Excursion: trip to Saaremaa10.00 – 13.00 Excursion: Kaali crater and Kuressaare13.00 – 14.30 Lunch In Kuressaare14.30 – 17.00 Excursion: Saaremaa and Muhu19.00 – 21.00 Arrival to the hotel and dinner Restaurant Senso (Radisson Blu Hotel Olümpia)21.00 Distribution of theory papers Radisson Blu Hotel Olümpia21.00 – 22.00 SKYPE meeting with Students Wednesday, 18 July 201207.00 – 09.00 Breakfast Radisson Blu Hotel Olümpia09.00 – 12.00 Free time12.00 – 13.00 Lunch Restaurant Senso (Radisson Blu Hotel Olümpia)13.00 – 19.00 Discussion of experimental problems Radisson Blu Hotel Olümpia Conference Centre19.00 – 20.30 Dinner Restaurant Senso (Radisson Blu Hotel Olümpia)20.30 ‐ ……… Translation of experimental problems Radisson Blu Hotel Olümpia Conference Centre Thursday, 19 July 201207.00 – 09.00 Breakfast Radisson Blu Hotel Olümpia08.00 – 12.00 Collection of marks from Leaders (theory) Online09.00 – 12.00 Free time7
  • 8. 12.00 – 13.00 Lunch Restaurant Senso (Radisson Blu Hotel Olümpia)13.00 – 17.00 Excursion: Tallinn18.00 – 19.30 Leaders group A: Transport to dinner19.30 – 22.30 Leaders group A: Dinner with Students22.30 Leaders group A: Transport to Tallinn Radisson Blu Hotel Olümpia17.30 Leaders group B: Transport to Tartu20.00 – 23.00 Leaders group B: Dinner with Students23.30 Leaders group A: Distribution of practical papers Radisson Blu Hotel Olümpia Friday, 20 July 2012 Tartu – the World Capital of Physics06.00 – 08.00 Breakfast Radisson Blu Hotel Olümpia and Hotel London08.00 Leaders group B: Distribution of practical papers Hotel London08.00 – 11.00 Leaders group B and Observers: Transport to Tartu11.00 – 17.00 Tartu ‐ the World Capital of Physics – public science activities13.00 – 14.00 Lunch In Tartu restaurants17.00 – 18.00 Lecture: Sir Harold Kroto (The 1996 Nobel Prize in Chemistry)18.00 – 20.00 Reception by Mayor of Tartu AHHAA Science Centre8
  • 9. 20.00 – 22.30 Transport to Tallinn22.30 Arrival at the hotel Radisson Blu Hotel Olümpia22.30 Collection of marks from Leaders (experiment) Online Saturday, 21 July 201207.00 – 09.00 Breakfast Radisson Blu Hotel Olümpia10.00 – 12.00 International Board Meeting Radisson Blu Hotel Olümpia Conference Centre11.00 Distribution of marks (theory and experiment) Online12.00 – 14.00 Lunch Restaurant Senso (Radisson Blu Hotel Olümpia)14.00 – 21.00 Moderation of theoretical papers Radisson Blu Hotel Olümpia Conference Centre19.00 – 20.30 Dinner Restaurant Senso (Radisson Blu Hotel Olümpia) Sunday, 22 July 201207.00 – 09.00 Breakfast Radisson Blu Hotel Olümpia09.00 – 12.00 Moderation of experimental papers Radisson Blu Hotel Olümpia Conference Centre12.00 – 14.00 Lunch Restaurant Senso (Radisson Blu Hotel Olümpia)14.00 – 17.00 Moderation of experimental papers Radisson Blu Hotel Olümpia Conference Centre17.00 – 19.00 International Board Meeting: Deciding final marks and medals Radisson Blu Hotel Olümpia Conference Centre19.00 – 23.00 Free time and dinner Monday, 23 July 201207.00 – 09.00 Breakfast Radisson Blu Hotel Olümpia09.00 – 13.00 Free time9
  • 10. 13.15 – 13.45 Walk to Closing Ceremony14. 00 – Closing Ceremony NOKIA Concert Hall17.0017.00 – 18.00 Walk back to the hotel Radisson Blu Hotel Olümpia18.30 – 01.00 Farewell Party The Tallinn Song Festival Grounds22.00 Round-the-clock transport back to the hotel Tuesday, 24 July 2012 Departure*The organizer reserves the rights to change the program.Tentative program for leaders and observers Sunday, 15 July 2012 Arrival and Registration in Radisson Blu Hotel Olümpia17.00 – 18.00 Departure from the hotel, transport to Open Air Museum18.00 – 21.00 Icebreaking Estonian Open Air Museum21.00 – 22.00 Arrival to the hotel10
  • 11. Monday, 16 July 201207.00 – 09.00 Breakfast Radisson Blu Hotel Olümpia09.15 – 09.45 Walk to Opening Ceremony10.00 – 12.00 Opening Ceremony NOKIA Concert Hall12.00 – 13.30 Welcome Banquet NOKIA Concert Hall13.30 – 14.30 Walk to the hotel14.30 – 19.00 International Board Meeting: Discussion of theoretical problems Radisson Blu Hotel Olümpia Conference Centre19.00 – 21.00 Dinner Restaurant Senso (Radisson Blu Hotel Olümpia)20.30 ‐ …….. Translation of theoretical problems Radisson Blu Hotel Olümpia Conference Centre Tuesday, 17 July 201206.00 – 07.30 Breakfast Radisson Blu Hotel Olümpia07.30 Departure from the hotel07.30 – 10.00 Excursion: trip to Saaremaa10.00 – 13.00 Excursion: Kaali crater and Kuressaare13.00 – 14.30 Lunch In Kuressaare14.30 – 17.00 Excursion: Saaremaa and Muhu19.00 – 21.00 Arrival to the hotel and dinner Restaurant Senso (Radisson Blu Hotel Olümpia)21.00 Distribution of theory papers Radisson Blu Hotel Olümpia11
  • 12. 21.00 – 22.00 SKYPE meeting with Students Wednesday, 18 July 201207.00 – 09.00 Breakfast Radisson Blu Hotel Olümpia09.00 – 12.00 Free time12.00 – 13.00 Lunch Restaurant Senso (Radisson Blu Hotel Olümpia)13.00 – 19.00 Discussion of experimental problems Radisson Blu Hotel Olümpia Conference Centre19.00 – 20.30 Dinner Restaurant Senso (Radisson Blu Hotel Olümpia)20.30 ‐ ……… Translation of experimental problems Radisson Blu Hotel Olümpia Conference Centre Thursday, 19 July 201207.00 – 09.00 Breakfast Radisson Blu Hotel Olümpia08.00 – 12.00 Collection of marks from Leaders (theory) Online09.00 – 12.00 Free time12.00 – 13.00 Lunch Restaurant Senso (Radisson Blu Hotel Olümpia)13.00 – 17.00 Excursion: Tallinn18.00 – 19.30 Leaders group A: Transport to dinner19.30 – 22.30 Leaders group A: Dinner with Students22.30 Leaders group A: Transport to Tallinn Radisson Blu Hotel Olümpia17.30 Leaders group B: Transport to Tartu12
  • 13. 20.00 – 23.00 Leaders group B: Dinner with Students23.30 Leaders group A: Distribution of practical papers Radisson Blu Hotel Olümpia Friday, 20 July 2012 Tartu – the World Capital of Physics06.00 – 08.00 Breakfast Radisson Blu Hotel Olümpia and Hotel London08.00 Leaders group B: Distribution of practical papers Hotel London08.00 – 11.00 Leaders group B and Observers: Transport to Tartu11.00 – 17.00 Tartu ‐ the World Capital of Physics – public science activities13.00 – 14.00 Lunch In Tartu restaurants17.00 – 18.00 Lecture: Sir Harold Kroto (The 1996 Nobel Prize in Chemistry)18.00 – 20.00 Reception by Mayor of Tartu AHHAA Science Centre20.00 – 22.30 Transport to Tallinn22.30 Arrival at the hotel Radisson Blu Hotel Olümpia22.30 Collection of marks from Leaders (experiment) Online Saturday, 21 July 201207.00 – 09.00 Breakfast Radisson Blu Hotel Olümpia10.00 – 12.00 International Board Meeting Radisson Blu Hotel Olümpia Conference Centre13
  • 14. 11.00 Distribution of marks (theory and experiment) Online12.00 – 14.00 Lunch Restaurant Senso (Radisson Blu Hotel Olümpia)14.00 – 21.00 Moderation of theoretical papers Radisson Blu Hotel Olümpia Conference Centre19.00 – 20.30 Dinner Restaurant Senso (Radisson Blu Hotel Olümpia) Sunday, 22 July 201207.00 – 09.00 Breakfast Radisson Blu Hotel Olümpia09.00 – 12.00 Moderation of experimental papers Radisson Blu Hotel Olümpia Conference Centre12.00 – 14.00 Lunch Restaurant Senso (Radisson Blu Hotel Olümpia)14.00 – 17.00 Moderation of experimental papers Radisson Blu Hotel Olümpia Conference Centre17.00 – 19.00 International Board Meeting: Deciding final marks and medals Radisson Blu Hotel Olümpia Conference Centre19.00 – 23.00 Free time and dinner Monday, 23 July 201207.00 – 09.00 Breakfast Radisson Blu Hotel Olümpia09.00 – 13.00 Free time13.15 – 13.45 Walk to Closing Ceremony14. 00 – Closing Ceremony NOKIA Concert Hall17.0017.00 – 18.00 Walk back to the hotel Radisson Blu Hotel Olümpia18.30 – 01.00 Farewell Party The Tallinn Song Festival Grounds22.00 Round-the-clock transport back to the hotel14
  • 15. Tuesday, 24 July 2012 Departure*The organizer reserves the rights to change the program.Steering CommitteeChairmanJanar Holm Estonian Ministry of Education and ResearchSecretaryViire Sepp Gifted and Talented Development Centre of University TartuMembersRait Toompere Archimedes FoundationPeeter Saari Estonian Academy of SciencesToomas Sõmera Estonian Information Technology FoundationÜlle Kikas Estonian Ministry of Education and Research15
  • 16. Kaido Reivelt Estonian Physical SocietyRaivo Stern National Institute of Chemical and BiophysicsJakob Kübarsepp Tallinn University of TechnologyJaan Kalda Tallinn University of Technology, The Institute of CyberneticsKristjan Haller University of TartuJaak Kikas University of TartuMarco Kim University of Tartu Institute of Physics16
  • 17. Head of the Academic CommitteeHead of Experimental Examination Academic CommitteeJaak Kikasjaak.kikas@ut.eeHead of Theoretical ExaminationJaan Kaldakalda@ioc.ee17
  • 18. Organizing CommitteeHead of the Organizing CommitteeEne Koitlaene.koitla@eitsa.eeProject ManagerMarily Hendriksonmarily.hendrikson@eitsa.ee18
  • 19. Formula sheetThis formula sheet has been originally compiled during the training sessions of the Estonian and Finnish teams (since 1994); actually, the students were the ones whorequested it. Now, since I decided to use it as a reference list for the on-line competition "Physics Cup – IPhO2012", I revised it trying to make it more easily understandable,and intend to continue doing so: please keep track of the version number, the current one is 29. Jan. 2012. Also, if you have any suggestions (eg. if some formulationsare not clear or you discover a mistake/typo), please let me know.Jaan Kalda, Academic Committee of IPhO-2012Frequently asked questionsQ: My country will not participate at IPhO-2012. Can I participate at the "Physics Cup – IPhO-2012"?A: Yes, you can. There are only two conditions: your 21st birthday should be later than 30th June 2012, and you should not be a university student.Q: How can I register for IPhO-2012?A: "Physics Cup – IPhO-2012" is not to be confused with IPhO-2012. In order to compete in IPhO-2012, you need to be selected for your national IPhO-team. Typically, thisinvolves participation in the national physics olympiad(s) and/or training camps with selection competitions.Q: Can two or more students form a team and solve together?A: No, this is an individual competition.Q: Can I use books, internet sources etc. for solving the competition problems?A: Yes, you can. This will teach you only a knowledge about physics, but you cannot find the solutions there (hopefully – at least for the majority of the problems).19
  • 20. Q: Can I ask the help of my teacher?A: Yes, as long as he/she gives you only a book or points out a chapter there. No, if he/she would be going to give you the solution itself.Q: Is my solution correct or wrong?A: According to the rules, this will be made known on the next Sunday – in order to give the contestants opportunity to find mistakes by themselves (there is also a penaltydifference, 0.9 vs 0.8 depending if the presence of an error was pointed out or the student found it himself/herself). But if you got "inner feeling" that everything is clear,then of course there are no mistakes. If not sure, check once again over (starting with a basic validation: are the dimensions correct, is the result OK for the special limitcases – some parameters are equal or zero, etc). Also, if you are not very sure about your solution, you better assume that it is not correct – the solutions dont get correctby lucky accidents.Q: How many awards will be given and what will be the awards?A: The number of results will depend on the level of the results; the awards will include the experimental sets of the practical problems of IPhO-2012.Q: I already sent a correct solution, but now I found a more elegant one, can I submit it?A: Yes, you can; if that elegant solution turns out to be the best one, youll receive a factor of e, but the time of arrival of that new solution will define the bonus factor dueto speed, and all the penalty factors earned with the previous solution(s) remain active. On the other hand, if it is still not the best solution, the speed factor of your firstsolution will be kept (ie. in that case the new solution would have no effect on your score).Q: Since the students can solve the problems at home, wouldnt it be very easy for them to cheat?A: Not very easy, because using reference materials is not considered to be a cheating (note that professional physicists can also use reference materials while doing theirresearch). The real cheating would be letting someone smarter than you solve the problem, and then presenting it as your own solution. However, only very few of you havesmart enough people nearby. And even if you happen to have such a smart person around, he (or she) has probably high enough ethical norms so he wouldnt solve theproblem for you. And even if hed be willing to do it, would you let it happen? I mean, would you be happy getting the first or second place knowing that you cheated and theothers did not? I am sure that the answer is no! So, let us assume that all the participants are nice people and there are no cheaters (there is even no need to mention thatsuch kind of cheating tends to get revealed sooner or later)!20
  • 21. Competition “Physics Cup – IPhO2012”Eligibility: age- and educational restrictions for the participants are the same as for IPhO-2012.Registration: please send e-mail to AC.IPhO2012@gmail.com, indicating: 1. Your given name; 2. Family name; 3. Date of birth; 4. E-mail address; 5. Full mailing address; 6. Your school; 7. Your physics teacherIt is recommended to register as soon as possible, prior to the publication of the first problem (18. Sept. 2011). However, you can also register at any later stage.Distribution of problems: At 2 pm (GMT) of the third Sunday of each month, between September 2011 and June 2012, a new problem is published at thehomepage of 43rd IPhO (www.ipho2012.ee) and also sent by e-mail to all the registered participants.Submitting the solutions: The contestants are asked to submit the answer as fast as they can, but not later than the publication time of the next problem. Theformulae can be submitted using the LaTeX symbols; these can be included directly into the e-mail text (e.g. m=m_0/sqrt{1-v^2/c^2}). Alternatively, these can bealso submitted as formulae in OpenOffice or MSWord files, or scanned (digitally photographed) images of a clear handwritten text. Also, the students are asked tosend their solutions (using one of the above mentioned formats), but this can be done with a delay of up to 48 hours (from the moment of submitting the answer).These solutions should contain as few text (in English) as possible (few words are typically enough), mostly using formulae and figures. When sending solutions,please use the subject line "Problem No 1" ("Problem No 2", etc), exactly as written here (but without quotation marks); in that case you will receive an auto-reply confirming that your solution has been received. If you do not want to receive an auto-reply, make sure the subject line does not contain (not necessarilybegin with) the text "Problem No 1" (for instance just drop "No" and write "Problem 1").Grading: Each problem costs 1.0 pts; this serves as a base score, which will be multiplied with factors corresponding to bonuses and penalties. Either a full or zerocredit is given; on a weekly basis (for the first three weeks on Sunday, for the fourth week on Thursday), the competitors are notified if their solutions are (a)21
  • 22. complete, (b) incomplete (e.g. some missing motivations), (c) with minor mistakes, or (d) incorrect. Cases (b)-(c) incur a penalty factor of 0.9 and the case (d) – 0.8.The students can send the corrected solutions – either upon finding an error or receiving the notification of incompleteness; each revision incurs a penalty factor of0.9. The first 10 correct answers (supplemented later with a complete solution) receive a bonus factor according to the formula k = 1.111-n, where n is the ordernumber. The best solution will receive a bonus factor of e (=2.718…) and will be published as the official solution at the web page. If there is no single solution,which is better than the others, the factor e may be distributed over several solutions of similar quality — in such a way that the product of all the factors equals stillto e. If, in addition to the best solutions, there are other good solutions which (due to certain reasons, eg. the usage of a significantly different approach) deservepublishing, these published alternative solutions will receive a bonus factor of 1.1. There is also an additional rule for those who send many incorrect solutionsbefore finding a correct one: if the product of penalty factors with the speed bonus factors gives a number which is smaller than e-2/3, the factor e-2/3 is used, instead.Example: a student submits an answer, discovers an error and therefore submits a revised answer (k1 = 0.9), together with the solution; upon being notified that thesolution has still a minor mistake (k2 = 0.9), he submits a third revision (k3 = 0.9) of the answer and the solution. This solution turns out to be the fifth correctsolution, so k4 = 1.16; later on, it is found that this is the best solution which will be published at the web-page (k5 = e). Thus, the overall score will be 0.93×1.16×e= 4.673…Publication of results: The names and results of half of the students with best scores are published at the web-page; the list is updated monthly.Distribution of awards: The awards are announced and handed over at the closing ceremony of the 43rd IPhO or sent to the mailing address (if the recipient is notpresent at the ceremony).In your solutions — what you may assume well-known and what not: things which are not in the formula sheet need to be motivated/derived. (You dont need tocheck – your typical high-school formulae are there.)Questions regarding the competition: e-mail to AC.IPhO2012@gmail.com.See also: Frequently asked questions.Problem No 0 – you will not receive points for this, but you can submit your answer and let us know how long time did it take for you to solve it.22
