Problems of the 9 th International Physics Olympiads (Budapest, Hungary, 1976)

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1 Problems of the 9 th International Physics Olympiads (Budapest, Hunary, 1976) Theoretical problems Problem 1 A hollow sphere of radius R = 0.5 m rotates about a vertical axis throuh its centre with an anular velocity of ω = 5 s -1. Inside the sphere a small block is movin toether with the sphere at the heiht of R/ (Fi. 6). ( = 10 m/s.) a) What should be at least the coefficient of friction to fulfill this condition? b) Find the minimal coefficient of friction also for the case of ω = 8 s -1. c) Investiate the problem of stability in both cases, ) for a small chane of the position of the block, β) for a small chane of the anular velocity of the sphere. R S N m mω Rsin R/ Fiure 6 Fiure 7 a) The block moves alon a horizontal circle of radius R sin. The net force actin on the block is pointed to the centre of this circle (Fi. 7). The vector sum of the normal force exerted by the wall N, the frictional force S and the weiht m is equal to the resultant: m ω Rsin. The connections between the horizontal and vertical components: mω Rsin = N sin S cos, m = N cos + S sin. The solution of the system of equations: ω Rcos S = m sin 1, 1

2 ω Rsin N = m cos +. The block does not slip down if ω Rcos 1 S 3 3 µ a = sin = = N ω Rsin 3 cos + In this case there must be at least this friction to prevent slippin, i.e. slidin down. ω Rcos b) If on the other hand > 1 some friction is necessary to prevent the block to slip upwards. m ω Rsin must be equal to the resultant of forces S, N and m. Condition for the minimal coefficient of friction is (Fi. 8): ω Rcos 1 S = sin = N ω Rsin cos = = µ b S N m mω Rsin Fiure 8 c) We have to investiate µ a and µ b as functions of and ω in the cases a) and b) (see Fi. 9/a and 9/b): 0.5 µ a ω = 5/s 0.5 µ b ω = 8/s ω < 5/s ω > 5/s 90 Fiure ω < 8/s Fiure ω > 8/s 90 In case a): if the block slips upwards, it comes back; if it slips down it does not return. If ω increases, the block remains in equilibrium, if ω decreases it slips downwards. In case b): if the block slips upwards it stays there; if the block slips downwards it returns. If ω increases the block climbs upwards -, if ω decreases the block remains in equilibrium. Problem The walls of a cylinder of base 1 dm, the piston and the inner dividin wall are

3 perfect heat insulators (Fi. 10). The valve in the dividin wall opens if the pressure on the riht side is reater than on the left side. Initially there is 1 helium in the left side and helium in the riht side. The lenths of both sides are 11. dm each and the temperature is 0 C. Outside we have a pressure of 100 kpa. The specific heat at constant volume is c v = 3.15 J/K, at constant pressure it is 11. dm 11. dm c p = 5.5 J/K. The piston is pushed slowly towards the dividin wall. When the valve 1 dm opens we stop then continue pushin slowly until the wall is reached. Find the work done on the piston by us. Fiure 10 The volume of 4 helium at 0 C temperature and a pressure of 100 kpa is.4 dm 3 (molar volume). It follows that initially the pressure on the left hand side is 600 kpa, on the riht hand side 100 kpa. Therefore the valve is closed. An adiabatic compression happens until the pressure in the riht side reaches 600 kpa (κ = 5/3). 100 V = 600, hence the volume on the riht side (when the valve opens): V = 3.8 dm 3. From the ideal as equation the temperature is on the riht side at this point pv T 1 = = 55K. nr Durin this phase the whole work performed increases the internal enery of the as: W 1 = (3.15 J/K) ( ) (55 K 73 K) = 1760 J. Next the valve opens, the piston is arrested. The temperature after the mixin has been completed: T = = 313K. 14 Durin this phase there is no chane in the enery, no work done on the piston. An adiabatic compression follows from = 15.0 dm 3 to 11. dm 3 : = T 3 11., hence T 3 = 381 K. The whole work done increases the enery of the as: W 3 = (3.15 J/K) (14 ) (381 K 313 K) = 3000 J. The total work done: W total = W 1 + W 3 = 4760 J. The work done by the outside atmospheric pressure should be subtracted: W atm = 100 kpa 11. dm 3 = 110 J. 3

