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- Lazo M. Manojlovi

*Table of contents : CoverHalf Title PageTitle PageCopyright PageAbout the AuthorTable of ContentsPrefacePhysical ConstantsProblemsSolutionsIndex*

NONSTANDARD PROBLEMS IN GENERAL PHYSICS WITH SOLUTIONS

NONSTANDARD PROBLEMS IN GENERAL PHYSICS WITH SOLUTIONS

Lazo M. Manojlovi

ARCLER

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www.arclerpress.com

Nonstandard Problems in General Physics with Solutions Lazo M. Manojlović

Arcler Press 2010 Winston Park Drive, 2nd Floor Oakville, ON L6H 5R7 Canada www.arclerpress.com Tel: 001-289-291-7705 001-905-616-2116 Fax: 001-289-291-7601 Email: [email protected] e-book Edition 2019 ISBN: 978-1-77361-622-3 (e-book) This book contains information obtained from highly regarded resources. Reprinted material sources are indicated and copyright remains with the original owners. Copyright for images and other graphics remains with the original owners as indicated. A Wide variety of references are listed. Reasonable efforts have been made to publish reliable data. Authors or Editors or Publishers are not responsible for the accuracy of the information in the published chapters or consequences of their use. The publisher assumes no responsibility for any damage or grievance to the persons or property arising out of the use of any materials, instructions, methods or thoughts in the book. The authors or editors and the publisher have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission has not been obtained. If any copyright holder has not been acknowledged, please write to us so we may rectify.

Notice: Registered trademark of products or corporate names are used only for explanation and identification without intent of infringement. © 2019 Arcler Press ISBN: 978-1-77361-490-8 (Hardcover) Arcler Press publishes wide variety of books and eBooks. For more information about Arcler Press and its products, visit our website at www.arclerpress.com

ABOUT THE AUTHOR

Dr. Lazo M. Manojlović was born in Vranje, Serbia, on April 30, 1972. He received the B.Sc.E.E. and M.Sc.E.E. degrees in electrical engineering from the Faculty of Electrical Engineering, University of Belgrade, Belgrade, Serbia, in 1996 and 2003, respectively, and the Ph.D. degree in electrical engineering from the Faculty of Technical Sciences, University of Novi Sad, Novi Sad, Serbia, in 2010. From 1997 to 1998, he was with the Institute Mihailo Pupin, Belgrade, Serbia, as a part of his extended studies, where he was engaged in research on the development of applicative software for text to speech synthesis of Serbian spoken language in the C programming language based on diphone concatenation. From 1998 to 2002, he was with the Military Institute of Technology, Belgrade, Serbia, where he was engaged in research on designing and testing several types of ultralownoise, wideband photodiode preamplifiers for quadrant photodetector and for tracking electronics circuits for analog signal processing in pulsed-laser tracking systems for laser guidance missiles. From 2002 and 2003, he was with Pupin Telecom DATACOM, Belgrade, Serbia, where he was involved in designing, installing, and commissioning of the first hybrid fiber coaxial network for the cable distribution systems in Belgrade. From 2003 to 2005, he was with the Vienna University of Technology, Vienna, Austria, where he was engaged in research on the development of highprecision piezoactuator- based positioning systems. From 2005 and 2008, he

was with Integrated Microsystems Austria GmbH, Wiener Neustadt, Austria, where he was engaged in research on the development of different types of fiber-optic sensors. He is currently with the Ministry of Telecommunications and Information Society of the Republic of Serbia, Belgrade, Serbia, as a Special Adviser to the Minister, and also with the Zrenjanin Technical College, Zrenjanin, Serbia, as a Professor. His research interests include interferometry, optical sensing systems, and laser rangefinders.

TABLE OF CONTENTS

Preface...........................................................................................................ix

Physical Constants..........................................................................................xi

Problems....................................................................................................... 1 Solutions...................................................................................................... 37 Index......................................................................................................... 273

PREFACE

Solving practical physical problems represents one of the best ways to understand the laws of physics. In order to get a better insight into the general laws of physics and how to apply them to the real-life problems, the students will find in this book the problems that were carefully selected. The book contains 100, mostly nonstandard, problems with detailed solutions covering the classical mechanics, classical electrodynamics, thermodynamics, optics, fluid mechanics, etc. A relatively large number of problems are new, original, and presented for the first time together with their detailed solutions. However, the reader may be familiar with some of the presented problems. The author looked for the problems that treat some well-known physical phenomena such as see wave formation, sandstorm forming, bursting of the soap bubble, origin of the planet rotation, estimating the thickness of the Earth’s crust, etc. Also, the author included some theoretical problems such as over a century-old Abraham– Minkowski dilemma treating the photon momentum within the dielectric medium. In order to avoid classical approach, which someone can find it bored, that the problems are strictly divided into the classical branches of the physics, the author placed the problems into a random order. In my opinion, this may be more natural way as a large number of problems may require knowledge from several physical branches. The presented solution of each problem is just one possibility among other possible solutions. The students are naturally strongly encouraged to find their own way in solving the problems.

PHYSICAL CONSTANTS

Gravitational constant

G = 6.674 × 10−11 N·kg−2·m2

Speed of light in vacuum

c = 299,792,458 m·s−1

Elementary charge

e = 1.60217662 × 10−19 C

Electron mass

me = 9.10938356 × 10−31 kg

Proton mass Boltzmann constant Planck constant Avogadro constant Gas constant Permittivity of vacuum Permeability of vacuum Mean radius of the Earth Mean Sun–Earth distance Mean density of the Earth Free-fall acceleration at the Earth’s surface

mp = 1.6726219 × 10−27 kg

kB = 1.38064852 × 10−23 J·K−1 h = 6.62607004 × 10−34 J·s

NA = 6.02214086 × 1023 mol−1 R = 8.3144598 J⋅mol−1⋅K−1

ε0 = 8.854187817 × 10−12 F⋅m−1 μ0 = 4π × 10−7 H⋅m−1 R = 6371 km

D = 1.49 × 108 km ρ = 5520 kg⋅m−3 g = 9.81 m⋅s−2

PROBLEMS

P1.

A billiard ball rolls without slipping on the billiard table, as it is depicted in the figure below. In order to keep the ball on the table, the table rim is surrounded by the sharp raised edge. Find the minimum value of the ratio of the table edge height and the ball radius as well as the minimum value of the friction coefficient between the ball and the table, so when a direct, head-on collision occurs between the ball and the table edge, there is no slipping of the ball. Consider the collision between the ball and the table edge absolutely elastic. The table surface and the table edge are covered with the same material.

P2.

The circus artist of mass m performs its artistic play by walking on the ball of mass M, whereby the ball rolls on the circus podium without slipping. If the friction coefficient between the acrobat shoes and the ball is μ1, and the friction coefficient between the ball and the podium is μ2, find the maximum acceleration that the acrobat can achieve by walking on the ball? The ball is considered homogeneous.

2 P3.

P4.

P5.

P6.

P7.

Nonstandard Problems in General Physics with Solutions

You may have noticed that when you throw a stone vertically down into the water, after the impact of the stone in the water, the drops of water are separated from the water surface and move vertically upward. Find what maximum height can reach the water drops if the stone of density ρ is dropped in water from the height h. The density of water is ρ0. One winter morning with lots of snow, children went to the nearby hill in order to play with the snowballs. During the play, one snowball accidentally came to one steep slant and began rolling down the slopes without slipping. As the snow was wet and sticky, it was picked up by the snowball while rolling, so that its mass was increasing along the slope. Find the snowball velocity at the end of the slope, if during the rolling down the slope, it crossed the altitude difference of h = 300 m. Consider that the slope of the slant is constant and that the snowball at the end of the slope was much larger than at its top. The amount of the snow picked by the snowball at any point of the slope is proportional to the square root of the snowball volume. The snow thickness is considered uniform. After jumping out of the plane and opening the parachute, the parachutist, after a certain time, reaches the stationary falling velocity. Find this velocity of the parachutist of mass m if the dome-shaped parachute of the diameter D has a small opening at the top of the parachute with the opening diameter d ( d D ). Consider the density of the air homogeneous and equal to ρ. Within a smooth and curved tube of length l, placed between two points A and B with mutual altitude difference h, there is a flexible homogenous rope of the same length l which is held in the point A, as presented in figure below. Find the acceleration of all the rope points at the instance when the rope is released.

A small body of mass m is located at the top of a homogeneous ball of mass M and radius R as it is presented in figure below. The body

Problems

3

can move on the ball surface without friction. The whole system is in the Earth’s gravity field with the gravitational acceleration g. If the body is lightly derived from a labile equilibrium position at the top of the ball, find the maximal velocity the ball achieves. Consider that the ball is rolling without slipping. The ratio of the ball mass and the small body mass is M / m = 70 / 27 .

P8.

A heavy, homogeneous chain is located at the top of the cube with the mass M and positioned along the middle line of the upper side of the cube, as shown in the figure below, where its length is smaller than the length of the cube edge. The chain can slip over the surface of the cube without friction, while the cube is on a horizontal rough surface, where the static coefficient of friction between the cube and the surface is equal to μ. The whole system is in the field of Earth’s gravity. Find the maximum possible mass of the chain in which the cube will not move if the system overcomes the effect of the Earth’s gravity. Also, find the maximum possible static friction coefficient that will keep cube motionless.

P9.

The infinitely long, thin wire is placed at an angle α relative to the horizontal plane. Around the wire, a very long, thin, flexible rope has been wrapped. The rope has mass that can be neglected in comparison with the masses of two weights that are attached at the rope ends, as it is presented in figure below. Both weights are at the beginning

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Nonstandard Problems in General Physics with Solutions

fixed and immobile. The weight masses are m1 and m2 ( m1 < m2 ). The whole system is placed in the gravitational field of the Earth. At one point, weights are released and the whole system is left to itself. Due to the friction between the rope and the wire weights, after certain amount of time, reach their terminal vertical velocity v with respect to the contact point between the wire and the rope. Find the velocity by which the point of contact of the rope and the wire slides down the wire after a sufficiently long time. Assume that the rope is sufficiently thick and cannot be folded into a diameter smaller than or equal to the wire diameter (the diameter of the wire is smaller than the diameter of the rope), so that the rope and the wire can be regarded as touching at just a single point.

P10. The propellant gas pressure in a liquid spray bottle is higher than the surrounding atmospheric pressure where the pressure difference is Δp. Determine the average size of drops of liquid that emerge from the bottle after spraying. The surface tension of the fluid in the bottle is σ. Assume that after dispersing, drops of liquid leave the bottle at a relatively low velocity. P11. A very high velocity shock wave, which propagates through the air with the overpressure Δp, hits a body with an average density ρ. Find the velocity of the body when the shock wave has passed the body if the shock wave velocity is u. Assume that there is an abrupt change of pressure between the shock wave and the surrounding air. P12. Using the results of the previous problem, find the velocity of a shock wave with the overpressure Δp that propagates through the gas with the density ρ. P13. The young scout got the task to burn a fire after a sunny day with a magnifying glass and a newspaper. Will the young scout perform the task successfully with a help of a magnifying glass of the diameter D

Problems

5

= 10 cm and the focal length f = 25 cm? The temperature of the surface of the Sun is T⨀ ≈ 5500°C while the temperature of self-inflation of the paper is T ≈ 230°C. It is considered that the surface of the Sun and the paper are behaving like absolute black bodies, and that paper acts as a very good thermal insulator. P14. In order to fully correct his distance vision, a shortsighted person wears contact lenses with the optical power of P = −2 diopters. If the diameter of the pupil is d = 5 mm, find the angular resolution of the distance vision of the shortsighted person when he doesn’t wear contact lenses. When a shortsighted person does not wear any corrective optical tool, he partially closes his eyes in order to get clear image of distant objects. Why is it so? P15. The sledding track profile between the starting point A and the endpoint B in a winter tourist center is shown in the figure below. Based on the graphic method, find the place on the track where the sled will stop if it starts to move from the starting point A with zero initial velocity. The friction coefficient between the sled and snow is μ = 0.25. Consider the radius of curvature of the track large enough so that the inertial forces have no effect on the final result. By using the graphical method, also find the place on the track where the sled will reach the maximum velocity. If the actual side length of a single square given on the diagram below, where the track profile is shown, is a = 10 m, estimate the maximum velocity the sled attained. How accurate is it possible to estimate the maximum velocity of the sled?

P16. Jeweler wants to measure the weight of a gold chain with an electronic analytical balance. Before dropping the chain into a balance plate, the jeweler held one end of the chain in his hand while the other end touched the plate. At one point, the jeweler let the chain fall to the plate. If the scale was adjusted by mistake to measure and display the maximum value, determine how much is jeweler erroneous in measuring the weight of the chain.

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Nonstandard Problems in General Physics with Solutions

P17. A small body is located at the top of a rough hemisphere of radius R. Find the minimum velocity that must be given to the body in the horizontal direction in order to fall from the hemisphere? The whole system is in the Earth’s gravity field of acceleration g. The coefficient of friction between the body and the hemisphere is μ.

P18. Have you ever wondered why the sky is dark at night? If we adopt the assumption of an infinite, eternal, and static Universe, as early astronomers deemed, it is easy to show that in this case, the night sky would shine with infinite shine. However, we are witnessing that this is not the case. It was this paradox that puzzled many astronomers from the 16th to the 18th centuries. The first who pointed out this paradox in 1610 was Johannes Kepler. Afterward, this paradox was dealt by Edmond Halley and Jean-Philippe Loys de Chéseaux in the 18th century. In 1823, Heinrich Wilhelm Matthias Olbers presented this paradox to the general public. The problem was then named after this German astronomer Olbers’ paradox. This simple paradox was the proof that the Universe cannot be regarded as infinite, eternal, and static. Since the age of the Universe is finite, find the age of the Universe if it is known from the astronomical measurements that the irradiation of the Universe due to star, diffuse galactic, and cosmic light is approximately E ≈ 40 nW/m2, while the estimated average density of stars in the Universe is approximately λ ≈ 3 × 10−59 1/m3. Consider that the average luminosity of stars in the Universe is approximately equal to the solar luminosity of L = 4.6 × 1025 W. The speed of light in vacuum is c = 300,000 km/s. Assume that the average star is of the same age as the Universe. P19. The cosmic probe of mass m = 500 kg, after leaving the sphere of influence of the Earth’s gravity field, continues to move along the same orbit around the Sun as the Earth at a distance of RE = 150 × 106 km. In order to start its journey toward the Mars’s orbit, the probe expands its solar sail. The solar sail surface area is S = 1000 m2 and it is composed of a highly reflective material that reflects almost completely the Sun’s light. Due to the pressure of light on the solar sail, the probe begins to gradually move away from the Earth’s orbit

Problems

7

through a spiral path. Find the optimal angle of the solar sail with respect to the Sun in order to reach the Mars’s orbit as fast as possible. If the Mars’s orbit is at a distance of RM = 230 × 106 km from the Sun, determine how long the probe will take to reach the Mars’s orbit if the solar irradiance on Earth’s surface is E = 1370 W/m2. The mass of the Sun is M = 2 × 1030 kg, while the gravitational constant is G = 6.67 × 10−11 Nm2/kg2. The speed of light in vacuum is c = 300,000 km/s. P20. Heron’s fountain is a hydraulic machine invented by the 1st-century AD inventor, mathematician, and physicist Heron of Alexandria. Heron studied the pressure of air and steam, described the first steam engine, and built toys that would spurt water, one of them known as Heron’s fountain. Various versions of Heron’s fountain are used today in physics classes as a demonstration of principles of hydraulics and pneumatics. A simple example of the Heron’s fountain is presented in figure below showing three water vessels A, B, and C placed at three different heights.

When the valve V is closed, the fountain doesn’t work. After opening the valve, the water, from the small vessel A that contains just a small amount of water, starts to flow into the vessel C. When the level of water starts to rise in the vessel C, the air pressure in it also starts to rise, thus raising the air pressure in the vessel B. Thereafter, the high air pressure in the vessel B pushes the water through the tube that connects vessels A and B, thus making the fountain working. In general case, the fountain will work as long as the water from the vessel B is not transferred into the vessel C. If the bottom of the vessel B is positioned at the height hB with respect to the bottom of the vessel C, find the condition that this height must fulfill in order to provide that

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Nonstandard Problems in General Physics with Solutions

all of the water of the vessel B be transferred into the vessel C, thus making the fountain working as long as possible. Consider both vessels B and C to be of the same shape thus having the same height H. All three vessels have the same cross-sectional area. Find the maximum height of the fountain water stream as well. Before opening the valve, the vessel B is full of water. P21. The asteroid belt consists of a very large number of asteroids of average density ρ that move along circular orbits around a massive star of mass M at a mean distance R, as shown in the figure below. The width of the asteroid belt w is much smaller than the mean distance of the asteroid belt from the star ( w R ). The total mass of the asteroids in the asteroid belt is m and it is much smaller than the mass of the star, whereby it can be assumed that the thickness of the asteroid belt is much smaller than its width, and that the asteroids are approximately equally distributed within the asteroid belt. Due to the mutual gravitational attraction between individual asteroids and local disturbances within the asteroid belt, the asteroids themselves began to make small clusters thus forming in this way the planet. Considering that the entire mass of the asteroid belt was involved in the planet formation and that after the formation of the planet, it is moving along a circular orbit around the star. In this case, find the rotation period of the planet around its axis, that is, the length of the day at the newly formed planet and the direction of rotation relative to the direction of rotation around the star of this formed planet if it is assumed that due to the internal pressure in the planet, there was no significant change in the density of the material of the planet relative to the mean density of the material asteroid belt.

P22. A spacecraft, after successfully completing the mission, returns to Earth. When returning to Earth, the spacecraft penetrates at a high velocity of v = 7.7 km/s in the upper, thin atmospheric layers. On this occasion, due to friction between the spacecraft and the air, the space-

Problems

9

craft is warmed up. Find the maximum temperature increase in the particular points of the spacecraft if the mean molar mass of the air is μ = 29 g/mol and the universal gas constant R = 8.314 J/(mol⋅K). It is considered that air is composed of diatomic molecules. P23. Find the minimum angular velocity of the basketball ball in order to rotate steadily at the top of the finger of the basketball player, as presented in figure below. The radius of the standard basketball ball is R = 12 cm, while the ball rotates in Earth’s gravitational field of acceleration g = 9.81 m/s2.

P24. In 1937, the famous English physicist Paul Dirac set up his hypothesis of large numbers in which he assumed that the universal gravitational

constant changes with time as G ( t ) ≈ G0τ / t , where G0 is the present day value of the universal gravitational constant and τ = 13.7 billions of years is the estimated age of the Universe. In order to experimentally check the accuracy of the Dirac hypothesis, it is necessary to have instruments of very high accuracy. Unfortunately, instruments and measuring techniques that have a satisfactory accuracy have not been developed yet. However, during several cosmic missions on the surface of the Moon, retro reflecting mirrors were installed, whose primary goal was to accurately measure the distance between Earth and Moon using lasers. During these measurements, it was found that the average distance between the Earth and the Moon is approximately equal to R ≈ 384,000 km, and that the Moon is moving away from Earth for ΔR ≈ 3.82 cm each year. The reason for this, according to today’s prevailing hypothesis, lies in tidal forces between the Earth and the Moon. If we assume that tidal forces are not the only reason for such Moon’s motion but also the influence and the variable value

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Nonstandard Problems in General Physics with Solutions

of the universal gravitational constant on it, is it possible, on the basis of these data, to reject the hypothesis that the universal gravitational constant changes with time in the manner that Dirac postulated? P25. It is well known that the mercury thermometer is used to measure body temperature. As the body temperature is usually higher than the ambient temperature after the measurement, an error will occur due to the rapid cooling of the thermometer. However, the construction of the thermometer is such that the sudden drop in temperature does not significantly affect the accuracy of the measurement. This is achieved by placing a very small radius tube between the mercury reservoir and the thermometer scale, as shown in the figure below. In the event of a sudden drop in temperature on the thermometer, this constriction leads to the interruption of the flow of mercury between the reservoir and the scale. In this way, the scale of the thermometer shows the maximum measured body temperature. Determine the maximum radius of constriction of the thermometer so that the mercury flow is interrupted when the temperature difference between the body and the environment is less than ΔT = 15°C. The coefficients of linear thermal expansion of mercury and glass are αM = 61 × 10−6 1/°C and αG = 8.5 × 10−6 1/°C, respectively. The surface tension of mercury is σ = 0.49 N/m, whereby the mercury does not wet the glass, while the specific heat capacity and density of mercury are c = 139 J/(kg·K) and ρ = 13600 kg/m3, respectively. The total area of the mercury reservoir is S = 1.5 cm2, while the coefficient of convective heat transfer between glass and air is equal to h = 100 W/(m2·K). It is considered that all glass parts of the heat meter are made of very thin glass, and that the glass in comparison with the mercury is a very bad heat conductor. The thermometer is made under evacuated atmospheric conditions so that it can be assumed that the gas pressure inside the thermometer is negligible.

Problems

11

P26. During the clear dark night, it is likely that each of you has noticed that the stars change their glow, that is, they twinkling. The reason for this is a very small angular size of a star that makes starlight relatively coherent. Due to the inhomogeneity of the atmosphere, there are various conditions for the propagation of light waves, so that their mutual interference causes the variable brightness of the star. On the other hand, observing the planets that can be seen by the naked eye, such as Venus, Mars, and Jupiter, similar twinkling cannot be observed. The relatively large angular size of the planets makes the light that comes from them incoherent, and thus there is no interference between the light waves. Considering that the opening of the pupil eye in very low light conditions is equal to d = 8 mm, and that the eye is sensitive to the wavelengths of light in the range from λL = 380 to λH = 770 nm, estimate the maximum angular size of the star in which it twinkles. Compare the result obtained with the angular sizes of individual planets and stars given in the table below. Planets

Stars

Celestial body

Venus

Mars

Jupiter

Sirius

Alpha Centauri

Betelgeuse

Angular size

9.6″–66″

3.5″–25″

30″–50″

0.006″

0.007″

0.05″–0.06″

P27. An infinitely large resistor network is shown in the figure below. Find the equivalent resistance between the endpoints of the net if the resistances of all resistors in the grid are identical and equal to R. What will be the equivalent resistance of the net if one of the resistors in the third column is removed, and what will be the equivalent resistance if one of the resistors is replaced by a short circuit in the same column?

P28. A motorcyclist rides a motorcycle at a constant velocity of v = 100 km/h. At one point, on the road in front of him, he sees the sharp curve of the radius R = 20 m. Since there is not enough time for braking, the motorcycle is trying to turn it at full speed. Determine whether the motorcycle will be successful in this turning of the motorcycle if the weight of the motorcycle with the motorcyclist is m = 150 kg, the moment of inertia of the wheels J = 10 kg∙m2, the radius

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Nonstandard Problems in General Physics with Solutions

of the wheels r = 30 cm, and the height of the center of gravity of the motorcycle with the motorcyclist from the road before entering the curve h = 1.25 m? Consider both wheels of the motorcycle identical as well as the friction coefficient between motorcycle wheels and the road is large enough. Assume that the weight of the motorcycle with the motorcyclist is much greater than the mass of the wheels, and that the center of gravity of the motorcycle with the motorcyclist during the ride does not change in relation to the wheels of the motorcycle. P29. An infinite triangular resistance grid is made from a uniform and homogeneous resistive wire, as shown in the figure below. The vertex of each smaller triangle is connected to the middle point of the side of the larger triangle. If R is the resistance of the side of the largest triangle, find the equivalent resistance between any two endpoints located in the vertexes of the largest triangle.

P30. Two small bodies with masses m1 and m2 are joined by a firm lightweight rod. The system is moving toward a wall where the angle between the rod and the line that is normal to the wall is θ = 60°, as presented in figure shown below. The velocity vector of the two-body system is perpendicular to the wall. Find the minimum ratio of body

masses m2 / m1 so that after the elastic collision of the first body with mass m1 with the wall, the second body with mass m2 does not collide with the wall.

Problems

13

P31. Suddenly while driving a car, the hubcap falls off from a car wheel. The motion of the car is uniform with constant velocity v. Find the minimum friction coefficient between the hubcap and the road if the hubcap continues to move without slipping after the impact with the road. The radius of the car wheel is R, while the radius of the hubcap is r ( r < R ). The hubcap is considered as a homogenous disk. The impact of the hubcap with the road is considered perfectly inelastic. The acceleration of Earth’s gravity field is g. P32. Waterwheel or saqiya was first introduced into the irrigation systems of ancient Egypt during the early reign of Ptolemy. The introduction of waterwheels into the irrigation systems has greatly improved the irrigation of fertile Egyptian fields, which led to the rapid expansion of Egyptian agriculture of that time. The principle of work is very simple and is shown in the picture below. Basically, the waterwheel is a water-driven water pump. In order to ensure a uniform rotation of the waterwheel on the perimeter of the waterwheel, there are evenly mounted drive paddles whose role is to drive the wheel when they find themselves in the water stream of the river. Also, watering vessels are evenly distributed along the perimeter of the waterwheel. By turning the wheel, vessels dive into the water and thus inflate, and when a full vessel reaches the highest point, it is emptied into an open tube that leads to the field. Determine the maximum capacity as well as the maximum power of this water pump if the water flow velocity is v and the height of the waterwheel is H. Assume that the force with which the water stream acts on the paddles is proportional to the square of the velocity of the paddle relative to the water flow and that the proportionality coefficient is equal to k. If the volume of the vessel is equal to V, determine the total number of evenly distributed vessels per circumference of the waterwheel in order to achieve the maximum pump capacity. The acceleration of the Earth’s gravity field is g, while the water density is equal to ρ. Assume that at any

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moment, there is only one paddle in the water. Ignore friction in the axis of the wheel.

P33. Around a homogenous massive cylinder, a lightweight flexible rope is winded as presented in figure below. One end of the rope is firmly attached to the cylinder, whereas on the other end of the rope, a small body of mass m has been attached. The whole system is in the Earth’s gravity field of acceleration g. Find the acceleration of the cylinder if there is no slipping between the cylinder and the floor. What is the minimum friction coefficient between the cylinder and the floor in this case?

P34. Two turtles participate in a friendly chase. They draw a large circle of radius R on the floor. One of the turtles starts out on the circle, whereas another at the center of the circle. They begin to run simultaneously. At all times, the first turtle runs along the circle at a speed v, while the second turtle runs directly toward the first turtle at a speed u ( u < v ). After a while, the second turtle notices that the distance between them is no longer changing. What is that constant distance?

Problems

15

P35. Find the maximum achievable coefficient of performance of a wind turbine. P36. A long slipway, inclined at an angle α to the horizontal plane, is fitted with many identical rollers. The distance between every two adjacent rollers is d. The rollers have horizontal axels and consist of rubbercovered cylinders each of mass m and radius r, as presented in figure shown below. A heavy body of mass M and length much greater than the distance between two adjacent rollers is released at the top of the slipway. Find the terminal velocity of the massive body. Air resistance and friction at the roller pivots are negligible.

P37. There is an interesting “optical illusion” made by two identical concave spherical mirrors facing each other as presented in figure shown below. If a small object is placed at the bottom of the subjacent mirror M2, an image of the object will be shown at the small opening positioned at the center of the upper mirror M1. So, virtually, an observer will see the object that hovers in the opening. In order to obtain such hovering image of an object, find the distance between the mirrors if the mirrors have the same focal length equal to f.

P38. When an inclined plane–parallel glass plate is positioned directly behind a positive thin lens, there are lateral (δ) and longitudinal (Δ) shifts of the lens focal point as presented in figure shown below. Find the longitudinal shift of the lens focal point (Δ) if the plate thickness is t and its refractive index n. The angle between the plate and the lens principle planes is φ.

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Nonstandard Problems in General Physics with Solutions

P39. A long, heavy, flexible rope with mass λ per unit length is stretched by a constant force T. A sudden movement causes a circular loop to form at the end of the rope, as presented in figure below. In a manner similar to that in which transverse waves propagate, the loop moves along the rope. Determine the velocity of the loop.

P40. Find the minimal velocity that should be given to a small body with mass m in order to cross over the hill of mass M as presented in figure shown below. The hill height is H. The friction between the small body, the hill, and the floor is negligible.

P41. A small homogenous ball is placed at the top of a rough hemisphere as presented in figure shown below. Find the angular position of the ball on the hemisphere when it starts to slip if the ball’s initial velocity is negligible? The whole system is in the Earth’s gravity field of acceleration g. The coefficient of friction between the ball and the hemisphere is μ.

Problems

17

P42. A metallic disk rotates around its pivot with the angular velocity Ω. Find the voltage that measures the voltmeter V, as presented in figure shown below, between the disk pivot and its circumference. The disk radius is R.

P43. The Crookes radiometer, also known as a light mill, consists of an airtight glass bulb, containing a partial vacuum. Inside are a set of vanes which are mounted on a spindle. The vanes rotate when exposed to light, with faster rotation for more intense light, providing a quantitative measurement of electromagnetic radiation intensity. The reason for the rotation was a cause of much scientific debate in 10 years following the invention of the device, but in 1879, the currently accepted explanation for the rotation was published. Today the device is mainly used in physics education as a demonstration of a heat engine run by light energy. It was invented in 1873 by the chemist Sir William Crookes as the by-product of some chemical research. In the course of very accurate quantitative chemical work, he was weighing samples in a partially evacuated chamber to reduce the effect of air currents, and noticed the weighings were disturbed when sunlight shone on the balance. Investigating this effect, he created the device named after him. In a typical design, inside the glass bulb of the radiometer, on a low friction spindle, is a rotor with several (usually four) vertical lightweight vanes spaced equally around the axis. The vanes are polished or white on one side and black on the other, as presented in figure shown below. Each vane is consisted of a thin mica plate with the plate thickness d. The vanes are located at the distance R from the spindle, which is much larger than the vane linear dimensions. Find the angular velocity of the rotor if the radiometer is illuminated with the diffuse light having the irradiance equal to E. The thermal con-

18

Nonstandard Problems in General Physics with Solutions

ductivity of mica is λ. The white or polished side of the vane is considered as an ideal reflector, whereas the black side of the vane is considered as an ideal absorber. The radiometer is placed in the environment with the ambient temperature T. Molar mass of the gas inside the radiometer bulb is μ, whereas the universal gas constant is equal to ℜ.

Consider the case when instead of a diffuse light, the radiometer is illuminated by a wide light beam, such as the direct sunlight. In order to estimate the angular velocity, take into consideration the following numerical values: plate thickness d = 1 mm, vane to spindle distance R = 2 cm, solar irradiance E = 1370 W/m2, mica thermal conductivity λ = 0.71 W/(m·K), ambient temperature T = 22°C, air molar mass μ = 29 g/mol, and universal gas constant ℜ = 8.314 J/(mol·K). P44. Determine the mean wavelength of water waves if the mean wind speed at the sea level is v. Consider the case when only a thin layer of air flow affects the wave formation. Compare the obtained result with the measurements shown in the table given below. Mean wind speed [km/h] 19 37 56 74 92

Mean wavelength [m] 8.5 33.8 76.5 136 212.2

(Source: http://en.wikipedia.org/wiki/Wind_wave). P45. By assuming that the light quantum energy is proportional to the frequency of an electromagnetic radiation, find the relation between the energy and the frequency of a photon. In order to determine this relation, consider the case when the moving transmitter T that emits

Problems

19

planar, unvarying, and monochromatic electromagnetic waves in the direction of the receiver R, as shown in the figure given below. The transmitter T and receiver R antennas are arranged in the way that each photon that is being emitted by the transmitting antenna of the transmitter is being received by the receiving antenna of the receiver, that is, the transmission system has the maximal possible efficiency.

P46. A semi-infinite resistors network is shown in the picture below. Determine the equivalent resistance between points A and B.

P47. Starting from the energy conservation law, find the relative Doppler shift Δλ/λ of the wavelength of a monochromatic light source moving at a velocity u relative to the observer. P48. Find the size of the bubbles that detach from the bottom of the bottle and move upwards after opening the bottle of the carbonated (sparkling) water. Why do some bubbles remain attached to the bottom bottle, what leads to such behavior? Consider that there is a clean carbon dioxide in the bubbles. The surface tension of water is σ, whereas the wetting angle is α. The density of water is equal to ρ. P49. The sound source is moving at a high speed away from an observer simultaneously broadcasting the sound record. How fast must be the sound source if the observer hears the same sound but played backward? P50. Find the total number of atoms of air in the Earth’s atmosphere if the Earth’s radius is R = 6400 km and the atmospheric pressure at the sea

20

Nonstandard Problems in General Physics with Solutions

level p = 105 Pa. Consider the atmosphere height much smaller than the Earth’s radius. Average molar mass of air is μ = 29 g/mol. P51. Find the average thickness of the Earth’s crust if the latent heat of melting of basalt (a common volcanic rock) is L = 4 × 105 J/kg. (Source: http://www.dartmouth.edu/~ears5/handouts/heat.html). P52. By electron transition from a higher to a lower energy level, the excited atom emits a photon of the corresponding frequency. Show that in the case when such an excited atom is moving at a certain nonrelativistic velocity and emits the photon in the direction of its movement, a photon frequency differs from the frequency of the photon emitted by a motionless atom for an amount that corresponds to Doppler shift. P53. After bursting the soap balloon, a very large number of small drops of soap are produced, as can be seen in the pictures below. Estimate the average radius of drops of soap and their number if the drops were caused by bursting the soap bubble of radius R and wall thickness d.

P54. Find the thickness of the wall of the soap bubble, which is constantly inflated, in the moment of its bursting if the surface tension of the soap is σ, and the tensile stress of the soap is p. P55. What is the diameter of the lens which was used to make the photo shown below? Although the photographic lenses are made of several optical components, they can be considered as ideal thin lenses.

Problems

21

P56. In the electric schematic, shown in figure below, all three movingcoil voltmeters are non-ideal and identical. The electromotive force of the ideal voltage generator is E = 9 V. The left voltmeter shows the voltage of U1 = 2 V. What rare the showings of other two voltmeters?

P57. A small pebble accidentally came into the interior of a car tire of radius R. Find the minimal velocity of the car so that the pebble rotates together with the tire. The friction coefficient between the pebble and the tire is μ. P58. The circuit shown in the diagram below contains an ideal battery and two resistors, R1 and R2. A voltmeter is used to measure the voltage across R1, then across R2, then across the battery. Its readings are, respectively, 2.0 V; 3.0 V; and 6.0 V. What are the actual voltages across the resistors?

22

Nonstandard Problems in General Physics with Solutions

P59. Determine the velocity of bursting of the soap membrane, as shown in the picture below where bursting of the soap bubble is shown, if the surface tension of the soap is σ, the soap density is ρ, and the thickness of the soap membrane is d.

P60. In the infinite circuit shown in the diagram below, batteries have electromotive forces ε1 and ε2 and the internal resistance r1 and r2. Find the electromotive forces and the internal resistance of the equivalent battery.

P61. A sandstorm occurs as the rapid air front encounters a sandy area, as shown in the picture below. Taking into account that the air pressure is equal to p and the air density is equal to ρ, determine the minimum wind velocity of the air front of altitude h that can cause the sandstorm.

P62. When observing the drop of water in the cloud, which during its fall through the cloud increases by collecting small droplets along its

Problems

23

path, it is determined that the motion of the drop is uniformly accelerated. Determine the acceleration of the drop if, at the initial moment, the size of the drop is very small. Neglect the air resistance. P63. Nowadays, the power consumption of all the energy sources used by the mankind is approximately ΔP = 4.5 × 1013 W, while the power of the Sun’s radiation reaching the surface of the Earth is P = 1.8 × 1017 W. Find the Earth’s temperatures rise due to the operation of all energy sources that are manmade? What is the maximum permissible power of all energy sources used by the mankind if, for ecological reasons, the maximum permissible increase in temperature on Earth is ΔTmax = 0.1°C? P64. The electric circuit consists of a battery, two identical voltmeters, and two identical ammeters, as shown in the figure below. Ammeters A1 and A2 show, respectively, I1 = 1.1 mA and I2 = 0.9 mA, while the voltmeter V2 shows U2 = 0.25 V. What does the voltmeter V1 show? What is the electromotive force of the battery?

P65. A very light cylindrical buoy with the cross-sectional area S and height h floats on the surface of the turbulent sea. In order to stay in a vertical position at the very bottom of the buoy, there is a very small in size but massive weight of mass m. What can be the maximum amplitude of sea waves of period τ so that the top of the buoy remains unsubmerged? Consider that the wavelength of the sea waves is much greater than the linear dimensions of the buoy, and that the period of small oscillations of the buoy in the sea water is larger than the period of sea waves. The seawater density is ρ. P66. A simple heat engine consists of two wheels around whose a thin strong wire has been tensed and wrapped, as it is presented in figure shown below. One of the wheel, with much larger radius, is constantly heated and kept at the constant temperature, whereas the other reaches in the steady state, the temperature that is lower than that of the heated wheel. At the initial moment, a small angular momentum is given to the hot wheel in the counterclockwise direction, which is at the same time loaded. From that moment, the engine starts to move

24

Nonstandard Problems in General Physics with Solutions

and after reaching the steady state, the velocity of the wire is v. If the coefficient of thermal expansion of the wire is ε and its specific heat capacity c, find the coefficient of performance of this simple heat engine. Consider that after leaving the wheel, the wire has the same temperature as the corresponding wheel.

P67. Find the positions of the front and rear wheels of the airplane so that the safest possible landing of the airplane has been achieved. The radius of gyration of the airplane along its pitch axis is ρ. The impacts of the wheels with the airport runaway is considered perfectly inelastic. P68. Determine the amount of the gunpowder of density ρ that should be filled into the cannon barrel of the cross-sectional area S and the length L, as shown in the picture below, so that the cannonball of mass m, when after almost instantaneous combustion of the gunpowder is fired from the cannon with the highest possible velocity. Take that the gunpowder after the combustion forms the gas that can be considered ideal and which molecules have complex structure. The mass of the cannon is considered much larger than the mass of the gunpowder and the mass of the cannonball. The given parameters are

related as follows ρ SL / m = 48 .

P69. In Greek mythology, Sisyphus was the king of Ephyra. He was punished for his self-aggrandizing craftiness and deceitfulness by being

Problems

25

forced to roll an immense boulder up a hill only for it to roll down when it nears the top, repeating this action for eternity, as shown in figure below.