  • 23. Physics solver’s mosaicWhat is needed to be able to solve problems so well that you could get a gold medal at IPhO? Is it enough to be just very gifted? Of course not, there are other students,who have solved a lot of problems – while you are thinking hard trying to "invent a bicycle", they are already writing the solution, because they had solved a similar problemearlier. Is it enough to solve a lot of problems and read a lot of problem solutions?Most often, no. Just solving or reading solutions, of course, will increase your technicalskills, but you also need to think over, what were the main ideas which made it possible to solve the problem, and take these ideas into your permanent arsenal; if you solvetoo many problems, you dont have time to think over. Is it possible to learn "the art of problem solving" and if yes then how? Well, 99% of the Olympiad problems aresolved using a rather limited set of ideas (for mathematics, that set is somewhat larger). So, if you acquire those ideas well enough – so that you can recognize themeven if they are carefully hidden – then the IPhO gold will be yours! Do not worry, no-one expects you to discover a solving technique which has been never seen before,because that would be an achievement worth of a Nobel Prize!Since we started the topic of Nobel Prize – is it enough to be the absolute winner of the IPhO to get, at a later stage, a Nobel Prize? (Each year, there is one Nobel Prize inphysics – similarly to the absolute winner of IPhO.) Of course, it is not; however, youll have better chances than anyone else. Becoming a great physicist requires severalcomponents, one of which is having brilliant problem solving skills (tested at IPhO). Another one is ability to make solvable models - formulate problems which can be solvedand which reflect important aspects of reality. Third component is ability to distinguish, which problems are important and which are not. You can be very skilful and smart,but if you study problems of marginal interest, no-one will pay attention to your research results. Finally, you need a considerable amount of luck. Indeed, that particularfield of physics in which you start your studies, eg. start making your PhD thesis, depends on somewhat random decisions – it is almost impossible to foresee, where are thebiggest scientific challenges after five or ten years. Also, in order to perfect yourself in regard of the above-mentioned three components, you need excellent supervisors andexcellent lab; although you have some freedom in choosing your supervisor and lab, you still need to be very lucky to find outstanding ones!I coined to name this section as "mosaic", because we shall describe here a set of solving techniques, fragments of the whole arsenal needed for a perfect problem solver.With a large number of pieces, the picture would become recognizable, but we need to start making it piece by piece … While some "tiles" will be useful for solving aspectrum of problems, other tiles are aimed to give more insight into certain physical concepts.Jaan Kalda, Academic Committee of IPhO-201223
  • 24. 1. Minimum or maximum?It is well-known that a system is stable at the minimum of its potential energy. But why? Why is a minimum different from a maximum? For Fermat principle it is clear:there is no longest optical path between two points – the ray could just go "zig-zag" -, but there is definitely one which is the shortest!The reason is simple – at an equilibrium state, the kinetic energy has always minimum (as long as masses are positive). What we actually do need for a stability isa conditional extremum of one conserved quantity (such as the net energy), under the assumption that the other conserved quantities are keptconstant (unconditional extremum is OK, too). Consider the motion of a body along x-axis and let us describe it on the phase plane, with coordinates x and p (themomentum). The overall energy is E =U(x)+p2/2m. Now, if we depict this energy as a surface in 3-dimensional space with coordinates x, p and E, the point describing the state ofthe system will move along the intersection line of that surface with a horizontal plane E=Const. At the minimum of U(x), with p=0, this intersection line would be just a singlepoint, because this is the lowest point of that surface. The near-by trajectories will be obtained if we ascend the horizontal plane a little, E =Emin+e, so that it no longer justtouches the surface, but cuts a tiny ellips from it. All the points of that trajectory (the ellips) are close to the equilibrium point, so the sate is, indeed, stable.It appears that a system can be stable also because of a conditional maximum of the net energy: while an unconditional extremum of the kinetic energy can only be aminimum, things are different for conditional extrema. Perhaps the simplest example is the rotation of a rigid body. Let us consider a rectangular brick with length a, width b,and thickness c (a>b>c). Let Ic be its moment of inertia for the axis passing its centre of mass and perpendicular to the (a,b)-plane; Ib and Ia are defined in a similar way. For ageneric case, the moment of inertia I will depend on the orientation of the rotation axis, but it is quite clear that Ic >= I >= Ia (it can be shown easily once you learn how to usetensor calculations). Now, let us throw the brick rotating into air and study the motion in a frame which moves together with the centre of mass of the brick (in that frame,we can ignore gravity). There are two conserved quantities: angular momentum L, and rotation energy K=L2/2I . We see that for a fixed L, the system has minimal energyfor I = Ic (axis is parallel to the shortest edge of the brick), and maximal energy for I = Ia (axis is parallel to the longest edge of the brick). You can easily check experimentallythat both ways of rotation are, indeed, stable! Not so for the axis parallel to the third edge… This phenomenon is demonstrated in a video made by NASA on the InternationalSpace Station.( http://mix.msfc.nasa.gov/abstracts.php?p=3873)Well, actually the rotation with the minimal energy is still a little bit more stable than that of with the maximal energy; the reason is in dissipation. If we try to represent themotion of the system in the phase space (as described above), we would start with touching a top of an hill with a horizontal plane E =Emax (so that the intersection is just apoint), but due to dissipation, the energy will decrease, E =Emin – e, and the phase trajectory would be a slowly winding-out spiral. So, while you are probably used to knowthat dissipation draws a system towards a stable state, here it is vice versa, it draws the system away from the stable state!24
  • 25. — Jaan Kalda, Academic Committee of IPhO-20122. Fast or slow?What is an adiabatic process? Most of the readers would probably answer that this is a process with a gas which is so fast that there is no heat exchange with thesurroundings. However, this is only a half of the truth, and actually the less important half. In fact, it is quite easy to understand that this is not entirely correct: consider acylinder, which is divided by a thin wall into two halves; one half is filled with a gas at a pressure p, and the other one is empty. Now, let us remove momentarily the wall:the gas from one half fills the entire cylinder. Since no external work is done (the wall can be removed without performing a work), the energy of the gas is preserved,hence, the temperature remains the same as it was at the beginning. Meanwhile, for an adiabatic process we would expect a decrease of temperature by a factor of 2γ-1: partof the internal energy is supposed to be spent on a mechanical work performed by the expanding gas. However, if the piston moves faster than the speed of sound, the gaswill be unable to catch up and push the piston. So, the adiabatic law was not followed because the process was too fast!It appears that the adiabatic law for thermodynamics has also a counterpart in classical mechanics – the conservation of the adiabatic invariant. For mechanical systems(oscillators) performing periodic motion, the adiabatic invariant is defined as the area of the closed curve drawn by the system in phase space(which is a graph wherethe momentum p is plotted as a function of the respective coordinate x), and is (approximately) conserved when the parameters of the system are changed adiabatically,ie. slowly as compared with the oscillation frequency. For typical applications, the accuracy of the conservation of the adiabatic invariant is exponentially good and canbe estimated as e-fτ, where f is the eigenfrequency of the oscillator, and τ is the characteristic period of the variation of the system parameters.How are related to each other (a) adiabatic invariant and (b) adiabatic process with a gas? The easiest way to understand this is to consider a one-dimensional motion of amolecule between two walls, which depart slowly from each other (Figure 1). Let us use the system of reference where one of the walls is at rest, and the other moves with avelocity u << v, where v is the velocity of the molecule (the interaction of the molecule with the walls is assumed to be absolutely elastic). One can say that such a moleculerepresents an oscillator with a slowly changing potential: the potential energy U(x) = 0 for 0<x< X (where X = a +ut)and otherwise, U(x) = ∞. The trajectory of themolecule in the phase space is a rectangle of side lengths Xand 2mv. So, the adiabatic invariant is 2mvX; hence, vX = Const. For a one-dimensional gas, thedistance Xbetween the walls plays the role of the “volume” V, and mv2/2=kT/2, hence v ~ T1/2 (here "~" means “is proportional to”). So, the adiabatic invariant can bewritten as V 2T = Const. On the other hand, from the adiabatic law for an ideal gas, we would expect TV γ-1 = Const. For the one-dimensional gas, the number of the degreesof freedom i = 1, hence γ = cp/cV = (i+2)/i =3, and TV2 = Const, ie. we can conclude that the adiabatic invariant and the adiabatic gas law give us exactly the same result!25
  • 26. How to prove that for an adiabatic forcing of an oscillator, the adiabatic invariant is conserved? Well, this is not a too simple mathematical task and thus we skip the proofhere (it can be found in good textbooks of theoretical mechanics). However, for a simple particular case of an elastic ball between two walls (see above), it can be done moreeasily. Indeed, with each impact with the departing wall, the speed of the ball is decreased by 2u, and this happens once per time interval t = 2X/v. So, the ball decelerateswith the rate of dv/dt = 2u/t =uv/X, hence dv/v= udt/X = –dX/X. Integrating this differential equation gives us directlyXv = Const.Conservation laws play a central role both for the physical processes, and for the physics as a science (cf “Minimum of Maximum”), and adiabatic invariant is not anexception. Perhaps the most important role of it is related to the quantum mechanics. Namely, during adiabatic processes, the system will not leave the stationary quantumstate it has taken (as long as the state itself does not disappear). To motivate this claim, let us consider a biatomic molecule, which can be modelled as an oscillator. Whentreating the process classically, the trajectory of a harmonic oscillator in the phase space is an ellips of surface areaJ = πp0x0, where p0 and x0 are the amplitudes of themomentum and coordinate. Note that p0 = mx0ω0, where ω0 is the circular eigenfrequency of the oscillator; therefore, the full energy of the oscillator (calculated as themaximal kinetic energy) is E = p02/2m = p0x0ω0/2 = J ω0/2π = J f0. Hence, the adiabatic invariant J = E/f0: during adiabatic processes, the oscillation energy is proportional tothe frequency. According to the quantum mechanics, the stationary energy levels of the oscillator are given byEn=hf0(n +1/2), where n is an integer representing the ordernumber of the energy level. Comparing the classical and quantum-mechanical results leads us to the conclusion that during adiabatic processes,n = Const: the systemwill remain at the stationary state of the same order number where it was(Figure 2) . (While it is not always completely correct to combine classical and quantum-mechanical results, classical mechanics is a macroscopic limit of the quantum mechanics and hence, the conservation laws of both theories need to be compatible.)Now, suppose our bi-atomic molecule is forced by an electromagnetic field in the form of an adiabatic pulse. In terms of classical mechanics we say that such a forcing isunable to pump energy into oscillations of the molecule, because the adiabatic invariant is conserved and hence, the energy of oscillations depends only on the currenteigenfrequency. In terms of quantum mechanics we’ll say exactly the same, but the motivation will be different: the adiabatic pulse contains no photons which are resonantwith the oscillator.26
  • 27. Another important role of the adiabatic invariant is protecting us from the cosmic radiation (in “collaboration” with the magnetic field of the Earth). It appears that the motionof a charged particle in a magnetic field can be represented as an Hamiltonian motion (we skip here the definition of the Hamiltonian motion as it would go too deeply intothe subject of theoretical mechanics), with a re-defined momentum. It appears also that with this new momentum (the so-called generalized momentum), the adiabaticinvariant of a gyrating (helicoidally moving) charged particle is its magnetic dipole moment (which is proportional to the magnetic flux embraced by the trajectory, hence thisflux is also conserved). So, if a charged particle moves helicoidally along magnetic field lines towards a stronger magnetic field, due to the conservation of its magneticmoment, the perpendicular (to the magnetic field) component of its velocity will increase. Owing to the conservation of its kinetic energy, the parallel component of thevelocity will decrease, and at a certain point, it becomes equal to zero: the particle is reflected back (Figure 3). This is exactly what happens with a majority of the chargedparticles approaching Earth along the field lines of its magnetic field.Adiabatic invariant has simple every-day applications, too. Suppose you try to carry a cup of coffee – this will be quite simple even if the cup is completely full. Now try thesame with a plate of soup – at least with full plate, this will be quite difficult! Finally, with a large full photographic tray, this will be nearly impossible! The reason is thatwhen you try to keep your hands motionless, they still move slightly, but the feedback from your vision allows you to correct the mistakes. The characteristic time-scale ofsuch a motion of hands is of the same order of magnitude as your reaction time, in the range of 0.2 – 0.4 s. This is to be compared with the reciprocal of the circulareigenfrequency ω0-1 of the water level oscillations. (ω0-1 differs from the full period T by 2π; ω0-1 serves as a better reference here, because the corrective motion of handsrepresents no more than a quarter of a full period of an oscillatory motion.) For a plate of depth h and length L, the smallest eigenfrequency can be estimated as thefrequency of standing waves of wavelength 2L (see also problem No 2 of IPhO-1984). The speed of shallow water waves is (gh)1/2, so that the eigenfrequency willbe f0 = (gh)1/2/2L. For a cup of coffee, the diameter and depth can be estimated as 7cm, hence the characteristic time scale of oscillations will be ω0-1 ≈ 0.03s; with respectto such oscillations, the hand motion is adiabatic – even if we apply our smallest estimate of 0.2s (note that counter-intuitively, here a slow reaction is better than a fast27
