last date stamped he 1 LIBRARY, ANGEL: BY E. R. HEDRICK PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF MISSOURI 0. D. KELLOGG ASSISTANT PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF MISSOURI GINN AND COMPANY BOSTON NEW YORK CHICAGO LONDON COPYRIGHT, 1909, BY E. R. HEDRICK AND O. D. KELLOGG ALL BIGHTS RESERVED 89.11 tEfte gtftcnautn jpreaa C.INN AM) COMPANY- PRO- PRIETORS BOSTON U.S.A. Sciences i A ?c H2>5 PKEFACE Wherever the teaching of mathematics to engineering students is discussed, and frequently in cases of other classes of students, the criticism which is almost without exception the most insistent is this : that the student leaves the course without adequate ability to apply his mathematical knowledge.* This means that he has not the faculty of taking a problem, giving it an analytic formu- lation, and interpreting the analytic results. It is an open question whether it is the duty of the teacher of mathematics, or of the teacher of the more technical work which involves mathematics, to supply the needed training, but usually the mathematician is glad at least to share the responsibility and to do whatever he can to make his work fruitful, fully conscious of the fact that if he can successfully make the contact of his subject with the problems of the laboratory, of the engineering office, and of other activities, he will thereby add immensely to the vitality and interest of his work. With such a motive, it has been the practice at the Uni- versity of Missouri to follow the course in sophomore calculus with several weeks in applications to mechanics, this being a sub- ject rich in the kind of material desired. The present book is a formulation of the work there attempted, and it is believed that the need at our institution which has called the book into being will make its appearance welcome to a large number of mathe- matical departments. Opinions will differ as to the subject-matter which such a book should contain. The authors were guided by the feeling that it was practice in applying calculus rather than a broad knowledge * See, for instance, the reports of the joint meeting of mathematicians and engineers held in Chicago, December, 1907, under the auspices of the Chicago Section of the American Mathematical Society. These reports appeared in Science during the ensuing year. iii iv PREFACE of mechanics that was desired, and that such an end would be hindered rather than helped by a wide diversity of subject-matter. It was felt that, when feasible, the student secures a better insight into a subject by developing a portion of the theory himself, and so " exercises " have been introduced which form a part of such development of the theory. The " problems " are the applications of the theory, usually to cases in which numerical data are given. They vary considerably in difficulty, but it is thought that they are sufficiently numerous to supply a good number of the proper grade for any given class ; and, furthermore, it is believed that a more difficult problem with a proper amount of elucidation in advance by the teacher will be of far more value to the student than a number of problems so well within his range of ability as to require very little study on his part concerning the method of attack. Some hints will be found in the text, and some sug- gestions are given among the answers at the end of the book. We have endeavored, by a judicious selection of the problems to which answers are given, to secure a safe mean between the evil of supplying detailed answers, which rob the student of his independence, and the evil of furnishing him with no check upon his work. We shall be most grateful for any corrections or sug- gestions concerning this or other aspects of the book. E. R. HEDRICK COLUMBIA, MISSOURI 0. D. KELLOGG CONTENTS CHAPTER I. INTRODUCTION SECTION PAGE 1. Mechanics. Subject-matter of mechanics 1 2. Units 1 3. Vectors 3 4. The fixing of vectors 3 5. Operations upon vectors 4 6. The fundamental relation between acceleration and force vectors 8 Analysis of Chapter I 11 CHAPTER II. STATICS 7. Statics. Subject-matter of statics 12 8. Equilibrium under the action of concurrent forces .... 12 9. Nonconcurrent forces ; moments 16 10. Couples 18 11. Equilibrium. General conditions for equilibrium of a rigid body 20 12. Centers of mass of systems of particles 26 13. Centers of mass of continuous bodies 28 14. Centers of mass of some bodies of special shape.; .... 30 Analysis of Chapter II . ..." 41 CHAPTER III. DYNAMICS OF A PARTICLE 15. Dynamics 42 16. Rectilinear motion ; concepts involved 42 17. Some special rectilinear motions 48 18. Motion of a particle in a plane 55 19. Some geometric properties of plane motion 57 20. The differential equations of plane motion 63 21. Some special plane motions 63 Analysis of Chapter III 77 vi CONTENTS CHAPTER IV. WORK AND ENERGY SECTION PAGE 22. Work .................. 78 23. Force fields ; work on curved paths ......... 79 24. Conservative fields ; potential energy ........ 80 2~>. Kinetic energy ............... 82 26. Conservation of energy ............. 83 Analysis of Chapter IV ............. 85 CHAPTER V. .MECHANICS OF RIGID BODIES 27. Instantaneous motion of ;i rigid body ........ 86 28. Pure translation ............... 89 29. Pure rotation ................ 90 30. Moments and products of inertia. Radius of gyration ... 92 31. Work and energy in the case of a rigid body ...... 98 32. Compound pendulum ; experimental determination of mo- ments of inertia ............. 102 33. General equations of motion of a rigid body ...... 104 Analysis of Chapter V ............. 107 AND A\SWEI!S .......... 109 INDEX . 115 CHAPTER I INTRODUCTION 1. Mechanics. Mechanics deals with the position or motion of bodies in space ; so that the ideas of space and time implied in motion, and of mass implied in body, are fundamental. These three concepts cannot be defined, for there is nothing simpler in terms of which we can define them. 2. Units. With each of the three concepts mentioned is asso- ciated the idea of quantity, the quantity being measured by com- parison with a conventional unit or standard quantity. In Great Britain, and for commercial purposes in the United States, the units of space, mass, and time generally employed are the foot, pound, and second respectively. The initials of these fundamental units are usually used to designate the system, " the F.P.S. system." In France, and for scientific purposes in the United States, the units are the centimeter, gram, and second respectively, giving " the C.G.S. system." The following table gives the important derived units of velocity, acceleration, and force. In engineering circles other units of mass are in use, with the result that the units of force are also changed. The " engineering " or " tech- nical" or "gravitational" units of mass and force are also given below.* * For further information concerning units, see Ziwet, Theoretical Mechanics ; Encyclopedia Britannica on "Weights and Measures." In the following we shall meet with other derived units, hut these will he defined as they arise. 1 INTRODUCTION QUAXTITT BRITISH OK F.P.S. SYSTEM FRENCH OB C.6.S. SYSTEM Velocity = time- rate of change of space . . One foot per second. One centimeter per second. Acceleration = time-rate of change of ve- locity . One foot per second per second. (The accelera- tion of a falling body near the earth's surface is fj = 32.2 of these units, nearly-) One centimeter per second per second. (The accelera- tion of a falling body near the earth's surface is g = 981 of these units, nearly.) Force = Mass x Acceleration One poundal : the force which, acting constantly throughout a second, will give to a pound of matter a unit acceleration, that is, increase the velocity by one foot per second. One dyne : the force which, acting constantly through- out a second, will give to a gram of matter a unit acceleration, that is, in- crease the velocity by one centimeter per second. The engineering units of mass and force in the two systems follow : <.M ANTITY BRITISH OR F.P.S. SYSTDM FRENCH m: r.<;.S. SYSTEM Mass .... The mass of a body weigh- ing g pounds. (To deter- mine the mass of a body in technical units, divide its weight in pounds by ,7=32.2.) The mass of a body weigh- ing (j kilograms. (To de- termine the mass of a body in technical units, divide its weight in kilograms by = 981.) Force = Mass x Acceleration One pound : the force which the earth exerts at its surface upon a body \\--ighing one pound. (One pound = g pound- als.) One kilogram : the force which the earth exerts at its surface upon a body weighing one kilogram. (One kilogram = 1000 g dynes.) One foot = 30.5 centimeters. One pound = .373 kilograms. One centimeter = .0328 feet. One kilogram = 2.68 pounds. VECTOES 3 I. PROBLEMS ON UNITS 1. A force equal to the weight of 30 gm. acts upon a mass of 2 kgm. What is the acceleration in C.G.S. and in F.P.S. units? Solution. The equation f = ma holds if f is measured in dynes and m in grams. It may be solved for a, thus giving the desired result. The weight of 30 gm. is a force of 30 g dynes, and 2 kgm. equals 2000 gm. Hence 30 fj = 2000 a, and a = 30 #/2000 = 30 x 981/2000, which is approxi- mately 14.7 cm. per sec. per sec. As one centimeter is .0328 in., this is 14.7 X .0328, or .482 ft. per sec. per sec. 2. A man walks 5 mi. per hour. What is his speed in F.P.S. units, and what in C.G.S. units? 