If so, find the minimum radius of the boulder that Sisyphus can push uphill if he is of an average height H = 1.75 m and wears the shoes that have relatively high friction coefficient between them and the hill μ = 0.9. What is the maximum angle of inclination of the hill that Sisyphus can overcome by pushing the boulder uphill? If boulder is made of a rock with the density of ρ = 2700 kg/m3 and it has the minimum possible radius, what force should be provided by Sisyphus in order to roll the boulder uphill that has the inclination angle that is equal to one and half of the maximum inclination angle of the hill? If Sisyphus cannot push the boulder uphill in the previous case, what force he must provide in order to keep the boulder motionless if the friction coefficient between the boulder and the hill is μ› = 0.3? Finally, decide if the myth of Sisyphus is physically realistic? P70. The Abraham–Minkowski dilemma lasts more than one century and refers to the value of the photon momentum in the dielectric. Namely, in 1908, Herman Minkowski defined the photon momentum in the = k nhν / c , where is the reduced Planck condielectric as p= M stant, k is the wave number, n is the refractive index of dielectric, h is the Planck constant, ν is the frequency, and c is the speed of light in vacuum, while in 1909, Max Abraham defined the photon momen-

mv = hν / ( nc ) , where m = hν / c 2 is tum in the dielectric as p= A the mass of the photon and v = c / n is the light velocity in the dielectric. Even after more than a century, scientists failed to solve the dilemma since they find confirmations for both assumptions. In order

26

(a)

Nonstandard Problems in General Physics with Solutions

to help scientists solve this dilemma, we will consider two following thought experiments. In the first thought experiment, we will observe a monochromatic light beam consisting of a very large number of evenly distributed photons per volume of the beam, which is reflected from two mirrors of a stable resonator located in the dielectric of the refractive index n, as shown in the figure below. All rays of the light beam are parallel and perpendicular to the mirrors. It is considered that at least one half of the photons are moving toward the left at each moment, while the other half of the photons to the right mirror. At one point, one of the mirrors starts to move very slowly parallel to the light beam. By observing the processes that are going to happen, determine the value of the photon momentum in the dielectric.

(b) In the second thought experiment, we will consider a single photon passing through a transparent dielectric block, where the block after the photon passing moves in the direction of the photon motion, as shown in the picture below. By observing the processes that are going to happen, determine the value of the photon momentum in the dielectric with the refractive index n.

P71. Determine the maximum speed at which the cannonball can be fired from the cannon barrel if the gunpowder with the combustion energy per unit mass λ is used. Consider that, when firing cannonballs, the

Problems

27

barrel gases are evenly distributed within the volume of the cannon barrel. P72. Determine the resistance between points A and C of the infinite large resistive network, shown in the figure below, if the resistance between points A and B is equal to RAB.

P73. Estimate the radius of curvature of the trajectory of the football ball if a football player kicks a ball off-center thus causing the ball to spin. The velocity of the ball is v, whereas its angular velocity of spinning is ω. The vectors of ball velocity and angular velocity are mutually perpendicular. P74. Find the position of the center of mass of the homogeneous semicircular disk of radius R by not using the apparatus of higher mathematics such as integral calculus. P75. By means of a resistive wire with constant resistance per unit length, an infinite square resistive net is made, as shown in the picture below. The vertices of each smaller square are connected to the middle of the side of the larger square. If R is the resistance of the side of the largest square, determine the equivalent resistance between the endpoints A and B as well as between the endpoints A and C.

28

Nonstandard Problems in General Physics with Solutions

P76. Determine the tensile stress of the material whose Yang’s modulus of elasticity is equal to E, while the Poisson’s ratio is equal to ν. P77. A liquid with the density ρ0 occupies an infinitely large volume. Into the liquid, two balls are immersed. The balls have the same radius r and are positioned at the mutual distance d. Find the gravitational force that acts on the balls if the balls densities are ρ1 and ρ2 and

where it is fulfilled ρ1 < ρ0 < ρ 2 . P78. Candlewick is made of a densely braided very thin string of diameter d, where the diameter of the wick is much larger than the diameter of the string, as presented in figure shown below. Determine the string density in the candlewick so that when the wick is lighted, the candle is shining with the strongest shine.

P79. A faucet aerator (or tap aerator) is often found at the tip of modern indoor water faucets. Aerators can be simply screwed onto the faucet head, creating a non-splashing stream and often delivering a mixture of water and air, as presented in the figure shown below. An aerator can prevent splashing, shape the water stream coming out of the faucet spout, produce a straight and evenly pressured stream, conserve water and reduce energy costs, reduce faucet noise, and increase perceived water pressure (often used in homes with low water pressure); sometimes described as a pressure regulator or flow regulator. When a single stream of water hits a surface, the water must go somewhere, and because the stream is uniform, the water will tend to go mostly in the same direction. If a single stream hits a surface which is curved, then the stream will conform to the shape and be easily redirected with the force of the volume of water falling. Adding the aerator does two things: reduces the volume of falling water which reduces the splash distance, and creates multiple “mini-streams” within the mainstream. Each mini-stream, if it were falling by itself, would splash or flow in a unique and different way when it hits the surface, as compared to the other mini-streams. Because they are all falling at the same time, the streams will splash in their own way but end up hit-

Problems

29

ting other splash streams. The resulting interference cancels out the majority of the splashing effect. Because the aerator limits the water flow through the faucet, water use is reduced as compared to the same time of flow without an aerator. In the case of hot water, because less water is used, less heat energy is used. Determine the minimum volumetric water flow rate that will produce the water stream aeration if the atmospheric pressure is p0 and the water density is ρ. The faucet aerator is constructed in such a way that input cross-sectional area of the aerator is S1, whereas the output cross-sectional area of the aerator is S2. What is the maximum air bubble radius if the water surface tension is σ? What is the maximum possible volumetric water savings?\

P80. A simple heat engine consists of evenly distributed n identical rods, where n is the even number, each of mass m, and length l around the common pivot, as presented in the figure shown below. The heat engine is divided into two equal parts where one of the parts of the engine is in the ambient with the temperature T1 and the other part is

in the ambient with the temperature T2 ( T1 > T2 ). The complete engine is in the Earth’s gravitational field with the acceleration g. At one moment, the engine is put into operation by rotating it with the angular velocity ω. Find the power and torque generated by this simple engine. Rods are made of the material with the coefficient of thermal expansion equal to α. Rods should be considered to have very low heat capacity.

30

Nonstandard Problems in General Physics with Solutions

P81. The slit electrostatic lens presented in the figure shown below, containing an aperture long in comparison with its width h, separates a region in which the electric field is E1 from a region in which the electric field is E2. A beam of charged particles focusing at a distance x to the left of the aperture is refocused at a distance y to the right of the aperture. If V is the voltage through which the particles were accelerated before reaching the lens show that the following is valid:

1 1 E2 − E1 + ≈ x y 2V , which represents the well-known lens equation. Use the following approximations in the analysis V E1 x, E2 y , and x, y h .

P82. If we let the large ball of mass M fall, and immediately after it, a small ball of mass m from the height h, as presented in figure shown below, determine the maximum height that the smaller ball can reach. In what proportion of the masses this will happen? Find what maximum height a small ball of mass m can reach if we repeat this procedure with three balls with the masses M1, M2, and m. In what mass relations will this happen? Find a solution to the general case when we have n balls. Consider ball collisions absolutely elastic.

P83. Find the pressure that monochromatic plane electromagnetic wave with the irradiance E exerts onto the vacuum–dielectric medium in-

Problems

31

terface surface that is covered with the antireflective coating (ARC) as it is depicted in the figure shown below. The index of refraction of the medium is n. By using the obtained result, find the photon momentum within the medium.

P84. When it falls, the chimney in most cases is broken into two parts. Why this happens and in which place? What condition must be fulfilled in order to do so? At what point in the chimney cross section, the fracture will occur? If the chimney does not break, at a certain moment, the chimney pivot point, located at the chimney bottom, will be detached from the ground. When will this happen? The height of the chimney is L, its mass m, the chimney outer radius is R, whereas the inner radius is r. The tensile strength of the material the chimney is made of is σ, whereas the chimney material shear strength is τ. P85. In a pinhole camera, the distance of the pinhole from the photographic plate is d = 10 cm. You want to take a picture of the sun in the visible spectrum (λ ≈ 500 nm). What diameter of the pinhole should you use in order to obtain the sharpest image? P86. Beads of equal mass m are strung at equal distances d on a long, horizontal wire, as it is presented in figure shown below. The beads are initially at rest but can move without friction. One of the beads is continuously accelerated (toward the right) by a constant force F. What are the speeds of the accelerated bead and the front of the “shock wave,” after a long time, if the collisions of the beads are considered completely inelastic and perfectly elastic?

P87. Beads with masses m1, m2, and m3 can move without friction along a long, horizontal wire, as it is presented in figure shown below, where

m2 m1 , m3 is valid. Find the maximum velocities of outer beads if

32

Nonstandard Problems in General Physics with Solutions

in the initial moment, the inner bead gains the velocity v. All collisions can be considered elastic.

P88. In what place in fencing with the sticks one need to hit them against each other, so that one does not feel the impact? The stick is held with one hand and its end. P89. What is the maximum height that the sack of sand can reach with the help of the plank of mass m1 and length l if at the other end of the plank from height H falls the same sack of sand as it is presented in figure shown below. The mass of the sacks of sand is m2.

P90. Two sacks of sand with the masses m1 and m2 are joined with a very light, flexible rope. The rope is squeezed through a small hole made in the table as presented in figure shown below. At one moment, the sack with the mass m2 is released. What will be the maximal height that this sack will reach again? The height of the table is H.

P91. Two thin, strong, and frictionless wires are joined in a way that they form the angle α between them, as presented in figure shown below. Around the wires, a light and flexible rope is wrapped. The rope length is l. Find the height difference h between the cross-sectional point of the wires and the rope if at one end of the rope, a heavy body is attached. The whole system is placed in the Earth’s gravitational

Problems

33

field with the acceleration g. What is the tension in the rope if the mass of the attached body is m?

P92. A uniform, homogenous, thin bar of mass m lies on the horizontal plane. What minimum possible force one must apply in order to move the bar? The friction coefficient between the bar and plane is μ. P93. Determine the equivalent resistance between points A and B of the infinite large resistive hexagonal network, shown in the figure below, if the resistance between every two adjacent points is R.

P94. The radius of curvature of a drop at its top point is R, as presented in figure shown below. What is the mass of the drop if the drop height is h and the radius of the contact surface between the drop and the plane is r? The drop liquid density is ρ, whereas its surface tension is σ, whereby the drop liquid does not wet the plane. What is the minimum radius of curvature of the drop? What should be the drop thickness in order to obtain maximum magnification of the small objects located on the plane surface if drop liquid is transparent with the index of refraction equal to n?

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Nonstandard Problems in General Physics with Solutions

P95. The surf skimmer is a toy used for sport at the beach. By jumping on a floating board, a player can skim along water smoothly for several meters, as presented in figure shown below. In order to provide successful skim, the board front is slightly curved, as can be seen in the figure below. If the board is curved in a way that the board front is raised for a small height H with respect to the rest of the board, determine the thickness of the water layer below the board if the skimming velocity is v. Consider the board to be approximately rectangular in shape having the area S. The skimmer mass is m, whereas the water density is ρ.

P96. Find the force that acts on one facet of a cube if the cube surface is uniformly charged with the surface charge density σ. The cube side length is a. What about if instead of a cube, we have a uniformly charged tetrahedron. P97. On a rough elevated plane with the elevation angle α lies a small body which is joined with a light, unstretchable string. The free end of the string is pulled through a small hole located on the plane. In an initial moment, the body lies on the plane in a way that the string is horizontal as it is presented in the figure shown below. At one moment, the strings start to be pulled very slowly, where the body in the moment of reaching the hole makes the trajectory in the form of a semicircle. Find the friction coefficient between the body and the plane.

P98. On the roof of the house that is inclined at an angle φ with respect to the horizon lays a lead sheet. The air temperature during the day rises thus reaching the maximum temperature of T1 and after that, during the night, falls down to the minimum temperature of T2. The sheet length is l. Find the distance which sheet passes on the roof during N

Problems

35

days. The friction coefficient between the sheet and the roof is μ and satisfies the condition µ < tan ϕ . The coefficients of linear thermal expansion of lead is α. P99. The science fiction writer R. A. Heinlein describes a “skyhook” satellite that consists of a long rope placed in orbit at the equator, aligned along a radius from the center of the Earth, and moving so that the rope appears suspended in space above a fixed point on the equator as presented in the figure shown below. The bottom of the rope hangs free just above the surface of the Earth. Assuming that the rope has uniform mass per unit length and that the rope is strong enough to resist breaking, find the length of the rope. The Earth’s radius is R.

P100. In a uniformly charged sphere with the radius R, a spherical hole is cut out having the radius r. Centers of the spheres are at the mutual distance d, as it is depicted in the figure shown below. The charge density of the sphere is ρ. Find the electric fields in both sphere centers. Proof that the electric field in the hole is uniform.

SOLUTIONS

S1.

The analysis will start from figure below where are depicted all relevant parameters for the analysis during the impact of the ball with the table edge. It will be taken into account that during the collision with the table edge, only the directions of the ball velocity v and angular velocity ω are changed, where their intensities remain unchanged. The reasons why it is so lies in the facts that there is no slipping of the ball during the collision with the table edge and the impact is absolutely elastic.

For the angle φ between the line segment that connects two points, that is, the ball center and the point of impact B and the horizontal line, the following is valid:

sinϕ=

h −1 R , (S1.1)

38

Nonstandard Problems in General Physics with Solutions

where h is the table edge height and R is the ball radius and where the following is satisfied: R < h < 2 R and consequently ϕ ∈ [ 0, π / 2] . As the ball rolls without slipping, immediately before the impact, the point B, which is placed at the ball circumference, has the total velocity with its horizontal component directed toward the table edge and its vertical component directed toward the table surface. Based on the direction of the velocity of the point B during the impact, there will be reaction forces of the table edge on the ball N 2 and T2 with the directions as it is depicted in the figure above, that is, in the opposite directions from the point B velocity. Due to the reaction forces between the ball and the table edge, there will be also reaction forces between the ball and the table surface in the point A (the point of the ball rotation in the moment of impact). The corresponding reaction forces N1 and T1 affect the ball motion during the impact in the way that is depicted in the figure above. It is important to mention that if the impact time duration τ is very short than the following, approximation can be taken into account: mg N1 , T1 , N 2 , T2 , where m is the ball mass and g is the gravitational acceleration, or the forces caused by the impact are much larger than the ball weight.

Based on the impact dynamics, the following equations are obtained:

N 2sinϕ = N1 + T2 cosϕ

, (S1.2)

2mv N 2 cosϕ + T2sinϕ − T1 =

τ

, and (S1.3)

2Jω

(T1 + T2 ) R = τ

, (S1.4)

where v is the ball velocity before and after the impact, J is the mo2 ment of inertia of the ball where J = 2mR / 5 is satisfied, and ω is the ball angular velocity where under condition that the ball rolls without slipping the following is obtained: ω = v / R . By substituting all these relations into equation (S1.4) and after the rearrangement, the following equation is obtained:

5 mv (T1 + T2 ) = 4 τ . (S1.5) By combining equations (S1.3) and (S1.5), the following equation is obtained:

Solutions

N 2 cosϕ + T2sinϕ − T= 1

5 (T1 + T2 ) 2 . (S1.6)

If we rearrange equations (S1.2) and (S1.6), the following system is obtained:

T= N 2sinϕ − N1 and (S1.7) 2 cosϕ

7T1 + ( 5 − 2sinϕ ) T2 = 2 N 2 cosϕ

39

. (S1.8)

The conditions that must be satisfied in order to avoid the ball slipping during the collision with the table edge are the following:

T1 ≤ µ N1

T ≤ µN

2 , where µ is the friction coefficient between and 2 the table and the ball. By combining these inequalities with equations (S1.7) and (S1.8), the following is obtained:

µ N 2 cosϕ ≥ N 2sinϕ − N1 and (S1.9)

7 µ N1 + ( 5 − 2sinϕ ) µ N 2 ≥ 2 N 2 cosϕ

S1.10)

By rearranging inequalities given in the expressions (S1.9) and (S1.10), we have:

N1 ≥ N 2 ( sinϕ − µ cosϕ )

N1 ≥

and (S1.11)

1 N 2 ( 2cosϕ + 2 µ sinϕ − 5µ ) 7µ . (S1.12)

One of the ways in which the inequalities (S1.11) and (S1.12) can be solved is that if their right sides are less than zero, then the inequalities (S1.11) and (S1.12) will be absolutely satisfied. In this case, we have:

sinϕ − µ cosϕ ≤ 0 and (S1.13) 2cosϕ + 2 µ sinϕ − 5µ ≤ 0 . (S1.14) Based on the inequality (S1.13), we have: µ ≥ tanϕ . If we rearrange inequality (S1.14), the following is obtained:

2cosϕ

. (

µ

+ 2sinϕ − 5 ≤ 0

. (S1.15)

40

Nonstandard Problems in General Physics with Solutions

The inequality (S1.15) will be satisfied if we instead of the friction coefficient put some other smaller quantity. If such inequality is satisfied, then the inequality (S1.15) is also satisfied. So, based on the inequalities µ ≥ tanϕ and (S1.15), we have:

2cosϕ + 2sinϕ − 5 ≤ 0 tanϕ . (S1.16) By rearranging the inequality (S1.16), the following is finally obtained:

sinϕ ≥

2 5 . (S1.17)

By combining the equation (S1.1) and the inequality (S1.17), finally for the required ratio of the table edge height and the ball radius, the following is obtained:

h 7 ≥ R 5 . (S1.18) Finally, according to the inequality (S1.18) for the minimal ratio of the table edge height and the ball radius, the following is obtained:

7 h = = 1.4 R min 5 , (S1.19) while for the friction coefficient, based on the inequalities µ ≥ tanϕ

and (S1.17), the following is obtained:

µ≥

2 21 , (S1.20)

so, finally, according to the inequality (S1.20) for the minimal friction coefficient, the following is obtained:

µ= min

2 ≈ 0.44 21 . (S1.21)

The solution of the inequalities (S1.11) and (S1.12), which is presented above, is one possible way to find the solution of these inequalities. This solution is not the most rigorous one but it provides the results that will certainly fulfill the required conditions. The second possible solution is presented below. In this second solution,

Solutions

41

the inequalities (S1.11) and (S1.12) can be observed in two possible ways. First, we will start from the inequality (S1.11). This inequality will be satisfied if the inequality, where instead of the N1 the smaller quantity is put, is satisfied. As this is smaller quantity, we can adopt the value from the right side of the inequality (S1.12). In this case, we have:

1 N 2 ( 2cosϕ + 2 µ sinϕ − 5µ ) ≥ N 2 ( sinϕ − µ cosϕ ) 7µ , (S1.22) where if this inequality is satisfied, then the inequality (S1.11) is certainly satisfied. By rearranging the inequality (S1.22), it becomes:

2 7µ sinϕ − + cosϕ ≤ 1 5µ 5 . (S1.23) If we now start from the inequality (S1.12), then this inequality will be certainly satisfied if it is satisfied for the smaller value from the inequality (S1.11). In this case, the inequality (S1.12) becomes:

N 2 ( sinϕ − µ cosϕ ) ≥

1 N 2 ( 2cosϕ + 2 µ sinϕ − 5µ ) 7µ , (S1.24)

where if this inequality is satisfied, then the inequality (S1.12) is certainly satisfied. By rearranging the inequality (S1.24), it becomes:

2 7µ sinϕ − + cosϕ ≥ −1 5µ 5 . (S1.25)

If we now take that for the parameter p the following is valid:

= p

2 7µ + 5µ 5 . (S1.26)

then the inequalities (S1.23) and (S1.25) become:

sinϕ − pcosϕ ≤ 1 and (S1.27)

sinϕ − pcosϕ ≥ −1 . (S1.28)

Based on the inequalities (S1.27) and (S1.28), we have:

sinϕ − 1 ≤ pcosϕ and (S1.29) sinϕ + 1 ≥ pcosϕ . (S1.30)

42

Nonstandard Problems in General Physics with Solutions

The inequality (S1.29) is certainly satisfied for every p and φ from ϕ ∈ [ 0, π / 2] the value set and p > 0 because in this case, the fol-

lowing is satisfied: sinϕ − 1 ≤ 0 and pcosϕ ≥ 0 . Further, we will observe only the inequality (S1.30). After the rearrangement, it becomes:

(1 + p ) sin ϕ + 2sinϕ + (1 − p ) ≥ 0 . (S1.31) 2

2

2

The nulls of the polynomial from the left side of the inequality (S1.31) are:

sinϕ1 =

p2 −1 , sinϕ2 = −1 p2 + 1 . (S1.32)

The second solution φ2 is omitted as ϕ ∈ [ 0, π / 2] and sinϕ ≥ 0 . As

1 + p 2 > 0 is satisfied, the quadratic function on the left side of the

inequality (S1.31) has the minimum, so the following condition is satisfied:

sinϕ ≥

p2 −1 p 2 + 1 . (S1.33)

The function, given on the right side of the inequality (S1.33), is an increasing function so if the inequality is satisfied for the maximal value of the parameter p, it will be satisfied for any other value of this parameter. Based on the expression (S1.26), it can be easily determined that for the maximal value of the parameter p is valid sinϕ ≥ 1 , which is, on the other side, impossible. From the problem statement, the minimal possible value of the ratio of the table edge height and the ball radius is to be found. This minimal possible value can be found when the parameter p has the minimal value. According to the expression (S1.26), it can be easily shown that for µ = 2 / 7,

p = 2 14 / 5 . the parameter p has the following minimal value: min Based on the previous analysis, the following must be satisfied:

sinϕ ≥

31 81 . (S1.34)

Solutions

43

By combining the expression (S1.1) and the inequality (S1.34) finally for the required ratio of the table edge height and the ball radius, the following is obtained:

h 112 ≥ R 81 . (S1.35) Finally, according to the inequality (S1.35) for the minimal ratio of the table edge height and the ball radius, the following is obtained:

ü h ≈ 1.3827 = 81 R ü , (S1.36) which is very close to the value shown in the expression (S1.19). This value is obtained if the following is satisfied:

= µ

2 ≈ 0.53 7 , (S1.37)

which also coincides with the condition given in the expression (S1.21).

At the very end, one can adopt the results obtained in the previous analysis as they provide the necessary and sufficient condition for the fulfillment of the conditions given in the problem statement, so finally one can take: ( h / R )min = 7 / 5 and µmin = 2 / 21 .

S2.

The analysis will start from figure below where are depicted all relevant parameters for the analysis of the circus artist movement on the ball. In order to stay on the ball, the artist must have the same acceleration as the ball. The contact between the artist shoe and the ball occurs at the point that is positioned at the ball surface where the angle between the line segment that connects the ball center and the contact point and the horizontal line is φ.

44

Nonstandard Problems in General Physics with Solutions

Based on the dynamics of the artist–ball system, the following system of equations is obtained:

= T1

( M + m ) a , (S2.1) ( M + m ) g , (S2.2)

= N1

Jα (T2 − T1 ) R = , (S2.3)

N 2 cosϕ − T2sinϕ = ma , and (S2.4)

N 2sinϕ + T2 cosϕ = mg , (S2.5)

where N1 and T1 are the reaction forces between the ball and the podium, N2 and T2 are the reaction forces between the artist shoe and the ball, a is the system acceleration, g is the gravitational acceleration, R is the ball radius, J is the moment of inertia of the ball where J = 2 MR 2 / 5 is satisfied, and α is the ball angular acceleration that satisfies α = a / R . The conditions that must be fulfilled in order to avoid the artist and the ball slipping during the movement are the following: T1 ≤ µ1 N1 and T2 ≤ µ2 N 2 . The first inequality and the equations (S2.1) and (S2.2) provide the following condition:

a ≤ µ1 g

Equation (S2.3) can be rearranged as:

2 T2 − T1 =Ma 5 , (S2.7)

. (S2.6)

where the relation for the ball moment of inertia and acceleration has been taken into account. By combining the relations (S2.1) and (S2.7), one obtains:

7M T= 2 1 + 5m

ma . (S2.8)

The last equation if combined with equations (S2.4) and (S2.5) gives the following system:

Solutions

7M N 2 cosϕ − 1 + 5m

7M N 2sinϕ + 1 + mg macosϕ = 5m .

ma masinϕ = , and

(S2.9) (S2.10)

This system can be solved in terms of system acceleration a and reaction force N2, so one obtains:

cosϕ g 7M 1+ + sinϕ 5m , and (S2.11)

7M 1 + 1 + sinϕ 5m N2 = mg 7M 1+ + sinϕ 5m . (S2.12)

a=

45

By substituting equation (S2.11) into equation (S2.8), the following is obtained:

7M 1 + cosϕ 5m T2 = mg 7M 1+ + sinϕ 5m . (S2.13)

Further, by taking into consideration that there is no slipping of the artist on the ball, that is, that the condition have:

T2 ≤ µ2 N 2 is fulfilled, we

7M 1 + cosϕ 5m µ2 ≥ 7M 1 + 1 + sinϕ 5m . (S2.14)

The relation (S2.6) and equation (S2.11) further provide the condition for ball rolling without slipping: cosϕ µ1 ≥ 7M 1+ + sinϕ 5m . (S2.15)

46

Nonstandard Problems in General Physics with Solutions

The relation (S2.11) and inequality (S2.14) provide:

7M 1 + 1 + sinϕ µ2 g 5m a≤ 7M 7M 1+ 1+ + sinϕ 5m 5m . (S2.16)

The right side of the last inequality has the maximal value for

sinϕ ≈ 1 , so inequality (S2.16) becomes: a≤

S3.

1+

7M 5m . (S2.17)

Finally, by taking into consideration the condition (S2.6), one obtains the following for the maximum achievable acceleration: amax

µ2 g

µg = min µ1 g , 2 7M 1+ 5m

. (S2.18)

A stone that is dropped from a certain height from the water surface reaches the velocity v slightly before the impact with the water occurs, which is given by:

v = 2 gh , (S3.1)

where g is the gravitational acceleration and h is the stone height from the water surface. During the stone–water collision, we have a rather complex impact mechanism. However, it can be modeled as presented in figure below.

Solutions

47

Slightly after the impact, the stone continues to move downward (sink) with the velocity v΄. During the stone sinking, effectively, we have that a certain amount of water moves upward with the same velocity. The effective mass m΄ of the water that moves upward is equal to the water mass that has the same volume as the sinking stone. Therefore, for this effective water mass, we have:

m' =

ρ0 m ρ , (S3.2)

where m is the mass of the stone. Apart from the stone sinking and the effective mass of water upward movement, during the impact, a certain mass of water is pushed by the stone that, due to the complex motion on the water surface, is repelled from the water surface in the form of droplets with the overall mass M that start to move upward from the water surface with the velocity u. According to the law of momentum conservation slightly before and after the impact, we have the following:

mv= mv′ − m′v′ + Mu , (S3.3)

where the last term on the left side of equation is presented with the positive sign as the droplets are moving downward before they have been repelled by the water surface. One can rearrange equation (S3.3) in the following form:

ρ M ´u =v − 1 − 0 v m ρ , (S3.4) where equation (S3.2) has been taken into account. Since u ≥ 0 is satisfied, the following condition must be fulfilled:

´v 1 ≤ v 1 − ρ0

ρ . (S3.5)

Consequently, by applying the energy conservation law slightly before and after the impact, one has the following:

mv 2 mv′2 m′v′2 Mu 2 = + + 2 2 2 2 . (S3.6)

48

Nonstandard Problems in General Physics with Solutions

One can further rearrange equation (S3.6) in the following form:

ρ M 2 u = v 2 − 1 + 0 v′2 m ρ , (S3.7)

where equation (S3.2) has been taken into account. It is obvious from the equation (S3.7) that the following condition must be also fulfilled:

´v ≤ v

1 1+

ρ0 ρ . (S3.8)

However, when the inequality (S3.8) is fulfilled the inequality (S3.5) is automatically fulfilled thus making the last inequality more rigorous. The combination of equations (S3.4) and (S3.7) further gives the following:

ρ 1 − 1 + 0 x 2 ρ u= v ρ0 1 − 1 − x ρ , (S3.9) where the parameter x is given by x = v′ / v and consequently must fulfill the inequality (S3.8). The droplets velocity has the extremum values for the following values of the parameter x: 1 1 1 x1,2 = ± − 2 ρ ρ0 1 + ρ0 1− 0 1 − ρ ρ ρ , (S3.10)

where these values have been obtained by taking the first derivative of the equation (S3.9) and finding the roots of such obtained function. In order to fulfill the condition (S3.8), only one solution may leave, that is: 1 1 1 x= − − 2 ρ0 ρ0 1 + ρ0 1− 1 − ρ ρ ρ . (S3.11)

Solutions

One can show that this value of parameter x fulfills the condition (S3.8) or: x=

1

ρ 1− 0 ρ

−

1 ρ0 1 − ρ

2

ρ 1− 0 ρ

−

1 1+

ρ0 ρ

≤

ρ 1+ 0 ρ 1− 2 ρ0 − 1 ρ

ρ 1+ 0 ρ

1

≤

1+

1

ρ 1− 0 ρ

+

ρ0 ρ

, (S3.12)

1 1+

2 ρ0 1 − ρ − − 1 1 ρ0 1+ ρ

ρ0 1 − ρ 1− 1+

1

ρ0 ρ

(S3.13)

that is obviously fulfilled. So, the solution (3.11) definitely fulfills the condition (S3.8). In order to test if this value of parameter x corresponds to the maximal value of the droplets velocity, one can make a simple test. Namely, for x = 0, we have u = v and for = x 1/ 1 + ρ0 / ρ , we have u = 0 , so, between these two limit values of the parameter x, there is no value of the droplets velocity that can, at the same time, take its minimum value and be greater than zero. Therefore, the value of the parameter x, given by equation (S3.11), corresponds to the maximum droplets velocity that is given by:

umax =

−

which by simple rearrangement further yields to: 1

49

ρ0 ρ

2

2

v

. (S3.14)

This maximum initial velocity of the droplets will bring the droplets to the maximum possible height above the water surface level. This maximum height is given by:

H max =

2 umax 2 g , (S3.15)

which finally gives:

50

Nonstandard Problems in General Physics with Solutions

H max

ρ 1+ 0 ρ 1 − 2 1 − ρ0 ρ =h

2 ρ0 1 − ρ 1 − 1 − ρ 1+ 0 ρ

ρ0 1 − ρ 1− 1+

ρ0 ρ

2

2

2

,

(S3.16)

where equation (S3.1) has been taken into account.

S4.

The small snowball at the top of the hill starts to roll downhill the slope of the hill without the slipping, as it is presented in the figure below. During the rolling, the snowball picks up the sticky snow thus increasing its size. At the end of the slope, the snowball crosses the altitude difference of h and gains the velocity at the end of the hill v.

According to the energy conservation law, the potential energy of the picked snow is transformed into the kinetic energy of the translator and rotational motion of the large snowball at the end of the slope. Therefore, one can write the following:

mv 2 J ω 2 + = mgÄh 2 2 , (S4.1)

where m is the snowball mass at the bottom of the hill, J is the moment of inertia of the snowball at the bottom of the hill where

J = 2mr 2 / 5 is satisfied, and where r is the snowball radius at the

Solutions

51

bottom of the hill, ω is the snowball angular velocity at the bottom of the hill for which is satisfied ω = v / r , g is gravitational acceleration, and Δh is the altitude difference between the picked snow center of gravity and the bottom of the hill. Equation (S4.1) thus can be rearranged by including all the presented relation in the following simpler form:

v=

10 gÄh 7 . (S4.2)

In order to find the velocity, one must find the altitude difference between the picked snow center of gravity and the bottom of the hill. As it was stated in the problem statement, the amount of the picked snow along the elementary slope path is proportional to the square root of the snowball volume. Therefore, the elementary volume ΔV of the snow that is picked along an elementary path of Δz is given by: 3 2

ÄV ∝ V Äz ∝ r Äz . (S4.3)

Due to the picked snow, the overall snowball radius will increase as:

4π 3 4π 3 r + ÄV = ( r + Är ) 3 3 , (S4.4)

where Δr is the snowball elementary radius increase for which, from equations (S4.3) and (S4.4), the following is valid:

r Är ∝ Äz (S4.5)

that after the integration gives:

r ∝ z 3 , (S4.6)

or how the snowball radius will increase along the path while rolling downhill. The snow thickness along the slope is uniform; thus for the width w of the snow trail that the snowball makes along the slope, one has:

2

3 2

= ÄV twÄz ∝ r Äz , (S4.7)

where t is the snow thickness. According to equation (S4.7), we have:

52

Nonstandard Problems in General Physics with Solutions 3 2

w ∝ r . (S4.8)

The equations (S4.6) and (S4.8) finally give the relation between the width of the snowball trail and the path length as:

w ∝ z , (S4.9)

or the snowball trail width linearly increases with the path that snowball passes along the slope. Therefore, according the equation (S4.7), the volume of the picked snow or equivalently the mass of the picked snow linearly increases with the snowball path length. Thus, the shape of the snowball trail is in the form of a triangle, as presented in figure below, where the center of gravity C of the picked snow is positioned, as in the case of any triangle, l / 3 from the triangle base (bottom of the hill) where l is the overall path length of the snowball (the equivalent triangle height).

Consequently, the altitude difference between the picked snow cenh / 3 . Finally, the ter of gravity and the bottom of the hill is ∆h = snowballs velocity is determined according to equation (S4.2) as:

v=

10 gh 21 , (S4.10)

or v ≈ 37.4 m/s by taking into account the exact numerical values. S5.

In order to find the parachutist terminal velocity after opening the parachute, we will consider a situation as presented in figure below. Namely, after reaching the steady-state velocity, the dome of the parachute is pressurized by a stream of air that flows into the parachute with the velocity that is equal to the parachutist falling velocity. Therefore, according to the Bernoulli’s principle, we have the following equation:

Solutions

p + ∆p +

53

1 2 1 ρ v = p + ρu 2 2 2 , (S5.1)

where p is the atmospheric pressure, Δp is the pressure difference between the pressure inside the dome of the parachute and the surrounding atmospheric pressure, v is the parachutist velocity, and u is the air stream velocity that leaves the parachute through the hole at the top of the parachute. A continuity equation gives the following relation between the input and output air stream:

Sv = su , (S5.2)

2 2 where S = D / ( 4π ) is the parachute base area and s = d / ( 4π ) is the area of the parachute hole.

Equations (S5.1) and (S5.2) give:

d v≈ D

2

2∆p

ρ , (S5.3)

where the relation d D has been taken into account. Due to the pressure difference below the parachute, there will be a drag force that will act on the parachute that is given by:

D2 ∆p (S5.4) 4π that is balanced by the parachutist weight, that is, F = mg . The comF ≈ S ∆p=

bination of the last three equations results in the parachutist terminal velocity which is finally given by:

54

S6.

Nonstandard Problems in General Physics with Solutions

v≈2

d2 D3

2π mg

ρ

. (S5.5)

After a very short period of time Δt, since releasing the rope, it has been moved from points A and B to the points A’ and B’, as it is presented in figure below. The distance that rope travels is consequently very short and given by AA′ = BB' = ∆l where ∆l l is fulfilled. Due to the change of the rope potential energy, we have the increase of its kinetic energy. The change of the rope potential energy is equivalent to the movement of small mass Δm, which is determined by the length Δl as:

∆l ∆m = m l , (S6.1)

where m is the rope mass, from the position AA’ to the position BB’.

Therefore, this small mass is shifted vertically for the distance that is equal to the altitude difference between the points A and B. The change in the rope potential energy is then given by:

∆Ep = ∆mgh

. (S6.2)

Consequently, the rope that has been held gains a small velocity thus giving the rise in the rope kinetic energy as:

∆Ek =

1 m∆v 2 2 , (S6.3)

where Δv is the rope velocity. According to the energy conservation ∆Ep = ∆Ek law, one has that by combining all three equations further gives:

∆v 2=

2gh ∆l l . (S6.4)

Solutions

55

Assuming that for this very short period of time after releasing the rope, the acceleration of all rope points can be considered constant; one has:

∆v = a∆t and (S6.5)

∆l=

1 a∆t 2 2 . (S6.6)

The combination of equations (S6.4), (S6.5), (S6.6), and finally gives the wanted rope acceleration as:

a=

h g l . (S6.7)

S7.

The analysis will start from figure below where are depicted all relevant parameters for the analysis of the small body movement on the ball surface. In order to develop a mathematical model of this complex movement, we will assume that the body of mass m is moved across the ball surface and is in the position at the ball surface where the angle between the line segment that connects the ball center and the contact point between the small body and the ball and the horizontal line is φ.

The altitude difference between the starting position of the body at the top of the ball and this particular point is given by:

∆h= R (1 − sinϕ )

. (S7.1)

At this particular point, the ball acceleration and velocity are A and v, respectively. The relative movement of the body with respect to the ball is determined by the relative radial acceleration ar′, relative tangential acceleration at′, and relative velocity u′. The relation between the relative radial acceleration and the relative velocity of the body is given by:

56

Nonstandard Problems in General Physics with Solutions

. (S7.2) Taking into consideration the ball and the body movements, the absolute vertical acceleration of the body is given by:

= ay at' cosϕ + ar 'sinϕ

, (S7.3)

whereas the absolute horizontal acceleration of the body is given by:

ax = at' sinϕ − ar' cosϕ − A

Consequently, the absolute vertical velocity of the body is given by:

uy = u'cosϕ

whereas the absolute horizontal velocity of the body is given by:

. (S7.4)

, (S7.5)

= ux u ′sinϕ − v . (S7.6)

Therefore, the absolute total velocity of the body is given by:

= u

ux2 + uy2

. (S7.7)

Based on the dynamic of the ball-body system, one can write the following equations:

ma= mg − Nsinϕ y

max = Ncosϕ

, (S7.8)

, (S7.9)

= MA Ncosϕ − T , and (S7.10) J α = TR , (S7.11)

where N is the reaction force between the ball and the body, T is the reaction force between the ball and the ground surface, J is the moment of inertia of the ball where J = 2MR 2 / 5 is satisfied, and α is the ball angular acceleration that satisfies α = A / R . The ball will continue to accelerate until the moment when the small body is detached from the ball surface. This will happen in the moment when the condition N = 0 has been reached. In this very moment, the ball will reach its maximal velocity. Consequently, according to the above-presented set of equations, one has ay = g , ax = 0 , and A = 0 fulfilled in the

Solutions

57

moment of the body detachment. In this particular case, equations (S7.3) and (S7.4) become:

at' cosϕ + ar' sinϕ = g , (S7.12)

at' sinϕ = ar' cosϕ

By solving equations (S7.12) and (S7.13), one obtains:

ar' = gsinϕ . (S7.14)

. (S7.13)

The combination of equations (S7.9), (S7.10), and (S7.11) further yields to:

max =

7 MA 5 , (S7.15)

which by taking into consideration that the small body initial velocity is zero and after the integration yields to:

mux =

7 Mv 5 . (S7.16)

Based on the energy conservation law, we have:

mg= ∆h

1 1 1 mu 2 + Mv 2 + J ω 2 2 2 2 , (S7.17)

where ω is the ball angular velocity for which ω = v / R is valid. This last equation simplifies into:

7 2mg ∆= h mu 2 + Mv 2 5 . (S7.18) The combination of equations (S7.1), (S7.5), (S7.6), (S7.7), (S7.16), and (S7.17) gives:

= 2mgR (1 − sinϕ ) mu'2cos 2ϕ +

7 7M Mv 2 1 + 5 5m

. (S7.19)

Further, from equations (S7.6) and (S7.16), one has:

sin = ϕ

v 7M 1 + u′ 5m

. (S7.20)

Also from equations (S7.2) and (S7.14), one also has:

58

Nonstandard Problems in General Physics with Solutions

u '2 sinϕ = gR . (S7.21)

The combination of the last two equations yields to:

= u′

3

7M gRv 1 + 5m

(S7.22)

and consequently to:

1 7M = sinϕ gR 1 + 5m gR

2

2 3 3 v . (S7.23)

Equations (S7.19), (S7.22), and (S7.23) give:

7M 1 + 5m gR

The last equation can be presented in the following form:

x3 − 3ax + 2 = 0 , (S7.25)

7M 1 1+ 3 2 7M 5m v − 3 1 + 5m gR

1 3

23 0 v +2= . (S7.24)

where: 1

7M 1 + 5m x= gR

7M 3 a= 1 + 5m . (S7.27)

3 23 v and (S7.26) 1

By taking into consideration the ball–body mass ratio of M / m = 70 / 27, the parameter a is a = 5 / 3 . For this particular value of the parameter a cubic equation (S7.25) becomes:

x3 − 5 x + 2 = 0 , (S7.28)

which can be presented in the following form:

Solutions

59

0 ( x − 2 ) ( x 2 + 2 x − 1) = . (S7.29)

x =2 = x 2 − 1 , and This equation has three real solutions: 1 , 2 x3 = − 2 − 1 . The variable x must have the positive value, so the third, negative solution cannot be the right one. Which one of the remaining two solutions is the right one can be determined according to the relation (S7.23) that can be also presented as:

sinϕ = ax . (S7.30)

As it must be fulfilled sinϕ ≤ 1 or equivalently x ≤ 1/ a = 3 / 5 = 0.6, the 2 − 1 ≈ 0.41 . Finally, only solution that fulfills this condition is x= one can find the maximal ball velocity as: S8.