  • 28. one). For a plate of H = 3cm and L = 25cm we get ω0-1 ≈ 0.15s – the hand motion is already not very adiabatic. Finally, for a photographic tray ofH = 3cm and L = 60cm, weobtain ω0-1 ≈ 0.35s, which is really difficult to handle.Finally, in the context of adiabaticity, it is interesting to analyse the IPhO problem about tides, which was posed in 1996 in Oslo (as Problem No 3). The problem is, indeed,very interesting: you are given a simplified model of a complex and important phenomenon, which, regardless of simplicity, gives you reasonable estimate and teachesvaluable physical concepts. Let us read its text and comment the model assumptions.In this problem we consider some gross features of the magnitude of mid-ocean tides on earth. We simplify the problem by making the following assumptions:(i) The earth and the moon are considered to be an isolated system,/a very reasonable assumption: even the effect of the Sun is small in the reference frame of Moon-Earth centre of mass, where the inertial force and Sun gravity cancel eachother out/(ii) the distance between the moon and the earth is assumed to be constant,/also reasonable: there are small variations, but nothing to worry about/(iii) the earth is assumed to be completely covered by an ocean,/this is definitely not the case, but at least the Pacific Ocean is very large; as a model, why not …/(iv) the dynamic effects of the rotation of the earth around its axis are neglected, and/Did you understand what they wanted to say? If not, you need to learn reading the problem texts! Well, it means that the forcing of the water by the Moon is to be assumedto be adiabatic (slow), so that the water level will take a quasi-equilibrium position (ie. equilibrium, where the equilibrium state changes slowly in time). The validity of thisassumption will be discussed below./(v) the gravitational attraction of the earth can be determined as if all mass were concentrated at the centre of the earth.Again, a perfectly reasonable assumption: the gravitational field of a sphere (assuming that the mass density depends only on the distance from the centre) is outside thesphere the same as that of a point mass. The departure of the Earths shape from a sphere is small, indeed.And so, is the tide forcing really adiabatic? We need to compare the period of forcing with the eigenfrequency, or, the speed of the "piston" with the speed of waves. Thespeed of the "piston" is the Earth perimeter divided by 24 h, ie. v = 460 m/s. The relevant wave is, in effect, a tsunami with the estimated speed of (gH)1/2 = 200 m/s (here,H = 4000 m is an estimate for the average ocean depth). So, the forcing is far from being adiabatic, we could say that the assumption (iv) is horribly wrong. On the otherhand, if we solve the problem according to these assumptions, we obtain for the tide amplitude h = 27 cm, which has at least a correct order of magnitude; why? Well,because for a typical resonance response curve, the response amplitude at a double eigenfrequency (which we would need as the "piston" speed is ca twice the wave speed)is of the same order of magnitude as that of a zero frequency (which is obtained in this Problem). Further, since the tidal motion of the water is by no means quasi-stationary, the ocean boundaries will play an important role. What will happen is very similar to the motion of tea in a cup, when you push the tea by a spoon: basinboundaries reflect the moving water, creating vortices and complex pattern of tidal heights. To conclude, we learned that the above tide model fails for water tides (providinga very rough estimate of the tidal height); perhaps it can be used somewhere else with a better accuracy? The answer is "yes, for the tides of the Earth crust "! Indeed, themantle thickness is of the order of few thousands km, which corresponds to almost ten-fold tsunami speed and makes the Moon as a "piston" reasonably adiabatic. Therelative crust deformation due to tidal movements is so small that the elastic response of the crust is also negligible: the result h = 27 cm is indeed very close to reality.—Jaan Kalda, Academic Committee of IPhO-201228
  • 29. 3. Force diagrams or generalized coordinates?Typically you are taught in high school that in order to solve problems with interacting bodies you need to draw force diagrams, and write down the force balance equations(based on Newton II law) for x and ycomponents (for three-dimensional problems, also the z-component). However, for problems which are more difficult than theelementary ones, this is typically far from being the simplest approach. Meanwhile, there is a very powerful method based on generalized coordinates, which provides in mostcases the easiest route to the solution. The basic idea of the method is as follows.Suppose the state of a system can be described by a single parameter , which we call the generalized coordinate (the method can be also applied with two or moreparameters, but this will complicate things, and in most cases, one parameter is perfectly enough). Then, what you need to do is to express the potential energy of thesystem in terms of , , and the kinetic energy in terms of , the time-derivative of : . Then, if there is no dissipation and external forces, the netenergy is conserved: . Upon taking time-derivative of this equality, we obtain , from where we can express the accelerationof the generalized coordinate:Note that most often, is constant, because the kinetic energy is proportional to , and plays the role of an effective mass . In some cases, it may happenthat depends also on and/or depends also on ; then, the above formula will not work, but the technique itself remains still applicable (cf. the example of rotatingspring below).In order to illustrate this method, let us start with a simple wedge problem. Consider a system where a ball of mass lays on a wedge of mass , and is attachedwith a weightless rope and pully to a wall as depicted in Figure; you are asked to find the acceleration of the wedge, assuming that all the surfaces are frictionless, and thereis a homogeneous gravity field .29
  • 30. When using the force diagram method, it would be a good idea to use the (non-inertial) reference frame associated with the wedge (introducing thereby the inertialforces and ), because otherwise, it would be difficult to write down equation describing the fact that the ball will remain on the inclined surface of the wedge.Here, however, we leave this for the reader as an exercise, and describe the state of the system via the displacement of the wedge. Then, the velocity of the wedge is ;the velocity of the ball with respect to the wedge is also , implying that the vertical component of the balls velocity is , and the horizontal componentis . Hence, we find thatUpon taking time derivative of this equation and cancelling out , we obtain an expression for the wedge acceleration:As another example, let us consider an old IPhO problem (5th IPhO in Sofia, 1971, Problem No 1). The set-up is quite similar to the previous problem, but there is nowall, there are two bricks instead of one ball, and the wedge has two inclined surfaces (see Figure); we ask again, what is the acceleration of the wedge.30
  • 31. You might think that the method does not work here, because there are two degrees of freedom: the wedge can slide on the table, and the bricks can slide with respect tothe wedge. However, if we make use of the conservation of the centre of mass (there are no external horizontal forces), we can express the displacement of the bricks(with respect to the wedge) via the displacement of the wedge :What is left to do, is to writesubstitute by , take time derivative of the full energy, and express . Well, there is some math do be done, but that is actually just an algebra. If you do it correctly, youobtain .A really simple example is provided by water level oscillations in U-tube. Let the water occupy length of the U-tube, and let us use the water level height (withrespect to the equilibrium level) as the generalized coordinate. For a state with , a water column of height from one arm has been lifted by an height differenceand moved into the other arm of the U-tube, which corresponds to the potential energy ; meanwhile, . So, upon applying our technique weobtain , which describes an harmonic oscillator of circular frequency .Actually, when in hurry and oscillation frequency is needed, two steps of the scheme (taking time derivative and writing the equation of motion) can be skipped. Indeed, foran harmonic oscillator, both and need to be quadratic in and , respectively, ie. should have form and , where and are constants;then, .Next, the technique can be used to analyse oscillations in simple rotating systems, such as, for instance, a system of two balls of mass , connected with a springof length and stiffness , rotating with angular momentum (which is perpendicular to the spring). Here, again, an additional (to the energy) conservation law (of angularmomentum) reduces the effective number of degrees of freedom down to one. Let us use the deformation of the spring as the generalized coordinate. Then,31
  • 32. This case is different in that the kinetic energy depends not only on , but also on ; in effect, the second term of the kinetic energy behaves as a potential one,and can be combined into an effective potential energy in the expression for the full energy. Following our technique,This equation of motion can be linearised around the state of equilibrium (such that for , the right-hand-side turns to zero), by introducing .Linearisation means approximating a non-linear function with a linear one, and is typically done by neglecting in the Taylor expansion quadratic and higher terms, ie. bysubstituting with ; this is legitimate if the argument varies in a narrow range, in this case for . As a result, we obtainwhich gives us immedieately the circular frequency of small oscillations, .What we did here can be also called a linear stability analysis (which is a very popular technique in physics). Indeed, it is easy to see that regardless of the parametervalues, the circular frequency is always a real number, ie. the circular trajectories of the balls are always stable (meanwhile, imaginary circular frequency would mean thatthe solution includes a component which grows exponentially in time, ie. the regular motion along the circular trajectory would be unstable).Note that almost exactly the same analysis which was done here for the rotating spring, was used in the"official" solution of the Problem 1 (subquestion 3) ofIPhO-2011. However, it appears that for the mentioned problem, this technique cannot be applied as easily: there is one mistake in the solution, and another one amongthe assumptions of the problem; for more details, see the mosaic tile "Are trojans stable?".Up til now we have dealt with problems where the task was to find an acceleration. What to do, if you are asked to find a force? For instance, a sphere and a wedge areplaced on two facing ramps as shown in Figure; all the surfaces are frictionless. Find the normal force between the wedge and the sphere.32
  • 33. Well, it would be very easy to find the acceleration of the ball (or that of the wedge) using the method of generalized coordinates (ball displacement can be used as thecoordinate). But once we know the acceleration, it is also easy to find the normal force between the wedge and the ball from the Newton II law! (The answeris .)The method of generalized coordinates is designed to work for dissipation-less systems.. However, in some cases it is also possible to take into account the friction.To illustrate this, let us modify the previous problem so that the right ramp remains frictionless, but the left ramp has high friction, so that the ball will rotate along it, andthe friction between the wedge and the ball is described by kinetic friction coefficient .The idea here is to "fix" the energy conservation law by adding the work performed by the friction force. Initally, such an equation will involve the normal force asa parameter, but it can be determined later: we express the normal force in the same ways as for the previous problem, and this will be the equation for finding .So, and ; the contact point leaves "traces" both on the wedge (of length ) and on the ball (of length ),corresponding to the net work of . So, the energy conservation law is written asfrom where .Now, assuming that we have heavy wedge, and the system moves leftwards, the Newton II law for the wedge can be written as33
  • 34. ,and hence, .As a final example illustrating this method, let us consider a somewhat more difficult problem, posed by W.H. Besant in 1859, and solved by Lord Rayleigh in 1917:in an infinite space filled with an incompressible liquid of density at pressure , there is a spherical "bubble" of radius , which has vacuum inside. Due to the pressure(far away, it is kept equal to ), the "bubble" starts collapsing; find the collaps time of the "bubble". Here we use the radius of the "bubble" as the generalized coordinate;there is no potential energy, but there is work done by the pressure, . What is left to do, is to express the kinetic energy of the fluid in terms of . Dueto the incompressibility of the fluid, the volume flux of liquid through any spherical surface of radius around the centre of the "bubble" is independent of: . So, the kinetic energy can be found asSo, the energy balance can be written asThis equation could be used to find the acceleration ; however, we need to know the collapse time; so we put , and express in terms of and :34
  • 35. Thus, we were able to obtain an answer, which contains a dimensionless integral: substituting allowed us to get rid of the dimensional quantities under theintegral (if possible, always use this technique to convert integrals into dimensionless numbers). This result could be left as is, since finding an integral is a task formathematicians. The mathematicians, however, have been up to the task: where denotes the gamma function. So, we can writeFinally, to close the topic of the generalized coordinates, it should be mentioned that this technique can be developed into generic theories – Lagrangian andHamiltonian formalisms, which are typically taught as a main component of the course of theoretical mechanics. In particular, the Hamiltonian formalism makes it possibleto prove the conservation of adiabatic invariant, as well as the KAM (Kolmogorov-Arnold-Mozer)theorem, as well as to derive conservation laws from the symmetryproperties of the Hamiltonian (orLagrangian) using the Noethers theorem. The Hamiltonian approach differs from what is described here by using the generalizedmomentum , instead of the generalized velocity . For the most typical cases when the kinetic energy is proportional to the square of the generalized velocity, one can justuse the effective mass (defined above): . Then, the expression for the full energy is considered as a function of and, , and is called the Hamiltonian; the equation of motion is written in the form of a system of equations, , . However, for thepractical application of problem solving, the simplified approach to the generalized coordinates provided above is just enough!—Jaan Kalda, Academic Committee of IPhO-201235
  • 36. 4. Are Trojans stable?To begin with, what are Trojans? (http://en.wikipedia.org/wiki/Trojan_%28astronomy%29) These are small celestial bodies which move together with two heavybodies (typically the sun and a planet) in such a way that (a) the relative position of the three bodies does not change (they rotate as if forming a solid body); (b) the motionof these small bodies is stable: small fluctuations in the relative position will not be amplified. It appears that for a two-body system, eg, the Sun and the Jupiter, there arefive points, where a small (third) body could move so that the condition (a) will be satisfied – the so called Lagrangian points(http://map.gsfc.nasa.gov/mission/observatory_l2.html), denoted by L1, L2, L3, L4 and L5. The first three of these lay at the same line with the Sun and Jupiter. Inaddition, as was shown in IPhO problem 1989-2, the condition (a) will be also satisfied, if the three bodies form an equilateral triangle; the respective points are denoted byL4 and L5. It appears that the Lagrangian points L1, L2 and L3 are always unstable, but the points L4 and L5 can be stable. In particular, for the Sun-Jupiter system, L4 andL5 are stable, and there are actually a considerable number of asteroids "trapped" into the vicinity of these points. These asteroids are named after the figures of the Trojanwar, which is why the satellites in Lagrangian points L4 and L5 are called the Trojans (the term is not limited to the Sun-Jupiter system).And so, the Trojans are stable by definition, and the title here is somewhat inaccurate; the actual question is, are the Lagrangian points L4 and L5 always stable? Thequestion is motivated by the Problem 1 (subquestion iii) of IPhO 2010 which made an attempt of studying the stability of L4 for a system consisting of two equal pointmasses (actually, small oscillations of a small body moving around L4). The official solution, concluded that the small body will oscillate, ie. the position is stable. However, acareful analysis shows that the stability of L4 and L5 is achieved only if the ratio of the two large masses is large enough – larger than , ie. for twoequal masses the equilibrium is unstable! So, what went wrong in the IPhO Problem 2010-1-iii? (http://map.gsfc.nasa.gov/media/ContentMedia/lagrange.pdf)To begin with, let us mention that stability for such a system is actually quite a surprising thing. Indeed, according to the Earnshawstheorem, (http://en.wikipedia.org/wiki/Earnshaw%27s_theorem) there are no stable equilibrium configurations for particles with Coulomb potential (gravitational potentialis identical to the Coulomb one). Indeed, if there were a point Pwhich is a stable equilibrium for positive charges, in the immediate vicinity of P, all the field lines need to bedirected towards P, because otherwise, a positive charge would escape from P along the outgoing field lines. This, however, would be in contradiction with the Gauss law fora small spherical neighbourhood of P: the flux of the force field needs to be negative (there are only incoming field lines), but equals strictly to zero for Coulomb potentials.Here we hope that L4 will be a stable equilibrium in the system of reference co-rotating with the two heavy masses; in that system, there is also the force field of thecentrifugal force. Unfortunately, centrifugal force is of no help, because it leads to the creation of field lines in vacuum, making the flux around P strictly positive (recall thatstability requires a negative flux). Now, let us recall that besides the gravitational and centrifugal forces, we have also the Coriolis force, which acts, however, only onmoving bodies. Hence, the stability can be created only by the Coriolis force!36