3. A mass of 25 gm. moves with an acceleration of 30 cm. per sec. per sec. What is the force acting, measured in C.G.S. and in F.P.S. units? 4. If sound travels 1000 ft. per sec. in air, what is its speed measured in C.G.S. units? 5. A boy throws a ball with a velocity of 50 m. per sec. W T hat is its speed in F.P.S. units? 6. A force of 6 Ib. acts upon a 3-lb. weight. What is the acceleration in F.P.S. and C.G.S. units? 3. Vectors. The units above apply primarily to bodies moving in a straight line. In the case of bodies not moving in a straight line, velocity, acceleration, and force cannot be characterized by one measurement alone, say of their magnitude ; but in addition their direction must be specified. They are examples of what are called vectors, and are usually represented geometrically by directed straight line segments. 4. The fixing of vectors. For the fixing of vectors in space three magnitudes are necessary. These may be selected in various ways ; we mention here only two of them. Let us denote the vector by V. It is to be noted then that V does not stand for a number in the ordinary sense, but for a set of numbers. The first way of fixing V is by its magnitude, v, essentially a positive num- ber, and the angles a, /3, and 7 which it makes with the coordinate axes (see Fig. 1). These three angles are not independent, for cos 2 a + cos 2 /3 + cos 2 7 = 1 ; so that V involves but three independ- ent magnitudes, v and two of the three angles a, ft, and 7. The relation given follows from the law for the diagonal of a cuboid : 1 = v 2 + v 2 + f 2 , whence, dividing by v 2 , we have FIG. 1 Another means of fixing V, and that usually used in analytic treatments of vectors, is by its projections on the axes, v x , v y , v z * If the initial point of the vector be placed at the origin, these pro- jections are simply the coordinates of the terminal points. Ex. 1. Express v x , r y , v z in terms of r, a, /3, and y, and conversely. Ex. 2. If our vector is required to lie in a fixed plane, two quantities suffice to fix it. (Jive two pairs of quantities analogous to the two sets given above for space, and give the relations between them. 5. Operations upon vectors. If an object move from a point A on a table to a point B on the table, while at the same time the table moves so that the point B passes to a point C fixed in the room, the body originally at A will arrive at C, and this no matter * If I stands for a given direction, upon a line with that direction. stands for the projection of the vector V VECTOR OPERATIONS 5 how the separate motions are carried out. Now the motion of the body relative to the table may be represented by a vector V v while the motion of the table may be represented by a vector F 2 , and the total effect also amounts to a vector F,. The last is said to be the o sum of the first two, and this illustrates the general definition : place the initial point of one vector upon the terminal point of the other ; the vector running from the free initial point to the free terminal point is called the sum, or resultant vector. Ex. 1. Show that F 3 will be the same whether the initial point of F 2 be put upon the terminal point of V 1 or the initial point of V 1 be put upon the terminal point of F 2 . Ex. 2. Show that another method of adding vectors, equivalent to the above, is to put the initial points of V 1 and F 2 together, to complete the parallelogram, and to draw the diagonal from the common initial point. This diagonal will be F 3 . Ex. 3. How would you obtain the sum of 3 vectors? of any given number of vectors ? Show that n vectors may be added in any order. On the other hand, a given vector may be resolved into two vectors whose sum or resultant it is. This may be done in an unlimited number of ways. Several especially important general ways are indicated in the following exercises. Ex. 4. Let there be given in a plane a vector F and two nonparallel lines Zj and 1 2 . Show that there are two vectors, V l parallel to Z t and F 2 parallel to Z 2 , whose sum is F. Give the construction for F x and F 2 . Is there more than one solution ? Ex. 5. Let there be given in space a vector F and three lines Z lf 1 2 , and Z 3 , not in the same plane or parallel. Show that there are three vectors, one parallel to l v one to Z 2 , and one to / 3 , whose sum is F. Is there more than one solution? To get a good figure, draw parallels to Zj, Z , and Z 3 through the origin of F, so that one looks into the solid angle formed. Ex. G. Let there be given a vector F and a J 7 :-^--'''' ~~-~^ V l vector F r Show how to construct a vector F 2 ^^ a * V a i such that F! + F 2 = F. " Ex. 7. Let F : and F 2 be two vectors whose resultant is F, and let -p IG their magnitudes be v l , v 2 , and v respectively ; let a v and a 2 be the positive angles between V : and F and between F 2 and F respectively ; 6 INTRODUCTION (a) Show that the magnitude v of the resultant is the square root of the sum of the squares of the magnitudes of the components increased by twice their product times the cosine of the angle (tij + a 2 ) between them, that is, v 2 = v? + r| + 2 v t v 2 cos (a l + a 2 ~). (b) Show that sine^/^ = sina 2 /u r (c) Show that tana 1 = [ 2 sin(a 1 + 2 )]/[ y i + v z cos(aj + 2 )1' Hint. The first two parts are essentially formulas from plane trigonom- etry. The third should be obtained from the figure by dropping a perpen- dicular from the tip of V to V t produced. These results are useful in finding direction and magnitude of resultants. Ex. 8. Take coordinate axes so that V lies on the a:-axis beginning with the origin. Find the projections on the axes of all the vectors, and thus prove analytically by the methods of Analytic Geometry the formulas of Ex. 7, (a) and (b). Hint. Show that v v^ cos a^ + i> 2 cos a 2 , = v 1 sin a 1 i> 2 sin a z ; then square and add. It should be noted that in the above statements concerning com- position and resolution of vectors it must be possible to regard the vectors as acting at the same point, or as concurrent. The fol- lowing important facts concerning operations with vectors should be clearly understood by the student. I. The projections upon each coordinate axis of the sum of two vectors is the sum of the projections upon that axis of the two com- ponent vectors. The same is true for the projections upon any line. Ex. 9. Prove the above theorem, making use of the " projection " the- orem of the Trigonometry or Analytic Geometry text-books. Draw your own figure. Ex. 10. Extend the above theorem to n vectors (X l , F 1? ZJ, (X 2 , F 2 , Z a ), .... (X n , Y,,, Z B ), with resultant (X, Y, Z). That is, show that X'= X l + X z + X 3 + + X n , etc. Such a sum is usually abbreviated in mathematical writing by 2A' i( the Greek sigma, 2, denoting a sum. II. The projections on each axis of k times a vector, where k is a number, is k times the projection upon the axis in question of the vector. The same is true of the projection upon any line. Ex. 11. This amounts to a definition of product of a vector % a number. Verify the fact that it holds when k is a positive integer and the multipli- cation is regarded as repeated addition. Show that in all cases the product vector has the same direction as the given one, and that its magnitude is k times that of the given one. VECTOR OPERATIONS If a vector vary with the time t, or any other parameter, its derivative may be defined, for we know how to subtract vectors and to divide by numbers. The notion of limit of a set of vectors will be sufficiently clear. The derivative will be found to have the property : III. The projection upon each axis of the derivative of a vector is the derivative of the projection upon the axes in question of the vector. The same is true of the projection upon any line. =1 Ex. 12. Working in two dimensions, draw the variable vector V for various values of , approaching 1 Q , and construct the appropriate difference vectors, multiplying their length by I/A/. The set of vectors thus obtained should have a limiting position, which we call the derivative vector. Find the magnitudes of the various approximating vectors and show that the limit of these magnitudes is -\/(do x /dt) 2 + (do y /dt) z . Next show that the limit of their slope is (W'' V /V/) = (di^./dt). Then noting that these are the magnitude and slopes of the vector whose projections are the derivatives of the projections of V, the truth of the above statement, III, is apparent. It is very important to note that the derivative vector in general will have a different direction from the given vector (see p. 57). We next consider a very useful question, namely,, given the pro- jections of a vector on the coordinate axes, supposed rectangular, to find the projections upon any given line. We consider the problem for vectors in a plane, where we shall need it most, and 8 INTRODUCTION leave the analogous considerations in space to the student to work out, or look up, as occasion arises. Let W denote a vector and / a direction, e.g. the direction of the positive ?