3 v = 2 −1 5

(

) 53 (

)

2 − 1 gR

. (S7.31)

In order to develop the mathematical model of the chain movement and its corresponding effect on the cube, we will observe the figure given below with all relevant parameters. Due to the Earth’s gravity influence, chain will accelerate with the acceleration given by:

a=

y g l , (S8.1)

where y is the chain length that hangs over the edge of the cube, l is the chain overall length, and g is the Earth’s gravity field acceleration.

The corresponding forces that act on the cube (or equivalently according to the third Newton’s law of motion on the chain) are given by:

Tx = m

l−y a l and (S8.2)

60

Nonstandard Problems in General Physics with Solutions

= Ty m

y ( g − a) l . (S8.3)

The overall vertical reaction force of the surface on the cube is given as the sum of the cube weight, weight of the chain that still lies on the top of the cube, and the vertical reaction force between the cube and the chain or:

N= Mg + m

l−y g + Ty l . (S8.4)

The combination of equations (S8.1), (S8.3), and (S8.4) gives the following:

N =( M + m ) g − mg

2

y l . (S8.5)

In order to keep cube motionless, the friction force must be equal to or larger than the acting horizontal force of the chain or:

Tx ≤ µ N

Equations (S8.1), (S8.2), and (S8.6) further give:

m≤

µM 2

y y − (1 − µ ) + − µ l l . (S8.7)

This condition must be fulfilled for any y. As the function in the denominator of the last inequality has the maximum value for:

y 1 = l 2 (1 − µ )

, (S8.8)

it is obvious that if the condition given by equation (S8.7) is satisfied

for this value of y / l , it will be satisfied for any other value of this parameter thus keeping the cube motionless if the following condition is satisfied:

m≤

. (S8.6)

4 µ (1 − µ )

1 − 4 µ (1 − µ )

M . (S8.9)

Solutions

The problem statement also asks to find the maximum possible static friction coefficient that will enable to keep cube motionless, that is, this value of friction coefficient will provide at least any real value of chain mass that will keep cube without motion. In order to find this value of friction coefficient, one can rearrange inequality (S8.9) in the following form:

m≤

61

1 − ( 2 µ − 1)

( 2µ − 1)

2

2

M . (S8.10)

It is obvious from the last inequality that the following condition must be fulfilled as the chain mass must be positive:

( 2µ − 1)

2

≤1

, (S8.11)

or equivalently:

−1 ≤ 2 µ − 1 ≤ 1 , (S8.12)

which finally gives:

0 ≤ µ ≤ 1 . (S8.13)

Therefore, the maximum value of the static friction coefficient is

S9.

After a certain period of time, masses m1 and m2 reach their terminal vertical velocity v, as presented in the figure below, where mass m2 goes down relative to the mass m1 and mass m1 goes up relative to the mass m2 as the following condition m1 < m2 is fulfilled. There are two reaction forces between the rope and the wire, namely the reac tion force N that is perpendicular to the wire and it is lying under an arbitrary angle θ with respect to the horizontal plane, as it is also presented in the figure below. Due to the friction between the rope

µmax = 1 .

and the wire, there is also a friction force T that is perpendicular to the normal reaction force and directed collinearly to the total velocity of the rope at the contact point between the rope and the wire.

62

Nonstandard Problems in General Physics with Solutions

In order to present the corresponding mathematical model of such complex movement, we will observe the figure below where are presented all relevant parameters for analyzing such movement in the plane where the rope is lying, that is, perpendicular to the horizontal plane. As there is no acceleration, the problem will be treated in the frame of classical statics. Therefore, we have the following equation with respect to the contact point A between the rope and the wire:

= m2 g T ′cosα + m1 g

, (S9.1)

where T′ is the friction force component normal to the wire.

For a plane that is perpendicular to the horizontal plane and which is collinear to the wire, as it is presented in figure below, one has the following set of equations:

Ncosα sinθ + T ′cosα cosθ + T ′′sinα = ( m1 + m2 ) g

and (S9.2) Nsinα sinθ + T ′sinα cosθ = T ′′cosα , (S9.3)

where T″ is the friction force component parallel to the wire.

Solutions

The equation (S9.3) can be reorganized as:

cosα N sin θ + T ′cosθ = T '' sinα . (S9.4)

The combination of equations (S9.2) and (S9.4) gives:

T ′′

cos 2α + T ′′sin= α sinα

( m1 + m2 ) g

,

(S9.5)

that after the rearrangements gives:

( m1 + m2 ) gsinα . (S9.6)

′′ T =

Equation (S9.1) also gives:

T′ =

( m2 − m1 ) g cosα

. (S9.7)

As the friction force T is collinear to the total velocity of the rope at the contact point between the rope and the wire, we have:

u T ′′ = v T ′ , (S9.8)

which finally gives:

u=

63

( m1 + m2 ) sin2α v 2 ( m2 − m1 ) . (S9.9)

S10. When the spray bottle valve is opened, the high pressured gas in the bottle pushes the liquid, which then passes through a tiny tube with high velocity. At the end of the tube, the liquid is scattered in a large number of small drops, as it is presented in the figure given below. The work that has been done by the gas inside the bottle is given by:

A =∆p∆V , (S10.1)

64

Nonstandard Problems in General Physics with Solutions

where ΔV is the volume of the ejected liquid.

According to the energy conservation law, this work has been done in order to scatter the ejected liquid in large number of small drops. Therefore, we have:

A = nSσ , (S10.2)

where n is the number of drops and S is their average surface, which is given by:

S = 4π r 2 , (S10.3)

where r is the average drop radius. The volume of the ejected liquid is divided into n small drops thus giving the following relation:

4 ∆V =π nr 3 3 . (S10.4) After substitution, one obtains the following relation for the average drop radius:

r=

3σ ∆p . (S10.5)

S11. We will assume that the shock wave propagates along the x-axis with the very high velocity u and crosses over the body, as it is presented in the figure below. Due to the pressure difference between the shock wave and the surrounding atmospheric pressure, there will be a net force that will be acting on the body. This force is equal to:

F ( x ) = ( p + ∆p − p ) S ( x ) = ∆pS ( x )

,

where p is the atmospheric pressure and sectional area of the position x.

(S11.1)

S ( x)

is the body cross-

Solutions

65

This force will be acting on the body during a very short period of time, which is given by:

∆x ∆t = u , (S11.2) where the body velocity is considered to be much smaller than the shock wave velocity. The change of the body momentum during this period of time is given as:

∆P = m∆v = F ( x ) ∆t

, (S11.3)

where m is the body mass and Δv is the body velocity change during this period of time. The overall body velocity after the shock wave has crossed over the body is given as a sum of all these elementary velocity changes, that is:

v = ∑ ∆v = ∑

F ( x ) ∆t m

. (S11.4)

By substituting equations (S11.1) and (S11.2) into equation:

v =∑

∆p ∆p S ( x ) ∆x = ∑ S ( x ) ∆x um um .

(S11.5)

The sum of terms S ( x ) ∆x along the body represents its volume. Therefore, equation (S11.5) transforms into:

v=

∆p V u m , (S11.6)

where V is the body volume. Bearing in mind that ρ = m / V is fulfilled, where ρ is the average body density, the following relation is finally obtained for the body velocity:

66

Nonstandard Problems in General Physics with Solutions

v=

∆p u ρ . (S11.7)

S12. Instead of a solid body that has been found under the influence of a shock wave in the previous problem, we will observe a small gas slab with the area S and length Δx that lies on the way of the shock wave propagation as it is presented in figure below. In order to cross over this gas slab the shock wave needs the following time:

∆x ∆t = u , (S12.1)

where u is the shock way velocity.

The force that acts on this small gas slab is constant as the slab has uniform cross section and it is given by:

F = ( p + ∆p − p ) S = ∆pS

, (S12.2)

where p is the surrounding gas pressure. The change of the momentum of the gas within the slab during the time that the shock wave needs to cross over the slab is given as:

∆P =∆mv =F ∆t ,

where Δm is the mass of the gas within the slab and v is the velocity of the gas at the moment when the shock wave crosses over the slab. The combination of last two equations gives:

∆mv = ∆pS ∆t , (S12.4)

which further after the substitution of equation (S12.1) in equation (S12.4) gives:

∆mv = ∆p

(S12.3)

∆xS u . (S12.5)

Solutions

The mass of the gas within the slab is:

∆m = ρ∆xS . (S12.6)

The last two equations further give:

v=

67

∆p u ρ . (S12.7)

As the gas slab, in the moment when the shock wave just crossed over the slab, is positioned at the shock wave wave front, it must have the same velocity as the shock wave, that is, v = u must be fulfilled. So, finally, having this in mind, according to the equation (S12.7), one has the following for the shock wave velocity:

u=

∆p

ρ . (S12.8)

S13. The total flux of the Sun is given by:

Φ =4π r 2σ T4

, (S13.1)

where r is the Sun radius and σ is the Stefan–Boltzmann constant. As the Sun radiates isotopically, the magnifying glass receives only a small portion of the total solar flux as it is presented in the figure below.

The flux captured by the magnifying glass is equal to:

= φc

2

D 2 Φ 4π R 2 , (S13.2)

π

where R is Earth distance from the Sun. As the distance between the Sun and the magnifying glass is very large, the image of the Sun is projected on the paper at the distance equal to the magnifying glass focal length. The image diameter is thus given by:

68

Nonstandard Problems in General Physics with Solutions

d=

2 fr R . (S13.3)

The captured flux falls on the paper thus heating it up. As the paper is very good isolator, there will be no radial heat transfer over the paper surface and the only part of the paper that will be heated is the illuminated part of the paper. Due to the energy conservation law, the flux that falls on the paper will be equal to the paper dissipated flux through the radiation. As the paper is very thin, it dissipates through its both sides thus giving the dissipated flux that is given by: 2

d φd = 2π σ TP4 2 , (S13.4)

where TP is the temperature of the illuminated part of the paper. The energy conservation law gives φc = φd , so we have: 2

D π 2 2 Φ =2π d σ T 4 P 4π R 2 2 . (S13.5) Taking into calculation equation (S13.1), one obtains the following from the last equation:

TP4 =

D2r 2 4 T 2d 2 R 2 . (S13.6)

Finally, by substituting equation (S13.3) into the last equation, one has:

TP =

D T 2 2 f . (S13.7)

Taking into account the numerical values of the parameters, we ob-

T T = 230℃ ,

tain TP ≈ 1900°C. Since P the paper will ignite for shore and the young scout will accomplish his task. S14. The analysis will start from figure given below where the human eye geometry of the shortsighted person together with the contact lens and all relevant parameters is depicted. It is well known that contact lenses are positioned close to the human eye surface. In this case,

Solutions

69

if we assume that the contact lens and the eye lens can be approximated as thin lenses, the focal length fCE of the optical system made of contact lens and eye lens can be written the following:

1 = f CE

1 1 + f CL fSE , (S14.1)

where fCL is the focal length of the contact lens for which is valid f CL = 1/ P = −50 cm , f is the focal length of the shortsighted perSE son eye lens, where due to the close positioning of the contact lens and the eye lens, the distance between these lenses is neglected. In order to fully correct the distance vision of the shortsighted person, the focal length of the optical system must be equal to the focal length fNE of the eye lens of the normal person, that is, the following must be

f

= f

NE . In this case, the parallel optical rays, which are fulfilled CE falling in the eye of the observer from the large distance, intersect in the focal plane that is positioned exactly at the surface of the retina of the eye. Therefore, the completely clear image of the distant objects is formed at the retina surface of the shortsighted person eye.

If we further observe in the analysis figure given below, where it is depicted the geometry of the parallel optical rays with all relevant parameters in the case when a shortsighted person doesn’t wear contact lenses, we will notice that parallel rays intersect in the focal plane in front of the retina. Therefore, instead of the clear image on the retina, we will have a blurred image as a consequence of the large light spot on the retina surface, which is formed by small objects placed at a relatively large distance from the eye (such objects form approximately parallel optical rays at the eye surface). The diameter of the light spot is D. Based on the optical rays’ geometry and the similarity of the triangles depicted in figure below, for the light spot diameter on the retina surface, we have the following relation:

70

Nonstandard Problems in General Physics with Solutions

d D = fSE f NE − fSE , (S14.2) from which for the light spot diameter is obtained:

f = D d NE − 1 fSE . (S14.3)

The parallel rays, whose are coming at the observer eye under very small mutual angle Δφ ( ∆ϕ 1 ), will form two adjacent light spots at the retina surface. The mutual distance between those two adjacent light spots is Δx. Based on the rays’ geometrical parameters depicted in figure below for the distance Δx, we have:

∆ϕ ≈ tan∆ϕ =

∆x → ∆x = f NE ∆ϕ f NE . (S14.4)

In order to be able to discriminate two adjacent objects, the light spots that are formed by the two adjacent objects on the retina surface of the shortsighted person eye must not overlap. In this case, the following condition must be fulfilled:

∆x ≥ D . (S14.5)

By substituting equations (S14.3) and (S14.4) in the equation (S14.5), we will obtain:

1 1 ∆ϕ ≥ d − fSE f NE

. (S14.6)

By starting from expression (S14.1) and taking into consideration

that f CL = 1/ P is fulfilled, expression (S14.1) becomes:

Solutions

71

1 1 − f CE fSE , (S14.7)

respectively, if the distance vision is fully corrected we have:

= P

f CE = f NE , so expression (S14.7) becomes: 1 1 = P − f NE fSE . (S14.8)

Finally, by substituting the expression (S14.8) in the inequality (S14.6), for the minimal angle Δφmin under which the eye of the shortsighted person can discriminate two adjacent objects, respectively, for the angular resolution of the shortsighted person distance vision when he doesn’t wear contact lenses, we have:

∆ϕmin = − Pd

. (S14.9)

By substituting the numerical values for the parameters, provided in the textual statement of the problem, we will finally obtain:

∆ϕmin ≈ 0.01 ≈ 34 '

. Just for the comparison, the angular resolution of the distance vision of the normal person is approximately 1′.

The reason why a shortsighted person partially closes his eyes when he doesn’t wear any corrective optical tool in order to get clear image of distant objects lies in the expression (S14.9). By partial closing the eyes, the effective size of the pupil is reduced. Therefore, the minimal angle under which the eye of the shortsighted person can discriminate two adjacent objects is also reduced. If d′ is the effective diameter of the pupil obtained by partial closing the eyes, where d ' < d is satisfied, for the new angular resolution of the shortsighted person, the following is obtained: ∆ϕ 'min ≈ − Pd ' < − Pd ≈ ∆ϕ min .

S15. As presented in figure below, the elementary work that has been done by the friction force over the elementary length of the sledding track ds = CC ' is given by:

dA = µ mgcosϕ ds , (S15.1)

where m is the mass of the sled, g is the Earth’s gravitational field acceleration, and φ is the slope of the track at a given point. As, ac-

2 cording to the problem statement, it is fulfilled mv / R mg , that

72

Nonstandard Problems in General Physics with Solutions

is, the inertial forces are much smaller than other forces, where v is the sled velocity and R is the sledding track radius of curvature, the influence of the inertial forces is omitted when calculating the elementary work of the friction force. Further, from the sledding track geometry, one has:

dx = cosϕ ds , (S15.2) where dx is the elementary horizontal displacement of the sled. The combination of the last two equations gives:

dA = µ mgdx , (S15.3) or, equivalently the total work of the friction force when the sled performs a horizontal shift equal to Δx is given by:

∆A = µ mg ∆x . (S15.4)

The sled will stop at a particular point on the track where the total change of the sled potential energy is, according to the energy conservation law, equal to the work done by the friction force. If the altitude difference between the start and stop point of the sled is Δy, then the overall change in the sled potential energy is:

∆E = mg ∆y . (S15.5)

The energy conservation law gives ∆E =∆A or:

µ mg ∆x= mg ∆y , (S15.6)

which finally gives:

∆y 1 = µ= 0.25= ∆x 4 . (S15.7) In order to graphically determine the place on the track where the sled will stop if it has moved from the starting point without initial velocity, one needs to draw a line from the starting point A where the angle between this line and the horizontal line is α, as presented

Solutions

73

1/ 4 . The point C where the sled in figure, that must satisfy tan will stop lies in the intersection between this line and the curve that represents the track profile.

In order to find the place on the track where the sled will reach the maximum velocity, we will partially use the graphical method. It is obvious that the sled will accelerate until the moment when the friction force overcomes the traction force. At a given point of the track, the traction force is given by:

F = mgsinϕ , (S15.8)

where φ is the slope of the tack at a given point. At this point, the friction force is given by:

T = µ mgcosϕ . (S15.9) For the particular point at the track where the friction force overtakes the traction force, the point from which the sled starts to decelerate is valid F = T , or equivalently:

mgsinα = µ mgcosα , (S15.10)

where α is the track slope at this particular point, which further gives: tanα = µ . (S15.11)

The corresponding point on the track where the sled reaches its maximum velocity can be found as a point on the track where the track slope is given by equation (S15.11), as presented in figure below, or when it is fulfilled:

dy =µ dx . (S15.12)

74

Nonstandard Problems in General Physics with Solutions

The point D at the above-presented diagram is the point on the track that fulfills this condition, that is, the tangent line at this curve point has the required slope. This very point has the vertical and the horizontal distances from the starting point A equal to Δy and Δx, respectively, as presented in figure above. According to the energy conservation law, one has:

mv 2 mg ∆= y + µ mg ∆x 2 , (S15.13)

where v is the sled maximum velocity. Equation (S15.13) further gives:

= v

2 g ( ∆y − µ∆x )

. (S15.14)

Based on the figure above, one may estimate ∆y ≈ 7 a and ∆x ≈ 12a , or if the actual side length of a single square given on the diagram below, where the track profile is shown, is a = 10 m , ∆y ≈ 70 m and ∆x ≈ 120 m . Taking into account these numerical values, one has the following value for the maximum sled velocity v ≈ 28 m / s .

The depicted track profile has a concave shape, thus having only a single point with a given slope. If the track has a more complex shape thus having more than one of such points, the maximum velocity will be reached for a point where the parameter ∆y − µ∆x has the maximum value. However, this point must have shorter horizontal distance from the starting point than the point where the sled will stop.

The error in estimating the maximum velocity can be found from the errors in measuring the vertical and the horizontal positions of the point where the sled reaches its maximum velocity. If the measurement errors are δy for the vertical and δx for the horizontal position

Solutions

75

of the corresponding point, the error in estimating the maximum velocity of the sled is given by:

= ∆v

∂v ∂v δy+ δx ∂ ( ∆y ) ∂ ( ∆x )

. (S15.15)

By substituting equation (S15.14) into equation (S15.15), one obtains:

= ∆v

(δ y − µδ x )

. (S15.16)

The actual measurement errors are δ y = ± a / 2 and δ x = ± a / 2 , so the maximum absolute error in estimating the maximum sled velocity is equal to:

∆v max =

or

ga (1 + µ ) 2v , (S15.17)

∆v max ≈ 2.2 m / s

.

S16. In the beginning, the jeweler holds the chain just above the balance plate, as it is depicted in the left part of the figure below. After releasing the chain, its motion is affected by the Earth’s gravity field. At a certain instant of time, a part of the chain of length x lies at the plate, as it is depicted in the right part of the figure below.

The rest of the chain attains the velocity v due to the chain free fall. Therefore, this velocity is given by:

v = 2 gx

, (S16.1)

where g is the Earth’s gravity field acceleration. The force that acts on the balance plate is equal to the sum of the weight of the chain part, which lies on the plate and the force, which is caused by the non-elastic impact of the incoming part of the chain, or equivalently:

76

Nonstandard Problems in General Physics with Solutions

x +F l , (S16.2)

= Q mg

where m is the chain mass, l is the chain length, and where the impact force F is given by:

F =v

∆m ∆t , (S16.3)

where ∆m / ∆t is the increase in mass of chain that lies on the plate in unit time, which is further given by:

∆m m ∆x m = = v ∆t l ∆t l . (S16.4) The combination of the last three equations gives:

= Q

m gx + v 2 l . (S16.5)

(

)

By substituting equation (S16.1) into equation (S16.5), one obtains:

Q = 3g

m x l . (S16.6)

The balance measures the mass that is given by:

Q x = 3m g l . (S16.7) Since 0 ≤ x ≤ l the equivalent mass, the balance measure is the one that is obtained for x = l , or: m='

m ' = 3m . (S16.8)

Therefore, the jeweler makes a significant error in measurement the chain mass. The measured mass is three times bigger than the real one.

S17. We will begin with the analysis from figure below, showing a small body that begins to move from the top of the hemisphere with the initial velocity v in the horizontal direction. At the moment when the line, which connects the center of the hemisphere and the small body, forms the angle θ with the vertical line, the velocity of the body is u. Based on the analysis of the forces acting on the body

Solutions

77

at that moment, we have the following for the normal force N with which the body acts on the hemisphere:

N mgcosθ − F , (S17.1) = where m is the mass of the small body and F the centrifugal force for which is valid:

mu 2 F= R , (S17.2)

where R is the radius of the hemisphere.

At that moment, the friction force that acts on the body is given by:

T = µ N . (S17.3)

Based on the energy conservation law, one has:

mv 2 mu 2 + mg ∆= h +A 2 2 , (S17.4)

where Δh is the altitude difference shown in the figure above and A is the friction force work. For the work of the friction force, it is valid: θ

A = ∫T (ϕ ) Rdϕ 0

, (S17.5)

T (ϕ ) where for the friction force at a given angle φ, based on equations (S17.1), (S17.2), and (S17.3), the following is valid:

mu 2 (ϕ ) = T (ϕ ) µ mg cos ϕ − R . (S17.6)

Substituting equation (S17.6) into (S17.5), we have:

78

Nonstandard Problems in General Physics with Solutions θ

= A µ mgRsinθ − 2 µ ∫W (ϕ ) dϕ

0

, (S17.7)

where W (ϕ ) is the kinetic energy of the body. The combination of equations (S17.3) and (S17.7) yields: θ

mv 2 + mg= ∆h W (θ ) + µ mgRsinθ − 2 µ ∫W (ϕ ) dϕ 2 0

where we also have:

∆h= R (1 − cosθ )

, (S17.8)

. (S17.9)

In this case, equation (S17.8) becomes:

mv 2 + mgR (1 − cosθ ) = W (θ ) + µ mgRsinθ − 2 µ ∫W (ϕ ) dϕ 2 0

The differentiation of the equation (S17.10) yields:

θ

W (θ ) −

(S17.10)

1 dW (θ ) 1 1 = mgR cosθ − sinθ 2 µ dθ 2 µ . (S17.11)

Equation (S17.11) represents the first-order differential equation, which solution will be obtained as a sum of complementary and particular solutions of this differential equation. For the complementary solution, one has:

WC (θ ) −

1 dWC (θ ) = 0 2 µ dθ , (S17.12)

which solution is given as:

WC (θ ) = Cexp ( 2 µθ )

, (S17.13)

where C is an arbitrary constant that will be found from the initial conditions. The particular solution of the differential equation will be found in the following shape:

W= asinθ + bcosθ P (θ )

. (S17.14)

Substituting equation (S17.14) into differential equation (S17.11), one obtains:

Solutions

79

b a 1 1 mgRcosθ − mgRsinθ a − cosθ + b + = sinθ 2 2 2 2 µ µ µ (S17.15)

From equation (S17.15), it is obvious that the following must be fulfilled:

a−

b 1 = mgR 2µ 2 and (S17.16)

b+

a 1 = − mgR 2µ 2µ . (S17.17)

The solutions of the above set of two equations are:

a=

2µ 2 − 1 mgR 4µ 2 + 1 and (S17.18)

3µ 2 b= − 2 mgR 4µ + 1 . (S17.19) The combination of equations (S17.13), (S17.14), (S17.18), and (S17.19) gives: W (θ ) = Cexp ( 2 µθ ) +

2µ 2 − 1 3µ 2 mgR cos θ − mgRsinθ 4µ 2 + 1 4µ 2 + 1 (S17.20)

The constant C will be determined from the initial condition, that is, from:

2µ 2 − 1 mv 2 W (θ = 0) = C+ 2 mgR = 4µ + 1 2 , (S17.21)

which further gives:

= C

mv 2 2 µ 2 − 1 − 2 mgR 2 4µ + 1 . (S17.22)

Therefore, the complete solution of the differential equation (S17.11) is given as: mv 2 2 µ 2 − 1 2µ 2 − 1 3µ 2 W (θ ) = mgR exp ( 2 µθ ) + 2 mgRcosθ − 2 mgRsinθ − 2 4µ + 1 4µ + 1 4µ + 1 2

(S17.23)

80

Nonstandard Problems in General Physics with Solutions

The minimum required velocity of the small body vmin, which is given to the body in the horizontal direction in order to fall from the hemisphere, will be found from the condition that the body reaches a point on the hemisphere where the friction force is equal to the tangential component of the body’s weight and at that moment has a zero velocity or zero kinetic energy. From this point, the tangential component of body weight will be greater than the friction force. Therefore, the body will continue to move and finally fall from the hemisphere. For the corresponding critical angle θC to be reached by the body is worth, the following is valid:

tanθ C = µ . (S17.24) According to the previous conclusions and expression (S17.23), the following is valid:

= vmin

{

}

2 gR 2 µ 2 − 1 + exp ( −2 µθ C ) 3µ tanθ C − 2 µ 2 − 1 cosθ C 4µ 2 + 1 (S17.25)

(

)

Substituting equation (S17.24) into (S17.25), one obtains finally the following:

= vmin

2 gR 2 2 µ − 1 + µ 2 + 1exp ( −2 µ arctanµ ) 2 4µ + 1 (S17.26)

v = gR , meaning that the body must be If µ → +∞, one has min given the velocity, which will provide that the centrifugal force overcomes the body weight, thus the body immediately leaves the hemisphere. S18. In order to find the age of the Universe relaying only on the given measured astronomical data, one must find the ratio between the irradiation of the Universe due to the star, diffuse galactic and cosmic light and the corresponding energy density due to the electromagnetic radiation. Therefore, we will observe an elementary volume

[ r , r + dr ] , elevation angles [θ ,θ + dθ ] , and dV limited to the radii [ϕ ,ϕ + dϕ ] , as depicted in figure below, where only azimuth angles a few of the total number dN of the photons contained in this elementary volume will directly reach the elementary surface, as the distribution of the photon wave vectors is uniform in solid angle.

Solutions

81

The elementary probability dp that the photon in the elementary volume dV will directly reach the elementary surface dS is equal to:

dp =

dΩ 4π , (S18.1)

where dΩ is the elementary solid angle at which the photon in the elementary volume dV sees the elementary surface dS. Each photon in the elementary ring limited with the radii [ r , r + dr ] and elevation angles [θ ,θ + dθ ] can directly reach the elementary surface with the same elementary probability as given in equation (S18.1), because every photon in this elementary ring “sees” the elementary surface with the same elementary solid angle.

The elementary number of photons dNR from the elementary ring with the elementary volume dVR that directly reach the elementary surface is equal to:

dN R = ndpdVR

, (S18.2)

where n is the photon density and where for the elementary volume of the elementary ring, one has:

dVR = 2π r 2 cosθ drdθ

. (S18.3)

The total number of photons dNS that directly reach the elementary surface from the elementary hemispherical shell in the time interval [t , t + dt ] , where r = ct , and where c is the speed of light in vacuum is equal to:

= dNS

dN ∫= R

dVS

1 ndSdr 4

, (S18.4)

where dVS is the elementary volume of the hemispherical thin shell and the elevation angle θ takes the values from the range [ 0, π / 2] .

82

Nonstandard Problems in General Physics with Solutions

The overall elementary energy dWS attained by the elementary surface dS during the time interval [t , t + dt ] is given by:

1 1 cε nd= Sdt cwdSdt 4 4 , (S18.5) where ε is the average photon energy and w = ε n is the energy den= dWS ε= dNS

sity of the electromagnetic radiation. For the irradiance E at the elementary surface, we have:

= E

dWS 1 = cw dSdt 4 . (S18.6)

The analysis considers a simple Universe model that is presented in figure below with all relevant parameters. Each photon that has been radiated throughout the Universe history is captured within its volume. The main principle behind this model is the energy conservation law, that is, stellar radiation fills the Universe volume with uniform energy density electromagnetic radiation. The total presentday electromagnetic energy W that is radiated by the stars and stored within the Universe volume is given by:

W = NLT , (S18.7)

where N is the total number of stars in the Universe and T its presentday age.

As a result of stellar radiation, the Universe is filled with a uniform electromagnetic energy density w. Since the Universe has been considered homogenous and isotropic, the electromagnetic energy density within the Universe volume has been considered to be uniform. Taking this into consideration, the electromagnetic energy P, stored within the Universe volume, is given by:

P = wV , (S18.8)

Solutions

83

where V is the Universe volume. According to equation (S18.6), the electromagnetic energy density can be obtained from the measured irradiance, that is:

w=

4E c . (S18.9)

According to the first principle, we have P = W , that is, the starsradiated and the Universe-stored electromagnetic energies have the same value. This fundamental principle in combination with equations (S18.7) and (S18.9) gives the following for the age of the Universe:

T=

4E cλ L , (S18.10)

where the average star density is λ = N / V . Substituting the numerical values, one obtains T ≈ 12.3 Gy as the Universe age.

S19. The light pressure on the probe solar sail is very small, so, the probe leaves the Earth’s orbit very gradually. Therefore, we can adopt that the probe orbit is almost circular, where the probe angular momentum is given as:

L = mvr , (S19.1)

where v is the probe velocity and r is the probe orbit radius. As the orbit is almost circular, we have also fulfilled:

mv 2 mM =G 2 r r , (S19.2)

where M is the solar mass. Due to the pressure of the light onto the sail, there will be net torque acting on the probe, which further causes the change of the angular momentum of the probe as:

dL =T dt , (S19.3)

where T is the torque. The combination of equations (S19.1) and (S19.2) gives:

L = m GMr , (S19.4)

84

Nonstandard Problems in General Physics with Solutions

so, in order to provide fastest possible journey of the probe the torque must be maximized. If the probe solar sail angle with respect to the radius vector is always fixed to θ, as presented in figure below, each photon that falls onto the sail will transfer the following momentum to the sail in the direction that is normal to the solar sail:

Äp = 2 psinθ , (S19.5)

where p is the photon momentum.

The overall momentum transferred to the solar sail during the time interval Δt is given by:

∆P =

∑∆p

i

i

= 2sinθ ∑ pi i

, (S19.6)

where Δpi is the transferred momentum of the i-th photon and pi is the i-th photon momentum. Therefore, the net force acting on the sail is given as:

= F

∆P ∑ pi = 2sinθ i ∆t ∆t . (S19.7)

Further, we have that the overall energy of the reflected photons during this period of time is:

∆W=

∑ε = ∑cp= c∑ p i

i

i

i

i

i

, (S19.8)

where εi is the energy of the i-th photon. The solar irradiance at the distance r from the Sun is thus given by:

E (r ) =

1 ∆W S ′ ∆t , (S19.9)

Solutions

85

where S’ is the effective surface of the sail seen from the Sun, or S ′ = Ssinθ . The combination of the last four equations yields:

F=

2S ′ 2S E ( r ) sinθ = E ( r ) sin 2θ c c . (S19.10)

The solar irradiance along the probe path is also given as:

R E (r ) = E E r . (S19.11)

2

The torque acting on the sail according to the geometrical parameters presented in the figure above is then given by:

T = rFcosθ . (S19.12)

Combining the last three equations, one has:

T=

2 SERE2 1 2 sin θ cosθ c r . (S19.13)

The torque reaches its maximum value for cosθ = 3 / 3 or equivalently for the optimal angle between the solar sail and the radius vecθ 54.7° . For this angle, the torque has its maximum tor equal to= value given by:

T=

4 3SERE2 1 9c r . (S19.14)

Now, when we have the maximal torque, we will go back on equation (S19.3) where it is described the probe movement. By substituting equation (S19.4) into equation (S19.3), one has:

1 1 dr m GM =T 2 r dt . (S19.15) By taking into account the maximal value of the torque, given in equation (S19.14), the last equation becomes:

r dr =

8 3SERE2 dt 9cm GM , (S19.16)

which after the integration gives:

86

Nonstandard Problems in General Physics with Solutions 3

8 3SERE2 t +C 9cm GM , (S19.17)

= r 2 (t )

where C is an arbitrary constant that will be found from the initial t 0= ) RE . So, according to the equation conditions, that is, from r (= (S19.17), the constant C takes the following value: 3

C = RE2 , (S19.18)

which finally yields to:

8SE 3RE 3 r= t ( t ) RE 1 + 9cm GM . (S19.19)

2

In order to find the time duration τ of the whole journey to the Mars’s r (τ ) = RM orbit, we have fulfilled: , which gives:

3 9cm GM RM 2 = τ − 1 8SE 3RE RE . (S19.20) Taking into account the numerical value, one obtains τ ≈ 60 years

as the journey duration.

S20. Analysis of the work of the Heron’s fountain will begin from the figure shown below where the relationships between the vessels are defined, as well as all relevant parameters for the analysis. All vessels (A, B, and C) have equal areas of the cross section S, while the vessels B and C have the same height H.

Solutions

87

In order to provide proper work of the Heron’s fountain, the folh > max {lA } lowing condition must be fulfilled: F , where hF is the height of the fountain tube in vessel A and lA the water level in the same vessel. At the very beginning, the water level in vessel A is equal to lAS, the level of water in vessel B is at its maximum level

l

l =0

=H

and equal to BS , while the vessel C is empty CS . Bottoms of the vessels A, B, and C are at the following heights in relation to

h =0

the lowest horizontal base: hA, hB, and C , respectively. At the initial moment, valve V is closed, while air pressure in hermetically sealed vessels B and C is equal to atmospheric pressure p0. By opening the valve, the water from the vessel A begins to flow through the connecting pipe connecting it to the vessel C. As a result, the level of water in the vessel C will rise as well as the air pressure in the vessels B and C, which are also interconnected by the pipe. As a result, the water level in the pipe connecting the vessels A and C will grow until it reaches the maximum level and starts to emerge from it. In this way, the Heron’s fountain is started up. In order to start up fountain at all, the air pressure in vessels B and C must be large enough to allow it. We consider that due to the thin walls of the vessels as well as the slow flow of water, the compression of air in the vessels B and C is isotherm process. Therefore, the following is valid for the pressure p in these vessels before the beginning of the work of the fountain:

= p0 SH pS ( H − lC )

, (S20.1)

where lC is the height of the water in the vessel C. This pressure should be high enough to expel the water so that it starts to exit from vessel B and enters vessel A. This will happen for some critical value of the lCC level of water in the vessel C, that is, the critical value of the pressure of air pC in vessels B and C. In order to continue after the start of the fountain process, at this critical moment of the start of the operation of the fountain, the condition of the continuous water flows from vessel A to vessel C must be met, that is, the following must be fulfilled:

p0 + ρ g ( hA + lAC ) ≥ pC + ρ glCC

, (S20.2)

where ρ is the density of the water, g is the Earth’s gravitational field acceleration, and lAC is water level in vessel A at the critical moment

88

Nonstandard Problems in General Physics with Solutions

of the start of the work of the fountain. As we are using an incompressible fluid (water) and as we are neglecting the volume of water

l

+l

= l

in the pipes, this must be fulfilled: AC CC AS (all three vessels have the same area of the cross section and the pipes have very small cross-sectional area). On the other hand, the level of water in the fountain tube has reached a maximum value, so the following is valid:

p0 + ρ g ( hA − hB + hF ) = pC + ρ gH

l

+l

. (S20.3)

= lAS is fulfilled and by com-

Taking into consideration that AC CC bining the last three equations, we have:

lCC ρg = ( hA − hB + hF − H ) H − lCC p0 and (S20.4)

lCC ρg ≤ ( hA + lAS − 2lCC ) H − lCC p0 . (S20.5)

2 p ≈ 105 Pa ρ = 103 kg / m3 As we have 0 , , and g = 9.81 m / s , we p / ( ρ g ) ≈ 10 m obtain 0 , which is much larger than all the heights in l / ( H − lCC ) 1 the fountain, the condition CC is then fulfilled, that

is, we have have:

lCC H . In this case, from the last two expressions, we

lAS ≥ hF − hB − H + 2lCC ≈ hF − hB − H . (S20.6) The necessary and sufficient condition to start up Heron’s fountain is then:

hB + H ≥ hF , (S20.7)

which is surely fulfilled, since the height of the fountain hose hF is relatively small.

In the following text, the work of the fountain after the start of work will be analyzed, as presented in figure below where the fountain is displayed at the arbitrary moment of its operation. In this case, the pressure p in the vessels B and C is:

Solutions

p0= SH pS ( 2 H − lB − lC ) ⇒ = p p0

89

H 2 H − lB − lC , (S20.8)

where it is also valid: lA + lB + lC = H + lAS ≈ H , as according to expressions (S20.6) and (S20.7), lAS can be very small, or much less than the height of the vessels H.

In order to flow the water from vessel A to vessel C, the following must be satisfied:

p0 + ρ g ( hA + lA ) ≥ p + ρ glC

. (S20.9)

On the other hand, based on Bernoulli’s law, for the velocity v of water sprinkling from the fountain, we have the following:

p0 + ρ g ( hA + hF ) +

ρv2 2

=p + ρ g ( hB + lB )

.