  • 37. Unfortunately, the Coriolis force is not included into the Syllabus of IPhO. (http://ipho.phy.ntnu.edu.tw/syllabus.html) Quite often, the usage of Coriolis force can beavoided, most typically by using non-rotating systems of reference (the origin can move along a circle, though), or studying only potential energies (Coriolis force does notperform work). Here, however, neither of these tricks can be used: the system of reference needs to rotate (because the net gravitational field is stationary only in such asystem), and as we saw, we cannot work with the potentials only, because the Coriolis force is needed to achieve the stability. The authors of the problem believed to havebeen found a work-around: assume that there is an approximate conservation of the angular momentum (with respect to the centre of mass O of the whole system, cf.Figure) of the small body, and apply the method of generalized coordinates: if the radial displacement from the equilibrium point L4 (or L5; marked in Figure as P) is usedas the coordinate, the tangential velocity can be expressed via the radial one , allowing us to write down the energy balance equation (recall that the Coriolisforce cancels out from that equation as it does not create any work). From that equation, one could immediately obtain the circular frequency of small oscillations. However,we have made two mistakes here. First, the angular momentum is not conserved, even not approximately. Indeed, angular momentum is conserved if the force field isrotationally symmetric. However, a superposition of the gravitational fields of two point masses has no such symmetry. Approximate conservation of would require thatsuch a symmetry is local: near L4, the curvature radius of the equipotential surface needs to be equal to the distance from the origin O; regrettably, this is also not thecase. Second, the gravitational energy depends not only on the radial coordinate , but also on the tangential displacement from L4; note that there is no way ofexpressing via , even the (non)conservation of the angular momentum is useless.37
  • 38. So, how to obtain a correct solution to this problem – what is the frequency of small oscillations around the Lagrangian point L4, assuming that the two heavy masses areequal? Well, we just need to follow the standard way of doing such things: first, we write down the equations of motion for both coordinates, and , and second,use linear approximation (which is valid for small displacements), ie. neglect the terms which involve second and higher powers of and ; when working with thegravitational potential, this corresponds to neglecting the terms with third and higher powers. In such a way, we obtain linear equations of motion. The third step is to findthe eigenfrequencies of that system of equations, ie. such values of that with a proper choice of , the equations will be satisfied with and . Ifthere is at least one eigenfrequency with a positive imaginary part then the system is unstable. On the other hand, if for all the eigenfrequencies , thesystem is stable (unless there is an eigenfrequency , in which case the linear analysis is not sufficient for proving stability). As afourth useful idea, let us notethat with more than one point mass, it is much more convenient tocalculate the gravitational potential, rather than the resultant gravitational force.According to what has been said, we need an expression for the Coriolis force. Of course, we could just take a ready formula, but it would be better to understand how itcan be obtained (if you are not interested, please skip this part). And so, consider a system of reference, which rotates around the origin with an angular velocity (thevector defines the rotation axis according to the corkscrew rule). Consider a point , which is motionless in the rotating system, and let us denote . In the labsystem of reference, the point moves with velocity , and when studying the direction of the velocity , one can see that . Now, if the pointmoves in the rotating system of reference with velocity (let us use to measure the time in the rotating system), then this additional velocity needs to be added towhat would have been for a motionless point:So, we can conclude that the time-derivatives of vectors in rotating and lab systems of reference are related via equalityThis is written in the form of an operator, which means that we can write any vector (eg or ) rightwards of all the three terms. In particular, we can apply this formula tothe right- and left-hand-sides of the equality :38
  • 39. Here we need to bear in mind that when taking derivatives of vectors and products of vectors, all the well-known rules can be applied; inparticular, and . We also need the rule for the double cross product, ; you canmemorize this equality by keeping in mind that the double product is a linear combination of the vectors from the inner braces, and that the sign + comes with the vectorfrom the middle position. And so, bearing in mind that and , and assuming that , we obtainLet us recall that is the acceleration of the point as seen in the lab system of reference, and is the same as seen in the rotating system of reference. Now, if is apoint mass , and there is an external force acting on , then and hence,ie. in the rotating system of reference, the body behaves as if there were additional forces: the Coriolis force , and the centrifugal force .Now we are finally ready to tackle the IPhO Problem 2010-1-iii. As mentioned above, the first step is writing down the potential energy in the rotating system ofreference (see Figure above):note that the last term corresponds to the potential energy of the centrifugal force. When working with this potential energy, we can forget about the constant part of it;additionally, we can also forget about the linear part, because it gives us the force, which is exactly zero: our point L4 has been chosen so as to provide an equilibrium.Owing to that equilibrium, we have also equality . We approximate the potential using the formula (which includes the first two termsof the Taylor expansion); keeping in mind that we obtainThis can be further simplified:39
  • 40. hence, the resultant force of the gravitational and centrifugal forces can be written as and . Component-wise, the Coriolisforce can be written as , . Finally, the equations of motion can be written asNow we can proceed with the final step, finding the eigenfrequencies. We look for the solutions in the form , , upon substituting theseexpressions we obtainThis is a quadratic equation for , which results in This can be brought to the formSo, we can conclude that due to the presence of an unstable solution ,the equilibrium point L4 (and L5) is not stable in the case of a binary gravitational system with two equal masses.Final notes. This mosaic tile is different from the others in that it is not motivated by a (more or less) universal problem solving technique or an important physical concept;instead, it is mainly aimed to clarify a single IPhO problem. The assumptions of physics contest problems dont need to be entirely correct. However, for physicists, it is veryimportant to be aware, how firm or loose are the assumptions of their study, and to which degree can the the conclusions of their study be affected by the mismatchbetween the assumptions and the real life. Studies based on wrong assumptions can be useful, but the fact that the assumptions are not valid needs to be emphasized. The40
  • 41. contest problems serve mostly educational purposes and are no different – if an invalid assumption is made, it should be clearly pointed out, and, if possible, explained whyan incorrect assumption was made. Of course, no-one is secured against accidental mistakes; in particular, the more interesting your newly invented problem is, the higherare the chances that there are some mistakes. Meanwhile, the IPhO problems serve as a well-tested pool of exercises, tested by the contestants and leaders of manycountries, and it is better to make sure that there are no unresolved issues in these problems. This is the reasoning which led to the current mosaic tile. Although we are notable to close here the list of all such problems (for instance, there are problems1988-2-iv and 2000-3-iv,v; you can let me know if you found out what is wrong there), morerecent problems get typically more attention.—Jaan Kalda, Academic Committee of IPhO-201241
  • 42. 5. Images or roulette?Suppose you are given a problem and you dont know how to really solve it; however, your intuition tells you that the answer is that and that (you can also call it "aneducated guess"). Is such an answer acceptable? If you are an experimentalist, it would be perfectly fine: who cares how you got your equipment working, as long as itworks! If you are a theoretical physicist, it would be not really good, but if you find some arguments to qualitatively motivate your answer, you can call it "a conjecture", andit will be better than nothing. If you are a mathematician, no-one will care about your guess-work. However, there are cases when a guess is as good as amethodically obtained result: when it is known (has been proved) that there is a unique solution to the problem, and you are able to show that your solution does,indeed, satisfy all the requirements. Such an approach is acceptable even for mathematicians! In physics, this method is mostly known as the method of electrical images.To begin with, let us consider the simplest and most classical problem on electrical images (the Problem 1): suppose there is an infinite conducting plane andat , there is charge . Find (a) the charge surface density of the induced charges at ; (b) the interaction force between the plane and thecharge; (c) the net charge induced at the conductor surface .Intuitively, it is quite clear that the problem is well-defined, ie. it should have a unique solution. Let us analyse it in mathematical details. First, the electrostatic field iseverywhere potential ( , or equivalently, ; just skip what is written in braces if you dont understand it) and second, in the half-space , exceptfor the point , it is source-free ( , or equivalently, ). These two conditions form a closed set of differential equations (in partial42
  • 43. derivatives; more specifically, owing to the first condition, the electric field can be expressed via an electrostatic potential, , due to the second condtition,). Now, in order to have a unique solution, we need appropriate boundary conditions (which correspond to initial conditions for ordinary differential equations) at theboundary of that region of space where we need to find the field (this region is marked with grey in the Fig.). In the case of our problem, the boundary consists of threeparts: (a) the conductor surface , where ; (b) the point occupied by the charge , around which ;(c) infinitely remoteregion where . Comparing (a) and (c) we can conclude that at the conductor surface, .Mathematicians have proved that the problem of finding a potential source-free field ( ) in a certain space region will have a unique solution, if each contiguousboundary segment of that space region has either (i) a fixed and known value of the potential , or (ii) a constant (but unknown) value of the potential,and a known total flux of the field . Note that boundary itself is excluded from the region where we need to find the field;however, the points of the boundary are at a zero-distance from the region.Using the mathematical terms, the formulation may seem somewhat obscure, but in physical terms, it is very simple: in order to have a unique solution to the problemof finding the electrostatic field in a certain region of space, the boundary can consist of two type of elements: (a) electrical charges, the values of which areknown, and (b) electrical conductors, for which one value out of two needs to be given (the other value will be found as a part of the solution): (i) the netcharge; (ii) the potential.Now, if we look back at our problem, it is easy to see that everything is fine: we know the potential of the infinite conducting plane (the net charge is yet to be found), andthe stand-alone charge is also known. What is left to do, is to construct such a field which will satisfy all the boundary conditions and is obtained as asuperposition of fields which are know to be potential and source free in our space region (then, the superposition will be too, owing to the superposition principle,which is valid for linear differential equations, such as ). As for the component-fields which we are going to use for the construction of the solution, we dont havemuch choice – we can use the fields of point charges, but the charges need to be placed outside the space region of interest, because the field of a point chargeis not source free at that point where the charge resides. If necessary, we can use also a homogeneous constant field, or the field of a homogeneously charged rod (the rodneeds to be outside the region).For the first problem, the task is easy: it is just enough to place one virtual charge at (blue in Fig.) to ensure that when superimposed to the fieldof the real charge at , the resultant potential is zero at the entire surface of the conductor [since we keep a charge at (red in Fig.), theboundary condition at that point is satisfied, too]. Note that in reality, there is no charge at : all the real charge is induced at the surface of the conductor,only the field in the region is as if there were a charge at . To sum up, at the electric field (we knew it from the verybeginning!), and at , the electric field is such as if there were a charge at .43
  • 44. In order to calculate the net charge induced at the surface of the conductor [question (c)], let us consider the flux of electric field through a very large sphere of divergingradius, centred around the charge (in Fig, orange circle ). In the region , the field is that of a dipole (the pair of red and blue charges in Fig), hence vanishesas . Meanwhile, the surface area of the sphere grows as ; therefore, the field flux is , ie. becomes zero for an infinite sphere (here means "proportional to").According to the Gauss law, this means that the sum of real charges inside the sphere is zero, which means that the net charge on the conductor surface must be equalto , to compensate the charge at . So, the real induced charge is equal to the image charge at . This is a universal result (consequenceof the Gauss law): the sum of image charges inside a conductor with a given potential equals to the net charge induced on the surface of that conductor.In order to answer the question (a), we consider a small cylindrical surface of a cross-section area and a negligible height, which is positioned at , coaxial withthe real and virtual charges (in Fig, green rectangle ). The electric field flux through that cylinder includes only the flux through that bottom of it which is turned towardsthe real charge (the side surfaces are small and the other bottom is inside the conductor, where ): . Note that stands for the charge inside thecylinder, and is the field at , ie. the superposition of the fields of the real and virtual charges. So, we finally obtain (we haveprefixed – to reflect the fact that the induced surface charge is of opposite sign.Finally, to answer the question (b), let us note that the electrostatic force acting on a charge depends only on the value of the electric field at the position of the charge(neglecting the field of the charge itself); since the field is such as if there were a charge at , the force must be also the same what would be if there werea charge at , ie. .Now it is becomes also clear, why the method is called the method of electrical images: plane conductor surfaces work as mirrors: we need to put virtual charges in theposition of those optical images of the real charges which would appear, if the conductor surfaces were mirrors. This is valid not only for a single infinite conductor surface,but also for configurations depicted in Figures below.44