/-axis. We denote by (I, W) the angle between the direction I and W measured from I FIG. 4 toward W, and similarly by (x, I) and (y, 1} the angle the line I makes with the axes measured from the axes to the line. Then, according to the definition of projection, the projection of W upon I is w l = w cos (/, W) = w cos [(x, W) (x, I)] w[cos(x, W)cos(x, /)+sin(.', /r)sm(#, I)] = w cos (x, W) cos (x, l}+ w cos (y, W} cos (y, I). But w cos(,/:, W} = u' x , and w cos(y, W) = w y . Hence we have IV. w t = w x cos (x, /) + w y cos (y, /), that is, the projection of a vector upon any line is the sum of the products of each projection upon an axis by the corresponding direction cosine of the line. 6. The fundamental relation between acceleration and force vectors. With the above information concerning vectors, we may proceed to consider motion other than in a straight line (cf. 3). Velocities, accelerations, and forces will then be vectors, and the fundamental equation F = mA is to be understood : the vector* F and A have the same direction, and the magnitude of F is m times the magnitude of A. Ex. Show that the above statement is equivalent to : the projection of F upon any line is m times the projection upon the same line of A. n. PROBLEMS ON VECTORS The following problems, as well as most of those in the book, should be worked with the aid of a careful figure. Measurement of the figure will then serve. as an approximate check on the work. In case of vectors in space the student can at least lay off the projections on straight lines. 1. A sailboat is sailing " into the wind," its course making an angle of 35 with the wind. The speed of the boat is 6 mi. per hr. and that of the wind is 20 mi. per hr. Find the speed and direction of the wind as it appears to a person on the boat. Solution. The forward velocity of the boat gives a relative backward component to the velocity of the wind. Our problem is to solve the tri- angle BA D, given AD = 20, BA = 6, and 4 BAD = 145. BD* = 6 2 + 20 2 + 2 x 6 x 20 xcos 35 = 436 + 240 x. 819 = 25.2, about. sin DBA = (20 sin 145)/25.2 = 20 x .574/25.2 = .456. DBA = 27. 1, about. Hence the apparent speed of the wind is 25.2 mi. an hour, and it apparently makes an angle 27.1 with the course of the boat. In some problems the formulas of the exercises in this chapter may prove useful, but the student had better rely upon his knowledge of trigo- nometry and his own ingenuity. 2. Two forces of 12 and 16 Ib. respectively act at an angle of 90. Find magnitude and direction of the resultant. 3. Find the resultant of two forces of the same magnitude f acting at an angle of 45. 4. Two men kick a football at the same instant. One kicks eastward at the rate of 71 ft. per sec., and the other northwest at the rate of 48 ft. per sec. Find the initial direction and the velocity of the ball. 5. The resultant of two forces is 9. One of them is 3 and the angle which it makes with the resultant is 60. What is the magnitude of the other force, and what the angle between it and the resultant? 6. Two forces are inclined to their resultant at angles of 120. How are the magnitudes of the forces related ? 7. A balloon rises 1 120 ft. per min. while the wind blows it horizontally 370 ft. per min. What is its velocity, and in which direction does it rise? (Use table of natural functions.) 10 INTRODUCTION 8 ? N i a CO B H o 1 o a as r. o * t- -0 a 2 ^3 00. 1 en N - a IH o -s | - * o C a 9 ^ J S . <<-! i S ^ 1 W l1 ^3 +3 tW * T o 90 1 10 o C - ^ n *-c o> < -- o oo 1 o T C b .5 ~^ rH ? 8 > e T . ~ 43 o UO B o r fe V> ^ C ^ O ia S o CO P3 1 ^' ^ 15 5 "I o - 00 o - < o C C ^ -f o o T * 1 * "o" " 1 d to 01 , "> ^ CO u h ^ = 8 T 1 1 ~ N ' 1 i ~ -: ff ^ I" ^ ~ 2 J5 K u . ^ e i M b c ^ 1 i j2 " 1 2 ^ 8 ^ ~ ^ C >i 2 s CO +=> B* 3 r" 1 * t- o -re i 1 e 1 s v4 -* C o fl 8 i r. . 1 - S" o Sh T 1 g ^ o> . E ~ 4 gq 1 c-i "K o 1 e C fe s 'no = e E = ^ -/: ; ~ t o - ? cos 150 o * , 1 -^> X ~-i JH 5 - -- 1 ci 1 I e c i e > e. fe "o S 1 8 - V" 1 1 i 1 T += .S - - i" H -re i s 73 -+3 < C ^ IO 3 .0 3 _r ;> i -i I r^ ;- i r - r H e t= c c 2 s V s - ^ ? ^a - - '- -r. 01 HM "3 c rt -| O N ~ ~ C = 8 1 I > 5 '-'- 1 1 1 ^ '-f. e s - i i 1> , C - = 8 8 06 c '^ ? e CO - c -^ I. IO 2 * 00 O +i - * O o , t - rCl ^ s a o c /!'>ii of rigid bodies is extremely useful in furnishing a simple approx- imation to what occurs in nature (see also p. 26). 12 CONCURRENT FORCES 13 If a number of vectors V v V 2 , , V n , are added by putting the initial point of each upon the terminal point of the preceding (cf. Ex. 3, 5), a polygonal line is formed which is called the vector polygon, or in particular, if the vectors are forces, the force polygon. The resultant force is the one which, starting from the initial point of the first force, closes the polygon. Ex. Show that if we add the n vectors T 7 lt F 2 , , V n in various orders, we obtain n\ force polygons, all of which have the same closing side. Thus we may state : A body can be at rest under the action of concurrent forces when, and only when, the force polygon is closed. For then, and then only, will the resultant be zero. Geometrically, this is the method -usually used in treating problems of equilib- rium. The analytic statement of the condition is found by pro- jecting the resultant upon the axes and referring to Ex. 10, 5. If X, Y, and Z denote the projections of the force F upon the axes, the conditions for equilibrium of concurrent forces takes the form =0. (1) HI. PROBLEMS ON EQUILIBRIUM OF CONCURRENT FORCES [Accompany each problem with a diagram. Each problem should be solved geometrically and analytically when possible.] 1. Show that the forces represented by the medians of a triangle are in equilibrium, the positive sense of each median being toward the vertex. The solution follows as an example of the method of attacking these problems. (a) (leometric proof. As the medians meet in a point, the forces are concurrent. We shall show that the force polygon, in this case a tri- angle, is closed. To do this we shift the force C'C par- allel to itself, so that its initial point C' falls upon the terminal point B of B'R. Then C'BDC is a parallelogram and CD = C'B = AC', C' being the mid-point of AB. To show the force polygon is closed we merely need to show that DB' is parallel with and equal to J '. 1. This will follow if we can show the two parallelograms AC'A'B' and B' FIG. 6 14 STATICS RA'DC, of -which A' A and DB' are corresponding diagonals, to be equal. But CD is equal to and parallel with AC', and so is B'A', for the segment joining the mid-points of two sides of a triangle is parallel with the third side and equal to one half of it. Finally AB' B'C, and hence the paral- lelograms are equal and the theorem is proven. (b) Analytic proof. If the coordinates of A, B, and C are respectively (z lf yj, (x 2 , y 2 ), (*, y g ), the coordinates of A' are \ (x 2 + z 3 ), |(y 2 + y,), of ', (:r 3 + xj, (y, + yj, of C', 1 (x 1 + ar g ), I (^ + y 2 ). The coniiionents or projections of the forces A' A, B'B, and C'C on the x-axis are therefore A\ + A 2 + A' 8 = *! + , + z, -$(2 *! + 2 a:, + 2 ar g ) = 0. Similarly, r x + T 2 + F 3 = 0, and the forces, having a vanishing resultant, are in equilibrium. In this problem the analytic treatment is briefer and more elegant, though this is by no means always the case. 2. Prove that three forces of magnitude 5, 6, and 12 can never be in equilibrium. 3. Let AB and CD be diameters of a circle. Three concurrent forces are represented in magnitude and direction by AB, DC, and 2 BD. Show they are in equilibrium. 4. Three concurrent forces are determined by their magnitudes and the angles their directions make with the axes as follows : F I; /=20, a = 30, = 60, y = 90. F 2 ; '/= 14, a = 90, = 45, y = 4r>. F 3 ; /= 10, a = 45, ft = 45, y = 90. Find the force which will hold them, in equilibrium. 5. Find the resultant of the forces given by their projections upon the axes: = = = 0. Three concurrent forces are in equilibrium. Show they lie in a plane. Show further that if their magnitudes are/ r /,,/ 3 , and the angles between them are a 23 , a sl , and a 12 , then ^ (Lamp's Tlieorem) ' smcr 23 smo- 31 smcr 12 . Hint. Consider the force triangle. 7. Let a system of forces be represented by OA , OB, OC, , OX. Show that if they are in equilibrium, the coordinates of are the averages of the corresponding coordinates of A, B, C, , A"; in other words, is the center of mass of a system of equal particles at the points A, B, C, -, N (see p. 27). 