(S20.10)

Due to the sprinkling velocity of the water from the fountain tube, the water reaches the maximum height Δh in relation to the top of the pipe of the fountain for which it is valid:

v2 ∆h = 2 g . (S20.11)

The combination of the last two equations gives:

= ∆h

p − p0 + hB + lB − hA − hF ρg . (S20.12)

Substituting equation (S20.12) into (S20.8), one obtains:

90

Nonstandard Problems in General Physics with Solutions

p0 lB + lC − H = ∆h + hB + lB − hA − hF ρ g 2 H − l − l B C . (S20.13)

l +l = H +l −l

AS A , the last equation becomes: As it is satisfied B C p ∆lA ∆h =− 0 +h +l −h −h ρ g H + ∆lA B B A F , (S20.14)

where ∆lA = lA − lAS is the change of the level of water in vessel A. Further, expression (S20.9) may be rearranged in the following way:

−

p0 ∆lA ≤ hA + lA − lC ρ g H + ∆lA . (S20.15)

The combination of the last two equations gives:

∆h ≤ hB + lA + lB − lC − hF . (S20.16)

As it is valid

∆h ≤ hB + H − 2lC + lAS − hF . (S20.17)

The maximum possible height that water may reach in fountain will

lA + lB + lC = H + lAS , the last expression becomes:

l =0

be possible for a minimum lC, or at the very beginning when C is fulfilled. In this case, the maximum possible height Δhmax of the water column in the fountain will be at the very beginning and will be given as:

∆hmax = hB + H + lAS − hF ≈ hB + H , (S20.18)

where lAS , hF hB + H was taken into account. Further, according to the expression (S20.14), the fountain will work for:

∆h =−

p0 ∆lA +h +l −h −h ≥ 0 ρ g H + ∆lA B B A F .

(S20.19)

Combining the last expression with expression (S20.15), we have:

hA + hF − hB − lB ≤ −

p0 ∆lA ≤ hA + lA − lC ρ g H + ∆lA , (S20.20)

or the following condition must be constantly fulfilled:

Solutions

hF − hB ≤ lA + lB − lC . (S20.21)

As

91

lA + lB = H + lAS − lC is fulfilled, the last expression becomes:

hF − hB ≤ H + lAS − 2lC

. (S20.22)

In order to make the fountain work as long as possible, it is necessary that almost all the water from vessel B be transferred into vessel C, that is, in the end, it will be

lB ≈ 0

and

lC ≈ H

, where on the basis of

∆l the expression (S20.15), it is obvious that A is much less than all other geometric parameters. It should be taken into account, on the

∆l < 0

basis of expression (S20.14), that A , meaning that the level of water in vessel A decreases. At the very end, the condition given by the expression (S20.22) must also be fulfilled:

hB − hF ≥ H + lAS

. (S20.23)

As it is fulfilled lAS , hF hB , H , we finally get the condition for providing the work of the fountain as long as all of the water is transferred from the vessel B to the vessel C:

hB ≥ H

. (S20.24)

If the condition given by the expression (S20.24) is fulfilled, at the very end, the situation in the fountain will be as shown in the figure below.

S21. This problem will be solved by using two fundamental conservation laws, that is, the conservation of momentum and the conservation

92

Nonstandard Problems in General Physics with Solutions

of angular momentum. The situations before and after planet formation, with all relevant parameters for the analysis, is presented in the figure shown below. Since the pressure inside the planet is not high enough to change the material density, one has the following:

m=

4π 3 r ρ 3 , (S21.1)

where r is the radius of newly formed planet.

The thickness of the asteroid belt is much smaller than its width. Therefore, one can define the surface density of the material (asteroids) within the asteroid belt as:

σ=

m 2π Rw , (S21.2)

where the condition w R was taken into account. The asteroid that rotates around the star at the distance RA, where we have

R − w / 2 ≤ RA ≤ R + w / 2 fulfilled, has the velocity v for which the A

following is valid:

mA vA2 m M = G A2 RA RA , (S21.3)

where mA is the mass of the corresponding asteroid and G is the universal gravitational constant. The asteroid velocity is thus given by:

vA =

GM RA

. (S21.4)

Due to the compensation effects of the gravitational attraction force and centrifugal force of the rotational movement, all asteroids are weightlessness. Therefore, one can take that the momentum of all asteroids is conserved in the momentum of the newly formed planet. The momentum of all asteroids is given as:

Solutions

PA =

93

R + w /2

∫

vA dmA

R − w /2

, (S21.5)

where the asteroid velocity vA is given by the equation (S21.4), and where is valid:

dmA = 2πσ RA dRA , (S21.6) is the asteroid mass in elementary asteroid belt. The solution of the integral given in equation (S21.5) is:

GM R , (S21.7) where the condition w R was taken into account. When the planet is formed its momentum is given by: PA ≈ m

= PP mv = mR= m P PÙ

GM RP

, (S21.8)

where RP is the planet orbital radius, vP its orbital velocity, and Ω its angular velocity of revolution around the star. According to the momentum conservation law, we have:

PP = PA

which gives:

RP ≈ R

, (S21.9) , (S21.10)

or the planet orbital radius is approximately equal to the asteroid mean distance from the star.

The angular momentum of all asteroids is given as:

LA =

R + w /2

∫

vA RA dmA

R − w /2

. (S21.11)

The solution of this integral gives: 2m GM LA = 5 Rw

5 5 2 2 w w R+ −R − 2 2 .

(S21.12)

94

Nonstandard Problems in General Physics with Solutions

The angular momentum of a newly formed planet is given by:

= LP mvP RP + J Pω , (S21.13)

where JP is the planet moment of inertia and ω its rotation angular velocity, where it was assumed that both vectors of revolution and rotation angular velocities have the same directions. According to the angular momentum conservation law, we have:

LP = LA , (S21.14)

which gives:

2m GM mvP= RP + J Pω 5 Rw

Bearing in mind that we have:

JP = = vP

2 2 mr 5 and (S21.16) GM GM ≈ RP R

, (S21.17)

equations (S21.15) becomes:

= ω

5 5 2 2 w w R+ R − − 2 2 . (S21.15)

GMR r2

5 5 w 2 w 2 5 R − 1 − − 1 + w 2 R 2 R 2 . (S21.18)

The function inside the square brackets can be developed in Taylor series as: 5

2 3 4 w 5 w 2 5 w 15 w 5 w 5 w 1+ 1 + = + + − + O 2 2 R 8 2 R 16 2 R 128 2 R 2R 2 R

(S21.19)

and

5

2 3 4 w 5 w 2 5 w 15 w 5 w 5 w 1 − = 1 − + − − + O 2 2 R 8 2 R 16 2 R 128 2 R 2R 2 R

(S21.20)

If we substitute equations (S21.19) and (S21.20) into equation (S21.18), one obtains:

Solutions

ω≈

5 w GMR 64 R

2

2

3

4πρ 3m . (S21.22)

The newly formed planet period of rotation is then given by:

=

2

5 GMR w 64r 2 R . (S21.21)

Finally, by combining equations (S21.1) and (S21.21), we have:

ω≈

95

2π

ω

≈

128π R GMR

3m πρ .

(S21.23)

The direction of the planet rotation is the same as the direction of planet revolution, so the planet has prograde motion.

S22. In order to find the maximum possible temperature that certain points at the spacecraft surface may reach when the spacecraft penetrates the high atmospheric layer with high velocity, we will assume that the air at this height is very cold. Therefore, due to the energy conservation law, we have:

mv 2 5 N = kB NT 2 2 , (S22.1)

where m is the molecule mass, N is the number of molecules that have been captured by a cavity at the spacecraft surface, kB is the Boltzmann’s constant, and T is the air temperature. It was assumed that the kinetic energy of molecules due to the spacecraft movement has been transformed into corresponding thermal energy of the molecules. The last equation may be rearranged into:

1 5 nµ v 2 = kB nN AT 2 2 , (S22.2)

where n is the amount of substances and NA is the Avogadro’s constant. Bearing in mind that:

kB N A = R

, (S22.3)

we have the following for the maximum temperature of the spacecraft surface:

96

Nonstandard Problems in General Physics with Solutions

T=

µv2

5 R . (S22.4) Substituting the numerical values, we obtain T ≈ 41, 000 K .

S23. The analysis will begin from figure given below, where the rotation and precession of basketball ball at the top of the basketball player finger with all relevant parameters are presented. In order to solve the problem, we will adopt certain assumptions. First, we will approximate the basketball ball to a homogeneous sphere whose entire mass is concentrated on its surface. Then the moment of inertia of the basketball ball for the axis passing through the center of the ball (axis of rotation of the ball) is valid:

J=

2 mR 2 3 , (S23.1)

where m is the mass of ball. Second, in order to prevent the basketball ball from slipping from its fingers, the ball is made of a material that has a high friction coefficient. Third, we consider that the basketball player holds his finger very firmly so that the finger can be considered immobile.

When the ball rotates with an angular velocity ω and it is placed at the top of the finger, it will perform a regular precession with an angular velocity Ω around the fixed point A at the top of the finger. Due to the symmetry in relation to the vertical axis passing through the top of the basketball player finger, that is, the point A, the direction of the angular velocity vector of the ball precession will be on that axis or in the direction of the vector of the Earth’s gravity field. The reason for such a precession is the large coefficient of friction between the ball and the finger, which will ensure that the motionless ballpoint remains at the tip of the finger (point A).

Solutions

97

The cause of the precession of the ball is the torque M, which is the result of the action of certain forces, that is, the weight Q of the ball and the inertial force (centrifugal forces) T acting in point C, which represents the center of gravity of the ball. For the torque acting on the ball is valid the following: M =R × Q + T , (S23.2)

(

)

where R is the position vector of the center of gravity of the ball C relative to the point of contact of the ball and finger A, where the direction of the vector of the torque is directed so that it enters the plane of the drawing at a given moment of time (at the moment when the axis of rotation and axis of the precession of the ball are in the plane of the drawing). Further, for the corresponding forces, and based on the analysis of above-presented figure, it is valid: Q = mg and (S23.3) 2 = T mR Ù sin ϕ ⋅ i r , (S23.4)

where g is the vector of the Earth’s gravity field acceleration, φ is the angle between the axis of the rotationand the axis of the precession of the ball, and ir is the unit vector ( ir = 1 ) in the radial direction perpendicular to the axis of the precession of the ball. Based on the expressions (S23.2), (S23.3), and (S23.4), as well as on the basis of the geometric relations between the individual parameters shown in figure above, the torque is given by: = M mR sin ϕ g + RΩ 2 cos ϕ ⋅ iφ , (S23.5)

(

)

98

Nonstandard Problems in General Physics with Solutions

where iφ is the unit vector ( iφ = 1 ) directed so that it enters the plane of the drawing at a particular instant of time (at the moment when the axis of the rotation and axis of the precession of the ball are in the level of the drawing). Based on the dynamics that describes the movement of the ball at the top of the finger, we have: dL M= dt , (S23.6)

where L is the total angular momentum of the ball for which it is valid: = LR = J ω is the angular momentum of the ball L LR + LP where to its rotation and LP is the angular momentum of the ball due to its precession. Since, on the basis of the last equation in a very short time interval Δt ( ∆t → 0 ), it will be fulfilled ∆L = M ∆t , therefore, at any given instant of time, the change in the angular momentum will be directed in the same direction as the torque M , or in this situa tion, perpendicular to the vector of the angular momentum L . Then according to the Pythagorean theorem, the following can be written: L ( t= + Ät )

2 ∆L ( t ) 2 2 L ( t ) + ∆L ( t ) ≈ L ( t ) + 2 L (t )

,

(S23.7)

where L ( t ) and L ( t + ∆t ) are the intensities of the angular momen

tum vectors at the time instants t and t + ∆t , respectively, and ∆L ( t ) the intensity of the angular momentum vector change. As ∆L ( t ) is the infinitesimal of the second order, the last equation becomes: 2

L ( t + ∆t ) ≡ L ( t ) = const.

, (S23.8)

that is, the intensity of the angular momentum vector is constant. Since the angular momentum change vector is perpendicular to the angular momentum vector and as the angular momentum is constant, the top of the angular momentum vector will move around the circle, that is, the ball will perform a precession. Therefore, the angular mo L mentum vector P of ball due to its precession must be constant. In this case, the equation (S23.6) becomes: dL dLR M = = dt dt . (S23.9)

Solutions

99

The geometry of the motion of the angular momentum vector is shown in figure below. As can be seen from figure below, the angular momentum vector rotates around the fixed point A (the basketball player fingertip) at angle φ relative to the horizontal plane. Based on the geometry shown in figure below and under the assumption

∆φ → 0 for the intensity of the change in the angular momentum ∆L =∆L ( t )

vector

, the following is valid:

∆= L Lsinθ∆φ , (S23.10) L = L (t )

where is the intensity of the angular momentum vector and Δϕ is the angle that the angular momentum vector transits for the infinitesimal time interval Δt.

By combining the expressions (S23.1), (S23.5), and (S23.10) and taking into account that it is satisfied L = J ω and L= M ∆t , where M= M ω=ω and , one obtains:

2 ∆φ g + RΩ 2 cosϕ = Rω 3 ∆t . (S23.11) As ∆φ / ∆t represents the angular velocity of rotation of the angular momentum vector, that is, the angular velocity of the precession of the ball, the following is valid ∆φ / ∆t =Ω, where we have Ω = Ω . Taking this into account, the last equation becomes:

2 R cos ϕΩ 2 − RωΩ + g = 0 3 . (S23.12) By solving the last equation, the following is obtained for the angular velocity of the precession of the ball:

100

Nonstandard Problems in General Physics with Solutions

Ω =ω

9g cosϕ Rω 2 3cosϕ . (S23.13)

1± 1−

In order to have a physically realistic solution for the angular velocity of the precession of the ball, that is, that the ball could be stabilized at the top of the basketball player finger, the basketball player must satisfy the following condition based on expression (S23.13):

ω ≥3

g cosϕ R . (S23.14)

In the case of a small angle φ ( ϕ 1 ), that is, when the axis of the rotation and axis of the precession of the ball almost coincide, we will have the strictest criterion for stable rotation of the ball at the top of the basketball player finger. Then the minimum angular velocity, which is necessary to be given to the basketball ball in order to steady rotates on the top of the basketball player finger, is given by:

g R . (S23.15)

ωmin = 3

By replacing the numerical values given in the problem statement, we obtain:

ωmin ≈ 27.1rad / s ≈ 259 rpm .

S24. As the Moon very gradually departures from Earth, we can write:

mv 2 ( t )

R (t )

= G (t )

mM R2 (t )

, (S24.1)

where m is the Moon’s mass, v is its orbiting velocity, and M is the Earth’s mass. According to the conservation of the Moon’s angular momentum, we also have:

mv ( t ) R ( t ) = L

, (S24.2)

where L is the constant, that is, it doesn’t depend on time. Therefore, according to the last two equations, we have:

G ( t )= R (t )

L2 = const. m2 M (S24.3)

Solutions

101

Based on the last equation, one has:

d G ( t ) R ( t ) = 0 dt , (S24.4)

which further gives:

R ( t )

R (t )

G (t )

. (S24.5)

= −

1 t

, (S24.6)

so, equation (S24.5) becomes:

R ( t )

G (t )

we also have:

G ( t )

G ( t )

= −

R (t )

=

1 t

. (S24.7)

The Moon–Earth distance thus changes as:

R (t ) = R

t

τ , (S24.8)

where R is the actual Moon–Earth distance and τ the present-day age of the Universe. According to the equation (S24.8), the distance between Moon and Earth increases for the following value:

∆R = R

∆t

τ , (S24.9)

each year where Δt is the period of 1 year. Taking the following val-

ues R ≈ 384, 000 km , ∆t =1 y , and τ ≈ 13.7 Gy into the account, one obtains ∆R ≈ 2.8 cm . Since this value is smaller than the measured one ( ∆R ≈ 3.82 cm ), it is possible that the Moon moves away from the Earth due to the change of gravitational constant.

S25. The analysis of the work of the mercury thermometer will begin from the figure shown below where the relationships between mercury reservoir, constriction, and capillary of the thermometer scale are defined, as well as all relevant parameters for analysis.

102

Nonstandard Problems in General Physics with Solutions

We will assume that the constriction of the capillary and the capillary of the scale are of the circular cross section with radiuses r and R, respectively, where r R is valid. After the measurement, the thermometer is released from the contact with the human body, thus being positioned in the environment with a lower temperature. As a result, it begins to cool down so that the temperature of the reservoir changes rapidly. We will assume that the rest of the thermometer, besides the reservoir, is cooling at a much lower speed. There are three reasons for this: (1). poor thermal conductivity of the glass, (2) very thin glass walls of the thermometer that slow down heat dissipation, and (3) the constriction of the capillary and the capillary of the thermometer scale are in the evacuated glass housing so that they very poorly conduct the heat from the constriction of the capillary and capillary of the thermometer scale to the glass thermometer housing. In addition, the very thin glass walls of the reservoir make it possible, despite the poor thermal conductivity of the glass, to quickly exchange heat with the environment. In this case, the volume of mercury outside the reservoir V and the rate of change of the volume of

mercury outside the reservoir V are related as: = V 3 (α M − α G ) VT , (S25.1) where T is the rate of the temperature change of the thermometer

reservoir. Due to the effect of intermolecular forces by reducing the volume of the mercury in the reservoir, they will start pulling the mercury out of the capillary of the thermometer scale to the reservoir. Then the following must be satisfied: V = S1v1 , (S25.2)

where S1 is the area of the scale capillary cross section for which is 2 valid S1 = π R with R being the capillary radius and v1 is the decay rate of the mercury on the scale. By combining expressions (S25.1)

Solutions

103

and (S25.2) for the rate of mercury decay on a scale, the following is valid:

v1 =

3 (α M − α G ) V T π R2 . (S25.3)

Since mercury is an incompressible fluid, the withdrawal of a certain amount of mercury from the scale capillary must follow the flow of the same amount of mercury through the constriction of the capil-

Sv =S v

2 2 , where v is the mercury lary. Therefore, it must be filled 1 1 2 velocity through the constriction and S2 is the area of the cross secS = π r2 tion of the constriction for which is valid 2 with r being the capillary constriction radius.

In the followings, we will move on to the analysis of the flow of mercury through the thermometer when the reservoir of the mercury starts to cool. In the analysis, we will look at the part of the system shown in the figure above. Using the energy conservation law for the system limited by the surfaces S1 and S2, the following is obtained:

1 1 ∆mv12 + F2 ∆x2 + p1S1∆x1 = ∆mv22 + p2 S2 ∆x2 2 2 , (S25.4)

where Δm is the mass of the observed elementary part of the mercury in the constriction of the capillary and the capillary of the scale for

which it is valid: ∆m = ρ S1∆x1 = ρ S 2 ∆x2 , where ρ is the density of mercury and Δx2 and Δx1 are the lengths of the observed elementary part of mercury in the constriction of the capillary and the capillary of the scale, respectively, F2 is the corresponding intermolecular force, and p1 is pressure in the capillary of the scale and p2 is the mean pressure in the capillary constriction along the path length Δx2, respectively. By further elaboration of the expression (S25.4), it becomes:

1 1 ∆mv12 + F2 ∆x2 + p1S1∆x1 = ∆mv22 + p2 S 2 ∆x2 2 2 ,(S25.5)

or equivalently:

1 2 F2 1 2 ρ v1 + + p= ρ v2 + p2 1 2 S2 2 . (S25.6)

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Nonstandard Problems in General Physics with Solutions

For the intermolecular force, it is valid:

F2 = 2π rσ . (S25.7)

where σ is the surface tension, while for pressure p1 in the capillary of the scale, it is valid:

p1 =

2σ RM . (S25.8)

where RM is the radius of the curvature of the meniscus for which

RM ≥ R is valid since the wetting angle is obtuse. By substituting

these values in expression (S25.6), it becomes:

1 1 1 2 ρ v1 + 2σ + 2 r RM

or since

1 2 2σ 1 2 ρ v1 + ≈ ρ v2 + p2 2 r 2 . (S25.10)

1 2 = ρ v2 + p2 2 ,

(S25.9)

r R ≤ RM :

Due to the viscous friction effect, the pressure along the capillary constriction will not be the same. The pressure difference Δp2 along the observed elementary part of the constriction of the capillary length Δx2 is equal to:

∆p= 2

8η v2 ∆x2 r2 , (S25.11)

where η is the viscous friction coefficient of mercury. In this case, the mean pressure

p2 =

p2 is:

p2 + ( p2 + ∆p2 ) 2

∆p2 = p2 +

4η v2 ∆x2 r2 . (S25.12)

Since the capillary of the scale has a much larger radius than the constriction of the capillary ( R r ), the velocity of the mercury flow in the capillary of the scale is much smaller than the velocity of

the stream in the constriction of the capillary ( v1 v2 ), and as the elementary pathway in the capillary scale is much shorter than the

Solutions

105

elementary path in the constriction of the capillary ( ∆x1 ∆x2 ), the pressure change along the elementary path in the case of a capillary scale is much smaller than the pressure change along the elementary path in the constriction of the capillary. Therefore, in expression (S25.4), instead of the mean pressure in the capillary of the scale, the pressure value on the external wall of the elementary path was adopted. By combining terms (S25.10) and (S25.12), we obtain:

4η v 1 2 2σ 1 2 ≈ ρ v2 + p2 + 2 2 ∆x2 ρ v1 + 2 r 2 r . (S25.13) Since the last member of the expression (S25.13) is infinitesimally small, by choosing arbitrarily small values for Δx2, the expression (S25.13) can be written as: 1 2 2σ 1 2 ρ v1 + ≈ ρ v2 + p2 2 r 2 . (S25.14)

Further, based on the continuity equation, it is valid: the last equation becomes:

S1v1 = S 2 v2 , so

2

1 2 2σ 1 2 S1 2 ρ v1 + ≈ ρ v1 v2 + p2 2 r 2 S2

which after the rearrangement gives:

r v1 ≈ R

The combination of equations (S25.3) and (S25.16) yields:

π

From the last equation, one obtains the following for the pressure p2:

3

2

, (S25.15)

2 2σ − p2 ρ r . (S25.16)

(α M − α G )VT ≈ r 2

2 2σ − p2 ρ r . (S25.17) 2

p2 ≈

2σ ρ 3 − 4 (α M − α G ) VT r 2r π . (S25.18)

In order to break the mercury flow in the constriction of the capillary, it must be valid: 3

π

p2 ≤ 0 so we have:

(α M − α G ) ρVT ≥ 2

σρ r 3

, (S25.19)

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Nonstandard Problems in General Physics with Solutions

or equivalently:

2π σρ r 3 mT ≥ 3 (α M − α G )

, (S25.20)

where m = ρV is the mass of mercury in the reservoir. Since the mercury exchanges heat with the environment through its thin glass reservoir, based on Newton’s law of cooling, we have:

hS ⋅ δ T mcT = , (S25.21)

where c is the specific heat capacity of mercury, h coefficient of convective heat transfer between glass and air, S total area of the mercury reservoir, and δT the temperature difference between the mercury in the reservoir and the environment. By combining the last two expressions, we obtain: 2

1 3hS r≤ (α M − α G ) δ T σρ 2π c . (S25.22) Since δ T ≤ ∆T is always satisfied due to cooling conditions, where ΔT is the difference in body temperature and environment, then the breakdown of the flow will occur on the very beginning of the cooling process (on the basis of expression (S25.18) and (S25.20)) (then the pressure p2 is the smallest), so we finally get: 3

2

1 3hS r≤3 (α M − α G ) ∆T σρ 2π c . (S25.23) By replacing the numerical values, it is obtained: r ≤ 14 µ m .

S26. On a clear, dark night, our eyes can see about 6000 or so stars in the sky. One of the most interesting phenomena of the clear night sky is why some stars twinkling and some not. In order to determine which kind of celestial bodies twinkle and which not, we will observe a star with the radius R located at a very large distance D from the observer. Two rays, which emerge from the point P located at the star surface, travel along two paths with the distance l1 and l2 until reach the observer eye. After the refraction at the eye lens, the rays impinge the retina at the point R. Due to the different ray paths, the turbulence of the atmosphere can cause random phase modulation of the wave fields which further lead to the random fluctuations at the

Solutions

107

observer retina known as the star twinkling. According to the geometry from figure shown below, we have: 2

l1 =

d Rd D + R − ≈ D 1 − 2 2 2 D and

l2 =

d Rd D + R + ≈ D 1 + 2 2 2 D ,

2

(S26.1)

2

2

(S26.2)

where d is the pupil diameter and where the following condition d R D were taken into consideration.

The optical path difference of these two rays is given by:

∆l = l2 − l1 ≈

Rd D . (S26.3)

Roughly speaking, the star will twinkle if the optical path difference is smaller than the coherence length ( ∆l < lC ) and will not if the optical path difference is larger than the coherence length ( ∆l > lC ). Therefore, if the star angular size, defined as:

ϕ≈

2R D , (S26.4)

is smaller than:

ϕ

v2 .

The power delivered to the wind turbine is given by:

P = Fv , (S35.2)

where v is the velocity of the air flow at the wind turbine. Similarly, the power delivered to the wind turbine according to the energy conservation law can be represented as:

= P

1 2 1 2 1 − mv 2 mv 2 2 . (S35.3)

From the last three equations, one obtains:

= v

1 ( v1 + v2 ) 2 . (S35.4)

Further, the mass flow rate can be represented as:

m = ρ Sv , (S35.5)

where ρ is the air density and S is the wind turbine cross-sectional area. Equations (S35.2) and (S35.5) further give:

= P ρ Sv 2 ( v1 − v2 )

, (S35.6)

which after the combining with equation (S35.4) gives: 2 = P ρ Sv 2 v1 − ( 2v − v= 1 ) 2 ρ Sv ( v1 − v )

. (S35.7)

The total available power of the air stream with the velocity v1, that is, the unobstructed wind velocity, over the area of the wind turbine of S, that is, when the air flow isn’t obstructed by the wind turbine, is given by:

Solutions

P0 =

131

1 ρ Sv13 2 . (S35.8)

Therefore, the coefficient of performance of a wind turbine is defined as:

2 P 4v ( v1 − v ) C= = P P0 v13 . (S35.9)

The coefficient of performance reaches its maximum value for v = ( 2 / 3) v1

, that is, when the air flow velocity at the wind turbine is equal to two-thirds of the wind velocity, so the maximum achievable coefficient of performance of a wind turbine is equal to:

CPmax =

16 = 0.59 27 . (S35.10)

S36. Depending on the way how the impacts between the heavy body and the rollers occur, we may distinguish two solutions. Since the rollers are covered with the rubber, there is a very high friction coefficient between the rollers and the body. Therefore, depending on the impact mechanism, we may say that the impact may be considered either elastic or perfectly inelastic.

In order to find the terminal velocity of the heavy body rolling down the slipway in the case of elastic impacts, we will refer to the figure shown below where it is presented the terminal motion of the body down the slipway. Moving from one to another roller, the body will cross over the distance that is equal to the distance between two adjacent rollers d. Therefore, the change in the body potential energy is given by:

∆EP = Mg ∆h , (S36.1)

where M is the body mass, g is the Earth’s gravity field acceleration, and Δh is the altitude change of the body, which is given by:

∆h = dsinα , (S36.2)

where α is the inclination angle of the slipway.

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Nonstandard Problems in General Physics with Solutions

Since there is no slipping between the body and the rollers (the rollers are covered by rubber) when the body touches the roller, it starts to rotate with the angular velocity that is given by:

ω=

v r , (S36.3)

where v is the body terminal velocity and r is the roller radius. Simply, according to the energy conservation law, the change in the body potential energy will be transferred to the kinetic energy of the roller rotation. Therefore, we can write:

1 ∆EP =J ω 2 2 , (S36.4) 2

where J is the roller moment of inertia given by J = mr / 2 with m being the roller mass. The combination of the last four equations gives:

Mgdsinα =

1 2 mv 4 . (S36.5)

From the last equation, one can finally obtain the terminal velocity of the body as:

v=2

M gdsinα m . (S36.6)

Now, if we consider perfectly inelastic impacts between the heavy body and the rollers, we may refer to the figure shown below where are presented the situation of the body movement slightly before and after the impact with the last roller.

Solutions

133

Between two impacts with the rollers, the heavy body slightly accelerates. Therefore, according to the energy conservation law, we may write:

1 1 2 Mv 2 += ∆EP M ( v + ∆v ) 2 2 , (S36.7)

where v is the body terminal velocity and Δv the change in the body velocity. Since the body mass is much larger than the mass of the roller beneath it, the corresponding change in the kinetic energy of the rollers is omitted. Taking into consideration that the distance between two adjacent rollers is very small and much smaller than the object dimensions, we have fulfilled ∆v v . In this case, equation (S36.7) transforms into:

∆v ≈

∆EP Mv . (S36.8)

According to the angular momentum conservation law in the steadystate situation, we have the following fulfilled:

M ( v + ∆v= ) r Mvr + J ω

, (S36.9)

where also it is valid ω = v / r . The last equation transforms into:

Jω ∆v = Mr . (S36.10)

Further, from equations (S36.8) and (S36.10), we have:

∆EP J ω ≈ v r . (S36.11)

By substituting equations (S36.1) and (S36.2) into the last equation, we have:

Mgdsinα 1 ≈ mv v 2 . (S36.12)

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Nonstandard Problems in General Physics with Solutions

So, finally from the last equation, we obtain the following for the body terminal velocity:

v≈ 2

M gdsinα m . (S36.13)

S37. If the distance between the mirrors is d and the object is small if compared with the mirror dimensions, as presented in figure given below, then the image that is formed by the upper mirror will be positioned at the distance:

q1 =

df d − f , (S37.1)

where f is the focal length of the mirrors (they are identical mirrors so both of them have the same focal length). The image of the upper mirror is at the same time the object of the subjacent mirror, where the position of the subjacent mirror object is given by:

p2= d − q1 . (S37.2)

The image of the subjacent mirror will be at the position: p f q2 = 2 p2 − f . (S37.3)

In order to see the object “hovering,” the observer must fulfill the following condition:

q2 = d . (S37.4)

The combination of the last four equations gives us:

d 2 − 4 fd + 3 f 2 = 0 , (S37.5)

where the solutions of this quadratic equation are given by:

Solutions

d=

( 2 ± 1) f

135

. (S37.6)

Simply by choosing the shorter distance d = f between the mirrors, that is, by positioning the mirrors at the distance that is equal to their focal length, one can obtain object “hovering” optical illusion with the smaller dimensions of the overall setup. Moreover, the “hovering” object will be of the same dimensions as the original one as the presented optical setup has no magnification (magnification is equal to unity).

S38. In order to calculate both shifts, a closer look of the optical system with all relevant geometric parameters has been presented in figure shown below. Based on the Snell’s law, we can write:

sinβ =

sinα n and (S38.1)

sin ( β + ∆β ) =

sin (α + ∆α ) n

, (S38.2)

where n is the plate refractive index, α, α + ∆α , β, and β + ∆β are the angles presented in figure shown below where for the paraxial rays is valid ∆α 1 and ∆β 1 .

After rearranging, equations (S38.1) and (S38.2) become:

∆β ≈

Geometry from the figure above gives the following relations:

AB =

cosα ∆α ncosβ . (S38.3) t

cos ( β + ∆β )

, (S38.4)

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Nonstandard Problems in General Physics with Solutions

AB' ABcos α − ( β + ∆β ) = , and (S38.5)

BB' ABsin α − ( β + ∆β ) = . (S38.6)

We also have:

BE = AA' + BB' , (S38.7)

BD' = BE − δ , (S38.8)

= ∆ AB' + D'F' − A'F , (S38.9)

D'F' ≈

BD' ∆α , and (S38.10)

A'F ≈

AA' ∆α . (S38.11)

Equations (S38.4), (S38.5), and (S38.6) give:

AB' = t

BB' = t

cos ( β + ∆β )

and (S38.12)

sin α − ( β + ∆β ) cos ( β + ∆β )

. (S38.13)

By combining equations (S38.7), (S38.8), and (S38.13), we have:

sin α − ( β + ∆β ) BD' = AA' + t −δ cos ( β + ∆β )

. (S38.14)

Further, the combination of the last eight equations gives: ∆≈t

cos α − ( β + ∆β )

cos α − ( β + ∆β ) cos ( β + ∆β )

+

t sin α − ( β + ∆β ) δ − ∆α ∆α cos ( β + ∆β )

.(S38.15)

Having in mind that the lateral displacement δ may be simply obtained from the standard case of a beam shift for a plane–parallel glass slab as:

Solutions

δ ≈ t 1 −

137

sinα n 2 − sin 2α , (S38.16) cosα

finally, by taking into account the first three equations for the longitudinal displacement, we have:

sin 2α ∆ ≈ t cosα + − 2 2 α n sin −

3 n 2 − sin 2α . (S38.17) n 2 cos 2α

(

)

The angle φ and the angle α are the angles with perpendicular sides, so it is valid α = ϕ , thus equation (S38.17) finally becomes:

sin 2ϕ ∆ ≈ t cosϕ + − 2 2 n sin ϕ −

3 n 2 − sin 2ϕ . (S38.18) n 2 cos 2ϕ

(

)

S39. The velocity of the loop may be found from the dynamic of a small arc representing a part of the loop. Since the lowest point of the loop is a part of the rope, which is fixed, the velocity of the lowest point of the loop has zero velocity. Therefore, one can consider that effectively the loop rolls without slipping over the rope. In order to find the velocity of the loop, an infinitesimally small arc of the rope will be observed, as presented in figure shown below. As effectively the arc rotates around its pivot, we have the force balance in the following way:

= F 2Tsin

∆θ ≈ T ∆θ 2 , (S39.1)

where F is the centrifugal force acting on a small arc, T is the rope force, and Δθ is an arc angle for which is valid ∆θ 1 .

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Nonstandard Problems in General Physics with Solutions

The centrifugal force is given by:

F=

∆mu 2 r , (S39.2)

where Δm is the arc mass, u is the loop velocity, and r the loop radius. The arc mass is given by:

∆m = λ r ∆θ , (S39.3)

where λ is the rope mass per unit length. From the last two equations, the centrifugal force is given by:

= F λu 2 ∆θ . (S39.4)

By combining this equation with equation (S39.1), one has:

λu 2 ∆θ ≈ T ∆θ , (S39.5)

which finally leads to the loop velocity of:

u≈

T

λ . (S39.6)

S40. According to the momentum conservation law, as presented in figure below, we have the following:

mv = mv′ + Mu , (S40.1)

where m is the small body mass, v is its initial velocity, v′ is the small body mass at the hilltop, M is the hill mass, and u is the hill velocity at the moment when the small body is at its top.

Also, according to the energy conservation law, one has the following:

mv 2 mv′2 Mu 2 = + + mgH 2 2 2 , (S40.2)

where g is the Earth’s gravity field acceleration and H is the hill height. By solving the last two equations in terms of the velocities v′ and u, we have:

Solutions

= v′

= u

v M 1+ m

M 2 gH 1− 2 1 m v

v 2 gH 1 ± 1 − 2 M v 1+ m

m 1 + M

m 1 + M

and

139

(S40.3)

. (S40.4)

Since v′ ≤ v and u ≤ ( m / M ) v must be fulfilled, we have the following solutions:

v=′

= u

v M 1+ m

M 2 gH 1− 2 1 + v m

m 1 + M

and

(S40.5)

v m 2 gH 1 − 1 − 2 1 + M v M 1+ m . (S40.6)

In order to cross over the hill, the following condition must be fulfilled: v′ ≥ u . (S40.7)

The combination of the last three equations then gives the following:

M 2 gH m 2 gH m 1 − 2 1 + ≥ − 1 − 2 1 + m v M v M . (S40.8)

This inequality is fulfilled for any value of the initial velocity of the small body for which the square root term gives the real value. Therefore, the following condition must be fulfilled:

1−

2 gH v2

m 1 + M

≥0 . (S40.9)

In order to cross over the hill, the velocity of the small body must be as follows:

m v ≥ 2 gH 1 + M

. (S40.10)

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Nonstandard Problems in General Physics with Solutions

The minimum velocity of the small body is thus given by:

m 2 gH 1 + M

= vmin

. (S40.11)

The minimum velocity of a small body may be found in a simpler way than the one presented above. As a matter of fact, we must find such initial velocity of a small body for which both the small body and the hill have the same velocities when the small body is positioned at the top of the hill as presented in figure below.

According to the momentum conservation law, we have:

mvmin =

whereas according to the energy conservation law, we have:

1 2 1 mvmin = ( m + M ) u 2 + mgH 2 2 . (S40.13)

( m + M ) u , (S40.12)

By solving the last two equations, we obtain the same solution as one presented in equation (S40.11).

S41. Having in mind that at the beginning, there is no slipping between the small ball and the hemisphere, the ball will reach a certain point at the hemisphere without slipping that is located at the angle θ with respect to the vertical, as presented in figure shown below. According to the energy conservation law, one has:

= mgh

1 2 1 mv + J ω 2 2 2 , (S41.1)

where m is the ball mass, g is the Earth’s gravity field acceleration, h is the ball altitude difference, v is the ball velocity, ω is the ball angular velocity, and J is the ball moment of inertia given by

J = 2mr 2 / 5 , with r being ball radius. Since there is no slipping, the

following is valid:

ω=

v r . (S41.2)

Solutions

141

Further, the altitude difference is given by:

h =( R + r )(1 − cosθ )

, (S41.3)

where R is the hemisphere radius.

The combination of the last three equations yields: = v

10 g ( R + r )(1 − cosθ ) 7 . (S41.4)

The dynamic of the ball on the hemisphere surface at this particular point is presented in figure shown below with all forces relevant to its motion in this particular time instance.

According to Newton’s second law of motion, we have:

= maθ mgsinθ − T

, (S41.5)

where aθ is the ball tangential acceleration and T is the friction force between the ball and the hemisphere. Similarly, we have:

J α = Tr , (S41.6)

where α is the ball angular acceleration given by:

α=

aθ r . (S41.7)

The last three equations give:

T=

2 mgsinθ 7 . (S41.8)

The normal reaction force N between the ball and the hemisphere, as presented in figure shown above, is determined by:

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Nonstandard Problems in General Physics with Solutions

N+F = mgcosθ , (S41.9)

where F is the centrifugal force acting on the ball that is given by:

F=

mv 2 R + r . (S41.10)

The combination of equations (S41.4), (S41.9), and (S41.10) gives:

1 mg (17cosθ − 10 ) 7 , (S41.11)

The ball is not slipping until the following condition is satisfied:

T ≤ µ N , (S41.12)

= N

with μ being the friction coefficient between the ball and the hemisphere. According to the equations (S41.8) and (S41.11) and the last inequality, the following is obtained:

2sinθ ≤ µ (17cosθ − 10 )

. (S41.13)

The ball will start to slip for the angle θS that satisfies the following equation:

= 2sinθS µ (17cosθS − 10 )

. (S41.14)

This equation may be transformed into the following quadratic equation:

( 289µ

The solutions of this quadratic equation are as follows:

2

)

+ 2 cos 2θS − 340 µ 2 cosθS + 100 µ 2 − 2 = 0 . (S41.15)

(

)(

4 289 µ 2 + 2 100 µ 2 − 2 340 µ 2 = cosθS 1± 1− 2 2 289 µ 2 + 2 340 µ 2

(

(

)

(S41.16)

After the rearrangement, the last equation becomes: = cosθS

)

)

378µ 2 + 4 170 µ 2 ± 1 289 µ 2 + 2 170 µ 2

. (S41.17)

It is easy to show mathematically that the following condition is fulfilled:

Solutions

143

378µ 2 + 4 170 µ 2 1 + ≤1 2 289 µ + 2 170 µ 2

As the ball starts to slip for a smaller angle θS of two values given

. (S41.18)

cosθ

S. in the equation (S41.17), we will choose the higher value of Therefore, the angle at which the ball starts to slip is finally given by:

170 µ 2 378µ 2 + 4 = θS arccos 1 + 2 170 µ 2 289 µ + 2 .

