39th International Physics Olympiad - Hanoi - Vietnam Theoretical Problem No. 1 /Solution. Solution

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1 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. /Slutin Slutin. The structure f the mrtar.. Calculating the distance TG The vlume f water in the bucket is V = = cm m. The length f the bttm f the bucket is d = L htan 6 = (. 74. tan 6 ) m =. 53m. (as the initial data are given with tw significant digits, we shall keep nly tw significant digits in the final answer, but we keep mre digits in the intermediate steps). The height c f the water layer in the bucket is calculated frm the frmula: ( 3 / b) V = bcd + b ctan 6 c = 3 c d V / + d Inserting numerical values fr V, b and d, we find c =.8m. When the lever lies hrizntally, the distance, n the hrizntal axis, between the rtatin axis and the center f mass f water N, is TG = ( m/ M)TH =.57m (see the figure belw). d c tan =., and TH a 6 474m R S P H N K T Answer: TG =.6m... Calculating the values f and. When the lever tilts with angle, water level is at the edge f the bucket. At that pint the water vlume is 3 3 m. Assume PQ < d. Frm gemetry V = hb PQ /, frm which P Q =.m. The assumptin PQ < d is bviusly satisfied ( d =.53m ). T cmpute the angle, we nte that tan = h/ QS= h/( PQ+ 3h). Frm this we find =.6.

2 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. /Slutin When the tilt angle is 3, the bucket is empty: = 3. h T G R N S P I Q.3. Determining the tilt angle f the lever and the amunt f water in the bucket m when the ttal trque μ n the lever is equal t zer Dente PQ = x(m). The amunt f water in the bucket is xhb m= ρwater = 9 x (kg). μ = when the trque cming frm the water in the bucket cancels ut the trque cming frm the weight f the lever. The crss sectin f the water in the bucket is the triangle PQR in the figure. The center f mass N f water is lcated at /3 f the meridian RI, therefre NTG lies n a straight line. Then: mg TN = Mg TG r m TN = M TG = 3.57 =.474 () Calculating TN frm x then substitute () : x x x TN = L+ a ( h 3 + ) = = which implies m TN = 9 x(.84 x/ 3) = 3x + 7.3x () S we find an equatin fr x : 3x + 7.3x =.474 (3) The slutins t (3) are x =.337 and x =.673. Since x has t be smaller than.53, we have t take x = x =.673 and m= 9x =.65kg. h tan = =. 436, r = x+ h 3 Answer: m =.6kg and =3.6.. Parameters f the wrking mde

3 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. /Slutin..Graphs f μ( ), ( t), and μ(t) during ne peratin cycle. Initially when there is n water in the bucket, =, μ has the largest magnitude equal t gm TG = = 4.64 N m. Our cnventin will be that the sign f this trque is negative as it tends t decrease As water flws int the bucket, the trque cming frm the water (which carries psitive sign) makes μ increase until μ is slightly psitive, when the lever starts t lift up. Frm that mment, by assumptin, the amunt f water in the bucket is cnstant. The lever tilts s the center f mass f water mves away frm the rtatin axis, leading t an increase f μ, which reaches maximum when water is just abut t verflw the edge f the bucket. At this mment = =.6. A simple calculatin shws that SI = SP + PQ / = / =.634m. TN = SI =.7644 m. 3 μ max = (. TN 3 TG) g cs.6. Therefre = ( ) 9. 8 cs. 6 =. 69 N m. μ =.7 N m. max As the bucket tilts further, the amunt f water in the bucket decreases, and when =, μ =. Due t inertia, keeps increasing and μ keeps decreasing. The bucket is empty when = 3, when μ equals 3 g TG cs 3 = 4. N m. After that keeps increasing due t inertia t ( μ = gm TGcs = 4.6cs N m ), then quickly decreases t ( μ = 4.6N m). belw On this basis we can sketch the graphs f ( t), μ() t, and μ( ) as in the figure 3

