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m a Lev Tarasov Aldina Tarasova Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S M I R P U B L I S H E R S M O S C O W
L E V T A R A S O V, A L D I N A T A R A S O V A Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S M I R P U B L I S H E R S M O S C O W
Translated from the Russian by Nicholas Weinstein. First published 1973. Revised from the 1968 Russian edition. This electronic version typeset in LATEX by Damitr Mazanav. Released on the web by http://mirtitles.org in 2020. Git Repository for obtaining the source files https://gitlab.com/mirtitles/tarasov-qasp/
Contents From The Editor Of The Russian Edition 7 Foreword 9 § 1 Can You Analyse Graphs Representing The Kinematics Of Straight-Line Motion? 13 § 2 Can You Show The Forces Applied To A Body? 21 § 3 Can You Determine The Friction Force? 31 § 4 How Well Do You Know Newton’s Laws Of Motion? 37 § 5 How Do You Go About Solving Problems In Kinematics? 51 Problems 59 § 6 How Do You Go About Solving Problems In Dynamics? 61 Problems 65 § 7 Are Problems In Dynamics Much More Difficult To Solve If Friction Is Taken Into Account? 67 Problems 73 § 8 How Do You Deal With Motion In A Circle? 77 Problems 89
4 § 9 How Do You Explain The Weightlessness Of Bodies? 91 Problems 95 § 10 Can You Apply The Laws Of Conservation Of Energy And Linear Momentum? 99 Problems 115 § 11 Can You Deal With Harmonic Vibrations? 119 Problems 126 § 12 What Happens To A Pendulum In A State Of Weightlessness? 127 § 13 Can You Use The Force Resolution Method Efficiently? 135 Problems 138 § 14 What Do You Know About The Equilibrium Of Bodies? 141 § 15 How Do You Locate The Centre Of Gravity? 149 Problems 155 § 16 Do you know Archimedes’ principle? 159 Problems 165 § 17 Is Archimedes’ Principle Valid In A Spaceship? 167 § 18 What Do You Know About The Molecular-Kinetic Theory Of Matter? 173 § 19 How Do You Account For The Peculiarity In The Thermal Expansion Of Water? 187 § 20 How Well Do You Know The Gas Laws? 189 § 21 How Do You Go About Solving Problems On Gas Laws? 203 Problems 211 § 22 Let Us Discuss Field Theory 215
5 § 23 How Is An Electrostatic Field Described? 221 § 24 How Do Lines Of Force Behave Near The Surface Of A Conductor? 231 § 25 How Do You Deal With Motion In A Uniform Electrostatic Field? 237 Problems 247 § 26 Can You Apply Coulomb’s Law? 249 Problems 256 § 27 Do You Know Ohm’s Law? 259 Problems 267 § 28 Can A Capacitor Be Connected Into A Direct-Current Circuit? 269 Problems 271 § 29 Can you compute the resistance of a branched portion of a circuit? 275 Problems 280 § 30 Why Did The Electric Bulb Burn Out? 283 Problems 288 § 31 Do You Know How Light Beams Are Reflected And Refracted? 293 Problems 299 § 32 How Do You Construct Images Formed By Mirrors And Lenses? 301 § 33 How Well Do You Solve Problems Involving Mirrors And Lenses? 313 Problems 318 Answers 321
From The Editor Of The Russian Edition It can safely be asserted that no student preparing for an entrance ex- amination in physics, for admission to an engineering institute has yet opened a book similar to this one. Employing the extremely lively form of dialogue, the authors were able to comprehensively discuss almost all the subjects in the syllabus, especially questions usually con- sidered difficult to understand. The book presents a detailed analysis of common mistakes made by students taking entrance examinations in physics. Students will find this to be an exceptionally clear and in- teresting textbook which treats of complicated problems from various viewpoints and contains a great many excellent illustrations promoting a deeper understanding of the ideas and concepts involved. The au- thors are lecturers of the Moscow Institute of Electronics Engineering and are well acquainted with the general level of training of students seeking admission to engineering institutes; they have years of experi- ence in conducting entrance examinations. The expert knowledge of the authors, in conjunction with the lively and lucid presentation, has made. This a very useful study guide for students preparing for physics examinations. Prof. G. Epifanov, D.Sc. (Phys. and Math.)
Foreword This book was planned as an aid to students preparing for an entrance examination in physics for admission to an engineering institute. It has the form of a dialogue between the author (the T E AC H E R:) and an in- quisitive reader (the S T U D E N T:). This is exceptionally convenient for analysing common errors made by students in entrance examinations, for reviewing different methods of solving the same problems and for discussing difficult questions of physical theory. A great many questions and problems of school physics are dealt with. Besides, problems are given (with answers) for home study. Most of the questions and problems figured in the entrance examinations of the Moscow Institute of Electronics Engineering in the years 1964- 66. An analysis of mistakes made by students is always instructive. Attention can be drawn to various aspects of the problem, certain fine points can be made, and a more thorough understanding of the fundamentals can be reached. Such an analysis, however, may prove to be very difficult. Though there is only one correct answer, there can be a great many incorrect ones. It is practically impossible to foresee all the incorrect answers to any question; many of them remain concealed forever behind the dis- tressing silence of a student being orally examined. Nevertheless, one can point out certain incorrect answers to definite questions that are heard continually. There are many questions that are almost inevitably answered incorrectly. This book is based mainly on these types of questions and problems. We wish to warn the reader that this is by no means a textbook em-
10 bracing all the items of the syllabus. He will not find here a systematic account of the subject matter that may be required by the study course in physics. He will find this text to be perhaps more like a freely told story or, rather, a freely conducted discussion. Hence, it will be of little use to those who wish to begin their study of physics or to sys- tematize their knowledge of this science. It was intended, instead, for those who wish to increase their knowledge of physics on the thresh- old of their examinations. Our ideal reader, as we conceive him, has completed the required course in school physics, has a good general idea of what it is all about, remembers the principal relationships, can cite various laws and has a fair knowledge of the units employed. He is in that “suspended” state in which he is no longer a secondary school student and has not yet become a full fledged student of an institute. He is eager, however, to become one. If this requires an extension of his knowledge in physics, our book can help him. Primarily, we hope our book will prove that memorizing a text- book (even a very good one) is not only a wearisome business, but indeed a fruitless one. A student must learn to think, to ponder over the material and not simply learn it by heart. If such an understand- ing is achieved, to some extent or other, we shall consider our efforts worthwhile. In conclusion, we wish to thank Prof. G. Epifanov without whose encouragement and invaluable aid this book could not have been written and prepared for publication. We also gratefully acknowledge the many helpful suggestions and constructive criticism that were made on the manuscript by Prof. V. A. Fabrikant, Associate-Prof. A. G. Chertov, and E. N. Vtorov, Senior Instructor of the Physics Department of the Moscow Power Engineering Institute. L. Tarasov A. Tarasova
Do not neglect kinematics! The question of how a body travels in space and time is of considerable interest, both from physical and practical points of view.
§ 1 Can You Analyse Graphs Representing The Kinematics Of Straight-Line Motion? T E AC H E R: You have seen graphs showing the dependence of the velocity and distance travelled by a body on the time of travel for straight-line, uniformly variable motion. In this connection, I wish to put the following question. Consider a velocity graph of the kind shown in Figure 1. On its basis, draw a graph showing the dependence of the distance travelled on time. S T U D E N T: But I have never drawn such graphs. T E AC H E R: There should be no difficulties. However, let us rea- son this out together. First we will divide the whole interval of time into three periods: 1, 2 and 3 (see Figure 1). How does the body travel in period 1? What is the formula for the distance travelled in this period?(1) B Fig. 1 where a is the acceleration of the body. TEACHER: Using the velo- city graph, can you find the accelerat ion? STUDENT: Yes. The acce- leration is the change in equals the ratio of length AC 1) velocity in unit time. It to length OCt TEACHER: Good. Now consider periods 2 and 3. STUDENT: In period 2 the body travels with uniform velo- city v acquired at the end of period 1. The formula for the dis- tance travelled is s=vt TEACHER: Youhave seengraphs showing the dependence of the velocity and distance travelled by a body on the time of travel for straight-line, uniformly va- riable motion. In this connection, I wish to put the following ques.. tion. Consider a velocity graph of the kind shown in Fig. I. On its basis, draw a graph showing the dependence of the distance tra- velled on time. STUDENT: But I have never drawn such graphs. TEACHER: There should be no difficulties. However, let us rea- son this out together. First we will divide the whole interval of time into three periods: 1,2 and 3 (see Fig. 1). How does the body travel in period 1? What is the formula for the distance travelled in this period? STUDENT: In period 1, the body has uniformly accelerated motion with no initial velocity. The formula for the distance travelled is of the form s (t) = § I. CAN YOU ANAL YSE GRAPHS REPRESENTING THE KINEMATICS OF STRAIGHT-LINE MOTION? Figure 1: On the basis of this graph can you draw a graph showing the dependence of distance travelled on time? S T U D E N T: In period 1, the body has uniformly accelerated mo- tion with no initial velocity. The formula for the distance trav-
14 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S elled is of the form spt q “ at 2 2 (1) where a is the acceleration of the body. T E AC H E R: Using the velocity graph, can you find the accelera- tion? S T U D E N T: Yes. The acceleration is the change in velocity in unit time. It equals the ratio of length AC to length OC . T E AC H E R: Good. Now consider periods 2 and 3. S T U D E N T: In period 2 the body travels with uniform velocity v acquired at the end of period 1. The formula for the distance travelled is s “ v t . T E AC H E R: Just a minute, your answer is inaccurate. You have forgotten that the uniform motion began, not at the initial in- stant of time, but at the instant t1. Up to that time, the body had already travelled a distance equal to at 2 1 2 . The dependence of the distance travelled on the elapsed time for period 2 is expressed by the equation spt q “ at 2 1 2 ` vpt ´ t1q (2) With this in mind, please write the formula for the distance travelled in period 3. S T U D E N T: The motion of the body in period 3 is uniformly de- celerated. If I understand it correctly, the formula of the distance travelled in this period should be spt q “ at 2 1 2 ` vpt2 ´ t1q ` vpt ´ t2q ´ a1pt ´ t2q2 2 where a1 is the acceleration in period 3. It is only one half of the acceleration a in period 1, because period 3 is twice as long as period 1.
C A N YO U A NA LY S E G R A P H S R E P R E S E N T I N G T H E K I N E M AT I C S O F S T R A I G H T-L I N E M O T I O N? 15 T E AC H E R: Your equation can be simplified somewhat: spt q “ at 2 1 2 ` vpt ´ t1q ´ a1pt ´ t2q2 2 (3) Now, it remains to summarize the results of equations (1), (2) and (3). S T U D E N T: I understand. The graph of the distance travelled has the form of a parabola for period 1, a straight line for period 2 and another parabola (but turned over, with the convexity facing upward) for period 3. Here is the graph I have drawn (Figure 2).travelled in period 3. STUDENT: The motion of the body in period 3 is uniformly decelerated. If I understand it correctly, the formula of the distance travelled in this period should be s (t) = + V (tz-t l ) + V (t-t z) _ adt;t z)2 Fig. 2 I I I I I I I I I I o -1-_ t3 t where a 1 is the acceleration in period 3. It is only one half of the acceleration a in period 1, because period 3 is twice as long as period 1. TEACHER: Your equation can be simplified somewhat: s(t)= +v(t-tl ) - adt;tz)Z (3) Now, it remains to summarize the results of equations (1), (2) and (3). STUDENT: I understand. The graph of the distance travelled has the form of a parabola for period 1, a straight line for s period 2 and another parabola (but turned over, with the con- vexity facing upward) for pe- riod 3. Here is the graph I have drawn (Fig. 2). TEACHER: There are two faults in your drawing: the graph of the distance travelled should have no kinks. It should be a smooth curve, i.e. the parabolas should be' tangent to the straight line. Moreover,' the vertex of the upper (inverted) parabola should correspond to the instant of time t 3' Here is a correct drawing of the graph (Fig. 3). STUDENT: Please explain it. Let us consider a portion of a distance-travelled vs timegraph (Fig. 4). The average velocity of the body in 12 Figure 2: Students’ incorrect graph showing distance covered s as a function of time t . T E AC H E R: There are two faults in your drawing: the graph of the distance travelled should have no kinks. It should be a smooth curve, i.e. the parabolas should be tangent to the straight line. Moreover, the vertex of the upper (inverted) parabola should correspond to the instant of time t3. Here is a correct drawing of the graph (Figure 3). S T U D E N T: Please explain it. T E AC H E R: Let us consider a portion of a distance-travelled vs time graph (Figure 4). The average velocity of the body in the interval from t to t ` ∆t equals spt ` ∆t q ´ spt q ∆t “ tan α where α is the angle between chord AB and the horizontal. To determine the velocity of the body at the instant t it is necessary
16 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C SThus v(t)=1im (4) l:1t -to 0 In the limit, the chord becomes a tangent to the distance- travelled vs time curve, passing through point A (see the dash- ed line in Fig. 4). The tangent of the angle this line (tangent s Fig. 3 s to Fig. 4 to the curve) makes with the horizontal is the value of the velocity at the instant t. Thus it is possible to find the velo- city at any instant of time from the angle of inclination of the tangent to the distance-travelled vs time curve at the corres- ponding point. But let us return to your drawing (see Fig. 2). It follows from your graph that at the instant of time t i (and at t 2) the velocity of the body has two different values. If we approach t, from the left, the velocity equals tan ai, while if we ap- proach it from the right the velocity equals tan a 2• According to your graph, the velocity of the body at the instant t 1 (and again at t 2) must have a discontinuity, which actually it has not (the velocity vs time graph in Fig. 1 is continuous). STUDENT: I understand now. Continuity of the velocity graph leads to smoothness of the distance-travelled vs time graph. 13 Figure 3: Corrected graph showing distance covered s as a function of time t . to find the limit of such average velocities for ∆t Ñ 0. Thus vpt q “ lim ∆t Ñ0 spt ` ∆t q ´ spt q ∆t (4) In the limit, the chord becomes a tangent to the distance-travelled vs time curve, passing through point A (see the dashed line in Figure 4). The tangent of the angle this line (tangent to the curve) makes with the horizontal is the value of the velocity at the instant t . Thus it is possible to find the velocity at any instant of time from the angle of inclination of the tangent to the distance- travelled vs time curve at the corresponding point. But let usto t + equals s (t (t) - t - an ex etween chord AB and the horizontal. ity of the body at the instant t it is limit of such average velocities for t)=1im (4) l:1t -to 0 becomes a tangent to the distance- e, passing through point A (see the dash- tangent of the angle this line (tangent s to Fig. 4 h the horizontal is the value of the nt t. Thus it is possible to find the velo- f time from the angle of inclination of the nce-travelled vs time curve at the corres- to your drawing (see Fig. 2). It follows t at the instant of time t i (and at t 2) the Figure 4: Corrected graph showing distance covered s as a function of time t . return to your drawing (see Figure 2). It follows from your graph that at the instant of time t1 (and at t2) the velocity of the body has two different values. If we approach t , from the
C A N YO U A NA LY S E G R A P H S R E P R E S E N T I N G T H E K I N E M AT I C S O F S T R A I G H T-L I N E M O T I O N? 17 left, the velocity equals tan α1, while if we approach it from the right the velocity equals tan α2. According to your graph, the velocity of the body at the instant t1 (and again at t2) must have a discontinuity, which actually it has not (the velocity vs time graph in Figure 1 is continuous). S T U D E N T: I understand now. Continuity of the velocity graph leads to smoothness of the distance-travelled vs time graph. T E AC H E R: Incidentally, the vertices of the parabolas should cor- respond to the instants of time 0 and t3 because at these instants the velocity of the body equals zero and the tangent to the curve must be horizontal for these points. Now, using the velocity graph in Figure 1, find the distance travelled by a body by the instant t2. S T U D E N T: First we determine the acceleration a in period 1 from the velocity graph and then the velocity v in period 2. Next we make use of formula (2). The distance travelled by the body during the time t2 equals spt2q “ at 2 1 2 ` vpt2 ´ t q T E AC H E R: Exactly. But there is a simpler way. The distance travelled by the body during the time t2 is numerically equal to the area of the figure OAB D under the velocity vs time graph in the interval O t2. Let us consider another problem to fix what we have learned. Assume that the distance-travelled vs time graph has kinks. This graph is shown in Figure 5, where the curved line is a parabola with its vertex at point A. Draw the velocity vs timegraph. S T U D E N T: Since there are kinks in the distance-travelled graph, there should be discontinuities in the velocity graph at the cor- responding instants of time (t1 and t2). Here is my graph (Fig- ure 6). T E AC H E R: Good. What is the length of BC ?
18 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C Sce travelled by the body during the time t 2 is numerically equal to the area of the figure OABD under the velocity vs time graph in the interval Ot 2. Let us consider another pro- blem to fix what we have learned. Assume that the distance-travelled os time graph has kinks. This graph is shown in Fig. 5, where the curved line is a para- bola with its vertex at point A. Draw the velocity timegraph. t B o v t s tz t3 Fig. 5 Fig. 6 STUDENT: Since there are kinks in the distance-travelled graph, there should be discontinuities in the 'velocity graph at the corresponding instants of time (t 1 and t 2) . Here is my graph (Fig. 6). TEACHER: Good. What is the length of BC? STUDEN.T: It is equal to tan a l (see Fig. 5). We don't, however, know the value of angle al. TEACHER: Nevertheless, we should have no difficulty in t determining the length of BC. Take notice that the distance travelled by the I'ody by the time t 3 is the same as if it had 14 Figure 5: Corrected graph with kinks showing distance travelled s as a function of time t .g the velocity graph in Fig. 1, find the distance body by the instant t 2. First we determine the acceleration a in period 1 ity graph and then the velocity v in period 2. use of formula (2). The distance travelled by g the time t 2 equals s(t 2)= T + v(t 2 - i t ) • Exactly. But there is a simpler way. The distan- the body during the time t 2 is numerically a of the figure OABD under the velocity vs e interval Ot 2. Let us consider another pro- hat we have learned. t the distance-travelled os time graph has kinks. shown in Fig. 5, where the curved line is a para- rtex at point A. Draw the velocity timegraph. t B o v t tz t3 Fig. 5 Fig. 6 Since there are kinks in the distance-travelled ould be discontinuities in the 'velocity graph nding instants of time (t 1 and t 2) . Here is my Good. What is the length of BC? It is equal to tan a l (see Fig. 5). We don't, the value of angle al. Nevertheless, we should have no difficulty in t e length of BC. Take notice that the distance e I'ody by the time t 3 is the same as if it had Figure 6: Velocity vs time graph for the function shown in Figure 5. S T U D E N T: It is equal to tan α1 (see Figure 5). We don’t, however, know the value of angle α1. T E AC H E R: Nevertheless, we should have no difficulty in deter- mining the length of BC . Take notice that the distance travelled by the body by the time t3 is the same as if it had travelled at uni- form velocity all the time (the straight line in the interval from t2 to t3 in Figure 5 is a continuation of the straight line in the interval from 0 to t1). Since the distance travelled is measured by the area under the velocity graph, it follows that the area of rectangle AD EC in Figure 6 is equal to the area of triangle ABC . Consequently, BC “ 2EC , i.e. the velocity at instant t2 when approached from the left is twice the velocity of uniform motion in the intervals from 0 to t1 and from t2 to t3.
The concept of a force is one of the basic physical concepts. Can you apply it with sufficient facility? Do you have a good understanding of the laws of dynamics?
§ 2 Can You Show The Forces Applied To A Body? S T U D E N T: Problems in mechanics seem to be the most difficult of all. How do you begin to solve them? T E AC H E R: Frequently, you can begin by considering the forces applied to a body. As an example, we can take the following cases Figure 7: (a) the body is thrown upward at an angle to the horizontal, (b) the body slides down an inclined plane, (c) the body rotates on the end of a string in a vertical plane, and (d) the body is a pendulum.(0) (d) Fig. 7 I 1 I \ I \ I \ I , / " -,'"... .",. (0), n c to b a p ca th th d th st (d D a these cases, and explain what STUDENT: Here is my draw is the weight of the body and second /.--..........., which fflJlJmJ/T//T/I/)7/l7!T/l, plane a the thi centrip in the the we T is th TEAC in all f rect dra · One clearly interac to show you mu interac in the racts w (Fig. 9 the we If we tion th § 2. CAN YOU SHOW THE FORCES APPLIED TO A BODY? Figure 7: A variety of different situations for bodies. For each case we have to find out the forces acting on the given body. Draw arrows showing the forces applied to the body in each of these cases, and explain what the arrows represent. S T U D E N T: Here is my drawing (Figure 8). In the first case, P is the weight of the body and F is the throwing force. In the second, P is the weight, F is the force which keeps the body sliding along the, plane and Ffr is the friction force. In the third, P is the weight, Fc , is the centripetal force and T is the tension in the string. In the fourth case, P is the weight, F is the restoring force and T is the tension in the string. T E AC H E R: You have made mistakes in all four cases. Here I have the correct drawing (Figure 9). One thing that you must under- stand clearly is that a force is the result of interaction between
22 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S bodies. Therefore, to show the forces applied to a body you must first establish what bodies interact with the given body. Thus, in the first case, only the earth interacts with the body by attract- ing it (Figure 9 (a)). Therefore, only one force, the weight P , is applied to the body. If we wished to take into consideration the resistance of the air or, say, the action of the wind, we would have to introduce additional forces. No “throwing force”, shown in your drawing, actually exists, since there is no interaction creating such a force.the action of the wind, we would ha forces. No "throwing force", shown exists, since there is no interaction STUDENT: But to throw a body, must be exerted on it. F /'lrP " , (a) (a) (e , l' I I I R (d) Fig. 8 TEACHER: Yes, that's true. Wh exert a certain force on it. In the dealt with the motion of the body after the force which imparted a flight to the body had ceased to act mulate" forces; as soon as the inter the force isn't there any more. STUDENT: But if only the weig why doesn't it fall vertically down along a curved path? TEACHER: It surprises you that tion of motion of the body does no tion of the force acting on it. This, 18 Figure 8: Student response for forces acting on bodies in different situations given in Figure 7. S T U D E N T: But to throw a body, surely some kind of force must be exerted on it. T E AC H E R: Yes, that’s true. When you throw a body you exert a certain force on it. In the case above, however, we dealt with the motion of the body after it was thrown, i.e. after the force which imparted a definite initial velocity of flight to the body had ceased to act. It is impossible to “accumulate” forces; as soon as the interaction of the bodies ends, the force isn’t there any more. S T U D E N T: But if only the weight is acting on the body, why doesn’t it fall vertically downward instead of travelling along a curved path? T E AC H E R: It surprises you that in the given case the direction of motion of the body does not coincide with the direction of the force acting on it. This, however, fully agrees with Newton’s second law. Your question shows that you haven’t given suffi- cient thought to Newton’s laws of dynamics. I intend to discuss this later (see § 4). Now I want to continue our analysis of the four cases of motion of a body. In the second case (Figure 9 (b)), a body is sliding down an inclined plane. What bodies are interact- ing with it? S T U D E N T: Evidently, two bodies: the earth and the inclined plane. T E AC H E R: Exactly. This enables us to find the forces applied to the body. The earth is responsible for the weight P , and the
C A N YO U S H OW T H E F O RC E S A P P L I E D TO A B O DY? 23 inclined plane causes the force of sliding friction Ffr and the force N ordinarily called the bearing reaction1. Note that you 1Also known as the normal force.entirely omitted force N in your drawing. S T U D E N T: Just a moment! Then the inclined plane acts on the body with two forces and not one?the action of the wind, we would have to introduce addi tiona! forces. No "throwing force", shown in your drawing, actually exists, since there is no interaction creating such a force. STUDENT: But to throw a body, surely some kind of force must be exerted on it. F /'lrP " , (a) (a) (e) I\\," .... ,I I " , l' I I I R (d) Fig. 8 Fig. 9 TEACHER: Yes, that's true. When you throw a body you exert a certain force on it. In the case above, however, we dealt with the motion of the body after it was thrown, i.e. after the force which imparted a definite initial velocity of flight to the body had ceased to act. It is impossible to "accu- mulate" forces; as soon as the interaction of the bodies ends, the force isn't there any more. STUDENT: But if only the weight is acting on the body, why doesn't it fall vertically downward instead of travelling along a curved path? TEACHER: It surprises you that in the given case the direc- tion of motion of the body does not coincide with the direc.. tion of the force acting on it. This, however, fully agrees with 18 Figure 9: For each case the correct drawing of forces acting on the bodies are made. Compare this with Figure 8. T E AC H E R: There is, of course, only one force. It is, however, more convenient to deal with it in the form of two component forces, one directed along the inclined plane (force of sliding friction) and the other perpendicular to it (bearing reaction). The fact that these forces have a common origin, i.e, that they are components of the same force, can be seen in the existence of a universal relation between Ffr and N : Ffr “ kN (5) where k is a constant called the coefficient of sliding friction. We shall deal with this relationship in more detail later (§ 3). S T U D E N T: In my drawing, I showed a sliding force which keeps the body sliding down the plane. Evidently, there is no such force. But I clearly remember hearing t.he term “sliding force” used frequently in the past. What can you say about this? T E AC H E R: Yes, such a term actually exists. You must bear in mind, however, that the sliding force, as you call it, is simply one of the components of the body’s weight, obtained when the weight is resolved into two forces, one along the plane and the other normal to it. If, in enumerating the forces applied to the body, you have named the weight, there is no reason to add the sliding force, one of its components. In the third case (Figure 9 (c)), the body rotates in a vertical plane. What bodies act on it? S T U D E N T: Two bodies: the earth and the string. T E AC H E R: Good, and that is why two forces are applied to the body: the weight and the tension of the string. S T U D E N T: But what about the centripetal force?
24 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S T E AC H E R: Don’t be in such a hurry! So many mistakes are made in problems concerning the motion of a body in a circle that I intend to dwell at length on this further on (§ 8). Here I only wish to note that the centripetal force is not some kind of additional force applied to the body. It is the resultant force. In our case (when the body is at the lowest point of its path), the centripetal force is the difference between the tension of the string and the weight. S T U D E N T: If I understand it correctly, the restoring force in the fourth case (Figure 9 (d)) is also the resultant of the tension in the string and the weight? T E AC H E R: Quite true. Here, as in the third case, the string and the earth interact with the body. Therefore, two forces, the tension of the string and the weight, are applied to the body. I wish to emphasize again that forces arise only as a result of interaction of bodies; they cannot originate from any “accessory” considerations. Find the bodies acting on the given object and you will reveal the forces applied to the object. S T U D E N T: No doubt there are more complicated cases than the ones you have illustrated in (Figure 7). Can we consider them? T E AC H E R: There are many examples of more complicated inter- action of bodies. For instance, a certain constant horizontal force F acts on a body as a result of which the body moves upward along an inclined surface. The forces applied to the body in this case are shown in (Figure 10).STUDENT: But what about the centripetal force? TEACHER: Don't be in such a hurry! So many mistakes are made in problems concerning the motion of a body in a circle that I intend to dwell at length on this further on (see § 8). Here I only wish to note that the centripetal force is not some kind of additional force applied to the body. It is the resultant force. In our case (when the body is at the lowest point of its path), the centripetal force is the difference between the tension of the string and the weight. STUDENT: If I understand it correctly, the restoring force in the fourth case (Fig. gel) is also the resultant of the tension in the string and the weight? . TEACHER: Quite true. Here, as in third case, the string and the earth interact with the body. Therefore, two forces, the tension of the string and the weight, are applied to the body. I wish to emphasize again that forces arise only as a result of interaction of bodies; they cannot originate from any "accessory" considerations. Find the bodies acting on the gi- ven object and you will reveal the forces applied to the object. STUDENT: No doubt there are more complicated cases than the ones you have illustrated in Fig. 7. Can we consider them? TEACHER: There are many examples of morecomplicated interaction of bodies. For instance, a certain constant hori- zontal force F acts on a body as a result of which the body mo- ves upward along an inclined surface. The forces applied to the body in this case are shown in Fig. 10. P Fig. 10 'I TI I J:W I I 'eI I " I0-_L_- + + p Fig. 11 Another example is the oscillation of an electrically char- ged pendulum placed inside a parallel-plate capacitor. Here we have an additional force Fe with which the field of the capacitor acts on the charge of the pendulum (Fig. 11). It is 20 Figure 10: A body on inclined plane and forces acting on it.
C A N YO U S H OW T H E F O RC E S A P P L I E D TO A B O DY? 25 Another example is the oscillation of an electrically charged pen- dulum placed inside a parallel-plate capacitor. Here we have an additional force Fe with which the field of the capacitor acts on the charge of the pendulum (Figure 11). It is obviously impos- sible to mention all the conceivable cases that may come up in solving problems.ion of the string and the weight, are applied hasize again that forces arise only as a result f bodies; they cannot originate from any nsiderations. Find the bodies acting on the gi- ou will reveal the forces applied to the object. o doubt there are more complicated cases than e illustrated in Fig. 7. Can we consider them? There are many examples of morecomplicated bodies. For instance, a certain constant hori- acts on a body as a result of which the body mo- ng an inclined surface. The forces applied to ase are shown in Fig. 10. P Fig. 10 'I TI I J:W I I 'eI I " I0-_L_- + + p Fig. 11 mple is the oscillation of an electrically char- laced inside a parallel-plate capacitor. Here tional force Fe with which the field of the n the charge of the pendulum (Fig. 11). It is Figure 11: Forces acting on an electrically charged pendulum inside a parallel plate capacitor. S T U D E N T: What do you do when there are several bodies in the problem? Take, for example, the case illustrated in Figure 12. T E AC H E R: You should clearly realize each time the motion of what bodies or combination of bodies you intend to consider. Let us take, for instance, the motion of body 1 in the example you proposed. The earth, the inclined plane and string AB inter- act with this body.T" J (C) p" N' (0) obviously impossible to mention all the conceivable cases that may come up in solving problems. STUDENT: What do you do when there are several bodies in the problem? Take, for example, the case illustrated in Fig. 12. TEACHER: You should clearly realize each time the motion of what bodies or combination of bodies you intend to consi- der. Let us take, for instance, the motion of body 1 in the example you proposed. The- earth, the inclined plane and string AB interact with this body. STUDENT: Doesn't body 2 interact with body 1? TEACHER: Only through string AB. The forces applied to body 1 are the weight P', force F,r of sliding friction, bearing reaction N' and the ten- (OJ sion T' of string AB (Fig. 13a). B C A 2 1 Fig. 12 Fig. 13 STUDENT: But why is the friction force directed to the left in your drawing? It would seem just as reasonable to have it act in the opposite direction. TEACHER: To determine the direction of the friction force, it is necessary to know the direction in which the body is travelling. If this has not been specified in the problem, we should assume either one or the other direction. In the given problem, I assume that body 1 (together with the whole system of bodies) is travelling to the right and the pulley is rotating clockwise. Of course, I cannot know this beforehand; the direction of motion becomes definite only after the corre- spond ing numerical values are substituted. If my assumption Figure 12: What are the forces acting on three bodies on balanced on an incline? S T U D E N T: Doesn’t body 2 interact with body 1? T E AC H E R: Only through string AB. The forces applied to body 1 are the weight P 1, force Ffr of sliding friction, bearing reaction N 1 and the tension T 1 of string AB (Figure 13 (a)).
26 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S S T U D E N T: But why is the friction force directed to the left in your drawing? It would seem just as reasonable to have it act in the opposite direction.T" J (C) p" N' (0) impossible to mention all the conceivable cases y come up in solving problems. NT: What do you do when there are several bodies lem? Take, for example, the case illustrated in ER: You should clearly realize each time the motion odies or combination of bodies you intend to consi- take, for instance, the motion of body 1 in the ou proposed. The- e inclined plane and B interact with this NT: Doesn't body 2 with body 1? ER: Only through B. The forces applied re the weight P', r of sliding friction, action N' and the ten- (OJ f string AB (Fig. 13a). B C A 2 1 Fig. 12 Fig. 13 NT: But why is the friction force directed to the left wing? It would seem just as reasonable to have it pposite direction. ER: To determine the direction of the friction force, ssary to know the direction in which the body is . If this has not been specified in the problem, we ume either one or the other direction. In the given assume that body 1 (together with the whole odies) is travelling to the right and the pulley is clockwise. Of course, I cannot know this beforehand; tion of motion becomes definite only after the corre- numerical values are substituted. If my assumption I shall obtain a negative value when I calculate the 21 Figure 13: Forces acting on the three bodies balanced on an incline as shown in Figure 12. T E AC H E R: To determine the direction of the friction force, it is necessary to know the direction in which the body is travelling. If this has not been specified in the problem, we should assume either one or the other direction. In the given problem, I assume that body 1 (together with the whole system of bodies) is travel- ling to the right and the pulley is rotating clockwise. Of course, I cannot know this beforehand; the direction of motion becomes definite only after the corresponding numerical values are sub- stituted. If my assumption is wrong, I shall obtain a negative value when I calculate the acceleration. Then I will have to as- sume that the body moves to the left instead of to the right (with the pulley rotating counterclockwise) and to direct the force of sliding friction correspondingly. After this I can derive an equa- tion for calculating the acceleration and again check its sign by substituting the numerical values.
C A N YO U S H OW T H E F O RC E S A P P L I E D TO A B O DY? 27 S T U D E N T: Why check the sign of the acceleration a second time? If it was negative when motion was assumed to be to the right, it will evidently be positive for the second assumption. T E AC H E R: No, it can turn out to be negative in the second case as well. S T U D E N T: I can’t understand that. Isn’t it obvious that if the body is not moving to the right it must be moving to the left? T E AC H E R: You forget that the body can also be at rest. We shall return to this question later and analyse in detail the complica- tions that arise when we take the friction force into considera- tion (see § 7). For the present, we shall just assume that the pulley rotates clockwise and examine the motion of body 2. S T U D E N T: The earth, the inclined plane, string AB and string C D interact with body 2. The forces applied to body 2 are shown in Figure 13 (b). T E AC H E R: Very well. Now let us go over to body 3. S T U D E N T: Body 3 interacts only with the earth and with string C D. Figure 13 (c) shows the forces applied to body 3. T E AC H E R: Now, after establishing the forces applied to each body, you can write the equation of motion for each one and then solve the system of equations you obtain. S T U D E N T: You mentioned that it was not necessary to deal with each body separately, but that we could also consider the set of bodies as a whole. T E AC H E R: Why yes; bodies 1, 2 and 3 can be examined, not sep- arately as we have just done, but as a whole. Then, the tensions in the strings need not be taken into consideration since they become, in this case, internal forces, i.e. forces of interaction be- tween separate parts of the item being considered. The system of the three bodies as a whole interacts only with the earth and the inclined plane.