8. A weight W upon an inclined plane whose angle with the horizon is /, may In- held at rest either by a force Q acting up the incline, or by a force P acting horizontally. Show that W = PQ/\ / P 2 + Q-, and cos i = Q/P. Is your result in poundals or pounds? CONCURRENT FORCES 15 . 9. Find the angles between three forces 2P, 3P, and 4P which are in equilibrium. 10. Two rafters meeting at an angle 60 in a vertical plane support a chandelier weighing 90 Ib. What is the force along each rafter? 11. A body of mass 140 Ib. is attached to the ends of two ropes of length 6 and 8 ft. respectively, which are fastened to a horizontal beam at two points 10 ft. apart. Find the tensions on the ropes. 12. A bar weighing 100 Ib. is suspended by chains passing from a ring to its ends. The chains make an angle of 45 with the vertical. Find the tension on the chains. Hint. Consider the bar replaced by two weights of 50 Ib. each at its end points and connected by a weightless rigid bar. 13. A body is kept at rest on a smooth inclined plane by two forces, each equal to half the weight of the body, the one acting horizontally and the other directly up the plane. Find the angle of inclination of the plane. 14. Two smooth rectangles have a horizontal edge in common, their faces being inclined to a horizontal plane at angles a t and a 2 respectively. Weights u\ and ?r 2 rest one on each plane, and are connected by a string running smoothly over the common horizontal edge. If the system is in equilibrium, find the ratio of ?/. < 1 to w 2 . Hint. Consider, say, the weight u^, remembering that the string trans- mits the force exerted upon w l by tc 2 in line with the plane on which w^ rests. Friction. When two bodies are in contact along plane surfaces, there is usually a pressure keeping them together and causing more or less of an interplay of the slight roughness of their surfaces. There thus arises a certain resistance to any force tending to make the surfaces slide over one another. This resistance is a force exerted by the protruding particles against each other, and has a direction opposite to that of the attempted motion and is called a frictional force. As long as there is no motion, the frictional force exactly balances the force tending to produce motion. But it is found that when the moving force exceeds a certain limit, the body does move, although still retanldl by the frictional force. Experiments justify as an approximation the assumption that this force is proportional to the force pressing the two bodies together, that is, to the normal com- ponent n of the resultant of the forces acting at the common surface. We have then the result (1) The frictional force has a direction opposite to that of the resultant of the other forces. (2) It has a magnitude y which, (a) when no motion takes place, is equal to the magnitude of this resultant : (b) when motion takes place, is proportional to the magnitude of the normal component of their resultant. In symbols f= p.n, where p. is a 16 STATICS proportionality factor, and is called the coefficient of friction. It depends upon the character of the surfaces in contact. The student should avoid feeling that the above is more than an approximation, and apply it only in the case of bodies sliding on one another. There are many other kinds of frictional forces, too numerous and varied to receive attention here. 15. Determine the frictional force which keeps a body of weight 1 Ib. at rest on a plane inclined at an angle of 30, the slipping point being not yet reached. 16. If a plane be tilted farther and farther, the normal component of the gravitational force acting upon a body on the surface of the plane diminishes, while the component tending to produce slipping increases. The angle of inclination e at which the slipping begins, is called the angle of friction. Prove p. = tane. 17. Suppose a body rest on a table and a force of magnitude h tends to move it. The magnitudes of the friction/" depend upon h. Draw the graph of f as a function of h up to and beyond the slipping point. What is the value of h at the slipping point ? 18. Let a be the angle of inclination of a plane, and f the frictional force tending to prevent or retard a body's sliding along the plane. Draw the graph of f as a function of a up to and beyond the angle of friction. Take e, say at 20. 19. A mass of 10 Ib. rests upon a table and can be just moved by a force of 3 Ib. acting horizontally. Find the coefficient of friction and the direction and magnitude of the resultant reaction of the plane. 20. A body of weight w rests on a rough horizontal table. If a force be applied with an upward component, this upward component will lessen the normal reaction and hence the friction. Show that the least force which will move the body makes an angle e with the horizontal and has a magni- tude w sin e. 21. Two w r eights of 10 and 20 Ib. lie upon a rough inclined plane con- nected by a string which passes around a pulley in the plane. Find the greatest inclination of the plane consistent with equilibrium of the system, the coefficient of friction being p. 22. How high can a particle rest in a hemispherical basin of radius r, the coefficient of friction being p. ? 9. Nonconcurrent forces; moments. We now turn to the consideration of a system of forces acting at various points of a rigid body whose lines of action do not all intersect in one point. Besides a tendency to translate there will, in general, be a tendency to rotate the body about some line in it. If there is to be equilib- rium, the tendency to rotate about any line whatever in the body MOMENTS 17 must be zero. We consider the question of measuring this tend- ency to rotate. Let us first examine a force whose line of action is perpendic- ular to the axis of rotation. The angle between two nonintersecting lines is defined to be the angle between two intersecting lines which are parallel, one with each of the given lines. The tendency of such a force to turn about the axis is found to be proportional to the magnitude / of the force and to the dis- tance p of its line of action from the axis, or the arm of the force as it is called ; in other words, the tendency is proportional to / p. The student will find the best substantiation of this empirical fact in the well-known properties of levers and balances. (See Mach, The Science of Mechanics, trans- lated by McCormack, Chicago, 1902.) This quantity / p is called the moment of the force about the axis. If a number of forces are acting, some tending to turn in one direction and others in the opposite, we fix on a positive direc- tion and give a positive sign to the moments of the forces tend- ing to turn in that direction, and a negative sign to the moments of those tending to turn in the opposite direction. Should the force F not be perpendicular to the axis, we resolve it into two components, F a parallel with the axis, and F n per- pendicular to the axis (see Fig. 8). The component F a evidently has no tendency to turn about a line parallel to it, so that the turning effect of F is that of F n - We therefore define the moment in this case to be f n -p, where / is the magnitude of F n and p the distance of its line from the axis. FIG. 7 18 STATICS When the forces studied all lie in one plane perpendicular to the axis the problem is usually regarded as a plane problem, and we speak of the moments of the forces about a point, namely, the point in which the axis pierces the plane. The following exercises are important parts of the theory of moments and should be thoroughly studied. Ex. 1. Let P be any point upon a force vector .For upon its line of action. Prove that the moment of F about a point O is OP times the projection f p , upon a line ~~~~~~ perpendicular to OP. Hi >if. Use the previous definition, and similarly of triangles. Ex. 2. Let OX be an axis, OP a perpendicular dropped upon it from any point P of a force vector F or of its line of action, and let f^ be the projec- tion of F on a line perpendicular to the plane of OX and OP. Calling OP =//, show that the monunt <>f ]' uliout OX is p'f p .. Ex. 3. From the preceding, together with I of o, prove that the moment of the resultant of several concurrent forces is the sum of the moments of the > >/; fl> fa s > ' > fny 5 /!, /2z> ' ' ' referring to I, 5, . Generalize to three parallel forces not all in the same plane. der the points where their lines cut a perpendicular plane, and let their coordinates be (x r y^), (z 2 , y 2 ), (x s , ?