4 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. /Slutin μ.7 N.m A 3.6 O E.6 B cs N.m D -4. N.m C -4.6 N.m F.. The infinitesimal wrk prduced by the trque μ( ) is dw = μ( ) d. The energy btained by the lever during ne cycle due t the actin f μ( ) is W = μ( ) d, which is the area limited by the line μ( ). Therefre is equal W ttal t the area enclsed by the curve (OABCDFO) n the graph μ( ). The wrk that the lever transfers t the mrtar is the energy the lever receives as it mves frm the psitin = t the hrizntal psitin =. We have Wpunding equals t the area f (OEDFO) n the graph gm TG sin = 4. 6sin (J). μ( ). It is equal t.3. The magnitudes f can be estimated frm the fact that at pint D the energy f the lever is zer. We have area (OABO) = area (BEDCB) Apprximating OABO by a triangle, and BEDCB by a trapezid, we btain: (/ ) = 4. [( 3.6) + ( 3)] (/ ), which implies = Frm this we find W = area (OEDFO) = Mg TG cs d = 46. sin 347. = 63. punding

5 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. /Slutin Thus we find Wpunding.6 J. μ 3. The rest mde The bucket is always verflwn with water. The tw branches f μ( ) in the vicinity f = crrespnding t increasing and decreasing cincide with each ther. The graph implies that = is a stable equilibrium f the mrtar Find the expressin fr the trque μ when the tilt angle is = +Δ ( Δ is small ). The mass f water in bucket when the lever tilts with angle is m= (/ ) ρbhpq, where PQ = h. A simple calculatin shws that tan tan 3 when increases frm t + Δ, the mass f water increases by Δ bh bh m = ρ ρ sin Δ sin Δ. The trque μ acting n the lever when the tilt is +Δ equals the trque due t Δm. We have μ =Δ m g TN cs( +Δ). TN is fund frm the equilibrium cnditin f the lever at tilting angle : TN = M TG / m= 3.57/.65 =.779m. We find at the end μ = 47. Δ N m 47 Δ N m Equatin f mtin f the lever d μ = where μ = 47 Δ, = +Δ, and I is the sum f mments I dt f inertia f the lever and f the water in bucket relative t the axis T. Here I is nt cnstant the amunt f water in the bucket depends n. When Δ is small, ne can cnsider the amunt and the shape f water in the bucket t be cnstant, s I is apprximatey a cnstant. Cnsider water in bucket as a material pint with mass.6 kg, a simple calculatin gives I = =.36.4 kg m. We have 47.4 d Δ Δ =. That is the equatin fr a harmnic scillatr with perid dt 5

6 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. /Slutin.4 τ = π = 3.7. The answer is therefreτ = 3. s Harmnic scillatin f lever (arund = ) when bucket is always verflwn. Assume the lever scillate harmnically with amplitude Δ arund =. At time t =, Δ =, the bucket is verflwn. At time dt the tilt changes by d. We are interested in the case d <, i.e., the mtin f lever is in the directin f decreasing, and ne needs t add mre water t verflw the bucket. The equatin f mtin is: Δ = Δ sin( πt / τ), therefre d( Δ ) = d = Δ ( π / τ)cs( πt/ τ) dt. Fr the bucket t be verflwn, during this time the amunt f water falling t the bucket shuld be at least bh ρ Δπ bh ρdt π t dm = d = cs sin τsin τ ; dm is πbh ρδ maximum at t =, dm = dt. τsin The amunt f water falling t the bucket is related t flw rate Φ ; dm =Φdt, πbh ρδ therefre Φ=. τ sin An verflwn bucket is the necessary cnditin fr harmnic scillatins f the lever, therefre the cnditin fr the lever t have harmnic scillatins with ampltude π/36 rad is Φ Φ with r S Φ =.3kg/s. Φ = πbh ρπ. 39kg/s 36τsin = 3.3 Determinatin f Φ If the bucket remains verflwn when the tilt decreases t.6, then the amunt f water in bucket shuld reach kg at this time, and the lever scillate harmnically with amplitude equal = 3 3Φ. The flw shuld exceed, therefre Φ = 3.3.7kg/s. This is the minimal flw rate fr the rice-punding mrtar nt t wrk. 6