(S41.19)

S42. The free electrons in the metallic disk are exposed to the centrifugal force due to the disk rotation, as presented in figure shown below. The mechanical, that is, the centrifugal, force acting on the electron at the distance r from the disk pivot is given by:

FM = mrÙ 2

, (S42.1)

where m is the electron mass and Ω is the angular velocity of the disk. Due to the mechanical force acting on the electrons, the electrons will be unevenly distributed along the disk volume thus creating the compensating electrical force.

The electrical force acting on the same electron is given by:

FE = eE ( r )

, (S42.2)

E (r ) where e is the elementary charge and is the electric field at this particular point where the electron is observed. Beside the electrical force, due to the presence of the moving charges, the magnetic force will be also present acting on each electron. However, since RΩ c is satisfied, with R being the disk radius and c being the

144

Nonstandard Problems in General Physics with Solutions

speed of light in vacuum, the influence of the magnetic field may be omitted. The electrons will be moved in the disk until the force balance has been achieved. Therefore, one has fulfilled in the steady state:

FE = FM . (S42.3)

According to the last three equations, the electric field is:

E (r ) =

m 2 Ù r e . (S42.4)

The voltage that measures the voltmeter is thus given by: R

U = ∫E ( r ) dr 0

, (S42.5)

which finally gives:

U=

m 2 2 RÙ 2e . (S42.6)

S43. One may wrongly conclude that the light pressure is responsible for the vanes rotation. As a matter of fact, the impinging photons do not carry sufficient momentum in order to rotate the rotor with the vanes. The reason why vane rotates lies in the thermodynamics of the Crookes radiometer. So, the Crookes radiometer may be treated as a heat engine. Therefore, the analysis of the Crooks radiometer that will be presented below relies on the thermodynamic principles.

Since the black side of the vane behaves as a black body, it absorbs all the radiation that impinges its surface. The flux captured by the black side of the vane is thus given by:

Φ =ES , (S43.1)

where E is the irradiance and S is the area of the vane surface. This amount of radiation flux is constant during the vane rotation as the illumination is diffuse. Due to the absorbed radiation, the black side of the vane is warmer than the polished one, as presented in figure shown below. Therefore, according to the energy conservation law, the corresponding temperature difference between the warmer (black) and the colder (polished) side of the vane is given by:

Solutions

Φ =λ S

145

TH − TC d , (S43.2)

where λ is the thermal conductivity of mica, TH is the temperature of the vane black side, TC is the temperature of the white or polished side of the vane, and d is the vane thickness. Since the gas in the radiometer bulb is under the partial vacuum, one may consider that there are no collisions between the molecules of air inside the bulb. Therefore, only collisions between the air molecules and the vanes as well as between the air molecules and the glass bulb occur. When reaching the bulb surface, the molecules collide with it and exchange the energy with the bulb walls. So, after the collision with it, the molecules gain the energy that is proportional to the bulb temperature, that is, to the ambient temperature of the surrounding environment. On average, the velocity of the molecule after colliding with the bulb walls and slightly before the collision with the vanes is thus given by:

1 2 1 mvx = kBT 2 2 , (S43.2)

where m is the mass of the molecule, vx is the molecule velocity component that is perpendicular to the vane surface, kB is the Boltzmann’s constant, and T is the ambient temperature.

When reaches either hot or cold side of the vane, the molecule exchanges its energy with the vane walls, thus gaining the energy that is proportional to the vane surface temperature. Therefore, the cor-

146

Nonstandard Problems in General Physics with Solutions

responding molecule velocity components that are perpendicular to the vane surfaces are given by:

1 2 1 mvxH = kBTH 2 2 , (S43.3) for the hot side where vxH is the corresponding molecule velocity component and

1 2 1 mvxC = kBTC 2 2 , (S43.4)

for the cold side where vxC is the corresponding molecule velocity

T >T

v

>v ,

C and consequently xH xC there will component. Since H be net force acting on the vanes, thus pushing them in the direction from the hot (black) side to the cold (white or polished) side of the vanes. Therefore, the rotor will accelerate until the moment when the force balance is reached. When the steady state has been reached, the velocity component of the molecule perpendicular to the vane surface after the collision of the molecule with the hot side of the vane will be given by:

' v= vxH − 2u xH

, (S43.5)

where u is the vane velocity, and the velocity component of the molecule perpendicular to the vane surface after the collision of the molecule with the cold side of the vane will be given by:

' v= vxC + 2u xC

. (S43.6)

Since the vane dimensions are much smaller than the distance between the spindle and the vanes, we have the following for the vane velocity:

u = Rω , (S43.7)

with R being the distance between the spindle axis and the vanes and ω being the rotor angular velocity that must be found. When the steady state has been reached, there will be no net force acting on the vanes; thus, the momentum change on both side of the vane of each molecule must be the same, giving the following condition:

(

)

(

m vx' H + vx = m vx' C + vx

) , (S43.8)

Solutions

or equivalently:

vx' H = vx' C

. (S43.9)

The combination of equations (S43.5), (S43.6), and (S43.9) yields to:

vxH − vxC = 4u . (S43.10) Further, by combining the last equation with equations (S43.3) and (S43.4), one obtains the following:

m TH − TC = 4u kB

. (S43.11)

Now, going back to the equation (S43.2), one may rearrange it in the following way: T −T λS TH + TC TH − TC Ö =λ S H C = d d . (S43.12)

(

)(

)

Since the captured radiant flux carries a relatively small amount of optical power, the temperature difference of the hot and cold sides of the vanes from the ambient temperature is very small in comparison with the ambient temperature one may take in the first approximation the following relation (S43.12) becomes:

T −T 2λ S S H C T = Ö λ= d d

(

TH − TC

TH ≈ TC ≈ T

. Therefore, equation

) . (S43.13)

The combination of equations (S43.1), (S43.11), and (S43.13) gives: E=

8λ d

mT u kB

. (S43.14)

One can express the molecular mass as:

m=

147

µ N A , (S43.15)

with μ being the molar mass of the bulb gas and NA being the Avogadro constant. The combination of the last two equations gives:

148

Nonstandard Problems in General Physics with Solutions

E=

8λ d

µT kB N A

u . (S43.16)

Taking into consideration that it is valid:

kB N A = R , (S43.17)

where ℜ is the universal gas constant, equation (S43.16) becomes:

E=

µT R

u

. (S43.18)

Taking into consideration equation (S43.7), we finally obtain the following for the angular velocity of the Crooks radiometer rotor:

ω=

8λ d

d 8λ R

R E µT . (S43.19)

According to the last equation, one can notice that the Crooks radiometer rotor angular velocity is directly proportional to the light irradiation.

In the case of a wide light beam, the absorbing sides of the vanes are illuminated only during a half of the period of the vane rotation. Moreover, the black side captured power during the vane rotation is proportional to cosθ with θ being the angle between the light beam and the normal to the vane. Therefore, only a fraction of the flux will be captured. This fraction is equal to: π /2

1 1 = C = cosθ dθ ∫ 2π −π /2 π

. (S43.20)

The corresponding angular velocity in the case of the illumination with the wide light beam is thus given by:

= ω ' C= ω

d 8πλ R

R E µT . (S43.21)

Taking into account the numerical values from the problem state-

′ ment one obtain ω ≈ 3.8 rad / s ≈ 36 rpm as the vane angular velocity.

Solutions

149

S44. The surface energy density of the water waves is given by:

E=

1 ρ0 gA2 2 , (S44.1)

where ρ0 is the water density, g is the Earth’s gravity field acceleration, and A is the wave amplitude. Similarly, for the momentum density of the water waves, we have:

P=

E cP , (S44.2)

where cP is the phase velocity of the water waves, which is further given by:

cP =

gλ 2π , (S44.3)

with λ being the wavelength of the water waves. In order to find the mean wavelength of the water waves, we will relay of figure shown below where it is presented the simple mechanism of water waves formation with all relevant parameters for the analysis. We will observe two slabs of moving air, that is, one over the sea part where the wind starts to blow and where there are no formed waves and another over the sea part where the water waves are fully formed. Therefore, we may say that according to the conservation laws, we can consider that the mass, the momentum, and the energy in both slabs are the same. According to the mass conservation law, the masses in both slabs are the same so we may write the following:

= m wlh = ρ wLH ρ , (S44.4)

where w is the slabs width, l and L are the slabs lengths, h and H are the slabs heights, and ρ is the air density. According to the last equation, we have:

lh = LH . (S44.5)

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Nonstandard Problems in General Physics with Solutions

Further, according to the simultaneous movement of the air from the first to the second slab, the air stream velocities in both slabs must fulfill the following condition:

l L = v u , (S44.6)

Ät=

where Δt is the time that wind needs in order to pass from one side to the other of both slabs, v is the wind velocity at the beginning, and u is the wind velocity over the waves.

The energy conservation law gives the following relation:

1 1 1 1 pwhl + mv 2 + mgh = pwHL + mu 2 + mgH + ELw 2 2 2 2 ,(S44.7)

where p is the air pressure and E is the surface energy density of the waves given by equation (S44.1). After taking into account equations (S44.1), (S44.4), (S44.5), and (S44.6), the last equation can be rearranged into:

(

)

(

)

lh ρ v 2 + gh= lh ρ u 2 + gH + ρ0 gA2 L

.

(S44.8)

The momentum conservation law further gives the following relation:

mv = mu + PLw . (S44.9)

After taking into account equations (S44.2), (S44.4), (S44.5), and (S44.6), the last equation can be rearranged into:

ρ0 gA2 L lhρ ( v − u ) = 2cP . (S44.10)

The combination of equations (S44.8) and (S44.10) further gives:

Solutions

ρ0 gA2 L

v + gh ) (= (v − u ) 2

ρ0 gA2 L

2

)

+ gH + ρ0 gA2 L

2cP

which can be further simplified into:

v 2 − u 2 + g ( h − H= ) 2cP ( v − u )

The combination of equations (S44.5) and (S44.6) provides:

H=

2cP

(u (v − u )

151

,(S44.11)

. (S44.12)

v h u , (S44.13)

which after substituting into equation (S44.12) gives:

v2 − u 2 −

gh ( v − u=) 2cP ( v − u ) u (S44.14)

that after simplifying, it gives:

v+u −

gh = 2cP u . (S44.15)

Since the wind velocity is just slightly changed when the waves are formed, we have approximately fulfilled the following:

u ≈ v , (S44.16)

thus equation (S44.15) becomes:

2v −

gh ≈ 2cP v . (S44.17)

If we take into account equation (S44.3), where the relation for the phase velocity is given, the last equation becomes:

2v −

gh gλ ≈2 v 2π , (S44.18)

which can be solved in order to finally give the relation between the wind parameters and corresponding water wave wavelength:

2π λ≈ g

2

gh v − 2v . (S44.19)

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Nonstandard Problems in General Physics with Solutions

Since only a thin layer of air flow affects the wave formation, the fol-

2 lowing is fulfilled gh v , that is, as it was stated in the problem, we finally have:

λ≈

2π v 2 g . (S44.20)

One can notice that this wavelength corresponds to the water wave with the phase velocity that is equal to the wind speed. In order to compare the theoretically obtained mean wavelengths of the water waves with the measured ones, the results are presented in the table shown below. Wind speed [km/h]

Measured wavelength [m]

Calculated wavelength [m]

19

8.5

17.8 (2.094 × Measured wavelength)

37

33.8

67.7 (2.003 × Measured wavelength)

56

76.5

155.0 (2.026 × Measured wavelength)

74

136

270.6 (1.990 × Measured wavelength)

92

212.2

418.3 (1.971 × Measured wavelength)

It is clear from the presented results that the theoretically obtained result is two times bigger than the measured one. This means that the thickness of an air flow that affects the wave formation is not so thin. Therefore, according to the equations (S44.19) and (S44.20), one can estimate the thickness of the wind layer that influences the formation of the water waves at open sea in the following way: 2

gh 2π v 2 2π ≈2 v − g g 2v , (S44.21) where it was taken into account that the real wavelength is two times smaller than the one presented in equation (S44.20). From the last equation, one can estimate the thickness of the air flow that takes a part in the wave formation from the following equation that has been obtained by simplifying equation (S44.21): 2

gh 1 1 − 2 ≈ 2 . (S44.22) 2v According to the equation (S44.22), the air flow thickness is given by:

Solutions

153

v2 h ≈ 2± 2 g . (S44.23)

(

)

Having in mind that according to equation (S44.18), one has the following condition fulfilled:

h≤

2v 2 g , (S44.24)

we finally obtain from equation (S44.23) and inequality (S44.24) the following result for the wind height that affects in the wave formation:

(

h ≈ 2− 2

)

v2 g . (S44.25)

S45. In order to determine the mathematical relationship between the photon energy and the frequency of the electromagnetic wave, we will consider the transmitter T that emits planar, unvarying, and monochromatic electromagnetic waves in the direction of the receiver R, as presented in figure shown below. The radiated power and angular frequency of the transmitter electromagnetic wave in the transmitter frame of reference are P and ω, respectively, and do not change with the passage of time. Since the transmitter power is constant, the transmitter photon emission rate is also constant and equals q in the transmitter frame of reference. The abovementioned three parameters P, q, and ω of the transmitter are related by the following relation:

P = qε (ω )

. (S45.1)

The transmitter T and receiver R antennas are arranged in the way that each photon that is being emitted by the transmitting antenna of the transmitter is being received by the receiving antenna of the receiver, that is, the transmission system has the maximal possible efficiency. If the transmitter starts to move from the receiver uni-

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Nonstandard Problems in General Physics with Solutions

formly with constant velocity u, as it is depicted in presented above, due to the Doppler effect, the receiving power and angular frequency of the electromagnetic wave in the receiver frame of reference are P′ and ω′, respectively. Due to the time dilation, caused by the motion of the transmitter in the receiver frame of reference, the number of photons per unit time that reaches the receiver is q′, where q ' ≠ q . In this case, the average time between the emission of two photons is

τ = 1/ q , while the average time between the capture of two photons is τ ' = 1/ q ' , where also τ ' ≠ τ is valid. The abovementioned three

parameters P′, q′, and ω′ of the receiver are related by the following relation:

P ' = q ' ε (ω ' )

Based on the Lorentz transformations, the following is valid for the angular frequency ω′ of the electromagnetic wave and the number of photons per unit time q′ that have been captured by the receiver:

ω'=ω

τ ′=

. (S45.2)

c −u c + u and (S45.3)

τ

u2 → q′= q 1 − 2 c u2 1− 2 c , (S45.4)

where c is the speed of light in vacuum. According to the energy conservation law in the receiver frame of reference, we have: P=′ P − Fu , (S45.5) where F is the braking force that must be applied to the transmitter in the opposite direction of its motion in order to provide the uniform motion of the transmitter with constant velocity, or the transmitter must not accelerate due to the repulsive force of the electromagnetic radiation that it emits. Further equation (S45.5) becomes:

q ' ε= (ω ') qε (ω ) − Fu

. (S45.6)

By combining equations (S45.3), (S45.4), and (S45.6), we have:

Solutions

155

u2 c −u ε ω qε (ω ) − Fu = 2 c c + u . (S45.7)

If the transmitter velocity is very close to the speed of light in vacu-

q 1−

um, that is, u → c , according to equation (S45.4), it is valid q ' → 0 , that is, the average time between the receiver and recipient of two ε (ω ' ) → ε ( 0 ) consecutive photons tends to infinity τ ′ → +∞ and ε (0) ,so regardless of the value of , equation. (S45.7) becomes:

F =q

ε (ω ) c

, (S45.8)

ε ( 0 ) < +∞ where it is assumed , while otherwise an indefinitely large energy would be needed to establish a time-invariant electromagnetic field even in a limited volume. Based on the third Newton’s law of motion, the force F is also the repulsive force of the electromagnetic radiation that acts on the transmitter. According to equation (S45.8), the photon momentum is given by: p (ω ) =

ε (ω ) c

, (S45.9)

which is the well-known relationship for the photon momentum.

The goal of the above-presented analysis was to find the photon momentum as a function of its energy. In the next step of the analysis, we will consider the case when the transmitter is moving steadily toward the receiver with constant velocity u, as shown in figure below. In order to overcome the repulsive force of the emitted electromagnetic radiation and to ensure uniform motion of the transmitter, the force F must be exerted on the transmitter in the direction of its motion. According to the energy conservation law in the receiver frame of reference, the following is valid: P=′ P + Fu . (S45.10)

156

Nonstandard Problems in General Physics with Solutions

Further, taking into consideration equation (S45.8), the following is valid for equation (S45.10):

q ' ε= (ω ') qε (ω ) + qε (ω )

By combining equations (S45.3), (S45.4), and (S45.11), we obtain the following functional equation:

ε ω

c+u c −u

u2 u 1 − 2 = ε (ω ) + ε (ω ) c c . (S45.12)

The last equation can be rearranged in the following way:

u c . (S45.11)

ε ω

c+u c −u

c+u ε (ω ) = c − u . (S45.13)

The functional equation, given by equation (S45.13), is of the following type:

ε ( aω ) = aε (ω )

, (S45.14)

a =( c + u ) / ( c − u ) where and obviously a ≥ 0 . In order to find the solution of this functional equation, we will assume that the solution could be found in the following form:

ε (ω )= ω ⋅ f (ω )

, (S45.15)

f (ω ) where is an arbitrary function. By substituting equation (S45.15) into equation (S45.14), we have:

aω ⋅ f ( aω ) =⋅ aω f (ω )

(S45.16)

that for ω > 0 further yields to:

f ( aω ) = f (ω )

. (S45.17)

Solutions

157

( )

As parameter a can take any positive value, the function must have the same, constant value for any argument, or f (ω ) = const and thus:

ε ( ω ) = ω

(S45.18)

or equivalently the photon energy–frequency relationship is given by:

ε = hν (S45.19)

where ν = 2π / ω is the frequency of the electromagnetic radiation and where constant h must be experimentally determined. Relationship given in equation (S45.19) represents the well-known relationship between the energy and the frequency of the photon, where h is −34

= h 6.626 ×10 Js that was also the well-known Planck constant first measured by Millikan a century ago. Equation (S45.18) conε ( 0 ) < +∞ firms our earlier assumption that , since according to ε 0 = 0 equation (S45.18), we have ( ) .

In order to show the uniqueness of ε (ω ) = ω as a solution to equaε ' (ω ) tion (S45.13), we will assume that there is another solution , ε= ' (ω ) g (ω ) ⋅ ε (ω ) which, in general, case can be represented as g (ω ) where is an arbitrary function of ω. This solution must also ε ' ( aω ) = aε ' (ω ) satisfy equation (S45.13); so, we have and cong ( aω ) ⋅ ε ( aω ) = ag (ω ) ⋅ ε (ω ) ε (ω ) sequently . As is the solution ε ( aω ) = aε (ω ) to equation (S45.13), we have , which further leads g ( aω ) = g (ω ) g (ω ) to . Therefore, function takes the same value g (ω ) = b for any value of ω, so, it has a constant value , or conseε ' (ω ) = bε (ω ) quently is satisfied, where b is an arbitrary but posiε ' (ω ) ε (ω ) tive constant. Therefore, both solutions and are of the same form, given by equation (S45.18), that is, they depend linearly on the frequency. Hence, the solution given by equation (S45.18)

158

Nonstandard Problems in General Physics with Solutions

is the unique solution to the functional equation given by equation (S45.13). S46. In order to find the equivalent resistance between two points in the semi-infinite resistors network, as presented in the problem statement, we will observe a resistors network as presented in figure given below that consists of two such semi-infinite networks.

The equivalent network of this resistors network, that is, the one given in figure above, is presented in figure shown below.

Therefore, we have the following relation:

ET RE = IT 2 , (S46.1)

where ET is the electromotive force of a test generator, IT is its current, and RE is the equivalent resistance of the network. Due to the symmetry, the voltage drops at each 2R resistor are the same, so the resistors network from the beginning of this analysis can be presented in the following form given in figure below, that is, one can short-circuit all nodes with the same potential.

Solutions

159

Since each two parallel connected resistors with the resistance 2R make one resistor with the resistance R, the above-presented infinite network may be rearranged in the network shown below, where the corresponding resistances between points A and B (or equivalently A′ and B′) is according to equation (S46.2) given by:

RAB =

RE 2 . (S46.2)

Therefore, in order to find the equivalent resistance, one must first find the resistance between two adjacent points of the infinite resistors network shown in figure above. Due to the symmetry and infiniteness of the resistors network, we may introduce an ideal current source into the point A as presented in figure shown below.

160

Nonstandard Problems in General Physics with Solutions

The potential difference between points A and B is given by: ' U AB =

RI T 4 , (S46.3)

where IT is the electric current of the test current source and where, due to the symmetry, was taken into consideration the current of the current source is divided into four identical part, as it is presented in figure shown above. Similarly, we may introduce another ideal current source into the point B as presented in figure shown below.

The potential difference between points A and B is given by: ″ U AB =

RI T 4 . (S46.4)

So, by superposition theorem, one can combine both current sources and obtain the circuit as presented in figure shown below.

According to the superposition theorem, the potential difference is thus given by:

' ″ U AB = U AB + U AB =

RI T 2 . (S46.5)

The resistance between points A and B is thus given by:

Solutions

R = AB

161

U AB R = IT 2 . (S46.6)

Finally, according to the equation (S46.2), the wanted equivalent resistance is given by:

= RE 2= RAB R . (S46.7)

S47. If we observe the transmitter T, which emits monochromatic electromagnetic radiation of frequency ν than if it moves with the velocity u and emits the radiation parallel to its motion, then the frequency ν′ of emitted radiation will differ due to the Doppler effect, as presented in figure shown below. According to the energy conservation law, we have:

ν + Fu = ν ' , (S47.1) nh nh

where n is the photon emission rate, h is the Planck constant, F is the repulsion force due to the photon emission, and u is the transmitter velocity.

The repulsion force can be calculated according to the generated momentum of radiated photons as:

F = n

hν ' c , (S47.2)

where c is the speed of light in vacuum. The combination of last two equations gives:

ν′ =

ν 1−

u c . (S47.3)

Since we have satisfied: ν ′= ν + ∆ν , (S47.4) where Δν is the frequency shift of the radiation and u c the combination the last two equations yields:

162

Nonstandard Problems in General Physics with Solutions

ν += ∆ν

u 1− c

u ≈ ν 1 + c , (S47.5)

which further gives:

∆ν

ν

ν

≈

u c . (S47.6)

Since λ = c / ν , with λ being the radiation wavelength, we have:

∆ν

ν

∆λ

λ , (S47.7)

where Δλ is the wavelength shift of the radiation. Finally, according to the equations (S47.6) and (S47.7), we have:

∆λ

= −

λ

≈−

u c , (S47.8)

for the relative Doppler shift.

S48. The bubble of carbon dioxide stays attached to the bottle bottom as long as the buoyancy is smaller than the adhesive force between the bubble and the bottle. The situation between the bubble and the bottle is presented in figure shown below with all relevant parameters for the analysis. At the moment when the bubble detaches from the bottle bottom, we have the following fulfilled:

F = Tsinα , (S48.1)

where F is the buoyancy, T is the adhesive force between the bubble and the bottle, and α is the wetting angle.

Solutions

163

The buoyancy is given by:

F = ρ gV , (S48.2)

where ρ is the water density, g is the Earth’s gravitational field acceleration, and V is the volume of the bubble, which is further given by:

= V

π

2 1 R 3 4 − (1 − cosα ) 3sin 2α + (1 − cosα ) 3 2 , (S48.3)

with R being the bubble radius.

Further, the adhesive force is given by:

T = 2πσ Rsinα , (S48.4)

where σ is the surface tension of water. The combination of all last four equations gives: π

3

ρ gR 3 4 −

According to the last equation, the radius of the detached bubbles is given by:

R = 2sinα

1 2 2πσ Rsin 2α (1 − cosα ) 3sin 2α + (1 − cosα ) = 2 .(S48.5)

3σ 1 ρ g 8 − (1 − cosα ) 3sin 2α + (1 − cosα )2 .(S48.6)

S49. If a sound source, that is, the speaker, is moving away from the observer, that is, the microphone, with high speed, as presented in figure shown below, the microphone will capture the sound signal that is given by:

l (t ) r (= t ) s t − c , (S49.1)

where s(t) is the sound signal broadcasted by the source, l(t) is the actual distance between the source and the observer, and c is the sound speed.

164

Nonstandard Problems in General Physics with Solutions

The actual distance between the source and the observer is further given by:

l ( t ) =l0 + u ( t − t0 )

, (S49.2)

where l0 is the distance at the time instance t0 and u is the source velocity. The combination of the last two equations gives:

l + u ( t − t0 ) r (= t ) s t − 0 c , (S49.3)

which may be simplified into:

ut − l u r ( t ) s 0 0 + 1 − t = c . (S49.4) c In order to hear the sound played backward, the received sound must be the function of time as given below:

r (= t ) s (τ − t )

, (S49.5)

where τ is an arbitrary time constant. So, according to the equations (S49.4) and (S49.5), one can see that the speed of sound source must be twice the speed of sound, that is:

u = 2c . (S49.6)

S50. If we observe a thin layer of air representing the Earth’s atmosphere that surrounds it, as presented in figure shown below, for the atmospheric pressure at the sea level, we have the following:

p=

mg S , (S50.1)

where m is the overall mass of the atmosphere, g is the Earth’s gravitational field acceleration, and S is the Earth’s surface overall area, which is further given by:

S = 4π R 2 , (S50.2)

Solutions

165

with R being the Earth’s radius.

According to the last two equations, the atmosphere mass is given by:

m=

4π R 2 p g . (S50.3)

Based on the known mass of the atmosphere and air average molar mass, we have the following for the number of molecules of air within the Earth’s atmosphere:

N=

m

µ

NA

, (S50.4)

where μ is the average molar mass of air and NA Avogadro constant. Based on equations (S50.3) and (S50.4), we have the following for the number of air molecules:

N=

4π N A R 2 p gµ . (S50.5)

Taking into account = N A 6.022 ×1023 1 / mol

the ,

following

numerical

R = 6400 km ,

values:

p = 105 Pa ,

g = 9.81 m / s 2 , and µ = 29 g / mol , we have obtained N ≈ 1.1×1044

molecules of air in the Earth’s atmosphere. S51. In order to find the Earth’s average crust thickness, we will refer to a box-shaped part of the Earth’s crust of average density ρ, box base width w, box base length l, and box height H, which at the same time represents the Earth’s thickness, where the crust thickness is defined as the altitude difference between the surface layer of the crust and the magma, as presented in figure shown below.

166

Nonstandard Problems in General Physics with Solutions

According to the virtual work principle, if we allow a small vertical movement of the crust box of ΔH, this small part of the crust has to be melted and transformed into magma. Therefore, according to the energy conservation law, we have the following:

∆mgH = ∆mL , (S51.1)

where g is the Earth’s gravitational field acceleration, L is the latent heat of melting of basalt, and Δm is the elementary mass of the melted basalt within the observed box of the Earth’s crust for which we have:

∆m = ρ wlH . (S51.2)

Since the Earth’s crust average thickness is not dependent on this elementary melted mass of the Earth’s crust, as equation (S51.1) shows, according to this equation, we finally have the following for the average crust thickness:

H=

L g . (S51.3)

Taking into account the problem specified numerical value of the

5 latent heat of melting of basalt of L= 4 × 10 J / kg and the Earth’s 2

gravitational field acceleration of g = 9.81 m / s , we have obtained H ≈ 41 km . This value very good corresponds to the real Earth’s crust thickness that, according to the source https://en.wikipedia.org/ wiki/Crust_(geology), lies in the range between 30 km and 50 km. S52. An excited atom that moves at the rate u emits a photon in the direction of its motion, as presented in figure shown below. According to the energy conservation law, we have:

Solutions

167

1 2 1 mu + h= mu '2 + hν ' ν 2 2 , (S52.1)

where m is the atom mass, hν is the energy of the excited state of the atom that corresponds to the energy of the emitted photon in the case when the atom is motionless with h being the Planck constant and ν being the frequency of the emitted photon in the case of the motionless atom, u’ is the atom velocity after emitting the photon, and ν› is the frequency of the emitted photon.

Similarly, according to the momentum conservation law, one has the following:

mu = mu '+

hν ′ c , (S52.2)

with c being the speed of light in vacuum. From the last equation, we have the following for the atom velocity after photon emission:

u′ = u −

hν ′ mc . (S52.3)

By substituting equation (S52.3) into the equation (S52.1), the following is obtained: 2

1 1 hν ′ hν ′ =+ hν mu 2 − m u − 2 2 mc .

The last equation can be rearranged into:

ν′ = ν+

(S52.4)

2 m 2 hν ′ u − m u − 2h mc , (S52.5)

which further gives:

∆ν = ν ′ −ν =

u hν ′ ν ′ 1 − c 2mcu , (S52.6)

with Δν being the frequency difference. Since it is fulfilled:

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Nonstandard Problems in General Physics with Solutions

mu 2 hν , (S52.7)

that is, the kinetic energy of the atom is much larger than the energy of the emitted photon, the last equation becomes:

u ∆ν = ν ′ −ν =ν ′ , c

(S52.8) which in the non-relativistic case u c becomes:

u ∆ν =ν c (S52.9)

that corresponds to the well-known Doppler shift.

S53. If we assume that r is the average radius of drops after bursting the soap bubble, then, according to the energy conservation law, we have:

8π R 2σ = 4π Nr 2σ ,

(S53.1)

where R is the bubble radius, σ is the soap surface tension, and N is the number of drops. This equation simplifies into:

2 R 2 = Nr 2 . (S53.2)

Since the soap volume before and after bursting of the bubble is the same, we consequently have the following:

4π R 2 d =

4π Nr 3 3 .

(S53.3)

The last equation simplifies into:

R2d =

1 3 Nr 3 . (S53.4)

By combining equations (S53.2) and (S53.4), we finally obtain the following results:

r=

3 d 2 , (S53.5)

for the average drop radius and

N=

8R 2 9d 2 , (S53.6)

for the total number of soap drops.

Solutions

169

S54. A small part of the soap bubble in the form of a cuboid is presented in figure shown below. The cuboid is chosen so that it is satisfied:

d w R , (S54.1)

with d being the height of the cuboid, that is, the thickness of the soap bubble, w being its width, and R being the bubble radius.

The force that stretches the part of the soap bubble is given by:

F = 2 wσ , (S54.2)

where σ is the soap surface tension. Therefore, the stress inside the soap bubble is given by:

τ=

F S , (S54.3)

where the surface S is given by:

S = wd . (S54.4)

According to the last three equations, we have:

τ=

2σ d . (S54.5)

The bubble will burst when the stress inside the bubble reaches the tensile stress of the soap, that is, when it is satisfied:

τ = p , (S54.6)

where p is the soap tensile stress. Finally, according to the equations (S54.5) and (S54.6), we have the following for the soap bubble wall thickness in the moment of its bursting:

d=

2σ p . (S54.7)

Nonstandard Problems in General Physics with Solutions

170

S55. The image formation of a ruler onto the image sensor of the camera is presented in figure shown below. According to the lens equation, we have:

1 1 1 + = p q f , (S55.1)

where p is the object distance, that is, the distance between the ruler and the camera lens, q is the image distance, and f is the focal length of a camera lens.

According to equation (S55.1), the image distance is given by:

q=

pf p − f . (S55.2)

The object and image dimensions P and Q, respectively, are related as the following relation shows:

P Q = p q . (S55.3) Based on equations (S55.2) and (S55.3), we have the following for the image dimension:

Q=

f P p − f . (S55.4)

From the photo in the problem statement, one can notice the bright circles that originate from distant bright point-like objects. So, in the case of these distant objects, we have the situation as presented in figure shown below. The parallel beams of this point-like distant objects form a bright circles on the image sensors, which diameter d is defined according to the following relation:

Solutions

D d = − ±

171

, (S55.5)

where D is the diameter of the camera lens.

By combining equations (S55.2) and (S55.5), we have

D=

p− f d f . (S55.6)

Equations (S55.4) and (S55.6) further give:

D=

P d Q . (S55.7)

According to the last equation and the picture shown above, one can say that the circle that has the same dimension as the bright circle, which radius is measured by the photographed ruler, represents at the same time the diameter of the lens, so based on the measurement performed in the photo presented above, we have D ≈ 1.6 cm .

Nonstandard Problems in General Physics with Solutions

172

S56. Since all three voltmeters are non-ideal and identical moving-coil voltmeters, they have finite and equal internal resistance RV. It is obvious from the electric schematic presented in figure shown below that, as the voltage generator is ideal and consequently has zero internal impedance, that the following is valid:

U1 + U 3 = E , (S56.1)

where U1 is the showing of the left voltmeter, U3 is the showing of the right voltmeter, and E is the electromotive force of the generator. So, the showing of the right voltmeter is given by:

U 3 =E − U1 =7 V .

(S56.2)

According to the first Kirchhoff’s law for the node where all three voltmeters are connected, we have the following:

I1 + I 2 = I 3 , (S56.3)

where electrical currents I1, I2, and I3 are the currents that flow through the voltmeter coils, as presented in figure above. Therefore, the current through the coil of the voltmeter in the center is thus given by:

±= ü

− . (S56.4)

By multiplying the last equation with the voltmeter internal resistance, we have:

R= RV I 3 − RV I1 . V I2

(S56.5)

Since it is fulfilled the following

U1 = RV I1 , U 2 = RV I 2 , and

U 3 = RV I 3 , with U being the measurement of the central voltmeter, 2

we have the following for the showing of the central voltmeter:

U 2 = U 3 − U1 = 5 V .

(S56.6)

Solutions

173

S57. In order to rotate together with the tire, as presented in figure shown below with all relevant parameters for the analysis, the forces acting on the pebble must fulfill the following conditions:

mv 2 N + mgsinα = R and (S57.1) T = mgcosα , (S57.2) where N is the normal reaction force of the tire acting on the pebble, m is the pebble mass, g is the Earth’s gravitational field acceleration, α is the angle between the center of tire–pebble line and the horizontal plane, v is the car velocity, R is the tire radius, and T is the friction force between the tire and the pebble.

The pebble will stay still with regard to the tire wall if the following condition is fulfilled:

T ≤ µ N , (S57.3) with μ being the friction coefficient between the tire and the pebble. The combination of the last three equations gives:

v2 ≥

gR

µ

( cosα + µsinα )

. (S57.4)

This condition must be fulfilled for any value of the angle α. Since the function of angle α, which is in the brackets of the last equation, has the maximum value for: tanα = µ , (S57.5) the following must be also fulfilled:

Nonstandard Problems in General Physics with Solutions

174

v 2 ≥ gR 1 +

1

µ 2 . (S57.6)

So, the minimal car velocity is:

= vmin

gR 1 +

1

µ2

. (S57.7)

S58. The actual voltages across the resistors and the ideal battery, according to the simple schematic shown below, are given by:

R1 U1 = E R + R 1 2 , (S58.1) R2 U2 = E R1 + R2 , and (S58.2)

U3 = E

, (S58.3)

where U1and U2 are the actual voltages across the resistors R1 and R2, respectively, and U3 is the voltage across the ideal battery with the electromotive force E.

Since a non-ideal voltmeter, with the internal resistance of RV, is used for the measurement of the corresponding voltages, we have the situation as presented in the electric schematics shown in the figure below.

Solutions

175

When the voltmeter is used for the measurement of the voltage across the resistor R1, we have, according to the schematic shown in the left part of the figure shown above, the following:

R1 RV R1 RV = U1' = E E R R + R R R + R R + R R 1 V 2 1 V 2 1 V 2 ,

(S58.4) ' 1

U = 2V where according to the problem statement, we have . Consequently, when the voltmeter is used for the measurement of the voltage across the resistor R2, we have, according to the schematic shown in the central part of the figure shown above, the following:

R2 RV R2 RV = U 2' = E E R R + R R R + R R + R R 2 V 1 1 V 2 1 V 2 , (S58.5) where according to the problem statement, we have last two equations give:

U 2' = 3 V

. The

U1' R1 = ' U 2 R2 . (S58.6) When the voltmeter is used for the measurement of the voltage across the ideal battery, we have, according to the schematic shown in the right part of the figure shown above, the following:

U' = E 3 , (S58.7) where according to the problem statement, we have

' U= E= 6 V 3

The combination of equations (S58.1) and (S58.6) gives:

.

Nonstandard Problems in General Physics with Solutions

176

R1 U1' R2 U 2' U1' = = U1 = E = E E 2.4 V ' ' ' R1 U + U U 1 1 2 1+ 1+ ' R2 U2 . (S58.8) The combination of equations (S58.2) and (S58.6) gives:

= U1

U 2' 1 1 = = E = E E 3.6 V ' ' ' R1 U + U U 1 1 2 1+ 1+ ' R2 U2 . (S58.9)

S59. By bursting, the soap membrane drags the rest of the soap already bursted with the constant velocity, that is, the velocity of the soap membrane bursting as it is presented in figure shown below. Therefore, according to the energy conservation law, we have the following:

1 2 1 mu + = ∆A ( m + ∆m ) u 2 2 2 , (S59.1) where m is the mass of the part of the dragged already bursted soap, u is the bursting velocity of the soap membrane, and ΔA is the work done by the molecular force of the membrane during the bursting of a small part of the soap membrane of the mass Δm. The energy conservation law, in the form presented in equation (S59.1), is valid only if there is no other energy loses such as heat generation.

