University of California, Berkeley Physics 7A Spring 2009 (Yury Kolomensky) SOLUTIONS TO PRACTICE PROBLEMS FOR THE FINAL EXAM

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1 1 University of Caifornia, Bereey Physics 7A Spring 009 (Yury Koomensy) SOLUIONS O PRACICE PROBLEMS FOR HE FINAL EXAM Maximum score: 00 points 1. (5 points) Ice in a Gass You are riding in an eevator hoding a gass of water with an ice cube foating in it. (a) (10 points) When the eevator is at rest, what fraction of the ice is submerged in water? (b) (15 points) As the eevator acceerates upward, you notice that the boc is dips beow the surface. What is the acceeration of the eevator? Use the density of water ρ w = 1 g/cm 3 and the density of ice ρ i = 0.9 g/cm Soution (a) his is a fairy simpe appication of Newton s and buoyancy aws. At rest, the acceeration of the ice cube is zero. here are two forces acting on the boc: buoyancy force F b = ρ w gv submerged, pointing up, and gravity mg, pointing down: F b mg = 0 ρ w gv submerged = mg Expressing the mass of the boc in terms of the tota voume of the boc and density: m = ρ i V we can sove for the fraction of the boc submerged in water: V submerged V = ρ i ρ w = 0.9 In other words, 90% of the boc (by voume) is submerged in water. (b)) In this case, the ice boc is acceerating upward (together with water and the eevator): F b mg = ma where F b = ρ i gv, since the entire voume is submerged in water. Again substituting m = ρ i V, we can sove for acceeration: a = ρ wgv ρ i gv ρ i V ( ) ρw = g 1 = 1.1 m/s. ρ i. (30 points) Hammer hrow at the Oympics Hammer hrow is an Oympic rac-and-fied event. he object is to throw a heavy stee ba (which used to be an actua hammer) for maximum distance. In men s events, the ba weighs 16 bs (m = 7.3 g). In the act of throwing, an athete quicy spins the ba in a circe on the end of a 4 ft ( = 1. m) wire hande before etting it fy. he word record in men s events beongs to Yuriy Sedyh of Soviet Union, who threw the hammer m in (a) (15 points) Ignoring air resistance and assuming the optima aunch ange, estimate the speed with which the hammer was aunched. (b) (15 points) Estimate the tension of the wire just before the ba was aunched, whie it was sti foowing a circuar trajectory. (a) Let s reca the inematics equation for the range of a projectie. If the ba is iced at ange θ over horizon and it ands at the same eevation, the ba traves the horizonta distance L = v sinθ g It immediatey foows, that if v is fixed, the maximum range L max = v g is achieved when θ = π/4 rad = 45. Pugging in L max = m, we can sove for veocity of the hammer:. v = L max g = 9 m/s. (1) (b) his part is actuay a bit ambiguous. I did not te you whether the hammer is rotating in the horizonta pane or at some ange to the horizon. If you watch

2 the video, it is cear that the hammer is actuay spun in a pane that is tited to the horizonta by about 45 o. his maes it easier to reease the ba at the proper ange. However, when the ba is reeased, the veocity vector is at 45 o to the horizonta, and the wire is in the horizonta pane. herefore, the wire tension provides centripeta acceeration, and the ony other force in the probem (gravity) provides tangentia acceeration. It wi be cear however, that the tangentia force is much smaer than the centripeta, and for the estimates, it can be ignored. he reationship between the force and the acceeration is given by the Newton s aw: = ma cp = m v where, the ength of the wire, is the radius of rotation. Pugging in the expression for v from Eq. (1), we compute = mg L max = 5 N. his force is about 80 times arger than the weight of the ba, in fact, it is even arger than the weight of the person! Ceary, compared to this force, the weight of the ba (and the additiona acceeration it may create) can be ignored. 3. (40 points) Haf-empty or haf-fu? Imagine a ta cyindrica gass of radius r = 3 cm, height h = 15 cm with wa thicness w = mm and bottom thicness W = 5 mm. Density of gass is ρ g =.5 g/cm 3. What configuration is most stabe against tipping over: (a) An empty gass (b) A gass fu of sugar (c) A gass haf-fu of sugar Assume the density of sugar ρ s = 1.6 g/cm 3. Hint: the most stabe configuration is the one with the owest center of gravity (center of mass). Why? 3. Soution You may now this intuitivey. Haf-fu gass is most stabe, because it has the owest center of mass. You can test this at home, but have a vacuum handy to safey pic up the pieces of gass. CM α R h CM How woud you quantify stabiity of an object? We did discuss it in ecture briefy. We can say that the object is in a stabe equiibrium if the gravitationa torques acting on it return it to the equiibrium position. As shoud be obvious from the picture above, the gass woud return into the vertica position if after it is tipped to the right, its center of mass is sti eft of the vertica ine drawn through the axis of rotation the edge of the base. For a given configuration empty, fu, or haf-fu gass, there is a maximum ange α max such that if the gass is tipped by ess than α max, it wi return to the vertica position, and wi tip over if it is tited by an ange arger than α max. It is cear from the picture then tan α max = r h CM where h CM is the height of the center of mass. he arger this ange α max, the more stabe the gass. he most stabe configuration woud be achieved when the center of mass of the gass is owest. Which of the three configurations has the owest center of mass? It shoud be intuitivey obvious that this woud be the haf-fu gass. On an exam, such answer woud be sufficient. But for educationa purposes, et s prove it with a bit of math. he gass density ρ g is the same in the sides and in the base, and the side was are fairy thin: thicness w r and w h. he position of the center of mass of the empty gass reative to the bottom is πrhwρ g h/ h e = πrhwρ g + πr Wρ g = h 1 ( ) = 6 cm () 1 + rw hw he center of mass is apparenty 1.5 cm beow the midde of the gass.

