Examination Cover Sheet Princeton University Undergraduate Honor Committee January 22, 2007 Course Name & Number: PHY 101 Professors: Galbiati, Kwon, McDonald Date: January 22, 2007 Time: 9 AM This examination is administered under the Princeton University Honor Code. Students should sit one seat apart from each other, if possible, and refrain from talking to other students during the exam. All suspected violations of the Honor Code must be reported to the Honor Committee Chair at honor@princeton.edu. The items marked with YES are permitted for use on this examination. Any item that is marked with NO or not mentioned may not be used and should not be in your working space. Assume items not on this list are not allowed for use on this examination. Please place items you will not need out of view in your bag or under your working space at this time. University policy does not allow the use of electronic devices such as cell phones, PDAs, laptops, MP3 players, iPods, etc. during examinations. Students may not wear headphones during an examination. вЂў вЂў вЂў вЂў Course textbooks: NO Course Notes: NO Other books/printed materials: NO Formula Sheet: YES, only the one available at the end of the exam booklet вЂў Comments on use of printed aids: NO вЂў Calculator: YES, but only for standard functions. Cannot use graphing functions or advanced functions, like equation solving Students may only leave the examination room for a very brief period without the explicit permission of the instructor. The exam may not be taken outside of the examination room. During the examination, the Professor or a preceptor will be available at the following location: outside the exam room. This exam is a timed examination. You will have 3 hours and 0 minutes to complete this exam. Write your name in capital letters in the appropriate space in the next page. Also, dont forget to write and sign the Honor Code pledge in the appropriate line on the next page: вЂњI pledge my honor that I have not violated the Honor Code during this examinationвЂќ 1 2 Your Name: PHYSICS 101 FINAL EXAM January 22, 2007 1 3 5 3 hours Please Circle your section 9 am Galbiati 2 10 am Kwon 11 am McDonald 4 12:30 pm McDonald 12:30 pm Kwon Problem Score 1 /20 2 /15 3 /12 4 /20 5 /20 6 /20 7 /20 8 /20 9 /15 Total /162 Instructions: When you are told to begin, check that this examination booklet contains all the numbered pages from 2 through 30. The exam contains 9 problems. Read each problem carefully. You must show your work. The grade you get depends on your solution even when you write down the correct answer. BOX your final answer. Do not panic or be discouraged if you cannot do every problem; there are both easy and hard parts in this exam. If a part of a problem depends on a previous answer you have not obtained, assume it and proceed. Keep moving and finish as much as you can! Possibly useful constants and equations are on the last page, which you may want to tear off and keep handy. Rewrite and sign the Honor Pledge: I pledge my honor that I have not violated the Honor Code during this examination. Signature 3 Problem 1 Miscellaneous Questions 1. (5 points) A marathon runner runs at a steady 15 km/hr. When the runner is 7.5 km from the finish, a bird begins flying from the runner to the finish at 30 km/hr. When the bird reaches the finish line, it turns around and flies back to the runner, and then turns around again, repeating the back-and forth trips until the runner reaches the finish line. How many kilometers does the bird travel? вЂў 10 km вЂў 15 km вЂў 20 km вЂў 30 km It takes 30 minutes for the runner to arrive at the finish line. In this time span, the bird covers a distance of 15 km . 