28 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S S T U D E N T: I should like to clear up one point. When I depicted the forces in Figure 13 (b) and (c), I assumed that the tension in string C D is the same on both sides of the pulley. Would that be correct? T E AC H E R: Strictly speaking, that’s incorrect. If the pulley is rotating clockwise, the tension in the part of string C D attached to body 3 should be greater than the tension in the part of the string attached to body 2. This difference in tension is what causes accelerated rotation of the pulley. It was assumed in the given example that the mass of the pulley can be disregarded. In other words, the pulley has no mass that is to be accelerated, it is simply regarded as a means of changing the direction of the string connecting bodies 2 and 3. Therefore, it can be assumed that the tension in string C D is the same on both sides of the pulley. As a rule, the mass of the pulley is disregarded unless otherwise stipulated. Have we cleared up everything? S T U D E N T: I still have a question concerning the point of applica- tion of the force. In your drawings you applied all the forces to a single point of the body. Is this correct? Can you apply the force of friction, say, to the centre of gravity of the body? T E AC H E R: It should be remembered that we are studying the kinematics and dynamics, not of extended bodies, but of mate- rial points, or particles, i.e. we regard the body to be of point mass. On the drawings, however, we show a body, and not a point, only for the sake of clarity. Therefore, all the forces can be shown as applied to a single point of the body. S T U D E N T: We were taught that any simplification leads to the loss of certain aspects of the problem. Exactly what do we lose when we regard the body as a material point? T E AC H E R: In such a simplified approach we do not take into ac- count the rotational moments which, under real conditions, may result in rotation and overturning of the body. A material point has only a motion of translation. Let us consider an example. Assume that two forces are applied at two different points of a body: F1 at point A and F2 at point B, as shown in Figure 14 (a).
C A N YO U S H OW T H E F O RC E S A P P L I E D TO A B O DY? 29B Fi' application of the force. In your drawings you applied all the forces to a single point of the body. Is this correct? Can you apply the force of friction, say, to the centre of gravity of the body? TEACHER: It should be remembered that we are studying the kinematics and dynamics, not of extended bodies, but of material points, or particles, i.e. we regard the body to be of point mass. On the drawings, however, (a) we show a body, and not a point, only for the sake of clarity. Therefore, all rz the forces can be shown as applied to a single point of the body. STUDENT: We were taught that any simplification leads to the loss of cer- tain aspects of the problem. Exactly what do we lose when we regard the Fz body as a material point? TEACHER: In such a simplified ap- r; proach we do not take into account . the rotational moments which, under FIg. 14 real conditions, may result in rota- tion and overturning of the body. A material point has only a motion of translation. Let us consider an example. Assume that two forces are applied at two different points of a body: FI at point A and F 2 at point B, as shown in Fig. 14a. Now let us apply, at point A, force equal and parallel to force F2 , and also force F; equal to F 2 but acting in the opposite direction (see Fig. 14b). SInce forces F; and F; counterbalance each other, their addi- 23 Figure 14: Forces acting on the three bodies on balanced on an incline shown in Figure 12. Now let us apply, at point A, force F 1 2 equal and parallel to force F2 , and also force F 2 2 equal to force F2 but acting in the opposite direction (see Figure 14 (b)). Since forces F 1 2 and F 2 2 counterbal- ance each other, their addition does not alter the physical aspect of the problem in any way. However, Figure 14 (b) can be in- terpreted as follows: forces F1 and F 1 2 applied at point A cause motion of translation of the body also applied to the body is a force couple (forces F2 and F 2 2 ) causing rotation. In other words, force F2 can be transferred to point A of the body if, at the same time, the corresponding rotational moment is added. When we regard the body as a material point, or particle, there will evidently be no rotational moment. S T U D E N T: You say that a material point cannot rotate but has only motion of translation. But we have already dealt with rotational motion - motion in a circle. T E AC H E R: Do not confuse entirely different things. The motion of translation of a point can take place along various paths, for instance in a circle. When I ruled out the possibility of rotational motion of a point I meant rotation about itself, i.e, about any axis passing through the point.
§ 3 Can You Determine The Friction Force? T E AC H E R: I should like to dwell in more detail on the calculation of the friction force in various problems. I have in mind dry sliding friction (sliding friction is said to be dry when there is no layer of any substance, such as a lubricant, between the sliding surfaces). S T U D E N T: But here everything seems to be quite clear.N § 3. CAN YOU DETERMINE THE FRICTION FORCE? TEACHER: I should like to dwell in more detail on the cal- culation of the friction force in various problems. I have in mind dry sliding friction (sliding fri- ction is said to be dry when there is no layer of any substance, such as a lubricant, between the sliding surfaces). But here everything seems to be quite clear. TEACHER: Nevertheless, many mistakes made in examinations are due to the inability to cal- culate the friction force. Consi- der the example illustrated in Fig. 15. A sled oi weight P is being pulled with a force F applied to a rope which makes an angle ex with the horizontal; the coefficient of friction is k. Find the force of sliding friction. How will you go about it? STUDENT: Why, that seems to be very simple. The friction force equals kP. N p P Fig. 15 Fig. 16 TEACHER: Entirely wrong. The force of sliding friction is equal, not to kP, but to kN, where N is the bearing reaction. Remember equation (5) from § 2. .STUDENT: But isn't that the same thing? TEACHER: In a particular case, the weight and the bearing reaction may be equal to each other, but, in general, they are entirely different forces. Consider the example I proposed. The forces applied to the body (the sled) are the weight P, bearing reaction N, force FIr of sliding friction and the ten- sion F of the rope (see Fig. 15). We resolve force F into its vertical (F sin ex) and horizontal (F cos ex) components. All 25 Figure 15: A sled being pulled with a rope. T E AC H E R: Nevertheless, many mistakes made in examinations are due to the inability to calculate the friction force. Consider the example illustrated in Figure 15. A sled of weight P is being pulled with a force F applied to a rope which makes an angle α with the horizontal; the coefficient of friction is k. Find the force of sliding friction. How will you go about it? S T U D E N T: Why, that seems to be very simple. The friction force equals kP .
32 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S T E AC H E R: Entirely wrong. The force of sliding friction is equal, not to kP , but to kN , where N is the bearing reaction. Remem- ber equation (5) from § 2. S T U D E N T: But isn’t that the same thing? T E AC H E R: In a particular case, the weight and the bearing reac- tion may be equal to each other, but, in general, they are entirely different forces. Consider the example I proposed. The forces applied to the body (the sled) are the weight P , bearing reaction N , force Ffr of sliding friction and the tension F of the rope (see Figure 15). We resolve force F into its vertical (F sin α) and horizontal (F cos α) components. All forces acting in the vertical direction counterbalance one another. This enables us to find the bearing reaction: N “ P ´ F sin α (6) As you can see, this force is not equal to the weight of the sled, but is less by the amount F sin α. Physically, this is what should be expected, because the taut rope, being pulled at an angle up- wards, seems to “raise” the sled somewhat. This reduces the force with which the sled bears on the surface and thereby the bearing reaction as well. So, in this case, Ffr “ kpP ´ F sin αq (7) If the rope were horizontal (α “ 0), then instead of equation (6) we would have N “ P , from which it follows that Ffr “ kP . S T U D E N T: I understand now. I never thought about this before. T E AC H E R: This is quite a common error of examinees who attempt to treat the force of sliding friction as the product of the coefficient of friction by the weight and not by the bearing reaction. Try to avoid such mistakes in the future. S T U D E N T: I shall follow the rule: to find the friction force, first determine the bearing reaction. T E AC H E R: So far we have been dealing with the force of sliding friction. Now let us consider static friction. This has certain
C A N YO U D E T E R M I N E T H E F R I C T I O N F O RC E? 33 specific features to which students do not always pay sufficient attention. Take the following example. A body is at rest on a horizontal surface and is acted on by a horizontal force F which tends to move the body. How great do you think the friction force will be in this case? S T U D E N T: If the body rests on a horizontal plane and force F acts horizontally, then N “ P . Is that correct? T E AC H E R: Quite correct. Continue. S T U D E N T: It follows that the friction force equals kP .N FORCE? various problems. I have in mind dry sliding friction (sliding fri- ction is said to be dry when there is no layer of any substance, such as a lubricant, between the sliding surfaces). But here everything seems to be quite clear. TEACHER: Nevertheless, many mistakes made in examinations are due to the inability to cal- culate the friction force. Consi- der the example illustrated in Fig. 15. A sled oi weight P is h a force F applied to a rope which makes an e horizontal; the coefficient of friction is k. of sliding friction. How will you go about it? hy, that seems to be very simple. The friction P. P ig. 15 Fig. 16 ntirely wrong. The force of sliding friction is P, but to kN, where N is the bearing reaction. ation (5) from § 2. ut isn't that the same thing? n a particular case, the weight and the bearing e equal to each other, but, in general, they fferent forces. Consider the example I proposed. ed to the body (the sled) are the weight P, n N, force FIr of sliding friction and the ten- pe (see Fig. 15). We resolve force F into its ex) and horizontal (F cos ex) components. All 25 Figure 16: Forces acting on a body at rest. T E AC H E R: You have made a typical mistake by confusing the forces of sliding and static friction. If the body were sliding along the plane, your answer would be correct. But here the body is at rest. Hence it is necessary that all forces applied to the body counterbalance one another. Four forces act on the body: the weight P , bearing reaction N , force F and the force of static friction Ffr (Figure 16). The vertical forces P and N counterbalance each other. So should the horizontal forces F and Ffr Therefore Ffr “ F (8) S T U D E N T: It follows that the force of static friction depends on the external force tending to move the body. T E AC H E R: Yes, that is so. The force of static friction increases with the force F . It does not increase infinitely, however. The
34 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S force of static friction reaches a maximum value: Fmax “ k0N (9) Coefficient k0 slightly exceeds coefficient k which characterizes, according to equation (5), the force of sliding friction. As soon as the external force F reaches the value k0N , the body begins to slide. At this value, coefficient k0 becomes equal to k, and so the friction force is reduced somewhat. Upon further increase of force F , the friction force (now the force of sliding friction) ceases to increase further (until very high velocities are attained), and the body travels with gradually increasing acceleration. The inability of many examinees to determine the friction force is disclosed by the following rather simple question: What is the friction force when a body of weight P is at rest on an inclined plane with an angle of inclination α? One hears a variety of incorrect answers. Some say that the friction force equals kP , and others that it equals kN “ kP cos α. S T U D E N T: I understand. Since the body is at rest, we have to deal with the force of static friction. It should be found from the condition of equilibrium of forces acting along the inclined plane. There are two such forces in our case: the friction force Ffr and the sliding force P sin α acting downward along the plane: Therefore, the correct answer is Ffr “ P sin α T E AC H E R: Exactly. In conclusion, consider the problem illus- trated in Figure 17. A load of mass m lies on a body of mass M ; the maximum force of static friction between the two is charac- terized by the coefficient k0 and there is no friction between the body and the earth. Find the minimum force F applied to the body at which the load will begin to slide along it.STUDENT: It follows that on the external force tendi TEACHER: Yes, that is s creases with the force F. It ever. The force of static fr Coefficient k o slightly excee zes, according to equation As soon as the external force begins to slide. At this valu k, and so the friction force is increase of force F, the fricti friction) ceases to increase f are attained), and the body acceleration. The inability the friction force is disclose question: what is the frictio is at rest on an 'inclined pl a? One hears a variety of i the friction force equals kP =kP cos a. STUDENT: I understand. S to deal with the force of st from the condition of equili inclined plane. There are t frictio ! 6 z t:;F P sin -g:::::Q : plane: ;;;;n» F/r=P Fig. 17 TEA consid Fig. 17. A load of mass m maximum force of static f racterized by the coefficient k the body and the earth. Find the body at which the load wi STUDENT: First I shall as small, so that the load will n two bodies will acquire the a= Figure 17: A load of mass m lies on a body of mass M . We have to find the minimum force F so that load begins to slide. S T U D E N T: First I shall assume that force F is sufficiently small, so that the load will not slide along the body. Then the two bodies will acquire the acceleration a “ F M ` m
C A N YO U D E T E R M I N E T H E F R I C T I O N F O RC E? 35 T E AC H E R: Correct. What force will this acceleration impart to the load? S T U D E N T: It will be subjected to the force of static friction Ffr by the acceleration. Thus Ffr “ ma “ F m M ` m It follows that with an increase in force F , the force of static friction Ffr also increases. It cannot, however, increase infinitely. Its maximum value is Ffr max “ k0N “ k0 m g Consequently, the maximum value of force F . at which the two bodies can still travel together as an integral unit is determined from the condition k0 m g “ F m M ` m This, then, is the minimum force at which the load begins to slide along the body. T E AC H E R: Your solution of the proposed problem is correct. I am completely satisfied with your reasoning.
§ 4 How Well Do You Know Newton’s Laws Of Motion? T E AC H E R: Please state Newton’s first law of motion. S T U D E N T: A body remains at rest or in a state of uniform mo- tion in a straight line until the action of other bodies compels it to change that state. T E AC H E R: Is this law valid in all frames of reference? S T U D E N T: I don’t understand your question. T E AC H E R: If you say that a body is at rest, you mean that it is stationary with respect to some other body which, in the given case, serves as the reference system, or frame of reference. It is quite pointless to speak of a body being in a state of rest or definite motion without indicating the frame of reference. The nature of the motion of a body depends upon the choice of the frame of reference. For instance, a body lying on the floor of a travelling railway car is at rest with respect to a frame of reference attached to the car, but is moving with respect to a frame of reference attached to the track. Now we can return to my question. Is Newton’s first law valid for all frames of reference? S T U D E N T: Well, it probably is. T E AC H E R: I see that this question has taken you unawares. Ex- periments show that Newton’s first law is not valid for all refer-
38 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S ence systems. Consider the example with the body lying on the floor of the railway car. We shall neglect the friction between the body and the floor. First we shall deal with the position of the body with respect to a frame of reference attached to the car. We can observe the following: the body rests on the floor and, all of a sudden, it begins to slide along the floor even though no action of any kind is evident. Here we have an obvious violation of Newton’s first law of motion. The conventional explanation of this effect is that the car, which had been travelling in a straight line and at uniform velocity, begins to slow down, because the train is braked, and the body, due to the absence of friction, continues to maintain its state of uniform straight-line motion with respect to the railway tracks. From this we can conclude that Newton’s law holds true in a frame of reference attached to the railway tracks, but not in one attached to a car being slowed down. Frames of reference for which Newton’s first law is valid are said to be inertial; those in which it is not valid are non-inertial. For most of the phenomena we deal with we can assume that any frame of reference is inertial if it is attached to the earth’s surface, or to any other bodies which are at rest with respect to the earth’s surface or travel in a straight line at uniform veloc- ity. Non-inertial frames of reference are systems travelling with acceleration (or deceleration), for instance rotating systems, ac- celerating or decelerating lifts, etc. Note that not only Newton’s first law of motion is invalid for non-inertial reference systems, but his second law as well (since the first law is a particular case of the second law). S T U D E N T: But if Newton’s laws cannot be employed for frames of reference travelling with acceleration, then how can we deal with mechanics in such frames? T E AC H E R: Newton’s laws of motion can nevertheless be used for non-inertial frames of reference. To do this, however, it will be necessary to apply, purely formally, an additional force to the body. This force, the so called inertial force, equals the product of the mass of the body by the acceleration of the reference
H OW W E L L D O YO U K N OW N E W TO N’S L AW S O F M O T I O N? 39 system, and its direction is opposite to the acceleration of the body. I should emphasize that no such force actually exists but, if it is formally introduced, then Newton’s laws of motion will hold true in a non-inertial frame of reference. I want to advise you, however, to employ only inertial frames of reference in solving problems. Then, all the forces that you have to deal with will be really existing forces. S T U D E N T: But if we limit ourselves to inertial frames of refer- ence, then we cannot analyse, for instance, a problem about a body lying on a rotating disk. T E AC H E R: Why can’t we? The choice of the frame of reference is up to you. If in such a problem you use a reference system attached to the disk (i.e. a non-inertial system), the body is con- sidered to be at rest. But if your reference system is attached to the earth (i.e. an inertial reference system), then the body is dealt with as one travelling in a circle. I would advise you to choose an inertial frame of reference. And now please state Newton’s second law of motion. S T U D E N T: This law can be written as F “ ma, where F is the force acting on the body, m is its mass and a - acceleration. T E AC H E R: Your laconic answer is very typical. I should make three critical remarks on your statement; two are not very im- portant and one is essential. In the first place, it is not the force that results from the acceleration, but, on the contrary, the ac- celeration is the result of the applied force. It is therefore more logical to write the equation of the law as a “ B F m (10) where B is the proportionality factor depending upon the choice of units of measurement of the quantities in equation (10). No- tice that your version had no mention of the proportionality factor B. Secondly, a body is accelerated by all forces applied to it (though some may counterbalance one another). Therefore,
40 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S in stating the law you should use, not the term “force”, but the more accurate term “resultant force”. My third remark is the most important. Newton’s second law establishes a relationship between force and acceleration. But force and acceleration are vector quantities, characterized not only by their numerical value (magnitude) but by their direction as well. Your statement of the law fails to specify the directions. This is an essential shortcoming. Your statement leaves out a vital part of Newton’s second law of motion. Correctly stated it is: the acceleration of a body is directly proportional to the resultant of all forces acting on the body, inversely proportional to the mass of the body and takes place in the direction of the resultant force. This statement can be analytically expressed by the formula ~a “ B ~F m (11) (where the arrows over the letters denote vectors). S T U D E N T: When in § 2 we discussed the forces applied to a body thrown upward at an angle to the horizontal, you said you would show later that the direction of motion of a body does not necessarily coincide with the direction of the force applied to it. You referred then to Newton’s second law.the change in velocity in unit tim -+ -+ are the velocity vectors VI and V 2 of stants of time t and The chan time is the vector ration is or, more rigorously, -+- L a(t)=lim &t-..O It follows that the acceleration vec tor which represents the change ficiently short interval of time. It Fig. 18 that the velocity vectors and the c can be oriented in entirely differen that, in the general case, the acceler are also differently oriented. Is tha STUDENT: Yes, now I understand. travels in a circle, the velocity of th a tangent to the circle, but its acce a radius toward the centre of rotatio celeration). TEACHER: Your example is quite return to relationship (11) and mak cisely the acceleration and not the in the direction of the applied for the acceleration and not the veloc magnitude of this force. On the oth body's motion at any given insta direction and magnitude of its velo (the velocity vector is always tangen Since the acceleration and velocit 32 Figure 18: Depicting change in velocity vectorially. T E AC H E R: Yes, I remember, and I think it would be quite appro- priate to return to this question. Let us recall what acceleration is. As we know, acceleration is characterized by the change in velocity in unit time. Illustrated in Figure 18 are the velocity vectors ~v1 and ~v2 of a body for two nearby instants of time t and t ` ∆t . The change in velocity during the time ∆t is the vector ∆ ~v “ ~v2 ´ ~v1. By definition, the acceleration is ~apt q – lim ∆t Ñ0 ∆ ~v ∆t (12) or, more rigorously, ~apt q “ lim ∆t Ñ0 ∆ ~v ∆t (13)
H OW W E L L D O YO U K N OW N E W TO N’S L AW S O F M O T I O N? 41 It follows that the acceleration vector is directed along vector ∆v, which represents the change in velocity during a sufficiently short interval of time. It is evident from Figure 18 that the veloc- ity vectors and the change in velocity vector can be oriented in entirely different directions. This means that, in the general case, the acceleration and velocity vectors are also differently oriented. Is that clear? S T U D E N T: Yes, now I understand. For example, when a body travels in a circle, the velocity of the body is directed along a tangent to the circle, but its acceleration is directed along a radius toward the centre of rotation (I mean centripetal acceleration). T E AC H E R: Your example is quite appropriate. Now let us return to relationship (equation (11)) and make it clear that it is pre- cisely the acceleration and not the velocity that is oriented in the direction of the applied force, and that it is again the acceleration and not the velocity that is related to the magnitude of this force. On the other hand, the nature of a body’s motion at any given instant is determined by the direction and magnitude of its veloc- ity at the given instant (the velocity vector is always tangent to the path of the body). Since the acceleration and velocity are different vectors, the direction of the applied force and the direction of motion of the body may not coincide in the general case. Consequently, the nature of the motion of a body at a given instant is not uniquely determined by the forces acting on the body at the given instant. S T U D E N T: This is true for the general case. But, of course, the direction of the applied force and the velocity may coincide. T E AC H E R: Certainly, that is possible. Lift a body and release it carefully, so that no initial velocity is imparted to it. Here the direction of motion will coincide with the direction of the force of gravity. If, however, you impart a horizontal initial velocity to the body then its direction of motion will not coincide with the direction of the gravity force; the body will follow a parabolic path. Though in both cases the body moves due to the action of the same force - its weight - the nature of its motion differs.
42 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S A physicist would say that this difference is due to the different initial conditions: at the beginning of the motion the body had no velocity in the first case and a definite horizontal velocity in the second.-+ -+- L\v a(t)=lim- (13) &t-..O It follows that the acceleration vector is di;ected along vec- tor which represents the change in velocity during a suf- ficiently short interval of time. It is evident from Fig. 18 Fig. 18 Fig. 19 that the velocity vectors and the change in velocity vector can be oriented in entirely different directions. This means that, in the general case, the acceleration and velocity vectors are also differently oriented. Is that clear? STUDENT: Yes, now I understand. For example, when a body travels in a circle, the velocity of the body is directed along a tangent to the circle, but its acceleration is directed along a radius toward the centre of rotation (I mean centripetal ac- celeration). TEACHER: Your example is quite appropriate. Now let us return to relationship (11) and make it clear that it is pre- cisely the acceleration and not the velocity that is oriented in the direction of the applied force, and that it is again the acceleration and not the velocity that is related to the magnitude of this force. On the other hand, the nature of a body's motion at any given instant is determined by the direction and magnitude of its velocity at the given instant (the velocity vector is always tangent to the path of the body). Since the acceleration and velocity are different vectors, 32 Figure 19: Trajectories of bodies with different initial velocities. Illustrated in Figure 19 are the trajectories of bodies thrown with initial velocities of different directions, but in all cases the same force, the weight of the body, is acting on it. S T U D E N T: Does that mean that the nature of the motion of a body at a given instant depends not only on the forces acting on the body at this instant, but also on the initial conditions? T E AC H E R: Exactly. It should be emphasized that the initial conditions reflect the prehistory of the body. They are the result of forces that existed in the past. These forces no longer exist, but the result of their action is manifested. From the philosophical point of view, this demonstrates the relation of the past to the present, i.e, the principle of causality. Note that if the formula of Newton’s second law contained the velocity and not the acceleration, this relationship of the past and present would not be revealed. In this case, the velocity of a body at a given instant (i.e. the nature of its motion at a given instant) would be fully determined by the forces acting on the body precisely at this instant; the past would have no effect whatsoever on the present.I I II' I __--L-_r" J .... ' ....... I II I I I I I I I I I I I I d :--- ....--t-----I ra) string. If it is deflected to tion and then released, it w a definite velocity is impart pendicular to the plane of travel in a circle at uniform ding upon the initial condi a plane (see Fig. 20a) , or circle act on the te STU ton's TEA stude forces reaso witho intera may That Figs. the s in th Actua applie the te p STU Fig. 20 the sa of dif on the nature of the moti starting point in determinin TEACHER: You have state is no need, however, to go t kinds of motion may be cau in Fig. 20), the numerical ces differ for the different k there will be a different res tion. Thus, for instance, in circle, the resultant force s oscillation in a plane, the r Figure 20: A ball hanging from a string can have different resultant motions depending on the initial conditions. I want to cite one more example illustrating the aforesaid. It is shown in Figure 20: a ball hanging on a string is subject to the action of two forces, the weight and the tension of the string. If it is deflected to one side of the equilibrium position and then released, it will begin to oscillate. If, however, a definite velocity is imparted to the ball in a direction perpendicular to the plane of deviation, the ball will begin to travel in a circle at uniform velocity. As you can see, depending upon the initial conditions, the ball either oscillates in a plane (see Figure 20 (a)), or travels at uniform velocity in a circle (see Figure 20 (b)). Only two forces act on it in either case: its weight and the tension of the string.
H OW W E L L D O YO U K N OW N E W TO N’S L AW S O F M O T I O N? 43 S T U D E N T: I haven’t considered Newton’s laws from this view- point. T E AC H E R: No wonder then that some students, in trying to de- termine the forces applied to a body, base their reasoning on the nature of motion without first finding out what bodies interact with the given body. You may recall that you did the same. That is exactly why, when drawing Figure 8 (c) and Figure 8 (d), it seemed to you that the sets of forces applied to the body in those cases should be different. Actually, in both cases two forces are applied to the body: its weight and the tension of the string. S T U D E N T: Now I understand that the same set of forces can cause motions of different nature and therefore data on the nature of the motion of a body cannot serve as a starting point in determining the forces applied to the body. T E AC H E R: You have stated the matter very precisely. There is no need, however, to go to the extremes. Though different kinds of motion may be caused by the same set of forces (as in Fig- ure 20), the numerical relations between the acting forces differ for the different kinds of motion. This means that there will be a different resultant applied force for each motion. Thus, for instance, in uniform motion of a body in a circle, the resultant force should be the centripetal one; in oscillation in a plane, the resultant force should be the restoring force. From this it follows that even though data on the kind of motion of a body cannot serve as the basis for determining the applied forces, they are far from superfluous. In this connection, let us return to the example illustrated in Figure 20. Assume that the angle α, between the vertical and the direction of the string is known and so is the weight P of the body. Find the tension T in the string when (1) the oscil- lating body is in its extreme position, and (2) when the body is travelling uniformly in a circle. In the first case, the resultant force is the restoring force and it is perpendicular to the string. Therefore, the weight P of the body is resolved into two components, with one component
44 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S along the resultant force and the other perpendicular to it (i.e. directed along the string). Then the forces perpendicular to the resultant force, i.e. those acting in the direction along the string, are equated to each other (see Figure 21 (a)). Thus T1 “ P cos α In the second case, the resultant force is the centripetal onep Fig. 21 Pcosa p fQ) In this connection, let us return to the example illustrated in Fig. 20. Assume that the angle ex, between the vertical and the direction of the string is known and so is the weight P of the body. Find the tension T in the string when (I) the oscillating body is in its extreme position, and (2) when the body is travelling uniformly in a circle. In the first case, the resultant force is the restoring force and it is perpendi- cular to the string. Therefore, the weight P of the body is resolved into two components, with one component along the resultant force and the other perpendicular to it (Le. directed along the string). Then the forces perpendicular to the re- suItant force, i. e. those acting in the direction along the string, are equated to each other (see Fig. 21a). Thus T 1 = P cos ex, In the second case, the resultant force is the centripetal one and is directed horizontally. Hence, the tension T 2 of the string should be resolved into a vertical and a horizontal force, and the forces perpendicular to the resultant force, i.e, the vertical forces, should be equated to each other (Fig. 2Ib). Then T 2 cos cx= P or p T=--2 cos ex As you can see, a knowledge of the nature of the body's mo- tion proved useful in finding the tension of the string. STUDENT: If I understand all this correctly, then, knowing the interaction of bodies, you can find the forces applied to one of them; if you know these forces and the initial con- ditions, you can predict the nature of the motion of the body (the magnitude and direction of its velocity at any instant). On the other hand, if you know the kind of motion of a body you can establish the relationships between the forces applied to it. Am I reasoning correctly? TEACHER: Quite so. But let us continue. I want to propose a comparatively simple problem relating to Newton's second 2* 35 Figure 21: Resolving the forces on a moving pendulum. and is directed horizontally. Hence, the tension T2 of the string should be resolved into a vertical and a horizontal force, and the forces perpendicular to the resultant force, i.e, the vertical forces, should be equated to each other (Figure 21 (b)). Then T2 cos α “ P or T2 “ P cos α As you can see, a knowledge of the nature of the body’s motion proved useful in finding the tension of the string. S T U D E N T: If I understand all this correctly, then, knowing the interaction of bodies, you can find the forces applied to one of them; if you know these forces and the initial conditions, you can predict the nature of the motion of the body (the magnitude and direction of its velocity at any instant). On the other hand, if you know the kind of motion of a body you can establish the relationships between the forces applied to it. Am I reasoning correctly?
H OW W E L L D O YO U K N OW N E W TO N’S L AW S O F M O T I O N? 45 T E AC H E R: Quite so. But let us continue. I want to propose a comparatively simple problem relating to Newton’s second law of motion. Two bodies, of masses M and m, are raised to the same height above the floor and are released simultaneous. Will the two bodies reach the floor simultaneously if the resistance of the air is the same for each of them? For slmplicity we shall assume that the air resistance is constant. S T U D E N T: Since the air resistance is the same for the two bodies, it can be disregarded. Consequently, both bodies reach the floor simultaneously. T E AC H E R: You are mistaken. You have no right to disregard the resistance of the air. Take, for example, the body of mass M . It is subject to two forces: the weight M g and the air resistance F . The resultant force is M g ´ F . From this we find the accelera- tion. Thus a “ M g ´ F M “ g ´ F M In this manner, the body of larger mass has a higher acceleration and will, consequently, reach the floor first. Once more I want to emphasize that in calculating the accel- eration of a body it is necessary to take into account I all the forces applied to it, i.e. you must find the resultant force. In this connection, the use of the term “driving force” is open to crit- icism. This term is inappropriate. In applying it to some force (or to several forces) we seem to single out the role of this force (or forces) in imparting acceleration to the body. As if the other forces concerned were less essential This is absolutely wrong. The motion of a body is a result of the action of all the forces applied to it without any exceptions (of course, the initial condi- tions should be taken into account). Let us now consider an example on Newton’s third law of mo- tion. A horse starts to pull a waggon. As a result, the horse and waggon begin to travel with a certain acceleration According to Newton’s third law, whatever the force with which the horse pulls the waggon, the waggon pulls back on the horse with ex- actly the same force but in the opposite direction. This being so,
46 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S why do the horse and waggon travel forward with an accelera- tion? Please explain. S T U D E N T: I have never given this any thought but I see no con- tradictions. The acceleration would be difficult to ex plain if the force with which the horse acts on the waggon was counterbal- anced by the force with which the waggon acts on the horse. But these forces cannot cancel each other since they are applied to different bodies: one to tho horse and the other to the waggon. Figure 22: Why do the horse and waggon travel forward with an acceleration? T E AC H E R: Your explanation is applicable to the case when the waggon is not harnessed to the horse. Then the horse pushes away from the waggon, as a result of which the waggon moves in one direction and the horse in the other. The case I proposed is entirely different. The horse is harnessed to the waggon. Thus they are linked together and travel as a single system. The forces of interaction between the horse and waggon that you mentioned are applied to different parts of the same system. In the motion of this system as a whole these forces can be regarded as mutually counterbalancing forces. Thus, you haven’t yet answered my question. S T U D E N T: Well, then I can’t understand what the matter is. Maybe the action here is not fully counterbalanced by the reac- tion? After all a horse is a living organism. T E AC H E R: Now don’t let your imagination run away with you. It was sufficient for you to meet with some difficulty and you are ready to sacrifice one of the principal laws of mechanics. To answer my question, there is no need to revise Newton’s third law of motion. On the contrary, let us use this law as a basis for our discussion. According to the third law. the interaction of the horse and the waggon cannot lead to the motion of this system as a whole (or,
H OW W E L L D O YO U K N OW N E W TO N’S L AW S O F M O T I O N? 47 more precisely, it cannot impart acceleration to the system as a whole). This being so there must exist some kind of supplemen- tary interaction. In other words at least one more body must participate in the problem in addition to the horse and waggon. This body, in the given case is the earth. As a result, we have three interactions to deal’ with instead of one, namely: (1) between the horse and the waggon (we shall denote this force by f0; (2) between the horse and the earth (force F ), in which the horse pushes against the ground; and (3) between the waggon and the earth (force f ) which is the friction of the waggon against the ground. All bodies are shown in Figure 22: the horse, the waggon and the earth and two forces are applied to each body. These two forces are the result of the interaction of the given body with the two others. The acceleration of the horse-waggon system is caused by the resultant of all the forces applied to it. There are four such forces and their resultant is F ´ f . This is what causes the acceleration of the system. Now you see that this acceleration is not associated with the interaction between the horse and the waggon. S T U D E N T: So the earth’s surface turns out to be, not simply the place on which certain events occur, but an active participant of these events. T E AC H E R: Your pictorial comment is quite true. Incidentally, if you locate the horse and waggon on an ideal icy surface, thereby excluding all horizontal interaction between this system and the earth, there will be no motion, whatsoever. It should be stressed that no internal interaction can impart acceleration to a system as a whole. This can be done only by external action (you can’t lift yourself by your hair, or bootstraps either). This is an important practical inference of Newton’s third law of motion.
If you know mechanics well, you can easily solve problems. The con- verse is just as true: if you solve problems readily, you evidently have a good knowledge of mechanics. Therefore, extend your knowledge of mechanics by solving as many problems as you can.
§ 5 How Do You Go About Solving Problems In Kinematics? T E AC H E R: Assume that two bodies are falling from a certain height. One has no initial velocity and the other has a certain initial velocity in a horizontal direction. Here and further on we shall disregard the resistance of the air. Compare the time it takes for the two bodies to fall to the ground. S T U D E N T: The motion of a body thrown horizontally can be re- garded as a combination of two motions: vertical and horizontal. The time of flight is determined by the vertical component of the motion. Since the vertical motions of the bodies are determined in both cases by the same data (same height and the absence of a vertical component of the initial velocity), the time of fall is the same for the two bodies. It equals b 2H {g , where H is the initial height. T E AC H E R: Absolutely right. Now let us consider a more com- plex case. Assume that both bodies are falling from the height H with no initial velocity, but in its path one of them meets a fixed plane, inclined at an angle of 45° to the horizontal. As a result of this impact on the plane the direction of the velocity of the body becomes horizontal (Figure 23). The point of impact is at the height h. Compare the times of fall of the two bodies. S T U D E N T: Both bodies take the same time to fall to the level of the inclined plane. As a result of the impact on the, plane one
52 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C SFig. 23 e time of fall is the same for the two bodies. It equals V2H/g, where H is the initial height. TEACHER: Absolutely right. Now let us consider a mor complex case. Assume that both bodies are falling from th height H with no initial velocity, but in its path one-of them meets a fixed plane, inclined at an angle of 45° to the horizon- tal. As a result of this impact on the plan the direction of the velocity of the bodybecomes horizontal (Fig. 23). The point o impact is at the height h. Compare the times of fall of the two bodies. STUDENT: Both bodies take the same time to fall to the level of the inclined plane. As a result of the impact on the , plane one of the bodies acquires a hori- zontal component of velocity. This hori- zontal component cannot, however, influence the vertical component of the body's motion. The- refore, it follows that in this case as well the time of fall should be the same for the bodies.' TEACHER: Your answer is wrong. You were right in saying that the horizontal component of the velocity doesn't influence the vertical motion of the body and, consequently, its time of fall. When the body strikes the inclined plane it not only 40 Figure 23: Two bodies falling from height H . One of them meets a fixed plane at height h. Problem is to compare the time of fall of the two bodies. of the bodies acquires a horizontal component of velocity. This horizontal component cannot, however, influence the vertical component of the body’s motion. Therefore, it follows that in this case as well the time of fall should be the same for the bodies. T E AC H E R: Your answer is wrong. You were right in saying that the horizontal component of the velocity doesn’t influence the vertical motion of the body and, consequently, its time of fall. When the body strikes the inclined plane it not only acquires a horizontal velocity component, but also loses the vertical component of its velocity, and this of course must affect the time of fall. After striking the inclined plane, the body falls from the height h with no initial vertical velocity. The impact against the plane delays the vertical motion of the body and thereby increases its time of fall. The time of fall for the body which dropped straight to the ground is b 2H {g ; that for the body striking the plane is b 2pH ´ hq{g ` b 2h{g . This leads us to the following question: at what h to H ratio will the time of fall reach its maximum value? In other words, at what height should the inclined plane be located so that it delays the fall most effectively? S T U D E N T: I am at a loss to give you an exact answer. It seems to me that the ratio h{H should not be near to 1 or to 0, because a ratio of 1 or 0 is equivalent to the absence of any plane what- soever. The inclined plane should be located somewhere in the middle between the ground and the initial point.