/ 3 ). Find first the point of application of the resultant of two of the forces and then bring in the third. EQUILIBRIUM 23 IV. PROBLEMS ON MOMENTS AND THE EQUILIBRIUM OF FORCES [In solving problems in equilibrium of forces the student should accom- pany each problem by a^ carefully drawn diagram in which are marked all forces acting on the body, or on each separate lo/ly in case more than one is involved, being careful to include " reactions " (pressures or tensions) at all points where bodies are in contact, tensions of strings, and so forth. Then he should apply the conditions for equilibrium to the body, or, in case of several, to each body and to the system as a whole. Geometric relations in the diagram should also be written down. The result will be a system of equations to be solved for the unknown forces. The number of equa- tions should agree with the number of unknowns if the problem is deter- minate. Notice that to a force in a plane correspond two unknowns, and in space three, though this number is diminished when either the magni- tude or direction or other similar data is given with respect to the force. In the following problems the forces all lie in one plane, and moments need only be taken with respect to one point.] 1. A uniform rod AB of length 21 and weight w rests with the end A against a smooth vertical wall, while to the lower end B is fastened a string BC of length 2 i, coming from a point C in the wall directly above A. If the system is in equi- librium, determine the angle ACB = 0. Show that the tension on the string T = '/ sec 0, and that the pres- sure against the wall P = iv tan 6. Sol n/ ion. We consider the forces acting on the bar. A s the wall is "smooth," the pressure between the bar and the wall is perpendicu- lar to the wall, for otherwise there would be slipping. The weight w may be con- sidered as acting directly downward at the mid-point of the bar, and the tension T acts along the string. Applying the first condi- tions of equilibrium, the sum of the horizontal components must vanish: P-rsin0=0; (1) and also the sum of the vertical components : -w+ Tecs = 0. (2) FIG. 12 24 STATICS Also, taking moments about A, -wxlsin+ Tx2lsin(ABC) = 0. (3) The unknowns are 6, P, T, and < and (ABC). We therefore need two more relations. These relations are geometric in character, and are (ABC) = -e, (4) sin d> sin and, from the law of sines, /^ ; = (5) & U I We now proceed to answer the questions proposed. Equation (2) solved for T gives T = w sec 0, one of the required results. This value substituted in (1) gives P = w tan 6, a second required result. To determine we sub- stitute the value of T, and the value of ABC given by (4) in equation (3), at the same time dividing by wl and expanding sin ( 0): sin + 2 sec 6 (sin < cos cos sin 6) = 0, or sin < + 2 sin < 2 cos < tan = 0, that is, tan (f> = 2 tan 0. Equation (5) may be written sin = - sin 0. Unless = 0, we may divide this equation by the preceding and obtain I cos

. 11. A weightless rod AB of length / can turn freely in a vertical plane about .1. A weight // is suspended from a point (' of the rod distant c from A. A string attached at B and making an angle 1-~>0 with the rod holds it in equilibrium in a horizontal position. Find the tension on the string and the magnitude of the pressure at A. 26 STATICS 12. A uniform thin rough rod passes under one peg and over a second higher one, its center being above the higher peg and distant from the higher and lower pegs a and b respectively. Show that if the rod is upon the point of motion, the coefficient of friction is /u, = (b a)tani/(b + a), where i is the inclination of the rod. For further problems in equilibrium the student is recommended to study the theory of balances and scales, such as freight and postal scales, also derricks and hoisting devices. Examples of the kind are not given here because of the space necessary to describe each apparatus. Moreover, it will be of great value to the student to formulate his own problem. 12. Centers of mass of systems of particles. We shall be con- cerned with a system of particles connected by weightless bars. A particle is the idea derived by imagining all the mass of a body concentrated at a point, whereas a weightless ~bar is to be thought of as a device for keeping the distance between two particles con- stant. Whereas such things do not exist in nature, problems are simplified by their use, and the results obtained, if not strictly faithful to actuality, are frequently in error by less than the errors of observation ; * moreover, they furnish a basis for more rigorous developments. An example of the latter use of such notions appears in the next section. Near the earth's surface the attraction of gravity furnishes an example of a set of parallel forces acting on the above-mentioned system with magnitudes proportional to the masses. We ask, What force will hold this set in equilibrium ? Let the particles have masses m v m 2 , , m n , and be situated at points (x v y v z^, (x 2 , 7/ 2 , 2 2 ), -, (x n , y n , z n ). Call the balancing force F, with pro- jections f x , f y , f z upon the coordinate axes, and with point of application (x, y, z). Applying the conditions I of 11, in which the balancing force is included, we have or fz - * As an illustration consider a symmetrical fly wheel with its cylindrical axle, which we shall suppose to protrude more on one side than on the other. The gravita- tional forces acting on the system may be considered as acting on a particle with the mass of the wheel situated at its geometric center and connected by a weightless bar to a particle with the weight of the axle situated at its geometric center. CENTEKS OF MASS 27 where M is the total mass of the system. This gives F by its components. Its point of application is involved in conditions II, into which we introduce the projections of F already determined : [y (My) - 0] + [y, (- m#) - 0] + [y a (- m,g] - 0] + = 0, [0 - x (,%)] + [0 - x, (- mM + [0 - x 2 (- m,g}} + . . = 0, [0-0] +[0-0] +[0-0] +--- = 0, or 11 = -= = > and x = My M M Thus the point of application is not determined, for the conditions of equilibrium will be satisfied for any z. The explanation is that the point of application of the balancing force may be shifted to any point of its line of action. Suppose we ask whether there is a point at which the body will balance even though turned in a different direction, or, what amounts to the same thing, when the parallel forces on the particles act in a different direction, say oppo- site to that of the #-axis. We should then have for each mass and conditions I would give us while conditions II would yield f & 7/1' f y/f*v f iy ^^ ni , ^'^ M M it is now x undetermined. If, however, we take the point "V x i m i ^ y i m i ^ z.rc x = > y = > z= M M M the body will balance at this point against a system of forces par- allel to either of the two axes considered, and, as may easily be proven, against any set of parallel forces with magnitudes vary- in'j as the masses. This point is called the center of mass. (The expressions centroid and center of gravity are also frequently used, but sometimes with slightly different meanings, so that we shall confine ourselves to the term given.) The center of mass is of 28 STATICS great importance in the mechanics of systems of particles and of bodies. Ex. Prove the assertion just made, that the system will balance at its center of mass for any direction of the parallel forces. Let the direction cosines of the direction of the forces be cos a, cos ft, and cos y. Conditions I give the components f x ,f y ,f z of the balancing force, and in conditions II it will be found that if the values given above for x, y, z are used, the coefficients of cos a, cos ft, and cos y vanish separately, so that the conditions are fulfilled no matter what these cosines be. Show that the magnitude of the balancing force is Mg. V. PROBLEMS ON THE CENTER OF MASS 1. A ball of 2 Ib. and one of 20 Ib. are fixed to the ends of a uniform bar 5 ft. long and of weight 5 Ib. Find the center of mass of the system. 2. A fly wheel weighing 2 T. rests upon an axle of 400 Ib., 3 ft. long, the plane of the wheel dividing the axle into two parts in the ratio of 1 to 3. Find the center of mass of the system. 3. Given two masses m t and i 2 situated at the points (z 1? y l , 2j) and (x 2 , y%, z 2 ) respectively, show by considering moments, that the center of mass divides the segment joining the masses in the ratio of ? 2 to m r Hence, applying a formula of Analytic Geometry, obtain the formulas for x, y, z for two masses. A third mass may now be added by considering the first two united in their center of mass, and applying the same method. Generalizing thus, obtain an independent proof of the formulas for z, y, z. 4. Show that the center of mass of three equal particles at the vertices of a triangle is at the intersection of the medians. 