  • 45. However, the analogy is not perfect, and does not work rigorously in the case of curved surfaces. It appears that in the case of spherical surfaces, the situation is actuallybetter than in the case of geometrical optics when the images are not perfect points, due to spherical abberations; this will be discussed in next paragraphs.Now, let us study the electric field created by a grounded conducting sphere of radius , together with a point charge at a distance from the centre of the sphere(Problem 2). Let us take the origin at the centre of the sphere; then the region where we need to find the electric field is , excluding the point where the external45
  • 46. charge resides. The boundary conditions are as follows: a fixed charge at , and at (which are good enough to provide a unique solution, cfcriteria given above). Solving this problem using the method of electrical images is possible owing to a non-trivial mathematical fact: for an arbitrary pair of positive annegative point charges, the surface of zero potential has the shape of a sphere (in the degenerate case of equal by moduli charges, the sphere has an inifiniteradius, ie. becomes a plane). So, we expect that for any , and (see Fig.), we can find such and that for any point on the sphere,In order to prove that this, indeed, the case, let us note that the green and orange triangles (in Fig. below) have one common angle, ie. the position of the image charge canbe chosen so that the green and orange triangles will be similar to each other. This will happen if the ratios of the respective sides of the green and orange triangles areequal, ie. if . Furthermore, due to the similarity of the triangles, ; combining this with the condition resultsin . Since these values of and are independent of and , the green and orange triangles remain similar (with the same similarity ratio), and the condition remains satisfied for any point on the sphere.This means that indeed, with an image charge , which is placed at thedistance from the centre of the sphere towards the external charge, all the boundary conditions are satisfied, ie. this configuration of charges produces the actualelectric field outside the sphere. Also, we can conclude that the net charge induced on the surface of the grounded sphere equals to .Now we can easily solve a problem of reversed geometry (Problem 3): suppose there is a charge inside a hollow conducting charge-less sphere of radius ; we ask, whatis the interaction force between the charge and the sphere, and what is the electric potential of the sphere.46
  • 47. For the inside region of the sphere, the boundary conditions are: (a) constant potential at the spere; (b) charge at the given point. It is easy to see that these conditionscan be satisfied with the charge placement from the previous problem, only the real and image charges swap the places. So, we need to put an image charge at thedistance from the centre of the sphere, with . So, the interaction force is .For the outside region of the sphere, the boundary conditions are: (a) constant potential at the spere; (b) net charge inside the sphere. It is easy to see that theseconditions will be satisfied with a field created by a point charge at the centre of the sphere (the respective fieldlines are depicted as blue dotted lines in Fig.). Due to theuniqueness of the solution, this is the actual field outside the sphere; hence, the potential of is given by the potential of a point charge, . A useful conclusion isthat the charge distribution inside a closed conducting vessel cannot be determined by outside observations.As a Problem 4, let us consider the interaction of an isolated electrically neutral conducting sphere with a point charge at distance from the centre of the sphere. Theonly difference from the Problem 2 is that the sphere is electrically neutral and isolated. From the solution of Problem 2 we know that putting image chargeat distance from the centre of the sphere yields a zero potential for any point on the sphere. Now we can add more image charges inside the sphere, but we needto keep the sphere surface equipotential. The only place we can put an image charge and satisfy this condition, is the centre of the sphere. On the other hand, the net chargeof the sphere is the sum of image charges; we already have the first image charge , hence, if we put a second image charge into the centre of the sphere, thiscondition will be satisfied, too!47
  • 48. Next, let us determine the surface charge density for a conducting cylinder, placed into a homogeneous electric field , which is perpendicular to the axis of the cylinder,also to used as the -axis (Problem 5). If there were a sphere instead of the cylinder, it could be studied as a limit case of Problem 4, when the point charge is at an infinitedistance , with charge equal to . In 2D geometry, however, point image charges (actually, homogeneously charged wires) are no good, because then thepotential is a logarithmic function of the distance, and for the system of two parallel wires, there are no equipotential surfaces with the shape of a cylinder.Solution to this problem, will be constructed step-by-step. First we find the electric field inside a dielectric cylinder of radius and homogeneous volume charge density ;we assume that the dielectric permeability . Writing the Gauss law for a coaxial cylindrical surface of radius and height , we obtain , from where .Here we have used vector notation to express the fact that the electric field is radial ( is a 2D vector laying in the -plane, pointing from the axis to the current point).The next step towards the solution is studying two cylinders of opposite volume charge densities , and finding the electric field inside the region where the cylindersoverlap (see Fig.). Using the last equation, we can write .48
  • 49. .The result is a constant vector, which means that the electric field inside the region of overlap is homogeneous. Now it becomes also clear, how to obtain the solution to theProblem 5: we take , and let and in such a way that the product of the two will remain constant,The result will be a homogeneous electric field inside the cylinder, and no volume charge ( and cancel out). The region where the cylinders dont overlap becomes verythin, so that the volume charge becomes a surface charge . The surface charge density , where is the thickness of the layer and is the polarangle between the electric field, and the radius vector pointing from the centre of the cylinder to the observation point, see Fig. Finally, what is left to do to obtain theelectric field in the case of the Problem 5, is to add an everywhere homogeneous electric field . Inside the cylinder, this cancels out the field created by the cylinders(but not outside). This field (the superposition of the homogeneous field and the fields of the two almost overlapping cylinders) satisfies all the boundary conditions. Infact, there is only one condition: near the outer surface of the cylinder, the perpendicular to the surface component of the field must vanish. This is, indeed, the case,because the tangential component of the electric field is zero inside the cylinder (due to our construction), and remains unchanged when the observation point crosses thethin surface charge layer (this is a consequence of the circulation theorem, written for a small loop embracing a piece of the surface charge). Note that here we actuallydidnt need virtual image charges: we managed to guess directly the surface charge distribution ;due to the uniqueness of the solution, this is the real surface charge distribution. Note that outside the cylinder, the field is the same as if there were an ideal 2D dipole.Indeed, this field is the superposition of the fields of two cylinders of opposite volume charge; outside the cylinders, these fields are the same what would be if there weretwo wires of linear charge density ; in our case , ie. the distance between the wires approaches zero, ie. we result in an ideal 2D dipole, the momentumof which is .49
  • 50. In order to close the topic of Problem 5, let us note that the technique applied here can be also used for a conducting sphere in an homogeneous electric field, as well as fora superconducting cylinder in an homogeneous magnetic field. In the case of magnetic field, we need to find the surface current distribution; we use the superposition of twocylinders with counter-directed currents densities ; for a convenient vectorial representation we can express the magnetic inductions produced by the cylinders by usingthe cross product as (the induction in the region of overlap will be ).The last problem (Problem 6) of this mosaic tile is the problem of electromagnetic wave propagation between two perfectly conducting parallel plates, in which case nearthe plates, the tangential electric field is always zero. Let the plates be at positions , and the let us try to find the form of waves, polarized along the -axis, andpropagating in the direction of the -axis. Similarly to what we did in the case of the electrostatic problems above, we try to construct the solution using the fields which areknown to satisfy the underlying equations. In fact, there is probably only one relevant field which we know – that of a plane electromagnetic wave:where denotes the unit vector along the -axis; the ratio of the circular frequency and wave vector equals to the speed of light, . Note that when weexpress as a complex number, we assume that the real, physically observable quantity is only the real part of it; we keep the complex part just for convenience: then wecan use exponential functions instead of sinusoidal functions, and exponential functions have very convenient mathematical properties. It should be emphasized that suchcomplex representation works well only as long as we deal with linear superpositions (because the real part of a sum equals to the sum of real parts), ie. we dont need todeal with nonlinearity. Luckily, the Maxwell equations in vacuum are linear.The idea is to construct a standing wave in -direction via a superposition of two plane waves:Now it is easy to see that the boundary condition at is satisfied, for instance, with :50
  • 51. It should be noted that unlike with the previous problems, solution of this boundary value problem is not unique: here, the number of standing waves across the slit cannotbe determined from the boundary conditions (it depends also on which wave source is placed in the slit. Nevertheless, since we have used valid solutions of the Maxwellequations for building our superposition, and the boundary conditions are satisfied, we can be sure that such a wave can, indeed, propagate in the slit.It is also useful to study the dispersion relationship of such a wave – which propagates in the waveguide formed by the two plates. To that end, let us recall that .What might seem surprising is that such a phase velocity exceeds the speed of light. However, energy (and information) transfer rate is determined by the groupvelocity . To find the group velocity, let us rewrite the dispersion relation in the form , and differentiate it: , hence .Finally, let us mention that such a method – finding the solution by guessing and using the uniqueness of the solution – is not limited to the problem of finding the electricfield. Another possibility, for instance, is the problem of finding the current distribution in conducting media. In particular, the equivalent scheme proposed in IPhO-2002problem No 2 (http://ipho.phy.ntnu.edu.tw/problems-and-solutions/2002/IPhO_2002_Theoretical%20Question%202.pdf) can be derived using this technique.—Jaan Kalda, Academic Committee of IPhO-201251
  • 52. Problem 0What is the diameter of the lens which was used to make the photo below?Remark: although the photographic lenses are made of several optical components, for many practical calculations – including this problem – they can be considered as idealthin lenses.52
  • 53. SolutionAlthough this was not a competition problem, two correct solutions have been submitted: by Dinis Cheian and Ivan Ivashkovskiy.The Ivans solution is very thorough – with error analysis (as should be for such a semi-experimental problem) and is presented below on two images. Note that hedefines n as the ratio of two lengths, him and l, which he can measure from the photo and thus calculate n. On the other hand, he finds expressions for him and l, and usingthese expressions shows that n =H/D. Knowing the object lenth H, he can find now D.In addition to that solution, I have two comments. First, the usage of the Newton formula would have simplified slightly the mathematics: x1x2 = F2,where x1 =d – F and x2 = f – F are the distances of the object and of the image from the respective focal planes (using Ivans notations, d and f are the respective distancesfrom the lens). So, the diameter of the circle of confusion l = Dx2 / F = FD / x1 = FD / (d – F). Second, pay attention to the main result: the diameter of the lens equals to thediameter of the circle of confusion created by a far-away dot-source, if measured by the image of an in-focus ruler. So, if you have a subject for your photo (eg.friends face for a portrait) and you want to blur the (far-away) background, the degree of the background blur is defined purely by the diameter of the lens: a tele-lens600mm/4 creates circles of confusion of the size of the face, leaving no visible details at the background; a point-and-shoot camera with a small sensor and 8mm/4 lenscreates almost no background blur: the diameter of the circles of confusion are smaller than the pupils of the portrait.Finally, let us analyse the benefits of a large camera: you have a full-frame DSLR, and your friend has a small point-and-shoot camera of a 4 times smaller sensor (in linearsize). Your friend takes a photo of something, zooming to the focal length of 13 mm and is using the full aperture of F/4 (ie. the focal-length-to-diameter ratio equals to 4).You want to obtain exactly the same result: in order to have the same angle of view (and perspective), you need to take the focal length equal to 13*4 = 52 (your50mm/1.4 standard prime lens works well). In order to have the same blur, you need to shut down the lens — down to the aperture F/16 (using the diaphragm you decreasethe effective diameter). On the other hand, if you take a photo at the full aperture of F/1.4, your friend would need the aperture of F/0.35, which is theoretically impossible(would violate the second law of thermodynamics). Meanwhile, if you want to have both sharp foreground and sharp background, the point-and-shoot is better: you can useF/22, which would correspond to F/88 for the DSLR. While theoretically this is possible, the smallest aperture is typically only F/32 (starting from ca F/16, diffraction startsdegrading the image).- Jaan Kalda – Academic Committe -53
  • 54. 54
  • 55. 55
  • 56. Problem No 1A ballistic missile is launched from the north pole, the target is at the latitude Phi (>0, if northern hemisphere, <0 otherwise). At which angle (with respect to the horizon)the missile needs to be launched in order to have minimal launch velocity of the missile?Remark: ballistic missile is like a stone: you give its initial velocity, and it moves due to inertia; you can neglect the air friction.NB! When submitting your answer, please use the subject line "Problem No 1".If you are not yet registered, you can still register and participate: just send the registration info (name, date of birth, etc.) together with your answer.Finally, what you may assume well-known and what not: things which are not in the formula sheet (http://www.ipho2012.ee/physicscup/formula-sheet/) need to bemotivated/derived. (You dont need to check – your typical high-school formulae are there.)56
  • 57. Results after Problem 1The list of the contest leaders after the first problem:Points Name Country School Physics teacher Preparatory School for Science & Technology of4.8156 Brahim Saadi Algeria Derradji Nasreddine Annaba2.5937 Dinis Cheian Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev2.5937 Mikhail Shirkin Russia Gymnasium of Ramenskoye P Petrova Elena Georgyevna2.1436 Ivan Tadeu Ferreira Antunes Filho Brazil Colégio Objetivo, Lins, São Paulo1.7538 Cristian Zanoci Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev1.6105 Jakub Supeł Poland 14th School of Stanisław Staszic, Warsaw Włodzimierz Zielicz1.6105 Nikita Sopenko Russia Lyceum No.14, Tambov Valeriy Vladimirovich Biryukov 1.1 Kohei Kawabata Japan Nada High School 1.1 Lars Dehlwes Germany Ohm-Gymnasium Erlangen Mr. Perleth 1.1 SZABÓ Attila Hungary Leőwey Klára High School, Pécs Simon Péter, Dr Kotek László 1 Ion Toloaca Moldova liceul "Mircea Eliade" Igor Iurevici Nemtov; Andrei Simboteanu 1 Jakub Šafin Slovak Pavol Horov Secondary, Michalovce Jozef Smrek 1 Jaan Toots Estonia Tallinn Secondary Science School Toomas Reimann 1 Luís Gustavo Lapinha Dalla Stella Brazil Colégio Integrado Objetivo, Barueri, Brazil Ronaldo Fogo 1 Lev Ginzburg Russia Advanced Educational Scientific Center, MSU, Moscow I.V. Lukjanov, S.N. Oks A.R. Zilberman, G.F. Lvovskaya, G.Z. 1 Alexandra Vasileva Russia Lyceum "Second School", Moscow Arabuly 1 Task Ohmori Japan Nada High School T.Hamaguchi 1 Ilie Popanu Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev 1 Sharad Mirani India Prakash Higher Secondary School Ruchi Sadana, Sunil Sharma0.9801 Mekan Toyjanow TurkmenistanTurgut Ozal Turkmen Turkish High School Halit Coshkun57