4 The work done on the piston by us: W = W total W atm = 3640 J. Problem 3 Somewhere in a lass sphere there is an air bubble. Describe methods how to determine the diameter of the bubble without damain the sphere. We can not rely on any value about the density of the lass. It is quite uncertain. The index of refraction can be determined usin a liht beam which does not touch the bubble. Another method consists of immersin the sphere into a liquid of same refraction index: its surface becomes invisible. A reat number of methods can be found. We can start by determinin the axis, the line which joins the centers of the sphere and the bubble. The easiest way is to use the tumbler-over method. If the sphere is placed on a horizontal plane the axis takes up a vertical position. The imae of the bubble, seen from both directions alon the axis, is a circle. If the sphere is immersed in a liquid of same index of refraction the spherical bubble is practically inside a parallel plate (Fi. 11). Its boundaries can be determined either by a micrometer or usin parallel liht beams. Alon the axis we have a lens system consistin, of two thick neative lenses. The diameter of the bubble can be determined by several measurements and complicated Fiure11 calculations. If the index of refraction of the lass is known we can fit a plano-concave lens of same index of refraction to the sphere at the end of the axis (Fi. 1). As ABCD forms a parallel plate the diameter of the bubble can be measured usin parallel liht beams. A C A R ϕ d r ψ B Fiure1 Fiure13 Focusin a liht beam on point A of the surface of the sphere (Fi. 13) we et a diverin beam from point A inside the sphere. The rays strike the surface at the other side and illuminate a cap. Measurin the spherical cap we et anle ϕ. Anle ψ can be obtained in a similar way at point B. From 4

5 r sin ϕ = and R + d sin ψ = r R d we have sinϕ sinϕ r = R, d = R. + sinϕ + sinϕ The diameter of the bubble can be determined also by the help of X-rays. X-rays are not refracted by lass. They will cast shadows indicatin the structure of the body, in our case the position and diameter of the bubble. We can also determine the moment of inertia with respect to the axis and thus the diameter of the bubble. Experimental problem The whole text iven to the students: At the workplace there are beyond other devices a test tube with 1 V electrical heatin, a liquid with known specific heat (c 0 =.1 J/ C) and an X material with unknown thermal properties. The X material is insoluble in the liquid. Examine the thermal properties of the X crystal material between room temperature and 70 C. Determine the thermal data of the X material. Tabulate and plot the measured data. (You can use only the devices and materials prepared on the table. The damaed devices and the used up materials are not replaceable.) Heatin first the liquid then the liquid and the crystalline substance toether two time-temperature raphs can be plotted. From the raphs specific heat, meltin point and heat of fusion can be easily obtained. Literature [1] W. Gorzkowski: Problems of the 1st International Physics Olympiad Physics Competitions 5, no pp6-17, 003 [] R. Kunfalvi: Collection of Competition Tasks from the Ist throuh XVth International 5

6 Physics Olympiads Roland Eötvös Physical Society in cooperation with UNESCO, Budapest, 1985 [3] A Nemzetközi Fizikai Diákolimpiák feladatai I.-XV. Eötvös Loránd Fizikai Társulat, Középiskolai Matematikai Lapok,

Physics 121k Exam 3 7 Dec 2012

Physics 121k Exam 3 7 Dec 2012 Answer each question and show your work. A correct answer with no supportin reasonin may receive no credit. Unless directed otherwise, please use =10.0 m/s 2. Name: 1. (15 points) An 5.0 k block, initially