3 3 he center of mass sugar in the fu gass is ocated at height h/. herefore, the center of mass of the gass and sugar is h f = h em g + h/m s M g + M s = 7.3 cm where M g = 177 g is the mass of the gass, and M s = 1060 g is the mass of the fu gass of sugar. he center of mass of the fu gass is above the center of mass of the empty gass. So the ony remaining contender is the haf-fu (or haf-empty ;-) gass. If M 1/ = 530 g is the mass of the haf of gass of sugar, then the center of mass of the haf-fu gass is h 1/ = M gh e + M 1/ h/4 M g + M 1/ = 4.3 cm herefore, our intuition that the haf-fu gass is the most stabe of three configurations was correct. 4. (35 points) Bird on a String A cothesine of ength = 10 m is strung horizontay between two trees to tension = 100 N. A bird of mass m = 1 g ands on the string exacty in the midde. Before catching the string, the bird had a downward veocity v = 1 m/s. As the bird hods on firmy to the string, it starts osciating in the vertica direction. (a) (15 points) Determine the period of the osciations. (b) (10 points) Find the owest and highest points of the osciations, reative to the horizonta string. (c) (10 points) What harmonics of the string does the bird excite? Are any of them audibe? 4. Soution (a) First of a, we sha prove that the motion of the bird on the string is osciatory. Osciations occur in a system dispaced from equiibrium, if there is a restoring force, proportiona to the dispacement. We ve covered a coupe of situations ie this in cass and on the Probem of the wee page. 0 y α mg So et s find the equiibrium position y 0 of the bird first. hat is the position where the net force acting on the bird is zero. We wi point Y axis upward, and count y vaues reative to the horizonta string. In other words, the equiibrium position of the bird wi be at negative y. here are three forces acting on the bird (see picture): two tension forces, which are equa in magnitude to and the gravity force mg. he baance of forces in the vertica direction impy sinα = mg he ange α is proportiona to the dispacement y 0 : sin α tan α = y 0 = y 0 So the tension force which baances the gravity force is F restoring = 4 y 0 = mg Bingo! his is just ie Hooe s aw: the force from the string F restoring is proportiona to y 0, and the effective spring constant is = 4 he equiibrium position is at y 0 = mg = 40 N/m = 5 cm. If the bird is dispaced from the equiibrium position to eevation y, it wi experience net force F net = 4 y mg = 4 (y y 0) = (y y 0 ) he bird s acceeration satisfies the osciator equation: a y = d y dt = m (y y 0) Just ie a mass on a vertica spring, the bird woud be osciating around y 0 with the anguar frequency ω = = 6.3 rad/sec m he period of osciations is = π ω = 1 sec