  • 58. 0.8733 Papimeri Dumitru Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev 0.81 Meylis Malikov TurkmenistanTurgut Ozal Turkmen Turkish High School Halit Coshkun 0,81 Bharadwaj Rallabandi India Narayana Jr. College, Basheer Bagh, India Vyom Sekhar Singh 0.81 Liara Guinsberg Brazil Colégio Integrado Objetivo, São Paulo, Brazil Ronaldo Fogo0.6561 Nadezhda Vartanian Russia Smolensk Pedagogical Lyceum Mishchenko Andrei AnatolievichThe list of those who completed successfully the problem, ordered according to the arrival time of the correct solution:1. Dinis Cheian (Moldova),2. Mikhail Shirkin (Russia),3. Ivan Tadeu Ferreira Antunes Filho (Brazil),4. Cristian Zanoci (Moldova),5. Brahim Saadi (Algeria).6. Nikita Sopenko (Russia)7. Jakub Supeł (Poland)8. Papimeri Dumitru (Moldova)9. Mekan Toyjanow (Turkemistan)10. Kohei Kawabata (Japan)11. Lars Dehlwes (Germany)58
  • 59. 12. Meylis Malikov (Turkemistan)13. Szabó Attila (Hungary)14. Jakub Šafin (Slovakia)15. Ion Toloaca (Moldova)16. Jaan Toots (Estonia)17. Bharadwaj Rallabandi (India)18. Luís Gustavo Lapinha Dalla Stella (Brazil)19. Lev Ginzburg (Russia)20. Liara Guinsberg (Brazil)21. Alexandra Vasileva (Russia)22. Task Ohmori (Japan)23. Ilie Popanu (Moldova)24. Nadezhda Vartanian (Russia)25. Sharad Mirani; (India)59
  • 60. The best solution: Brahim Saadi.Besides, there was one almost correct solutions and 23 wrong solutions. The overall number of registered participants: 186.———-NB! Dont forget to try solving Problem No 2, the faster ones will get more points!Jaan Kalda – Academic Committee of IPhO-201260
  • 61. SolutionI hope you found the first problem to be interesting. I am fond of it myself, because (a) it extends nicely the widely-known fact that 45 degrees is the optimal throwingangle; (b) is exactly solvable regardless of seeming difficulty; (c) optimal solution is technically quite simple; (d) the answer has a real practical relevance. According to theresults, it was not a simple one; still the difficulty was perhaps close to optimal – it kept the most of you busy the whole month, and still 13% of you were able to solve it.The second problem is probably at least as difficult (maybe the third one will be simpler).Regardless of the relative difficulty, there was one of you, for whom it was not difficult enough, and who solved a more generic star-war-problem: the launching site andtarget are at different distances from the centre of Earth. This is not a cosmetic change, because the problem becomes non-symmetric. Solution-wise, it adds one more step,which makes use of the geometric property of hyperbolas. For a professional physicist, it is really important to be able to figure out if the problem (or model) under study canbe made more generic while maintaining solvability, so for this problem, we have a clear winner of the best solution. (However, I am inclined to think that for the major partof the contest problems, it will be impossible to make such nice, non-cosmetic generalizations, which would justify giving the award of the best solution.)And so, the award for the best solution, a bonus factor e, goes to Brahim Saadi.The second-best solution, the one made by Mikhail Shirkin, is actually the one I had in mind when giving you the problem. Mikhail has written it down in a very laconicway: for a research paper, this writing style is definitely not appropriate, but for the solution of an Olympiad problem, it is just perfect! So, I am giving him a bonus factorof 1.1.Another factor of 1.1 goes to the solution of Szabó Attila, which represents a good style of research papers: the model assumptions are clearly formulated, formulae areput into correct sentences, text is understandable even for people who are not very well-prepared.There are two more solutions which I found useful to show you (both will also get a factor of 1.1). The first one is of Lars Dehlwes, who had made a useful graph of howthe minimal velocity depends on the latitude of the target (note that small differences in the required velocity imply actually a large economy, because the fuel mass of themissile depends exponentially on the terminal velocity, cf. Eq IV-21 of the latest formula sheet).The second one is of Jakub Supeł, who was not the only one to derive the formula E=-GMm/2a, but was among the first ones to do so, and did it nicely in LaTeX. Whileactually you did not need to derive this formula as it is in my formula sheet, it is useful for you to know, how it is done.61
  • 62. Many of you (majority, in fact) used the "brute force" approach: express the launch velocity (or the square of it) via some parameter (eg. launch angle or ellipticity of theorbit) and find the minimum from the condition that the derivative is zero. This was definitely a difficult way of doing it, but I was quite amazed by your technical skills!There was not a single mistake in very long mathematical manipulations, ending up in perfectly correct results! (Though, some final answers were left non-simplified.)Last but not least, what is the lesson of this problem? First, quite often, problems on extrema can be solved without taking derivatives, geometrically, which is typicallya much simpler way. Second, when solving the problems put on the Keplers laws,the geometrical and optical properties of ellips (see Eq XII-7 on the formula sheet; by theway, these properties are connected with each other via the Fermats principle) are always very useful. Third, expression for the full energy, E=-GMm/2a is extremelyhandy (Id like to call it the Keplers fourth law — but it was not derived by Kepler.)And one more thing: one of you, Comoglio Lorenzo, pointed out that there is a satellite simulation written in Java, if you want to play with trajectories, have a look.Jaan Kalda – Academic Committee of IPhO-2012———————Best solutions.Brahim Saadi:62
  • 63. 63
  • 64. 64
  • 65. 65
  • 66. 66
  • 67. Mikhail Shirkin:67
  • 68. Szabó Attila:68
  • 69. Lars Dehlwes:69
  • 70. Jakub Supeł:70
  • 71. Problem No 2Consider a brick-shaped ferromagnetic of relative magnetic permeability µ >> 1, which has dimensions 2a × 2a × a and a narrow slit of width d and depth a sawed into it asshown in Figure. You may assume that µd>> a >> d. A circular loop of diameter a and inductance L, made of a superconducting material, is put into that slit; the loopcarries electric current I. What is the mechanical work A needed to be done in order to pull the loop out of the slit and move it to a large distance from the ferromagnetic?Remarks: (a) the inductance of the loop equals to L when it is far away from the ferromagnetic. (b) The current in the loop equals to I when the loop is inside the slit. (c) Youmay assume that the hysteresis of the ferromagnetic is negligible, and µ is constant (independent of B).71
  • 72. Results after Problem 2The list of the contest leaders after the second problem:Points Name Country School Physics teacher Preparatory School for Science & Technology of5.6695 Brahim Saadi Algeria Derradji Nasreddine Annaba4.9694 SZABÓ Attila Hungary Leőwey Klára High School, Pécs Simon Péter, Dr Kotek László4.5041 Nikita Sopenko Russia Lyceum No.14, Tambov Valeriy Vladimirovich Biryukov4.4353 Ivan Tadeu Ferreira Antunes Filho Brazil Colégio Objetivo, Lins, São Paulo2.7363 Jakub Šafin Slovak Pavol Horov Secondary, Michalovce Jozef Smrek2.6785 Lars Dehlwes Germany Ohm-Gymnasium Erlangen Mr. Perleth2.5937 Dinis Cheian Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev2.5937 Mikhail Shirkin Russia Gymnasium of Ramenskoye Petrova Elena Georgyevna2.435 Ilie Popanu Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev1.9801 Luís Gustavo Lapinha Dalla Stella Brazil Colégio Integrado Objetivo, Barueri, Brazil Ronaldo Fogo A.R. Zilberman, G.F. Lvovskaya, G.Z.1.9703 Alexandra Vasileva Russia Lyceum "Second School", Moscow Arabuly1.81 Ion Toloaca Moldova liceul "Mircea Eliade" Igor Iurevici Nemtov; Andrei Simboteanu1.7643 Papimeri Dumitru Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev1.7538 Cristian Zanoci Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev1.6105 Jakub Supeł Poland 14th School of Stanisław Staszic, Warsaw Włodzimierz Zielicz1.1 Kohei Kawabata Japan Nada High School1 Jaan Toots Estonia Tallinn Secondary Science School Toomas Reimann1 Lev Ginzburg Russia Advanced Educational Scientific Center, MSU, Moscow I.V. Lukjanov, S.N. Oks1 Sharad Mirani India Prakash Higher Secondary School Ruchi Sadana, Sunil Sharma1 Task Ohmori Japan Nada High School T.Hamaguchi72
  • 73. 0.9801 Mekan Toyjanow TurkmenistanTurgut Ozal Turkmen Turkish High School Halit Coshkun0.81 Bharadwaj Rallabandi India Narayana Jr. College, Basheer Bagh, India Vyom Sekhar Singh0.81 Liara Guinsberg Brazil Colégio Integrado Objetivo, São Paulo, Brazil Ronaldo Fogo0.81 Meylis Malikov TurkmenistanTurgut Ozal Turkmen Turkish High School Halit Coshkun0.6561 Nadezhda Vartanian Russia Smolensk Pedagogical Lyceum Mishchenko Andrei AnatolievichPoints for Problem No 2:Pts Name Country3,8694 SZABÓ Attila Hungary2,8935 Nikita Sopenko Russia2,2917 Ivan Tadeu Ferreira Antunes Filho Brazil1,7363 Jakub Šafin Slovak1,5785 Lars Dehlwes Germany1,435 Ilie Popanu Moldova0,9801 Luís Gustavo Lapinha Dalla Stella Brazil0,9703 Alexandra Vasileva Russia0,891 Papimeri Dumitru Moldova0,8539 Brahim Saadi Algeria0,81 Ion Toloaca MoldovaCorrect solutions (ordered according to the arrival time):1. Szabó Attila (Hungary).73
  • 74. 2. Ivan Tadeu Ferreira Antunes Filho (Brazil)3. Nikita Sopenko (Russia)4. Jakub Šafin (Slovak)5. Lars Dehlwes (Germany)6. Ilie Popanu (Moldova)7. Brahim Saadi (Algeria)8. Alexandra Vasileva (Russia)9. Luís Gustavo Lapinha Dalla Stella (Brazil)10. Papimeri Dumitru (Moldova)11. Ion Toloaca (Moldova)Overall number of registered particpants: 204 (from 38 countries).During the first week, only one correct solution has been submitted (by Szabó Attila). So, the problem was judged to be very difficult, and after the first week, few hintswere published:"You need to understand how to calculate fields using the circulation theorem and Gauss law; relevant formulas (from the formula sheet) are IX-2, IX-3 and IX-6; I alsorecommend studying the formulae VIII-8, VIII-9, VIII-13, IX-27, IX-28, and IX-29. And finally – this is a common mistake – the inductance of a coil is not always multipliedby a factor of when you supply it with a ferromagnetic core (in fact, it is multiplied by only for very specific cases); the inductance will depend on the geometry of theferromagnetic!"74
  • 75. During the second week, two more correct solutions were received: by Ivan Tadeu Ferreira Antunes Filho and Nikita Sopenko. So, another hint was added:"In particular, it would be helpful to study the magnetic field created by electric transformers with closed (for instance, toroidal) ferromagnetic cores; Wikipedia is not toouseful (there is no calculation of B), except for the figure for leakage flux (which is small/negligible, if µ is large)."During the first half of the third week, the correct solution of Jakub Šafin was received. After that, on 3rd Nov, another amendment to the hints was made, so that thefinal wording was as follows:"you need to understand how to calculate fields using the circulation theorem and Gauss law; relevant formulae (from the formula sheet) are IX-2, IX-3 and IX-6; I alsorecommend studying the formulae VIII-8, VIII-9, VIII-13, IX-27, IX-28, and IX-29. In particular, it would be helpful to study the magnetic field created by electrictransformers with closed (for instance, toroidal) ferromagnetic cores; suggested reading from Wikipedia: figure for leakage flux (which is small, if µ is large)(http://en.wikipedia.org/wiki/File:Transformer_flux.gif); how to deal with magnetic circuits. (http://en.wikipedia.org/wiki/Magnetic_circuit) Please bear in mind thatyou are not supposed to copy directly formulae from the latter article, because the shape of our ferromagnetic brick differs from a simplified model of a closed-coretransformer; instead, it should be considered as a reading which helps you understand, what is going on with the B-field in our case, and how to correctly apply thecirculation theorem.————————Jaan Kalda – Academic Committee of IPhO-201275
  • 76. SolutionI expected this problem to be of the same difficulty level as the Problem No 1. However, it turned out to be more difficult – probably because typically in high schools,magnetism is not taught as well as mechanics. On the other hand, the problem, indeed, tests the knowledge of several things: (a) the property of magnetic materials to"attract" the magnetic field lines; (b) Amperes law; (c) Gauss law; (d) the property of superconducting loops to conserve the magnetic flux; (e) the energy of magneticfields.Regarding the distribution of the award for the best solution: all the first three solutions are very good, with different strong points (which will be commented below). Idecided to distribute the award between these three evenly, giving a small bias to Szabó Attila, who was the only one to solve the problem without any hints. So, the bonusfactors are e0.4, e0.3 and e0.3. The next three solutions are also partially published – due to different reasons – and receive bonus factors of 1.1.Let us start with the solution by Szabó Attila, which is almost perfect, including all the required components: (a) noting that the dominant part of the magnetic flux is keptinside the ferromagnetic (either using energy-based arguments, or applying Amperes law — as is done here); (b) showing that B has the same order of magnitude bothinside the slit and in the ferromagnetic — using the Gauss law; (c) recognizing that inside the ferromagnetic, B is not homogeneous and hence, the contribution of thesegment residing inside the ferromagnetic brick to the circulation integral of the Amperes law can only be estimated (and not calculated precisely); (d) applying theAmperes law to show that inside the slit region surrounded by the current loop, B is homogeneous, and calculating the value of that B; (e) calculating the initial energy — asis done here, or via magnetic field energy; (f) applying the flux conservation law for the superconducting loop to calculate the final energy; (g) calculating work as adifference of energies. If there is anything to be desired then it would be a motivation that in the slit, B is perpendicular to the plane of the loop (it is only stated as a fact,without motivation). Note that there is a typo in his text – instead of should be .76
  • 77. 77
  • 78. The next solution is that of Nikita Sopenko. As compared with the first solution, it includes a proof of the formula (which is not mandatory as it is covered bythe formula sheet). Further, his solution does not require a proof that in the slit, B is perpendicular to the current loop, because what is used here is only the perpendicularcomponent of B (which enters both into the Amperes law and the expression for the magnetic flux). Finally, he has nicely and explicitly shown the continuity of B at the slitboundary using the Gauss law (though, he does miss explicit proof that majority of the flux resides inside the ferromagnetic).78
  • 79. 79
  • 80. The solution of Ivan Tadeu Ferreira Antunes Filho is provided here as a .pdf file (it is too long to present page-by-page) (http://www.ipho2012.ee/wp-content/uploads/physics%20cup2%20-%20magnetic%20circuits-1.pdf); it differs from the first two solutions in that the approach is based on the concept of reluctance.This is not as clear physically as the approach based on the direct application of the Amperes law (in particular leaving open the question of why B is homogeneous in theslit); however, Ivan does manage to keep things correct (providing first a theoretical motivation of the method, and then calculating the reluctances in a correct way). Thereason why he does receive a bonus is not motivated by his method, but by the fact that he does study, what will happen if L becomes so large (when made of a very thinwire) that the expression in the braces would become negative. In particular, he shows that then, the solution needs to be modified, and the work would be still positive.Also, he applies the formula for the loop inductance to estimate if it is realistic to have such large values of L which would be comparable to the initial inductance of the loop(surrounded by the ferromagnetic); the answer is "not really". Note that intuitively, all the other solutions just imply that the ferromagnetic makes the initial inductancemuch larger than the inductance L of the stand-alone loop.Next, the solution of Jakub Šafin; what is worth highlighting, is his way of motivating, why in the slit, B is perpendicular to the plane of the current loop (whilemathematically not as clear and correct as the magnetic field line refraction law described by Ilie Popanu, see below, intuitively and qualitatively these are very usefularguments):"Now, we only need to find the initial inductance M. For large , fieldlines of B are attracted to the ferromagnetic; inside the slit, therefore, these fieldlines try to escape fromthe slit, so theyll be perpendicular to the loop (we can think of it as a deformation of magnetic field of the loop in vacuum). (This wont hold perfectly for fieldlines close tothe edge of the loop, though, because theyre curved, but for large , this is negligible.)"As mentioned above, the initial energy can be also calculated via the energy density of the magnetic field (for this method, it is important to understand that themagnetic field fills only the circular sub-region of the slit surrounded by the current loop). The first one to do so was Lars Dehlwes:80