H OW D O YO U G O A B O U T S O LV I N G P RO B L E M S I N K I N E M AT I C S? 53 T E AC H E R: Your qualitative remarks are quite true. But you should find no difficulty in obtaining the exact answer. We can write the time of fall of the body as t “ d 2H g ´?1 ´ x ` ?x ¯ where x “ h H Now we find the value of x at which the function t pxq is a maxi- mum. First we square the time of fall. Thus t 2 “ 2H g ˆ 1 ` 2 b p1 ´ xq x ˙ If the time is maximal, its square is also maximal. It is evident from the last equation that t 2 is a maximum when the function y “ p1 ´ xq x is a maximum. Thus, the problem is reduced to finding the maximum of the quadratic trinomial y “ ´x2 ` x “ ´ ˆ x ´ 1 2 ˙2 ` 1 4 This trinomial is maximal at x “ 1 2 . Thus, height h should be one half of height H . Our further discussion on typical procedure for solving prob- lems in kinematics will centre around the example of a body thrown upward at an angle to the horizontal (usually called the elevation angle). S T U D E N T: I’m not very good at such problems. T E AC H E R: We shall begin with the usual formulation of the problem: a body is thrown upward at an angle of α, to the hori- zon with an initial velocity of v0. Find the time of flight T , maximum height reached H and the range L. As usual, we first find the forces acting on the body. The only force is gravity. Consequently, the body travels at uniform ve- locity in the horizontal direction and with uniform acceleration g in the vertical direction. We are going to deal with the verti- cal and horizontal components of motion separately, for which
54 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S purpose we resolve the initial velocity vector into the vertical (v0 sin α,) and horizontal (v0 cos α,) components. The horizon- tal velocity component remains constant throughout the flight while the vertical component varies as shown in Figure 24. LetSTUDENT.: I'm not very good at such problems. TEACHER: We shall begin with the usual formulation of the problem: a body is thrown upward at an angle of ex, to the ho- rizon with: an initial velocity of Vo. Find the time of flight T, maximum height reached H and the range L. As usual, we first find the forces acting on the body. The only force is gra- vity. Consequently, the body travels at uniform velocity in the .horizontal direction and with uniform acceleration g .in the vertical direction. We are going to deal with the vertical and horizontal components of motion separately, for which Fig. 24 purpose we resolve the initial velocity vector into the verti- cal (vo sin cx,) and horizontal (vo cos ex,) components. The ho- rizontal velocity component remains constant throughout the flight while the vertical component varies as shown in Fig. 24. Let us examine the vertical component of the motion. The time of flight T= T i +T 2, where T 1 is the time of ascent (the body travels vertically with uniformly decelerated motion) and T 2 is the time of descent (the body travels vertically downward with uniformly accelerated motion). The vertical velocity of the body at the highest point of its trajectory (at the instant t=T 1) is obviously equal to zero. On the other hand, this velocity can be expressed by the formula showing the dependence of the velocity of uniformly decelerated motion on time. Thus we obtain 0= Vo sin a-gT l or T _ 0 0 sin Ct 1- g When T 1 is known we can obtain H T · 05 sin 2 Ct =vo ISllla-- 2- = 2g 42 (14) ( 15) Figure 24: A body thrown with an angle α to the horizon. The problem is to find the time of flight, maximum height reached and the range of the body. us examine the vertical component of the motion. The time of flight T “ T1 ` T2, where T1 is the time of ascent (the body travels vertically with uniformly decelerated motion) and T2 is the time of descent (the body travels vertically downward with uniformly accelerated motion). The vertical velocity of the body at the highest point of its trajectory (at the instant t “ T1) is obviously equal to zero. On the other hand, this velocity can be expressed by the formula showing the dependence of the velocity of uniformly decelerated motion on time. Thus we obtain 0 “ v0 sin α ´ g T1 or T1 “ v0 sin α g (14) When T1 is known we can obtain H “ v0T1 sin α ´ g T 2 1 2 “ v2 0 sin2 α 2g (15) The time of descent T2 can be calculated as the time a body falls from the known height H without any initial vertical velocity: T2 “ d 2H g “ v0 sin α g
H OW D O YO U G O A B O U T S O LV I N G P RO B L E M S I N K I N E M AT I C S? 55 Comparing this with equation (14) we see that the time of de- scent is equal to the time of ascent. The total time of flight is T “ T1 ` T2 “ 2 v0 sin α g (16) To find the range L, or horizontal distance travelled, we make use of the horizontal component of motion. As mentioned before, the body travels horizontally at uniform velocity. Thus L “ pv0 cos αq T “ v2 0 sin 2α g (17) It can be seen from equation (17) that if the sum of the angles at which two bodies are thrown is 90° and if the initial velocities are equal, the bodies will fall at the same point. Is everything clear to you so far? S T U D E N T: Why yes, everything seems to be clear. T E AC H E R: Fine. Then we shall add some complications. Assume that a horizontal tail wind of constant force F acts on the body. The weight of the body is P . Find, as in the preceding case, the time of flight T , maximum height reached H , and range L. S T U D E N T: In contrast to the preceding problem, the horizontal motion of the body is not uniform; now it travels with a hori- zontal acceleration of a “ pF {P q g . T E AC H E R: Have there been any changes in the vertical compo- nent of motion? S T U D E N T: Since the force of the wind acts horizontally the wind cannot affect the vertical motion of the body. T E AC H E R: Good. Now tell me which of the sought for quantities should have the same values as in the preceding problem. S T U D E N T: These will evidently be the time of flight T and the height H . They are the ones determined on the basis of the vertical motion of the body. They will therefore be the same as in the preceding problem.
56 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S T E AC H E R: Excellent. How about the range? S T U D E N T: The horizontal acceleration and time of flight being known, the range can be readily found. Thus L “ pv0 cos αq T ` aT 2 2 “ v2 0 sin 2α g ` 2F P v2 0 sin2 α g T E AC H E R: Quite correct. Only the answer would best be written in another form: L “ v2 0 sin 2α g ˆ 1 ` F P tan α ˙ (18) Next we shall consider a new problem: a body is thrown at an angle α to an inclined plane which makes the angle β with the horizontal (Figure 25). The body’s initial velocity is v0. Find the distance L from the point where the body is thrown to the point where it falls on the plane." STUDENT: The horizontal acceleration and time of flight being known, the range can be readily found. Thus L- ( ) T + aT2 _ sin 2a + 2F sln 2 a - vo cos 2 - g P g TEACHER: Quite correct. Only the answer would best be written in another form: L= (1 + tan « ) (18) Next we shall consider a new problem: a body is thrown at an angle ex to an inclined plane which makes the.angle with the horizontal (Fig. 25). The body's initial velocity is V o• LJo Find the distance L from the point where the body is thrown to the point where it falls on the plane. STUDENT: I once made an at- tempt to solve such a problem but failed. TEACHER: Can't you see any Fig. 25 similarity between this problem and the preceding one? STUDENT: No, I can't. TEACHER: Let us imagine that the figure for this problem is turned through the angle so that the inclined plane becomes horizontal (Fig. 26a). b Then the force of gravity is no () longer vertical. Now we resolve it into a vertical (P cos p) and (O) a horizontal (P sin component. You can readily see now that we have the preceding problem again, in which the 44 Figure 25: A body thrown with an angle α to an inclined plane. The problem is to find the distance at which the body will land on the inclined plane. S T U D E N T: I once made an attempt to solve such a problem but failed. T E AC H E R: Can’t you see any similarity between this problem and the preceding one? S T U D E N T: No, I can’t. T E AC H E R: Let us imagine that the figure for this problem is turned through the angle β so that the inclined plane becomes horizontal (Figure 26 (a)). Then the force of gravity is no longer vertical. Now we resolve it into a vertical (P cos β) and a horizontal (P sin β) component.
H OW D O YO U G O A B O U T S O LV I N G P RO B L E M S I N K I N E M AT I C S? 57 Figure 26: Rotating the Fig- ure 25 through angle β gives us a problem similar to the one we have solved in the previous section. You can readily see now that we have the preceding problem again, in which the force P sin β plays the role of the force of the wind, and P cos β the role of the force of gravity. Therefore we can find the answer by making use of equation (18) provided that we make the following substitutions: P sin β for F , P cos β for P , and g cos β for g Then we obtain L “ v2 0 sin 2α g cos β p1 ` tan β tan αq (19) At β “ 0, this coincides with equation (17). Of interest is another method of solving the same problem. We introduce the coordinate axes O x and Oy with the origin at the point the body is thrown from (Figure 26 (b)). The inclined plane is represented in these coordinates by the linear function y1 “ ´x tan β and the trajectory of the body is described by the parabola y2 “ ax2 ` b x in which the factors a and b can be expressed in terms of v0, α and β. Next we find the coordinate xA of the point A of intersec- tion of functions y1 and y2 by equating the expressions for these
58 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S functions ´x tan β “ ax2 ` b x From this it follows that xA “ ˆ tan β ` b ´a ˙ Then we can easily find the required distance L “ xA cos β “ ´ tan β ` b a cos β (20) It remains to express factors a and b in terms of v0, α and β. For this purpose, we examine two points of the parabola - B and C (see Figure 26 (b)). We write the equation of the parabola for each of these points: y2C “ ax2 C ` b xC y2B “ ax2 B ` b xB + The coordinates of points C and B are known to us. Conse- quently, the preceding system of equations enables us to de- termine factors a and b . I suggest that in your spare time you complete the solution of this problem and obtain the answer in the form of equation (19). S T U D E N T: I like the first solution better. T E AC H E R: That is a matter of taste. The two methods of solution differ essentially in their nature. The first could be called the “physical” method. It employs simulation which is so typical of the physical approach (we slightly altered the point of view and reduced our problem to the previously discussed problem with the tail wind). The second method could be called “mathemati- cal”. Here we employed two functions and found the coordinates of their points of intersection. In my opinion, the first method is the more elegant, but less general. The field of application of the second method is substan- tially wider. It can, for instance, be applied in principle when the profile of the hill from which the body is thrown is not a straight
H OW D O YO U G O A B O U T S O LV I N G P RO B L E M S I N K I N E M AT I C S? 59 line. Here, instead of the linear function y1, some other func- tion will be used which conforms to the profile of the hill. The first method is inapplicable in principle in such cases. We may note that the more extensive field of application of mathematical methods is due to their more abstract nature. PROBLEMS 1. Body A is thrown vertically upward with a velocity of 20 m{s. At what height was body B which, when thrown at a horizontal velocity of 4 m{s at the same time body A was thrown, collided with it in its flight? The horizontal distance between the initial points of the flight equals 4 m. Find also the time of flight of each body before the collision and the velocity of each at the instant of collision. 2. From points A and B, at the respective heights of 2 m and 6 m, two bodies are thrown simultaneously towards each other: one is thrown horizontally with a velocity of 8 m{s and the other, downward at an angle of 45° to the horizontal and at an initial velocity such that the bodies collide in flight. The horizontal distance between points A and B equals 8 m. Calculate the initial velocity v0 of the body thrown at an angle of 45°, the coordinates x and y of the point of collision, the time of flight t of the bodies before colliding and the velocities vA and vB of the two bodies at the instant of collision. The trajectories of the bodies lie in a single plane. 3. Two bodies are thrown from a single point at the angles α1 and α2 to the horizontal and at the initial velocities v1 and v2, respectively. At what distance from each other will the bodies be after the time t ? Consider two cases: (1) the trajectories of the two bodies lie in a single plane and the bodies are thrown in opposite directions, and (2) the trajectories lie in mutually perpendicular planes. 4. A body falls from the height H with no initial velocity. At the height h it elastically bounces off a plane inclined at an angle of 30° to the horizontal. Find the time it takes the body to reach the ground. 5. At what angle to the horizontal (elevation angle) should a body of weight P be thrown so that the maximum height reached is equal to the range? Assume that a horizontal tail wind of constant force F acts on the body in its flight.
60 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S 6. A stone is thrown upward, perpendicular to an inclined plane with an angle of inclination α. If the initial velocity is v0, at what distance from the point from which it is thrown will the stone fall? 7. A boy 1.5 m tall, standing at a distance of 15 m from a fence 5 m high, throws a stone at an angle of 45° to the horizontal. With what minimum velocity should the stone be thrown to fly over the fence?
§ 6 How Do You Go About Solving Problems In Dynamics? T E AC H E R: In solving problems in dynamics it is especially im- portant to be able to determine correctly the forces applied to the body (see § 2). S T U D E N T: Before we go any further, I wish to ask one question. Assuming that I have correctly found all the forces applied to the body, what should I do next? T E AC H E R: If the forces are not directed along a single straight line, they should be resolved in two mutually perpendicular directions. The force components should be dealt with separately for each of these directions, which we shall call “directions of resolution”. We can begin with some practical advice. In the first place, the forces should be drawn in large scale to avoid confusion in resolv- ing them. In trying to save space students usually represent forces in the form of almost microscopic arrows, and this does not help. You will understand what I mean if you compare your drawing (Figure 8) with mine (Figure 9). Secondly, do not hurry to re- solve the forces before it can be done properly. First you should find all forces applied to the body, and show them in the draw- ing. Only then can you begin to resolve some of them. Thirdly, you must remember that after you have resolved a force you should “forget” about its existence and use only its components. Either the force itself, or its components, no compromise.
62 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S S T U D E N T: How do I choose the directions of resolution? T E AC H E R: In making your choice you should consider the nature of the motion of the body. There are two alternatives: (1) the body is at rest or travels with uniform velocity in a straight line, and (2) the body travels with acceleration and the direction of acceleration is given (at least its sign). In the first case you can select the directions of resolution arbi- trarily, basing (or not basing) your choice on considerations of practical convenience. Assume, for instance, that in the case illus- trated in Figure 10 the body slides with uniform velocity up the inclined plane. Here the directions of resolution may be (with equal advantage) either vertical and horizontal (Figure 27 (a)) or along the inclined plane and perpendicular to it (Figure 27 (b)).p F p horizontal (Fig. 27a) or along the inclined plane and perpen- dicular to it (Fig. 27b). After the forces have been resolved, the algebraic sums of the component forces for each direction of resolution are equated to zero (remember that we are still dealing with the (6) Fig. 27 motion of bodies without acceleration). For the case illust- rated in Fig. 27a we can write the system of equations NCOsa.-F/rsina.-p=O} (21) F-F/rcosa.-N sina.= 0 The system of equations for the case in Fig. 27b is N -P cos a.-F sina.= 0 } F/r+Psinex-Fcosex=O (22) STUDENT: But these systems of equations differ from each other. TEACHER: They do but, nevertheless, lead to the same re- sults, as can readily be shown. Suppose it is required to find the force F that will ensure the motion of the body at uniform velocity up along the inclined plane. Substituting equation (5) into equations (21) we obtain Figure 27: Two ways of resolving the forces applied to a body in motion. After the forces have been resolved, the algebraic sums of the component forces for each direction of resolution are equated to zero (remember that we are still dealing with the motion of bod- ies without acceleration). For the case illustrated in Figure 27 (a) we can write the system of equations N cos α ´ Ff r sin α ´ P “ 0 F ´ Ff r cos α ´ N sin α “ 0 + (21)
H OW D O YO U G O A B O U T S O LV I N G P RO B L E M S I N DY NA M I C S? 63 The system of equations for the case in Figure 27 (b) is N ´ P cos α ´ F sin α “ 0 Ff r ` P sin α ´ F cos α “ 0 + (22) S T U D E N T: But these systems of equations differ from each other. T E AC H E R: They do but, nevertheless, lead to the same results, as can readily be shown. Suppose it is required to find the force F that will ensure the motion of the body at uniform velocity up along the inclined plane. Substituting equation (5) into equations (21) we obtain N pcos α ´ k sin αq ´ P “ 0 F ´ N pk cos α ` sin αq “ 0 + From the first equation of this system we get N “ P cos α ´ k sin α which is substituted into the second equation to determine the required force. Thus F “ P ˆ k cos α ` sin α cos α ´ k sin α ˙ Exactly the same answer is obtained from equations (22). You can check this for yourself. S T U D E N T: What do we do if the body travels with acceleration? T E AC H E R: In this case the choice of the directions of resolution depends on the direction in which the body is being accelerated (direction of the resultant force). Forces should be resolved in a direction along the acceleration and in one perpendicular to it. The algebraic sum of the force components in the direction perpendicular to the acceleration is equated to zero, while that of the force components in the direction along the acceleration is equal, according to Newton’s second law of motion, to the product of the mass of the body by its acceleration.
64 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S Let us return to the body on the inclined plane in the last prob- lem and assume that the body slides with a certain acceleration up the plane. According to my previous remarks, the forces should be resolved as in the case shown in Figure 27 (b). Then, in place of equations (22), we can write the following system N ´ P cos α ´ F sin α “ 0 F cos α ´ Ff r ´ P sin α “ ma “ P ˆ a g ˙ , /. /- (23) Making use of equation (5), we find the acceleration of the body a “ ˆ g p ˙ pF cos α ´ pP cos α ` F sin αq k ´ P sin αq S T U D E N T: In problems of this kind dealing with acceleration, can the forces be resolved in directions other than along the acceleration and perpendicular to it? As far as I understand from your explanation, this should not be done. T E AC H E R: Your question shows that I should clear up some points. Of course, even in problems on acceleration you have a right to resolve the forces in any two mutually perpendicular di- rections. In this case, however, you will have to resolve not only the forces, but the acceleration vector as well. This method of solution will lead to additional difficulties. To avoid unnecessary complications, it is best to proceed exactly as is advised. This is the simplest course. The direction of the body’s acceleration is always known (at least its sign), so you can proceed on the basis of this direction. The inability of examinees to choose the di- rections of force resolution rationally is one of the reasons for their helplessness in solving more or less complex problems in dynamics. S T U D E N T: We have only been speaking about resolution in two directions. In the general case, however, it would probably be more reasonable to speak of resolution in three mutually perpendicular directions. Space is actually three-dimensional. T E AC H E R: You are absolutely right. The two directions in our discussions are explained by the fact that we are dealing
H OW D O YO U G O A B O U T S O LV I N G P RO B L E M S I N DY NA M I C S? 65 with plane (two-dimensional) problems. In the general case, forces should be resolved in three directions. All the remarks made above still hold true, however. I should mention that, as a rule, two-dimensional problems are given in examinations. Though, of course, the examinee may be asked to make a not- too-complicated generalization for the three-dimensional case. PROBLEMS 8. A body with a mass of 5 kg is pulled along a horizontal plane by a force of 3 N1 applied to the body at an angle of 30° to the horizontal. The 1Originally unit of kgf is used, here after we will use the unit of Newtons denoted by N. coefficient of sliding friction is 0.2. Find the velocity of the body 10 s after the pulling force begins to act, and the work done by the friction force during this time. 9. A man pulls two sleds tied together by applying a force of F “ 12 N to the pulling rope at an angle of 45° to the horizontal (Figure 28). The masses of the sleds are equal to m1 “ m2 “ 15 kg. The coefficient of friction between the runners and the snow is 0.02. Find the acceleration of the sleds. the tension of the rope tying the sleds together, and the force with which the man should pull the rope to impart uniform velocity to the sleds.avoid unnecessary complications, it is best to proceed exactly as 1 advised. This is the simplest course. The direction of the body's acceleration is always known (at least its sign), so you can proceed on the basis of this direction. The inability of examinees to choose the directions of force resolution rationally is one of the reasons for their helplessness in solv- ing more or less 'complex problems in dynamics. STUDENT: We have only been speaking about resolution in two directions. In the general case, however, it would pro- bably be more reasonable to speak of resolution in three mutually perpendicular directions. Space is actually three- dimensional. TEACHER: You are absolutely right. The two directions in our discussions are explained by the fact that we are dealing with plane (two-dimensional) problems. In the general case, forces should be resolved in three directions. All the remarks made above still hold true, however. I should mention that, as a rule, two-dimensional problems are given in examina- tions. Though, of course, the examinee may be asked to make a not-too-complicated generalization for the three-dimensional case. 1 PROBLEMS 8. A body with a mass of 5 kg is pulled along a horizontal plane by a force of 3 kgf applied to the body at an angle of 30° to the horizontal. The coefficient of sliding friction is 0.2. Find the velocity of the body 10 seconds after the pulling force begins to act, and the work , done by the friction force during this time. 9. A man pulls two sleds tied together by applying a force of F= 12 kgf to the pulling rope at an angle of 450 to the horizontal (Fig.28). The masses of the sleds are equal to ml=m2= 15 kg. The coefficient of friction between the runners and the snow is 0.02. Find the acceleration of the sleds, Fig. 28 Fig. 29 the tension of the rope .tying the sleds together, and the force with which the man should pull the rope to impart uniform velocity to the sleds. 10. Three equal weights of a mass of 2 kg each are hanging on a string passing over a fixed pulley as shown in Fig. 29. Find the acceleration of the system and the tension of the string connecting weights 1 and 2. 51 Figure 28: Sled being pulled by a rope with an angle 45°. See Problem 9. 10. Three equal weights of a mass of 2 kg each are hanging on a string passing over a fixed pulley as shown in Figure 29. Find the acceleration of the system and the tension of the string connecting weights 1 and 2.necessary complications, it is best to proceed exactly ed. This is the simplest course. The direction of the celeration is always known (at least its sign), so proceed on the basis of this direction. The inability nees to choose the directions of force resolution y is one of the reasons for their helplessness in solv- or less 'complex problems in dynamics. NT: We have only been speaking about resolution in tions. In the general case, however, it would pro- more reasonable to speak of resolution in three perpendicular directions. Space is actually three- onal. HER: You are absolutely right. The two directions in ussions are explained by the fact that we are dealing e (two-dimensional) problems. In the general case, ould be resolved in three directions. All the remarks ve still hold true, however. I should mention that, two-dimensional problems are given in examina- ough, of course, the examinee may be asked to make a complicated generalization for the three-dimensional 1 MS y with a mass of 5 kg is pulled along a horizontal plane by a kgf applied to the body at an angle of 30° to the horizontal. The t of sliding friction is 0.2. Find the velocity of the body 10 seconds lling force begins to act, and the work , fricti
66 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S 11. Calculate the acceleration of the weights and the tension in the strings for the case illustrated in Figure 30. Given: α “ 30°, P1 “ 4 N, P2 “ 2 newton, and P3 “ 8 N. Neglect the friction between the weights and the inclined plane.11. Calculate the acceleration of the weights and the tension In the strings for the case illustrated in Fig. 30. Given: a=30°, P1=4 kgf, P2= =2 kgf, and Ps=8 kgf. Neglect the friction between the weights and the inclined plane. Fig. 30 P, Fig. 31 12. Consider the system of weights shown in Fig. 31. Here Pl= 1 kgf, P2=2 kgf, Ps=5 kgf, P4=0.5 kgf, and a=30°. The coefficient of friction between the weights and the planes equals 0.2. Find the acceleration of the set of weights, the tension of the strings and the force with which weight P4 presses downward on weight Pa. Figure 30: A system of three masses on an incline. See problem 11. 12. Consider the system of weights shown in Figure 31. Here P1 “ 1 NP2 “ 2 N, P5 “ 8 N and P4 “ 0.5 N, and α “ 30°. The coefficient of friction between the weights and the planes equals 0.2. Find the acceleration of the set of weights, the tension of the strings and the force with which weight P4 presses downward on weight P3.11. Calculate the acceleration of the weights and the tension In the strings for the case illustrated in Fig. 30. Given: a=30°, P1=4 kgf, P2= =2 kgf, and Ps=8 kgf. Neglect the friction between the weights and the inclined plane. Fig. 30 P, Fig. 31 12. Consider the system of weights shown in Fig. 31. Here Pl= 1 kgf, P2=2 kgf, Ps=5 kgf, P4=0.5 kgf, and a=30°. The coefficient of friction between the weights and the planes equals 0.2. Find the acceleration of the set of weights, the tension of the strings and the force with which weight P4 presses downward on weight Pa. Figure 31: A system of three masses on an incline. See problem 12
§ 7 Are Problems In Dynamics Much More Diffi- cult To Solve If Friction Is Taken Into Account? T E AC H E R: Problems may become much more difficult when the friction forces are taken into account. S T U D E N T: But we have already discussed the force of friction (§ 3). If a body is in motion, the friction force is determined from the bearing reaction (Ff r “ kN ); if the body is at rest, the friction force is equal to the force that tends to take it out of this state of rest. All this can readily be understood and remembered. T E AC H E R: That is so. However, you overlook one important fact. You assume that you already know the answers to the following questions: (1) Is the body moving or is it at rest? (2) In which direction is the body moving (if at all)? If these items are known beforehand, then the problem is com- paratively simple. Otherwise, it may be very complicated from the outset and may even require special investigation. S T U D E N T: Yes, now I recall that we spoke of this matter in § 2 in connection with our discussion concerning the choice of the direction of the friction force. T E AC H E R: Now I want to discuss this question in more detail. It is my firm opinion that the difficulties involved in, solving prob- lems which take the friction force into account are obviously
68 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S underestimated both by students and by certain authors who think up problems for physics textbooks. Let us consider the example illustrated in Figure 10. The angle of inclination α of the plane, weight P of the body, force F and the coefficient of friction k are given. For simplicity we shall assume that k0 “ k (where k0 is the coefficient determining the maximum possible force of static friction). It is required to determine the kind of motion of the body and to find the acceleration. Let us assume that the body slides upward along the inclined surface. We can resolve the forces as shown in Figure 27 (b) and make use of the result obtained for the acceleration in § 6.Thus a “ g P rF cos α ´ P sin α ´ pP cos α ` F sin αq ks (24) It follows from equation (24) that for the body to slide upward along the inclined plane, the following condition must be com- plied with: F cos α ´ P sin α ´ pP cos α ` F sin αq k ě 0 This condition can be written in the form F ě P ˆ k cos α ` sin α cos α ´ k sin α ˙ or F ě P ˆ k ` tan α 1 ´ k tan α ˙ (25) We shall also assume that the angle of inclination of the plane is not too large, so that p1 ´ k tan αq ą 0 or tan α ă 1 k (26) We shall next assume that the body slides downward along the inclined plane. We again resolve all the forces as in Figure 27 (b) but reverse the friction force. As a result we obtain the fall owing expression for the acceleration of the body a “ g P rP sin α ´ F cos α ´ pP cos α ` F sin αq ks (27)
A R E P RO B L E M S I N DY NA M I C S M U C H M O R E D I F F I C U LT TO S O LV E I F F R I C T I O N I S TA K E N I N TO AC C O U N T? 69 From equation (27) it follows that for the body to slide down- ward along the inclined plane, the following condition must be met: P sin α ´ F cos α ´ pP cos α ` F sin αq k ě 0 This condition we write in the form F ď P ˆ sin α ´ k cos α cos α ` k sin α ˙ or F ď P ˆ tan α ´ k 1 ` k tan α ˙ (28) In this case, we shall assume that the angle of inclination of the plane is not too small, so that ptan α ´ kq ą 0, or tan α ą k (29) Combining conditions (25), (26), (28) and (29), we can come to the following conclusions: (1) Assume that the condition k ă tan α ă 1 k holds good for an inclined plane. Then: (a) if F ą P ˆ k ` tan α 1 ´ k tan α ˙ , the body slides upward with an acceleration that can be determined by equation (24); (b) if F “ P ˆ k ` tan α 1 ´ k tan α ˙ , the body slides upward at uniform velocity or is at rest; (c) if F ă P ˆ tan α ´ k 1 ` k tan α ˙ , the body slides downward with an acceleration that can be determined by equation (27); (d) if F “ P ˆ tan α ´ k 1 ` k tan α ˙ , the body slides downward with uniform velocity or is at rest; (e) if P ˆ tan α ´ k 1 ` k tan α ˙ ă F ă P ˆ k ` tan α 1 ´ k tan α ˙ , the body is at rest.
70 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S Note that upon increase in force F from P ˆ tan α ´ k 1 ` k tan α ˙ to P ˆ k ` tan α 1 ´ k tan α ˙ , the force of static friction is gradually reduced from kpP cos α ` F sin αq to zero; then, after its direction is reversed, it increases to the value kpP cos α ` F sin αq. While this goes on the body remains at rest. (2) Now assume that the inclined plane satisfies the condition 0 ă tan α ď k then: (a) if F ą P ˆ k ` tan α 1 ´ k tan α ˙ , the body slides upward with an acceleration that can be determined by equation (24); (b) if F “ P ˆ k ` tan α 1 ´ k tan α ˙ , the body slides upward at uniform velocity or is at rest; (c) if F ă P ˆ k ` tan α 1 ´ k tan α ˙ , the body is at rest; no downward motion of the body along the inclined plane is possible (even if force F vanishes). (3) Finally, let us assume that the inclined plane meets the condition tan α ě 1 k then: (a) if F ă P ˆ tan α ´ k 1 ` k tan α ˙ , the body slides downward with an acceleration that can be determined by equation (27); (b) if F “ P ˆ tan α ´ k 1 ` k tan α ˙ , the body slides down ward with uniform velocity or is at rest; (c) if F ą P ˆ tan α ´ k 1 ` k tan α ˙ , the body is at rest; no upward motion of the body along the inclined plane is possible. On the face of it, this seems incomprehensible because force F can be increased indefinitely! The inclination of the plane is so
A R E P RO B L E M S I N DY NA M I C S M U C H M O R E D I F F I C U LT TO S O LV E I F F R I C T I O N I S TA K E N I N TO AC C O U N T? 71 large, however, that, with an increase in force F, the pressure of the body against the plane will increase at an even faster rate. S T U D E N T: Nothing of the kind has ever been demonstrated to us in school. T E AC H E R: That is exactly why I wanted to draw your attention to this matter. Of course, in your entrance examinations you will evidently have to deal with much simpler cases: there will be no friction, or there will be friction but the nature of the motion will be known beforehand (for instance, whether the body is in motion or at rest). However, even if one does not have to swim over deep spots, it is good to know where they are. S T U D E N T: What will happen if we assume that k “ 0? T E AC H E R: In the absence of friction, everything becomes much simpler at once. For any angle of inclination of the plane, the results will be: § at F ą P tan α, the body slides upward with the accelera- tion a “ g P pF cos α ´ P sin αq (30) § at F “ P tan α, the body slides with uniform velocity (upward or downward) or is at rest; § at F ă P tan α, the body slides downward with an accelera- tion a “ g P pP sin α ´ F cos αq (31) Note that the results of equations (30) and (31) coincide with an accuracy to the sign. Therefore, in solving problems, you can safely assume any direction of motion, find a and take notice of the sign of the acceleration. If a ą 0, the body travels in the direction you have assumed; if a ă 0, the body will travel in the opposite direction (the acceleration will be equal to |a|). Let us consider one more problem. Two bodies P1 and P2 are connected by a string running over a pulley. Body P1 is on an
72 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S inclined plane with the angle of inclination α and coefficient of friction k; body P2 hangs on the string (Figure 32). Find the acceleration of the system.Assume that the system is moving from left to right. Con.. sidering the motion of the system as a whole, we can write the following equation for the acceleration: (32) Assuming now that the system moves from right to left, we obtain . (33) should be met. Equation (33) irn- plies that for motion from right to left the necessary condition is 1 sina-kcosa Fig. 32 We will carry out an investigation for the given a, and k values, varying the ratio p=P 21Pt • From equation (32) it follows that for motion from left to right, the condition I" sin ce-j-e cos c Here an additional condition is required: the angle of in- clination should not be too small, i.e. tan a,>k. If tan then the system will not move from right to left, however large the ratio p may be. If tan a>k, the body is at rest provided the following ine- quality holds true: 'I 1 sin a+k cosa < p < sin a-k cos a If, instead, tan then the body is at rest at 1 p ;» sin ce-j-e cos c STUDENT: And what will happen if we change the angle ex or the coefficient k? TEACHER: I leave investigation from this point of view to you as a home assignment (see Problems Nos. 13 and 14). 57 Figure 32: Two bodies con- nected via string on an inclined plane with friction. The prob- lem is to find the acceleration of the system. Assume that the system is moving from left to right. Consid- ering the motion of the system as a whole, we can write the following equation for the acceleration: a “ g ˆ P2 ´ P1 sin α ´ P1k cos α P1 ` P2 ˙ (32) Assuming now that the system moves from right to left, we obtain a “ g ˆ P1 sin α ´ P2 ´ P1k cos α P1 ` P2 ˙ (33) We will carry out an investigation for the given α, and k values, varying the ratio p “ P2{P1. From equation (32) it follows that for motion from left to right, the condition p ď 1 sin α ` k cos α should be met. Equation (33) implies that for motion from right to left the necessary condition is p ě 1 sin α ´ k cos α Here an additional condition is required: the angle of inclination should not be too small, i.e. tan α ą k. If tan α ď k, then the system will not move from right to left, however large the ratio p may be. If tan α ą k, the body is at rest provided the following inequality holds true: 1 sin α ` k cos α ă p ă 1 sin α ´ k cos α
A R E P RO B L E M S I N DY NA M I C S M U C H M O R E D I F F I C U LT TO S O LV E I F F R I C T I O N I S TA K E N I N TO AC C O U N T? 73 If, instead, tan α ď k, then the body is at rest at p ą 1 sin α ` k cos α S T U D E N T: And what will happen if we change the angle α or the coefficient k? T E AC H E R: I leave investigation from this point of view to you as a home assignment (see Problems Nos. 13 and 14). PROBLEMS 13. Investigate the problem illustrated in Figure 32 assuming that the angle α of inclination of the plane and the ratio p “ P2{P1 are given, and assigning various values to the coefficient k. 14. Investigate the problem illustrated in Figure 32, assuming that the coeffi- cient of friction k and the ratio p “ P2{P1 are given and assigning various values to the angle α of inclination of the plane. For the sake of simplicity, use only two values of the ratio: p “ 1 (the bodies are of equal weight) and p “ 1{2 (the body on the inclined plane is twice as heavy as the one suspended on the string).
Motion in a circle is the simplest form of curvilinear motion. The more important it is to comprehend the specific features of such motion. You can see that the whole universe is made up of curvilinear motion. Let us consider uniform and non-uniform motion of a material point in a circle, and the motion of orbiting satellites. This will lead us to a discussion of the physical causes of the weightlessness of bodies.