Equation (S59.1) may be simplified into:

∆A =

1 ∆mu 2 2 . (S59.2)

Solutions

177

The work ΔA is further given by:

= 2σ w∆l , (S59.3) ∆A where σ is the soap surface tension, w is the observed membrane part width, and Δl is the length of the bursted soap membrane. In the equation (S59.3), it was assumed w d , with d being the membrane thickness. Similarly, the mass of the bursted membrane is given by: ∆m= ρ wd ∆l , (S59.4) with ρ being the soap density. The combination of the last three equations yields to the bursting velocity of:

u=2

σ ρ d . (S59.5)

If there is heat generated by the process of soap membrane bursting, one must rely on Newton’s second law of motion that gives the following:

F ∆t =∆ ( mu )

, (S59.6)

where F is the force caused by the surface tension of the soap membrane and Δt is the time needed for bursting the soap membrane of the length Δl. The corresponding force is given by: F = 2σ w , (S59.7) whereas the corresponding time is given by:

∆l ∆t = u . (S59.8) Since the bursting velocity is constant, equation (S59.6) becomes: F ∆t = u ∆m . (S59.9) The combination of equations (S59.4), (S59.7), (S59.8), and (S59.9) gives:

σ= w u ρ wd ∆l

, (S59.10)

which is finally simplified into the following bursting velocity:

Nonstandard Problems in General Physics with Solutions

178

u=

2σ ρ d . (S59.11)

S60. The equivalent schematic of the infinite circuit from the problem statement is shown in the left diagram given below, where the electromotive forces and the internal resistance of the equivalent battery are E and R, respectively. According to Norton’s theorem, this circuit is equivalent to the circuit shown in the right diagram given below, where the voltage between nodes A and B, as well as the resistance between the same nodes when the sources of the electromotive forces are shut down must be the same in both circuits. Therefore, we have the following fulfilled:

R= r1 +

r2 R r2 + R , (S60.1)

where r1 and r2 are the internal resistances of the corresponding batteries.

Equation (S60.1) transforms into the following quadratic equation:

R 2 − r1 R − r1r2 = 0 . (S60.2)

The solutions of this quadratic equation are:

= R

4r 1 r1 1 ± 1 + 2 2 r1

. (S60.3)

Since the resistance must be positive, we have:

= R

4r 1 r1 1 + 1 + 2 2 r1

. (S60.4)

Similarly, for the electromotive forces, we have:

Solutions

E = ε1 + ε 2 + r2

179

E − ε2 r2 + R . (S60.5)

This equation simplifies into:

r E= ε 2 + ε1 1 + 2 R . (S60.6) By substituting the value for R from the equation (S60.4) into the equation (S60.6), we finally obtain:

r2 r1 E= ε 2 + ε1 1 + 2 4r 1+ 1+ 2 r1

. (S60.7)

S61. In order to determine the minimum wind velocity of the air front that can cause the sandstorm, we will rely on the figure shown below where it is presented the simple mechanism of sandstorm formation with all relevant parameters for the analysis. We will observe two slabs of moving air and the mix of air and send, that is, one over the part of the ground where the wind starts to blow and where there are no sand grains in the air and another over the part of the ground where the sand grains are mixed with the air. Therefore, we may say that according to the conservation laws, we can consider that the mass of air, the momentum, and the energy in both slabs are the same but taking into account the acting forces due to the air pressure. According to the mass conservation law, the masses of air in both slabs are the same so we may write the following:

= = ρ wLH ρ , (S61.1) m wlh where w is the slabs width, l and L are the slabs lengths, h and H are the slabs heights, and ρ is the air density. According to the last equation, we have: lh = LH . (S61.2)

Nonstandard Problems in General Physics with Solutions

180

Further, according to the simultaneous movement of the air from the first to the second slab, the air stream velocities in both slabs must fulfill the following condition:

∆t =

l L = v u , (S61.3)

where Δt is the time that wind needs in order to pass from one side to the other of both slabs, v is the wind velocity at the beginning, and u is the wind velocity of the sandy air. The energy conservation law gives the following relation: 1 1 1 1 pwhl + mv 2 + = mgh pwHL + ( m + ∆m ) u 2 + ( m + ∆m ) gH 2 2 2 2 , (S61.4)

where p is the air pressure and Δm is the mass of the sand in the observed slab. After taking into account last four equations, the last equation can be rearranged into:

ρ v 2 + ρ gh=

( ρ + ∆ρ ) u 2 + ( ρ + ∆ρ ) gh

v u , (S61.5)

where the following equation has been taken into account:

= wLH ∆ρ , (S61.6) ∆m with Δρ being the density of sand grains in the sandstorm. The momentum conservation law further gives the following relation:

pwh∆t − pwH ∆= t

( m + ∆m ) u − mv .

(S61.7)

After taking into account equations (S61.1), (S61.2), (S61.3), and (S61.6), the last equation can be rearranged into:

Solutions

181

∆ρ u p vu 1 + = ρ u v − ρ . (S61.8) We will take into consideration that in the moment of the sandstorm formation, we approximately have 0 ≤ ∆ρ ρ and u = v − ∆v with ∆v v being obviously fulfilled. In this case, the last equation becomes:

∆v ∆ρ 1 ≈ ρ 1− p v ρ v 2 . (S61.9) Equation (S61.5) can be further rearranged into:

gh ∆ρ ∆ρ 2 2 v = 1+ u −v u − 1 + u ρ ρ

, (S61.10)

If we take that in the moment of the sandstorm formation, we have ∆ρ ρ and u = v − ∆v with ∆v v , the last equation becomes:

gh ∆ρ ∆v ∆ρ ∆v + −2 ≈ 2 v ρ v . (S61.11) v ρ −

By substituting equation (S61.9) into equation (S61.11), we have:

( ρ v ) + ( p − 2ρ gh ) ( ρ v ) + pρ gh ≈ 0 . (S61.12) 2

2

2

The solutions of this quadratic equation are: 2

p p pgh v ≈ gh − ± gh − − 2ρ 2ρ ρ . 2

(S61.13)

In order to have real solutions, one has the following conditions fulfilled:

gh −

p ≥0 2ρ and (S61.14) 2

p pgh ≥0 gh − − 2 ρ ρ . (S61.15)

182

Nonstandard Problems in General Physics with Solutions

Since the last condition is stricter, we must fulfill: 2

p pgh gh − ≥ 2ρ ρ . (S61.16) This inequality may be rearranged into: 2

p 2p h − h+ ≥0 g g 2 ρ ρ . 2

(S61.17)

In order to fulfill the above condition, the height of the air front must fulfill the following:

3 p h ≤ 1 − 2 ρ g and (S61.18) 3 p h ≥ 1 + 2 ρ g . (S61.19) The second condition, given by the inequality (S61.19), is unreal, since in this case, the height of the air front is larger than the height of the Earth’s atmosphere. Therefore, the maximal height of the air front that can initiate the sandstorm is given by:

3 p hmax ≈ 1 − 2 ρ g . (S61.20) For example, for the standard atmosphere of

p ≈ 105 Pa ,

ρ ≈ 1.2 kg / m3 and the Earth’s gravity field acceleration of g = 9.81 m / s 2 , the maxima height is hmax ≈ 1.14 km .

Finally, according to the equation (S61.13), the minimum velocity at which the sandstorm starts its formation is given by: 2

p p pgh v ≈ gh − − gh − − 2ρ 2ρ ρ .(S61.21)

Solutions

183

S62. The momentum change of the drop during its fall through the cloud is given by:

d ( mv ) = mgdt

, (S62.1)

with m being the drop mass, v being the drop velocity, and g being the Earth’s gravity field acceleration. Since the mass of the drop increases by collecting the small droplets along its path through the cloud, the increase of the drop mass during the elementary time is given by: dm = k ρ Svdt , (S62.2) that is, the drop collects all small droplets in the elementary volume dV = Svdt , where S is the drop cross-sectional area, ρ is the water density, and k is the volumetric concentration of droplets in the cloud, that is, the ratio between the volume of the droplets and the total volume containing the droplets. Since the cross-sectional area of the drop is given by: 2 S = π r , (S62.3)

where r is the drop radius and the elementary mass increase of the drop is given by: 2 dm = 4πρ r dr . (S62.4)

The combination of the last three equations yields:

dr k k dz = = v 4 dt , (S62.5) dt 4 where the drop velocity is given by v = dz / dt . Since the initial drop size is negligible, according to the last equation, we have:

r=

k z 4 , (S62.6)

with z being the drop path length. Equation (S62.1) can be further developed into:

mgdt . vdm + mdv = Mass of the drop is given by:

(S62.7)

Nonstandard Problems in General Physics with Solutions

184

m=

4π 3 ρr 3 . (S62.8)

By substituting equations (S62.4) and (S62.8) into equation (S62.7), we obtain:

3v

dr dv +r = rg dt dt .

(S62.9)

Since v = at is valid where a is the drop acceleration, we have further:

3at

dr + ra = rg dt . (S62.9)

Taking into account relation (S62.6), the last equation becomes:

3at

dz + za = zg dt . (S62.10)

For the drop path, we have:

z=

1 2 at 2 , (S62.11)

so, equation (S62.10) becomes:

1 1 2 3a 2t 2 + a 2t 2 = at g 2 2 . (S62.12) From the last equation, one simply obtains for the drop acceleration:

a=

g 7 . (S62.13)

S63. The temperature of the Earth’s surface only due to the Sun’s radiation is given by:

P = εσ ST , (S63.1) 4

where P is the power of the Sun’s radiation reaching the surface of the Earth, ε is the Earth’s surface emissivity, σ is the Stefan–Boltzmann constant, S is the Earth’s surface area, and T is the absolute temperature of the Earth’s surface. Due to the action of all manmade energy sources, there will be an increase in Earth’s surface temperature ΔT for which the following is valid:

Solutions

P= + ∆P εσ S (T + ∆T )

185

4

, (S63.2)

where ΔP is the overall power of all manmade energy sources. Since

∆P P, we have fulfilled ∆T T , so equation (S63.2) may be

simplified into:

4∆T P + ∆P ≈ εσ ST 4 1 + T

, (S63.3)

By substituting equation (S63.1) into the last equation, we have:

∆T ≈

∆P T 4 P . (S63.4)

Taking into account the numerical values for the corresponding parameters, we obtain the following rise of the Earth’s surface temperature: ∆T ≈ 0.018°C , where for the average Earth’s surface temperature, the value of T ≈ 15°C = 288 K is taken into account. In order to keep the rise of the Earth’s surface temperature below 0.1°C, we obtain the following for the maximum permissible power of all energy sources used by the mankind:

∆Pmax ≈ 4 P

∆Tmax T ,

or numerically

(S63.5)

∆Pmax ≈ 2.5 ×1014 W

.

S64. As shown on the schematic presented in the figure shown below, the current that flows through the voltmeter V2 is, according to the first Kirchhoff’s law, given by:

I 3 = I1 − I 2 = 0.2 mA

.

(S64.1)

The resistance of the voltmeter V2 is thus given by:

R= V2

U2 = 1.25 kΩ I3 .

(S64.2)

Since the voltmeters are identical, we have the following for the resistance of the voltmeter V1:

R= R= 1.25 kΩ . V1 V2

(S64.3)

Nonstandard Problems in General Physics with Solutions

186

The resistance of the ammeter A2 is given by:

R= A2

U2 = 278 Ω I2 . (S64.4)

Since the ammeters are identical, we have the following for the resistance of the ammeter A1:

R= R= 278 Ω A1 A2

. (S64.5)

The showing of the voltmeter V1 is then given by:

= U R= 1.375 V . (S64.6) V1 I1 1 Finally, the electromotive force of the battery is given by:

E =U1 + RA1 I1 + U 2 =1.93 V

.

(S64.7)

S65. The dynamics of a buoy that is affected by the sea waves with all relevant parameters for the analysis is presented in the figure shown ¨

below. The vertical acceleration y of the buoy is given by: ¨

y mg − F , (S65.1) m= where F is the buoyancy, m is the buoy (i.e., inner weight) mass, and g is the Earth’s gravity field acceleration. The buoyancy is given by:

= F ρ gS ( h − ∆h ) , (S65.2) where ρ is the sea water density, S is the buoy cross-sectional area, h is the buoy height, and Δh is the height of the buoy part that sticks out the water.

Solutions

187

Since the whole mass of the buoy is concentrated at the bottom of the buoy, where a small in dimension weight is placed, the center of mass of the buoy is thus located at the bottom of the buoy. So, we have fulfilled: h= y + x + ∆h , (S65.3) where y is the position of the buoy center of mass with respect to the average sea level and x is the actual wave height that is given by:

= x X 0 cos (ωt + θ ) , (S65.4) with X0 being the wave amplitude, ω being the wave angular frequency, and θ being the initial phase of the wave. Further, the combination of the first three equations gives: ¨

mg − ρ gSy − ρ gSx . (S65.5) m y = The differential equation given with the last expression represents the second-order linear differential equation, which solution may be represented in the following form:

y= y0 + Y0 cos (ωt + ϕ )

. (S65.6)

By combining the last three equations, one obtains: mY0ω 2 cos= (ωt + ϕ ) ρ gSY0 cos (ωt + ϕ ) + ρ gSX 0 cos (ωt + θ ) − mg + ρ gSy0

(S65.7)

X ≡Y ≡0

.

0 In the case when there are no waves, we have 0 , that is, there are no oscillations of the buoy caused by the wave, so the last equations becomes:

−mg + ρ gSy0 = 0

, (S65.8)

Nonstandard Problems in General Physics with Solutions

188

that is, parameter y0 represents the position of the bottom of the buoy with respect to the sea level, and it is given by:

y0 =

m ρ S . (S65.9)

In this case, equation (S65.7) becomes:

(

)

Y0 mω 2 − ρ gS cos = (ωt + ϕ ) ρ gSX 0 cos (ωt + θ )

. (S65.10)

It is obvious that from the last equation, ϕ ≡ θ must be fulfilled in order to satisfy the last equation for any time instance. So, the buoy oscillation amplitude is thus given by:

Y0 =

ρ gS X0 mω 2 − ρ gS . (S65.11)

Therefore, the equation of the buoy movement, that is, the position of the buoy bottom (center of mass) with respect to the average sea level is thus given by:

ρ gS m y= + X 0 cos (ωt + θ ) 2 ρ ω ρ S m − gS . (S65.12) Further, by combining equations (S65.3), (S65.4), and (S65.12), we obtain the following for the height of the buoy part that sticks out the water:

∆h = h −

m mω 2 − X0 cos (ωt + θ ) ρS mω 2 − ρ gS . (S65.13)

In order to keep the top of the buoy unsubmerged, we must fulfill ∆h ≥ 0 for any time instance, so we have:

1 g X 0 ≤ ( ρ Sh − m ) − 2 ρ S mω . (S65.14) Having in mind that the wave period is given by:

τ=

2π

ω , (S65.15)

the maximum wave amplitude is given by:

Solutions

189

1 gτ 2 X 0max = ( ρ Sh − m ) − 2 ρ S 4π m . (S65.16) Since the period of small oscillation of the buoy in the water is given by:

T=

4π 2 m ρ Sg , (S65.17)

the maximal wave amplitude is then given by:

m τ 2 X 0max = h − 1 − ρ S T 2 ,

(S65.18)

which can be further rewritten as:

τ2 X= Ä h 1 − 2 0max T , (S65.19) where Δh is the height of the part of the buoy the sticks out of the water when there are no waves. S66. When in the initial moment, an angular momentum is given to the hot wheel (the wheel m with the temperature T1) in the counterclockwise direction, then, due to the heating of the wire that comes from the cold wheel (the wheel with the temperature T2 in the steady state), there will be a thermal dilatation of the wire in the moment when it touches the wheel. The instantaneous dilatation of the wire and friction between the wire and the hot wheel will further cause different forces in the wire thus causing the continuous movement, that is, the rotation of the wheels in the counterclockwise direction, as it is presented in figure shown below.

Since the hot wheel is loaded, the mechanical power caused by the wire dilatation is transferred to the hot wheel due to the friction be-

Nonstandard Problems in General Physics with Solutions

190

tween the wire and the wheel. Therefore, the power of this simple heat engine is given by:

= P

( F1 − F2 ) Rω , (S66.1)

where F1 and F2 are the forces in the hot and cold part of the wire, respectively, R is the hot wheel radius, and ω is the angular velocity of the hot wheel. Typically, coefficient of thermal expansion of the wire is very small. Therefore, we can write the following:

ω≈

v1 v2 v ≈ ≈ R R R , (S66.2)

where v1 and v2 are the velocities of the hot and cold part of the wire, respectively, that are slightly different due to the heating and cooling of the wire when it is in the contact with the wheel that are at different temperatures. Now, if we observe the dynamics of this simple heat engine, which is presented in the figure shown below, according to Newton’s second law of motion, we can write:

( F1 − F2 ) ∆t =∆m ( v1 − v2 ) ,

(S66.3)

where F1 and F2 are the forces in the hot and cold part of the wire, respectively, Δm is the elementary mass of the wire, and v1 and v2 are the velocities in the hot and cold part of the wire, respectively.

The combination of the last three equations gives:

P≈

∆m v ( v1 − v2 ) ∆t . (S66.4)

Further, we have:

v1 =

∆l1 ∆t and (S66.5)

Solutions

v2 =

191

∆l2 ∆t , (S66.6)

with Δl1 and Δl2 being the length of the elementary mass of the wire Δm in the hot and cold part of the wire, respectively. Due to the thermal expansion of the wire, we have:

∆l1 − ∆ = l2 ε (T1 − T2 ) ∆l

, (S66.7)

where ε is the coefficient of thermal expansion of the wire, where due to the small coefficient of thermal expansion of the wire,

∆l1 ≈ ∆l2 ≈ ∆l

into:

P≈

is valid. Now equation (S66.4) can be rearranged

∆m ε (T1 − T2 ) v 2 ∆t , (S66.8)

∆l / ∆t was taken into account. For the elementary mass, where v = it is valid: ∆m = λ∆l , (S66.9) where λ is the mass per unit length of the wire, so the power transferred to the hot wheel is given by:

P ≈ λε (T1 − T2 ) v 3

. (S66.10)

In order to find the coefficient of performance of this simple machine, we must find the power provided to the wire in order to heat it from the temperature T2 to the temperature T1. This power is given by:

= P0

∆m c (T1 − T2 ) ∆t , (S66.11)

with c being the specific heat capacity of the wire. Equation (S66.11) may be simplified into:

= P λ vc (T1 − T2 ) 0 . (S66.12) Finally, the coefficient of performance of this simple machine is given by:

Nonstandard Problems in General Physics with Solutions

192

P ε 2 = η ≈ v P c . (S66.13) 0 It is interesting to note that the coefficient of performance doesn’t depend on the temperature difference of the wheels. S67. During the lending, the vertical velocity of the airplane is typically much smaller than its horizontal velocity. In the moment of impact between the airplane rear wheels and the airport runaway, the pitch angle of the airplane is typically very small, as it is presented in the figure shown below. Since the impact between the airplane rear wheels and the runaway is considered perfectly inelastic, we have the following equations valid:

= P m (v − u ) and (S67.1) Pr = J ω , (S67.2) where P is the momentum transferred to the airplane during the impact, m is the airplane mass, v and u are the vertical velocities of the airplane before and after the impact, respectively, r is the horizontal distance between the rear wheels and the center of mass of the air-

plane, J is the airplane moment of inertia given by J = mρ , where ρ is the radius of gyration of the airplane along its pitch axis, and ω is the angular velocity of the airplane. 2

Since the impact is perfectly inelastic the velocity of the rear wheels after the impact is zero, so the following must be fulfilled: u = ω r . (S67.3)

The combination of the last three equations yields:

ω =v

r ρ + r 2 . (S67.4) 2

Solutions

193

One may consider that after the inelastic impact, the airplane rotates around the pivot point that is positioned at the airplane rear wheels. In order to provide the fastest possible contact of the front wheels with the runaway after the impact of the rear wheels, that is, the shortest possible lending time of the airplane and consequently the safest possible landing, the angular velocity of the airplane must take its maximum possible value, which is, according to the last equation, obtained for: r = ρ , (S67.5)

or the distance between the rear wheels and the airplane center of mass must be the same as the airplane radius of gyration. Further, we will consider, also, the perfectly inelastic impact of the front wheels with the runaway. The impact dynamic is presented in figure shown below. Similar to the previous case, we have the following fulfilled:

P1 + P2 = mu and (S67.6)

P2 d − Pr J ω , (S67.7) 1 =

with P1 and P2 being the momentums transferred to the airplane during the impact from the rear and front wheel, respectively, u is the airplane center of mass velocity, d is the horizontal distance between the airplane center of mass and the front wheels, r is, as earlier mentioned, the horizontal distance between the rear wheels and the center of mass of the airplane, and ω is the angular velocity of the airplane. As the airplane rotates around rear wheels, we also have fulfilled equation (S67.3).

Nonstandard Problems in General Physics with Solutions

194

After rearranging last two equations, we have the following for the momentum transferred from the rear wheels to the airplane during the impact of the front wheels.

P1 = mω

rd − ρ 2 r + d . (S67.8)

In order to keep airplane stable during the landing and thus to provide safest possible airplane landing, during the impact of the front wheels, the reaction force between the rear wheels and the runaway must not be changed. In this way, there will be no change in the traction conditions of the rear wheels. So, the following must be fulfilled:

P1 ≡ 0 . (S67.9)

This last condition will be fulfilled if the distance between the front wheels and the airplane center of gravity fulfills the following condition:

d=

ρ2 r . (S67.10)

Since according to the equation (S67.5) we have r = ρ , finally we obtain the same result as for the rear wheels: d = ρ . (S67.11) Therefore, in order to provide the safest possible landing of the airplane, one must position the front and rear wheels of the airplane from its center of gravity at the distances that are equal to the airplane radius of gyration along its pitch axis. S68. At the beginning of the analysis, we will assume that the gunpowder is filled into the cannon barrel, so that it occupies the length l of the cannon barrel, as presented in the left part of the figure shown below. The mass of the gunpowder is then given by:

M = ρ Sl , (S68.1) where ρ is the gunpowder density and S is the cross-sectional area of the barrel. After the instantaneous combustion, the gunpowder is transformed into the ideal gas that occupies the same volume as the gunpowder itself. Therefore, the internal energy of such a gas is given by:

Solutions

195

U = M λ , (S68.2) where λ is the gunpowder combustion energy per unit mass. This energy is equal at the same time to: U = 3nRT , (S68.3) where n is the amount of substance of the gas obtained by the gun-

powder combustion that is also given by n = M / µ , with μ being molar mass of the gunpowder, R is the universal gas constant, and T is the absolute temperature of the gas. From that moment, the gas expands very quickly, so just before leaving the barrel, the gas occupies the whole barrel as presented in the right part of the figure shown below. As the expansion is very rapid, we can assume that the gas expansion is adiabatic. Therefore, we have:

T ( Sl )

γ −1

= T ' ( SL )

γ −1

, (S68.4)

where T′ is the temperature of the gas just before the cannonball leaves the barrel, L is the barrel length, and γ is the constant for which is valid γ = 4 / 3 since the molecule of the gunpowder gas has complex structure, thus having 6 degrees of freedom.

In this case, equation (S68.4) becomes: 1

l 3 T '=T L . (S68.5) The internal energy of the gas in the moment just before firing the cannonball is given by: U ' = 3nRT ' . (S68.6) According to the energy conservation law, we have:

1 U= U ′ + mv 2 + W 2 , (S68.7)

Nonstandard Problems in General Physics with Solutions

196

where m is the mass of the cannonball, v is the cannonball velocity, and W is the kinetic energy of the gunpowder gas, since the center of gravity of the gas rapidly moves. In order to find this kinetic energy of the gas, we will observe the figure shown below where it is presented the situation just before the cannonball firing. If we assume that the gas concentration is uniform within the barrel, then the velocity of the elementary gas slab at the position x in the barrel is given by:

u=

x v L . (S68.8)

The elementary kinetic energy of the gas slab is then given by:

1 dW = u 2 dm 2 , (S68.9) where the elementary mass of the gas slab is given by:

dm =

M dx L . (S68.10)

The corresponding kinetic energy of the gunpowder gas is then given by: L

W = ∫dW

0

. (S68.11)

The combination of the last four equations gives:

W=

1 Mv 2 6 . (S68.12)

Equation (S68.7) then becomes:

Solutions

197

1 1 3nRT = 3nRT '+ mv 2 + Mv 2 2 6 , (S68.13) By substituting the corresponding values for each parameter, the last equation becomes: 1 3 l 18nRT 1 − L 2 v = 3 m + Sl ρ . (S68.14)

According to the equations (S68.1), (S68.2), and (S68.3), the temperature of the combusted gunpowder is given by

T=

λρ Sl 3nR . (S68.15)

The combination of the last two equations gives: 1 3 l 6λρ Sl 1 − L v2 = 3m + ρ Sl . (S68.16) One can notice from the last equation that for l = 0 and l = L , we have v = 0 . Therefore, there is a certain value of the parameter l for which the cannonball gains maximum velocity. This value can be found by finding the first derivative of the last equation and equaling it with zero. If we do so, we obtain the following equation:

4 0 , (S68.17) az + 12 z − 9 =

where it is valid:

a=

ρ SL m and (S68.18) 1

l 3 z = L . (S68.19) Since ρ SL / m= a= 48, equation (S68.17) becomes:

Nonstandard Problems in General Physics with Solutions

198

4 0 . (S68.20) 48 z + 12 z − 9 =

The last equation may be simplified as:

0 ( 2 z − 1) 2 z ( 4 z 2 + 2 z + 1) + 3 =

. (S68.21)

The only physically realistic solution is:

z=

1 2 , (S68.22)

or:

l=

L 8 . (S68.23)

The mass of the gunpowder that provides maximum cannonball velocity is then given by:

= M

1 = ρ SL 6m 8 . (S68.24)

S69. Since Sisyphus mass is much smaller than the mass of rock that he pushes uphill, Sisyphus must pose his body so that his body line is perpendicular to the rock surface, as it is presented in figure shown below. Otherwise, there will be net torque acting on his body that eventually will overturn him. If the Sisyphus body angle with respect to the hill surface is α, then the following is valid from the geometry:

sin α =

R R + H , (S69.1)

where R is the rock radius and H is the Sisyphus height.

As Sisyphus pushes the rock slowly uphill, the following must be valid: F sin α = N and (S69.2)

Solutions

199

F cos α = T , (S69.3) where F is the force that is produced by Sisyphus, for which is valid

F mg , with m being Sisyphus mass and g being the Earth’s gravi-

tational field acceleration, and N and T are the vertical and horizontal reaction forces between Sisyphus legs and the hill, respectively. In order to avoid slipping of Sisyphus legs, the following condition must be fulfilled: T ≤ µ N , (S69.4) where μ is the friction coefficient between Sisyphus legs and the hill. The last three equations give: cot α ≤ µ . (S69.5) The last equation can be rearranged into:

sin α ≥

1 1+ µ 2

. (S69.6)

The combination of equations (S69.1) and (S69.6) finally gives:

R≥

H 1 + µ 2 −1

. (S69.7)

Having in mind that from the problem statement we have H = 1.75 m

and µ = 0.9, we obtain the following value for the minimal radius of the rock that Sisyphus can push uphill

Rmin ≈ 5.1 m

following must be fulfilled:

Mg ( R − H sin α ) sin β ≤ MgR cos α cos β

, (S69.8)

where M is the rock mass and β is the angle of inclination of the hill. The minimum rock radius is achieved for:

sin α =

1 1+ µ 2

. (S69.9)

Taking the minimum value of the rock radius into account and the corresponding value for the angle α, the last inequality becomes: tan β ≤ µ . (S69.10)

Nonstandard Problems in General Physics with Solutions

200

So, the maximal angle of the hill must be

β max ≈ 42° .

When Sisyphus pushes the rock uphill, we have the situation as presented in figure shown below. Since the rock moves uphill very slowly, we have following fulfilled:

′ F cos α − Mg sin β − T = Ma and

(S69.11)

′

T R = J ϕ , (S69.12) where a is the uphill acceleration of the rock, T’ the corresponding 2 horizontal reaction force between the rock and the hill, J = 2 MR / 5

is the rock moment of inertia, and ϕ = a / R angular acceleration of the rock since there is no slipping of the rock.

According to the last two equations, we have:

7 = F cos α M g sin β + a 5 . (S69.13) The minimum pushing force in order to move rock uphill is obtained for a g , so the last equation becomes:

F ≈ Mg

sin β cos α . (S69.14)

In the case of the minimum rock radius and half of the maximum inclination angle of the hill, we have:

F ≈ Mg 1 +

β sin max µ 2 1

2

. (S69.15)

Taking into consideration that the rock mass is related to its radius, we have:

Solutions

F≈

201

4π 1 β 3 ρ Rmin g 1 + 2 sin max µ 3 2 . (S69.16)

By substituting the corresponding parameters, one obtains the following force that must be produced by Sisyphus F ≈ 7.9 MN . Since the pushing force is enormous, it is obvious that the myth of Sisyphus is just a myth. In order to not allow sliding the rock downhill, we will have the situation as presented in figure shown below. Therefore, the following set of equations is valid:

F cos α Mg sin β − T ′ and (S69.17) = = N ′ + F sin α , (S69.18) Mg cos β

where N’ is the normal reaction force between the rock and the hill. The friction force between the rock and the hill is also given by:

′ ′ T = µ ' N , (S69.19) with μ› being the friction coefficient between the rock and the hill. The minimal force that must be provided by Sisyphus in order to not allow sliding the rock downhill is according to the last three equations given by:

F = Mg

sin β − µ ′ cos β cos α − µ ′ sin α , (S69.20)

In the case of minimal rock radius and half of the maximum inclination angle of the hill, we have:

β sin max 4π 2 3 = F ρ Rmin g 1+ µ 2 3

β max − µ ′ cos 2 µ − µ′

. (S69.21)

202

Nonstandard Problems in General Physics with Solutions

By substituting the corresponding parameters, one obtains the following force that must be produced by Sisyphus F ≈ 2.6 MN , which still represents an enormous force. S70. Inside the resonator, there are in total N ( N 1 ) photons and since the light is monochromatic, each of them carries the energy of ε = hν , where h is the Planck constant and ν is the frequency of the photons. Due to the photon reflections from the mirrors, there is net force exerted on the mirrors. At one point, the one of the mirrors starts to move parallel to the light beam with very small but constant velocity u ( u c , where c is the speed of light in vacuum), as presented in figure shown below. After a very long period of time Δt ( ∆t l / c , where l is the length of the resonator, that is, this period of time is much larger than the photon round-trip time within the resonator), the resonator will be longer for a very small length Δl ( ∆l l ) for which is valid ∆l = u ∆t , as presented in figure shown below. Because the adiabatic expansion of the resonator has been made, due to the law of conservation of energy, we have:

∆A + ∆W = 0 , (S70.1) where ΔA is the elementary work that is carried out by the force acting on the moving mirror, and ΔW is the elementary change of the total energy contained in the cavity, for which the following is valid: ∆A = F ∆l and (S70.2) ∆W = N ∆ε , (S70.3) where F is the force acting on the moving mirror and Δε is the energy change of the photons after the expansion.

Solutions

203

After the reflection from the moving mirror, the change of the momentum Δp of the photon is perpendicular to the moving mirror and it is given by:

2 p , (S70.4) ∆p = where p is the corresponding photon momentum in the dielectric. During the motion of the mirror for a period of time Δt for each photon, there will be in total q reflections from the moving mirror for which the following is valid:

= q

∆t c∆t = τ 2nl , (S70.5)

where τ = 2nl / c is the photon round-trip time within the resonator and n is the refractive index of the medium. Therefore, the force excreted on the moving mirror is given by:

F = Nq

∆p ∆t . (S70.6)

The combination of the last three equations gives:

F=N

cp nl . (S70.7)

Due to the photon reflection from the slowly moving mirror, the photon frequency will be changed as the result of the Doppler effect. After the reflection, the frequency ν′ of the photon is equal to: ν ′= ν + Äν , (S70.8) where ∆ν ν is the change in frequency of the photon for which

is also valid ∆ν ν / q , that is, the mirror velocity and observing time are chosen in order to fulfill this condition. This frequency change is, according to the Doppler effect, equal to:

∆ν =−

2u 2nu ν =− ν c c n . (S70.9)

Since each photon changes its energy for the following value when reflects from the moving mirror: ∆ε ' =h∆ν , (S70.10)

Nonstandard Problems in General Physics with Solutions

204

where the total energy change of each photon during the observed time Δt is given by ∆ε = q∆ε ' , the overall change of the elementary change of the total energy is given by: ∆W = Nq∆ε ' . (S70.11) Taking into consideration equations (S70.5), (S70.9), (S70.10), and (S70.11), we obtain:

∆W = −N

u ∆t hν l . (S70.12)

The combination of equations (S70.2), (S70.7), and (S70.12) gives:

cp ∆l − hν u ∆t =0 n . (S70.13) Since ∆l = u ∆t , we finally have:

= p

nhν = np = pM 0 c , (S70.14)

p = hν / c

where 0 is the corresponding photon momentum in vacuum. Equation (S70.14) shows that Minkowski variant of the photon momentum is obtained in the presented thought experiment. Regarding the second thought experiment when photon enters the transparent block, then, according to the momentum conservation, one has the following:

= p0 Mu + p , (S70.15) p = hν / c

where 0 is the corresponding photon momentum in vacuum, M is the mass of the transparent block, u is its velocity when the photon is inside the block, and p is the photon momentum in the dielectric.

Solutions

205

The photon will be in the block for the following period of time:

τ=

l nl = c c n , (S70.16)

with l being the block length and n its refractive index. The corresponding shift of the block is then given by: ∆l =τ u. (S70.17) During the period of time when the photon is in the block, the center of mass of the system will be shifted at the position that is given by:

M ∆l + ml ∆xCM = , M + m (S70.18) where m is the corresponding photon mass in vacuum. Since it is fulfilled M m , the last equations becomes:

∆xCM =∆l +

m l. M (S70.19)

For the period of time when the photon is inside the block, the same photon shifts for the following length: L = cτ , (S70.20) which at the same time corresponds to the shift of the center of mass of a single photon. According to the momentum conservation law, we have:

mL= M ∆xCM .

(S70.21)

Nonstandard Problems in General Physics with Solutions

206

The combination of the last three equations gives: mcτ = M ∆l + ml. (S70.22) By substituting equations (S70.16) and (S70.17) into the last equation, one obtains:

= mn M

n u + m. c (S70.23)

The velocity of the block u one can obtain from equation (S70.15) as:

u=

p0 − p M . (S70.24)

The combination of last two equations gives:

mn=

n ( p0 − p ) + m. c (S70.25)

Since it is valid:

p0 = mc, (S70.26)

equation (S70.25) becomes:

np0= n ( p0 − p ) + p0 .

(S70.27)

After the rearrangement of the last equation, one finally obtains:

= p

p0 = pA . n (S70.28)

Equation (S70.28) shows that Abraham variant of the photon momentum is obtained in the presented thought experiment. It seems that we fail in clarification of the more than a century old dilemma as we obtain both momentums in two independent thought experiments.

Solutions

207

S71. The solution of this problem is similar to the solution of the problem P68. According to the calculated speed of the cannonball that is given by the equation (S68.16), we have the following for the cannonball speed: 1 3 l 6λρ Sl 1 − L v2 = 3m + ρ Sl , (71.1)

where λ is the gunpowder combustion energy per unit mass, l is the length of the part of the cannon barrel that is filled with the gunpowder, ρ is the gunpowder density, S is the cross-section area of the barrel, m is the cannonball mass, and L is the barrel length. One can note from the last equation that the smaller the mass of the cannonball, the bigger the speed of its firing. So, in the case of small mass of the cannonball, the corresponding speed is: 1 3 l 2 = v 6λ 1 − L .

(71.2)

It is further obvious that the longer the barrel length, the higher can-

nonball speed. So, in the case of L l , the maximum speed has been achieved. Finally, the maximum speed at which the cannonball can be fired from the cannon barrel is given by:

vmax = 6λ

. (71.3)

S72. The equivalent resistance between points A and C as presented in figure shown below can be found in the following way:

RAC =

ET I T , (72.1)

where ET is the electromotive force of the test generator and IT its corresponding current.

Nonstandard Problems in General Physics with Solutions

208

Since the infinite resistor network is symmetric, we can introduce two independent current test generators, as presented in figures shown below and find the corresponding voltage between the points A and C. In this case, the equivalent resistance between points A and C is:

RAC =

U AC I T , (72.2)

where UAC is the voltage between corresponding points.

According to the equivalent schematics shown above, the voltage between the points A and C is given by:

(

)

U = R I1' + I 3' + I1″ + I 3″ AC , (72.3) I' I' I″ I″ where currents 1 , 3 , 1 , and 3 are marked on the schematics shown above. Due to the symmetry of the network, we have the following relation fulfilled:

Solutions

' ″ I= I= 1 3

209

IT 4 , (72.4)

= I ' 2 I 2' + I 3' 1 and (72.5)

″ I= I1″ + 2 I 2″ 3

. (72.6)

The combination of the last four equations gives:

(

)

U AC = R I T − 2 I 2' − 2 I 2″ .

(72.7)

The symmetry of the network also gives:

I 3' = I1″ ' 2

I =I

and (72.8)

″ 2

. (72.9)

Therefore, equation (72.7) becomes:

(

)

U = R I T − 4 I 2' AC . (72.10) Similarly, in order to find the resistance between the points A and B, we can introduce two independent current test generators, as presented in figures shown below and find the corresponding voltage between the points A and B. In this case, the equivalent resistance between points A and B is:

RAB =

U AB I T , (72.11)

where UAB is the voltage between corresponding points.

According to the equivalent schematics shown above, the voltage between the points A and B is given by:

Nonstandard Problems in General Physics with Solutions

210

(

)

U = R I1' + I 2' + I1''' + I 2''' AB ,

(72.12)

I ' I ' I ''' I ''' where currents 1 , 2 , 1 , and 2 are marked on the schematics shown above. Due to the symmetry of the network, we have the following relation fulfilled:

' ''' I= I= 1 2

I 2' = I1'''

IT 4 and (72.13)

. (72.14)

The combination of the last three equations gives:

I = U AB R T + 2 I 2' 2 . (72.15) Further, by combining equations (72.11) and (72.15), we have:

I R= R T + 2 I 2' AB I T 2 (72.16) that further yields:

I 2' = I T

2 RAB − R 4 R . (72.17)

By substituting this value into equation (72.10), one obtains:

U= 2 ( R − RAB ) I T AC . (72.18) Finally, according to the equations (72.2) and (72.18), we obtain the following resistance between the points A and C:

R= 2 ( R − RAB ) AC

. (72.19)

S73. Due to the Magnus effect, there will be net force acting on the ball that will further induce curved ball trajectory instead of a straight line. The reason for this is different air velocities at the opposite ballpoints that will further cause different air pressures at that points and at the end the net centripetal force that will push the ball toward the center of the ball revolution. Since the radius of curvature of the ball is large enough, we can, in the simplified analysis, neglect the centrifugal force of such ball motion. Moreover, since the area of the

Solutions

211

ball surface, where the velocities of the ballpoints due to its spinning are the highest is the largest, we can look for such a ball motion where two points at the opposite ball sides have the same total velocities, as it is presented in figure shown below. If we assume that the ball makes a revolution around the point C, as presented in figure below, the angular velocity of the ball revolution is given by:

Ù=

v R , (73.1)

where v is the ball velocity after kicking it and R is the radius of curvature of ball trajectory.