  • 81. 81
  • 82. Finally, Ilie Popanu proved that in the slit, B is perpendicular to the slit using the refraction law for the magnetic field lines:82
  • 83. Problem No 3Determine or estimate the net heat flux density between two parallel plates at distance from each other, which are at temperatures and , respectively. The spacebetween the plates is filled with a monoatomic gas of molar density and of molar mass . You may use the following approximations:1. The gas density is so low that the mean free path ;2. .3. When gas molecules bounce from the plates, they obtain the temperature of the respective plates (for instance, this will happen if they are absorbed/bound for a shorttime by the molecules of the plate, and then released back into the space between the plates).4. You may neglect the black body radiation.5. "Estimate" means that the numeric prefactor of your expression does not need to be accurat83
  • 84. Results after Problem 3The list of the contest leaders after the third problem:Points No 1 No 2 No 3 Name Country School Physics teacher 7,5632 1,1 3,8694 2,5937 SZABÓ Attila Hungary Leőwey Klára High School, Pécs Simon Péter, Dr Kotek László Preparatory School for Science & Technology 5,6695 4,8156 0,8539 Brahim Saadi Algeria Derradji Nasreddine of Annaba 5,5041 1,6105 2,8935 1 Nikita Sopenko Russia Lyceum No.14, Tambov Valeriy Vladimirovich Biryukov 5,4546 1 1,7363 2,7183 Jakub Šafin Slovak Pavol Horov Secondary, Michalovce Jozef Smrek Ivan Tadeu Ferreira 5,4353 2,1436 2,2917 1 Brazil Colégio Objetivo, Lins, São Paulo Antunes Filho 5,2722 1,1 1,5785 2,5937 Lars Dehlwes Germany Ohm-Gymnasium Erlangen Mr. Perleth Igor Iurevici Nemtov; Andrei 4,1679 1 0,81 2,3579 Ion Toloaca Moldova liceul "Mircea Eliade" Simboteanu A.R. Zilberman, G.F. Lvovskaya, G.Z. 3,919 1 0,9703 1,9487 Alexandra Vasileva Russia Lyceum "Second School", Moscow Arabuly Luís Gustavo Lapinha Dalla 3,7517 1 0,9801 1,7716 Brazil Colégio Integrado Objetivo, Barueri, Brazil Ronaldo Fogo Stella 3,3137 2,5937 0,72 Dinis Cheian Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev 3,3062 1 1,435 0,8712 Ilie Popanu Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev 3,0746 1,6105 1,4641 Jakub Supeł Poland 14th School of Stanisław Staszic, Warsaw Włodzimierz Zielicz 2,8424 0,8733 0,891 1,0781 Papimeri Dumitru Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev 2,7105 1,1 1,6105 Kohei Kawabata Japan Nada High School 2,5937 2,5937 Mikhail Shirkin Russia Gymnasium of Ramenskoye P Petrova Elena Georgyevna 2,4738 1,7538 0,72 Cristian Zanoci Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev 2 1 1 Jaan Toots Estonia Tallinn Secondary Science School Toomas Reimann 1,81 0,81 1 Bharadwaj Rallabandi India Narayana Jr. College, Basheer Bagh, India Vyom Sekhar Singh 1,6561 0,6561 1 Nadezhda Vartanian Russia Smolensk Pedagogical Lyceum Mishchenko Andrei Anatolievich84
  • 85. 1,1 1,1 Krzysztof Markiewicz Poland XIV Highschool in Warsaw Robert Stasiak Advanced Educational Scientific Center, MSU, 1 1 Lev Ginzburg Russia I.V. Lukjanov, S.N. Oks Moscow 1 1 Sharad Mirani India Prakash Higher Secondary School Ruchi Sadana, Sunil Sharma 1 1 Task Ohmori Japan Nada High School T.Hamaguchi 0,9801 0,9801 Mekan Toyjanow Turkmenistan Turgut Ozal Turkmen Turkish High School Halit Coshkun 0,8712 0,8712 Petar Tadic Montenegro Gimnazija ,,Stojan Cerovic" Niksic Ana Vujacic 0,81 0,81 Liara Guinsberg Brazil Colégio Integrado Objetivo, São Paulo, Brazil Ronaldo Fogo 0,81 0,81 Meylis Malikov Turkmenistan Turgut Ozal Turkmen Turkish High School Halit Coshkun 0,72 0,72 Rajat Sharma India Pragati Vidya Peeth,Gwalior Mr. Rakesh Ranjan 0,5648 0,5648 Lorenzo Comoglio Italy Liceo Scientifico del Cossatese e Valle Strona Chiara BandiniPoints for Problem No 3: 2,7183 Jakub Šafin 2,5937 SZABÓ Attila 2,5937 Lars Dehlwes 2,3579 Ion Toloaca 1,9487 Alexandra Vasileva Luís Gustavo Lapinha 1,7716 Dalla Stella 1,6105 Kohei Kawabata 1,4641 Jakub Supeł 1,1 Krzysztof Markiewicz 1,0781 Papimeri Dumitru 1 Nikita Sopenko Ivan Tadeu Ferreira 1 Antunes Filho85
  • 86. 1 Jaan Toots 1 Bharadwaj Rallabandi 1 Nadezhda Vartanian 0,8712 Ilie Popanu 0,8712 Petar Tadic 0,72 Dinis Cheian 0,72 Cristian Zanoci 0,72 Rajat Sharma 0,5648 Lorenzo ComoglioCorrect solutions (ordered according to the arrival time; best solutions in bold):1. Lars Dehlwes (Germany)2. Ion Toloaca (Moldova)3. Szabó Attila (Hungary).4. Alexandra Vasileva (Russia)5. Luís Gustavo Lapinha Dalla Stella (Brazil)6. Kohei Kawabata (Japan)7. Jakub Supeł (Poland)8.Papimeri Dumitru (Moldova)9. Ilie Popanu (Moldova)86
  • 87. 10. Petar Tadic (Montenegro)11. Cristian Zanoci (Moldova)12. Dinis Cheian (Moldova)13. Bharadwaj Rallabandi (India)14. Nadezhda Vartanian (Russia)15. Nikita Sopenko (Russia)16. Jakub Šafin (Slovak)17. Krzysztof Markiewicz (Poland)18. Ivan Tadeu Ferreira Antunes Filho (Brazil)19. Jaan Toots (Estonia)20. Comoglio Lorenzo (Italy)21. Rajat Sharma (India)Overall number of registered participants: 214 (from 38 countries).Before the beginning of the final week, the following hints were given:Few hints: youll be ready to tackle the problem as soon as you understand how the basic formulae of the kinetic theory (formula sheet X-9) are derived;Wikipedia has goodenough coverage. And, of course you need the definition of the net heat flux density: it is the difference between the incoming and outgoing thermal energies per unit timeand unit surface area. Also, please pay attention that for a molecule, the round trip time (between the plates) is dominated by the period when its velocity is small. This is87
  • 88. similar to what happens in a road reconstruction region: the distance between the cars in seconds (the number of cars per minute) remains the same as it was in high-speedregions, hence, the distance in meters will be much smaller (and the density of the cars will be respectively higher).————————-Jaan Kalda – Academic Committee of IPhO-201288
  • 89. SolutionIn the problem text, it was stated that all numeric prefactors are considered to be acceptable, and thus, the problem was graded generously. Actually, there were only twocompletely correct answers (by Szabó Attila and Jakub Šafin), and two answers which were also flawless – except that instead of the correct ,approximation was used (Petar Tadic and Krzysztof Markiewicz).The most common mistake was not noticing that unlike in the case of a normal gas, both for the hot "faction" and cold "faction", the molecules move only in one direction.Hence, the ready formulae, such as are two times smaller than needed, and the Maxwell velocity distribution function should be also multiplied by two. Othertypical mistakes were that instead of the projection or , the modulus of the vector ( or ) was used. Meanwhile, the modulus should be usedwhen calculating the transferred energy, but in some solutions, there was , instead.The best solutions were judged to be those of Szabó Attila and Jakub Šafin. However, Szabó Attila sent first an approximate solution, which he later corrected – lateenough to lose his bonus points due to speed. If the best solution bonus would have been divided between these two, Szabó Attila would have got less points than whentaking into account his speed bonus. Therefore, he was given his speed bonus, and additionally a double 1.1-factor-bonus – for using both his originally submitted and therevised solutions on this web page. And so, the best solution bonus goes entirely to Jakub Šafin; Petar Tadic and Krzysztof Markiewicz both receive a 1.1-factor-bonus.Finally, Lorenzo Comoglio recieves also a bonus of 1.1: he made a very nice visualization of the process.There are two ways of calculating the frequency of collisions: (a) using the round-trip time, and (b) calculating first the densities of both "factions" of molecules (hot andcold).89
  • 90. The solution of Szabó Attila follows method (a).90
  • 91. The solution of Jakub Šafin is based on calculating the densities of "factions". Also, he makes a very useful analysis of the results.91
  • 92. 92
  • 93. 93
  • 94. 94
  • 95. The solution of Krzysztof Markiewicz:95
  • 96. The solution of Petar Tadic:96
  • 97. Finally, the initial solution of Szabó Attila: while incorrect, the idea itself is very nice, and the mistake is well hidden; so I judged it to be useful to display the first page,and analyse, why the prefactor will be wrong, if calculated in such a way.97
  • 98. Notice the nice trick of introducing and arranging the molecules according to the values of . Unfortunately, the trick does not work here: re-arranging the order of themolecule speeds does introduce false correlations. In such a way, we create molecules which are always faster than average, and the ones which are slower than average;the amount of transported heat is defined by after the hot wall, and the round-trip time is defined by after the cold wall; so, relatively large amount of heat would betransported in relatively shorter time, and therefore, the average of the product of the transported heat with the collision frequency would not be equal to the product of therespective averages; however, it would be equal if these two quantities were uncorrelated, as is actually the case!98
  • 99. Problem No 4There are three point masses m, 2m and 3m, each of which is fixed to a weightless rod; the three rods are of equal length L and are fixed to each other via a connector,which allows a free rotation (frictionless, torque-less) of the rods with respect to each other (so that the angles between the rods will change). Initially the angle between therods is 120o and the system is motionless; all the rods and point masses lay on the same plane. The heaviest mass (3m) is hit so that it obtains instantaneously a velocity v0,perpendicular to the rod to which it is fixed to and coplanar to the other rods. Determine the accelerations of all three point masses immediately after the point mass 3m washit. Remark: there is no gravity field, the system can be thought to be in weightlessness, or on frictionless horizontal surface.99
  • 100. Results after Problem 4The list of the contest leaders after the fourth problem:Points Name Country School Physics teacher 10,157 SZABÓ Attila Hungary Leőwey Klára High School, Pécs Simon Péter, Dr Kotek László 7,6476 Nikita Sopenko Russia Lyceum No.14, Tambov Valeriy Vladimirovich Biryukov 6,8666 Lars Dehlwes Germany Ohm-Gymnasium Erlangen Martin Perleth 6,7301 Jakub Šafin Slovak Pavol Horov Secondary, Michalovce Jozef Smrek Ivan Tadeu Ferreira 6,0833 Brazil Colégio Objetivo, Lins, São Paulo Antunes Filho 6,0244 Ilie Popanu Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev Preparatory School for Science & Technology 5,6695 Brahim Saadi Algeria Derradji Nasreddine of Annaba 5,1968 Jakub Supeł Poland 14th School of Stanisław Staszic, Warsaw Włodzimierz Zielicz Igor Iurevici Nemtov; Andrei 5,0589 Ion Toloaca Moldova liceul "Mircea Eliade" Simboteanu A.R. Zilberman, G.F. Lvovskaya, G.Z. 4,919 Alexandra Vasileva Russia Lyceum "Second School", Moscow Arabuly 4,1057 Dinis Cheian Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev 3,9424 Papimeri Dumitru Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev 3,8105 Kohei Kawabata Japan Nada High School 3,7997 Nadezhda Vartanian Russia Smolensk Pedagogical Lyceum Mishchenko Andrei Anatolievich Luís Gustavo Lapinha Dalla 3,7517 Brazil Colégio Integrado Objetivo, Barueri, Brazil Ronaldo Fogo Stella 3,4738 Cristian Zanoci Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev 3,4205 Bharadwaj Rallabandi India Narayana Jr. College, Basheer Bagh, India Vyom Sekhar Singh 2,5937 Mikhail Shirkin Russia Gymnasium of Ramenskoye P Petrova Elena Georgyevna 2,1 Krzysztof Markiewicz Poland XIV Highschool in Warsaw Robert Stasiak100
  • 101. 2 Jaan Toots Estonia Tallinn Secondary Science School Toomas Reimann 1,4641 Hideki Yukawa Japan Nada high school 1,3896 Petar Tadic Montenegro Gimnazija ,,Stojan Cerovic" Niksic Ana Vujacic 1,21 Midhul Varma India Vidyadham Junior, Hyderabad Manikanta Kumar 1 Task Ohmori Japan Nada High School T.Hamaguchi 1 Sharad Mirani India Prakash Higher Secondary School Ruchi Sadana, Sunil Sharma Advanced Educational Scientific Center, MSU, 1 Lev Ginzburg Russia I.V. Lukjanov, S.N. Oks Moscow 0,9801 Mekan Toyjanow Turkmenistan Turgut Ozal Turkmen Turkish High School Halit Coshkun 0,81 Meylis Malikov Turkmenistan Turgut Ozal Turkmen Turkish High School Halit Coshkun 0,81 Liara Guinsberg Brazil Colégio Integrado Objetivo, São Paulo, Brazil Ronaldo Fogo 0,792 Ulysse Lojkine France Lycée Henri IV, Paris M. Lacas 0,72 Rajat Sharma India Pragati Vidya Peeth,Gwalior Mr. Rakesh Ranjan 0,5648 Ng Fei Chong Malaysia SMJK Chung Ling, Penang 0,5648 Lorenzo Comoglio Italy Liceo Scientifico del Cossatese e Valle Strona Chiara BandiniPoints for Problem No 4: 2,7183 Ilie Popanu 2,5937 SZABÓ Attila 2,1436 Nikita Sopenko 2,1436 Nadezhda Vartanian 2,1222 Jakub Supeł 1,6105 Bharadwaj Rallabandi 1,5944 Lars Dehlwes 1,4641 Hideki Yukawa 1,2755 Jakub Šafin 1,21 Midhul Varma 1,1 Papimeri Dumitru101
  • 102. 1,1 Kohei Kawabata 1 Alexandra Vasileva 1 Cristian Zanoci 1 Krzysztof Markiewicz 0,891 Ion Toloaca 0,792 Dinis Cheian 0,792 Ulysse Lojkine Ivan Tadeu Ferreira 0,648 Antunes Filho 0,5648 Ng Fei Chong 0,5184 Petar TadicCorrect solutions (ordered according to the arrival time; best solutions in bold):1. Szabó Attila (Hungary)2. Jakub Supeł (Poland)3. Nadezhda Vartanian (Russia)4. Nikita Sopenko (Russia)5. Lars Dehlwes (Germany)6. Ilie Popanu (Moldova)7. Jakub Šafin (Slovak)8. Bharadwaj Rallabandi (India)102
  • 103. 9. Hideki Yukawa (Japan)10. Midhul Varma (India)11. Dinis Cheian (Moldova)12. Ion Toloaca (Moldova)13. Kohei Kawabata (Japan)14. Cristian Zanoci (Moldova)15. Ng Fei Chong (Malaysia)16. Alexandra Vasileva (Russia)17. Papimeri Dumitru (Moldova)18. Krzysztof Markiewicz (Poland)19. Ivan Tadeu Ferreira Antunes Filho (Brazil)20. Petar Tadic (Montenegro)21. Ulysse Lojkine (France)The number of incorrect solutions: 13The overall number of registered participants: 238 from 41 countries103
  • 104. For the last two weeks, a small hint was given: it is helpful to consider the motion of the balls in the connector frame of reference. For the last three days, relatively detailedhints were given: (1) Note that in the lab system of reference, there is only one force applied to each of the balls: the rod tension. Due to the Newton II law, once you knowthe tensions, you can obtain immediately the accelerations. (2) Force balance for the connector allows you to find, how the tensions in different rods are related to eachother, ie. to express T2 and T3 in terms of T1. (3) In order to advance further with the solution, it is helpful to consider the motion of the balls in the connector frame ofreference, where they perform circular motions: the radial (centripetal) acceleration is caused by the the tension in rod, together with the force of inertia; the tangentialacceleration is caused only by the force of inertia (because there is no bending stress in the rods). So, you have three equations (the force balance for each of the balls,projected onto the direction of the respective rod), and three unknowns (two components of the connector acceleration, and the tension T1. This system can be solvedgeometrically, arithmetically using trigonometric functions, or performing symbolic vectorial calculations; the length of the solution depends on the route you choose.104
  • 105. SolutionThis problem turned out to be a really good one, because the contestants came up with so many different solutions. This time, let us start with the solution, and at the endwell count the points. The first step towards the solution is showing that all the rod tensions are equal; this is an unavoidable step, unless you derive the therelationship between the ball accelerations from the conservation of linear momentum – as was done by Ulyss Lojkine:Another way of avoiding that first step is to apply Lagrangian formalism, as was done by Lars Dehlwes, Papimeri Dumitru, Cristian Zanoci and Dinis Cheian. This is a "bruteforce" approach, which is definitely a solid way of doing it, but be prepared for long calculations, where a single small mistake can invalidate your results (to be on a safeside, dont forget to check the absence of mistakes by checking the validity of the conservation laws!). In order to give you an idea about the amount work needed for thisapproach, here is a link to the Papimeri Dumitrus solution.While most of the contestants, indeed, used the fact that the rod tensions are equal, very few cared to show it properly; the clearest (and simple) proof is provided by IonToloaca:105
  • 106. The next step is showing that initially, the connector is at rest; while most of the contestants implicitly assumed it (perhaps assuming it to be obvious), very few caredto prove it. This is, again, best done by Ion Toloaca:106