§ 8 How Do You Deal With Motion In A Circle? T E AC H E R: I have found from experience that questions and problems concerning motion of a body in a circle turn out to be extremely difficult for many examinees. Their answers to such questions contain a great many typical errors. To demonstrate this let us invite another student to take part in our discussion. This student doesn’t know what we have discussed previously. We shall conditionally call him “ST U D E N T B” (the first student will hereafter be called “ST U D E N T A”). Will ST U D E N T B please indicate the forces acting on a satellite, or sputnik, in orbit around the earth? We will agree to neglect the resistance of the atmosphere and the attraction of the moon, sun and other celestial bodies. S T U D E N T B: The satellite is subject to two forces: the attraction of the earth and the centrifugal force. T E AC H E R: I have no objections to the attraction of the earth, but I don’t understand where you got the centrifugal force from. Please explain. S T U D E N T B: If there were no such force, the satellite could not stay in orbit. T E AC H E R: And what would happen to it? S T U D E N T B: Why, it would fall to the earth.
78 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S T E AC H E R: (turning to ST U D E N T A): Remember what I told you before! This is a perfect example of an attempt to prove that a certain force exists, not on the basis of the interaction of bodies, but by a backdoor manoeuvre - from the nature of the motion of bodies. As you see, the satellite must stay in orbit, so it is necessary to introduce a retaining force. Incidentally, if this centrifugal force really did exist, then the satellite could not remain in orbit because the forces acting on the satellite would cancel out and it would fly at uniform velocity and in a straight line. S T U D E N T A: The centrifugal force is never applied to a rotating body. It is applied to the tie (string or another bonding member). It is the centripetal force that is applied to the rotating body. S T U D E N T B: Do you mean that only the weight is applied to the satellite? T E AC H E R: Yes, only its weight. S T U D E N T B: And, nevertheless, it doesn’t fall to the earth? T E AC H E R: The motion of a body subject to the force of gravity is called falling. Hence, the satellite is falling. However, its “falling” is in the form of motion in a circle around the earth and there- fore can continue indefinitely. We have already established that the direction of motion of a body and the forces acting on it do not necessarily coincide (see § 4). S T U D E N T B: In speaking of the attraction of the earth and the centrifugal force, I based my statement on the formula GmM r 2 “ mv2 r (34) where the left-hand side is the force of attraction (m=mass of the satellite, M =mass of the earth, r =radius of the orbit and G=gravitational constant), and the right-hand side is the cen- trifugal force (v=velocity of the satellite). Do you mean to say that this formula is incorrect?
H OW D O YO U D E A L W I T H M O T I O N I N A C I RC L E? 79 T E AC H E R: No, the formula is quite correct. What is incorrect is your interpretation of the formula. You regard equation (34) as one of equilibrium between two forces. Actually, it is an expression of Newton’s second law of motion F “ ma (34a) where F “ GmM r 2 and a “ ˆ v2 r ˙ is the centripetal acceleration. S T U D E N T B: I agree that your interpretation enables us to get along without any centrifugal force. But, if there is no centrifugal force, there must at least be a centripetal force. You have not, however, mentioned such a force. T E AC H E R: In our case, the centripetal force is the force of attrac- tion between the satellite and the earth. I want to underline the fact that this does not refer to two different forces. By no means. This is one and the same force. S T U D E N T B: Then why introduce the concept of a centripetal force at all? T E AC H E R: I fully agree with you on this point. The term “cen- tripetal force”, in my opinion, leads to nothing but confusion. What is understood to be the centripetal force is not at all an in- dependent force applied to a body along with other forces. It is, instead, the resultant of all the forces applied to a body travelling in a circle at uniform velocity. The quantity mv2{r is not a force. It represents the product of the mass m of the body by the centripetal acceleration v2{r . This acceleration is directed toward the centre and, consequently, the resultant of all forces, applied to a body travelling in a circle at uniform velocity, is directed toward the centre. Thus, there is a centripetal acceleration and there are forces which, added together, impart a centripetal acceleration to the body. S T U D E N T B: I must admit that this approach to the motion of a body in a circle is to my liking. Indeed, this motion is not a static case, for which an equilibrium of forces is characteristic, but a dynamic case.
80 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S S T U D E N T A: If we reject the concept of a centripetal force, then we should probably drop the term “centrifugal force” as well, even in reference to ties. T E AC H E R: The introduction of the term “centrifugal force” is even less justified. The centripetal force actually exists, if only as a resultant force. The centrifugal force does not even exist in many cases. S T U D E N T A: I don’t understand your last remark. The centrifu- gal force is introduced as a reaction to the centripetal force. If it does not always exist, as you say, then Newton’s third law of motion is not always valid. Is that so? T E AC H E R: Newton’s third law is valid only for real forces deter- mined by the interaction of bodies, and not for the resultants of these forces. I can demonstrate this by the example of the conical pendulum that you are already familiar with (Figure 33).The quantity mv2/r is not a force. It represents the product of the mass m of the body by the centripetal accelerationv 2/r. This acceleration is. directed toward the centre and, conse- quently, the resultant of all forces, applied to a body tra- velling in a circle at uniform velocity, is directed toward the centre. Thus, there is a centripetal acceleration and there are forces which, added together, impart a centripetal acce- leration to the- body. STUDENT B: I must admit that this approach to the motion of a body in a circle is to my liking. Indeed, this motion is not a static case, for which an equilibrium of forces is cha- racteristic, but a dynamic case. STUDENT A: If we reject the concept of a centripetal force, then we should pr.obably drop the term "centrifugal force" as well, even in reference to ties. TEACHER: The introduction of the term "centrifugal force" is even less justified. The centripetal force actually exists, if only as a resultant force. The centrifugal force does not even exist in many cases. STUDENT A: I don't understand your last remark. The cent- rifugal force is introduced as a reaction to the centripetal force. If it does not always exist, as you say, then Newton's third law of motion is not always valid. Is that so? I : 7' I I .,,----t-- -I \, ....__ • centripeta/, force \ centrifugal I " force , I , I , I r; Fig. 33 Fig. 34 TEACHER: Newton's third law is valid only for real forces determined by the interaction of bodies, and not for the resultants of these forces. I can demonstrate this by the example of the conical pendulum that you are already familiar with (Fig. 33). The ball is subject to two forces: the weight P and the tension T of the string. These forces, taken together, provide the centripetal acceleration of the ball, and their sum is called the centripetal force. Force P is due to the inter- action of the ball with the earth. The reaction of this force is force P 1 which is applied to the earth. Force T results 62 Figure 33: A conical pendulum. The body attached to the string performs circular motion in the horizontal plane. During this motion the solid that it traces has the shape of a cone. The ball is subject to two forces: the weight P and the tension T of the string. These forces, taken together, provide the cen- tripetal acceleration of the ball, and their sum is called the cen- tripetal force. Force P is due to the interaction of the ball with the earth. The reaction of this force is force P1 which is applied to the earth. Force T results from interaction between the ball and the string. The reaction of this force is force T1 which is applied to the string. If forces P1 and T1 are formally added together we obtain a force which is conventionally understood to be the centrifugal force
H OW D O YO U D E A L W I T H M O T I O N I N A C I RC L E? 81 (see the dashed line in Figure 33). But to what is this force ap- plied? Are we justified in calling it a force when one of its com- ponents is applied to the earth and the other to an entirely differ- ent body-the string? Evidently, in the given case, the concept of a centrifugal force has no physical meaning. S T U D E N T A: In what cases does the centrifugal force exist? T E AC H E R: In the case of a satellite in orbit, for instance, when only two bodies interact the earth and the satellite. The cen- tripetal force is the force with which the earth attracts the satel- lite. The centrifugal force is the force with which the satellite attracts the earth. S T U D E N T B: You said that Newton’s third law was not valid for the resultant of real forces. I think that in this case it will be invalid also for the components of a real force. Is that true? T E AC H E R: Yes, quite true. In this connection I shall cite an example which has nothing in common with rotary motion. A ball lies on a floor and touches a wall which makes an obtuse angle with the floor (Figure 34).the mass m of the body by the centripetal accelerationv 2/r. This acceleration is. directed toward the centre and, conse- quently, the resultant of all forces, applied to a body tra- velling in a circle at uniform velocity, is directed toward the centre. Thus, there is a centripetal acceleration and there are forces which, added together, impart a centripetal acce- leration to the- body. STUDENT B: I must admit that this approach to the motion of a body in a circle is to my liking. Indeed, this motion is not a static case, for which an equilibrium of forces is cha- racteristic, but a dynamic case. STUDENT A: If we reject the concept of a centripetal force, then we should pr.obably drop the term "centrifugal force" as well, even in reference to ties. TEACHER: The introduction of the term "centrifugal force" is even less justified. The centripetal force actually exists, if only as a resultant force. The centrifugal force does not even exist in many cases. STUDENT A: I don't understand your last remark. The cent- rifugal force is introduced as a reaction to the centripetal force. If it does not always exist, as you say, then Newton's third law of motion is not always valid. Is that so? I : 7' I I .,,----t-- -I \, ....__ • centripeta/, force \ centrifugal I " force , I , I , I r; Fig. 33 Fig. 34 TEACHER: Newton's third law is valid only for real forces determined by the interaction of bodies, and not for the resultants of these forces. I can demonstrate this by the example of the conical pendulum that you are already familiar with (Fig. 33). The ball is subject to two forces: the weight P and the tension T of the string. These forces, taken together, provide the centripetal acceleration of the ball, and their sum is called the centripetal force. Force P is due to the inter- action of the ball with the earth. The reaction of this force is force P 1 which is applied to the earth. Force T results 62 Figure 34: A ball on the floor touching a wall at an obtuse angle with the floor. Let us resolve the weight of the ball into two components: per- pendicular to the wall and parallel to the floor. We shall deal with these two components instead of the weight of the ball. If Newton’s third law were applicable to separate components, we could expect a reaction of the wall counter balancing the com- ponent of the weight perpendicular to it. Then, the component of the weight parallel to the floor would remain unbalanced and the ball would have to have a horizontal acceleration. Obviously, this is physically absurd. S T U D E N T A: So far you have discussed uniform motion in a circle. How do you deal with a body moving nonuniformly in a circle? For instance, a body slides down from the top of a vertically held hoop. While it slides along the hoop it is moving in a circle. This cannot, however, be uniform motion because the velocity of the body increases. What do you, do in such cases?
82 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S T E AC H E R: If a body moves in a circle at uniform velocity, the resultant of all forces applied to the body must be directed to the centre; it imparts centripetal acceleration to the body. In the more general case of nonuniform motion in a circle, the resultant force is not directed strictly toward the centre. In this case, it has a component along a radius toward the centre and another component tangent to the trajectory of the body (i.e. to the circle). The first component is responsible for the centripetal acceleration of the body, and the second component, for the so-called tangential acceleration, associated with the change in velocity. It should be pointed out that since the velocity of the body changes, the centripetal acceleration v2{r must also change. S T U D E N T A: Does that mean that for each instant of time the centripetal acceleration will be determined by the formula a “ v2{r , where v is the instantaneous velocity? T E AC H E R: Exactly. While the centripetal acceleration is constant in uniform motion in a circle, it varies in the process of motion in nonuniform motion in a circle. S T U D E N T A: What does one do to find out just how the velocity v varies in nonuniform rotation? T E AC H E R: Usually, the law of conservation of energy is resorted to for this purpose. Let us consider a specific example. Assume that a body slides without friction from the top of a vertically held hoop of radius R. With what force will the body press on the hoop as it passes a point located at a height h cm below the top of the hoop? The initial velocity of the body at the top of the hoop equals zero. First of all, it is necessary to find what forces act on the body. S T U D E N T A: Two forces act on the body: the weight P and the bearing reaction N . They are shown in Figure 35. T E AC H E R: Correct. What are you going to do next?
H OW D O YO U D E A L W I T H M O T I O N I N A C I RC L E? 83c' of a vertically held hoop of radius R. With what force will (a) the body press on the hoop as it passes a point located at a height h em below the top of Fi.g. 35 Fig. 36 the hoop? The initial velocity of the body at the top of the hoo equals zero. First of all, it is necessary to find what force act on the body. STUDENT A: Two forces act on the body: the weight P an the bearing reaction N. They are shown in Fig. 35. TEACHER: Correct. What are you going to do next? 64 Figure 35: Resolution of forces acting on a body in nonuniform circular motion. S T U D E N T A: I’m going to do as you said. I shall find the resultant of these two forces and resolve it into two components: one along the radius and the other tangent to the circle. T E AC H E R: Quite right. Though it would evidently be simpler to start by resolving the two forces applied to the body in the two directions instead of finding the resultant, the more so because it will be necessary to resolve only one force - the weight. S T U D E N T A: My resolution of the forces is shown in Figure 35. T E AC H E R: Force P2 is responsible for the tangential acceleration of the body, it does not interest us at present. The resultant of forces P1 and N causes the centripetal acceleration of the body, i.e. P1 ´ N “ mv2 R (35) The velocity of the body at the point we are interested in (point A in Figure 35) can be found from the law of conservation of energy P h “ mv2 2 (36) Combining equations (35) and (36) and taking into consideration that P1 “ P cos α “ P ˆ R ´ h R ˙ , we obtain P R pR ´ hq ´ N “ ˆ 2P h R ˙ ,
84 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S The sought-for force with which the body presses on the hoop is equal, according to Newton’s third law, to the bearing reaction N “ P ˆ R ´ 3h R ˙ (37) S T U D E N T B: You assume that at point A the body is still on the surface of the hoop. But it may fly off the hoop before it gets to point A. T E AC H E R: We can find the point at which the body leaves the surface of the hoop. This point corresponds to the extreme case when the force with which the body presses against the hoop is reduced to zero. Consequently, in equation (37), we assume N “ 0 and solve for h, i.e, the vertical distance from the top of the hoop to the point at which the body flies off. Thus h0 “ R 3 (38) If in the problem as stated the value of h complies with the condition h ă h0 then the result of equation (37) is correct; if, instead, h ě h0, then N “ 0. S T U D E N T A: As far as I can make out, two physical laws, equa- tions (35) and (36), were used to solve this problem. T E AC H E R: Very good that you point this out. Quite true, two laws are involved in the solution of this problem: Newton’s second law of motion [see equation (35)] and the law of conser- vation of energy [see equation (36)]. Unfortunately, examinees do not always clearly understand just which physical laws they employ in solving some problem or other. This, I think, is an essential point. Take, for instance, the following example. An initial velocity v0 is imparted to a body so that it can travel from point A to point C . Two alternative paths leading from A to C are offered (Figure 36 (a), (b)). In both cases the body must reach the same height H , but in different ways. Find the minimum initial veloc- ity v0 for each case. Friction can be neglected.
H OW D O YO U D E A L W I T H M O T I O N I N A C I RC L E? 85 S T U D E N T B: I think that the minimum initial velocity should be the same in both cases, because there is no friction and the same height is to be reached. This velocity can be calculated from the law of conservation of energy m g H “ mv2 0 2 from which v0 “ a2g Hc' a=v where v is the instantaneous velocity? " TEACHER: Exactly. While the centripetal acceleration is constant in uniform motion in a circle, it varies in the process of motion in nonuniform motion in a circle. STUDENT A: What does one do to find out just how the velo- city v varies in nonuniform rotation? TEACHER: Usually, the law of conservation of energy is resorted to for this purpose. Let us consider a specific example. A ssume that a body slides without friction from the top B of a vertically held hoop of radius R. With what force will (a) the body press on the hoop as it passes a point located at a height h em below the top of Fi.g. 35 Fig. 36 the hoop? The initial velocity of the body at the top of the hoop equals zero. First of all, it is necessary to find what forces act on the body. STUDENT A: Two forces act on the body: the weight P and the bearing reaction N. They are shown in Fig. 35. TEACHER: Correct. What are you going to do next? 64 Figure 36: A body travels from point A to point C via two different paths. The problem is to find the minimum initial velocity in the two cases. T E AC H E R: Your answer is wrong. You overlooked the fact that in the first case, the body passes the upper point of its trajectory when it is in a state of rotational motion. This means that at the top point B (Figure 36 (a)) it will have a velocity v1 determined from a dynamics equation similar to equation (35). Since the problem involves the finding of a minimum, we should consider the extreme case when the pressure of the body on its support at point B is reduced to zero. Then only the weight will be acting on the body and imparting to it the centripetal acceleration. Thus m g “ mv2 1 R “ 2mv2 1 H (39) Adding to the dynamics equation (39) the energy equation mv2 0 2 “ mv2 1 2 ` m g H (40) we find that the minimum initial velocity equals. b 5g H {2. In the second case, the body may pass the top point at a velocity in- finitely close to zero and so we can limit ourselves to the energy equation. Then your answer is correct. S T U D E N T B: Now I understand. If in the first case the body had no velocity at point B, it would simply fall off its track. T E AC H E R: If in the first case the body had the initial velocity v0 “ a2g H as you suggested, it would never reach point B but would fall away from the track somewhat earlier. I propose that you find the height h of the point at which the body would fall away from the track if its initial velocity was v0 “ a2g H . S T U D E N T A: Please let me try to do this problem.
86 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S(42) STUDENT B: Now I understand. If in the first case the body had no velocity at point B, it would simply fall off its track. TEACHER: If in the first case the body had the initial velocity vo=V2gH as you suggested, it would never reach point B but would fall away from the track somewhat earlier. J'propose that you find the height h of the point at which the body would fall away from the B A. track if its initial velocity was vo=V2gH. STUDENT A: Please let me try to do this problem. TEACHER: Certainly. STUDENT A: At the point the body drops off the track the bearing rea.. Fig. 37 ction is evidently equal to zero. Therefore, only the weight acts on the body at this point. We can resolve the weight into two components, one along the radius (mg cos a) and the other perpendicular to the radius (mg sin a) as shown in Fig. 37 (point A is the point at which the body falls away from the track). The component along the radius imparts a centri- petal acceleration to the body, determined by the equation mv; mgcos a= -R (41) where V2 is the velocity of the body at point A. To find it we can make use of the energy equation 2· 2 mV2 + h mvo -2- mg =-2- Combining the dynamics (41) and energy (42) equations, tak- ing into consideration that cos a= (h-R)/R, we obtain mg(h-R)= from which n-. 6g After substituting the final result is 5 h=r;H 3* (43) 67 Figure 37: A body moving on a circular track. The problem is to find the point at which it would fall away from the track if the initial velocity is v0 “ a2g H . T E AC H E R: Certainly. S T U D E N T A: At the point the body drops off the track the bearing reaction is evidently equal to zero. Therefore, only the weight acts on the body at this point. We can resolve the weight into two components, one along the radius (m g cos α) and the other perpendicular to the radius (m g sin α) as shown in Figure 37 (point A is the point at which the body falls away from the track). The component along the radius imparts a centripetal acceleration to the body, determined by the equation m g cos α “ mv2 2 2 (41) where v2 is the velocity of the body at point A. To find it we can make use of the energy equation mv2 2 2 ` m g h “ mv2 0 2 (42) Combining the dynamics (41) and energy (42) equations, taking into consideration that cos α “ ph ´ Rq{R we obtain m g ph ´ Rq “ mv2 0 ´ 2m g h from which h “ 2v2 0 ` g H 6g (43) After substituting v2 0 “ 2g H the final result is h “ 5 6 H
H OW D O YO U D E A L W I T H M O T I O N I N A C I RC L E? 87 T E AC H E R: Entirely correct. Note that you can use equation (43) to find the initial velocity v0 for the body to loop the loop. For this we take h “ H in equation (43). Then H “ 2v2 0 ` g H 6g From this we directly obtain the result previously determined v0 “ c 5g H 2 S T U D E N T A: Condition (43) was obtained for a body falling off its track. How can it be used for the case in which the body loops the loop without falling away? T E AC H E R: Falling away at the very top point of the loop actually means that the body does not fall away but passes this point, continuing its motion in a circle. S T U D E N T B: One could say that the body falls away as if only for a single instant.(C) rbJ N(c NCOSd N(ko Frr sind / STU --- Ffr d' two .... Fff' COSd known TEA We do unkno The u po p 38 easily 19. first e STUDENT A: After dividing we cos a+ko sin a ko cos a- sin a which we solve for the required k _ (j)2R cos cx 0- gcosa-(j) perpendicular to the acceleration cussed in § 6. STUDENT A: Now I understand. in the horizontal and vertical dire The vertical components of the fa) anothe ponen the b N c FJr Tak Ffr=k we can the fo TEACHER: It is evident from eq tion (g cos ex- (J)2R s should hold true. This condition form tan « < Figure 38: A body lies on a ro- tating inclined plane. Problem is to find the maximum value of static friction coefficient at which body remains on the plane without sliding off. T E AC H E R: Quite true. In conclusion I propose the following problem (Figure 38 (a)). A body lies at the bottom of an in- clined plane with an angle of inclination α. This plane rotates at uniform angular velocity ω about a vertical axis. The distance from the body to the axis of rotation of the plane equals R. Find the minimum coefficient k0 (I remind you that this coefficient characterizes the maximum possible value of the force of static friction) at which the body remains on the rotating inclined plane without sliding off. Let us begin as always with the question: what forces are applied to the body? S T U D E N T A: Three forces are applied to the body: the weight P , bearing reaction N , and the force of friction Ff r . T E AC H E R: Quite correct. It’s a good thing that you didn’t add the centripetal force. Now what are you going to do next?
88 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S S T U D E N T A: Next, I shall resolve the forces in the directions along the plane and perpendicular to it as shown in Figure 38 (b). T E AC H E R: I’ll take the liberty of interrupting you at this point. I don’t like the way you have resolved the forces. Tell me, in what direction is the body accelerated? S T U D E N T A: The acceleration is directed horizontally. It is centripetal acceleration. T E AC H E R: Good. That is why you should resolve the forces horizontally (i.e. along the acceleration) and vertically (i.e. per- pendicular to the acceleration). Remember what we discussed in § 6. S T U D E N T A: Now I understand. The resolution of the forces in the horizontal and vertical directions is shown in Figure 38 (c). The vertical components of the forces counterbalance one an- other, and the horizontal components impart acceleration to the body. Thus N cos α ` Ff r sin α “ P Ff r cos α ´ N sin α “ mv2 R , /. /- Taking into consideration that Ff r “ k0N , ˆ v2 R ˙ “ ω2R and m “ ˆ P g ˙ , we can rewrite these equations in the form N pcos α ` k0 sin αq “ P N pk0 cos α ´ sin αq “ P ω2Rg + S T U D E N T B: You have only two equations and three unknowns: k0, P and N . T E AC H E R: That is no obstacle. We don’t have to find all three unknowns, only the coefficient k0. The unknowns P and N can be easily eliminated by dividing the first equation by the second.
H OW D O YO U D E A L W I T H M O T I O N I N A C I RC L E? 89 S T U D E N T A: After dividing we obtain cos α ` k0 sin α k0 cos α ´ sin α “ g ω2R which we solve for the required coefficient k0 “ ˜ ω2R cos α ` g sin α g cos α ´ ω2R sin α ¸ (44) T E AC H E R: It is evident from equation (44) that the condition ´ g cos α ´ ω2R sin α ¯ ą 0 should hold true. This condition can also be written in the form tan α ă g ω2R (45) If condition (45) is not complied with, no friction force is capable of retaining the body on the rotating inclined plane. PROBLEMS 15. What is the ratio of the forces with which an army tank bears down on the middle of a convex and of a concave bridge? The radius of curvature of the bridge is 40 m in both cases and the speed of the tank is 45 km{h. 16. A body slides without friction from the height H “ 60 cm and then loops the loop of radius R “ 20 cm (Figure 39). Find the ratio of the forces with which the body bears against the track at points A, B and C .If condition (45) is not complied with, no friction force is capable of retaining the body on the rotating inclined plane. PROBLEMS 15. What is the ratio of the forces with which an army tank bears down on the middle of a convex and of a concave bridge? The radius of curvature of the bridge is 40 m in both cases and the speed of the tank is 45 km per hr. 16. A body slides without friction from the height H=60 em and then loops the loop of radius R=20 ern (Fig. 39). Find ,the ratio of the :t; forces with wnich the body bears against the track at points A, Band C. 17. A body can rotate in A a vertical plane at the end of a string of length R. What Fig. 39 horizontal velocity should be imparted to the body in the top position so that the tension of the string in the bottom position is ten times as great as the weight of the body? 18. Calculate the density of the substance of a spherical planet if a satellite rotates about it with a period T in a circular orbit at a distance from the surface of the planet equal to one half of its radius R. The gravitat ional constant is denoted by G. Figure 39: A body slides from a given height and then loops on a circular track. Problem is to find ratio of forces which the body bears against the track at points A, B and C . See Problem 16. 17. A body can rotate in a vertical plane at the end of a string of length R. What horizontal velocity should be imparted to the body in the top posi- tion so that the tension of the string in the bottom position is ten times as great as the weight of the body?
90 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S 18. Calculate the density of the substance of a spherical planet if a satellite rotates about it with a period T in a circular orbit at a distance from the surface of the planet equal to one half of its radius R. The gravitational constant is denoted by G. 19. A body of mass m can slide without friction along a trough bent in the form of a circular arc of radius R. At what height h will the body be at rest if the trough rotates at a uniform angular velocity ω (Figure 40) about a vertical axis? What force F does the body exert on the trough?of the bridge is 40 m in both cases and the speed of the tank is 45 km per hr. 16. A body slides without friction from the height H=60 em and then loops the loop of radius R=20 ern (Fig. 39). Find ,the ratio of the :t; forces with wnich the body bears against the track at points A, Band C. 17. A body can rotate in A a vertical plane at the end of a string of length R. What Fig. 39 horizontal velocity should be imparted to the body in the top position so that the tension of the string in the bottom position is ten times as great as the weight of the body? 18. Calculate the density of the substance of a spherical planet if a satellite rotates about it with a period T in a circular orbit at a distance from the surface of the planet equal to one half of its radius R. The gravitat ional constant is denoted by G. Fig. 40 Fig. 41 19. A body of mass m can slide without friction along a trough bent in the form of a circular arc of radius- R. At what height h will the body be at rest if the trough rotates at a uniform angular veloci- ty ro (Fig. 40) about a vertical axis? What force F does the body exert on the trough? 20. A hoop of radius R is fixed vertically on the floor. A body slides without friction from the top of the hoop (Fig. 41). At what distance 1 from the point where the hoop is fixed will the body fall? Figure 40: A body slides with- out friction on a uniformly rotating circular trough. Prob- lem is to find the force the body exerts on the trough and the height at which the body will be at rest. See Problem 19. 20. A hoop of radius R is fixed vertically on the floor. A body slides without friction from the top of the hoop (Figure 41). At what distance l from the point where the hoop is fixed will the body fall?ondition (45) is not complied with, no friction force able of retaining the body on the rotating inclined EMS hat is the ratio of the forces with which an army tank bears down iddle of a convex and of a concave bridge? The radius of curvature idge is 40 m in both cases and the speed of the tank is 45 km per hr. 16. A body slides without friction from the height H=60 em and then loops the loop of radius R=20 ern (Fig. 39). Find ,the ratio of the forces with wnich the body bears against the track at points A, Band C. 17. A body can rotate in A a vertical plane at the end of a string of length R. What Fig. 39 horizontal velocity should be imparted to the body in the tion so that the tension of the string in the bottom position es as great as the weight of the body? 8. Calculate the density of the substance of a spherical planet if ite rotates about it with a period T in a circular orbit at a e from the surface of the planet equal to one half of its radius gra vitat ional constant is denoted by G. Fig. 40 Fig. 41 A body of mass m can slide without friction along a trough e form of a circular arc of radius- R. At what height h will be at rest if the trough rotates at a uniform angular veloci- ig. 40) about a vertical axis? What force F does the body exert ugh? hoop of radius R is fixed vertically on the floor. A body ithout friction from the top of the hoop (Fig. 41). At what e 1 from the point where the hoop is fixed will the body fall? Figure 41: A body slides without friction form top of the hoop. Problem is to find distance l from the point where the hoop is fixes and the body will fall. See Problem 20.
§ 9 How Do You Explain The Weightlessness Of Bodies? T E AC H E R: How do you understand the expression: At the equa- tor of a planet, a body weightless than at the poles? S T U D E N T B: I understand it as follows. The attraction of a body by the earth is less at the equator than at the poles for two rea- sons. In the first place, the earth is somewhat flattened at the poles and therefore the distance from the centre of the earth is somewhat less to the poles than to the equator. In the second place, the earth rotates about its axis as a result of which the force of attraction at the equator is weakened due to the centrifugal effect. S T U D E N T A: Please make your last remark a little clearer. S T U D E N T B: You must subtract the centrifugal force from the force of attraction. S T U D E N T A: I don’t agree with you. Firstly, the centrifugal force is not applied to a body travelling in a circle. We already discussed that in the preceding section (§ 8). Secondly, even if such a force existed it could not prevent the force of attraction from being exactly the same as if there was no rotation of the earth. The force of attraction equals GmM {r 2 and, as such, does not change just because other forces may act on the body. T E AC H E R: As you can see, the question of the “weightness of bodies” is not as simple as it seems at first glance. That’s why it
92 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S is among the questions that examinees quite frequently fail to an- swer correctly. As a matter of fact, if we agree on the definition that the “weight of a body” is the force with which the body is attracted by the earth, i.e. the force GmM {r 2, then the reduc- tion of weight at the equator should be associated only with the flattening at the poles (or bulging at the equator). S T U D E N T B: But you cannot disregard rotation of the earth! T E AC H E R: I fully agree with you. But first I wish to point out that usually, in everyday life, the “weight of a body” is under- stood to be, not the force with which it is attracted to the earth, and this is quite logical, but the force measured by a spring bal- ance, i.e. the force with which the body bears against the earth. In other words, the bearing reaction is measured (the force with which a body bears against a support is equal to the bearing reaction according to Newton’s third law). It follows that the expression “a body weighs less at the equator than at the poles” means that at the equator it bears against its support with a lesser force than at the poles. Let us denote the force of attraction at the poles by P1 and at the equator by P2, the bearing reaction at the poles by N1 and at the equator by P2. At the poles the body is at rest, and at the equator it travels in a circle. Thus P1 ´ N1 “ 0 P2 ´ N2 “ mac p where ac p is the centripetal acceleration. We can rewrite these equations in the form N1 “ P1 N2 “ P2 ´ mac p + (46) Here it is clear that N2 is less than N1 since, firstly, P2 is less than P1 (from the effect of the flattening at the poles) and, secondly, we subtract from P2 the quantity mac p (the effect of rotation of the earth).
H OW D O YO U E X P L A I N T H E W E I G H T L E S S N E S S O F B O D I E S? 93 S T U D E N T B: So, the expression “a body has lost half of its weight” does not mean that the force with which it is attracted to the earth (or any other planet) has been reduced by one half? T E AC H E R: No, it doesn’t. The force of attraction may not change at all. This expression means that the force with which the body bears against its support (in other words, the bearing reaction) has been reduced by one half. S T U D E N T B: But then it follows that I can dispose of the “weight- ness” of a body quite freely. What can prevent me from digging a deep pit under the body and letting it fall into the pit together with its support? In this case, there will be no force whatsoever bearing on the support. Does that mean that the body has com- pletely “lost its weight”? That it is in a state of weightlessness? T E AC H E R: You have independently come to the correct conclu- sion. As a matter of fact, the state of weightlessness is a state of fall of a body. In this connection I wish to make several remarks. I have come across the interpretation of weightlessness as a state in which the force of attraction of the earth is counterbalanced by some other force. In the case of a satellite, the centrifugal force was suggested as this counterbalancing force. The statement was as follows: the force with which the satellite is attracted by the earth and the centrifugal force counter balance each other and, as a result, the resultant force applied to the satellite equals zero, and this corresponds to weightlessness. You now understand, of course, that such a treatment is incor- rect if only because no centrifugal force acts on the satellite. Incidentally, if weightlessness is understood to be a state in which the force of attraction is counterbalanced by some other force, then it would be more logical to consider a body weightless when it simply rests on a horizontal plane. This is precisely one of the cases where the weight is counterbalanced by the bearing reaction! Actually, no counter balancing of the force of attraction is required for weightlessness. On the contrary, for a body to become weightless, it is necessary to provide conditions in which no other force acts on it except attraction. In other words, it is
94 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S necessary that the bearing reaction equal zero. The motion of a body subject to the force of attraction is the falling of the body. Consequently, weightlessness is a state of falling, for example the falling of a lift in a mine shaft or the orbiting of the earth by a satellite. S T U D E N T A: In the preceding section (§ 8) you mentioned that the orbiting of a satellite about the earth is none other than its falling to the earth for an indefinitely long period of time. T E AC H E R: That the motion of a satellite about the earth is falling can be shown very graphically in the following way. Imagine that you are at the top of a high mountain and throw a stone horizontally. We shall neglect the air resistance. The greater the initial velocity of the stone, the farther it will fall. Figure 42 (a) illustrates how the trajectory of the stone gradually changes with an increase in the initial velocity. At a certain velocity v1 the trajectory of the falling stone becomes a circle and the stone becomes a satellite of the earth. The velocity v1 is called the circular orbital velocity. It is found from equation (34) v1 “ c GM r (47)If the radius r of the satellite's orb equal to the radius of the earth then STUDENT A: What will happen if the mountain top we continue to in TEACHER: The stone will orbit th elongated ellipse (Fig. 42b). At a trajectory of the stone becomes ceases to be a satellite of the earth. the e to ca xima is ab ST v, ned as a the i reach the s In th that How the w (6) TE Weig the f sun. ST lJ2 weig locat stella with in of so TE Fig. 42 ST to m weightlessness as a state of falling A parachutist also falls, but he h associated with weightlessness. TEACHER: You are right. Weig kind of falling. Weightlessness is the motion of a body SUbject only I have already mentioned that for a 74 Figure 42: Trajectory of a body around Earth with different initial velocities. If the radius r of the satellite’s orbit is taken approximately equal to the radius of the earth then v1 « 8 km{s. S T U D E N T A: What will happen if in throwing a stone from the mountain top we continue to increase the initial velocity? T E AC H E R: The stone will orbit the earth in a more and more elongated ellipse (Figure 42 (b)). At a certain velocity v2 the trajectory of the stone becomes a parabola and the stone ceases to be a satellite of the earth. The velocity v2 is called the escape velocity. According to calculations, v2 « 11 km{s. This is about ?2 times v1. S T U D E N T A: You have defined the state of weightlessness as a state of fall. However, if the initial velocity of the stone reaches the escape velocity, the stone will leave the earth. In this case,
H OW D O YO U E X P L A I N T H E W E I G H T L E S S N E S S O F B O D I E S? 95 you cannot say that it is falling to the earth. How, then, can you interpret the weightlessness of the stone? T E AC H E R: Very simply. Weightlessness in this case is the falling of the stone to the sun. S T U D E N T A: Then the weightlessness of a spaceship located somewhere in interstellar space is to be associated with the falling of the ship in the gravitational field of some celestial body? T E AC H E R: Exactly. S T U D E N T B: Still, it seems to me that the definition of weightless- ness as a state of falling requires some refinement. A parachutist also falls, but he has none of the sensations associated with weightlessness. T E AC H E R: You are right. Weightlessness is not just any kind of falling. Weightlessness is the so-called free fall, i.e. the motion of a body subject only (!) to the force of gravity. I have already mentioned that for a body to become weightless it is necessary to create conditions under which no other force, except the force of attraction, acts on the body. In the case of the fall of a parachutist, there is an additional force, the resistance of the air. PROBLEMS 21. Calculate the density of the substance of a spherical planet where the daily period of rotation equals 10 hours, if it is known that bodies are weightless at the equator of the planet.