4 2. (5 points) From what minimum height h does the car need to start to make it around the loop? Ignore friction. h R вЂў 3 R 2 вЂў 2R вЂў 5 R 2 вЂў Not enough info By application of the law of conservation of the total energy, the kinetic energy at the top of the loop is: 12 mv 2 = mg(h в€’ 2R). If the car barely makes it around the loop, the contact force vanishes on the very top of the loop. At that point, the centripetal force 2 is provided entirely by gravity. Therefore: mv R = mg. Combining the two equations: mg(h в€’ 2R) = h = 25 R 1 mgR 2 5 3. (5 points) ItвЂ™s a very hot summer in Princeton and the air conditioner is broken. Your roommate decides that a possible relief might come from the refrigerator, and proposes to leave the door of the refrigerator open (with the windows closed, of course). This will: вЂў decrease the temperature in the room вЂў increase the temperature in the room вЂў will not affect the temperature in the room It will increase the temperature in the room . Every cycle, the refrigerator removes from the room a quantity of heat QC , from the chilled side of the fridge which is open to the room, and exhausts the quantity QH , through the vent and to the same room. The net balance is that the room gets a positive quantity of heat: QH в€’ QC = W where W is the work needed to run the refrigerator through one cycle. 6 4. (5 points) A satellite orbiting around the Earth passes nearby the International Space Station. Operators from the Space Station double the mass of the satellite by adding a payload with scientific instruments. The radius of the orbit of the satellite can stay constant if its speed: вЂў increases by a factor of 8 вЂў increases by a factor of 4 вЂў increases by a factor of 2 вЂў stays constant вЂў decreases by a factor of 2 The centripetal force acting on satellites is the gravitational force due to the central body, around which the satellite is orbiting: mv 2 mM =G 2 R R where m is the mass of the satellite, M is the mass of the central object, and R is the radius of the orbit. The equation can be solved for R: R=G M v2 The radius depends on the velocity of the satellite, not on its mass. Therefore the radius stays constant if the speed stays constant 7 Problem 2 A copper flask (emissivity = 1) of radius 10 cm contains liquid argon at its boiling point for a pressure of 1 atm, i.e. 87 K. The flask is enclosed in a steel vessel which has a uniform temperature of 200 K. The flask is held to the top of the vessel by a teflon neck, whose cross-sectional area if 1 cm2 and whose length is 10 cm. The top of the teflon neck is in good thermal contact with the steel vessel, and the bottom of the neck is at the temperature of the copper flask. The flask is open to the atmosphere outside the steel vessel (pressure 1 atm). Teflon Neck Copper Flask Stainless Vessel a.) (5 points) What is the net thermal power adsorbed by the vessel due to the radiant heat? You may neglect here the radiant power from the neck. Here we call the area of the spherical flask A = 4ПЂR2 . The net power adsorbed by the flask through radiation is given by: 4 4 в€’ Tflask = 1 В· 5.67 Г— 10в€’8 Prad = ПѓA Tvessel W В· 4ПЂ(0.1 m)2 В· 2004 в€’ 874 K4 m2 K4 Prad = 10.98 W 8 b.) (5 points) What is the thermal power adsorbed by the vessel due to conduction along the neck? The thermal conductivity of teflon is 0.2 W mв€’1 Kв€’1 . The area through which the heat is conducted is B = 1 cm2 = 10в€’4 m2 . The thermal power adsorbed by conduction is: Pcond = kBв€†T 0.2 W mв€’1 Kв€’1 В· 10в€’4 m2 В· 113 K = L 0.1 m Pcond = 0.023 W c.) (5 points) What is the rate at which the argon inside the flask is boiling? The heat of vaporization of argon is 6.43 kJ molв€’1 . The molar mass of argon is 40 g. Quote your answer in units of grams/seconds. The total thermal power adsorbed is: P = Prad + Pcond = 11 W The rate of evaporation is: 11 J 1 mol g Г— Г— 40 = 0.068 g/s s 6430 J mol 9 Problem 3 a.) (6 points) Two cars with masses of 1,500 kg each, both traveling at a speed of 100 km/hr, collide in a head-on collision and stick together. Find the change in entropy of the Universe if the temperature is 27 degrees Celsius. This is the вЂњclassicвЂќ head-on collision. Due to momentum conservation, the two cars stop altogether. The final kinetic energy is zero. The initial kinetic energy, suddenly, becomes вЂњunavailableвЂќ for work and is degraded into heat or other forms (work required for deformation) not available for performing work. The change in entropy is of course positive, as with any irreversible process. 