  • 107. The final and most difficult step is solving the force balance problem in the connectors frame of reference(which is non-inertial, moving with an acceleration ),where the masses obtain centripetal acceleration due to the joint effect of the inertial force ( ) and the rod tension. This can be done geometrically; an elegant wayof doing it is once again provided by Ion Toloaca:107
  • 108. 108
  • 109. It can be also done by solving an algebraic system of equations, as was, for instance, done by Ng Fei Chong:109
  • 110. Alternatively, you can solve a trigonometric system of equation; example is contributed by Kohei Kawabata:110
  • 111. Finally, you can treat the force balance vectorially, and this is the advertised shortest solution! How it can be done is demonstrated by Ilie Popanu:111
  • 112. His writing, however, includes some redundant lines, so let us do it once again. Let us introduce unit vectors along the rods, . Then, the Newton second law, asprojected to the direction of the rod, can be written aswhere are the ball masses ( ) and – the velocities ( ). Adding up these three equations and bringing before the braces,we end up withNow, since , we obtain immediatelySo, what is the lesson from here? First, using vector calculus saves often a lot of mathematical work. Second, in order to preserve the symmetry of theproblem, it may be helpful to use over-determined vector basis (two vectors would have been enough, but we used three, ).It should be noted that another person to apply efficiently vector calculus was Jakub Šafin, who actually solved a more general problem, when the masses and anglesbetween the rods are arbitrary, and all the masses move.112
  • 113. This (and even more general) result can be actually obtained quite easily by generalizing the vector approach explained above. To begin with, we need togeneralize the relationship between the rod tensions to the case when the angles are not equal. Bharadwaj Rallabandi has actually dug out an appropriate theorem, whichis nothing more than a sine theorem for the triangle of vectors (representing the force balance for the connector):So, we can write , where is the diameter of the circumcircle of the triangle of vectors and is the angle between the two rodsexcluding the i-th one. Similarly to what we did above, the force balance for the balls is written asand we want to eliminate from this system of equations. We can use the same trick as above (adding equations), if we make use of the sine theorem, expressed in vectorform as . So, we multiply the equations by and add up, to obtain113
  • 114. Finally, the acceleration of the j-th body is given by Before we proceed to counting the points, there are nice and very original solutions contributed by Nikita Sopenko andHideki Yukawa. First, the solution of Nikita Sopenko:114
  • 115. Here he actually manages to solve the problem without using a non-inertial system of reference, and applies a nice trick of finding the projection of a vector to an axis whileknowing its projection to two other directions. And finally, the solution of Hideki Yukawa:115
  • 116. Few comments are needed here. First, I added a red arrow indicating the acceleration of the connector in the s reference frame; in that frame of reference, theconnector rotates, so the projection of its acceleration to the rod direction is , which happens to be the height of the triangle in his figure. Finally, the height equals tothe sum of the connectors distances to the sides of the triangle because the triangle area can be represented as the sum of the areas of the three smaller triangles, formedby the connector and the triangle vertices.And thats it, we are ready to count the points. The results are impressive, there are many candidates for the best solution: Ilie Popanu, Jakub Šafin, Hideki Yukawa,Nikita Sopenko, Ion Toloaca. One possibility would be to divide the bonus between them, but then there would be a problem: Ilie Popanu submitted his first solution quiteearly, but his short version was sent when there was no speed bonus left; as a result, for him a partial best-solution-bonus would be smaller than his original speed bonus.So I decided to give the full bonus of the best solution to Ilie Popanu; everyone else (mentioned in bold text above) will receive a bonus of 1.1.—Jaan Kalda, Academic Committee of IPhO-2012116
  • 117. Problem No 5Determine all the eigenfrequencies (=natural frequencies) of the circuit shown in Figure. You may assume that all the capacitors and inductances are ideal, and that thefollowing strong inequalities are satisfied: , and . Note that your answers need to be simplified according to these strong inequalities.117
  • 118. Results after Problem 5The list of the contest leaders after the fifth problem:Points Name Country School Physics teacher 12,725 SZABÓ Attila Hungary Leőwey Klára High School, Pécs Simon Péter, Dr Kotek László 11,535 Nikita Sopenko Russia Lyceum No.14, Tambov Valeriy Vladimirovich Biryukov 8,168 Ilie Popanu Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev 7,7004 Jakub Šafin Slovak Pavol Horov Secondary, Michalovce Jozef Smrek 7,5883 Lars Dehlwes Germany Ohm-Gymnasium Erlangen Martin Perleth Ivan Tadeu Ferreira 6,0833 Brazil Colégio Objetivo, Lins, São Paulo Antunes Filho 5,8911 Papimeri Dumitru Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev Igor Iurevici Nemtov; Andrei 5,8689 Ion Toloaca Moldova liceul "Mircea Eliade" Simboteanu Preparatory School for Science & 5,6695 Brahim Saadi Algeria Derradji Nasreddine Technology of Annaba 5,5407 Dinis Cheian Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev 5,1968 Jakub Supeł Poland 14th School of Stanisław Staszic, Warsaw Włodzimierz Zielicz 5,0844 Cristian Zanoci Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev A.R. Zilberman, G.F. Lvovskaya, 4,919 Alexandra Vasileva Russia Lyceum "Second School", Moscow G.Z. Arabuly 3,8105 Kohei Kawabata Japan Nada High School 3,7997 Nadezhda Vartanian Russia Smolensk Pedagogical Lyceum Mishchenko Andrei Anatolievich Luís Gustavo Lapinha Dalla 3,7517 Brazil Colégio Integrado Objetivo, Barueri, Brazil Ronaldo Fogo Stella 3,4205 Bharadwaj Rallabandi India Narayana Jr. College, Basheer Bagh, India Vyom Sekhar Singh 2,5937 Mikhail Shirkin Russia Gymnasium of Ramenskoye P Petrova Elena Georgyevna 2,1 Krzysztof Markiewicz Poland XIV Highschool in Warsaw Robert Stasiak118
  • 119. 2 Jaan Toots Estonia Tallinn Secondary Science School Toomas Reimann 1,903 Petar Tadic Montenegro Gimnazija ,,Stojan Cerovic" Niksic Ana Vujacic Liceo Scientifico del Cossatese e Valle 1,8606 Lorenzo Comoglio Italy Chiara Bandini Strona 1,6214 Ng Fei Chong Malaysia SMJK Chung Ling, Penang 1,4641 Hideki Yukawa Japan Nada high school 1,21 Midhul Varma India Vidyadham Junior, Hyderabad Manikanta Kumar 1 Teoh Yee Seng Malaysia Hong Siang Ean, Loh Pei Yee 1 Task Ohmori Japan Nada High School T.Hamaguchi 1 Sharad Mirani India Prakash Higher Secondary School Ruchi Sadana, Sunil Sharma Advanced Educational Scientific Center, 1 Lev Ginzburg Russia I.V. Lukjanov, S.N. Oks MSU, Moscow 0,9801 Mekan Toyjanow Turkmenistan Turgut Ozal Turkmen Turkish High School Halit Coshkun 0,81 Meylis Malikov Turkmenistan Turgut Ozal Turkmen Turkish High School Halit Coshkun Colégio Integrado Objetivo, São Paulo, 0,81 Liara Guinsberg Brazil Ronaldo Fogo Brazil 0,792 Ulysse Lojkine France Lycée Henri IV, Paris M. Lacas 0,72 Rajat Sharma India Pragati Vidya Peeth,Gwalior Mr. Rakesh Ranjan Bosnia- 0,5648 Selver Pepić Fourth Gymnasium Ilidža, Sarajevo Rajfa Musemić HerzegovinaPoints for Problem No 5: 3,8876 Nikita Sopenko 2,5678 SZABÓ Attila 2,1436 Ilie Popanu 1,9487 Papimeri Dumitru119
  • 120. 1,6105 Cristian Zanoci 1,435 Dinis Cheian 1,2958 Lorenzo Comoglio 1,0567 Ng Fei Chong 1 Teoh Yee Seng 0,9703 Jakub Šafin 0,81 Ion Toloaca 0,7217 Lars Dehlwes 0,5648 Selver Pepić 0,5134 Petar TadicCorrect solutions (ordered according to the arrival time; best solutions in bold):1. Szabó Attila (Hungary)2. Nikita Sopenko (Russia)3. Ilie Popanu (Moldova)4. Papimeri Dumitru (Moldova)5. Dinis Cheian (Moldova)6. Cristian Zanoci (Moldova)7. Ng Fei Chong (Malaysia)8. Lorenzo Comoglio (Italy)120
  • 121. 9. Jakub Šafin (Slovak)10. Lars Dehlwes (Germany)11. Petar Tadic (Montenegro)12. Ion Toloaca (Moldova)13. Teoh Yee Seng (Malaysia)14. Selver Pepić (Bosnia-Herzegovina)(The list is ordered according to the arrival time)Also, there are two incorrect solutions and two solutions, which are based on correct principles, but contain some mistakes.Overall number of registered participants: 247 from 42 countries.—————-Since the number of correct solutions was relatively low, the following hints were given by the end of the third week:- study the formula sheet (section V-3) to get an idea of how many eigenfrequencies you need to find; if in difficulties counting the number of oscillators, take it equal to thenumber of degrees of freedom (how many independent loop currents are needed for a superposition to represent arbitrary current distribution of the circuit);- make use of the strong inequalities at as early stage as possible, to simplify the mathematical task;- study the section VIII of the formula sheet; particularly useful are pts. 11, 5, 3, 2 (though, depending on your approach, you may not need all of these formulae).- in order to find the natural frequencies, you can write down the system of differential equations, and based on that system, write down the characteristic equation, thesolutions of which are the natural frequencies. However, note that you can avoid writing down the differential equations. Instead, you can make use of the concept of current121
  • 122. and voltage resonances. So, for a voltage resonance, there will be a non-zero voltage amplitude U between two nodes A and B, even if there is no current flowing into thenode A (and from the node B). Indeed, if the impedance between the nodes A and B is , we can write |U| =|I||Z|, hence a non-zero U is compatible with I=0if . Therefore, natural frequencies can be found as the solutions of the equation .——————-The solutions of Problem 5 will be uploaded on 21st February.122
  • 123. SolutionThose of you who know (a) how to find eigenvalues of a system of linear differential equations, and (b) solve electrical bridge circuits using node potential or loop currentmethods had an option of solving this problem with a "brute force": write down a complete set of differential equations for four unknown functions (currents or charges; four,because this is the number of degrees of freedom here), write the characteristic function for eigenfrequencies, and find the asymptotic solutions of it. This may sound easy,but mathematical work associated with this approach it is quite large; besides, the required mathematical techniques are typically taught only during university courses (thiswas supposed to be a problem solvable within the knowledge of the IPhO syllabus). Also, the characteristic equation for the squared frequency leads to a cubic algebraicequation here (actually a fourth order equation, but with one root being equal to zero), which cannot be solved directly. Well, there are the Cardanos formulae, which,however, lead to the casus irreducibilis (real roots are expressed via complex numbers in such a way that it cannot be simplified algebraically), the asymptotic simplificationof which is a very-very difficult task. Thats why the right procedure for this brute force approach is to apply asymptotic simplifications (small-root-approximations and large-root-approximations) to the equation itself, and then solve the resulting lower-order equations; in any case, there is a lot of math involved. Mathematical complexity of thebrute force approach was actually intentional, the aim was to promote the high- and low-frequency simplifications of electrical circuits. For physicists, using suchsimplifications is a very important skill: even if you manage solving the problem by brute force, you cant grasp the qualitative features of the circuit unless you learn tosimplify it for different frequency ranges; in particular, this is why the best solution awards were not given to brute-force-solutions.The bonus for the best solutions was divided equally between two contestants – Lorenzo Comoglio and Nikita Sopenko. Lorenzo Comoglio was able to simplify the circuitcorrectly on the same day when the problem was published; however, he had not been taught how to solve systems of linear differential equations. Because of that, it tookhim a lot of time to correct the mathematical part. The award is given him because he was the only one who was able to simplify the electrical circuit before any hints. Belowis his final version.123
  • 124. 124
  • 125. So, he obtains the natural frequencies as frequencies at which the impedance of an AC-circuit turns to zero. The reason why this method works can be explained as follows.Consider two arbitrary nodes of the circuit: one of these will be an input node for a vanishing AC current , the other node will be for the output of the same current. Ifthere are free oscillations taking place in the circuit, we have still zero input current, but non-zero voltage oscillations between the two nodes. Due to Ohms law, ,hence the free oscillations require . If we apply this equation to the Lorenzos high-frequency circuit, we havewhere we have denotedNote that we have kept here only the two positive roots. The result, while seemingly different, is actually identical to the one of Lorenzo.Finally, what would have happened if another pair of nodes were chosen for writing the impedance? There would have been a slightly different expression for the impedance,but the final result would have been, of course, the same!Next solution is that of Nikita Sopenko. Instead of solving the circuit problem, he decided to consider an equivalent mechanical problem. This can be done because ifproperly matched, an LC-circuit and a system of springs and masses are described by exactly the same system of differential equations. The basic rule is that a masscorresponds to an inductance, a spring – to a capacitor, and a displacement – to a relocation of a charge. His asymptotic simplification of the mechanical oscillator isessentially equivalent to Lorenzos simplification of the electrical circuit.125
  • 126. 126
  • 127. Now let us have a look on the solution of Jakub Šafin. He simplified the circuit in the same way as Lorenzo did, but instead of considering current resonance (whennegligible input current into a node leads to large loop currents), he studied the voltage resonance. To that end, a single wire of the circuit is chosen, and cut broken; as aresult of cutting, two endpoints of the wire are formed, which are considered as input terminals of an AC voltage with zero amplitude, . This is compatible with theoriginal unbroken circuit, too, because in that case the cut points coincide and hence, have the same potential. In the case of free oscillations, there is still a current in thecircuit, which is compatible with only if , which is the condition for the voltage resonance. Jakubs solution is given here as his original .pdf file.Another solution based on the simplified circuits is provided by Ng Fei Chong. However, he does not make use of the resonance phenomenon; instead, he finds the naturalfrequencies via characteristic equation of a system of two differential equations.127
  • 128. 128
  • 129. Next solution is an example of the "brute force" approach (cf. the introduction paragraph) – as was perfectly executed by Szabó Attila; it is provided here as his original .pdffile. Note that most of those contestants who tried this method failed initially – because when simplifying the determinant and/or a fourth order algebraic equation, someneglected terms were actually important. However, Attila got everything right at his first attempt.Finally, one of the contestants, Selver Pepić, took an "experimentalists approach": he extended the circuit with small resistances and a voltage source, coined numericalvalues, and calculated resonance curves; the result is given below.129
  • 130. 130
  • 131. Problem No 6There is a solid metal sphere of radius , which is cut into two parts along a plane in such a way that the outer surface of the smaller part of the sphere is . The cutsurfaces are coated with a negligibly thin insulating layer, and the two parts are put together so that the original shape of the sphere is restored.Initially the sphere iselectrically neutral. Then the smaller part of the sphere is given a positive electric charge ; the larger part of the sphere remains neutral. Find (a) the charge distributionalong the sphere; (b) the electrostatic interaction force between the two pieces of the sphere; (c) the electrostatic energy of the sphere.Acknowledgment: This problem is contributed by Gyula Honyek.131
  • 132. Intermediate conclusionAs of 26th February, correct solutions of Problem 6 have been submitted by:1. Szabó Attila (Hungary)2. Nikita Sopenko (Russia)3. Lars Dehlwes (Germany)4. Kohei Kawabata (Japan)5. Kai-Chi Huang (Taiwan)6. Petar Tadic (Montenegro)7. Bruno Bento Barros de Araújo (Brazil)8. Ilie Popanu (Moldova)The number of registered participants: 252 from 43 countries.————————Next update: 4th March 2012132