The role of the physical laws of conservation can scarcely be overesti- mated. They constitute the most general rules established by mankind on the basis of the experience of many generations. Skillful application of the laws of conservation enables many problems to be solved with comparative ease. Let us consider examples concerning the laws of conservation of energy and momentum.
§ 10 Can You Apply The Laws Of Conservation Of Energy And Linear Momentum? T E AC H E R: To begin with I wish to propose several simple prob- lems. The first problem: Bodies slide without friction down two in- clined planes of equal height H but with two different angles of inclination α1 and α2. The initial velocity of the bodies equals zero. Find the velocities of the bodies at the end of their paths. The second problem: We know that the formula expressing the final velocity of a body in terms of the acceleration and distance travelled v “ ?2as refers to the case when there is no initial velocity. What will this formula be if the body has an initial velocity v0? The third problem: A body is thrown from a height H with a horizontal velocity of v0. Find its velocity when it reaches the ground. The fourth problem: A body is thrown upward at an angle a to the horizontal with an initial velocity v0. Find the maximum height reached in its flight. S T U D E N T A: I shall solve the first problem in the following way. We first consider one of the inclined planes, for instance the one with the angle of inclination α1. Two forces are applied to the body: the force of gravity P and the bearing reaction N1. We
100 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S resolve the force P into two components, one along the plane (P sin α1) and the other perpendicular to it (P cos α1). We then write the equations for the forces perpendicular to the inclined plane P cos α1 ´ N1 “ 0 and for the forces along the plane P sin α1 “ Pa1 g where a1 is the acceleration of the body. From the second equa- tion we find that a1 “ g sin α1. The distance travelled by the body is H { sin α1. Next, using the formula mentioned in the second problem, we find that the velocity at the end of the path is v1 “ ?a1 s1 “ d 2g H sin α1 sin α1 “ a2g H Since the final result does not depend upon the angle of inclina- tion, it is also applicable to the second plane inclined at the angle α2. To solve the second problem I shall make use of the well known kinematic relationships v “ v0 ` at s “ v0 t ` at 2 2 From the first equation we find that t “ pv ´ v0q{a. Substituting this for t in the second equation we obtain s “ v0pv ´ v0q a ` a 2 pv ´ v0q2 a2 or 2sa “ 2v0v ´ 2v2 0 ` v2 ´ 2v0v ` v2 0 from which 2sa “ v2 ´ v2 0 . The final result is v “ b 2as ` v2 0 (48)
C A N YO U A P P LY T H E L AW S O F C O N S E RVAT I O N O F E N E RG Y A N D L I N E A R M O M E N T U M? 101 To solve the third problem, I shall first find the horizontal v1 and vertical v2 components of the final velocity. Since the body travels at uniform velocity in the horizontal direction, v1 “ v0. In the vertical direction the body travels with acceleration g but has no initial velocity. Therefore, we can use the formula v2 “ a2g H . Since the sum of the squares of the sides of a right triangle equals, the square of the hypotenuse, the final answer is v “ b v2 1 ` v2 2 “ b v2 0 ` 2g H (49) The fourth problem has already been discussed in § 5. It is nec- essary to resolve the initial velocity into the horizontal (v0 cos α) and vertical (v0 sin α) components. Then we consider the vertical motion of the body and, first of all, we find the time t1 of ascent from the formula for the dependence of the velocity on time in uniformly decelerated motion pvv “ v0 sin α ´ g t q, taking into account that at t “ t1 the vertical velocity of the body vanishes. Thus v0 sin α ´ g t1 “ 0, from which t1 “ pv0{g q sin α. The time t1 being known, we find the height H reached from the formula of the dependence of the distance travelled on time in uniformly decelerated motion. Thus H “ v0 t1 sin α ´ g t 2 1 2 “ v2 0 2g sin α T E AC H E R: In all four cases you obtained the correct answers. I am not, however, pleased with the way you solved these prob- lems. They could all have been solved much simpler if you had used the law of conservation of energy. You can see for yourself. First problem. The law of conservation of energy is of the form m g H “ mv2{2 (the potential energy of the body at the top of the inclined plane is equal to its kinetic energy at the bottom). From this equation we readily find the velocity of the body at the bottom v “ a2g H
102 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S Second problem. In this case, the law of conservation of energy is of the form mv2 0 {2 ` mas “ mv2{2, where mas is the work done by the forces in imparting the acceleration a to the body. This leads immediately to v2 0 ` 2as “ v2 or, finally, to v “ b 2as ` v2 0 Third problem. We write the law of conservation of energy as m g H ` mv2 0 2 “ mv2{2. Then the result is v “ b 2g H ` v2 0 Fourth problem. At the point the body is thrown its energy equals mv2 0 {2. At the top point of its trajectory, the energy of the body is m g H ` mv2 1 {2. Since the velocity v1 at the top point equals v0 cos α, then, using the law of conservation of energy mv2 0 2 “ m g H ` mv2 0 2 cos2 α we find that H “ ˜ v2 0 2g ¸ ´ 1 ´ cos2 α ¯ or, finally H “ ˜ v2 0 2g ¸ sin2 α S T U D E N T A: Yes, it’s quite clear to me that these problems can be solved in a much simpler way. It didn’t occur to me to use the law of conservation of energy. T E AC H E R: Unfortunately, examinees frequently forget about this law. As a result they begin to solve such problems by more cumbersome methods, thus increasing the probability of errors. My advice is: make more resourceful and extensive use of the law of conservation of energy. This poses the question: how skillfully can you employ this law?
C A N YO U A P P LY T H E L AW S O F C O N S E RVAT I O N O F E N E RG Y A N D L I N E A R M O M E N T U M? 103 S T U D E N T A: It seems to me that no special skill is required; the law of conservation of energy as such is quite simple. T E AC H E R: The ability to apply a physical law correctly is not de- termined by its complexity or simplicity. Consider an example. Assume that a body travels at uniform velocity in a circle in a horizontal plane. No friction forces operate. The body is subject to a centripetal force. What is the work done by this force in one revolution of the body? S T U D E N T A: Work is equal to the product of the force by the distance through which it acts. Thus, in our case, it equals pmv2{Rq2πR “ 2πmv2, where R is the radius of the circle and m and v are the mass and velocity of the body. T E AC H E R: According to the law of conservation of energy; work cannot completely disappear. What has become of the work you calculated? S T U D E N T A: It has been used to rotate the body. T E AC H E R: I don’t understand. State it more precisely. S T U D E N T A: It keeps the body on the circle. T E AC H E R: Your reasoning is wrong. No work at all is required to keep the body on the circle. S T U D E N T A: Then I don’t know how to answer your question. T E AC H E R: Energy imparted to a body can be distributed, as physicists say, among the following “channels”: (1) increasing the kinetic energy of the body; (2) increasing its potential energy; (3) work performed by the given body on other bodies; and (4) heat evolved due to friction. Such is the general principle which not all examinees understand with sufficient clarity. Now consider the work of the centripetal force. The body travels at a constant velocity and therefore its kinetic energy is not increased. Thus the first channel is closed. The body travels in a horizontal plane and, consequently, its potential energy is not changed. The second channel is also closed. The given body does
104 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S not perform any work on other bodies, so that the third channel is closed. Finally, all kinds of friction have been excluded. This closes the fourth and last channel. S T U D E N T A: But then there is simply no room for the work of the centripetal force, or is there? T E AC H E R: As you see, none. It remains now for you to declare your position on the matter. Either you admit that the law your this your of conservation of energy is not valid, and then all troubles are gone, or you proceed from the validity of law and then . . . . However, try to find the way out of difficulties. S T U D E N T A: I think that it remains to conclude that the cen- tripetal force performs no work whatsoever. T E AC H E R: That is quite a logical conclusion. I want to point out that it is the direct consequence of the law of conservation of energy. S T U D E N T B: All this is very well, but what do we do about the formula for the work done by a body? T E AC H E R: In addition to the force and the distance through which it acts, this formula should also contain the cosine of the angle between the direction of the force and the velocity A “ F s cos α. In the given case, cos α “ 0. S T U D E N T A: Oh yes, I entirely forgot about this cosine. T E AC H E R: I want to propose another example. Consider com- municating vessels connected by a narrow tube with a stopcock. Assume that at first all the liquid is in the left vessel and its height is H (Figure 43(a)). Then we open the stopcock and the liquid flows from the left into the right vessel. The flow ceases when there is an equal level of H {2 in each vessel (Figure 43(b)). Let us calculate the potential energy of the liquid in the initial and final positions. For this we multiply the weight of the liquid in each vessel by one half of the column of liquid. In the initial
C A N YO U A P P LY T H E L AW S O F C O N S E RVAT I O N O F E N E RG Y A N D L I N E A R M O M E N T U M? 105 position the potential energy equalled P H {2, and in the final one it is ˆ P 2 ˙ ˆ H 4 ˙ ` ˆ P 2 ˙ ˆ H 4 ˙ “ ˆ P H 4 ˙ .Fig. 43 (C) TEACHER: That is quite point out that' it is the dir servation of energy. STUDENT B: All this is v the formula for the work do TEACHER: In addition to which it acts, this formula fa) the a force In th ST got a TE er ex vesse with all th its h open f rom fl ow le vel Let of th posit weigh one the in equal is Thus energy of the liquid turns ou initial state. Where has one STUDENT A: I shall attem potential energy PH/4 could Figure 43: Problem with potential energy of a liquid column in a communicating vessel. Thus in the final state, the potential energy of the liquid turns out to be only one half of that in the initial state. Where has one half of the energy disappeared to? S T U D E N T A: I shall attempt to reason as you advised. The poten- tial energy P H 4 could be used up in performing work on other bodies, on heat evolved in friction and on the kinetic energy of the liquid itself. Is that right? T E AC H E R: Quite correct. Continue. S T U D E N T A: In our case, the liquid flowing from one vessel to the other does not perform any work on other bodies. The liquid has no kinetic energy in the final state because it is in a state of rest. Then, it remains to conclude that one half of the potential energy has been converted into heat evolved in friction. True, I don’t have a very clear idea of what kind of friction it is. T E AC H E R: You reasoned correctly and came to the right con- clusion. I want to add a few words on the nature of friction. One can imagine that the liquid is divided into layers, each char- acterizing a definite rate of flow. The closer the layer to the walls of the tube, the lower its velocity. There is an exchange of molecules between the layers, as a result of which molecules with a higher velocity of ordered motion find themselves among molecules with a lower velocity of ordered motion, and vice versa. As a result, the “faster” layer has an accelerating effect on the “slower” layer and, conversely, the “slower” layer has a decelerating effect on the “faster” layer. This picture allows us to speak of the existence of a peculiar internal friction between the layers. Its effect is stronger with a greater difference in the velocities of the layers in the middle part of the tube and near the walls. Note that the velocity of the layers near the walls is influenced by the kind of interaction
106 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S between the molecules of the liquid and those of the walls. If the liquid wets the tube then the layer directly adjacent to the wall is actually stationary. S T U D E N T A: Does this mean that in the final state the tempera- ture of the liquid should be somewhat higher than in the initial state? T E AC H E R: Yes, exactly so. Now we shall change the conditions of the problem to some extent. Assume that there is no interac- tion between the liquid and the tube walls. Hence, all the layers will flow at the same velocity and there will be no internal fric- tion. How then will the liquid flow from one vessel to the other? S T U D E N T A: Here the potential energy will be reduced owing to the kinetic energy acquired by the liquid. In other words, the state illustrated in Figure 43(b) is not one of rest. The liquid will continue to flow from the left vessel to the right one until it reaches the state shown in Figure 43(c). In this state the potential energy is again the same as in the initial state Figure 43(a). T E AC H E R: What will happen to the liquid after this? S T U D E N T A: The liquid will begin to flow back in the reverse direction, from the right vessel to the left one. As a result, the levels of the liquid will fluctuate in the communicating vessels. T E AC H E R: Such fluctuations can be observed, for instance, in communicating glass vessels containing mercury. We know that mercury does not wet glass. Of course these fluctuations will be damped in the course of time, since it is impossible to completely exclude the interaction between the molecules of the liquid and those of the tube walls. S T U D E N T A: I see that the law of conservation of energy can be applied quite actively. T E AC H E R: Here is another problem for you. A bullet of mass m, travelling horizontally with a velocity v0, hits a wooden block of mass M , suspended on a string, and sticks in the block. To
C A N YO U A P P LY T H E L AW S O F C O N S E RVAT I O N O F E N E RG Y A N D L I N E A R M O M E N T U M? 107 what height H will the block rise, after the bullet hits it, due to deviation of the string from the equilibrium position (Figure 44)?Fig. :44 this state the potential energy is again the same as in the initial state (Fig. 43a). TEACHER: What will happen to the liquid after this? STUDENT A: The liquid will begin to flow back in the reverse direction, from the right vessel to the left one. As a result, the levels of the liquid will fluctuate in the commu- nicating vessels. TEACHER: Such fluctuations can be observed, for instance, in communicating glass vessels containing mercury. We know that mercury does not wet glass. Of course these fluctuations will be damped in the course of time, since it is impossible to completely exclude the interaction between the molecules of the liquid and those of the tube walls. STUDENT A: I see that the law of conservation of energy can be applied quite actively. TEACHER: Here is another prob- lem for you. A bullet of mass m, travellinghorizontally with a velocity v 0, hits a wooden block of mass M, suspended on a string, and sticks in the block. To what height H will the block rise, after the bullet hits it, due to devia- tion of the string from the equilibrium position (Fig. 44)? STUDENT A: We shall denote by Vt the velocity of the block with the bullet immediately after the bullet hits the block. To find this velocity we make use of the law of conservation of energy. Thus from which 2 2 m;o = (m + M) (50) VI = Vo y- (51) This velocity being known, we find the sought-for height H by again resorting te the law of conservation of energy 2 (m+ M) gH = (m + M) (52) Equations (50) and (52) can be combined mv2 (m+M)gH=T 83 Figure 44: A bullet hits a block of wood and sticks in it. The problem is to find the height to which the block will rise after the impact. S T U D E N T A: We shall denote by v1 the velocity of the block with the bullet immediately after the bullet hits the block. To find this velocity we make use of the law of conservation of energy. Thus mv2 o 2 “ pm ` M q v2 1 2 (50) from which v1 “ vo c m m ` M (51) This velocity being known, we find the sought-for height H by again resorting to the law of conservation of energy pm ` M qg H “ pm ` M q v2 1 2 (52) Equations (50) and (52) can be combined pm ` M qg H “ mv2 0 2 from which H “ m pm ` M q v2 0 2g (53) T E AC H E R: (to ST U D E N T B) What do you think of this solution? S T U D E N T B: I don’t agree with it. We were told previously that in such cases the law of conservation of momentum is to be used.
108 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S Therefore, instead of equation (50) I would have used a different relationship mv0 “ pm ` M qv1 (54) (the momentum of the bullet before it hits the block is equal to the momentum of the bullet and block afterward). From this it follows that v1 “ v0 m pm ` M q (55) If we now use the law of conservation of energy (52) and substi- tute the result of equation (55) into (52) we obtain H “ ˆ m m ` M ˙2 v2 0 2g (56) T E AC H E R: We have two different opinions and two results. The point is that in one case the law of conservation of kinetic energy is applied when the bullet strikes the block, and in the other case, the law of conservation of momentum. Which is correct? (to ST U D E N T A): What can you say to justify your position? S T U D E N T A: It didn’t occur to me to use the law of conservation of momentum. T E AC H E R: (to ST U D E N T B): And what do you say? S T U D E N T B: I don’t know how to substantiate my position. I remember that in dealing with collisions, the law of conservation of momentum is always valid, while the law of conservation of energy does not always hold good. Since in the given case these laws lead to different results, my solution is evidently correct. T E AC H E R: As a matter of fact, it is indeed quite correct. It is, however, necessary to get a better insight into the matter. A collision after which the colliding bodies travel stuck together (or one inside the other) is said to be a “completely inelastic collision”. Typical of such impacts is the presence of permanent set in the colliding bodies, as a result of which a certain amount of heat is evolved. Therefore, equation (50), referring only to the kinetic energy of bodies, is inapplicable. In our case, it is
C A N YO U A P P LY T H E L AW S O F C O N S E RVAT I O N O F E N E RG Y A N D L I N E A R M O M E N T U M? 109 necessary to employ the law of conservation of momentum (54) to find the velocity of the box with the bullet after the impact. S T U D E N T A: Do you mean to say that the law of conservation of energy is not valid for a completely inelastic collision? But this law is universal. T E AC H E R: There is no question but that the law of conserva- tion of energy is valid for a completely inelastic collision as well. The kinetic energy is not conserved after such a collision. I specifically mean the kinetic energy and not the whole energy. Denoting the heat evolved in collision by Q, we can write the following system of laws of conservation referring to the com- pletely inelastic collision discussed above mv0 “ pm ` M qv1 mv2 0 2 “ pm ` M qv2 1 2 ` Q , /. /- (57) Here the first equation is the law of conservation of momentum, and the second is the law of conservation of energy (including not only mechanical energy, but heat as well). The system of equations (57) contains two unknowns: v1 and Q. After de- termining v1 from the first equation, we can use the second equation to find the evolved heat Q Q “ mv2 0 2 ´ pm ` M qm2v2 0 2pm ` M q2 “ mv2 0 2 ˆ 1 ´ m m ` M ˙ (58) It is evident from this equation that the larger the mass M , the more energy is converted into heat. In the limit, for an infinitely large mass. M , we obtain Q “ mv2 0 {2, i.e. all the kinetic energy is converted into heat. This is quite natural: assume that the bullet sticks in a wall. S T U D E N T A: Can there be an impact in which no heat is evolved? T E AC H E R: Yes, such collisions are possible. They are said to be “perfectly elastic”. For instance, the impact of two steel balls can be regarded as perfectly elastic with a fair degree of approxima- tion. Purely elastic deformation of the balls occurs and no heat
110 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S is evolved. After the collision, the balls return to their original shape. S T U D E N T A: You mean that in a perfectly elastic collision, the law of conservation of energy becomes the law of conservation of kinetic energy? T E AC H E R: Yes, of course. S T U D E N T A: But in this case I cannot understand how you can reconcile the laws of conservation of momentum and of energy. We obtain two entirely different equations for the velocity after impact. Or, maybe, the law of conservation of momentum is not valid for a perfectly elastic collision. T E AC H E R: Both conservation laws are valid for a perfectly elastic impact: for momentum and for kinetic energy. You have no rea- son to worry about the reconciliation of these laws because after a perfectly elastic impact, the bodies fly apart at different veloc- ities. Whereas after a completely inelastic impact the colliding bodies travel at the same velocity (since they stick together), after an elastic impact each body travels at its own definite velocity. Two unknowns require two equations. Let us consider an example. Assume that a body of mass m travelling at a velocity v0 elastically collides with a body of mass M at rest. Further assume that as a result of the impact the incident body bounces back. We shall denote the velocity of body m after the collision by v1 and that of body M by v2. Then the laws of conservation of momentum and energy can be written in the form mv0 “ M v2 ´ mv1 mv2 0 2 “ mv2 2 2 ` mv2 1 2 , /. /- (59) Note the minus sign in the first equation. It is due to our assump- tion that the incident body bounces back. S T U D E N T B: But you cannot always know beforehand in which direction a body will travel after the impact. Is it impossible for
C A N YO U A P P LY T H E L AW S O F C O N S E RVAT I O N O F E N E RG Y A N D L I N E A R M O M E N T U M? 111 the body m to continue travelling in the same direction but at a lower velocity after the collision? T E AC H E R: That is quite possible. In such a case, we shall obtain a negative velocity v1 when solving the system of equations (59). S T U D E N T B: I think that the direction of travel of body m after the collision is determined by the ratio of the masses m and M . T E AC H E R: Exactly. If m ă M , body m will bounce back; at m “ M , it will be at rest after the collision; and at m ą M , it will continue its travel in the same direction but at a lower velocity. In the general case, however, you need not worry about the direction of travel. It will be sufficient to assume some direction and begin the calculations. The sign of the answer will indicate your mistake, if any. S T U D E N T B: We know that upon collision the balls may fly apart at an angle to each other. We assumed that motion takes place along a single straight line. Evidently, this must have been a special case. T E AC H E R: You’ are right. We considered what is called a central collision in which the balls travel before and after the impact along a line passing through their centres. The more general case of the off-centre collision will be dealt with later. For the time being, I’d like to know if everything is quite clear. S T U D E N T A: I think I understand now. As I see it, in any colli- sion (elastic or inelastic), two laws of conservation are applicable: of momentum and of energy. Simply the different nature of the impacts leads to different equations for describing the conserva- tion laws. In dealing with inelastic collisions, it is necessary to take into account the heat evolved on impact. T E AC H E R: Your remarks are true and to the point. S T U D E N T B: So far as I understand it, completely elastic and perfectly inelastic collisions are the two extreme cases. Are they always suitable for describing real cases?
112 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S T E AC H E R: You are right in bringing up this matter. The cases of collision we have considered are extreme ones. In real colli- sions some amount of heat is always generated (no ideally elastic deformation exists) and the colliding bodies may fly apart with different velocities. In many cases, however, real collisions are described quite well by means of simplified models: completely elastic and perfectly inelastic collisions. Now let us consider an example of an off-centre elastic collision. A body in the form of an inclined plane with a 45° angle of inclination lies on a horizontal plane. A ball of mass m, flying horizontally with a velocity v0, collides with the body (inclined plane), which has a mass of M . As a result of the impact, the ball bounces vertically upward and the body M begins to slide without friction along the horizontal plane. Find the velocity with which the ball begins its vertical travel after the collision. (Figure 45). Which of you wishes to try your hand at this problem?(60) Fig. 45 m STUDENT B: Allow me to. Let us denote the sought-for ve· locity of the ball by VI and that of body M by V2' Since the collision is elastic, I have the right to assume that the kinetic energy is conserved. Thus 22M 2 mvo = mVl 222 I need one more equation, for which I should evidently use the law of conservation of momentum. I shall write it in the form mvO=Mv2+mvt (61) True, I'm not so sure about this last equation because velocity VI is at right angles to velocity V2' • TEACHER: Equation (60) is correct. Equation (61) is in- correct, just as you thought. You should remember that the law of conservation of momentum is a 111 vector equation, since the momentum is a vector quantity having the 'same direction as the velocity. True enough, when all the velocities are directed along a single straight line, the vector ","77fi77h'?77J.'777;"7j'J;i"];1 equation can be replaced by a scalar one. That is precisely what happened when we discussed central collisions. In the general case, however, it is necessary to resolve all velocities in mutually perpendicular directions and to write the law of conservation of momentum for each of these directions separately (if the problem is considered in a plane, the vector equation can be replaced by two scalar equations for the projections of the momen- tum in the two mutually perpendicular directions). For the given problem we can choose the horizontal and vertical directions. For the horizontal direction, the law of conservation of momentum is of the form (62)mvo= MV 2 From equations (60) and (62) we find the velocity v1 = VO V M M m STUDENT B: What do we do about the vertical direction? TEACHER: At first sight, it would seem that the law of conservation of momentum is not valid for the vertical dire- Figure 45: A ball collides with another body at rest in the form of an inclined plane. The problem is to find the velocity of the ball after the impact. S T U D E N T B: Allow me to. Let us denote the sought-for velocity of the ball by v1 and that of body M by v2. Since the collision is elastic, I have the right to assume that the kinetic energy is conserved. Thus mv2 0 2 “ mv2 1 2 ` mv2 2 2 (60) I need one more equation, for which I should evidently use the law of conservation of momentum. I shall write it in the form mv0 “ M v2 ` mv1 (61) True, I’m not so sure about this last equation because velocity v1 is at right angles to velocity v2.
C A N YO U A P P LY T H E L AW S O F C O N S E RVAT I O N O F E N E RG Y A N D L I N E A R M O M E N T U M? 113 T E AC H E R: Equation (60) is correct. Equation (61) is incorrect, just as you thought. You should remember that the law of con- servation of momentum is a vector equation, since the momen- tum is a vector quantity having the same direction as the velocity. True enough, when all the velocities are directed along a single straight line, the vector equation can be replaced by a scalar one. That is precisely what happened when we discussed central col- lisions. In the general case, however, it is necessary to resolve all velocities in mutually perpendicular directions and to write the law of conservation of momentum for each of these directions separately (if the problem is considered in a plane, the vector equation can be replaced by two scalar equations for the pro- jections of the momentum in the two mutually perpendicular directions). For the given problem we can choose the horizontal and vertical directions. For the horizontal direction, the law of conservation of momentum is of the form mv0 “ M v2 (62) From equations (60) and (62) we find the velocity v1 “ v0 c M ´ m m S T U D E N T B: What do we do about the vertical direction? T E AC H E R: At first sight, it would seem that the law of conserva- tion of momentum is not valid for the vertical direction. Actu- ally it is. Before the impact there were no vertical velocities; after the impact, there is a momentum mv1, directed vertically up- wards. We can readily see that still another body participates in the problem: the earth. If it was not for the earth, body M would not travel horizontally after the collision. Let us denote the mass of the earth by MC and the velocity it acquires as a result of the impact by vC. The absence of friction enables us to treat the interaction be- tween the body M and the earth as taking place only in the
114 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S vertical direction. In other words, the velocity vC of the earth is directed vertically downwards. Thus, the participation of the earth in our problem doesn’t change the form of equation (62), but leads to an equation which describes the law of conservation of momentum for the vertical direction mv1 ´ MCvC “ 0 (63) S T U D E N T B: Since the earth also participates in this problem it will evidently be necessary to correct the energy relation (60). T E AC H E R: Just what do you propose to do to equation (60)? S T U D E N T B: I wish to add a term concerning the motion of the earth after the impact mv2 0 2 “ mv2 1 2 ` M v2 2 2 ` MCv2 C 2 (64) T E AC H E R: Your intention is quite logical. There is, however, no need to change equation (60). As a matter of fact, it follows from equation (63) that the velocity of the earth is vC “ mv1 MC Since the mass MC is practically infinitely large, the velocity vC of the earth after the impact is practically equal to zero. Now, let us rewrite the term MCv2 C{2 in equation (64) to obtain the form pMCvCqvC{2. The quantity MCvC in this product has, according to equation (63), a finite value. If this value is multiplied by zero (in the given case by vC) the product is also zero. From this we can conclude that the earth participates very pe- culiarly in this problem. It acquires a certain momentum, but at the same time, receives practically no energy. In other words, it participates in the law of conservation of momentum, but does not participate in the law of conservation of energy. This circumstance is especially striking evidence of the fact that the laws of conservation of energy and of momentum are essentially different, mutually independent laws.
C A N YO U A P P LY T H E L AW S O F C O N S E RVAT I O N O F E N E RG Y A N D L I N E A R M O M E N T U M? 115 PROBLEMS 22. A body with a mass of 3 kg falls from a certain height with an initial ve- locity of 2 m{s, directed vertically downward. Find the work done to overcome the forces of resistance during 10 s if it is known that the body acquired a velocity of 50 m{s at the end of the 10 s interval. Assume that the force of resistance is constant. 23. A body slides first down an inclined plane at an angle of 30° and then along a horizontal surface. Determine the coefficient of friction if it is known that the body slides along the horizontal surface the same distance as along the inclined plane. 24. Calculate the efficiency of an inclined plane for the case when a body slides off it at uniform velocity. 25. A ball of mass m and volume V drops into water from a height H , plunges to a depth h and then jumps out of the water (the density of the ball is less than that of water). Find the resistance of the water (assuming it to be constant) and the height h1 to which the ball ascends after jumping out of the water. Neglect air resistance. The density of water is denoted by ρw . 26. A railway car with a mass of 50 tons, travelling with a velocity of 12 km{h, runs into a flatcar with a mass of 30 tons standing on the same track. Find the velocity of joint travel of the railway car and flatcar directly after the automatic coupling device operates. Calculate the distance travelled by the two cars after being coupled if the force of resistance is 5 per cent of the weight. 27. A cannon of mass M , located at the base of an inclined plane, shoots a shell of mass m in a horizontal direction with an initial velocity v0. To what height does the cannon ascend the inclined plane as a result of recoil if the angle of inclination of the plane is α and the coefficient of friction between the cannon and the plane is k? 28. Two balls of masses M and 2M are hanging on threads of length l fixed at the same point. The ball of mass M is pulled to one side through an angle of α and is released after imparting to it a tangential velocity of v0 in the direction of the equilibrium position. To what height will the balls rise after collision if: (a) the impact is perfectly elastic, and (b) if it is completely inelastic (the balls stick together as a result of the impact)? 29. A ball of mass M hangs on a string of length l . A bullet of mass m, flying horizontally, hits the ball and sticks in it. At what minimum velocity must the bullet travel so that the ball will make one complete revolution in a vertical plane?
116 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S 30. Two wedges with angles of inclination equal to 45° and each of mass M lie on a horizontal plane (Figure 46). A ball of mass m (m ! M ) drops freely from the height H . It first strikes one wedge and then the other and bounces vertically upward. Find the height to which the ball bounces. Assume that both impacts are elastic and that there is no friction between the wedges and the plane.ball is less than that of water it to be constant) and the hei out of the water. Neglect ai by P'W. 26. A railway car with a of 12 km per hr, runs into a same track. Find the velocity _--om direc I " opera : two c I tance I 2 , of an tl: : a hor I To w : ned p I inclin I fricti 28 . ing FIg. 46 point throu imparting to it a tangential rium position. To what heigh impact is perfectly elastic, a stick together as a result of th 29. A ball of mass M han flying horizontally, hits the city must the bullet travel so tion in a vertical plane? 90 Figure 46: A ball strikes one of the two edges after being dropped from a height. First it strikes the second wedge and then goes vertically up. The problem is to find the height to which the ball rises. 31. A wedge with an angle of 30° and a mass M lies on a horizontal plane. A ball of mass m drops freely from the height H , strikes the wedge elastically and bounces away at an angle of 30° to the horizontal. To what height does the ball ascend? Neglect friction between the wedge and the horizontal plane.
The world about us is full of vibrations and waves. Remember this when you study the branch of physical science devoted to these phenom- ena. Let us discuss harmonic vibrations and, as a special case, the vi- brations of a mathematical pendulum. We shall analyse the behaviour of the pendulum in non-inertial frames of reference.
§ 11 Can You Deal With Harmonic Vibrations? T E AC H E R: Some examinees do not have a sufficiently clear un- derstanding of harmonic vibrations. First let us discuss their definition. S T U D E N T A: Vibrations are said to be harmonic if they obey the sine law: the deviation x of a body from its equilibrium position varies with time as follows x “ Asinpωt ` aq (65) where A is the amplitude of vibration (maximum deviation of the body from the position of equilibrium), ω is the circular frequency (ω “ 2πT , where T is the period of vibration), and α is the initial phase (it indicates the deviation of the body from the position of equilibrium at the instant of time t “ 0). The idea of harmonic vibrations is conveyed by the motion of the projection of a point which rotates at uniform angular velocity ω in a circle of radius A (Figure 47).Fig. 47 § II. CAN YOU DEAL WITH HARMONIC VIBRATIONS? .frequency (ro=2njT, wher and a is the initial phase body from the position o time t=O). The idea of ha the point lar v (Fig. STU tion know of t direc libriu reced Harm which the restoring force tion x of the body from t Such a force is said to be " TEACHER: I am fully sa tions. In the first case, ha the basis of how they occur of their cause. In other w Figure 47: Simple harmonic motion and its relations. S T U D E N T B: I prefer another definition of harmonic vibrations. As is known, vibrations occur due to action of the restoring force, i.e. a force directed toward the position of equilibrium and increasing as the body recedes from the equilibrium position. Harmonic vibrations are those in which the restoring force F is proportional to the deviation x of the body from the equilibrium position. Thus F “ k x (66) Such a force is said to be “elastic”.
120 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S T E AC H E R: I am fully satisfied with both proposed definitions. In the first case, harmonic vibrations are defined on the basis of how they occur; in the second case, on the basis of their cause. In other words, the first definition uses the space-time (kine- matic) description of the vibrations, and the second, the causal (dynamic) description. S T U D E N T B: But which of the two definitions is preferable? Or, maybe, they are equivalent? T E AC H E R: No, they are not equivalent, and the first (kinematic) is preferable. It is more complete. S T U D E N T B: But whatever the nature of the restoring force, it will evidently determine the nature of the vibrations. I don’t understand, then, why my definition is less complete. T E AC H E R: This is not quite so. The nature of the restoring force does not fully determine the nature of the vibrations. S T U D E N T A: Apparently, now is the time to recall that the nature of the motion of a body at a given instant is determined not only by the forces acting on the body at the given instant, but by the initial conditions as well, i.e. the position and velocity of the body at the initial instant. We discussed this in § 4. T E AC H E R: Absolutely correct. With reference to the case being considered this statement means that the nature of the vibrations is determined not only by the restoring force, but by the con- ditions under which these vibrations started. It is evident that vibrations can be effected in various ways. For example, a body can be deflected a certain distance from its equilibrium position and then smoothly released. It will begin to vibrate. If the begin- ning of vibration is taken as the zero instant, then from equation (65), we obtain α “ π{2, and the distance the body is deflected is the amplitude of vibration. The body can be deflected different distances from the equilibrium position, thereby setting different amplitudes of vibration. Another method of starting vibrations is to impart a certain ini- tial velocity (by pushing) to a body in a state of equilibrium. The
C A N YO U D E A L W I T H H A R M O N I C V I B R AT I O N S? 121 body will begin to vibrate. Taking the beginning of vibration as the zero point, we obtain from equation (65) that α “ 0. The amplitude of these vibrations depends upon the initial velocity imparted to the body. It is evidently possible to propose innu- merable other, intermediate methods of exciting vibrations. For instance, a body is deflected from its position of equilibrium and, at the same time, is pushed or plucked, etc. Each of these methods will set definite values of the amplitude A and the initial phase α of the vibration. S T U D E N T B: Do you mean that the quantities A and α do not depend upon the nature of the restoring force? T E AC H E R: Exactly. You manipulate these two quantities at your own discretion when you excite vibrations by one or another method. The restoring force, i.e. coefficient k in equation (66), determines only the circular frequency ω or, in other words, the period of vibration of the body. It can be said that the period of vibration is an intrinsic characteristic of the vibrating body, while the amplitude A and the initial phase α depend upon the external conditions that excite the given vibration. Returning to the definitions of harmonic vibrations, we see that the dynamic definition contains no information on either the amplitude or initial phase. The kinematic definition, on the contrary, contains information on these quantities. S T U D E N T B: But if we have such a free hand in dealing with the amplitude, maybe it is not so important a characteristic of a vibrating body? T E AC H E R: You are mistaken. The amplitude is a very important characteristic of a vibrating body. To prove this, let us consider an example. A ball of mass m is attached to two elastic springs and accomplishes harmonic vibrations of amplitude A in the horizontal direction Figure 48. The restoring force is determined by the coefficient of elasticity k which characterizes the elastic properties of the springs. Find the energy of the vibrating ball. S T U D E N T A: To find the energy of the ball, we can consider its
122 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C Sties. .. STUDENT B: aut if we have such a free hand in dealing with the amplitude, maybe it is not so important a characteristic of a vibrating body? TEACHER: You are mistaken. The amplitude is a very im- portant characteristic of a vibrating body. To prove this, let us consider an example. A ball of mass m is attached to two o Fig. 48 Fig. 49 elastic springs and accomplishes harmonic vibrations of ampli- tude A in the horizontal direction (Fig. 48). The restoring force is determined by the coefficient of elasticity k which characterizes the elastic properties of the springs. Find the energy of the vibrating ball. STUDENT A: To find the energy of the ball, we can consider its position of extreme deflection (x=A). In this position, the velocity' of the ball equals zero and therefore its total energy is its potential energy. The latter can be determined as the work done against the restoring force F in displacing the 95 Figure 48: Simple harmonic motion and its relations. position of extreme deflection (x “ A). In this position, the velocity of the ball equals zero and therefore its total energy is its potential energy. The latter can be determined as the work done against the restoring force F in displacing the ball the distance A from its equilibrium position. Thus W “ F A (67) Next, taking into account that F “ kA, according to equation (66), we obtain W “ kA2 T E AC H E R: You reasoned along the proper lines, but committed an error. Equation (67) is applicable only on condition that the force is constant. In the given case, force F varies with the distance, as shown in Figure 49. The work done by this force over the distance x “ A is equal to the hatched area under the force curve. This is the area of a triangle and is equal to kA2 2 . Thus W “ kA2 2 (68) Note that the total energy of a vibrating body is proportional toB: Do you mean that the quantities A and a do on the nature of the restoring force? : Exactly. You manipulate these two quantities scretion when you excite vibrations by one or od. The restoring force, i. e. coefficient k in 6), determines only the circular frequency <U or, s, the period of vibration of the body. It can be e period of vibration is an intrinsic characteris- ibrating body, while the amplitude A and the e a depend upon the external conditions that en vibration. to the definitions of harmonic vibrations, we dynamic definition contains no information on mplitude or initial phase. The kinematic defini- ntrary , contains information on these quanti- .. B: aut if we have such a free hand in dealing with e, maybe it is not so important a characteristic ng body? : You are mistaken. The amplitude is a very im- racteristic of a vibrating body. To prove this, let example. A ball of mass m is attached to two o Fig. 48 Fig. 49 gs and accomplishes harmonic vibrations of ampli- e horizontal direction (Fig. 48). The restoring ined by the coefficient of elasticity k which Figure 49: Force varying with distance.