5. The vertices of a square carry weights 1, 1, 1, and 2. Determine the position of the center of mass. 13. Centers of mass of continuous bodies. If equal volumes taken from all parts of a body weigh the same, the body is said to have uniform density, and the density is denned to be the constant ratio of mass to volume. If, however, there is no such constant ratio, we have recourse to the method of derivatives, saying first, the average density of a portion of a body is the mass of that por- tion divided by its volume; and secondly, the density at a point is the limit of the average density of a volume, including the point, as the dimensions of the including volume approach zero. Thus for a nonuniform body the density will usually vary from point CONTINUOUS BODIES 29 to point, that is, will be a function of x, y, and z, for which we shall use the notation d = B (x, y, z). Turning now to the problem of determining the centers of mass of continuous bodies, we imagine the body split up into cuboids by planes parallel with the coordinate planes. The mass of one of the cuboids will be approximately the volume multiplied by the den- sity S (x, y, z), say at the mid-point of the cuboid. We have, there- fore, for the ^-coordinate of the center of mass of the system of cuboids, according to the preceding paragraph, approximately the summations being extended over all cuboids containing masses, and no others. We obtain now the center of mass of the given continuous body by passing to the limit x = M and similarly M .)/ M where f*f*f* M = lim ^ ^ ^ 8 (x, y, z) AA?/A^ = I I I S(x, y,z) dxdydz J JJ is the total mass of the body. In all the integrals the limits of integration are determined by the surfaces bounding the body, just as in the volume problems of the Integral Calculus. Ex. Show that the formulas of 12 and 13 hold also for oblique axes, except that M no longer admits the interpretation of being the mass. 30 STATICS VI. PROBLEMS ON CENTERS OF MASS OF CONTINUOUS BODIES Unless the contrary is specified, the body is to be assumed homogeneous. 1. Find the center of mass of a hemisphere of radius r whose density varies as the distance from the plane surface. Solution, Choosing the plane as xy-plane with origin at the center of the sphere, we see at once by symmetry that x = y = 0, whereas by the above formulas, since S(x, y, z) = kz, k r +r r + \^-^ r Vr *-* 2 -* , z = I I ; _ I fdzdydx, M J-r i/-Vr2-al J f+r / + Vr2 a* ^.Vrl & y* . and M = k I I . _ I zdzdydx, Jr J -\Jrt- Xi /0 whence, as the student may verify by evaluating these integrals, ibrr 5 far* 8r M = - and 8 15 8 15 2. Determine the center of mass of a homogeneous tetrahedron. Hint. Oblique axes should be used. The bounding planes may then be written x = 0, y = 0, z = 0, x/a + y/b + z/c 1. >. 3. Find the center of mass of a homogeneous hemisphere. 4. Find the center of mass of that half of the ellipsoid x 2 /a 2 + y 2 /b 2 + z 2 /c 2 = 1 which lies to one side of the plane x = 0. 5. Find the center of mass of that part of the paraboloid x 2 /a 2 + y 2 /b 2 = 2 z/c which lies below z = h. 6. Find the center of mass of a plate in the form of a sector of a circle of radius r and angle 2 a. In particular, put a = Tr/2 and get the centroid of a semicircular plate. For a = TT the centroid should lie at the geometric center. Check the result in this way. 14. Centers of mass of some bodies of special shapes. The formulas of the preceding sections take on simpler forms in special cases. Some of these we here mention, leaving most of the devel- opments to the student. (a) Cylinders, or bodies bounded by a cylindrical surface and two parallel planes, the density being the same along all lines parallel to the elements of the cylindrical surface. The last means that the density depends, if we take the z-axis in the direction of the elements,, on x and y only, i.e. d = 8 (x, y). If the j;7/-plane be taken halfway between the parallel bases, we see, by symmetry, that 2=0, and furthermore we may carry out one integration in the other formulas : CENTERS OF MASS 31 I I I x8 (x, y] dzdxdy ~2 M h I I xS (x, y) dxdy M J/ = li\ I 8 (x, y} dxdy, li\ \ y%(x,y)dxdy and similarly, y = The form of the integrals suggests " the center of mass of an area," and this expression is sometimes used, though its real meaning should be kept in mind. The quantity h 8 (x, y) is sometimes called the surface density. Low cylinders are occasionally called plates. (b) Straight wires. Take the x-axis through the wire. Then 8 (x) means the " linear density " (that is, the mass of a piece of the wire divided by its length, or, in case this quotient varies, its limit). The student should show that where J/ = I 8(x)dx. (c) Bodies of revolution, in which the density is the same at all points of any plane perpendicular to the axis of revolution. These are a simple generalization of the preceding, in which 8 (,/:) means the mass of a plate bounded by two planes perpendicular to the axis and a distance A# apart, divided by A;, or, in case this quan- tity varies, 8(x) is its limit as A 0. In particular, if the volume density is constant, say /.', then 8 (./) is k times the area of a cross section a distance x from the origin. The same formulas of course hold. 32 EXAMPLE. Consider problem 1 of the preceding paragraph. Taking the ar-axis as the axis of revolution of the hemisphere, we have 8 (x) = den- sity times the area of a cross section = z(Vr 2 x 2 ) 2 . Whence The student will notice how much more simply the integrals are evaluated by this method whenever it applies. (d) Bodies with an axis such that there is a set of parallel planes which cut from the body plates whose centers of mass lie upon the axis. Taking the axis for x-axis, and the other axes in a plane T FIG. 13 parallel with the set of planes mentioned (the axes may be oblique ; compare the exercise at the end of 13), these bodies admit of a treatment like wires. Then 8(x) means the mass of the body between two of the planes of the set, corresponding to x and to x + A#, divided by Ax, or, if this ratio varies, its limit. The same formulas again hold. EXAMPLE. Any homogeneous cunt-. rir/Jtt or oblit/ue, has its center of mass ~ the distance from its vertex to its base. Planes parallel with the base make CENTERS OF MASS 33 similar sections whose areas vary with the square of the distance from the vertex. We have, therefore, 8(x) = kx 2 , whence C l k I x 3 dx o r l k I x 2 dx Jo (e) Surfaces of revolution. Since a surface has no thickness, mate- rial surfaces are idealizations, though they are nearly realizable in the case of bodies made of thin metal. We shall suppose the mate- rial of constant thickness and density. Then, the a?-axis being chosen to coincide with the axis of revolution, and the meridian curve being given by y =f(x), we have Ac = VAor + A?/ 2 for a chord of the meridian curve, 2 ir(y + Ay/2) Ac for the surface formed by revolving the chord about the axis (see the section on surfaces of bodies of revolution in the Integral Calculus), so that if d is the surface density, 2 vr(y + Ay/ 2) Vl + (A?//A^) 2 A d is the mass of such a surface. Thence and, by symmetry, y = 0, z = 0. Compare this with the derivation of the formula in calculus for the area of a surface of revolution. It need scarcely be pointed out that the dis- tinction between this case and case (c) is that here a shell is meant, whereas in case (c) the solid bounded by such a shell was intended. (f) Curved wires. The equations of the curve which give the form of the wire should be put into parameter form, say, x =f(t), y f j (t], z = li (t}. Then we have for a chord 34 STATICS Heiice the mass of a short piece of the arc is "We therefore have, remembering that lim As/Ac = 1, X , 9(W)^f"(t) + tf* (t) + h'*(t)dt and similarly, M z = where If possible, the length of the curve should be taken as parameter, for then the radical reduces to (g) Bodies whose bounding surfaces are simpler in some special coordinate system may be treated by the use of such a system. We illustrate by a plate bounded by a cylindric surface whose generating CENTERS OF MASS 35 curve is given in polar coordinates (see the section on areas bounded by curves given in polar coordinates in the Integral Calculus). The formulas become (cf. case (a)) CCpa(\ cos = 0. | Trer - 1 7T// 3 a s ^-b s If the student desires to avoid negative masses, he may use the formula x = (nijXj + i 2 a; 2 )/Af, where M is the mass of the ichole body, and x the ^-coordinate of its center of mass. The resulting equation should then be solved for x t . Note that m l = M m z . VI. PROBLEMS ON CENTERS OF MASS OF CONTINUOUS BODIES (Continued) 7. Find the center of mass of a straight wire of length I whose density varies (a) as the distance from one end ; (b) as the square of the distance from one end ; (c) as the nth power of the distance from one end ; (d) as (/ 2 -x 2 ); (e) as sin(7rz//); (f) as sin(-!rx/2l). 8. Find the center of mass of a plate in the shape of a quarter ellipse bounded by semimajor and semiminor axes. 9. Find the center of mass of a plate bounded by the parabola y z = 4ax and the chord x = h. Particular case where the chord is the latus rectum (h = a). 