4 4 (b) he extremes can be most easiy found from energy conservation. At the owest and highest points, the inetic energy of the bird is zero. he initia energy of the bird is E 0 = mv Suppose the maximum/minimum position is y, reative to the horizonta (unperturbed) string. At that position, the tota energy of the bird is a sum of the gravitationa potentia energy mgy and the spring potentia energy is y /. From energy conservation mv = mgy + y his quadratic equation can be soved for y: y = mg ± m g + mv which has two vaues: y max = 5 cm, y min = 55 cm. (c) his is a itte more interesting. Normay, for the cosed-cosed boundary conditions, a string woud have the foowing mode frequencies: f = v, = 1,,3,... (3) L However, not a of these modes wi be excited by hitting the string in the midde. When we the bird ands in the midde of a string, it creates an anti-node there, a pace where the string has argest ampitude. So the ony modes from Eq. (3) that wi be excited are the modes which have an anti-node in the string midde. hese are ony odd modes, i.e. = 1,3,5,.... Hence, the tones we wi hear wi have frequencies f n = (n + 1) = (n + 1) Hz, n = 0... For n > 10, these frequencies are in the audibe range (but barey). he other way to get the same picture woud be to say that the string has to satisfy 3 boundary conditions: cosed on each end, and open in the midde. So if you consider haf of the string, it wi have cosed-open (node-antinode) conditions. hese conditions produce the tones given above (ony odd harmonics of 1 Hz). Yet another, reay coo (to some), way to arrive at the same answer (many ways to sin this cat) is to recognize that the shape of the string, when the bird ands on it, is trianguar. We are trying to decompose it into harmonics of a standing wave on a string, which are sine waves. his is nown as a Fourier decomposition, and foows pretty standard toos of advanced cacuus (EE majors wi earn this the hard way in upper division). One can show [1] that the Fourier series of a triange wave contains ony odd harmonics. 5. (40 points) Leay Centrifuge A horizonta pipe AB of ength is rotating with a constant anguar veocity ω around the vertica axis OO (see picture beow). he pipe is fied with some incompressibe iquid to ength h <. he end A is open to air, and the end B is cosed, but has a very sma hoe. Find the veocity with which the iquid eas out of the hoe. 5. Soution he veocity of the water is given by the Bernoui s equation. he pressure on the right of the hoe is P a, the atmospheric pressure. he pressure on the eft of the hoe is P a +P, the sum of the atmospheric pressure and the hydrostatic pressure created by the iquid. he veocity of the iquid is zero on the right of the hoe (reative to the pipe) and the veocity on the right is v. Bernoui s equation says P a + P = P a + ρv and therefore, the water exists with the veocity P v = ρ (4) Now we need to find the hydrostatic pressure P. We do it simiary to how we derived the hydrostatic pressure. Let s consider a thin ayer of iquid (sma width dx) at the distance x from the axis of rotation. a cp P+dP 0 h x P dx

5 5 his ayer experiences centripeta acceeration a cp = ω x, produced by the difference in hydrostatic pressure on the right and on the eft from it dp : dma cp = dpa where dm = ρadx is the mass of the ayer, and A is the cross sectiona area of the pipe. his gives us an expression for dp : dp = ρω xdx which a simpe differentia equation soved by direct integration. he pressure at the far right end of the pipe is P() P = h = ρω h dp = ρω [ h 1 h ] xdx Pugging this into Eq. (4), we find the veocity of the iquid streaming from the hoe v = ωh h 1 6. (30 points) Foghorns in the Bay wo tugboats, moving toward each other amost headon in the Oaand Harbor, sound foghorns at the same time. Each tugboat is moving with the speed of 10 nots (5 m/s) and the frequency of the foghorn sound is 00 Hz. What woud the captains of each boat hear? Assume the speed of sound in air v = 343 m/s (at air temperature of 0 o C). 6. Soution his is an appication of two effects we studied: the Dopper effect and the beating. First, et s determine what frequencies woud be heard by the captain of one of the boats. Let s denote the natura frequency of the horn f 0 = 00 Hz. he captain and his own boat move with the same veocity, so the captain woud observe the sound of his own horn at the same frequency f 1 = f 0. he sound from the other boat, on the other hand, woud be shifted in frequency. he detector (captain) is moving with veocity v D = V = 5 m/s toward the source of sound. At the same time, the source is moving with veocity v S = V = 5 m/s towards the boat. he genera formua that describes the Dopper effect is the foowing: f detected = f 0 v ± v D v v S he sign assignment is as foows. If the source (or detector) is moving aong the direction of the wave, we pic the minus sign in the denominator (numerator). If the source (or detector) are moving opposite to the wave, we pic the pus sign. he other way to thin about it is to remember this. If the source and the detector are moving toward each other, the detected frequency is higher than the origina. If they are moving away from each other, the detected frequency is ower (this is simiar to the red shift observed in the ight from distant stars, which are moving away from us due to the expansion of the Universe). Putting this a in pace, we get the frequency of the sound from the other boat: f = f 0 v + V v V f 0 ( 1 + V ) v (5) he ast expression is an approximation vaid for V v. Eq. (5) predicts that the sound from another boat wi be heard by captain at frequency ( f = f V ) 06 Hz. v His own boat s horn wi be heard at the origina frequency f 1 = f 0 = 00 Hz. he two waves wi interfere, and the resuting sound woud be a pitch of f = 0.5(f 1 + f ) = 03 Hz moduated by beat frequency f beat = f f 1 = 6 Hz. [1] See for exampe

Module 22: Simple Harmonic Oscillation and Torque

Module 22: Simple Harmonic Oscillation and Torque Modue : Simpe Harmonic Osciation and Torque.1 Introduction We have aready used Newton s Second Law or Conservation of Energy to anayze systems ike the boc-spring system that osciate. We sha now use torque