C A N YO U D E A L W I T H H A R M O N I C V I B R AT I O N S? 123 the square of the amplitude of vibration. This demonstrates what an important characteristic of a vibrating body the amplitude is. If 0 ă x ă A, then the total energy W is the sum of two components-the kinetic and potential energies W “ kA2 2 “ mv2 2 ` k x2 2 (69) Equation (69) enables the velocity v of the vibrating ball to be found at any distance x from the equilibrium position. My next question is: what is the period of vibration of the ball shown in Figure 48? S T U D E N T B: To establish the formula for the period of vibration it will be necessary to employ differential calculus. T E AC H E R: Strictly speaking, you are right. However, if we si- multaneously use the kinematic and dynamic definitions of har- monic vibrations we can manage without differential calculus. As a matter of fact, we can’ conclude from Figure 47, which is a graphical expression of the kinematic definition, that the velocity of the body at the instant it passes the equilibrium position is v1 “ ωA “ 2πA T (70) Using the result of equation (68), following from the dynamic definition, we can conclude that velocity v1 can be found from the energy relation mv2 1 2 “ kA2 2 (71) (at the instant the ball passes the equilibrium position the entire energy of the ball is kinetic energy). Combining equations (70) and (71), we obtain 4π2A2 m T 2 “ kA2, from which T “ 2π c m k (72) As mentioned previously, the period of vibration is determined fully by the properties of the vibrating system itself, and is inde- pendent of the way the vibrations are set up.
124 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S S T U D E N T A: When speaking of vibrations we usually deal, not with a ball attached to springs, but with a pendulum. Can the obtained results be generalized to include the pendulum? T E AC H E R: For such generalization we must first find out what, in the case of the pendulum, plays the role of the coefficient of elasticity k. It is evident that a pendulum vibrates not due to an elastic force, but to the force of gravity. Let us consider a ball (called a bob in a pendulum) suspended on a string of length l . We pull the bob to one side of the equilibrium position so that the string makes an angle α, (Figure 50) with the vertical. Two forces act on the bob: the force of gravity m g and the tension T of the string. Their resultant is the restoring force. As is evident from the figure, it equals m g sin α.from the energy relation kA2 -2-=-2- (71) (at the instant the ball passes the equilibrium position the entire energy of the ball is kinetic energy). Combining equa- tions (70) and (71), we obtain 4n 2A 2m/T2=kA 2, from which T=2n (72) As mentioned previously, the period of vibration is determined fully by the properties of the vibrating system itself, and is independent of the way the vib- rations are set up. STUDENT A: When speaking of vibrations we usually deal, not with a ball attached to springs, but with a pendulum. Can the obtained results be generalized to include the pendulum? TEACHER: For such generali- c zation we must first find out I ,\ what, in the case of the pendu- \ lurn, plays the role of the coef- \ ficient of elasticity k. It is evident I , \ that a pendulum vibrates not due \ to an elastic force, but to the force ,,- mi'costr of gravity. Let us consider a ball (called a bob in a pendulum) Fig. 50 suspended on a string of length 1. We pull the bob to one side of the equilibrium position so that the string makes an angle (1, (Fig. 50) with the vertical. Two forces act on the bob: the force of gravity mg and the tension T of the string. Their resultant is the restoring force. As is evident from the figu- re, it equals mg sin (1,. STUDENT A: Which of the lengths, AB or AC, should be considered the deflection of the pendulum from the equilib- rium position (see Fig. 50)? . TEACHER: We are analysing the harmonic vibrations of a pendulum. For this it is necessary that the angle of maximum deviation of the string from the equilibrium position be very small ex 1 (73) Figure 50: Analysing the motion of a pendulum. S T U D E N T A: Which of the lengths, AB or AC , should be con- sidered the deflection of the pendulum from the equilibrium position (see Figure 50)? . T E AC H E R: We are analysing the harmonic vibrations of a pendu- lum. For this it is necessary that the angle of maximum deviation
C A N YO U D E A L W I T H H A R M O N I C V I B R AT I O N S? 125 of the string from the equilibrium position be very small α ! 1 (73) (note that here angle α is expressed in radians; in degrees, angle α should, in any case, be less than 10°). If condition (73) is com- plied with, the difference between the lengths AB and AC can be neglected AB “ l sin α « AC “ l tan α Thus your question becomes insignificant. For definiteness, we can assume that x “ AB “ l sin α. Then equation (66) will take the following form for a pendulum m g sin α “ k l sin α (66a) from which k “ m g l (74) Substituting this equation into equation (72), we obtain the formula for the period of harmonic vibrations of a pendulum T “ 2π d l g (75) We shall also take up the question of the energy of the pendu- lum. Its total energy is evidently equal to m g h, where h is the height to which the bob ascends at the extreme position (see Figure 50). Thus W “ m g h “ m g l p1 ´ cos αq “ 2m g l sin2 α 2 (76) Relationship (76) is evidently suitable for all values of angle α. To convert this result to relationship (68), it is necessary to satisfy the condition of harmonicity of the pendulum’s vibrations, i.e. inequality (73). Then, sin α can be approximated by the angle α expressed in radians, and equation (76) will change to W « 2m g l ˆ α 2 ˙2 “ m g l ˆ α2 2 ˙
126 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S Taking equation (74) into consideration, we finally obtain W “ k ˆ pl αq2 2 ˙ « k pABq2 2 which is, in essence, the same as equation (68). S T U D E N T B: If I remember correctly, in previously studying the vibrations of a pendulum, there was generally no requirement about the smallness of the angle of deviation. T E AC H E R: This requirement is unnecessary if we only deal with the energy of the bob or the tension of the string. In the given case we are actually considering, not a pendulum, but the motion of a ball in a circle in a vertical plane. However, if the prob- lem involves formula (75) for the period of vibrations, then the vibration of the pendulum must necessarily be harmonic and, consequently, the angle of deviation must be small. For instance, in Problem No. 33, the condition of the smallness of the angle of deviation of the pendulum is immaterial, while in Problem No. 34 it is of vital importance. PROBLEMS 32. A ball accomplishes harmonic vibrations as shown in Figure 48. Find the ratio of the velocities of the ball at points whose distances from the equilibrium position equal one half and one third of the amplitude. 33. A bob suspended on a string is deflected from the equilibrium position by an angle of 60° and is then released. Find the ratio of the tensions of the string for the equilibrium position and for the maximum deviation of the bob. 34. A pendulum in the form of a ball (bob) on a thin string is deflected through an angle of 5°. Find the velocity of the bob at the instant it passes the equilibrium position if the circular frequency of vibration of the pendulum equals 2{s.
§ 12 What Happens To A Pendulum In A State Of Weightlessness? T E AC H E R: Suppose we drive a nail in the wall of a lift and sus- pend a bob on a string of length l tied to the nail. Then we set the bob into motion so that it accomplishes harmonic vibrations. Assume that the lift ascends with an acceleration of a. What is the period of vibration of the pendulum? S T U D E N T A: When we go up in a lift travelling with acceleration, we feel a certain increase in weight. Evidently, the pendulum should “feel” the same increase. I think that its period of vibra- tion can be found by the formula T “ 2π d l g ` a (77) I cannot, however, substantiate this formula rigorously enough. T E AC H E R: Your formula is correct. But to substantiate it we will have to adopt a point of view that is new to us. So far we have dealt with bodies located in inertial frames of reference only, avoiding non-inertial frames. Moreover, I even warned you against employing non-inertial frames of reference (§ 4). Be that as it may, in the present section it is more convenient to use just this frame of reference which, in the given case, is attached to the accelerating lift. Recall that in considering the motion of a body of mass m in a non-inertial frame of reference having an acceleration a, we
128 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S must, on purely formal grounds, apply an additional force to the body. This is called the force of inertia, equal to ma and acting in the direction opposite to the acceleration. After the force of inertia is applied to the body we can forget that the frame of reference is travelling with acceleration, and treat the motion as if it were in an inertial frame. In the case of the lift, we must apply an additional force ma to the bob. This force is constant in magnitude and its direction does not change and coincides with that of the force of gravity m g . Thus it follows that in equation (75) the acceleration g should be replaced by the arithmetical sum of the accelerations pg ` aq. As a result, we obtain the formula (77) proposed by you. S T U D E N T A: Consequently, if the lift descends with a downward acceleration a, the period of vibration will be determined by the difference in the accelerations pg ´ aq, since here the force of inertia rna is opposite to the gravitational force. Is that correct? T E AC H E R: Of course. In this case, the period of vibration of the pendulum is T “ 2π d l g ´ a (78) This formula makes sense on condition that a ă g . The closer the value of the acceleration a is to g , the greater the period of vibration of the pendulum. At a “ g , the state of weightlessness sets in. What will happen to the pendulum in this case? S T U D E N T A: According to formula (78), the period becomes infinitely large. This must mean that the pendulum is stationary. T E AC H E R: Let us clear up some details of your answer. We started out with the pendulum vibrating in the lift. All of a sudden, the lift breaks loose and begins falling freely downward (we neglect air resistance). What happens to the pendulum? S T U D E N T A: As I said before, the pendulum stops. T E AC H E R: Your answer is not quite correct. The pendulum will indeed be stationary (with respect to the lift, of course) if at the
W H AT H A P P E N S TO A P E N D U LU M I N A S TAT E O F W E I G H T L E S S N E S S? 129 instant the lift broke loose the bob happened to be in one of its extreme positions. If at that instant the bob was not at an extreme position it will continue to rotate at the end of the string in a vertical plane at a uniform velocity equal to its velocity at the instant the accident happened. S T U D E N T A: I understand now. T E AC H E R: Then make a drawing illustrating the behaviour of a pendulum (a bob attached to a string) inside a spaceship which is in a state of weightlessness. S T U D E N T A: In the spaceship, the bob at the end of the string will either be at rest (with respect to the spaceship), or will ro- tate in a circle whose radius is determined by the length of the string (if, of course, the walls or ceiling of the spaceship do not interfere). T E AC H E R: Your picture is not quite complete. Assume that we are inside a spaceship in a state of weightlessness. We take the bob and string and attach the free end of the string so that nei- ther walls nor ceiling interfere with the motion of the bob. After this we carefully release the bob. The ball remains stationary. Here we distinguish two cases: (1) the string is loose, and (2) the string is taut.Fig. 51 so that neither walls nor ceiling in the bob. After this we carefully r remains stationary. Here we disti string is loose, and (2) the string case (position 1 in Fig. 5Ia)..We Vo to the bob. As a result, the bo line at uniform velocity until the tion 2, Fig. 5Ia). At this instant, will act on the bob in the same manner as the reaction of a wall acts on a ball bouncing off it. As a result, the direction of travel of the bob will change abruptly and it will then again travel at uniform (position 3, Fig. 5Ia). In this pecul rule of the equality of the angles o should be valid. Nowconsider the sec string taut and then carefully relea case, the bob will remain stationary leased (position 1, Fig. 5Ib). Then we to the bob in a direction perpend result the bob begins to rotate in a The plane of rotation is determined ctor of the velocity imparted to th Let us consider the following pr 1 with a bob at one end is attached t 102 Figure 51: Anaysing the motion of a pendulum in a spaceship. Consider the first case (Position 1 in Figure 51 (a)). We impart a certain velocity v0 to the bob. As a result, the bob will travel in a straight line at uniform velocity until the string becomes taut (Position 2 in Figure 51 (a)). At this instant, the reaction of the string will act on the bob in the same manner as the reaction of a wall acts on a ball bouncing off it. As a result, the direction of travel of the bob will change abruptly and it will then again travel at uniform velocity in a straight line (Position 3 in Fig- ure 51 (a)). In this peculiar form of “reflection” the rule of the equality of the angles of incidence and reflection should be valid. Now consider the second case: we first stretch the string taut and then carefully release the bob. As in the first case, the bob will remain stationary in the position it was released (Position 1 in
130 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S Figure 51 (b)). Then we impart a certain velocity v0 to the bob in a direction perpendicular to the string. As a result the bob begins to rotate in a circle at uniform velocity. The plane of rotation is determined by the string and the vector of the velocity imparted to the bob.g Fig. 52Fig. 51 remains stationary. Here we distinguish two cases: (I) the string is loose, and (2) the string is taut. Consider the first case (position 1 in Fig. 5Ia)..We impart a certain velocity Vo to the bob. As a result, the bob will travel in a straight line at uniform velocity until the string becomes taut (posi- tion 2, Fig. 5Ia). At this instant, the reaction of the string will act on the bob in the same manner as the reaction of a wall ta) acts on a ball bouncing off it. As a result, the direction of travel of the bob will change abruptly and it will then again travel at uniform velocity in a straight line (position 3, Fig. 5Ia). In this peculiar form of "reflection" the rule of the equality of the angles of incidence and reflection should be valid. Nowconsider the secondcase: we first stretch the string taut and then carefully release the bob. As in the first case, the bob will remain stationary in the position it was re- leased (position 1, Fig. 5Ib). Then we impart a certain velocity V o to the bob in a direction perpendicular to the string. As a result the bob begins to rotate in a circle at uniform velocity. The plane of rotation is determined by the string and the ve- ctor of the velocity imparted to the bob. . Let us consider the following problem. A string of length 1 with a bob at one end is attached to a truck which slides with- 102 Figure 52: Anaysing the motion of a pendulum moving with an accelerated frame at an angle to earth’s gravity. Let us consider the following problem. A string of length l with a bob at one end is attached to a truck which slides without friction down an inclined plane having an angle of inclination α (Figure 52 (a)). We are to find the period of vibration of this pendulum located in a frame of reference which travels with a certain acceleration. However, in contrast to the preceding problems with the lift, the acceleration of the system is at a certain angle to the acceleration of the earth’s gravity. This poses an additional question: what is the equilibrium direction of the pendulum string? S T U D E N T A: I once tried to analyse such a problem but became confused and couldn’t solve it. T E AC H E R: The period of vibration of a pendulum in this case is found by formula (75) except that g is to be replaced by a certain effective acceleration as in the case of the lift. This acceleration (we shall denote it by geff ) is equal to the vector sum of the acceleration of gravity and that of the given system. Another matter to be taken into account is that in the above mentioned sum, the acceleration vector of the truck should appear with the reversed sign, since the force of inertia is in the direction opposite to the acceleration of the system. The acceleration vectors are shown in Figure 52 (b), the acceleration of the truck being equal to g sin α. Next we find geff geff “ b g 2 eff x ` g 2 eff y “ b pg sin α cos αq2 ` pg ´ g s i n2αq2 “ g cos α (79) from which T “ 2π d l g cos α (80)
W H AT H A P P E N S TO A P E N D U LU M I N A S TAT E O F W E I G H T L E S S N E S S? 131 S T U D E N T A: How can we determine the equilibrium direction of the string? T E AC H E R: It is the direction of the acceleration geff . On the basis of equation (79) it is easy to see that this direction makes an angle α with the vertical. In other words, in the equilibrium position, the string of a pendulum on a truck sliding down an inclined plane will be perpendicular to the plane. S T U D E N T B: Isn’t it possible to obtain this last result in some other way? T E AC H E R: We can reach the same conclusion directly by consid- ering the equilibrium of the bob with respect to the truck. The forces applied to the bob are: its weight m g , the tension T of the string and the force of inertia ma (Figure 53). We denote the angle the string makes with the vertical by β.mg- Fig. 53 Next we resolve all these forces tal directions and then write the for the force components in eac T sin ma cos Taking into consideration that a= tem of equations (81) in the form T sin = mg sin After dividing one equation by the cot = co Thus, angles p and ex turn out the equilibrium direction of the p cular to the inclined plane. STUDENT B: I have followed yo and come to the conclusion that I when, in about th lite, I in and the Simply, red to th ched to rifugal f being the nertial f to the sa not in d is a prob forces of force of TEACHER: Such an approach permissible. However, in referring § 8, you did not consider it to be simply trying to think up somethi falling to the earth. Moreover, in was no necessity for passing over tached to the satellite: the physica more clearly demonstrated withou force of inertia. My previous ad no special need, do not employ a n '104 Figure 53: Anaysing the motion of a pendulum. Next we resolve all these forces in the vertical and horizontal directions and then write the conditions of equilibrium for the force components in each of these directions. Thus T cos β ` ma sin α “ m g T sin β “ ma cos α + (81) Taking into consideration that a “ g sin α, we rewrite the system of equations (81) in the form T cos β “ m g p1 ´ sin2 αq T sin β “ m g sin α cos α + After dividing one equation by the other we obtain cot β “ cot α Thus, angles β and α turn out to be equal. Consequently, the equilibrium direction of the pendulum string is perpendicular to the inclined plane. S T U D E N T B: I have followed your explanations very closely and come to the conclusion that I was not so wrong after all when, in
132 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S answer to your question about the forces applied to a satellite, I indicated the force of gravity and the centrifugal force (see § 8). Simply, my answer should be referred to the frame of reference attached to the satellite, and the centrifugal force is to be un- derstood as being the force of inertia. In a non-inertial frame of reference attached to the satellite, we have a problem, not in dy- namics, but in statics. It is a problem of the equilibrium of forces of which one is the centrifugal force of inertia. T E AC H E R: Such an approach to the satellite problem is permissi- ble. However, in referring to the centrifugal force in § 8, you did not consider it to be a force of inertia. You were simply trying to think up something to keep the satellite from falling to the earth. Moreover, in the case you mention, there was no necessity for passing over to a frame of reference attached to the satellite: the physical essence of the problem was more clearly demonstrated without introducing a centrifugal force of inertia. My previous advice is still valid: if there is no special need, do not employ a non-inertial frame of reference.
The laws of statics are laws of equilibrium. Study these laws carefully. Do not forget that they are of immense practical importance. A builder without some knowledge of the basic laws of statics is inconceivable. We shall consider examples illustrating the rules for the resolution of forces. The subsequent discussion concerns the conditions of equilib- rium of bodies, which are used, in particular, for locating the centre of gravity.
§ 13 Can You Use The Force Resolution Method Efficiently? T E AC H E R: In solving mechanical problems it is frequently nec- essary to resolve forces. Therefore, I think it would be useful to discuss this question in somewhat more detail. First let us recall the main rule: to resolve a force into any two directions it is necessary to pass two straight lines through the head and two more through the tail of the force vector, each pair of lines being parallel to the respective directions of resolution. As a result we obtain a parallelogram whose sides are the components of the given force. This rule is illustrated in Figure 54 in which force F is resolved in two directions: AA1 and BB1.THE FORCE RESOLUTION METHOD EFFICIENTLY? to me us a f is lin mo vec par of tai are for Fig. 54 in which .force F is re and BB r- Let us consider seve solution is the common approa rated in Fig. 55: we have two i X B 'A, A B1 F Fig. 54 from the middle of a string. The make angles of a 1 and CX 2 wit strings is subject to greater te STUDENT A: I can resolve t same drawing in directions p strings. From this resolution 106 Figure 54: Illustration for resolution of forces resulting in a parallelogram. Let us consider several problems in which force resolution is the common approach. The first problem is illustrated in Figure 55: we have two identical loads P suspended each from the middle of a string. The strings sag due to the loads and make angles of α1 and α2 with the horizontal. Which of the strings is subject to greater tension?§ 13. CAN YOU USE THE FORCE RESOLUTION METHOD EFFICIENTLY? TEACHER: In solving mecha- nical problems it is frequently necessary to resolve forces. The- refore, I think it would be useful to discuss this question in so- mewhat more detail. First let us recall the main rule: to resolve a force into any two directions it is necessary to pass two straight lines through the head and two more through the tail of the force vector, each pair of lines being parallel to the respective directions of resolution. As a result we .ob- tain a parallelogram whose sides are the components of the given force. This rule is illustrated in Fig. 54 in which .force F is resolved in two directions: AA 1 and BB r- Let us consider several problems in which force re- solution is the common approach. The first problem is illust- rated in Fig. 55: we have two identical loads P suspended each X B 'A, A B1 F Fig. 54 p Fig. 55 from the middle of a string. The strings sag due to the loads and make angles of a 1 and CX 2 with the horizontal. Which of the strings is subject to greater tension? STUDENT A: I can resolve the weight of each load on the Figure 55: Anaysing the motion of a pendulum. S T U D E N T A: I can resolve the weight of each load on the same drawing in directions parallel to the branches of the strings. From this resolution it follows that the tension in the string is T “ P 2 sin α Thus, the string which sags less is subject to greater tension.
136 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S T E AC H E R: Quite correct. Tell me, can we draw up the string so tightly that it doesn’t sag at all when the load is applied? S T U D E N T A: And why not? T E AC H E R: Don’t hurry to answer. Make use of the result you just obtained. S T U D E N T A: Oh yes, I see. The string cannot be made so taut that there is no sag. The tension in the string increases with a decrease in angle α. However strong the string, it will be broken by the tension when angle α becomes sufficiently small. T E AC H E R: Note that the sagging of a string due to the action of a suspended load results from the elastic properties of the string causing its elongation. If the string could not deform (elongate) no load could be hung from it. This shows that in construction engineering, the strength analysis of various structures is closely associated with their capability to undergo elastic deformations (designers are wont to say that the structure must “breathe”). Exceedingly rigid structures are unsuitable since the stresses developed in them at small deformations may prove to be ex- cessively large and lead to failure. Such structures may even fail under their own weight. If we neglect the weight of the string in the preceding problem, we can readily find the relationship between the angle α of sag of the string and the weight P of the load. To do this we make use of Hooke’s law for elastic stretching of a string or wire (see Problem No. 35). Consider another example. There’ is a Russian proverb, “a wedge is driven out by a wedge” (the English equivalent being “like cures like”). This can be demonstrated by applying the method of force resolution (Figure 56 (a)). Wedge 1 is driven out of a slot by driving wedge 2 into the same slot, applying the force F . Angles α and β are given. Find the force that acts on wedge 1 and enables it to be driven out of the slot. S T U D E N T A: I find it difficult to solve this problem.
C A N YO U U S E T H E F O RC E R E S O LU T I O N M E T H O D E F F I C I E N T LY? 137 T E AC H E R: Let us begin by resolving force F into components in the horizontal direction and in a direction perpendicular to side AB of wedge 2. The components obtained are denoted by F1 and F2 (Figure 56 (b)). Component F2 is counterbalanced by the reac- tion of the left wall of the slot; component F1 equal to F { tan α, will act on wedge 1. Next we resolve this force into components in the vertical direction and in a direction perpendicular to the side C D of wedge 1. The respective components are F3 and F4 (Figure 56 (c)). Component F4 is counterbalanced by the reaction of the right wall of the slot, while component F3 enables wedge 1 to be driven out of the slot. This is the force we are seeking. It can readily be seen that it equals F1 tan β “ F ˆ tan α tan β ˙components in the vertical direction and pendicular to the side CD of wedge 1. Th nents are F8 and F4. (Fig. 56c). ComponentF by the reaction of the right wall of the nent F 3 enables wedge 1 to be driven out the force we are seeking. It can readily be A tan F1 tan p= F--tan ex. Let us now consider a third example, illu Two weights, PI and P 2, are suspended fr the portion of the string between them (Q) angle P (an and the ten of the strin T c D) . This the preced wedges. fa) F (C) icD Fig. 56 STUDENT A: First I shall resolve the w components in the directions AB and Be ( resolution we find that TAB=P 1/sin a a Thus we have already found the tension the string. Next.I shall resolve the weight 108 Figure 56: Anaysing the motion of a pendulum. Let us now consider a third example, illustrated in (Figure 57 (a)).components in the vertical direction and in a direction per- pendicular to the side CD of wedge 1. The respective compo- nents are F8 and F4. (Fig. 56c). ComponentF. is counterbalanced by the reaction of the right wall of the slot, while compo- nent F 3 enables wedge 1 to be driven out of the slot. This is the force we are seeking. It can readily be seen that it equals A tan F1 tan p= F--tan ex. Let us now consider a third example, illustrated in Fig. 57a. Two weights, PI and P 2, are suspended from a string so that the portion of the string between them is horizontal. Find (Q) angle P (angle a,. being known) and the tension in each portion of the string (TAB, T BC and T c D) . This example resembles the preceding one with the wedges. fa) F (C) icD Fig. 56 Fig. 57 STUDENT A: First I shall resolve the weight PI into force components in the directions AB and Be (Fig. 57b). From this resolution we find that TAB=P 1/sin a and T BC=P 1/tan a, Thus we have already found the tension in two portions of the string. Next.I shall resolve the weight P 2 into components 108 Figure 57: Anaysing the motion of a pendulum. Two weights, P1 and P2, are suspended from a string so that the portion of the string between them is horizontal. Find angle β (angle α being known) and the tension in each portion of the string (TAB , TBC and TC D ). This example resembles the preceding one with the wedges. S T U D E N T A: First I shall resolve the weight P1 into force com- ponents in the directions AB and BC (Figure 57 (b)). From this resolution we find that TAB “ P1 sin α and TBC “ P1 tan α Thus we have already found the tension in two portions of the string. Next I shall resolve the weight P2 into components in the directions BC and C D (Figure 57 (c)). From this resolution we can write the equations: TBC “ P2 tan β and TC D “ P2 sin β Equating the values for the tension in portion BC of the string obtained in the two force resolutions, we can write P1 tan α “ P2 tan β
138 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S from which β “ arctan ˆ P2 tan α P1 ˙ Substituting this value into the equation for TC D we can find the tension in portion C D of the string. T E AC H E R: Is it really so difficult to complete the problem, i.e. to find the force TC D ? S T U D E N T A: The answer will contain the sine of the arctan β, i.e. TC D “ P2 sin ˆ arctan ˆ P2 tan α P1 ˙˙ T E AC H E R: Your answer is correct but it can be written in a simpler form if sin β is expressed in terms of tan β. As a matter of fact sin β “ tan β a1 ` tan2 β Since tan β “ tan α ˆ P2 P1 ˙ we obtain TC D “ P1 tan α d 1 ` ˆ P2 P1 ˙2 tan2 α ST U D E N T B: I see that before taking an examination in physics, you must review your mathematics very thoroughly. T E AC H E R: Your remark is quite true. PROBLEMS 35. An elastic string, stretched from wall to wall in a lift, sags due to the action of a weight suspended from its middle point as shown in Figure 55. The angle of sag α equals 30° when the lift is at rest and 45° when the lift travels with acceleration. Find the magnitude and direction of acceleration of the lift. The weight of the string is to be neglected.
C A N YO U U S E T H E F O RC E R E S O LU T I O N M E T H O D E F F I C I E N T LY? 139 36. A bob of mass m “ 100 g is suspended from a string of length l “ 1 m tied to a bracket as shown in Figure 58 (α “ 30°). A horizontal velocity of 2 m{s is imparted to the bob and it begins to vibrate as a pendulum. Find the forces acting in members AB and BC when the bob is at the points of maximum deviation from the equilibrium position.sics, you must review y TEACHER: Your remar PROBLEMS \ I' 1\. \ Fig. 58 A 35. An elastic string, stretch the action of a weight suspende The angle of sag a equals 30° w travels and di weight 36. from a as show city of begins acting i the poi librium Figure 58: Anaysing the mo- tion of a pendulum, Problem 36.
§ 14 What Do You Know About The Equilibrium Of Bodies? T E AC H E R: Two positions of equilibrium of a brick are shown in Figure 59. Both equilibrium positions are stable, but their degree of stability differs. Which of the two positions is the more stable?THE EQUILIBRIUM OF BODIES? (Q) (6) the t stable STU positio TEA STU of gra to the TEA STU bearin in the TEACHER: And this isn't all ei consider the equilibrium of two b lelepiped with a square base (a) and a right circular cylinder (Fig. 60a). Assume that the parallelepiped and cylinder are of the same height H and have bases of the same area S. In this case, the cen- tres of gravity of the bodies are at the same height and, in addition, they have bea- ring surfaces of the same area. Their degrees of stabi- (0) lity, however, are different. Fig. 59 The measure of the stability of a is the energy that must be expend the given state of the body. 110 Figure 59: Which brick is more stable? S T U D E N T A: Evidently, the position of the brick in Figure 59(a). T E AC H E R: Why? S T U D E N T A: Here the centre of gravity of the brick is nearer to the earth’s surface, T E AC H E R: This isn’t all. S T U D E N T B: The area of the bearing surface is greater than in the position shown in Figure 59 (b). T E AC H E R: And this isn’t all either. To clear it up, let us consider the equilibrium of two bodies: a rectangular parallelepiped with a square base and a right circular cylinder Figure 60 (a). Assume that the parallelepiped and cylinder are of the same height H and have bases of the same area S. In this case, the centres of gravity of the bodies are at the same height and, in addition, they have bearing surfaces of the same area. Their degrees of stability, however, are different.
142 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S The measure of the stability of a specific state of equilibrium is the energy that must be expended to permanently disturb the given state of the body.IES? (6) stable? STUDENT A: Evidently, the position of the brick in Fig. 59a. TEACHER: Why? STUDENT A: Here the centre of gravity of the brick is nearer to the earth's surface, TEACHER: This isn't all. STUDENT B: The area of the bearing surface is greater than in the position shown in Fig. 59b. HER: And this isn't all either. To clear it up, let us r the equilibrium of two bodies: a rectangular paral- d with a square base (a) ht circular cylinder a). Assume that the epiped and cylinder same height H bases of the same In this case, the cen- avity of the bodies same height and, ion, they have bea- aces of the same eir degrees of stabi- (0) r-----,.----r----r---_ wever, are different. Fig. 59 Fig. 60 sure of the stability of a specific state of equilibrium rgy that must be expended to permanently disturb n state of the body. Figure 60: Comparing equilib- rium of two bodies, which is more stable? S T U D E N T B: What do you mean by the word “permanently”? T E AC H E R: It means that if the body is subsequently left to itself, it cannot return to the initial state again. This amount of energy is equal to the product of the weight of the body by the height to which the centre of gravity must be raised so that the body cannot return to its initial position. In the example with the parallelepiped and cylinder, the radius of the cylinder is R “b S{π and the side of the parallelepiped’s base is a “ ?S. To disturb the equilibrium of the cylinder, its centre of gravity must be raised through the height Figure 60 (b) h1 “ dˆ H 2 ˙2 ` R2 ´ H 2 To disturb the equilibrium of the parallelepiped, its centre of
W H AT D O YO U K N OW A B O U T T H E E Q U I L I B R I U M O F B O D I E S? 143 gravity must be raised Figure 60 (b) h2 “ dˆ H 2 ˙2 ` ˆ a 2 ˙2 ´ H 2 Since ˆ a{2 R ˙ “ ?πS 2?S “ ?π 2 ă 1 it follows that h2 ă h1 Thus, of the two bodies considered, the cylinder is the more stable. Now I propose that we return to the example with the two positions of the brick. S T U D E N T A: If we turn over the brick it will pass consecutively from one equilibrium position to another. The dashed line in Figure 61 shows the trajectory described by its centre of gravity in this process. To change the position of a lying brick its centre of gravity should be raised through the height h1 expending an energy equal to m g h1 and to change its upright position, the centre of gravity should be raised through h2, the energy expended being m g h2. The greater degree of stability of the lying brick is due to the fact that m g h1 ą m g h2 (82)STUDENT B: What do you m TEACHER: It means that if itself, it cannot return to the of energy is equal to the produ the height to which the centr that the body cannot return example with the parallelepip the cylinder is R=VS/n and base is a=VS. To disturb th centre of gravity must be rais h1 = 1/ 2 To disturb the equilibrium o of gravity must be raised (F h2 = 1f( Since (a/2)/R=VnS/2VS=V Thus, of the two bodies cons stable. Now I propose that we retu positions of the brick. STUDENT A: If we turn ove tively from one equilibrium p line desc this of a shou hh e mgh h Fig. 61 posit uld b expended being mgh 2• The gr lying brick is due to the f mgh; TEACHER: At last you've s greater stability of the lying STUDENT B: But it is evid depend upon the height of th level and on the area of the b Figure 61: Trajectory described by the centre of gravity of a brick when turning over. T E AC H E R: At last you’ve succeeded in substantiating the greater stability of the lying position of a body. S T U D E N T B: But it is evident that the heights h1 and h2 depend upon the height of the centre of gravity above floor level and on the area of the base. Doesn’t that mean that in discussing the degree of stability of bodies it is correct to compare the heights of the centres of gravity and the areas of the bases? T E AC H E R: Why yes, it is, but only to the extent that these quan- tities influence the difference between the heights h1 and h2. Thus, in the example with the parallelepiped and cylinder, the comparison of the heights of the centres of gravity and the areas of the bases is insufficient evidence for deciding which of the
144 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S bodies is the more stable. Besides, I wish to draw your attention to the following. Up till now we have tacitly assumed that the bodies were made of the same material. In this case, the inequal- ity (82) could be satisfied by observing the geometric condition h1 ą h2. In the general case, however, bodies may be made of different materials, and the inequality (82) may be met even when h1 ă h2 owing to the different densities of the bodies. For example, a cork brick will be less stable in the lying position than a lead brick in the upright position. Let us now see what conditions for the equilibrium of bodies you know. S T U D E N T A: The sum of all the forces applied to a body should equal zero. In addition, the weight vector of the body should fall within the limits of its base. T E AC H E R: Good. It is better, however, to specify the conditions of equilibrium in a different form, more general and more con- venient for practical application. Distinction should be made between two conditions of equilibrium: First condition: The projections of all forces applied to the body onto any direction, should mutually compensate one an- other. In other words, the algebraic sum of the projections of all the forces onto any direction should equal zero. This condition enables as many equations to be writ- ten as there are independent directions in the problem: one equation for a one-dimensional problem, two for a two-dimensional problem and three for the general case (mutually perpendicular directions are chosen). Second condition (moment condition): The algebraic sum of the moments of the forces about any point should equal zero. Here, all the force moments tending to turn the body about the chosen point in one direction (say, clockwise) are taken with a plus sign and all those tending to turn the body in the opposite direction (counterclockwise) are taken with a minus sign. To specify the moment condition, do the following:
W H AT D O YO U K N OW A B O U T T H E E Q U I L I B R I U M O F B O D I E S? 145 (a) establish all forces applied to the body; (b) choose a point with respect to which the force moments are to be considered; (c) find the moments of all the forces with respect to the chosen point; (d) write the equation for the algebraic sum of the moments, equating it to zero. In applying the moment condition, the following should be kept in mind: (1) the above stated condition refers to the case when all the forces in the problem and their arms are in a single plane (the problem is not three-dimensional), and (2) the algebraic sum of the moments should be equated to zero with respect to any point, either within or outside the body. It should be emphasized that though the values of the separate force moments do depend upon the choice of the point - with respect to which the force moments are considered), the algebraic sum of the moments equals zero in any case. To better under- stand the conditions of equilibrium, we shall consider a specific problem. A beam of weight P1 is fixed at points B and C (Figure 62 (a)). At point D, a load with a weight of P2 is suspended from the beam. The distances AB “ a, BC “ 2a and C D “ a. Find the reactions NB and NC at the two supports. Assume that the reactions of the supports are directed vertically. As usual, first indicate the forces applied to the body. S T U D E N T A: The body in the given problem is the beam. Four forces are applied to it: weights P1 and P2 and reactions NB and NC . T E AC H E R: Indicate these forces on the drawing.