1 km m 1 hr Wunavailable = KEi = 2Г— mv 2 = 1, 500 kg 100 1000 2 hr km 3600 s в€†S = Wunavailable 1.16 Г— 106 J = = 3,860 J/K T 300 K 2 = 1.16Г—106 J 10 b.) (6 points) The surface of the Sun has a temperature of 7,500 K and the surface of the Earth is at 290 K. Find the change in entropy of the Universe per each 1,000 J of energy that are transferred, by radiant heat, from the Sun to the Earth. The adsorption of heat by the Earth causes an entropy increase. The release of heat from the Sun causes an entropy decrease. The net balance is an entropy increase, since the heat flows from a hot to a colder source. The process, of course, in not reversible. в€†S = 1, 000 J 1, 000 J в€’ = 3.31 J/K 290 K 7, 500 K 11 Problem 4 Suppose that there is a cylindrical tunnel bored along the diameter of the Earth through its center. At one end (entrance) of the tunnel, a small ball of mass m = 1 kg is dropped from rest. In this problem, assume that the Earth is a solid sphere with uniform mass density. r R a.) (4 points) What is the mass density of the Earth? ME = 6.0 Г— 1024 kg RE = 6.4 Г— 106 m Therefore, ПЃE = ME ME 3 3 = 3 = 5.5 Г— 10 kg/m VE (4ПЂ/3)RE 12 b.) (4 points) Find the gravitational force on the ball when it is at a distance r from the center of the Earth. The gravitational force at this position is determined as if all the mass inside the sphere of radius r (see the figure) is concentrated at the center. It is also known that the mass outside the sphere of radius r does not affect any net gravitational force on the ball. Mass of the Earth inside the radius r is: M (r < RE ) = 4 3 ПЂr ПЃE 3 Therefore the gravitational force at r on a ball of mass m is: F (r) = в€’G ПЃE (4ПЂ/3)r3 m = в€’GПЃE (4ПЂ/3)mr r2 where the (в€’) sign indicates the direction of the force is toward the center. The numerical value for GПЃE (4ПЂ/3) is 1.5 Г— 10в€’6 N kgв€’1 mв€’1 , to be used later. c.) (4 points) The force found in part (b) implies that the ball moves in simple harmonic oscillation. How long does it take for the ball to reach the other end of the tunnel? Note that the force found in (b) can be written in the form F = в€’kr, where k = m В· 1.5 Г— 10в€’6 N kgв€’1 mв€’1 . This is the force for simple harmonic osciallation. Then, the angular frequency П‰ for the oscillation can be found as: П‰= k = m m В· 1.5 Г— 10в€’6 N kgв€’1 mв€’1 = 1.2 Г— 10в€’3 sв€’1 m The period T of the oscillation is: T = 2ПЂ = 5.1 Г— 103 s П‰ The time в€†t for the ball released from the entrance (r = RE ) to reach the other end is: в€†t = T /2 = 2.5 Г— 103 s 13 d.) (4 points) What is the speed of the ball when it passes the center of the Earth? In a simple harmonic oscillation, the energy E = (1/2)kx2 + (1/2)mv 2 is conserved. Comparing the energy at the entrance and at the center, we have: 1 2 1 kR = mv 2 , 2 E 2 where v is the speed of the ball at the center. Therefore, k v= RE = П‰RE = 7.9 Г— 103 m/s m e.) (4 points) Now we release the ball from rest at a distance r = R/2 from the center. (DonвЂ™t ask how we could get there to begin with.) For this case, what are the period of oscillation and the speed of the ball at the center of the Earth? In a simple harmonic oscillation, the period does not depend on the amplitude of the motion. Therefore, the period T remains the same: T = 5.1 Г— 103 s On the other hand, reduing the amplitude to half (RE в†’ RE /2) results in the change in total energy, hence the change in the speed at the center. v= k RE RE =П‰ = 4.0 Г— 103 m/s m 2 2 14 Problem 5 A ball of mass m = 0.1 kg is attached to one end of a wire (of length = 1 m), the other end being fastened to a fixed point. Initially the ball is released from rest when the angle between the tightly stretched wire and the vertical is 60в—¦ . It swings down and strikes a block (of mass M = 0.2 kg) initially at rest on a horizontal frictionless surface. Assume that the air resistance is negligible and the collision is elastic. m M a.) (5 points) During the downward swing of the ball (before collision), are any of the following observables conserved. Circle the observable(s) which are conserved and provide a very brief explanation for your answer. вЂў the momentum of the ball вЂў the kinetic energy of the ball вЂў the total energy of the ball The gravitation acts as an external force on the ball. Therefore, the momentum of the ball is not conserved. In the downward swing the gravitaional potential energy decreases, resulting in an increase of the kinetic energy. Therefore, the kinetic energy of the ball is not conserved. The total energy of the ball is conserved. Ignoring the air resistance (and any other non-conservative forces), the conserved energy is equal to the sum of the gravitational potential energy and the kinetic energy. 