Two opposite ballpoints A and B have the following total velocities:

vA= Ù ( R + r ) − ω r

and

vB= Ù ( R − r ) + ω r

, (73.3)

(73.2)

where r is the ball radius and ω the angular velocity of the ball spinning. In order to have same air pressures on both ball sides, the following must be fulfilled:

vA = vB . (73.4)

The last three equations give:

Ù ( R + r ) − ω r= Ù ( R − r ) + ω r

, (73.5)

or equivalently after the simplification: Ù = ω . (73.6) By substituting equation (73.1) into the last equation, we finally obtain:

R=

v

ω . (73.7)

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212

Since the radius of trajectory curvature of the ball does not depend on the ball radius, we will now observe two centrally symmetric points A′ and B′ on the ball surface, that is, two arbitrary points positioned on two sides of the ball diameter as presented in figure shown below. According to the geometry, the corresponding velocity components of these points are as follows:

vAy ′ = Ù ( R + ρ cos ϕ ) − ωρ cos ϕ

vAx ′ = ωρ sin ϕ

, (73.9)

v = Ù ( R − ρ cos ϕ ) + ωρ cos ϕ y B′

v = −ωρ sin ϕ x B′

where

y A′

v

and

, (73.8)

and (73.10)

, (73.11) x vA′

are the y and x components of the velocity of vy vx point A′, respectively, B′ and B′ are also the y and x components of the velocity of point B′, respectively, ρ is the radius of the spinning trajectory of the centrally symmetric points A′ and B′ on the ball surface, and φ is the angle of the corresponding points with respect to the radius vector of the ball revolution.

The total velocities of these points are given by:

= vA′

(v ) + (v )

= vB′

(v ) + (v )

y A′

y B′

2

2

x A′

x B′

2

and

(73.12)

2

. (73.13)

In order to have same air pressures on both ball sides, the following must be fulfilled:

vA′ = vB′ . (73.14)

Solutions

213

The last three equations give:

(v ) + (v ) = (v ) + (v ) y A′

2

x A′

2

2

y B′

x B′

2

. (73.15)

By substituting equations (73.8), (73.9), (73.10), and (73.1) into the last equations, we obtain the following relation: 2

2

Ù ( R + ρ cos ϕ ) − ωρ cos ϕ + (ωρ sin ϕ = ) Ù ( R − ρ cos ϕ ) + ωρ cos ϕ + ( −ωρ sin ϕ ) 2

(73.16)

2

After the simplification, equation (73.16) becomes: 2

2

ÙR + ( Ù − ω ) ρ cos ϕ = ÙR − ( Ù − ω ) ρ cos ϕ . (73.17) Further, the last expression may be reduced to:

4ΩR ( Ω − ω ) ρ cos ϕ = 0

. (73.18)

If Ù = ω is satisfied, that is, if the angular velocities of ball spinning and revolution are the same, for any given point at the ball surface, that is, for an arbitrary chosen ρ and φ, we can find centrally symmetric point with the same total velocity. Therefore, the corresponding air pressures at these points will be the same, thus causing opposite forces with the same intensities that are mutually canceled. So, finally, we can conclude that the angular velocity of the ball revolution will be the same as the angular velocity of its rotation (spinning) and the corresponding radius of curvature of ball trajectory will be given by the equation (73.7). Since in this approximated analysis there is no net force acting on the ball, yet the ball trajectory has, although large, finite radius of the curvature, the presented analysis is valid if the centripetal acceleration of the ball is relatively small. The centripetal acceleration of the ball is thus given by:

v2 = a = vω R , (73.19) and the analysis is valid if the angular velocity of ball spinning is small.

214

Nonstandard Problems in General Physics with Solutions

S74. In order to find the position of the center of mass of the homogeneous semicircular disk of radius by not using the apparatus of higher mathematics such as integral calculus that is usually employed, we will rely on a virtual work principle as presented in figure shown below. We will assume that the disk is positioned in the Earth’s gravitational field with the acceleration equal to g. The disk surface is parallel to the vector of gravitational field acceleration, whereas the disk diam-

eter is also parallel to this vector ( BB ' g ). Now, we will perform a small perturbation of the disk position by rotating it around its center A for a very small angle Δφ ( ∆ϕ 1 ). In this way, the center of mass of the disk is shifted from the point CM to the point CM′

' ∆ϕ . The where the corresponding angle is given by ∠CMACM = position of the center of mass of the disk is defined as x = ACM . In order to perform this perturbation, the following mechanical work must be done:

∆A= mgx∆ϕ , (74.1) where m is the mass of the disk.

Equivalently, this mechanical work is carried out in order to move a small triangle ABC from its initial position to a new position where we have the triangle AB'C' . Although these triangles are not real triangles, in fact small circular sectors, due to the very small central angle Δφ, they can be considered triangles. From the geometry, we have the following fulfilled for the corresponding angles:

∆ϕ and (74.2) ∠ABC = ∆ϕ . (74.3) ∠AB'C' =

Solutions

215

The equivalent change of the potential energy of the disk is equal to: ∆E= 2 gh∆m , (74.4) where h is the position of the triangle center of mass from its vertex A and Δm is the triangle mass. For every triangle, we have:

2 H 3 , (74.5)

h=

where H is the height of the triangle, which in our case is equal to H = R , with R being the disk radius. The mass of the triangle is given by:

∆= m

1 σ R 2 ∆ϕ 2 , (74.6)

with σ being the disk mas per unit area that is further given by:

σ=

m 1 π R2 2 . (74.7)

The last four equations give:

= ∆E

4 mgR∆ϕ 3π .

(74.8)

According to the energy conservation law, we have: ∆E =∆A . (74.9) By substituting equations (74.1) and (74.8) into the last equation, one obtains:

4 mgR∆ϕ= mgx∆ϕ 3π , (74.10) which after the rearrangement finally gives the position of the center of mass:

x=

4 R 3π . (74.11)

216

Nonstandard Problems in General Physics with Solutions

S75. Because of the symmetry of the infinite resistance network, it may be equivalently presented as it is shown in the left part of the figure shown below where RE represents a certain equivalent resistance. Due to the linearity and scaling effect of the smaller square that is positioned at the midpoints of the largest square, one can equivalently represent the same infinite resistance network as it is presented in the right part of the figure shown below, where R is the resistance of the side of the largest square.

Based on the Y–Δ transform, the schematic shown in the right part of the figure shown above may be presented as it is shown in figure below. The following is valid for the corresponding resistance:

R R , 2 2 R1 = = R R RE + + 2 2 2

2 4

R2 2 R + RE and

R RE , 1 RRE 2 2 = R2 = R R RE 2 2 R + RE + + 2 2 2 .

(75.1)

(75.2)

Solutions

217

The simplified schematic of the schematic shown above is given in the figure shown below. Between every two points at the largest square vertices, the resistance must be the same. So, according to the equivalent schematic shown at the beginning of this solution, we have fulfilled the following for the resistance between points A and B:

RE ⋅ 3RE 3 = RAB R= = RE E 3RE R + 3 R 4 E E .

(75.3)

Similarly, from the above-presented schematic, we have:

3 RAB = R1 + ( 2 R2 ) 3 ( 2 R2 ) + R1 = 2 R1 + R2 2 . (75.4) According to the last two equations, we have:

3 3 R = 2 R1 + R2 E 2 . (75.5) 4 After substituting equations (75.1) and (75.2) into the last equation, we have:

3 = RE 4

2 2

R2 3 RRE + 2 R + RE 4 2 R + RE .

(75.6)

After rearranging the last equation, one obtains the following quadratic equation:

3RE2 + 3R

(

)

2 − 1 RE − 2 2 R 2 = 0 .

(75.7)

The solutions of this quadratic equation are:

RE = R

−3

(

)

2 − 1 ± 27 + 6 2 6

.

(75.8)

Nonstandard Problems in General Physics with Solutions

218

Since the resistance must be positive, the value of the equivalent resistance RE is given by:

= RE

1 R 27 + 6 2 − 3 6

(

)

2 −1 . (75.9)

So, according to the equation (75.3) and to the last equation, we have the following resistance between the points A and B:

= RAB

1 R 27 + 6 2 − 3 8

(

)

2 − 1 ≈ 0.589 R . (75.10)

Similarly, according to the equivalent schematic shown at the beginning of this solution, we have fulfilled the following for the resistance between points A and C:

RAC = 2 RE 2 RE = RE =

1 R 27 + 6 2 − 3 6

(

)

2 − 1 ≈ 0.786 R . (75.11)

S76. When a force pulls a long bar, as presented in figure shown below, the bar will elongate due to the applied stress. The stress is defined as:

σ=

F S , (76.1)

with F being the force and S the area of the bar cross section. Due to the applied stress, there is a resulting strain defined as the bar length change Δl divided by the bar length l, or:

εx =

∆l l . (76.2)

The bar elongates in the direction of the pulling force and contracts at the same time in the direction perpendicular to the force, or for the corresponding strains, the following is valid:

Solutions

εy =

∆t t and (76.3)

εz =

∆w w , (76.4)

219

where Δw and Δt are the bar width and thickness change, respectively, of the corresponding bar width w and thickness t. In the case of relatively low stresses, the changes of the bar dimensions are proportional to the applied stress and the ratio of stress to strain for the bar is an elastic constant called the Young’s modulus E, that is:

E=

σ ε x , (76.5)

or there is a linear relationship between the stress and the strain. At the same time, we have the dimensions changes perpendicular to the direction of the applied force. Poisson’s ratio defines the ratio of transverse strain to axial strain, where for small values of the corresponding strains, we have this ratio defined as the amount of transversal expansion (compression) divided by the amount of axial compression (expansion):

∆t ∆l

∆w ∆l . (76.6)

ν= − = −

The corresponding area of the cross section of the stretched bar is given by:

S=

( w + ∆w )( t + ∆t ) . (76.7)

By combining the last two equations and equation (76.2), one has:

S=

( w −ν∆l )( t −ν∆l ) ≈ wt (1 − 2νε x ) , (76.8)

where the higher order components are omitted. So, equations (76.1), (76.2), (76.5), and (76.8) further give:

Eε x ≈

σ (1 − 2νε x )

. (76.9)

The last equation may be rearranged into the following quadratic equation:

Nonstandard Problems in General Physics with Solutions

220

2 Eνε x2 − Eε x + σ ≈ 0

. (76.10)

The solutions of the above-presented quadratic equation are:

εx ≈

1± 1− 4ν

8νσ E . (76.11)

According to the solution, for the strain given above if the following condition is fulfilled:

8νσ ≥1 E (76.12) there will be no real solutions of the strain. This means that the bar will keep on stretching until it breaks. So, according to the last inequality, the tensile stress is given by:

σT =

E 8ν . (76.13)

S77. Since the liquid occupies an infinite large volume, that is, the border of the liquid is at the infinite distance from the balls, we have the situation as it is presented in figure below with all relevant parameters for the analysis. In order to find the gravitational force acting on the balls, one must take into consideration the gravitational influence of the surrounding liquid.

The effect of the liquid may be mathematically modeled as two balls of liquid but with the negative density −ρ0 that occupy the same volume as the balls as it is presented in the figure shown below. Therefore, the gravitational attraction between the balls is given by:

F =G

( m1 − m0 )( m2 − m0 ) d2

,

(77.1)

Solutions

221

where G is the universal gravitational constant, d is the distance between the balls, m1 is the mass of the first ball, m2 is the mass of the second ball, and m0 is the mass of the expelled water. The corresponding masses are given as follows:

m0 =

4π 3 r ρ0 3 , (77.2)

m1 =

4π 3 r ρ1 3 , and (77.3)

m2 =

4π 3 r ρ2 3 , (77.4)

with r being the ball’s radius.

The combination of the last four equations gives:

= F

4π r 3 G ( ρ1 − ρ0 )( ρ 2 − ρ0 ) 3 d2 , (77.5)

or since it is valid

ρ1 < ρ0 < ρ 2 , the last equations becomes

4π r 3 F= − G ( ρ0 − ρ1 )( ρ 2 − ρ0 ) 3 d2 , (77.6) so as F < 0 is valid, we have instead of the gravitational attraction the gravitational repulsion of the balls. S78. In order to find the average distance between the strings within the candlewick which will provide the strongest shine of the candle, we will observe the candlewick cross section as it is presented in figure shown below with all relevant parameters for the analysis, where we will assume that the candlewick has circular cross section.

Nonstandard Problems in General Physics with Solutions

222

Due to the generated heat from burning the paraffin or wax from the candle, the paraffin or wax that is below the melted layer is melting and rising along the candlewick due to the surface tension of the melted liquid. The overall force that acts on the melted liquid from the candle is then given by: F = π dσ N , (78.1) where d is the string diameter, σ is the surface tension of the melted liquid, and N is the total number of the strings within the candlewick. The equivalent pressure is given by:

p=

F S , (78.2)

where S is the melted liquid surface, that is, the effective surface between the stings that is given by: 2

d = S π R −π N 2 , 2

(78.3)

with R being the candlewick radius. The last three equations give:

p=

dσ N

2

d R −N 2 . 2

(78.4)

Due to the resulting pressure, the melted liquid will rise with the velocity that may be determined according to the Bernoulli’s principle in the following way:

p=

1 2 ρv 2 , (78.5)

Solutions

223

with ρ being the melted liquid density and v the rising melted liquid velocity. The equivalent mass flow of the melted liquid is then given by: m = ρ Sv . (78.6) At the same time, in the steady state, this amount of the melted liquid will be also burned in the unit of time. So, in order to maximize the candle shine, one must maximize this mass flow of the melted liquid. Taking into consideration previous equations, the burned mass of the melted liquid per unit of time is thus given by:

1 = m π 2σρ Nd R 2 − Nd 2 4 . (78.7) In order to find the maximum mass flow, we must find the corresponding solution for the following equation:

dm =0 dN , (78.8) which will give the optimal number of the strings within the candlewick as:

N opt =

2R2 d 2 . (78.9)

Finally, the optimal string density within the candlewick that will provide the highest possible shine of the candle is given by:

n=

N opt

π R 2 , (78.10)

or:

n=

2 π d 2 . (78.11)

In order to check if this string density is physically achievable, we will try to find the maximal possible string density. According to the figure shown below, where are presented the string locations that will provide the highest possible string density, the corresponding maximum string density is thus given by:

224

Nonstandard Problems in General Physics with Solutions

3 2 3 6 = nü = 2 1 d 3 3d d⋅ 2 2 . (78.12)

n= / n π 3 / 3 > 1, the obtained string density is physically Since max achievable. S79. In order to analyze the working principle of the faucet aerator, we will rely on the figure shown below where the aerator is presented in its generic form with all relevant parameters for the analysis. We will take into consideration the water streams before and after mixing with the air. So, according to the energy conservation law, we have the following relation valid: 1 1 4π 3 2 r N v2 + p0 S 2 x2 ρ S1 x1v12 + p0 S1 x1= 4π r 2 Nσ + ρ S2 x2 − 2 2 3 , (79.1)

where ρ is the water density, S1 and S2 are the water streams crosssectional areas before and after the mixing with the air, respectively, x1 and x2 are their corresponding widths of the analyzed water stream slabs, v1 and v2 are the corresponding water stream velocities, p0 is the atmospheric pressure, r is the average radius of the air bubbles in the mixed water stream, N is the number of the bubbles in the observed water stream slab, and σ is the water surface tension.

Solutions

225

Due to the simultaneous motion of the observed water stream slabs, we have the following fulfilled:

x1 x2 = v v2 . (79.2) 1 According to the mass conservation law of the slabs, we have the following:

4π 3 = r N ρ S1 x1 ρ S 2 x2 − 3 , (79.3) where the air mass has been considered negligible. The combination of equations (79.1) and (79.2) gives: S1 x2

p p v13 v1 σ 4π 3 2 + 2 0 S1 x2 = r N v2 + 2 0 S 2 x2 8π r 2 N + S 2 x2 − v2 v2 ρ ρ 3 ρ .(79.4)

Also, the combination of equations (79.2) and (79.3) gives:

v1 4π 3 ρ S= ρ S2 x2 − r N 1 x2 v 3 . (79.5) 2 The last equation may be rearranged into:

4π 3 r Nv2 3 x2 = S 2 v2 − S1v1 . (79.6) By substituting this value into the equation (79.4), we have the following:

(

S1v1 v12 − v22 S 2 v2 − S1v1

) + 2p

6σ = ρ ρ r . (79.7) 0

The average radius of the air bubbles is according to the last equation given by:

r=

6σ

1 ρ S1v1 v − v22

(

2 1

S 2 v2 − S1v1

) + 2p

0

ρ , (79.8)

where the following condition is obviously fulfilled:

Nonstandard Problems in General Physics with Solutions

226

v S 1< 1 < 2 v2 S1 . (79.9) Further, according to Newton’s second law of motion, we have the following fulfilled:

p0 ( S 2 − S1 )= τ ρ S1 x1 ( v1 − v2 )

, (79.10)

where τ is the time that water stream needs to move from one to another position, that is, where the following is valid:

τ=

x1 v1 , (79.11)

so equation (79.10) becomes:

p0 ( S 2 −= S1 ) ρ S1v1 ( v1 − v2 )

. (79.12)

From the last equation, one can find the aerated water stream velocity as:

v2= v1 −

p0 ( S 2 − S1 )

ρ S1v1

. (79.13)

By substituting equation (79.13) into equation (79.8), we have the following:

r=

6σ p0

1 p (S − S ) ( S2 − S1 ) 2v1 − 0 2 1 ρ S1v1 p (S − S ) S 2 v1 − 0 2 1 − S1v1 ρ S1v1

+2 . (79.14)

If we take into account that the volumetric flow rate of the water is given by:

±=

1 1

, (79.15)

equation (79.14) becomes:

r=

ρ Q 2 − p0 S1S2 6σ p0 4 ρ Q 2 − p0 S1 ( 3S 2 − S1 )

. (79.16)

Solutions

227

In order to aerate the water stream, the air bubbles must have physically real value, that is, r > 0 must be fulfilled. Therefore, the following conditions must be fulfilled:

ρ Q 2 − p0 S1S2 > 0 and (79.17)

ü ρ Q 2 ℵp0 S1 ( S 2 S1 )

,

(79.18)

or equivalently:

ρ Q 2 > p0 S1S 2 and (79.19) 3S − S ρ Q 2 > p0 S1 2 1 4 . (79.20)

It is obvious that the first condition, given by the inequality (79.19), is more stringent; therefore, for aerating the water stream, the water volumetric flow rate must have the following minimum value:

Qmin =

p0 S1S 2

ρ

. (79.21)

According to the equation (79.16), the maximum air bubble radius is ρ Q 2 p0 S1S2 and it is equal to: achieved for

rmax =

3σ 2 p0 . (79.22)

Now, in order to find air bubble concentration, we will refer to the equation (79.6) that can be rearranged in the following form:

3 ( S 2 v2 − S1v1 ) N N = = n= 4π r 3 S 2 v2 , (79.23) S 2 x2 V2 where V2 is the volume of the aerated water stream observed slab. The volumetric part of the air bubbles in the aerated water stream is given by:

ε=

4π 3 rn 3 , (79.24)

which after combining with the equations (79.15) and (79.23) gives:

Nonstandard Problems in General Physics with Solutions

228

ε = 1−

Q S 2 v2 . (79.25)

Taking into consideration equation (79.13), the last equation becomes:

ε = 1−

ρ S1Q 2

ρ S2Q 2 − p0 S1S2 ( S2 − S1 )

. (79.26)

The maximum possible water savings is obtained for ρ S2Q 2 p0 S1S2 ( S2 − S1 ) , or for the water stream volumetric flow rate that fulfills:

Q

p0 S1S 2 ( S 2 − S1 )

ρ S2

, (79.27)

and the maximum possible water savings is:

ε = 1−

S1 S 2 . (79.28)

S80. Since the rods of this simple heat engine have very small heat capacities, they reach the local ambient temperature the moment they come into it. So, as presented in figure shown below, the moment the rods get into the ambient with the local temperature, they change their lengths due to the thermal expansion or contraction. The change in the rod lengths is given by:

= ∆l α l (T1 − T2 ) , (80.1) where l is the rod length, α is the thermal expansion coefficient of

T >T

2 ) are the local ambient temperatures. the rods, and T1 and T2 ( 1 Since the pivot of this simple heat engine is firmly fixed, due to the change in rod lengths, the moment the upper rod enters the colder ambient (from T1 to T2); there will be change in its potential energy, that is, there will be decrease in the potential energy of the upper rod in the following way:

∆E12 =

1 mg ∆l 2 , (80.2)

Solutions

229

where m is the mass of the rod and g is the Earth’s gravitation field acceleration. Similarly, by entering the warmer ambient of the lower rod (i.e., from T2 to T1), there will be also decrease in the potential energy of the lower rod in the following way:

∆E21=

1 mg ∆l 2 . (80.3)

The total decrease in the potential energy of this simple heat engine in the moment when rods change the ambient temperature is thus given by:

∆E = ∆E12 + ∆E21 = mg ∆l .

(80.4)

Since two new rods change their temperature every Δt seconds, which is given by:

2π ∆t = nω , (80.5)

where n is the total number of rods in the heat engine and ω is the angular velocity of the heat engine, the overall power of this engine is thus given by:

P=

By combining the last three equations, we have:

= P

∆E ∆t . (80.6) 1 nmgω∆l 2π , (80.7)

which further by substituting equation (80.1) into the last equation gives:

1 = P nmgωα l (T1 − T2 ) 2π .

(80.8)

230

Nonstandard Problems in General Physics with Solutions

Due to the conservation of energy, we also have: P = M ω , (80.9)

where M is the torque generated by this simple heat engine. Finally, the combination of the last two equations gives:

1 = M nmgα l (T1 − T2 ) 2π .

(80.10)

V E x, E y

1 2 , it is a good approximation to take S81. Since it is valid the trajectory of a charged particle to be a straight line except in the immediate vicinity of the slit. Here the particle is acted upon by a transverse electric field which bends the trajectory. This transverse electric field may be found by use of the equation ∇ ⋅ E = 0 at the center of the slit. Therefore, we have: ∂Ex E2 − E1 ≈ ∂x d , (81.1)

where d is the lens thickness. Expanding Ez in the neighborhood of z = 0 , we may write:

Ez ≈ kz

, (81.2) where k is the proportionality factor. Thus ∇ ⋅ E = 0 implies:

E2 − E1 d . (81.3) A particle of charge q entering the slit at height z is therefore acted upon by a force: z Fz ≈ −q ( E2 − E1 ) d . (81.4) k ≈−

The force acts for a following time:

d v , (81.5) where v is the velocity of the particle. The net momentum given is then: z ∆= p Fzτ ≈ −q ( E2 − E1 ) v . (81.6)

τ=

Solutions

231

According to the trajectory geometry shown in the figure below, this deflects the trajectory by the angle:

δ= −

∆p p , (81.7)

where p is the particle momentum. The particle will, therefore, intersect the horizontal axis at a point:

y=

z

β , (81.8)

or equivalently:

1 β = y z . (81.9) From the triangle geometry, we have: β= δ − α and (81.10)

α=

z x . (81.11)

Combining the last three equations, we have:

1 1 q ( E2 − E1 ) + = x y pv . (81.12) Since we have:

1 2 mv = qV 2 and (81.13) p = mv , (81.14)

Nonstandard Problems in General Physics with Solutions

232

where m is the particle mass, equation (81.12) finally becomes:

1 1 E2 − E1 + = 2V . (81.15) x y S82.

We are going to observe the situation when a big ball is bounced from the surface and then collides with a smaller ball. According to the momentum conservation law, we have the following:

Mv − mv = mu + Mv1

, (82.1)

where M is the mass of the bigger ball, m is the mass of the smaller ball, v is their mutual velocity slightly before the impact with the surface, u is the smaller ball velocity after the impact, and v1 is the bigger ball velocity after the impact. The ball velocity just before the impact with the surface is given by:

v = 2 gh , (82.2)

where g is the Earth’s gravitational field acceleration and h is the balls height. The velocity of the bigger ball after the collision is according to the equation (82.1) given by:

m v1 = v − (v + u ) M . (82.3) Further, according to the energy conservation law, we have: 1 1 1 1 Mv 2 + mv 2 = Mv12 + mu 2 2 2 2 2 . (82.4) Substituting equation (82.3) into the equation (82.4), we have the following for the small ball velocity after the impact:

m M u=v m 1+ M . (82.5) The last expression has no maximum for any particular value of the mass ratio m / M . However, it is clearly seen that it increases when the mass ratio m / M decreases, and in the end for m / M → 0, we have maximum small ball velocity: 3−

umax = 3v , (82.6)

Solutions

233

which finally leads to the maximum possible height of:

H max = 9h (82.7)

that the small ball may reach. In the case when we have three balls, we set the following equations:

M 1v − M 2 v = M 1v1 + M 2 v2 and (82.8) M v − mv = mu + M 2 v3 , (82.9) 2 2

based on the momentum conservation law for both collisions, where M1 and M2 are the corresponding masses of the bigger balls with their velocities v1 and v2 after the first collision between these balls, m is the mass of the small ball, v is the velocity of all three balls just before the impact with the surface, u is the velocity of small ball after the collision between small ball and the ball with the mass M2, and v3 is the final velocity of the ball with the mass M2. Further according to the energy conservation law, we have the set of following equations:

1 1 1 1 M 1v 2 + M 2 v 2 = M 1v12 + M 2 v22 2 2 2 2 and 1 1 1 1 M 2 v22 + mv 2 = mu 2 + M 2 v32 2 2 2 2 .

(82.10) (82.11)

By using the previous procedure, we obtain the velocity v2 as:

M2 M1 v2 = v M 1+ 2 M 1 . (82.12) For the velocity v3, we obtain the following value: m v3 = v2 − (v + u ) M2 . (82.13) 3−

By solving the presented set of the equations, we obtain the following for the velocity of the small ball after its collision with the ball with the mass M2:

Nonstandard Problems in General Physics with Solutions

234

M 23 − 2 M1 m +1− M M2 1+ 2 M1 u=v m 1+ M2 . (82.14) From here it is clearly seen that we get the highest velocity u for M 2 / M1 → 0 m / M2 → 0 and :

umax = 7v

, (82.15)

which finally leads to the maximum possible height of:

H max = 49h

(82.16)

that the small ball may reach. So, the upper ball should always be far lighter from the bottom one, in order to get the maximum possible height that the ball at the top may reach. In the case of a four balls M1, M2, M3, and m, we would have:

M 23 − 2 M1 M +1− 3 M M2 1+ 2 M1 m +1− v M3 M3 1+ M2 u=v m 1+ M3 , (82.17) with the maximum velocity: u = max

( 2 ( 2 ⋅ 3) + 1) v=

15v

. (82.18)

Therefore, we can introduce the proportionality factor kn that relates the highest possible velocity that is obtained with n balls with the highest possible velocity obtained with the n−1 balls in the following way:

Solutions

235

= k 2kn −1 + 1 , (82.19) n where the highest possible velocity with n balls is thus given by:

unmax = kn v

. (82.20)

The proportionality factor kn from the equation (82.19) can be rearranged into: n−2

= k n 2 n − 2 k 2 + ∑ 2 n − 2 −i i =1 , (82.21) for n ≥ 3 . The last equation may be rearranged into: n

kn = 3 ⋅ 2 n − 2 + ∑ 2i − 3 i =3 , (82.22) which is also valid for n ≥ 3 . After summation, we obtain: k= 2n − 1 , (82.23) n which is valid for n ≥ 2 . So, the maximum velocity is given by:

u = nmax

(2

n

)

−1 v

, (82.24) which is obviously also valid for n = 1 and the corresponding maximum height the ball at the top may reach is given by: S83.

H nmax =

(2

n

)

2

−1 h

. (82.25)

Typically, the plane electromagnetic wave can be considered as the uniform photon flux, so it will be taken into consideration in the analysis that uniform photon flux arrives perpendicularly at the vacuum–medium interface, as it is depicted in figure shown below. As it was mentioned in the problem statement, there is an ARC between the vacuum and medium, so every photon that arrives at the vacuum–medium interface will be transferred from the vacuum into the medium. According to the geometry in the figure shown below, all photons that are contained within the slab ABCD will be shifted into the slab A′B′C′D′ after time interval: ∆z n∆z ′ ∆t= = c c , (83.1)

236

Nonstandard Problems in General Physics with Solutions

where Δz is the thickness of the slab ABCD, Δz′ is the thickness of the slab A′B′C′D′, c is the vacuum speed of light, and n is the medium index of refraction. Bearing in mind that the dielectric medium is lossless and that there are no photon reflections at the vacuum–medium interface as well as no photon scattering within the medium volume, virtually all photons that are contained within the small slab AA′DD′ will be transferred into the small slab BB′CC′. Therefore, the corresponding vacuum and medium photon densities q and q′, respectively, are given with:

N q= A∆z and (83.2) N q' = A∆z ′ , (83.3) where N is the number of photons within each small slab and A is the surface of the slab side parallel to the vacuum–medium interface plane. The corresponding time-averaged vacuum and medium electromagnetic energy densities w and w′, respectively, are given by: w = ε q and (83.4)

w ' = ε q ' , (83.5) where ε = ω is the photon energy, is the reduced Planck constant, and ω is the angular frequency of the electromagnetic wave.

Taking into account last four, the ratio of these two densities is given with:

Solutions

237

w′ q′ ∆z = = w q ∆z ′ . (83.6) One can obtain the same result as one given by equation (83.6) if we simply apply the energy conservation law to the photons contained in the slab ABCD during their movement. As there are no external forces exerted on the box surfaces AD (A′D′) and BC (B′C′) and thus there is no mechanical work done over the distances Δz and Δz′, the energy stored in the slab ABCD must be the same as the energy stored in the box A′B′C′D′. Furthermore, this leads to the conclusion that the energy content of the small slab AA′DD′ is the same as the energy content of the small slab BB′CC′. If we take into account that the movement of the slab sides from AD to A′D′ and from BC to B′C′ occurs simultaneously and lasts for the time period ∆t =∆z / c =n∆z ′ / c , we have ∆z = n∆z ′ / c , which leads us further to:

q′ =n q . (83.7) The same result can be also obtained by analyzing the electromagnetic wave propagation from the vacuum through the ARC to the medium by taking into account the influence and the characteristics of the ARC. Having in mind that there are no losses, the timeaveraged vacuum Poynting vector S must be the same as the timeaveraged medium Poynting vector S′, that is, S ' = S . The relation between the time-averaged Poynting vector and the time-averaged electromagnetic energy density is given by S = vw , where v is the corresponding medium speed of light. Taking into consideration the vacuum and dielectric medium speed of light, we have S = cw and S ' = cw′ / n that further leads to the same relation for the energy densities ratio as it is given by equations (83.6) and (83.7). In order to find the intensity and direction of the pressure exerted on the dielectric surface exposed to electromagnetic radiation, we will assume that force F acts across the dielectric surface A in the same direction as the z-axis as it is presented in figure shown below. Based on the virtual work principle, we will consider the situation where, due to the radiation force action, the medium surface is moved along the virtual displacement δz, as presented in figure shown below. Ac-

Nonstandard Problems in General Physics with Solutions

238

cording to the energy conservation law, we have: Fäz = äW , (83.8) where Fäz is the virtual work done by the radiation force and δW is the virtual total change of energy in the system for which is valid:

äW = ε ( q ' q ) Aäz ( w − w ') Aäz =−

. (83.9)

By combining equations (83.7), (83.8), and (83.9), we have the following for the pressure acting on the vacuum–medium interface plane:

F P= = −ε q ( n − 1) A . (83.10) Bearing in mind that the irradiance of the incoming electromagnetic radiation is given by: E = cε q , (83.11) the pressure is then given by:

P= −

E ( n − 1) c

. (83.12)

The resultant pressure (force) is negative meaning that electromagnetic radiation pulls the dielectric surface toward the vacuum. Therefore, in compliance with Newton’s third law, there is a force with the same amplitude but opposite direction that acts on the photons in a way to “push” them when in the medium thus increasing their momentum. In order to calculate the photon momentum in medium, we will observe the photons contained in the slab with the thickness Δz and surface A. One side of the slab is located at the vacuum–me-

Solutions

239

dium interface, as it is depicted in figure shown below. All photons contained in the slab will be shifted in medium after time interval ∆t =∆z / c =n∆z ′ / c . The total momentum change Δp is equal to:

∆= p N ( p '− p )

, (83.13)

where N= qA∆z is the total number of photons in the slab, p = E / c is the photon momentum in vacuum, and p′ is the corresponding photon momentum in medium. For the total momentum change, we have: ∆p = F ' ∆t , (83.14) where F ′ = − F is the force that acts on the photons.

Finally, by combining the last three equations, we obtained the following for the photon momentum in medium:

nE c , (83.15) which represents the Minkowski version of photon momentum. The identical result can be obtained if we observe the moving transmitter of plane, monochromatic electromagnetic waves, which are submerged into the dielectric medium with the index of refraction n, as it is presented in figure shown below. The transmitter T is moving toward the receiver R of the electromagnetic radiation with very small uniform velocity u ( u c ). In the transmitter frame of reference, the emitted power and angular frequency of the electromagnetic radiation are P and ω, respectively. Power P and angular frequency ω are related as P = ω r , where r is the transmitter photon emission rate. The receiver is constructed in such a way that it collects every photon emitted by the transmitter. p' =

240

Nonstandard Problems in General Physics with Solutions

Due to the Doppler effect, the received power of the electromagnetic radiation is P′, where P ' ≠ P and received angular frequency is ω′ for which is valid:

nu = ω ' ω 1 + c . (83.16)

Having in mind that transmitter velocity is very small as compared with the speed of light, we can neglect the relativistic effects that can influence the photon reception rate and thus the power P′ and anguP′ = ω ′r ′ ω ′r , where the lar frequency ω′ are also related as= photon receiver reception rate r′ is equal to the photon transmitter emission rate r, that is, r ′ = r . According to the energy conservation law in the receiver frame of reference, we have: P=′ P + Fu , (83.17) where F is the force exerted on the transmitter in order to compensate the repulsion force caused by the photon emission. By combining last two equations, we have:

nω r c (83.18) that also leads us to the conclusion that the medium photon momentum has the Minkowski form. S84. According to the mathematical model of the chimney falling dynamics, which is presented in the figure shown below with all relevant parameters for the analysis, we have the following valid: ¨ 1 J ϕ = mgL sin ϕ 2 , (84.1) F=

where J is the corresponding chimney moment of inertia given by:

1 J = mL2 3 , (84.2)

Solutions

241 ¨

with m being the chimney mass and L its length, and where ϕ is the

chimney angular acceleration, g is the Earth’s gravitational field acceleration, and φ is the angle between the chimney and the vertical line.

These two equations give the angular acceleration of the chimney as: ¨

ϕ=

3g sin ϕ 2L . (84.3)

According to the energy conservation law, one can obtain the angular velocity of the chimney as:

3g = ϕ 2 (1 − cos ϕ ) L . (84.4) The corresponding stretching stress that one can found in the chimney cross section AB is given by: 1 m ( y ) ϕ 2 + rC ( y ) − m ( y ) g cos ϕ ,ϕ ) σ ( y= S , (84.5) where y is the position of the observed chimney cross section AB from the pivot,

m( y)

is the mass of the upper part of the chimney

(from the observed cross section to the chimney top),

C( y)

rC ( y )

is the

position of the center of mass of the upper part of the chimney from the pivot, and S is the chimney cross-sectional area given by:

Nonstandard Problems in General Physics with Solutions

242

(

)

= S π R2 − r 2 , (84.6) with R being the outer and r being the inner radius of the chimney. The mass of the upper part of the chimney is given by: y m (= y ) m 1 − L , (84.7) whereas the position of the center of mass the chimney from the pivot is given by:

C( y)

of the upper part of

1 rC = ( y) (L + y) 2 . (84.8) The last four equations give: mg

3y2

2y

− cos ϕ + 5cos ϕ − 3 σ ( y, ϕ ) = 2 (1 − cos ϕ ) − L 2π ( R 2 − r 2 ) L

. (84.9)

In order to have chimney stretching according to the last equation, the following condition must be fulfilled:

≥

2 cos 3 (1 cos

)

− . (84.10)

Let us calculate the torque at a chimney cross section at a distance y, observed from the chimney pivot. In order to do so, we will observe separately the lower and the upper chimney part. Therefore, we have the following fulfilled: ¨ y y − M ( y, ϕ ) + m g sin ϕ + τ ( y, ϕ ) Sy = J ( y )ϕ L 2 and (84.11) ¨

= τ ( y, ϕ ) S m ( y ) g sin ϕ − m ( y ) ϕ rC ( y ) , where

M ( y, ϕ )

is the corresponding torque,

J ( y)

(84.12)

τ ( y, ϕ )

is the cor-

responding shear stress, and is the moment of inertia of the lower part of the chimney given by:

1 y J ( y ) = m y2 3 L . (84.13)

Solutions

243

Taking into account above-presented equations, we have the following for the torque

M ( y, ϕ )

:

2

y 1 mgy 1 − sin ϕ 4 L . (84.14) This is the torque acting on the lower part of the chimney and the opposite is the one from the one shown in the figure above for the upper part of the chimney. For the upper part, we see that the stress due to the torque and due to the longitudinal force will be summed at point A and subtracted at point B, as presented in figure shown below. Therefore, the fracture occurs at point A (the lower side of the chimney cross section during the fall) if the cause of the fracture is the tensile stress. According to the equation (84.14), the torque has the maximum value for: 1 y= L 3 , (84.15) M= ( y, ϕ )

which represents one of the critical points where fracture can occur, of course with the addition of stress due to longitudinal forces. Further, for the second moment of inertia of the chimney cross section

π

(

)

= JR R4 − r 4 4 . (84.16) Due to the acting torque, there is a stress present in the chimney cross section, as presented in figure shown below that depends on the radial distance ρ from the chimney center in the following form:

σ M ( y, ϕ , ρ ) =

M ( y, ϕ ) JR

ρ . (84.17)

The total stress acting on the chimney cross section is given by:

σ T ( y= , ϕ , ρ ) σ ( y, ϕ ) + σ M ( y, ϕ , ρ )

.

(84.18)

The total stress has its extremum values at the points A and B, that is, for ρ = R that are given by:

Nonstandard Problems in General Physics with Solutions

244

σ T ( y, ϕ ) = −

2 3 y 2 (1 − cos ϕ ) 2 y cos ϕ mgR sin ϕ mg y − + 5cos ϕ − 3 ± y 1 − 4 4 L2 L 2π R 2 − r 2 L π R −r

(

)

(

)

(84.19)

,

where the positive sign stands for point A and the negative sign stands for point B. The condition that the stretching fracture is reached must be:

max {σ T ( y, ϕ )} = σ , (84.20)

where σ is the tensile strength of the material the chimney is made of.

Let us now calculate the shear stress in the corresponding chimney cross section according to the equation (84.12), which after the rearrangement gives:

= τ ( y, ϕ )

mg sin ϕ y 3y 1 − 1 − 2 2 L 4π R − r L

(

)

.