146 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C Stl» (C) A J3 B Fig. 62 c C .D Figure 62: A beam with suspended loads, the problem is to find the reactions of the supports. S T U D E N T A: But I don’t know whether the reactions are directed upward or downward. T E AC H E R: Assume that both reactions are directed upward. S T U D E N T A: Well, here is my drawing (Figure 62 (b)). Next I can specify the first condition of equilibrium by writing the equation NB ` NC “ P1 ` P2 T E AC H E R: I have no objection to this equation as such. However, in our problem it is simpler to use the second condition of equi- librium (the moment condition), employing it first with respect to point B and then to point C . S T U D E N T A: All right, I’ll do just that. As a result I can write the
W H AT D O YO U K N OW A B O U T T H E E Q U I L I B R I U M O F B O D I E S? 147 equations with respect to point B: aP1 ´ 2aNc ` 3aP2 “ 0 with respect to point C : 2aNB ´ aP1 ` aP2 “ 0 + (83) T E AC H E R: Now you see: each of your equations contains only one of the unknowns. It can readily be found. S T U D E N T A: From equations (83) we find NB “ pP1 ´ P2q 2 (84) NB “ pP1 ` 3P2q 2 (85) T E AC H E R: Equation (85) always has a positive result. This means that reaction NC is always directed upward (as we assumed). Equation (84) gives a positive result when P1 ą P2, negative when P1 ă P2 and becomes zero when P1 “ P2. This means that when P1 ă P2, reaction NB is in the direction we assumed, i. e. upward (see (Figure 62 (b)); that when P1 ă P2, reaction NB is downward (see (Figure 62 (c)); and at P1 “ P2 there is no reaction NB .
§ 15 How Do You Locate The Centre Of Gravity? T E AC H E R: In many cases, examinees find it difficult to locate the centre of gravity of a body of system of bodies. Is everything quite clear to you on this matter? S T U D E N T A: No, I can’t say it is. I don’t quite understand how you find the centre of gravity in the two cases shown in Fig- ure 63 (a) and Figure 64 (a).Fig.Fig. 63 § 15. , HOW DO YOU LOCATE THE CENTRE OF GRAVITY? TEACHER: In many cases, exa- minees find it difficult to locate the centre of gravity of a body or system of bodies. Is everything quite clear to you on this matter? STUDENT A: No, I can't say it is. I don't quite understand how you find the centre of gra- vity in the two cases shown in Figs. 63a and 64a. TEACHER: All right.· In the first case it is convenient to di- vide the plate into two rectangles as shown by the dashed line in Fig. 63b. The centre of gravity of rectangle 1 is at point A; the weight of this rectangle is pro- portional to its area and is equal, as is evident from the figure, to 6 units (here the weight is conditionally measured in square centimetres). The centre of gravity of rectangle 2 is at point B; ( Q) the weight of this rectangle is equal to 10 units. Next we project the points A and B on the coordinate axes Ox and Oy; these projections are denoted by A 1 and B 1 on the x-axis and by A 2 and B 2 on the y-axis. Then we consider the "bars" A 181 115 Figure 63: Problem is to find the centre of gravity of the given body. T E AC H E R: All right. In the first case it is convenient to divide
150 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S the plate into two rectangles as shown by the dashed line in Figure 63 (b). The centre of gravity of rectangle 1 is at point A; the weight of this rectangle is proportional to its area and is equal, as is evident from the figure, to 6 units (here the weight is conditionally measured in square centimetres). The centre of gravity of rectangle 2 is at point the weight of this rectangle is equal to 10 units. Next we project the points A and B on the coordinate axes Ox and Oy ; these projections are denoted by A1 and B1 on the X -axis and by A2 and B2 on the Y -axis. Then we consider the “bars” A1B1 and A2B2, assuming that the masses are concentrated at the ends of the “bars”, the mass of each end being equal to that of the corresponding rectangle (see Figure 63 (b). As a result, the problem of locating the centre of gravity of our plate is reduced to finding the centres of gravity of “bars” A1B1 and A2B2 The positions of these centres of gravity will be the coordinates of the centre of gravity of the plate. But let us complete the problem. First we determine the location of the centre of gravity of “bar” A1B1 using the well-known rule of force moments (see Figure 63 (b)): 6x “ 10p2 ´ xq then, x “ 5 4 cm Thus, the X -coordinate of the centre of gravity of the plate in the chosen system of coordinates is X “ p1 ` xq cm “ 9 4 cm In a similar way we find the centre of gravity of “bar” A2B2: 6y “ 10p1 ´ yq from which it follows that y “ 5 8 cm. Thus the Y -coordinate of the centre of gravity of the plate is Y “ p1.5 ` yqcm “ 17 8 cm S T U D E N T A: Now I understand. That is precisely how I would go about finding coordinate X of the centre of gravity of the
H OW D O YO U L O C AT E T H E C E N T R E O F G R AV I T Y? 151 plate. I was not sure that coordinate Y could be found in the same way. T E AC H E R: Let us consider the second case, shown in Figure 64 (a).Fig.ig. 63 Figs. 63a and 64a. TEACHER: All right.· In the first case it is convenient to di- vide the plate into two rectangles as shown by the dashed line in Fig. 63b. The centre of gravity of rectangle 1 is at point A; the weight of this rectangle is pro- rea and is equal, as is evident from the figure, he weight is conditionally measured in square he centre of gravity of rectangle 2 is at point B; his rectangle is equal to 10 units. Next we ints A and B on the coordinate axes Ox and Oy; ions are denoted by A 1 and B 1 on the x-axis and 2 on the y-axis. Then we consider the "bars" A 181 115 Figure 64: Problem is to find the centre of gravity of the given body. Two approaches are available. For instance, instead of the given circle with one circular hole, we can deal with a system of two bodies: a circle with two symmetrical circular holes and a cir- cle inserted into one of the holes (Figure 64 (b)). The centres of gravity of these bodies are located at their geometric centres. Knowing that the weight of the circle with two holes is propor- tional to its area, i.e. ˆ πR2 ´ 2πR2 4 ˙ “ πR2 2 and that of the small circle is proportional to its area πR2{4, we reduce the problem to finding the point of application of the resultant of the two parallel forces shown below in Figure 64 (b). We denote by x the distance from the sought-for centre of grav- ity to the geometric centre of the large circle. Then, according to
152 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S Figure 64 (b), we can write ˆ πR2 4 ˙ ˆ R 2 ´ x ˙ “ πR2 2 x from which x “ R 6 There is another possible approach. The given circle with the hole can be replaced by a solid circle (with no hole) plus a circle located at the same place where the hole was and having a neg- ative weight (i.e. one acting upward) (Figure 64 (c)) which will compensate for the positive weight of the corresponding portion of the solid circle. As a whole, this arrangement corresponds to the initial circle with the circular hole. In this case, the problem is again reduced to finding the point of application of the resultant of the two forces shown at the bottom of Figure 64 (c). According to the diagram we can write: πR2 x “ ˆ πR2 4 ˙ ˆ R 2 ` x ˙ from which, as in the preceding case, x “ R 6 . S T U D E N T A: I like the first approach better because it does not require the introduction of a negative weight. T E AC H E R: In addition, I want to propose a problem involving locating the centre of gravity of the system of loads shown in Figure 65 (a). We are given six loads of different weights (P1, P2, . . . , P6) arranged along a bar at equal distances a from one another. The weight of the bar is neglected. How would you go about solving this problem? S T U D E N T A: First I would consider two loads, for instance, P1 and P2, and find the point of application of their resultant. Then I would indicate this resultant (equal to the sum P1 ` P2) on the drawing and would cross out forces P1 and P2, from further consideration. Now, instead of the six forces, only five would remain. Next, I would find the point of application of the resultant of another pair of forces, etc. Thus, by consecutive operations I would ultimately find the required resultant whose point of application is the centre of gravity of the whole system.
H OW D O YO U L O C AT E T H E C E N T R E O F G R AV I T Y? 153N Fig. 65 (11) e can write: nR2x:r=(nR2/4) (R/2+x), from which, as in the precedin case, x=R/6.STUDENT A: I like the first approach better because it doe not require the introduction of a negative weight. TEACHER: In addition, I want to propose a problem invol ving locating the centre of gravity of the system of loads show t a) a in Fig. 65a. We are given si loads of different weights (P h P 2, ... , p6)' arranged along a ba at equal distances a from on P6 another. The weight of the bar isneglected. How would you go about solving this problem? STUDENT A: First I would con- sider two loads, for instance, PI and P 2, and find the point of application of their resultant. c Then I would indicate this resul- A tant (equal to the sum PI +P 2) on the drawing and would cross PG out forces P 1 and P 2, from fur- ther consideration. Now, instead of the six forces, only five would remain. Next, I· would find the point of application of the resul- tant of another pair of forces, etc. Thus, by consecutive ope- rations I would ultimately find the required resultant whose point of application is the centre of gravity of the whole system. TEACHER: Though your method of solution is absolutely correct, it is by far too cumbersome. I can show you a much more elegant solution. We begin by assuming that we are sup- porting the system at its centre of gravity (at point B in Fig. 65b). STUDENT B (interrupting): But you don't yet know the loca- tion of the centre of gravity. How do you know that it is between the points of application of forces P3 and P4 ? TEACHER: It makes no difference to me where exactly the centre of gravity is. I shall not take advantage of the fact that in Fig. 65b the centre of gravity turned out to be bet- ween the points of application of forces Pa and P4. So we 117 Figure 65: Problem is to find the centre of gravity of the bodies in the given configuration. T E AC H E R: Though your method of solution is absolutely cor- rect, it is by far too cumbersome. I can show you a much more elegant solution. We begin by assuming that we are sup porting the system at its centre of gravity (at point B in Figure 65 (b). S T U D E N T B: (interrupting): But you don’t yet know the location of the centre of gravity. How do you know that it is between the points of application of forces P3 and P4? T E AC H E R: It makes no difference to me where exactly the cen- tre of gravity is. I shall not take advantage of the fact that in Figure 65 (b) the centre of gravity turned out to be between the points of application of forces P3 and P4. So we assume we are supporting the system at its centre of gravity. As a result, the bar is in a state of equilibrium. In addition to the six forces, one more force - the bearing reaction N - will act on the bar. Since the bar is in a state of equilibrium, we can apply the conditions of equilibrium (see § 14). We begin with the first condition of
154 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S equilibrium for the projection of all the forces in the vertical direction N “ P1 ` P2 ` P3 ` P4 ` P5 ` P6 (86) Then we apply the second condition (moment condition), considering the force moments with respect to point A in Fig- ure 65 (b) (i.e. the left end of the bar). Here, all the forces tend to turn the bar clockwise, and the bearing reaction tends to turn it counterclockwise. We can write N pABq “ aP2 ` 2aP3 ` 3aP4 ` 4aP5 ` 5aP6 (87) Combining conditions (86) and (87), we can find the length AB, i.e. the required position of the centre of gravity measured from the left end of the bar AB “ aP2 ` 2aP3 ` 3aP4 ` 4aP5 ` 5aP6 P1 ` P2 ` P3 ` P4 ` P5 ` P6 (88) S T U D E N T A: Yes, I must admit that your method is much sim- pler. T E AC H E R: Also note that your method of solving the problem is very sensitive to the number of loads on the bar (the addition of each load makes the solution more and more tedious). My solution, on the contrary, does not become more complicated when loads are added. With each new load, only one term is added to the numerator and one to the denominator in equation (88). S T U D E N T B: Can we find the location of the centre of gravity of the bar if only the moment condition is used? T E AC H E R: Yes, we can. This is done by writing the condition of the equilibrium of force moments with respect to two different points. Let us do precisely that. We will consider the condition for the force moments with respect to points A and C (see Fig- ure 65 (b)). For point A the moment condition is expressed by equation (87); for point C , the equation will be N p5a ´ ABq “ aP5 ` 2aP4 ` 3aP3 ` 4aP2 ` 5aP1 (89)
H OW D O YO U L O C AT E T H E C E N T R E O F G R AV I T Y? 155 Dividing equation (87) by (89) we obtain AB 5a ´ AB “ aP2 ` 2aP3 ` 3aP4 ` 4aP5 ` 5aP6 aP5 ` 2aP4 ` 3aP3 ` 4aP2 ` 5aP1 From which AB paP5 ` 2aP4 ` 3aP3 ` 4aP2 ` 5aP1` aP2 ` 2aP3 ` 3aP4 ` 4aP5 ` 5aP6q “ 5a paP2 ` 2aP3 ` 3aP4 ` 4aP5 ` 5aP6q or AB ˆ 5a pP1 ` P2 ` P3 ` P4 ` P5 ` P6q “ 5a paP2 ` 2aP3 ` 3aP4 ` 4aP5 ` 5aP6q Thus we obtain the same result as in equation (88). PROBLEMS 37. Locate the centre of gravity of a circular disk having two circular holes as shown in Figure 66. The radii of the holes are equal to one half and one fourth of the radius of the disk.Dividing equation (87) by (89) we obtain AB _ aP 2+ 2aP3+3aP 4 +4aPs+5aP 6 5a- AB aPs+2aP 4+ 3aP3+4aP 2+ 5aP! From which AB (aPft + 2aP4 + 3aPa+4aP2+ 5aPl +aP2+ 2aPa+ +3aP 4 +4aP s + 5aP6 = 5a (aP2+2aPJ + 3aP4 +4aPs + 5aP6) PROBLEM 37. Locate the centre of gravity of a circular disk having two circular holes as shown in Fig. 66. The radii of the holes are equal to one half and. one fourth of the radius of the disk.Fig. 66 or AB x5a (Pt + P2+P 3 + P4 + P s + P6)= 5a(aP2 + 2aP3 + + 3aP4 + 4aP5 + 5aP6) Thus we obtain the same result as in equation (88). Figure 66: Problem is to find the centre of gravity of the given body.
Archimedes’ principle does not usually draw special attention. This is a common mistake of students preparing for physics exams. Highly inter- esting questions and problems can be devised on the basis of this prin- ciple. We shall discuss the problem of the applicability of Archimedes’ principle to bodies in a state of weightlessness.
§ 16 Do you know Archimedes’ principle? T E AC H E R: Do you know Archimedes’ principle? S T U D E N T A: Yes, of course. The buoyant force exerted by a liquid on a body immersed in it is exactly equal to the weight of the liquid displaced by the body. T E AC H E R: Correct. Only it should be extended to include gases: a buoyant force is also exerted by a gas on a body “immersed” in it. And now can you give a theoretical proof of your statement? S T U D E N T A: A proof of Archimedes’ principle? T E AC H E R: Yes. S T U D E N T A: But Archimedes’ principle was discovered directly as the result of experiment. T E AC H E R: Quite true. It can, however, be derived from simple energy considerations. Imagine that you raise a body of volume V and density ρ to a height H , first in a vacuum and then in a liquid with a density ρ0. The energy required in the first case equals ρgV H . The energy required in the second case is less because the raising of a body of volume V by a height H is accompanied by the lowering of a volume V of ρ the liquid by the same height H . Therefore, the energy expended in the second case equals ρgV H ´ ρ0 gV H Regarding the subtrahend ρ0 gV H as the work done by a certain force, we can conclude that, compared with a vacuum, in a liquid
160 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S an additional force F “ ρ0 gV H acts on the body making it easier to raise. This force is called the buoyant force. Quite obvi- ously, it is exactly equal to the weight of the liquid in the volume V of the body immersed in the liquid. (Note that we neglect the energy losses associated with friction upon real displacements of the body in the liquid.) h S P P ` ρ0 g h Figure 67: Buoyant force for Archimedes’ principle. Archimedes’ principle can be deduced in a somewhat different way. Assume that the body immersed in the liquid has the form of a cylinder of height h and that the area of its base is S (Fig- ure 67). Assume that the pressure on the upper base is p. Then the pressure on the lower base will equal p ` ρg h. Thus, the difference in pressure on the upper and lower bases equals ρ0 g h. If we multiply this difference by the area S of the base, we obtain the force F “ ρg hS which tends to push the body up- ward. Since hS “ V , the volume of the cylinder, it can readily be seen that this is the buoyant force which appears in Archimedes’ principle. S T U D E N T A: Yes, now I see that Archimedes’ principle can be arrived at by purely logical reasoning. T E AC H E R: Before proceeding any further, let us recall the condi- tion for the floating of a body. S T U D E N T A: I remember that condition. The weight of the body should be counterbalanced by the buoyant force acting on the body in accordance with Archimedes’ principle.
D O YO U K N OW A RC H I M E D E S’ P R I N C I P L E? 161 T E AC H E R: Quite correct. Here is an example for you. A piece of ice floats in a vessel with water. Will the water level change when the ice melts? S T U D E N T A: The level will remain unchanged because the weight of the ice is counterbalanced by the buoyant force and is there- fore equal to the weight of the water displaced by the ice. When the ice melts it converts into water whose volume is equal to that of the water that was displaced previously. T E AC H E R: Exactly. And now let us assume that there is, for instance, a piece of lead inside the ice. What will happen to the water level after the ice melts in this case? S T U D E N T A: I’m not quite sure, but I think the water level should reduce slightly. I cannot, however, prove this. T E AC H E R: Let us denote the volume of the piece of ice together with the lead by V , the volume of the piece of lead by v, the volume of the water displaced by the submerged part of the ice by V1 the density of the water by ρ0, the density of the ice by ρ1 and the density of the lead by ρ2. The piece of ice together with the lead has a weight equal to ρ1 g pV ´ vq ` ρ2 g v This weight is counterbalanced by the buoyant force ρ0 gV t . Thus ρ1 g pV ´ vq ` ρ2 g v “ ρ0 gV t (90) After melting, the ice turns into water whose volume V2 is found from the equation ρ1 g pV ´ vq “ ρ0 gV2 Substituting this equation into (90) we obtain ρ0 gV2 ` ρ2 g v “ ρ0 gV1 From which we find that the volume of water obtained as a result of the melting of the ice is V2 “ V1 ´ v ρ2 ρ1 (91)
162 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S Thus, before the ice melted, the volume of water displaced was V1. Then the lead and the water from the melted ice began to occupy the volume pV2 ` vq. To answer the question concerning the water level in the vessel, these volumes should be compared. From equation (91) we get V2 ` v “ V1 ´ v ρ2 ´ ρ0 ρ0 (92) Since ρ2 ą ρ0 (lead is heavier than water), it can be seen from equation (92) that pV2 ` vq ă V1. Consequently, the water level will reduce as a result of the melting of the ice. Dividing the difference in the volumes V1 ´ pV2 ` vq by the cross-sectional area S of the vessel (assuming, for the sake of simplicity, that it is of cylindrical shape) we can find the height h by which the level drops after the ice melts. Thus h “ v ˆ ρ2 ´ ρ0 ρ0S ˙ (93) Do you understand the solution of this problem? S T U D E N T A: Yes, I’m quite sure I do. T E AC H E R: Then, instead of the piece of lead, let us put a piece of cork of volume v and density ρ3 inside the ice. What will happen to the water level when the ice melts? S T U D E N T A: I think it will rise slightly. T E AC H E R: Why? S T U D E N T A: In the example with lead the level fell. Lead is heavier than water, and cork is lighter than water. Consequently, in the case of cork we should expect the opposite effect: the water level should rise. T E AC H E R: You are mistaken. Your answer would be correct if the cork remained submerged after the ice melted. Since the cork is lighter than water it will surely rise to the surface and float. Therefore, the example with cork (or any other body lighter than water) requires special consideration. Using the result of
D O YO U K N OW A RC H I M E D E S’ P R I N C I P L E? 163 equation (91), we can find the difference between the volume of the water displaced by the piece of ice together with the cork, and that of the water obtained by the melting of the ice. Thus V1 ´ V2 “ v ˆ ρ3 ρ0 ˙ (94) Next we apply the condition for the floating of the piece of cork: ρ3v “ ρ0v1 (95) where v1 is the volume of the part of the cork submerged in water. Substituting this equation into (94), we obtain V1 “ V2 ` v1 Thus the volume of water displaced by the piece of ice is ex- actly equal to the sum of the volume of water obtained from the melted ice. and the volume displaced by the submerged por- tion of the floating piece of cork. So in this case the water level remains unchanged. S T U D E N T A: And if the piece of ice contained simply a bubble of air instead of the piece of cork? T E AC H E R: After the ice melts, this bubble will be released. It can readily be seen that the water level in the vessel will be exactly the same as it was before the ice melted. In short, the example with the bubble of air in the ice is similar to that with the piece of cork. S T U D E N T A: I see that quite interesting question and problems can be devised on the basis of Archimedes’ principle. T E AC H E R: Unfortunately, some examinees don’t give enough attention to this principle when preparing for their physics ex- aminations. Let us consider the following example. One pan of a balance carries a vessel with water and the other, a stand with a weight suspended from it. The pans are balanced (Fig- ure 68 (a)). Then the stand is turned so that the suspended weight is completely submerged in the water. Obviously, the state of equilibrium is disturbed since the pan with the stand becomes
164 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S lighter (Figure 68 (b)). What additional weight must be put on the pan with the stand to restore equilibrium?Fig. 68 8, (0) TEACHER: You are mistaken. You would do well to recall Newton's third law of motion. According to this law, the force with which the water in the vessel acts on the submerged we- ight is exactly equal to the force with which the submerged weight acts on the water in the opposite direction. Consequent- ly, as the weight of the pan with the stand reduces, the weight of the pap with the vesselincreases. Therefore, to restore equilib- rium, a weight equal to 2P should be added to the pan with the stand. STUDENT A: I can't quite understand your reasoning. After all, the interaction of the submerged weight and the water in no way resem- bles the interaction of two bodies in mechanics. TEACHER: The field of application of Newton's third law is not limited to mechanics. The expression "to every action there is an equal and opposite reaction" refers to a great many kinds of interaction. We can, however, apply a different line of reasoning in our case, one to which you will surely have no objections. Let us deal with the stand with the weight and the vessel with the water as part of a single system whose total weight is obviously the sum of the weight of the left pan and that of the right pan. The total weight of the system should not change due to interaction of its parts with one another. Hence, if as the result of interaction the weight of the right pan is decreased by P, the weight of the left pan must be increased by the same amount (P). Therefore, after the weight is submerged in the vessel with water, the difference between the weights of the left and right pans should be 2P. PROBLEM 38. A vessel of cylindrical shape with a cross-sectional area S is filled with water in which a piece of ice, containing a lead ball, floats. The vo- lume of the ice together with the lead ball is V and 1/20 of this volume is above the water level. To what mark will the water level in the vessel reduce a Her the ice melts? The densities of water, ice and lead are assu- med to be known. Figure 68: What additional weight must be put on the pan with the stand to restore equilibrium? S T U D E N T A: The submerged weight is subject to a buoyant force equal to the weight of the water of the volume displaced by the submerged weight (we denote this weight of water by P ). Consequently, to restore equilibrium, a weight P should be placed on the pan with the stand. T E AC H E R: You are mistaken. You would do well to recall New- ton’s third law of motion. According to this law, the force with which the water in the vessel acts on the submerged weight is ex- actly equal to the force with which the submerged weight acts on the water in the opposite direction. Consequently, as the weight of the pan with the stand reduces, the weight of the pan with the vessel increases. Therefore, to restore equilibrium, a weight equal to 2P should be added to the pan with the stand. S T U D E N T A: I can’t quite understand your reasoning. After all, the interaction of the submerged weight and the water in no way resembles the interaction of two bodies in mechanics. T E AC H E R: The field of application of Newton’s third law is not limited to mechanics. The expression “to every action there is an equal and opposite reaction” refers to a great many kinds of in- teraction. We can, however, apply a different line of reasoning in our case, one to which you will surely have no objections. Let us
D O YO U K N OW A RC H I M E D E S’ P R I N C I P L E? 165 deal with the stand with the weight and the vessel with the water as part of a single system whose total weight is obviously the sum of the weight of the left pan and that of the right pan. The total weight of the system should not change due to interaction of its parts with one another. Hence, if as the result of interaction the weight of the right pan is decreased by P , the weight of the left pan must be increased by the same amount (P ). Therefore, after the weight is submerged in the vessel with water, the difference between the weights of the left and right pans should be 2P . PROBLEMS 38. A vessel of cylindrical shape with a cross-sectional area S is filled with water in which a piece of ice, containing a lead ball, floats. The volume of the ice together with the lead ball is V and 1/20 of this volume is above the water level. To what mark will the water level in the vessel reduce after the ice melts? The densities of water, ice and lead are assumed to be known.
§ 17 Is Archimedes’ Principle Valid In A Space- ship? T E AC H E R: Is Archimedes’ principle valid in a spaceship when it is in a state of weightlessness? S T U D E N T A: I think it is not. The essence of Archimedes’ prin- ciple is that due to the different densities of the body and the liquid (of equal volumes, of course), different amounts of work are required to raise them to the same height. In a state of weight- lessness, there is no difference in these amounts of work since the work required to lift a body and that required to lift an equal volume of the liquid is equal to zero. We can reach the same conclusion if we consider the pressure of the liquid on a body submerged in it because the buoyant force is due to the difference in the pressures exerted on the bottom and top bases on the body. In a state of weightlessness, this difference in pressure vanishes and, with it, the buoyant force. I may add that in a state of weightlessness there is no difference between “up” and “down” and so it is impossible to indicate which base of the body is the upper and which the lower one. Thus, in a state of weightlessness, no buoyant force acts on a body submerged in a liquid. This means that Archimedes’ princi- ple is not valid for such a state. S T U D E N T B: I don’t agree with the final conclusion of S T U- D E N T A. I am sure that Archimedes’ principle is valid for a state
168 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S of weightlessness. Let us reason more carefully. We shall not pass over directly to a state of weightlessness, but begin with a lift travelling with a certain acceleration a which is in the same direction as the acceleration g of gravity. Assume that a ă g . It is easy to see that in the given case a body submerged in a liquid will be subject to the buoyant force F “ ρ0pg ´ aqV (96) and the weight of the liquid of a volume displaced by the body is also equal to ρ0pg ´ aqV . Thus, the buoyant force is still equal to the weight of the liquid displaced by the body, i.e. Archimedes’ principle is valid. Next we will gradually increase the accelera- tion a, approaching the value of g . According to equation (96), the buoyant force will be gradually reduced, but simultaneously and in exactly the same way, the weight of a volume of liquid equal to the volume of the body will also be reduced. In other words, as acceleration a approaches acceleration g , Archimedes’ principle will continue to be valid. In the limit a “ g a state of weightlessness sets in. At this the buoyant force becomes zero, but so does the weight of the liquid displaced by the body. Con- sequently, nothing prevents us from stating that Archimedes’ principle is valid for a state of weightlessness as well. I wish to illustrate my argument by the following example. Let us suppose that a piece of cork floats in a vessel with water. According to equation (95) the ratio of the volume of the piece of cork sub- merged in the water to the total volume of the piece is equal to the ratio of the density of cork to the density of water. Thus v1 v “ ρ3 ρ0 (97) Next, we suppose that this vessel is in a lift and the lift begins to descend with a certain acceleration a. Since this does not change the densities of cork and water, equation (97) holds. In other words, in the motion of the lift with acceleration, the position of the piece of cork with reference to the water level remains the same as in the absence of acceleration. Obviously, this condition will not change in the limiting case when a “ g and we reach a state of weightlessness. In this way, the position of the piece of
I S A RC H I M E D E S’ P R I N C I P L E VA L I D I N A S PAC E S H I P? 169 cork with respect to the water level, determined by Archimedes’ principle, turns out to be independent of the acceleration of the lift. In this case no distinction can be made between the presence and absence of weightlessness. T E AC H E R: I should say that both of your arguments are well sub- stantiated. However, I must agree with Student A: Archimedes’ principle is not valid for a state of weightlessness. S T U D E N T B: But then you must refute my proofs. T E AC H E R: That’s just what I’ll try to do. Your arguments are based on two main points. The first is that at an acceleration a ă g a body is buoyed up in the liquid in a manner fully complying with Archimedes’ principle. The second is that this statement must hold for the limiting case as well, when a “ g , i.e. a state of weightlessness is reached. I have no objection to the first point, but I don’t agree with the second. S T U D E N T B: But you can’t deny that the piece of cork remains in the same position in a state of weightlessness as well! And its position directly follows from Archimedes’ principle. T E AC H E R: Yes, that’s true. The piece of cork actually does re- main in the same position in a state of weightlessness as well. However, in this state its position with respect to the surface of the liquid is no longer a result of Archimedes’ principle. Push it deep into the water with your finger and it will remain sus- pended at the depth you left it. On the other hand, if there is even the smallest difference pg ´ aq , the piece of cork will come up to the surface and float in the position determined by Archimedes’ principle. Thus, there is a basic difference between weightlessness and the presence of even an insignificant weight- ness. In other words, in passing over to a state of weightlessness, at the “very last instant” there occurs an abrupt change, or jump, that alters the whole situation qualitatively. S T U D E N T B: But what is this jump due to? Where did it come from? In my reasoning, acceleration a smoothly approached acceleration g .
170 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S T E AC H E R: This jump is related to the fact that at a “ g , a certain symmetry appears: the difference between “up” and “down” disappears, which, incidentally, was very aptly pointed out by Student A. If the difference pg ´ aq is infinitely small, but still not equal to zero, the problem contains a physically defined direction “upward”. It is precisely in this direction that the buoy- ant force acts. However, at a “ g , this direction disappears, and all directions become physically equivalent. That’s what I mean by a jump. The destruction or the appearance of symmetry always occurs with jump.
Basically, modern physics is molecular physics. Hence it is especially im- portant to obtain some knowledge of the fundamentals of the molecular- kinetic theory of matter, if only by using the simplest example of the ideal gas. The question of the peculiarity in the thermal expansion of water is discussed separately. The gas laws will be analysed in detail and will be applied in the solution of specific engineering problems.
§ 18 What Do You Know About The Molecular- Kinetic Theory Of Matter? T E AC H E R: One of the common examination questions is: what are the basic principles of the molecular-kinetic theory of matter? How would you answer this question? S T U D E N T A: I would mention the two basic principles. The first is that all bodies consist of molecules, and the second, that the molecules are in a state of chaotic thermal motion. T E AC H E R: Your answer is very typical: laconic and quite incom- plete. I have noticed that students usually take a formal attitude with respect to this question. As a rule, they do not know what should be said about the basic principles of the molecular-kinetic theory, and explain it away with just a few general remarks. In this connection, I feel that the molecular-kinetic theory of matter should be discussed in more detail. I shall begin by mentioning the principles of this theory that can be regarded as the basic ones. 1. Matter has a “granular” structure: it consists of molecules (or atoms). One gram-molecule of a substance contains NA “ 6 ˆ 1023 molecules regardless of the physical state of the substance (the number NA is called Avogadro’s number). 2. The molecules of a substance are in a state of incessant thermal motion.
174 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S 3. The nature of the thermal motion of the molecules de- pends upon the nature of their interaction and changes when the substance goes over from one physical state to another. 4. The intensity of the thermal motion of the molecules depends upon the degree to which the body is heated, this being characterized by the absolute temperature T . The theory proves that the mean energy e of a separate molecule is proportional to the temperature T . Thus, for instance, for monoatomic (single-atom) moleculese e “ 3 2 kT (98) where k “ 1.38 ˆ 10´16 erg/deg is a physical constant called Boltzmann’s constant. 5. From the standpoint of the molecular-kinetic theory, the total energy E of a body is the sum of the following terms: E “ Ek ` Ep ` U (99) where Ek is the kinetic energy of the body as a whole, Ep is the potential energy of the body as a whole in a certain external field, and U is the energy associated with the thermal motion of the molecules of the body. Energy U is called the internal energy of the body. Inclusion of the internal energy in dealing with various energy balances is a characteristic feature of the molecular-kinetic theory. S T U D E N T B: We are used to thinking that the gram-molecule and Avogadro’s number refer to chemistry. T E AC H E R: Evidently, that is why students taking a physics exam- ination do not frequently know what a gram-molecule is, and, as a rule, are always sure that Avogadro’s number refers only to gases. Remember: a gram-molecule is the number of grams of a substance which is numerically equal to its molecular weight (and by no means the weight of the molecule expressed in grams, as some students say); the gram-atom is the number of grams of a substance numerically equal to its atomic weight; and Avogadro’s
W H AT D O YO U K N OW A B O U T T H E M O L E C U L A R-K I N E T I C T H E O RY O F M AT T E R? 175 number is the number of molecules in a gram-molecule (or atoms in a gram-atom) of any substance, regardless of its physical state. I want to point out that Avogadro’s number is a kind of a bridge between the macro and micro characteristics of a substance. Thus, for example, using Avogadro’s number, you can express such a micro characteristic of a substance as the mean distance between its molecules (or atoms) in terms of the density and molecular (or atomic) weight. For instance, let us consider solid iron. Its density is ρ “ 7.8 g{cm3 and atomic weight A “ 56. We are to find the mean distance between the atoms in iron. We shall proceed as follows: in A g of iron there are NA atoms, then in 1 g of iron there must be NA{A atoms. It follows that in 1 cm3 there are ρNA{A atoms. Thus each atom of iron is associated with a volume of A{ρNAcm3. The required mean distance between the atoms is approximately equal to the cube root of this volume x « 3 d A ρNA “ 3 d 56 7.8 ˆ 6 ˆ 1023 cm « 2 ˆ 10´8 cm S T U D E N T B: Just before this you said that the nature of the ther- mal motion of the molecules depends upon the intermolecular interaction and is changed in passing over from one physical state to another. Explain this in more detail, please. Ep e1 r0 A B Figure 69: Dependence of the potential energy Ep of interaction of the molecules on the distance r between their centres. T E AC H E R: Qualitatively, the interaction of two molecules can be described by means of the curve illustrated in Figure 69. This curve shows the dependence of the potential energy Ep of inter- action of the molecules on the distance r between their centres. At a sufficiently large distance between the molecules the curve Ep pr q asymptotically approaches zero, i.e. the molecules practi- cally cease to interact. As the molecules come closer together, the curve Ep pr q turns downward. Then, when they are sufficiently close to one another, the molecules begin to repulse one another and curve Ep pr q turns upward and E continues to rise (this re- pulsion means that the molecules cannot freely penetrate into each other). As can be seen, the Ep pr q curve has a characteristic minimum. S T U D E N T B: What is negative energy?