15 b.) (5 points) Find the velocity (magnitude and direction) of the ball just before and just after the collision. вЂў Just before collision: letвЂ™s v1 be the velocity of the ball just before collision. mgh = mg (1 в€’ cos(60в—¦ )) = v1 = 1 mv 2 2 1 2g (1 в€’ cos(60в—¦ )) = 3.13 m/s The direction is to the right. вЂў Just after collision: Since the ball makes an elastic collision with the block, both the total linear momentum and the total kinetic energy are conserved just before and just after the collision. One way to answer this problem is to solve the simultaneous equations arising from linear momentum and kinetic energy conservations. Another way is to work in the center-of-mass (CM) frame where the velocity vectors simply change its sign before and after the collision. Velocity of the CM: vCM = mv1 /(m + M ) = 1.04 m/s Let v1CM and V1CM be the veolcities of the ball and the block, respectively, measured in the CM frame, just before collision: v1CM = v1 в€’ vCM = 2.09 m/s V1CM = V1 в€’ vCM = в€’1.04 m/s Just after collision, the velocity vectors flip sign (denoted with a subscript вЂњ2вЂќ): v2CM = в€’v1CM = в€’2.09 m/s V2CM = в€’V1CM = +1.04 m/s Coming back to the laboratory frame, the velocity of the ball becomes: v2 = v2CM + vCM = -1.04 m/s The (в€’) sign indicates that the ball moves to the left after the collision. 16 c.) (5 points) If the ball is in contact with the block for a time of 0.001 s, what is the average force exerted on the ball? Let F be the average force exerted on the ball during collision. m(v2 в€’ v1 ) 0.1 kg В· (в€’1.04 m/s в€’ 3.13 m/s) в€†p = = = - 417 N в€†t в€†t 0.001 s The (в€’) sign indicates that the force (on the ball) is directed to the left. F = d.) (5 points) How high (measuring from the point of collision) will the ball reach after the collision? mgh = h= 1 mv 2 2 2 v22 = 5.56 Г— 10в€’2 m 2g 17 Problem 6 Playing a baseball game, you are catching a ball (of mass m = 0.15 kg) with an initial speed of 40 m/s (в‰€ 90 miles per hour) thrown by the pitcher in a horizontal direction. a.) (3 points) If the ball moves 18 m horizontally before being caught by you, how much will it fall down due to vertical gravitational force? You may neglect air resistance. Time interval Оґt for the ball to move 40 m horizontally: Оґt = 18 m = 0.45 s 40 m/s Vertical fall Оґy during this time is: Оґy = 1 g В· Оґt2 = 0.99 m 2 b.) (3 points) What is the vertical component of its velocity just before it is caught? vy = в€’g В· Оґt = 4.41 m/s 18 c.) (4 points) How much work must be done on the ball to stop it? (Do not forget the vertical component of the velocity.) The work needed to stop the ball is equal to the kinetic energy of the ball just before being stopped. 1 1 W = в€†KE = 0 в€’ mv 2 = в€’ (0.15 kg) (40 m/s)2 + (4.41 m/s)2 = - 121 J 2 2 where we ignored the work done by the gravitational force during the stopping process. The (в€’) sign indicates that the work was done to decrease the kinetic energy of the object. d.) (3 points) If the ball is stopped in 2.0 cm with a constant deceleration, what is the average force acting on the ball while it is brought at rest? Suppose that you apply a force in the opposite direction to the final velocity of the ball. Then, the work (which you exert on the ball) can be written as W = F Оґs, where F is the average force you exert on the ball, and Оґs is the stopping distance. Then, W F = = в€’6.07 Г— 103 N Оґs The (в€’) sign indicates that the force was applied in the opposite direction to the displacement of the ball. 19 e.) (3 points) What is the average power transferred to the ball during the stopping process in part (d)? Assuming that the ball is stopped in a constant deceleration, the average velocity during the stopping process is obtained by averaging the initial and final velocity: v= 1 0+ 2 (40 m/s)2 + (4.41 m/s)2 = 