(84.21)

From the last equation, it is obvious the bigger y the smaller

τ ( y, ϕ )

, so the maximum shear stress is for y = 0 , or:

τ max =

mg sin ϕ 4π R 2 − r 2

(

) . (84.22)

Solutions

245

In order to have a shear fracture, it must be satisfied:

τ max = τ , (84.23)

where τ is the chimney material shear strength. However, the above function presented in equation (84.21) has an extremum at the following point:

y=

2 L 3 . (84.24)

At this point, we have the maximum shear stress, but in the opposite direction, which is three times smaller than the one at the bottom of the chimney. Therefore, most likely, due to shear to fracture, it will come at a lower base. Based on the figure at the beginning of the solution of this problem, let’s find the chimney reaction force that is parallel to the chimney axis in the following way:

1 = N mg cos ϕ − mϕ 2 L 2 . (84.25) Taking into account equation (84.4), we have: 3g = ϕ 2 (1 − cos ϕ ) L . (84.26) The last two equations yield:

1 = N mg ( 5cos ϕ − 3) 2 . (84.27) Similarly, for the chimney reaction force that is normal to the chimney axis, we have the following: ¨ 1 = N ⊥ mg sin ϕ − mL ϕ 2 . (84.28) By substituting equation (84.3) into the last equation, we have the following:

N⊥ =

1 mg sin ϕ 4 . (84.29)

The vertical reaction force at the chimney pivot point is given by:

= N N cos ϕ + N ⊥ sin ϕ y ,

(84.30)

246

Nonstandard Problems in General Physics with Solutions

which after substituting the corresponding values for the reaction forces, one has:

1 1 = cos ϕ ( 5cos ϕ − 3) + mg sin 2 ϕ Ny 2 4 . (84.31) The chimney will detach from the ground at the moment when Ny = 0 has been reached. The corresponding critical angle is thus given by: 1 = ϕC cos −1 = 70.5° 3 . (84.32) S85. Consider sa point-like source of light P on the Sun as it is presented in figure shown below. The image of this source is spread over a region of linear dimension given by: λd ∆x ≈ D + D , (85.1) where D is the pinhole diameter, λ is the light source wavelength, and d is the distance of the pinhole from the photographic plate (film).

The first term of the equation (85.1) is due to the spread of rays over the pinhole opening. In addition, due to the diffraction, there is a spread in angle of approximately λ / D upon passing through the opening, and hence a further increase in the width of the image of about λ d / D . The total width of the image of the corresponding point has minimum for the following value of the pinhole diameter: D ≈ λ d , (85.2) which for λ = 500 nm and d = 10 cm gives D ≈ 220 µ m .

Solutions

247

S86. In the case of completely inelastic collisions between the beads, we will refer to the second Newton’s law of motion, where we have:

F=

∆ ( mv ) ∆t

. (86.1)

Since the speed of the beads, that is, the speed of the shock wave is constant, we have v = const. , so the last equation becomes:

F =v

∆m ∆t . (86.2)

The overall mass of the moving beads increases for the mass of an individual bead mass m every period of time that is equal to ∆t =d / v . So, the last equation becomes:

m m 2 = F v= v d d v . (86.3) Therefore, the speed of the beads in the case of the inelastic collisions is given by:

Fd v= m . (86.4) In the case of perfectly elastic collisions between the beads, we will refer to the energy conservation law, where we have: 1 1 Mu 2 + Fd= ( M + m) u2 2 2 , (86.5) where M is the overall mass of the moving beads and u their speed. Finally, according to the last equation, the speed of the beads in the case of the perfectly elastic collisions is given by:

u=

2 Fd m . (86.6)

Nonstandard Problems in General Physics with Solutions

248

S87.

The maximum velocities of the outer beads are reached when the velocity of the inner bead is smaller than the velocities of the outer beads. In this case, the inner bead cannot collide with the outer beads and thus cannot transfer its momentum to the outer beads. Therefore,

v < v ,v

1 3 we have the situation as presented in figure below where 2 must be fulfilled, with v1, v2, and v3 being the terminal velocities of beads with the masses m1, m2, and m3, respectively. According to the momentum conservation law, we have:

m2 v = m2 v2 + m3v3 − m1v1

Since it is valid simplify into:

.

m2 m1 , m3

and

(87.1)

v2 < v1 , v3 ,

the last equation may

m2 v ≈ m3v3 − m1v1 . (87.2)

Similarly, according to the energy conservation law, we have:

1 1 1 1 m2 v 2 = m2 v22 + m3v32 + m1v12 2 2 2 2 , (87.3) which also may be simplified into:

m v 2 ≈ m3v32 + m1v12 2 . (87.4) Based on equations (87.2) and (87.4), one can find the corresponding velocities as:

m2 m3 v1 ≈ v m1 ( m1 + m3 ) and

(87.5)

m1m2 v3 ≈ v m3 ( m1 + m3 ) . (87.6)

Solutions

S88.

249

We will assume that during the fencing when the impact between the stick occurs, there is a net transfer of momentum P between the sticks at the impact place, as presented in figure shown below. Simultaneously, we will assume that during the fencing, there is a net impact transfer P′ between the stick and the hand holding it at one end. According to the momentum conservation law, we have: mv , (88.1) P − P′ = where m is the stick mass and v is the velocity of the stick center of mass. The velocity of the center of mass of the stick is then given by: P − P′ v= m . (88.2)

Also, based on the angular momentum conservation law, we can write:

l l P x − + P′ = Jω 2 2 , (88.3) 2

where l is the stick length, x is the position of the impact, J = ml /12 is the stick moment of inertia, and ω is the stick angular velocity. The angular velocity of the stick is then given by:

ω=

6 P ( 2 x − l ) + 6 P′l ml 2

. (88.4)

In order to avoid that one who holds the stick at its end does not feel the impact, the hand must stay at the place, that is, the velocity of the stick at the end that is held by hand must be zero. Therefore, for the hand velocity, we have:

Nonstandard Problems in General Physics with Solutions

250

l vH =− v ω = 0 2 . (88.5) The combination of equations (88.2), (88.4), and (88.5) gives: 3x = P′ P 1 − 2l . (88.6) Further, the hand must not feel any force, so one must also fulfill the following condition: P′ = 0 . (88.7) This is achieved for the following position of the sticks impact:

2 x= l 3 . (88.8) S89.

Figures below show the situation just before (left) and after (right) the impact of the falling sack of sand with the plank. The velocity of the falling of the sack just before the impact is given by:

v = 2 gH , (89.1) where g is the Earth’s gravitational field acceleration and H the falling height of the sack.

According to the angular momentum conservation law, one has the following: l l l m2 v = m2u + m2u + J ω 2 2 2 , (89.2)

Solutions

251

where m1 is the mass of the plank, m2 is the mass of the sand sacks, J = m1l 2 / 12 l is the plank length, u is the sacks velocity, is the plank moment of inertia, and ω is the plank angular velocity. Since the sack of sand falls onto the plank, one can take into consideration that the impact is perfectly inelastic, so the following condition is fulfilled: l u =ω 2 . (89.3) Based on the last two equations, we have: 3m2 u= v m + 6 m 1 2 . (89.4) The height that the second sack will reach is given by:

u2 h= 2 g . (89.5) The combination of equations (89.1), (89.4), and (89.5) finally gives: 2

3m2 h=H m1 + 6m2 . (89.6) S90. At the moment when the sack with the mass m2 touches the ground, we will have the situation as presented in the figure shown below. The velocity of both sacks will be found on the basis of the energy conservation law in the following way: 1 1 m= m1v 2 + m2 v 2 2 gH 2 2 , (90.1) where v is the corresponding velocity, m1 and m2 masses of the sand sacks, g is the Earth’s gravitation field acceleration, and H is the height of the table. The impact velocity of the sacks is according to the last equation given by:

v=

2m2 gH m1 + m2

. (90.2)

Nonstandard Problems in General Physics with Solutions

252

Since the sacks are full of sand, the impact of the sack with the mass m2 is perfectly inelastic. Therefore, after the impact, we have the situation as presented in the figure shown below. According to the moment conservation law, we have the following:

m= 1v

( m1 + m2 ) u , (90.3)

where u is the velocity of the sacks after the impact. This velocity is according to the last equation given by: m1 u= v m + m 1 2 . (90.4)

The height h that the sack with mass m2 will reach, which situation is presented in the figure shown below, will be found on the basis of the energy conservation law as:

1 1 m1u 2 + m2u 2 = m2 gh 2 2 .

(90.5)

Solutions

253

By substituting equation (90.4) into the last equation, one has:

m12 v2 h= m2 ( m1 + m2 ) 2 g .

(90.6)

Finally, by taking into account equation (90.2), we have: 2

m1 h= H m1 + m2 . (90.7) S91. In order to analyze this problem, we will refer to the figure shown below where it is presented the position of the rope wrapped around the wires with all relevant parameters for the analysis. Since the wires are frictionless, only possible reaction force must be normal to the wires. Therefore, the resultant force of both tension forces in the rope at the point of contact C must be normal to the wire or AC ⊥ CD . This is achieved if the angles satisfy the following condition: ∠DCE . (91.1) ∠BCD = Also, from the triangle geometry, we have:

π

α

∠ACB = − 2 2 , (91.2) where α is the angle between the wires.

Also, from the geometry, we have:

Nonstandard Problems in General Physics with Solutions

254

π

∠ACB + ∠BCD = 2 . (91.3) From the last three equations, we have:

α

∠BCD = ∠DCE = 2 , (91.4) or equivalently due to the symmetry, the following: ∠CEB = α . (91.5) ∠BCE = From the triangle ∆BCE geometry, we also have:

π

α

∠BEC = − 2 2 . (91.6) Further, according to the law of sines, we have:

BC BE = sin ∠BEC sin ∠BCE . (91.7) By substituting equations (91.5) and (91.6) into the last equation, one has: α α 2sin cos sin α sin α 2 2 = = BC = BC = BE BC α α π α sin − 2 2

cos

2

cos

2

. (91.8)

As the rope length is l, we have: BC + BE + CE = BC + 2BE = l , (91.9) since due to the symmetry, we have BE = CE . The combination of the last two equations yields:

BC =

l 1 + 4sin

α

2 . (91.10) The triangle ∆ABC geometry gives:

BC tan = 2 2 h , (91.11)

α

Solutions

255

where h is the wanted height difference. Based on the last two equations, we have: l h= α α 2 tan 1 + 4sin 2 2 . (91.12) The rope tension will be found according to the forces acting on the point E of the triangle ∆BCE in the following way:

∠BEC 2T cos = mg 2 , (91.13) where T is the rope tension, m is the body mass, and g is the Earth’s gravitational field acceleration. Taking into account equation (91.6), one finally has for the rope tension: mg T= π α 2 cos − 4 4 . (91.14) S92.

The minimum possible force will be achieved if the force is applied perpendicularly to the bar, in the horizontal plane at the bar end, as it is presented in the figure shown below. We will assume that when the acting force starts to increase at the moment when the bar starts to move, the bar will rotate around the center C that is positioned at the distance y from the free bar end. Further, as the bar is homogenous and uniform, we will assume that the distribution of the reaction force between the plane and the bar is also uniform. Therefore, we can write the following for the moment when the bar starts to rotate: l

F (l − y ) = ∫xdT 0

, (92.1)

where F is the acting force, l is the bar length, x is the acting position of the elementary friction force, and dT is the elementary friction force given by: dT = µ gdm , (92.2) where μ is the friction coefficient, g is the Earth’s gravitation field acceleration, and dm is the elementary mass given by:

Nonstandard Problems in General Physics with Solutions

256

m dm = dx l , (92.3) where dx is the elementary bar length.

Taking all this into consideration, equation (92.1) becomes:

F = (l − y )

µ gm

y x x xdx d + ∫ ∫ 0 0 .

l

l− y

(92.4)

By solving the integrals in the last equation, one obtains:

F=

2 µ gm y + ( l − y )

2l

l−y

2

. (92.5)

In order to find the minimum possible force, we need to find the solution of the following equation:

dF =0 d y , (92.6) or equivalently: 2 2 0 . (92.7) 2 y − 4ly + l =

The solutions of this quadratic equation are:

2 y= 1 ± l 2 . (92.8) Since y ≤ l , we have only one solution: 2 y= 1 − l 2 . (92.9)

Solutions

257

The minimum possible force will be obtained by substituting the value for y, given by the last equation, into the equation (92.5):

(

)

= F 2 − 1 µ gm . (92.10) S93. In order to find the equivalent resistance of the infinite hexagonal resistor network, we will introduce a test current source I connected between the point A and the ground that is located in the infinity of the network, as presented in the figure shown below. Due to the symmetry, this current is divided into three equal parts, thus giving the following result: 1 I'= I 3 , (93.1) where I′ is the corresponding current flowing out of the point A into three different directions, as presented in figure below.

Further, also due to the symmetry, which can be seen from the figure above, each current I′ is further divided into two equal parts thus giving:

1 1 = I″ = I' I 2 6 , (93.2) where I″ is the corresponding current that also flows into the point B. Therefore, the potential difference between the points A and B in this case is given by: 1 1 1 ' U AB = RI ′ + RI ′′ = RI + RI = RI 3 6 2 . (93.3) Due to the superposition theorem, in order to get the overall potential difference between the points A and B, one must also introduce

Nonstandard Problems in General Physics with Solutions

258

symmetrically the test current source I that flows that flows out the point B toward the point located in the infinity of the network, as presented in figure shown below. The current division in this case is symmetrical to one shown in the figure above, and it is presented in figure below. Similar to the previous case, the potential difference between the points A and B in this case is given by:

1 1 1 ″ U AB = RI ′ + RI ′′ = RI + RI = RI 3 6 2 . (93.4)

The overall voltage difference is thus due to the superposition theorem given by: ' ″ U = U AB + U AB = RI . (93.5) AB

Finally, the equivalent resistance is given by:

U AB = R I . (93.6) S94. The pressure at the contact between the drop and the plane is given by: 2σ = p ρ gh + R , (94.1) R = AB

where ρ is the liquid density, g is the Earth’s gravitational field acceleration, h is the drop height, σ is the surface tension of the drop liquid, and R is the drop radius at its top point. This pressure is equal to the following pressure:

p=

mg S , (94.2)

Solutions

259

where m is the drop mass and S is the area of the contact surface between the drop and the plain, which is given by: 2 S = π r , (94.3)

where r is the radius of the contact surface between the drop and the plane. Finally, based on the last three equations, we have:

2σ = m π r2 ρh + gR . (94.4) The smallest radius of the drop is obtained at the bottom of the drop, that is, at the contact point between the drop and the plane surface, where we have the following for the corresponding pressure: σ σ p= + r rmin , (94.5) where rmin is the corresponding minimum drop radius, for which is according to the equations (94.1) and (94.5), the following valid:

rmin =

1

ρ gh 2 1 + − σ R r . (94.6)

In order to find the magnification of a small object that is located on the contact surface between the drop and the plane, that is, at the plane surface, we will refer to the figure shown below where it is presented the drip in a form of a plano-convex lens. The focal length of such a plano-convex lens is given by:

f =

R n − 1 , (94.7)

where n is the index of refraction of the drop liquid. Since, we have the plano-convex lens, one of the principal planes (H′) is located at the vertex of the lens convex side. The other principal plane (H″) is located at the distance p from the lens plane side. This is, at the same time, the distance of the object O, which is located at the surface of the plane. This distance is given by:

= p

fh ( n − 1) h = nR n . (94.8)

Nonstandard Problems in General Physics with Solutions

260

According to the lens equation, we have:

1 1 1 + = p q f , (94.9) where q is the image distance. Taking into account the last three equations, one has:

hR q= ( n − 1) h − nR .

(94.10)

The magnification of the drop is thus given by:

M=

q = p

nR ( n − 1) h − nR

. (94.11)

In order to have maximum magnification, one can theoretically obtain infinite magnification if the denominator at the right side of the expression, given by the equation (94.11), is identically equal to zero, which is obtained for:

n h= R n − 1 . (94.12) S95. In order to find the water layer thickness, we will refer to the figure shown below where it is presented the dynamics of the surf skimmer with all relevant parameters for the analysis. According to the volume flow rate conservation law, one can write the following relation: = = whu , (95.1) Q wHv where w is the board width, H is the height of the front of the board, h is the water layer thickness, v is the skimmer velocity, and u is the water layer velocity.

Solutions

261

The corresponding mass flow rate is given by: m = ρ Q , (95.2) where ρ is the water density. Further, according to the second Newton’s law of motion, one can write:

= F m ( u − v ) , (95.3) where F is the force acting on the water and consequently acting on the board. The force, which is normal to the board and produces such force on the water layer, is given by: F N= sin θ , (95.4) where θ is the angle of the board with respect to the horizontal line, which is given by:

H −h = sin θ ≈θ l , (95.5) where l is the board length. This normal force is a component of the skimmer weight thus giving the following relation: = N mg cos θ ≈ mg , (95.6) where m is the skimmer mass and g is the Earth’s gravitational field acceleration. If we take into consideration equation (95.3) and substitute all relevant parameters given by the other equations, we have:

H = F ρ wHv 2 − 1 h . (95.7) Further by combining the last equation, the equations (95.5) and (95.6) with the equation (95.4), we obtain the following: H mg ≈ ρ wlv 2 h . (95.8)

Nonstandard Problems in General Physics with Solutions

262

Finally, by taking into account that the board is approximately rectangular in shape, its area is then given by: S ≈ wl , (95.9) thus giving the water layer thickness of:

ρ HSv 2 h≈ mg . (95.10) S96. The force that acts on one facet of the charged cube is given by: = F ∬σ E = ⋅ dS ⋅ σ ∬E ⋅ dS A A , (96.1) where E is the electric field vector at the surface of one cube facet of area A, dS is the elementary surface vector, and σ is the surface charge density of the cube. According to the Gauss’s law, we can write the following for the electric flux through the closed surface S that just covers the charged cube:

, (96.2)

where E is the electric field vector at the surface S that represents also the cube that just covers the charged cube, dS is the elementary

surface vector, Q is the overall cube charge, and ε0 is the vacuum permittivity. The cube charge is given by:

Q = 6σ a , (96.3) 2

with a being the cube side length and σ being the surface charge density of the cube. Due to the symmetry, the electric flux through a single cube facet is thus given by:

1 Q σ a2 = φ = Ö = 6 6ε 0 ε 0 . (96.4) This electric flux may be divided into two portions as follows: φ = Ö′ + φ ' , (96.5) with Φ′ being the electric flux part through the observed cube facet from other five facets of the cube and ϕ′ being the electric flux part through the observed cube facet that originates from this charged

Solutions

263

cube facet. In order to find the electric flux through the cube facet that originates only from this facet, we will observe a closed surface that just covers this facet. For the electric flux through the closed surface S′ that also has cuboid shape, which just covers the cube facet, we can write the following:

(96.6)

where ϕ′ is the electric flux through the outer side of the observed cube facet and ϕ″ isthe electric flux through the inner side of the ob-

served cube facet, E ' is the electric field vector originating from the observed cube facet at the position of the observed closed surface, dS ' is the elementary surface vector of the closed surface S′, and Q′ is the charge of the observed cube facets given by:

Q ' = σ a . (96.7) 2

Since due to the symmetry we have φ ′ = φ ′′ , then, according to the equations (96.7) and (96.7), we have:

σ a2 φ′ = 2ε 0 . (96.8) For this flux, it is also valid:

= φ′

∬E ⋅ dS A

, (96.9)

E where is the electric field vector at the surface of one cube facet of area A, dS is the elementary surface vector. Finally, by combining equations (96.1), (96.8), and (96.9), we obtain the acting force in the following form:

F=

σ 2a2 2ε 0 . (96.10)

It is very simple to show in an analogous way that in the case of a tetrahedron, the corresponding force acting on a single facet is given by:

264

Nonstandard Problems in General Physics with Solutions

3σ 2 a 2 8ε 0 . (96.11) S97. The starting position of the body is presented in the figure shown below. The string that is pulled through the small hole H on the inclined plane is pulled very slowly. Therefore, the inertial forces in the analysis of the body motion may be omitted. In the beginning, the string lies horizontally. When the motion of the body begins the trajectory of the body is semicircular until it reaches the hole. The normal reaction of the elevated plane on the body, that is, the body weight projection onto the Y′ axis is given by: N = mg cos α , (97.1) F=

where m is the body mass, g is the Earth’s gravitational field acceleration, and α is the plane angle of inclination.

In order to analyze the body motion, we will refer to the figure given below where it is presented the body trajectory in the elevated plane. One can notice that if omitting the inertial forces, there are three forces acting on the body. The friction force which is tangential on the trajectory at the actual body position is given by: F = µ N , (97.2) with μ being the friction coefficient between the body and the plane.

Solutions

265

The forces that act on the body must be in balance, thus along Xaxis, we have: T cos ϕ = F sin θ , (97.3) where T is the string tension, φ is the angle between the string and the horizontal line (X-axis), and θ is the angle between the radius vector of the actual body position with respect to the center of the semicircular trajectory (point C) and the horizontal line (X-axis). Based on the triangle geometry, we have: θ = 2ϕ . (97.4) Regarding the force components along Y-axis, we have the following relation:

sin α T sin ϕ + F cos θ . mg=

(97.5)

By substituting string tension from equation (97.3) and the angular relation from equation (97.4) into the last equation, the following is obtained: mg sin α = F . (97.6) Then, after substituting equations (97.1) and (97.2) into the last equation, one finally obtains: µ = tan α . (97.7) S98. In order to find the solution of this problem, we will assume that the weight distribution of the sheet onto the roof is uniform. When the temperature decreases during the night from the maximum temperature T1 down to the minimum temperature T2, we have the situation as presented in figure shown below. Due to the sheet weight component and the friction forces caused by the sheet shrinkage, there will be a set of sheet points that are motionless, or the parts of the sheet above and below of these motionless points will move in different directions. Due to the force balance, since the sheet movement is very slow, we will have the following: mg sin ϕ + F ″ = F ' , (98.1)

where m is the sheet mass, g is the Earth’s gravitational field acceleration, φ is the roof angle, F″ is the friction force of the lower sheet part, and consequently, F′ is the friction force of the upper sheet part

Nonstandard Problems in General Physics with Solutions

266

These friction forces are given by: F ′ = µ N ' and (98.2)

′′ µ N ″ , (98.3) F= where μ is the friction coefficient between the sheet and the roof and N′ and N″ are the normal reaction forces of the roof onto the lower and upper parts of the sheet, respectively, that are further given by:

l−x mg cos ϕ l and (98.4) x N ′′ = mg cos ϕ l , (98.5) N′ =

with l being the sheet length and x being the position of the sheet motionless points as presented in the figure shown above, and according to the last five equation, it is given by:

= x

l ( µ − tan ϕ ) 2µ . (98.6)

Due to the temperature decrease, each part of the sheet (upper and lower) will be shorter for the following lengths:

∆l =' α ( l − x )(T1 − T2 )

and (98.7)

= ∆l ″ α x (T1 − T2 ) , (98.8) with α being the lead coefficient of thermal expansion. The center of mass of the sheet will be thus shifted down, that is, slipped down from the position C1 to the position C2 thus passing the distance that is given by: 1 ∆s12= ( ∆l ′ − ∆l ″) 2 . (98.9)

Solutions

267

Taking into account the corresponding parameter values, we have:

= ∆s12

l α (T1 − T2 ) tan ϕ 2µ . (98.10)

The opposite situation occurs, that is, when the temperature increases during the day from the minimum temperature T2 up to the maximum temperature T1. The corresponding situation is presented in figure shown below. However, although the situation is opposite, the center of mass of the sheet will be shifted down for the same distance, that is, the following is valid:

∆s21 = ∆s12 . (98.11)

So, the overall shift of the center of mass of the sheet during a whole day will be given by:

∆s = ∆s21 + ∆s12 =

l

µ

α (T1 − T2 ) tan ϕ

.

(98.12)

Since the same shift of the sheet center of mass occurs at each of in total N days, the overall shift of the sheet is thus given by: s= N ∆s (98.13) that after substituting equation (98.12) into the last equation finally gives:

= s

Nl

µ

α (T1 − T2 ) tan ϕ

. (98.14)

Nonstandard Problems in General Physics with Solutions

268

S99.

In order to remain in orbit, the centripetal force acting on the “satellite” due to its orbital motion must exactly balance the gravitational attraction, or equivalently:

R +l

R +l

R

R

2 ∫ rω dm =

M

∫G r

dm

2

, (99.1)

where R is the Earth’s radius, l is the rope length, r is the actual position from the center of the Earth of an elementary mass dm of the rope, ω is the angular velocity of the Earth’s rotation, G is the universal gravitational constant, and M is the mass of the Earth. Since the mass of the rope per unit length, we also have: dm = λ dr , (99.2) with λ being the rope mass per unit length. The last two equations give: R +l

R +l

M

∫ rω λdr = ∫ G r R

2

2

R

λ dr

. (99.3)

By solving the integrals in the last equation, one obtains the following quadratic equation:

1 2 1 2 1 2 GM − ω ( R + l ) − R= R R + l , (99.4) 2 which after the rearrangement gives:

2GM Rω 2 . (99.5) The solutions of the above-given quadratic equation are:

( R + l )( 2 R + l ) =

3 1 2GM l= − R± − 2R2 9R2 + 4 2 R 2 2 ω .

(99.6)

Since only positive lengths are physically real, we have:

1 8GM l= R 1 + 3 2 − 3 2 Rω . (99.7) Further, as the Earth’ gravitational field acceleration is given by: GM g= 2 R , (99.8)

Solutions

269

and the rate of rotation of the Earth is given by:

2π T , (99.9) with T being the day length, the equation becomes:

ω=

1 2 gT 2 l= R 1+ 2 − 3 2 π R .

(99.10)

The Earth’s radius is approximately equal to R = 6400 km , so the 8 rope length is l ≈ 1.44 × 10 m .

S97.

In order to find the electric field in the charged sphere center, one must find the equivalent arrangement of the charge distribution that due to the superposition theorem make the same electric field distribution. Therefore, instead of a sphere with a hole, one can equivalently use two uniformly charged spheres, one bigger at the position of the sphere with the hole having the same charge density ρ and the another smaller filling the hole and having the charge density −ρ. Therefore, the electric field in the center of the sphere isn’t affected by the bigger sphere but just by the smaller charged sphere. The electric field in the sphere center is directed toward the hole and, according to the situation that is presented in the figure shown below, we have the following for the corresponding electric field:

4π 3 ρr 1 3 ρr3 = E1 = 4πε 0 d 2 3ε 0 d 2 , (100.1) with ε0 being the vacuum permittivity, ρ being the charge density, r being the hole radius, and d being the distances between the centers of the sphere and the hole.

270

Nonstandard Problems in General Physics with Solutions

In order to find the electric field in the hole center, we have the equivalent situation as presented in the figure shown below. The only charge that affects the electric field is the charged sphere located in the center of the bigger sphere, having the same radius as the distance between the centers of the sphere and the hole. Therefore, the electric field is given by:

4π ρd 3 ρ 1 3 d = E2 = 2 4πε 0 d 3ε 0 , (100.2) where the direction of the field is as presented in the figure shown below.

In order to find the electric field in an arbitrary point within the hole, we will observe the situation given in the figure shown below. Point P, in which we want to find the electric field, is located at the distance r1 from the sphere center S and at the distance r2 from the hole center H. So, only charges that are located within the spheres having the same radiuses r1 for the charge density ρ and r2 for the charge density −ρ influence the electric field in that particular point. The electric fields directions are as presented in the figure shown below and have the following intensities:

4π 3 ρr ρ r1 1 3 1 = E1 = 2 4πε 0 r1 3ε 0 and (100.3) 4π 3 ρr 1 3 2 ρ r2 = E2 = 4πε 0 r22 3ε 0 . (100.4) The overall field at the point P is according to the law of cosines given by:

Solutions

E=

E12 + E22 + 2 E1 E2 cos ϕ

271

, (100.5)

with φ being the angle between electric field vectors.

By substituting equations (100.3) and (100.4) into the last equation, one obtains:

ρ

= E 3ε 0

r12 + r22 + 2r1r2 cos ϕ

. (100.6)

The angle ∠SPH is simply given by:

π ϕ . (100.7) ∠SPH =− So, by combining the last two equations, one obtains:

ρ

= E 3ε 0

r12 + r22 − 2r1r2 cos ∠SPH

.

(100.8)

Further, also according to the law of cosines for the triangle ÄSPH, we have:

d=

r12 + r22 − 2r1r2 cos ∠SPH

. (100.9)

Finally, by combining the last two equations, we have: ρd E= 3ε 0 , (100.10) which has constant value regardless of the position of the point P within the hole

INDEX A absolute black bodies 5 absolute error 75 acceleration 1, 2, 3, 6, 9, 13, 14, 16, 23, 29, 33 altitude 2, 22 Ammeters 23 angular acceleration 44, 56, 121, 125, 141, 200, 241 angular frequency 153, 154, 187, 236, 239, 240 angular momentum 23 Angular momentum conservation law 94, 133, 249, 250 angular position 16 angular resolution 5, 71 angular size 11 angular velocity 9, 17, 18, 27, 29, 37, 38, 51, 57, 93, 94, 96, 99, 100, 112, 117, 120, 121, 122, 132, 140, 143, 146, 148, 190, 192, 193, 211, 213, 229, 241, 249, 251, 268

antireflective coating (ARC) 31 arbitrary function 156, 157 arc mass 138 asteroids 8 astronomers 6 astronomical data 80 atmosphere mass 165 atmospheric absorption 107 atmospheric layer 95 Atmospheric layers 8 atmospheric pressure 4, 19, 29, 53, 64, 87, 164, 224 average molar mass 165 axial compression 219 B Bernoulli’s law 89 Bernoulli’s principle 52, 222 billiard table 1 buoyancy 162, 163, 186 C capillary 105

274

Nonstandard Problems in General Physics with Solutions

capillary radius 102 center of gravity 51, 52, 97, 111, 194, 196 center of mass 117, 118, 124, 187, 188, 192, 193, 205, 214, 215, 241, 242, 249, 266, 267 center of the circle 14 central voltmeter 172 centrifugal force 77, 80, 92, 111, 137, 138, 142, 143, 210 centripetal force 210, 268 chimney 31 circular loop 16 Circumference 38, 120, 124 classical statics 62 coefficient of thermal expansion 24, 29, 190, 191, 266 coherence length 107, 108 Complex movement 55, 62, 111 conservation laws 91, 149, 179 cosmic light 6, 80 cross-sectional area 8, 23, 24, 29 cuboid 169, 263 D degrees of freedom 195 diameter 2, 4, 5, 20, 28, 31 diatomic molecules 9 dielectric 25, 26, 30 differential equation 78, 79, 187 dilatation 189 Dirac hypothesis 9 distance 5, 6, 7, 8, 9, 14, 15, 17, 18, 28, 30, 31, 34, 35 Doppler effect 154, 161, 203, 240 Doppler shift 19, 20 E Earth’s crust 165, 166

Earth’s gravity field 3, 6, 13, 14, 16 elastic collision 12 electric circuit 23 electric current 160 electric field 30, 35 electric field vector 262, 263 electric flux 262, 263 electromagnetic radiation 17, 18 electromagnetic radiation intensity 17 electromagnetic wave 153, 154, 235, 236, 237 electromotive force 21, 23, 158, 172, 174, 186, 207 electromotive forces 178 electron 143 electron mass 143 elementary probability 81 elementary surface 80, 81, 82, 262, 263 Energy conservation law 47, 50, 54, 57, 64, 68, 72, 74, 77, 82, 95, 103, 117, 130, 132, 133, 138, 140, 144, 150, 154, 155, 161, 166, 168, 176, 180, 195, 215, 224, 232, 233, 237, 238, 240, 241, 247, 248, 251, 252 energy density 80, 82, 83, 149, 150, 237 equivalent mass flow 223 equivalent resistance 11, 12, 19, 27, 33, 108, 109, 110, 113, 114, 116, 158, 159, 161, 207, 208, 209, 216, 218, 257, 258 F faucet aerator 224 free electrons 143 frequency 18, 20, 25

Index

frequency shift 161 friction 1, 3, 4, 5, 6, 8, 12, 13, 14, 15, 16, 17, 21, 25, 31, 33, 34, 35 friction coefficient 1, 3, 5, 12, 13, 14, 21, 25, 33, 34, 35, 39, 40, 61, 96, 104, 121, 122, 128, 131, 142, 173, 199, 201, 255, 264, 266 Friction force 60, 61, 62, 63, 71, 72, 73, 77, 80, 120, 121, 141, 173, 201, 255, 264, 265 G gas concentration 196 generators 208, 209 geometry 68, 69, 72, 99, 107, 198, 212, 214, 231, 235, 253, 254, 265 Graphically determine 72 graphic method 5 gravitational acceleration 38, 44, 46, 51 gravitational attraction 92, 220, 221, 268 gravitational constant 7, 9, 10, 92, 101, 221, 268 gravitational field 4, 9, 29, 32 gravitational force 28 H head-on collision 1 heat conductor 10 heat transfer 10 hemisphere 6, 16 hemisphere of radius 6 Heron’s fountain 7 homogeneous sphere 96 homogenous disk 13

275

horizontal line 37, 43, 55, 72, 261, 265 Horizontal plane 61, 62, 99, 173, 255 horizontal velocity 56, 192 I ideal gas 194 image sensors 170 inclination angle 25 inelastic 13, 24, 31 inertial forces 72, 264 Infinite triangular resistance grid 12 initial velocity 49, 57, 72, 76, 113, 138, 139, 140 instantaneous combustion 24 integral calculus 27 interference 11, 29 Internal resistance 172, 174, 178 K kinetic energy 50, 54, 78, 80, 95, 132, 133, 168, 196 Kirchhoff’s law 172, 185 L labile equilibrium 3 lateral displacement 136 lens 15, 20, 30 lens equation 170, 260 lens focal point 15 light energy 17 linear velocity 120 longitudinal force 243 Lorentz transformations 154 M magnetic field 144 magnification 135, 259, 260

276

Nonstandard Problems in General Physics with Solutions

Magnus effect 210 mass 1, 2, 3, 6, 7, 8, 9, 12, 14, 15, 16, 18, 20, 23, 24, 25, 26, 27, 29, 30, 31, 32, 33, 34, 35 mass conservation law 149, 179, 225 mathematical model 55, 59, 62, 240 Maximum velocity 73, 74, 75, 197, 234, 235 mercury thermometer 10 microphone 163 Minimum value 1 Minimum velocity 6 moment of inertia 11, 38, 44, 50, 56, 94, 96, 112, 117, 121, 125, 132, 140, 192, 200, 240, 242, 243, 249, 251 momentum 47, 65, 66, 83, 84, 91, 92, 93, 94, 98, 99, 100, 112, 116, 117, 129, 133, 138, 140, 144, 146, 149, 150, 155, 161, 167, 179, 180, 183, 189, 192, 194, 203, 204, 205, 206, 230, 231, 232, 233, 238, 239, 240, 248, 249, 250 momentum conservation 47, 93, 94, 117, 133, 138, 140, 150, 167, 180, 204, 205, 232, 233, 248, 249, 250 monochromatic electromagnetic waves 19 monochromatic light 19, 26 moving charges 143 mutual angle 70 N Newton’s law of cooling 106 Newton’s second law of motion

141, 177, 190, 226 Newton’s third law 238 non-ideal voltmeter 174 O optical illusion 15 optical power 5 orbit 6, 7, 8, 35 orbiting velocity 100 oscillations 23 P Parachutist terminal velocity 52, 53 paraffin 222 parallel beams 170 parallel optical rays 69 parallel rays 69, 70 parameter 41, 42, 48, 49, 58, 60, 74, 157, 188, 197, 267 pebble 21 perimeter 13 perturbation 214 photographic lenses 20 photon 18, 19, 20, 25, 26, 31 photon emission 153, 161, 167, 239, 240 photon emission rate 153, 161, 239 photopic vision 108 Planck constant 25, 157, 161, 167, 202, 236 pneumatics 7 Poisson’s ratio 28, 219 potential energy 50, 54, 72, 124, 131, 132, 215, 228, 229 Poynting vector 237 probe 83, 84, 85 propagation of light waves 11

Index

Q quadratic equation 116, 126, 127, 134, 142, 178, 181, 217, 219, 220, 256, 268 quantum energy 18 R radiant flux 147 radiation 68, 80, 82, 107, 108, 144, 154, 155, 157, 161, 162, 184, 237, 238, 239, 240 radiometer 17, 18 radius 1, 2, 5, 6, 9, 10, 11, 13, 14, 15, 17, 19, 20, 21, 23, 24, 25, 27, 28, 29, 31, 33, 35 radius of curvature 5, 27, 33, 72, 111, 210, 211, 213 radius of gyration 24 random phase modulation 106 refractive index 15, 25, 26, 135, 203, 205 relative motion 129 relative radial acceleration 55 Relative tangential acceleration 55 relative velocity 55 Relevant parameter 37, 43, 55, 59, 62, 68, 69, 82, 86, 92, 96, 101, 111, 122, 125, 149, 162, 173, 179, 186, 220, 221, 224, 240, 253, 260, 261 resistance 11, 12, 15, 19, 22, 23, 27, 33 resistor 11 resonator 202, 203 S self-inflation 5 shear strength 245 Sisyphus mass 198, 199

277

slopes 2 Small snowball 50 solar luminosity 6 specific heat capacity 10, 24, 106, 191 starlight spectrum 107 static coefficient of friction 3 steady-state velocity 52 Stefan–Boltzmann constant 67, 184 strain 218, 219, 220 stress 169, 218, 219, 220, 241, 242, 243, 244, 245 string density 223, 224 superposition theorem 160, 257, 258, 269 surface charge density 34 surface density 92 Surface temperature 145, 184, 185 surface tension 4, 10, 19, 20, 22, 29, 33 Surface tension 104, 163, 168, 169, 177, 222, 224, 258 T tangent line 74 tensile stress 20, 28, 169, 220, 243 terminal velocity 52, 53, 120, 131, 132, 133, 134 terminal vertical velocity 4 thermal conductivity 17, 18, 102, 145 thermodynamics 144 thermometer 10 thermometer scale 10 time dilation 154 torque 83, 84, 85, 97, 98, 111, 112, 122, 123, 124, 198, 230, 242, 243 traction force 73

278

Nonstandard Problems in General Physics with Solutions

transmission system 153 transmitter 18, 19 transversal expansion 219 transverse waves 16 triangle 12 U Uniform electromagnetic energy 82 Uniform energy density electromagnetic radiation 82 universal gas constant 9, 18 V vane linear dimensions 17 velocity 2, 3, 4, 5, 6, 8, 9, 11, 12, 13, 15, 16, 17, 18, 19, 20, 21, 22, 24, 25, 27, 29, 32, 34 vertex 12

vertical velocity 56, 61, 120, 192 virtual work principle 124, 166, 214, 237 voltage 17, 21, 30 voltmeter 17, 21, 23 volumetric concentration 183 W water density 13, 29, 34 Water stream 122, 123, 224, 225, 226, 227, 228 wavelength 108, 149, 151, 152, 162, 246 wavelength range 108 wind turbine 129, 130, 131 Y Yang’s modulus of elasticity 28