176 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S T E AC H E R: As we know, energy can be measured from any value. For instance, we can measure the potential energy of a stone from ground level of the given locality, or we can measure it from sea level, it makes no difference. In the given case, the zero point corresponds to the energy of interaction between molecules separated from each other at an infinitely large dis- tance. Therefore, the negative energy of the molecule means that it is in a bound state (bound with another molecule). To “free” this molecule, it is necessary to add some energy to it to increase the energy of the molecule to the zero level. Assume that the molecule has a negative energy e1 (see Figure 69). It is evident from the curve that in this case the molecule cannot get farther away from its neighbour than point B or get closer than point A. In other words, the molecule will vibrate between points A and B in the field of the neighbouring molecule (more precisely, there will be relative vibration of two molecules forming a bound system). In a gas molecules are at such great distances from one another on an average that they can be regarded as practically nonin- teracting. Each molecule travels freely, with relatively rare col- lisions. Each molecule participates in three types of motion: translatory, rotary (the molecule rotates about its own axis) and vibratory (the atoms in the molecule vibrate with respect to one another). If a molecule is monoatomic it will have only transla- tory motion. In a crystal the molecules are so close together that they form a single bound system. In this case, each molecule vibrates in some kind of general force field set up by the interaction of the whole collective of molecules. Typical of a crystal as a common bound system of molecules is the existence of an ordered three- dimensional structure - the crystal lattice. The lattice points are the equilibrium positions of the separate molecules. The molecules accomplish their complex vibratory motions about these positions. It should be noted that in some cases when molecules form a crystal, they continue to retain their individ- uality to some extent. In these cases, distinction is to be made between the vibration of the molecule in the field of the crystal
W H AT D O YO U K N OW A B O U T T H E M O L E C U L A R-K I N E T I C T H E O RY O F M AT T E R? 177 and the vibration of the atoms in the separate molecules. This phenomenon occurs when the binding energy of the atoms in the molecules is substantially higher than the binding energy of the molecules themselves in the crystal lattice. In most cases, however, the molecules do not retain their individuality upon forming a crystal so that the crystal turns out to be made up, not of separate molecules, but of separate atoms. Here, evidently, there is no intramolecular vibration, but only the vibration of the atoms in the field of the crystal. This, then, is the min- imum amount of information that examinees should possess about atomic and molecular thermal motions in matter. Usually, when speaking about the nature of thermal motions in matter, examinees get no farther than saying it is a “chaotic motion”, thus trying to cover up the lack of more detailed knowledge of thermal motion. S T U D E N T B: But you haven’t said anything about the nature of the thermal motions of molecules in a liquid. T E AC H E R: Thermal motions in a liquid are more involved than in other substances. A liquid occupying an intermediate position between gases and crystals exhibits, along with strong particle interaction, a considerable degree of disorder in its structure. The difficulty of dealing with crystals, owing to the strong inter- action of the particles, is largely compensated for by the existence of an ordered structure-the crystal lattice. The difficulty of deal- ing with gases owing to the disordered position of the separate particles is compensated for by a practically complete absence of particle interaction. In the case of liquids, however, there are both kinds of difficulties mentioned above with no corre- sponding compensating factors. It can be said that in a liquid the molecules, as a rule, completely retain their individuality. A great diversity of motions exists in liquids: displacement of the molecules, their rotation, vibration of the atoms in the molecules and vibration of the molecules in the fields of neighbouring molecules. The worst thing is that all of these types of motion cannot, strictly speaking, be treated separately (or, as they say, in the pure form) because there is a strong mutual influence of the motions.
178 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S S T U D E N T B: I can’t understand how translational motion of the molecule can be combined with its vibration in the fields of neighbouring molecules. T E AC H E R: Various models have been devised in which attempts were made to combine these motions. In one model, for in- stance, it was assumed that the molecule behaves as follows: it vibrates for a certain length of time in the field set up by its neighbours, then it takes a jump, passing over into new sur- roundings, vibrates in these surroundings, takes another jump, etc. Such a model is called the “jump-diffusion model”. S T U D E N T B: It seems that is precisely the way in which atoms diffuse in crystals. T E AC H E R: You are right. Only remember that in crystals this process is slower: jumps into a new environment occur consider- ably more rarely. There exists another model according to which a molecule in a liquid behaves as follows: it vibrates surrounded by its neighbours and the whole environment smoothly travels (“floats”) in space and is gradually deformed. This is called the “continuous-diffusion model”. S T U D E N T B: You said that a liquid occupies an intermediate position between crystals and gases. Which of them is it closer to? T E AC H E R: What do you think? S T U D E N T B: It seems to me that a liquid is closer to a gas. T E AC H E R: In actuality, however, a liquid is most likely closer to a crystal. This is indicated by the similarity of their densities, specific heats and coefficients of volume expansion. It is also known that the heat of fusion is considerably less than the heat of vaporization. All these facts are evidence of the appreciable similarity between the forces of inter-particle bonding in crystals and in liquids. Another consequence of this similarity is the existence of elements of ordered arrangement in the atoms of a liquid. This phenomenon, known as “short-range order”, was established in X -ray scattering experiments.
W H AT D O YO U K N OW A B O U T T H E M O L E C U L A R-K I N E T I C T H E O RY O F M AT T E R? 179 S T U D E N T B: What do you mean by short-range order? T E AC H E R: Short-range order is the ordered arrangement of a certain number of the nearest neighbours about any arbitrarily chosen atom (or molecule). In contrast to a crystal, this ordered arrangement with respect to the chosen atom is disturbed as we move away’ from it, and does not lead to the formation of a crystal lattice. At short distances, however, it is quite similar to the arrangement of the atoms of the given substance in the solid phase. Shown in Figure 70 (a) is the long-range order for a chain of atoms. It can be compared with the short-range order shown in Figure 70 (b). (a) (b) Figure 70: Long range order of crystals and short range order of liquids. The similarity between liquids and crystals has led to the term “quasi-crystallinity” of liquids. S T U D E N T B: But in such a case, liquids can evidently be dealt with by analogy with crystals. T E AC H E R: I should warn you against misuse of the concept of quasi-crystallinity of liquids and attributing too much impor- tance to it. Firstly, you must keep in mind that the liquid state corresponds to a wide range of temperatures, and the structural- dynamic properties of liquids cannot be expected to be the same (or even approximately the same) throughout this range. Near the critical state, a liquid should evidently lose all similarity to a solid and gradually transform to the gaseous phase. Thus, the concept of quasi-crystallinity of liquids may only be justified somewhere near the melting point, if at all. Secondly, the na- ture of the intermolecular interaction differs from one liquid to another. Consequently, the concept of quasi-crystallinity is not equally applicable to all liquids. For example, water is found to
180 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S be a more quasi-crystalline liquid than molten metals, and this explains many of its special properties (see § 19). S T U D E N T B: I see now that there is no simple picture of the thermal motions of molecules in a liquid. T E AC H E R: You are absolutely right. Only the extreme cases are comparatively simple. Intermediate cases are always complex. S T U D E N T A: The physics entrance examination requirements include the question about the basis for the molecular-kinetic theory of matter. Evidently, one should talk about Brownian motion. T E AC H E R: Yes, Brownian motion is striking experimental evi- dence substantiating the basic principles of the molecular-kinetic theory. But, do you know what Brownian motion actually is? S T U D E N T A: It is thermal motion of molecules. T E AC H E R: You are mistaken; Brownian motion can be observed with ordinary microscopes! It is motion of separate particles of matter bombarded by molecules of the medium in their thermal motion. From the molecular point of view these particles are macroscopic bodies. Nevertheless, by ordinary standards they are extremely small. As a result of their random uncompensated collisions with molecules, the Brownian particles move con- tinuously in a haphazard fashion and thus move about in the medium, which is usually some kind of liquid. S T U D E N T B: But why must the Brownian particles be so small? Why don’t we observe Brownian motion with appreciable parti- cles of matter such as tea leaves in a glass of tea? T E AC H E R: There are two reasons for this. In the first place, the number of collisions of molecules with the surface of a particle is proportional to the area of the surface; the mass of the particle is proportional to its volume. Thus, with an increase in the size R of a particle, the number of colIisions of molecules with its sur- face increases proportionally to R2, while the mass of the particle
W H AT D O YO U K N OW A B O U T T H E M O L E C U L A R-K I N E T I C T H E O RY O F M AT T E R? 181 R2 R3 y R 0 Figure 71: Effect of surface and volume relationships. which is to be displaced by the collision increases in propor- tion to R3. Therefore, as the particles increase in size it becomes more and more difficult for the molecules to push them about. To make this clear, I plotted two curves in Figure 71: y “ R2 and y “ R3. You can readily see that the quadratic relationship predominates at small values of R and the cubic relationship at large values. This means that surface effects predominate at small values of R and volume effects at large values. In the second place, the Brownian particle must be very small since its collisions with molecules are uncompensated, i.e. the number of collisions from the left and from the right in unit time should differ substantially. But the ratio of this difference in the number of collisions to the whole number of collisions will be the greater, the less the surface of the particle. S T U D E N T A: What other facts substantiating the molecular- kinetic theory are we expected to know? T E AC H E R: The very best substantiation of the molecular- kinetic theory is its successful application in explaining a great number of physical phenomena. For example, we can give the explana- tion of the pressure of a gas on the walls of a vessel containing it. The pressure p is the normal component of the force F acting on
182 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S unit area of the walls. Since F “ m ˆ ∆v ∆t ˙ “ ∆pmvq ∆t (100) to find the pressure we must determine the momentum trans- mitted to a unit area of the wall surface per unit time due to the blows with which the molecules of the gas strike the walls. Assume that a molecule of mass m is travelling perpendicular to a wall with a velocity v. As a result of an elastic collision with the wall, the molecule reverses its direction of travel and flies away from the wall with a velocity of v. The change in the momentum of the molecule equals ∆pmvq “ m∆v “ 2mv. This momentum is transmitted to the wall. For the sake of simplicity we shall assume that all the molecules of the gas have the same velocity v and six directions of motion in both directions along three coordinate axes (assume that the wall is perpendicular to one of these axes). Next, we shall take into account that in unit time only those molecules will reach the wall which are at a distance within v from it and whose velocity is directed toward the wall. Since a unit volume of the gas contains N {V molecules. in unit time 1 6 ˆ N V ˙ v molecules strike a unit area of the wall surface. Since each of these molecules transmits a momentum of 2mv, as a result of these blows a unit area of the wall surface receives a momentum equal to 2mv 1 6 ˆ N V ˙ v. According to equation (100). this is the required pressure p. Thus p “ 2 3 N V mv2 2 (101) According to equation (98), we can replace the energy of the molecule mv2 2 by the quantity 3 2 kT [in reference to the transla- tional motion of molecules, equation (98) is valid for molecules with any number of atoms]. After this, equation (101) can be rewritten as pV “ N kT (102) Note that this result was obtained by appreciable simplification of the problem (it was assumed, for instance, that the molecules
W H AT D O YO U K N OW A B O U T T H E M O L E C U L A R-K I N E T I C T H E O RY O F M AT T E R? 183 of the gas travel with the same velocity). However, theory shows that this result completely coincides with that obtained in a rigorous treatment. Equation (102) is beautifully confirmed by direct measure- ments. It is good proof of the correctness of the concepts of the molecular-kinetic theory which were used for deriving equation (102). Now let us discuss the phenomena of the evaporation and boiling of liquids on the basis of molecular-kinetic conceptions. How do you explain the phenomenon of evaporation? S T U D E N T A: The fastest molecules of liquid overcome the attrac- tion of the other molecules and fly out of the liquid. T E AC H E R: What will intensify evaporation? S T U D E N T A: Firstly, an increase in the free surface of the liquid, and secondly, heating of the liquid. T E AC H E R: It should be remembered that evaporation is a two- way process: while part of the molecules leave the liquid, another part returns to it. Evaporation will be the more effective the greater the ratio of the outgoing molecules to the incoming ones. The heating of the liquid and an increase of its free surface intensify the escape of molecules from the liquid. At the same time, measures can be taken to reduce the return of molecules to the liquid. For example, if a wind blows across the surface of the liquid, the newly escaped molecules are carried away, thereby reducing the probability of their return. That is why wet clothes dry more rapidly in the wind. If the escape of molecules from a liquid and their return compen- sate each other, a state of dynamic equilibrium sets in, and the vapour above the liquid becomes saturated. In some cases it is useful to retard the evaporation process. For instance, rapid evap- oration of the moisture in bread is undesirable. To prevent fast drying of bread it is kept in a closed container (bread box, plastic bag). This impedes the escape of the evaporated molecules, and
184 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S a layer of saturated vapour is formed above the surface of the bread, preventing further evaporation of water from the bread. Now, please explain the boiling process. S T U D E N T A: The boiling process is the same as evaporation, but proceeds more intensively. T E AC H E R: I don’t like your definition of the boiling process at all. I should mention that many examinees do not understand the essence of this process. When a liquid is heated, the solubil- ity of the gases it contains reduces. As a result, bubbles of gas are formed in the liquid (on the bottom and walls of the vessel). Evaporation occurs in these bubbles, they become filled with saturated vapour, whose pressure increases with the temperature of the liquid. At a certain temperature, the pressure of the sat- urated vapour inside the bubbles becomes equal to the pressure exerted on the bubbles from the outside (this pressure is equal to the atmospheric pressure plus the pressure of the layer of water above the bubble). Beginning with this instant, the bubbles rise rapidly to the surface and the liquid boils. As you can see, the boiling of a liquid differs essentially from evaporation. Note that evaporation takes place at any temperature, while boiling occurs at a definite temperature called the boiling point. Let me remind you that if the boiling process has begun, the temperature of the liquid cannot be raised, no matter how long we continue to heat it. The temperature remains at the boiling point until all of the liquid has boiled away. It is evident from the above discussion that the boiling point of a liquid is depressed when the outside pressure reduces. In this. connection, let us consider the following problem. A flask contains a small amount of water at room temperature. We begin to pump out the air above the water from the flask with a vacuum pump. What will happen to the water? S T U D E N T A: As the air is depleted, the pressure in the flask will reduce and the boiling point will be depressed. When it comes down to room temperature, the water will begin to boil.
W H AT D O YO U K N OW A B O U T T H E M O L E C U L A R-K I N E T I C T H E O RY O F M AT T E R? 185 T E AC H E R: Could the water freeze instead of boiling? S T U D E N T A: I don’t know. I think it couldn’t. T E AC H E R: It all depends upon the rate at which the air is pumped out of the flask. If this process is sufficiently slow, the water should begin to boil sooner or later. But if the air is exhausted very rapidly, the water should, on the contrary, freeze. As a re- sult of the depletion of the air (and, with it, of the water vapour), the evaporation process is’ intensified. Since in evaporation the molecules with the higher energies escape from the water, the remaining water will be cooled. If the air is exhausted slowly, the cooling effect is compensated for by the transfer of heat from the outside. As a result the temperature of the water remains con- stant. If the air is exhausted very rapidly, the cooling of the water cannot be compensated by an influx of heat from the outside, and the temperature of the water begins to drop. As soon as this happens, the possibility of boiling is also reduced. Continued rapid exhaustion of the air from the flask will lower the temper- ature of the water to the freezing point, and the unevaporated remainder of the water will be transformed into ice.
§ 19 How Do You Account For The Peculiarity In The Thermal Expansion Of Water? T E AC H E R: What are the peculiarities of the thermal expansion of water? T E AC H E R: When water is heated from 0 °C to 4 °C its density increases. It begins to expand only when its temperature is raised above 4 °C. T E AC H E R: How do you explain this? S T U D E N T A: I don’t know. Figure 72: Explaining pecu- liarity of thermal expansion of water. T E AC H E R: This distinctive feature of water is associated with its atomic structure. Molecules of water can interact only in one way: each molecule of water can add on only four neighbouring molecules whose centres then form a tetrahedron (Figure 72). This results in a friable, lace-like structure indicative of the quasi- crystallinity of water. Of course, we can speak of the structure of water, as of any other liquid, only on a short-range level (see § 18). With an increase in the distance from a selected molecule this order will undergo gradual distortion due to the bending and rupture of intermolecular bonds. As the temperature is raised, the bonds between the molecules are ruptured more frequently, there are more and more molecules with unoccupied bonds fill- ing the vacancies of the tetrahedral structure and, consequently, the degree of quasi-crystallinity is reduced.
188 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S The above-mentioned lace-like structure of water as a quasi- crystalline substance convincingly explains the anomaly of the physical properties of water, in particular, the peculiarity of its thermal expansion. On one hand, an increase in temperature leads to an increase in the mean distances between the atoms and molecule due to the intensification of intramolecular vibrations, i. e. the molecules seem to “swell” slightly. On the her hand, an increase in temperature breaks up the lace-like structure of water which, naturally, leads to a more dense packing of the molecules themselves. The first (vibrational) effect should lead to a reduc- tion in the density of water. This is the common effect causing the thermal expansion of solids. The second effect, that of struc- ture breakup, should, on the contrary, increase the density of water as it is heated. In heating water to 4 °C, the structural effect predominates and the density of water consequently in- creases. Upon further heating, the vibrational effect begins to predominate and therefore the density of water is reduced.
§ 20 How Well Do You Know The Gas Laws? T E AC H E R: Please write the equation for the combined gas law. S T U D E N T A: This equation is of the form pV T “ p0V0 T0 (103) where p, V and T are the pressure, volume and temperature of a certain mass of gas in a certain state, and p0, V0 and T0 are the same for the initial state. The temperature is expressed in the absolute scale. S T U D E N T B: I prefer to use an equation of a different form pV “ m μ RT (104) where m is the mass of the gas, μ is the mass of one gram- molecule and R is the universal gas constant. T E AC H E R: Both versions of the combined gas law are correct. (To S T U D E N T B) You have used the universal gas constant. Tell me, how would you compute its value? I don’t think one can memorize it. S T U D E N T B: To compute R, I can use equation (103), in which the parameters p0, V0 and T0 refer to a given mass of gas but taken at standard conditions. This means that p0 “ 76 cm Hg (cm of mercury column), T0 “ 273 K and V0 “ ˆ m μ ˙ ˆ 22.4 l,
190 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S since a gram-molecule of any gas at standard conditions occupies a definite volume equal to 22.4 l. The ratio m{μ is evidently the number of gram-molecules contained in the given mass of the gas. Substituting these values in equation (103) we obtain pV “ ˆ m μ ˙ T ˆ 76 cm Hg ˆ 22.4 l 273 K ˙ Comparing this with expression (104) we find that R “ 6.2 pcm Hgq litres deg . T E AC H E R: I purposely asked you to do these calculations in order to demonstrate the equivalence of expressions (103) and (104). Unfortunately, examinees usually know only equation (103) and are unfamiliar with (104), which coincides with equa- tion (102) obtained previously on the basis of molecular-kinetic considerations. From a comparison of equations (102) and (104) it follows that pm{μqR “ N k. Then R “ N ˆ m μ ˙ k “ NAk (105) Thus the universal gas constant turns out to be the product of Avogadro’s number by Boltzmann’s constant. Next, we shall see whether you can use the equation of the combined gas law. Please draw a curve showing an isobaric process, i.e. a process in which the gas pressure remains constant, using coordinate axes V and T . S T U D E N T A: I seem to recall that this process is described by a straight line. T E AC H E R: Why recall? Make use of equation (104). On its basis, express the volume of the gas as a function of its temperature. S T U D E N T A: From equation (104) we get V “ ˆ m μ ˙ ˆ R p ˙ T (106) T E AC H E R: Does the pressure here depend upon the temperature?
H OW W E L L D O YO U K N OW T H E GA S L AW S? 191 S T U D E N T A: In the given case it doesn’t because we are dealing with an isobaric process. T E AC H E R: Good. Then the product ˆ m μ ˙ ˆ R p ˙ in equation (106) is a constant factor. We thus obtain a linear dependence of the volume of the gas on its temperature. Examinees can usually depict isobaric ( p=const), isothermal (T =const) and isochoric (V =const) processes in diagrams with coordinate axes p and V . At the same time they usually find it difficult to depict these processes with other sets of coordinate axes, for instance V and T or T and p. These three processes are shown in Figure 73 in different sets of coordinate axes. Isobar Isochore Isotherm Isochore Isotherm Isobar Isotherm Isobar Isochore P V T T PV 0 00 Figure 73: The Gas Laws. S T U D E N T B: I have a question concerning isobars in a diagram with coordinate axes V and T . From equation (106) and from the corresponding curve in Figure 73 we see that as the tempera- ture approaches zero, the volume of the gas also approaches zero. However, in no case can the volume of a gas become less than the total volume of all its molecules. Where is the error in my reasoning? T E AC H E R: Equations (102), (103), (104) and (106) refer to the so-called ideal gas. The ideal gas is a simplified model of a real gas in which neither the size of the molecules nor their mutual attraction is taken into consideration. All the curves in Figure 73 apply to such a simplified model, i.e. the ideal gas.
192 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S S T U D E N T B: But the gas laws agree well with experimental data, and in experiments we deal with real gases whose molecules have sizes of their own. T E AC H E R: Note that such experiments are never conducted at extremely low temperatures. If a real gas has not been excessively cooled or compressed, it can be described quite accurately by the ideal gas model. Note also that for the gases contained in the air (for instance, nitrogen and oxygen), these conditions are met at room temperatures and ordinary pressures. S T U D E N T B: Do you mean that if we plot the dependence of the volume on the temperature in an isobaric process for a real gas, the curve will coincide with the corresponding straight line in Figure 73 at sufficiently high temperatures but will not coincide in the low temperature zone? T E AC H E R: Exactly. Moreover, remember that on a sufficiently large drop in temperature a gas will be condensed into a liquid. S T U D E N T B: I see. The fact that the curve of equation (106) in Figure 73 passes through the origin, or zero point, has no physical meaning. But then maybe we should terminate the curve before it reaches this point? T E AC H E R: That is not necessary. You are just drawing the curves for the model of a gas. Where this model can be applied is an- other question. Now I want to propose the following. Two isobars are shown in Figure 74 in coordinate axes V and T : one corresponds to the pressure p1 and the other to the pressure p2. Which of these pressures is higher? S T U D E N T A: Most likely, p2 is higher than p1. V T00 p1 p2 Figure 74: Which pressure is higher? T E AC H E R: You answer without thinking. Evidently, you decided that since that isobar is steeper, the corresponding pressure is higher. This, however, is entirely wrong. The tangent of the angle of inclination of an isobar equals ˆ m μ ˙ ˆ R p ˙ according to equation (106). It follows that the higher the pressure, the less the angle of inclination of the isobar. Thus, in our case, p2 ă
H OW W E L L D O YO U K N OW T H E GA S L AW S? 193 p1 We can reach the same conclusion by different reasoning. Let us draw an isotherm in Figure 74 (see the dashed line). It intersects isobar p2 at a higher value of the gas volume than isobar p1. We know that at the same temperature, the pressure of the gas will be the higher, the smaller its volume. This follows directly from the combined gas law [see equation (103) or (104)]. Consequently, p2 ă p1. S T U D E N T A: Now, I’m sure I understand. T E AC H E R: Then look at Figure 75 which shows two isotherms (the coordinate axes are p and V ) plotted for the same mass of gas at different temperatures, T1 and T2 Which is the higher temperature? p V0 T1 T2 Figure 75: Which pressure is higher?S T U D E N T A: First I shall draw an isobar (see the dashed line ( ) in Figure 75). At a constant pressure, the higher the temperature of a gas, the larger its volume. Therefore, the outermost isotherm T2 corresponds to the higher temperature. T E AC H E R: Correct. Remember: the closer an isotherm is to the origin of the coordinates p and V , the lower the temperature is. S T U D E N T B: In secondary school our study of the gas laws was of much narrower scope than our present discussion. The com- bined gas law was just barely mentioned. Our study was re- stricted to Boyle and Mariette’s, Gay-Lussac’s and Charles’ laws. T E AC H E R: In this connection, I wish to make some remarks that will enable the laws of Boyle and Mariotte, Gay-Lussac and Charles to be included in the general scheme. Boyle and Mari- otte’s law (more commonly known as Boyle’s law) describes the dependence of p on V in an isothermal process. The equation for this law is of the form p “ constant V (107) where the constant “ ˆ m μ ˙ RT .
194 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S Gay-Lussac s law describes the dependence of p on T in an isochoric process. The equation of this law is p “ constant T (108) where the constant “ ˆ m μ ˙ R V . The law of Charles describes the dependence of V on T in an isobaric process. Its equation is V “ constant T (109) where the constant “ ˆ m μ ˙ R p . [Equation (109) evidently repeats equation (106).] I will make the following remarks con- cerning the above-mentioned gas laws: (1) All these laws refer to the ideal gas and are applicable to a real gas only to the extent that the latter is described by the model of the ideal gas. (2) Each of these laws establishes a relationship between some pair of parameters of a gas under the assumption that the third parameter is constant. (3) As can readily be seen, each of these laws is a corollary of the combined gas law [see equation (104)] which es- tablishes a relationship between all three parameters regardless of any special conditions. (4) The constants in each of these laws can be expressed, not in terms of the mass of the gas and the constant third pa- rameter, but in terms of the same pair of parameters taken for a different state of the same mass of the gas. In other words, the gas laws can be rewritten in the following form p “ p0V0 V (107a) p “ p0 T0 T (108a) V “ V0 T0 T (109a)
H OW W E L L D O YO U K N OW T H E GA S L AW S? 195 S T U D E N T A: It seems I have finally understood the essence of the gas laws. T E AC H E R: In that case, let us go on. Consider the following example. A gas expands in such a manner that its pressure and volume comply with the condition pV 2 “ constant (110) We are to find out whether the gas is heated or, on the contrary, cooled in such an expansion. S T U D E N T A: Why must the temperature of the gas change?T o STUDENT A: It seems I have finally of the gas laws. TEACHER: In that case, let us go on example. A gas expands insuch a man volume comply with the condition pV2 = const We are to find out whether the gas is rary, cooled in such an expansion. ST.UDENT A: Why must the tempera TEACHER: If the temperature remain mean that the gas expands according p Fig. 76 Mariotte [equation (107)]. For an isoth while in our case the dependence of p o ture: pce(I/V2). STUDENT .A: Maybe I can try to p The curves will be of the shape shown TEACHER: That's a good idea. Wha STUDENT A: I seem to understand n tracing the curve p ce (1 IV2) toward g will gradually pass over to isotherm closer to the origin, i.e, isotherms 148 Figure 76: Comparing the isochores for p9 1 V 2 and p9 1 V . T E AC H E R: If the temperature remained constant, that would mean that the gas expands according to the law of Boyle and Mariotte [equation (107)]. For an isothermal process p9 1 V , while in our case the dependence of p on V is of a different nature: p9 1 V 2 ). S T U D E N T A: Maybe I can try to plot these relationships? The curves will be of the shape shown in Figure 76. T E AC H E R: That’s a good idea. What do the curves suggest? S T U D E N T A: I seem to understand now. We can see that in trac- ing the curve p9 1 V 2 toward greater volumes, the gas will gradu- ally pass over to isotherms that are closer and closer to the origin, i.e, isotherms corresponding to ever-decreasing temperatures. This means that in this expansion process the gas is cooled. T E AC H E R: Quite correct. Only I would reword your answer. It is better to say that such a gas expansion process is possible only provided the gas is cooled. S T U D E N T B: Can we reach the same conclusion analytically? T E AC H E R: Of course. Let us consider two states of the gas: p1, V1, T1 and p2, V2, T2. Next we shall write the combined
196 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S gas law [see equation (104)] for each of these states p1V1 “ m μ RT1 p2V2 “ m μ RT2 We can write the given gas expansion process, according to the condition, in the form p1V 2 1 “ p2V 2 2 Substituting the two preceding equations of the gas law in the last equation, we obtain m μ RT1V1 “ m μ RT2V2 After cancelling the common factors we find that T1V1 “ T2V2 (111) From this equation it is evident that if the gas volume is, for example, doubled, its temperature (in the absolute scale) should be reduced by one half. S T U D E N T A: Does this mean that whatever the process, the gas parameters ( p, V and T ) will be related to one another in each instant by the combined gas law?gas laws can be rewritten in the following form Povo P=y- (107a) p= T T (IOBa) (I09a) o STUDENT A: It seems I have finally understood the essence of the gas laws. TEACHER: In that case, let us go on. Consider the following example. A gas expands insuch a manner that its pressure and volume comply with the condition pV2 = const (110) We are to find out whether the gas is heated or, on the cont- rary, cooled in such an expansion. ST.UDENT A: Why must the temperature of the gas change? TEACHER: If the temperature remained constant, that would mean that the gas expands according to the law of Boyle and p Fig. 76 Fig. 77 Mariotte [equation (107)]. For an isothermal process pce(l/V), while in our case the dependence of p on V is of a different na- ture: pce(I/V2). STUDENT .A: Maybe I can try to plot these relationships? The curves will be of the shape shown in Fig. 76. TEACHER: That's a good idea. What do the curves suggest? STUDENT A: I seem to understand now. We can see that in tracing the curve p ce (1 IV2) toward greater volumes, the gas will gradually pass over to isotherms that are closer and closer to the origin, i.e, isotherms corresponding to ever- 148 Figure 77: Work done by a gas. T E AC H E R: Exactly. The combined gas law establishes a relation- ship between the gas parameters regardless of any conditions whatsoever. Now let us consider the nature of the energy ex- change between a gas and its environment in various processes. Assume that the gas is expanding. It will move back all bodies restricting its volume (for instance, a piston in a cylinder). Con- sequently, the gas performs work on these bodies. This work is not difficult to calculate for isobaric expansion of the gas. As- sume that the gas expands isobarically and pushes back a piston of cross-sectional area S over a distance ∆l (Figure 77). The pres- sure exerted by the gas on the piston is p. Find the amount of work done by the gas in moving the piston: A “ F ∆l “ p pSq∆l “ ppS∆l q “ ppV2 ´ V1q (112)
H OW W E L L D O YO U K N OW T H E GA S L AW S? 197 where V1 and V2 are the initial and final volumes of the gas. The amount of work done by the gas in nonisobaric expansion is more difficult to calculate because the pressure varies in the course of gas expansion. In the general case, the work done by the gas when its volume increases from V1 to V2 is equal to the area under the ppV q curve between the ordinates V1 and V2. The amounts of work done by a gas in isobaric and in isothermal expansion from volume V1 to volume V2 are shown in Figure 78 by the whole hatched area and the crosshatched area, respectively. The initial state of the gas is the same in both cases. Thus, in expanding, a gas does work on the surrounding bodies at the expense of part of its internal energy. The work done by the gas depends upon the nature of the expansion process. Note also that if a gas is compressed, then work is done on the gas and, consequently, its internal energy increases.The pressure exerted by the amount of work done by the Fig. 78 p where V 1 and V 2 are the init The amount of work done by is more difficult to calculate course of gas expansion. In t the gas when its volume incre area under the p (V) curve be The amounts of work done by mal expansion from volume Fig. 78 by the whole hatched respec gas is Thu on the of part done nature o also t v work quentl The not th between a gas and the medi expansion a gas does a cert refore, loses an amount of hand, however, as follows in § 18 [see equation (98)], gas in an isothermal proces energy U remains unchanged determined by the thermal m the mean energy of the mole perature T). The question is perform the work in the give STUDENT B: Evidently, t from the outside. TEACHER: Correct. In this' that a gas exchanges energy w two channels: by doing work volume of the gas, and by h 150 Figure 78: Work done by a gas. The performance of work, however, is not the only method of energy exchange between a gas and the medium. For example, in isothermal expansion a gas does a certain amount of work A and, therefore, loses an amount of energy equal to A. On the other hand, however, as follows from the principles enumerated in § 18 [see equation (98)], a constant temperature of the gas in an isothermal process should mean that its internal energy U remains unchanged (let me remind you that U is determined by the thermal motion of the molecules and that the mean energy of the molecules is proportional to the temperature T ). The question is: what kind of energy is used to perform the work in the given case? S T U D E N T B: Evidently, the heat transmitted to the gas from the outside. T E AC H E R: Correct. In this manner, we reach the conclusion that a gas exchanges energy with the medium through at least two channels: by doing work associated with a change in the volume of the gas, and by heat transfer. The energy balance can be expressed in the following form ∆U “ Q ´ A (113)
198 Q U E S T I O N S A N D A N S W E R S I N S C H O O L P H Y S I C S where ∆U is the increment of internal energy of the gas charac- terized by an increase in its temperature, Q is the heat transferred to the gas from the surrounding medium, and A is the work done by the gas on the surrounding bodies. Equation (113) is called the first law of thermodynamics. Note that it is universal and is applicable, not only to gases, but to any other bodies as well. S T U D E N T B: To sum up, we may conclude that in isothermal expansion, all the heat transferred to the gas is immediately converted into work done by the gas. If so, then isothermal processes cannot take place in a thermally insulated system. T E AC H E R: Quite true. Now consider isobaric expansion of gas from the energy point of view. S T U D E N T B: The gas expands. That means that it performs work. Here, as can be seen from equation (106), the temperature of the gas is raised, i.e. its internal energy is increased. Conse- quently, in this case, a relatively large amount of heat must be transferred to the gas: a part of this heat is used to increase the internal energy of the gas and the rest is converted into the work done by the gas. T E AC H E R: Very good. Consider one more example. A gas is heated so that its temperature is increased by ∆T . This is done twice: once at constant volume of the gas and then at constant pressure. Do we have to expend the same amount of heat to heat the gas in both cases? S T U D E N T A: I think so. S T U D E N T B: I would say that different amounts are required. At constant volume, no work is done, and all the heat is expended to increase the internal energy of the gas, i.e. to raise its tempera- ture. In this case Q1 “ C1∆T (114) At constant pressure, the heating of the gas is inevitably asso- ciated with its expansion, so that the amount of work done is A “ ppV ´ V1q. The supplied heat Q2 is used partly to increase