20.1 m/s Then the average power P transferred to the ball is: P = F v = 1.22 Г— 105 W f.) (4 points) Suppose that the same amount of power as obtained in part (e) is applied to 1 kg of water at 20в—¦ C. How long will it take to heat this water to 90в—¦ C? Let M be the amount of water. Q = CM в€†T = 4186 J kgв€’1 в—¦ Cв€’1 Г— 1 kg Г— 70 в—¦ C = 2.93 Г— 105 J The time в€†t required to transfer this amount of heat with the power obtained in part (e) is: Q в€†t = = 2.40 s P 20 Problem 7 Water emerges from a vertical faucet, whose diameter is D = 1 cm, with an initial downwards velocity of v0 = 1 m/s. Thereafter, gravity accelerates the water, and since water is incompressible the diameter of the water column must decrease so as to keep the mass flux ПЃvA constant. a.) (5 points) What is the diameter d of the column at a distance h = 30 cm below the outlet of the faucet? The velocity v of the water after falling through height h is given by: v 2 = v02 + 2gh = (1 m/s)2 + 2 В· 9.8 m sв€’2 В· 0.3 m = 6.88 m2 /s2 so that v = 2.62 m/s. The flow of incompressible water with mass density ПЃ has a constant flux of mass, ПЃvA, where A is the area of the flow. Hence, v0 Atop = vAbottom , or ПЂv0 D2 /4 = ПЂvd2 /4, and the diameter d of the flow at the bottom is: d=D v0 /v = 1 В· 1/2.62 = 0.62 cm 21 b.) (5 points) Water emerges from a hole in the side of a water tank at a height h = 5 m below the water level. What is the velocity v of the water just after it exits the hole in the tank? What is the pressure P1 inside the tank at the height of the hole? You may ignore the viscosity of water in this situation, so that energy is conserved. The tank is large enough that the vertical velocity v0 of water in the tank is negligible. Also, you may ignore the small difference in the air pressure at the top of the tank and at the hole. A solution can be based directly on conservation of energy, or on BernoulliвЂ™s law, which is based on conservation of energy. One view is that the water that emerges from the hole has disappeared from the top of the tank, so that it has gained gravitational potential energy PE = mgh. This energy is converted to its kinetic energy KE = mv 2 /2. Thus: v= в€љ 2gh = 2 В· 9.8 m/s В· 5 m = 9.89 m/s Alternatively, we can invoke BernoulliвЂ™s law that P +ПЃgh+ПЃv 2 /2 is constant for smooth fluid flow in which viscosity can be ignored. Assuming also that the pressure P is the same at the top of the tank and just outside the hole and that the velocity of the water is negligible at the top of the tank, we obtain the relation gh = v 2 /2, which leads to the boxed answer for v. The pressure inside the tank at the height of the hole follows from another application of BernoulliвЂ™s law in which v = 0 in both locations. That is: P1 = P0 + ПЃgh = 100, 000 Pa + 1000 kg mв€’3 В· 9.8 m sв€’2 В· 5 m = 149,000 Pa 22 c.) (5 points) The velocity of the water increases as it falls after emerging from the hole of area A in the tank, and the area of the water jet decreases in a manner related to part a of this problem. Careful observation reveals that, in addition, the area of the water jet decreases rapidly for a short time just after it exits the hole, before gravity has any effect on the velocity. The area of the jet after it has converged is A0 , as shown in the figure. To understand this effect, you will make an analysis in parts c and d using the concept of momentum flux, О¦p , i.e. the total momentum carried by all the water particles that crosses a surface perpendicular to the flow per unit time. Give an expression for the momentum flux О¦p across area A0 in terms of the velocity v of the jet across that area (which is the velocity that you found in part b). Hint: keep in mind that the mass flow is ПЃAv. What is the relation between mass and momentum? Momentum = mass times velocity, so momentum flux = mass flux times velocity, and, recalling parts a and b: О¦p = ПЃv 2 A0 = 2ПЃghA0 23 d.) (5 points) Continue the analysis of the horizontal, converging flow of water out of a hole of area A by considering a tube of area A within the tank, together with the volume outside the tank that tapers to an area A0 , as shown in the figure. The pressure in the tank at the