10. Find the center of mass of three faces of a cube regarded as plates, (a) when the faces all meet in a point ; (b) when they do not. PEOBLEMS ON CENTERS OF MASS 37 11. Find the center of mass of the cone formed by revolving y mx about the ar-axis, between the vertex and the base plane x = h. 12. Find the center of mass of the body between the paraboloid formed by revolving ?/ 2 = 4 ax about its axis and the plane x = h. 13. Find the same when the density varies with I/a:. 14. Find the center of mass of the homogeneous hemisphere by method of case (c). 15. Find the center of mass of half of the ellipsoid of revolution obtained by revolving x 2 /a 2 + y*/b 2 1 about the x-axis, the bounding plane being the plane- a: = 0. 16. Find the center of mass of the plate bounded by the a>axis and an arch of the sine curve y = sin x. 17. Find the center of mass of a regular tetrahedron composed of four equilateral triangular plates. Hint. Use oblique axes and regard the plates as concentrated at their centers of mass. 18. Find the center of mass of a solid homogeneous tetrahedron. Hint. Take origin at a vertex and a;-axis through the point of intersec- tion of the medians, i.e. at the centroid of the opposite face. Use method of case (d). 19. Find the center of mass of a hemispherical bowl of radius r. 20. Find the center of mass of a conical surface formed by revolving y = mx about the ar-axis, between the planes x = a and x b. 21. Find the center of mass of the wire bent into the form of a circular arc of radius r and angle 2 a. Check by putting a = TT. Hint. Polar coordinates advised. 22. Find the center of mass of a wire bent into the form of a cardioid p = a(l cos0). Hint. Use half angles. 23. Find the center of mass of a spring x = a cost, y = asint, z amt, the spring to be ended by the planes z = and z = h. Show that the cen- troid lies on the axis only when the spring makes an integral number of complete turns. 24. Find the center of mass of the wire p = e a6 from 6 = to = 2 -n. 25. From a right circular cylinder of height h and base of radius r is cut a cone of the same base and altitude. Find the center of mass of the remaining body. 26. Find the center of mass of a hemispherical shell bounded by a plane and two concentric spheres of radii a and b about one of its points. 27. Show that the center of mass of the body between two right cir- cular cones with same vertex, axis, and base plane, but different angles, coincides with the center of mass either cone would have if solid. 38 STATICS A A 1 C \ Y > A I Y I i 1 V FIG. 14 d > 28. Find the center of mass of a plate in the form of a segment of a circle of radius r, subtending an angle 2 a at the center. 29. By adding the proper triangle to the segment, get the center of mass of the sector and thus check the results of problems 6 and 28. 30. Find the center of mass of the body formed by revolving the oval of y 2 = x(x- a)(b - x), (6 14 37. Find the center of mass of the plate bounded by half the lemniscate p- = n 2 cos 2 0. 38. Find the center of mass of the plate bounded by one loop of p = a sin 2 0. 39. Show that the center of mass of the surface of a zone of a sphere is halfway between the bounding planes. 40. Find the center of mass of an arc of the hypocycloid x! + yl = a? (or x = a cos 3 *, y = a sin 3 /) between two successive cusps. 41. Find the center of mass of half a right circular cone, the dividing plane passing through the axis. 42. Find the center of mass of the plates indicated in Figs. 14-17. 43. Find the center of mass of a wire in the form of a cat- enary y = 2 (e-r/a -|- e-*/<) from x to x = a. 44. Find the center of mass of a wire given by the equa- tions y = | Va^/a, z = x 2 /4 a (or x = at 2 , y = 2 at s /3, c = a? 4 /4), from the origin to the point (a, 2 a/3, a/4). 45. Find the center of mass of a plate in the form of a sec- tor of the logarithmic spiral p = be md bounded by the radii = and = a. 46. Find the center of mass of the search-light reflector obtained by revolving the parabola y z = 4 ax about the x-axis, and bounded by the plane x = a. 47. Find the center of mass of a spherical wedge bounded by two planes meeting at an angle 2 a and a sphere of radius r with center on their line of intersection. Hint. Split up into plates normal to edge of the wedge and use problem 6. -i -- FIG. 17 40 STATICS 48. Prove the first theorem of Pappus : The area of the surface gener- ated by revolving a plane curve y =f(x) between the ordinates x = a and x = b about the z-axis is equal to the product of the length of the arc and the length of the path described by the rotation about the axis of the arc. 49. Prove the second theorem of Pappus : The volume of the body \vithin the surface of the above problem is the area under the curve times the length of the path described by the rotation about the axis of the center of mass of the area. Hint. Use double integrals. 50. Find by the theorems of Pappus the surface and volume of the torus, formed by rotating about an axis a circle of radius a whose center is a distance b from the axis (b>a). 51. Find the center of mass of the part included in a sphere of radius r and a right circular cone whose elements make an angle a with its axis, and whose vertex is at the center of the sphere. 52. Find the center of mass of the body bounded by a sphei-e of radius r and a right circular cone with vertex at the center of the sphere, and whose elements make an angle a with its axis. Verify by putting a = Tr/2 and comparing with the result of problem 14. 53. Find the center of mass of a spherical cap bounded by a sphere of radius r and a plane distant rcosa from the center. Verify as in the pre- ceding problem. 54. Find the center of mass of the part remaining of a hemisphere of radius r after removing the part contained within a circular cylinder of radius a, and whose axis coincides with the axis of the circle forming the edge of the hemisphere. 55. Find the center of mass of the plate bounded by the x-axis and an arch of the cycloid x = a(0 sin 0), y = a(l cos 9). 56. Treat problems 4 and 5 by the method of case (d). 57. Find the center of mass of a parabolic wire y 2 = 4 ax bounded by the latus rectum. The answer also holds for the right cylindric surface of which the wire is generatrix, and which is bounded by two planes perpen- dicular to the elements. 58. Find the center of mass of the part remaining of a sphere after removing the part contained within a cone with axis a diameter and ver- tex on the surface, the elements of the cone making an angle a with its axis. 59. From the problems on areas and volumes of surfaces of revolution in your Calculus text, find by means of the theorems of Pappus the center of mass of the corresponding curves and areas (wires and plates). ANALYSIS 41 ANALYSIS OF CHAPTER H 1. Subject-matter of statics. 2. Equilibrium denned. 8. The two conditions for equilibrium of concurrent forces. 4. Analytic statement of the condition for equilibrium of concurrent forces. 5. Assumptions regarding frictional forces. 6. Definition of the moment of a force about an axis ; its value for different positions of the axis. 7. Definition of couple. Constancy of the moment of a couple when the axis is shifted parallel to itself. 8. The shifting of the point of application of a force by the intro- duction of a couple. 9. Conditions for the equilibrium of nonconcurrent forces. 10. Analytic statement of the first conditions for equilibrium. 11. Analytic statement of the second conditions for equilibrium. 12. Definition of center of mass and its property with respect to a system of parallel forces proportional to the masses of the parts. 13. Formulas for the centers of mass of a system of particles and of continuous bodies. CHAPTER III DYNAMICS OF A PARTICLE 15. Dynamics. Dynamics treats of the effect of forces in produc- ing motion. We shall here limit ourselves principally to the motion of particles in a plane, for we thereby obtain methods applicable to the majority of interesting problems and avoid unnecessary com- plications of treatment. Furthermore, we start with the simplest kind of motion in a plane, namely, motion in a straight line. 16. Rectilinear motion; concepts involved. Take as origin a convenient point on the line, and call the distance OP of the particle at P from 0, s ; then the fundamental notions involved in the motion are the distance s, the time t, the velocity v, and the acceleration a.* To these should be added the mass m of the body and the force / acting upon it. As, however, the mass is constant and / = ma, we have really to deal essentially only with the four quantities above mentioned, and the problems of straight-line motion are usually concerned with a relation between two or more of these. We proceed to consider the more important cases that may arise, and to indicate how they should be treated. (a) A relation given between distance and time, say s =f(t). This we shall call a complete description of the motion, because the * Velocity. If the motion is uniform, a distance traversed, $. 