height of this tube is P1 , while the pressure outside the tank is P0 . The horizontal force on the water in this tube is its area times the pressure difference between its left and right ends. But also, the force equals the rate of change of momentum in the tube, which equals the momentum flux across the right end of the tube, if we choose the length of this tube so that its volume equals the volume of water that leaves the tank in 1 sec. Combine this fact with your results from parts b and c to find the reduced area A0 of the flow (a few cm outside the hole). IF we knew the velocity of the water vH at the hole, then we could use the law of mass conservation, v0 A0 = vH A. However, we do not know vH , so we need another approach. As hinted, a second approach is based on momentum flux, rather than energy conservation. The horizontal force on tube is F = A(P1 в€’ P0 ) and NOT AP1 в€’ A0 P0 since air pressure acts on the curved part of the jet as well as on its flat end such that the total horizontal force is just P0 times the total vertical area of the jet outside the hole, which area is simply the area of the hole = A. The horizontal force equals the momentum flux, so A(P1 в€’ P0 ) = ПЃv 2 A0 , using part c. Also, from part b we have that P1 = P0 + ПЃgh and v 2 = 2gh. Altogether, AПЃgh = 2ghA0 , so that: A0 = A/2 This analysis was first made by Borda in 1766. 24 Problem 8 A hexagonal pencil of mass m = 5 g has edges of width a = 3 mm. When placed with its axis horizontal on an inclined plane of angle Оё to the horizontal, the motion (if any) of the pencil is in general a combination of rolling and sliding. a.) (5 points) If the coefficient of static friction between the pencil and the plane is Вµ = 0.3, what is the largest angle Оёmax for which the pencil will remain at rest if placed on the plane at rest? The forces on the pencil when at rest are shown in the figure to the right. The force of gravity has components mg cos Оё and mg sin Оё perpendicular and parallel to the inclined plane, respectively. The normal force N balances the perpendicular component of the force of gravity, so that N = mg cos Оё. The force of static friction point up the slope and has magnitude F = ВµN = Вµmg cos Оё. The pencil does not slide so long as the force of friction exceeds the component of the force of gravity parallel to the slope. The critical condition is therefore, mg sin Оёmax = F = Вµmg cos Оёmax , or Вµ = tan Оёmax , and hence, Оёmax = tanв€’1 Вµ = tanв€’1 0.3 = 0.29 rad = 16.7в—¦ . 25 b.) (5 points) In the rest of the problem the angle of the inclined plane is less than Оёmax , so the pencil moves only if it is given an initial velocity. The pencil then rolls in a sequence of 1/6 turns during each of which the edge of the pencil in contact with the plane serves as the instantaneous axis of rotation. At the end of each 1/6 turn a face of the pencil collides with the inclined plane. Energy is lost during this collision, and so there exist a вЂњterminal velocityвЂќ of the pencil in which case the energy lost in each collision just equal the potential energy change of the pencil during the previous 1/6 turn. Give an expression for the change in potential energy, в€†PE, during each 1/6 turn of the pencil. As seen in the figure to the right, during a 1/6 turn, the center of mass of the pencil moves distance a parallel to the inclined plane, and hence the CM falls by height a sin Оё. Thus, в€†PE = mga sin Оё. 26 c.) (5 points) A simple model of the collision at the end of each 1/6 turn of the pencil between the pencil and the plane is that a fixed fraction of the kinetic energy gets lost during the collision. For the pencil and the plane under consideration, the fraction of kinetic energy lost in each collision is 50%. Give an expression for the change of kinetic energy, в€†KE, during a collision in terms of the velocity v of the CM just before the collision. I = 17ma2 /12 is the moment of inertia of a hexagonal pencil about one of its edges. The angular velocity just before the collision is П‰ = v/a. We know the moment of inertia I about the instantaneous axis of rotation, so the total kinetic energy is simply KE = IП‰ 2 /2. We do NOT add mv 2 /2 to this; the energy of the CM motion is already included. If we want to write KE = mv 2 /2 + I П‰ 2 /2, then I must be the moment of inertia about the center of mass, which is ICM = 7ma2 /12. Then, since v = aП‰, we find that KE = ma2 П‰ 2 /2 + (7ma2 /12)П‰ 2 /2 = 17ma2 /12)П‰ 2 /2 = IП‰ 2 /2. Since 1/2 of this KE is lost during the collision, the change of kinetic energy during a collision is: в€†KE = IП‰ 2 /4 = Iv 2 /4a2 = 17mv 2 /48 27 d.) (5 points) During вЂњsteadyвЂќ rolling of the pencil down the plane the kinetic energy lost in each collision with the plane equals the change in potential energy experienced each 1/6 turn. What is the velocity v at the end of a 1/6 turn for this вЂњsteadyвЂќ motion when the angle Оё of the inclined plane is 15в—¦ ? From parts b and c, в€†KE = Iv 2 /4a2 = 17mv 2 /48 = в€†PE = mga sin Оё So: v= 4mga3 sin Оё/I = 48ga sin Оё/17 = 48 В· 9.8 В· 0.003 В· 0.26/17 = 0.15 m/s = 15 cm/s 28 Problem 9 a.) (9 points) A tube of length L = 1 m is closed at one end, open at the other, and rotates with uniform angular velocity about an axis perpendicular to the closed end, as shown in the figure, which motion excites resonant vibrations in the air inside the tube. What is the lowest frequency heard by an observer in the plane of the rotation, if the period of rotation is 1 sec? The lowest frequency vibration of a tube of length L that has one end closed and one end open has wavelength О» = 4L = 4 m. The frequency f of the vibration is given by f = vs /О» = (343 m/s)/(4 m) = 85.8 Hz. However, because the open end of the tube is moving, an observer at rest hears Doppler-shifted frequencies. The lowest frequency heard occurs when the end of the tube is moving away from the observer. The velocity of the end of the tube is vs = П‰L = 2ПЂL/T = 2ПЂ m/s for length L = 1 m and period T = 1 s. The observed frequency fo is given by the appropriate Doppler formula: fo = f 1+vs /v = 85.6 1+2ПЂ/343 = 84.1 Hz 29 b.) (6 points) When a guitar string is plucked, its initial waveform is triangular, as shown in the upper figure below. The subsequent motion of the string is periodic with a period T . Sketch the waveforms at times T /4 and T /2 in the blank grids below Recall that a standing wave is equivalent to the sum of a left-moving traveling wave and a right moving traveling wave: 2 cos(kx) cos(П‰t) = cos(kx в€’ П‰t) + cos(kx + П‰t). A similar decomposition holds for the triangular wave on the guitar string. The spatial period of the wave is 2L if the length of the string is L, so the left- and right-moving component waves at time t = 0 both have the form of the dotted curve in the figure. 30 At time t = T /4 the left and right moving triangular waves are as shown by the dotted lines in the upper figure below. Adding these curves in the physical region of length L, we obtain the desired waveform. The same goes for the time t = T /2. 31 POSSIBLY USEFUL CONSTANTS AND EQUATIONS You may want to tear this out to keep at your side x = x0 + v0 t + 12 at2 F = ВµFN s = rОё p = mv П‰ = П‰0 + О±t KE = 21 IП‰ 2 W = F s cos Оё 2 ac = vr П„ = F sin Оё Ifull cylinder = 21 mr2 Ifull sphere = 52 mr2 F = Y A в€†L L0 P V = nRT Q = mL V Wisotherm = nRT ln Vfi Q = kA в€†T t L v = v0 + at F = в€’kx v = rП‰ F в€†t = в€†p P E = 12 kx2 KE = 12 mv 2 L = IП‰ П‰ 2 = П‰02 + 2О±в€†Оё П‰ = k/mв€љ 2ПЂr3/2 = T GM Ihollow cylinder = mr2 в€†L = О±L0 в€†T P1 V1Оі = P2 V2Оі P V = nkT Q = cmв€†T Оі = CP /CV Q = ПѓT 4 At v= v = О»f v= ОІ = (10 dB) log II0 f0 = fs f0 = fs / 1 в€“ sin Оё = DО» vs v в€љ ОіkT m Bad ПЃ fn = n sin Оё = v2 = v02 + 2ax F = в€’G Mr2m a = rО± xcm = M1tot i xi mi P E = mgh Wnc = в€†KE + в€†P E П„ = IО± в€†Оё = П‰0 t + 12 О±t2 П‰ = g/l I = mi ri2 Q = ПЂR4 в€†P/8О·L Q = Av P + ПЃgh + 21 ПЃv 2 = const в€†S = в€†Q T в€†U = Q в€’ W U = 32 nRT v= F в€љ m/L v = YПЃ v 1В± v0 1в€“ vvs v 2L 1.22 О»d f0 = fs (1 В± fn = n v0 ) v v 4L Monatomic: Diatomic: CV = 3R/2 CV = 5R/2 CP = 5R/2 CP = 7R/2 R = 8.315 J/K/mol u = 1.66 Г— 10в€’27 kg Пѓ = 5.67 Г— 10в€’8 W/m2 /K4 NA = 6.022 Г— 1023 molв€’1 k = 1.38 Г— 10в€’23 J/K Mearth = 6.0 Г— 1024 kg Mmoon = 7.4 Г— 1022 kg 0в—¦ C = 273.15 K Rearth = 6.4 Г— 106 m Rmoon = 1.7 Г— 106 m 1 kcal = 4186 J GNewton = 6.67 Г— 10в€’11 Nm2 /kg2 vsound = 343 m/s