2 s\, is proportional to the time consumed, t% 1\. The velocity is then defined as the constant ratio v = (*2 K\)/(tz t\) = As/A. If the motion is not uniform, this ratio depends upon both t\ and tz, and is called the average velocity during the interval t\ to to- By the velocity at the time t we mean the limit of the average velocity over an interval starting with t as the interval is shortened indefinitely, i.e. w = lira As/A<= cls/dt. Acceleration. Force tends to change "motion," i.e. to change velocity. The rate of change is called acceleration. If the motion is "uniformly accelerated," (v 2 v\)/(t = dx/dt. Whence ds/dt = (fc/2 m) / 2 and s = (k/Gm)t s + c 2 . But as s = when t 0, we have c 2 = 0, whence s = (k/G ?)< 3 , the required relation. (c) A relation given between acceleration (or force) and distance, say, a = f(s). This also requires two integrations, which may be carried out as follows : Write the relation dv/dt =f($), and multiply 44 DYNAMICS OF A PAETICLE by 2 v on the left and by its equivalent 2 (ds/dt) on the right, obtaining 2 v(dv/df) = 2f(s)(ds/dt). Whence, integrating with regard to t, we have v 2 = C < 2f(s)ds + c l . To carry out the second integration, solve for v, obtaining whence dt = = = and t = I , + e, ) ds If, after determining c 1 and c 2 by the auxiliary conditions given, and carrying out the integrations, the resulting equation is easily solvable for s, we have our complete description of case (a). Other- wise we have a, v, and t all expressed as functions of s, which may be considered as giving a parameter representation of the motion. EXAMPLE. A body falls toward the origin under a force inversely pro- portional to the square of the distance away. If v approaches as s approaches GO, and if s for t 0, determine the motion. Considering the body to move on the positive part of the line, the force will be nega- tive, as it is toward the origin. A proportionality factor giving simple results is km/I. Then, as/= Jan/2 s 2 , and ma =f, we have a = dc/dt = - l- 2 /2 s 2 . Then, as above, 2 i- (dc/dt) = - (F/* 2 ) (ds/df), or v 2 =k 2 /* + q. But v = as s = GO, hence = + c l and v 2 = k~/s, whence r = k/\ * The sign to choose is the negative one, as the body is falling toward the origin. Hence v = ds/dt = k/\s and 'Vsds = kdt. Hence | /= kt + c a , where c 2 = by the auxiliary conditions. Thus the motion is given by / 3 JtAi s = I I V 2 ) Among some interesting questions that may be attached to the above problem are : (1) Has the body fallen from an infinite distance in finite time? (2) With what velocity does it reach the origin? (3) Does the solution given hold for positive or negative at ds at ds dv , rvdv we have v ~j~~J \ v >> whence I - = s + c. The formula a = v (dv/ds) is a very useful one. (e) A relation given connecting acceleration (or force), velocity, and distance. In the more interesting problems this relation is linear in a, v, and s, and has constant coefficients, say, a + bv'+ cs = Q or d' 2 s/dt~ + b (ds/dt) + cs = 0, for the integration of which the student is referred to text-books on the calculus or on differential equations. Graphic representations of a motion, (a) On a straight line. Plot the points corresponding to a succession of values of t differing by equal increments. The direction of motion may be indicated by arrows parallel to the line. The closeness together of the points gives an indication of the speed. Jf the motion returns upon itself, draw a number of parallel lines and on each one plot the motion between two of its turning points. Jf in doing this we imagine the points actually marked at equal intervals of time, the point of the pencil has a motion which is a picture of the required motion. (b) Using the plane. Draw a <-axis perpendicular to the s-axis and plot s =/(/). If now a slit be cut in a piece of paper, and the plot be put behind the slit, so that the slit is parallel with the s-axis, and if the plot be drawn with uniform speed downward in the direction of the negative -axis, the point of the plot which shows through the slit will appear with the required motion. The maxima and minima of s =f(() will be the turn- ing points, and the speed is given by the slope of the curve with respect to the ^-axis. In discussing the motion the student should use one of these methods, and it is highly valuable for him to become acquainted with both (see Fig. 18). 46 DYNAMICS OF A PARTICLE VII. PROBLEMS ON RECTILINEAR MOTION In the first few of the following problems the " complete description " is given and the motion should be discussed. A discussion should include such points as the following : (a) Times and points where the motion stops, i.e. where the velocity vanishes. (b) The points reached by the motion ; e.g. in s P only the points to the right of the origin, with the origin. (c) Direction of the motion between stops (judged by the sign of v). (d) The number of times each part of the line is traversed, e.g. in the above example the positive half of the line is traversed twice, the negative half not at all. (e) Direction of the acceleration (i.e. of the force). (f) Tendency of the motion for large negative and for large positive values of t. It may be that not all these characteristics will be of interest in a given motion, whereas special motions will have other characteristics of special interest, which should be pointed out. The student should in advance get clearly in mind the meanings of sign of the velocity and acceleration. This may be done by answering the question, What is characteristic in the following four types of motion : (l)v>0, a>0; (2) w>0, a<0; (3) v<0, a>0; (4) v<0, a<0? Solution. v = 3t 2 -12t = Zt(t-4). a = 6 t - 12 = 6 (t - 2). (a) The motion ceases for t = and t = 4, that is, at the points s = 10 and s = 22. (b) All points are reached, because a cubic equation t s G t 2 + 10 = s will have a real root, no .matter what the value of s. (c) The direction of motion depends upon the sign of v, which depends upon the signs of its factors. For t<0, v>Q, and the motion is forward to s = 10. For 0<4, v<0, and the motion is backward to 22. For 40, and the motion is forward always thereafter. (d) The points between 22 and + 10 are traversed three times, other points once. (e) The acceleration is backward till t = 2, at the point s = 6, then forward. (f) The farther back in time we go, the farther to the left was the point (because for large negative t, s is large and negative), and there is no point to the left of which the moving point has not been. The velocity is positive for negative t, but decreasing as time progresses ; that is, the motion is a slower and slower forward motion. For larger and larger positive t we find s increasing without limit and v also ; that is, the point moves forward beyond any point whatever with an always increasing velocity. Graphic representations of the motion are shown in Fig. 18. PROBLEMS ON RECTILINEAR MOTION 47 ;- 16 t 2 - 32 1- 10. : U + S. 2. * 3. s ^. 5 ^ i O t JL 5. s = \/t. 6. * = l/(* 2 + l). 7. * = l/(* s -l). 8. s = 6sin<. 9. s = b sin (/.-< + e). 10. * = g'. 11. = e-. Can a point actually have the motion given by problems 20 and 21? 12. s = \ogt. 13. s = e~ t sint. 14. s = e- 1 /' 2 . 15. * = &(*- I) 8 . 16. s = 6-tanf. 17. * = |(ew + e^*'). 18. * = |("-e-*'). 19. s = 6(e* - e~ *)/(<*' + e~ kt ). 20. < = s s -6s-l. 21. t = s + sins. FIG. 18 In the next problems find the " complete description," determining the constants of integration by the auxiliary conditions given. If the motion is not recognized as one of the above types, it should also be discussed in the same way. Tell in all cases the force involved. 48 DYNAMICS OF A PARTICLE 22. v - kt ; s = 1 when t = 0. 23. v = ks ; s = s when / = ( . 24. a = kv ; v = 2, s = 1 when < = 0. 25. a = kt; s = when < = 0, s = when t = 1. 26. a = k*s ; i' = 0, s = a when * = 0. 27. a = k*$ ; r = ka, s = when = 0. 28. a = k 2 s ; w = i? , s = when t = t . 29. a = A: 2 /* 2 ; t? approaches as 5 approaches co, s = s when J = 0. Show that in this problem the body can only move on the positive side of the origin. 30. a = 2 6u + cs ; obtain the general solution and discuss for i 2 c = 0. 17. Some special rectilinear motions, (a) The inclined plane. Let * denote the angle of inclination of the plane to the horizon. Then, resolving the force of gravity into two components, perpen- dicular and parallel to the plane respectively, the parallel compo- nent, mg cos i, is alone effective in producing motion. The further development of this topic is left to the student in the exercises and problems to follow. Ex. 1. Determine the motion of a body sliding from rest down an inclined plane of inclination i. Ex. 2. Show that bodies starting simultaneously to slide down various chords of a vertical circle from the highest point of the circle will all reach the circumference at the same time. (b) Simple harmonic motion. Let a body be attracted toward a fixed center with a force varying with its distance away from the center.* Let the proportionality factor be ml? ; then, as the sign of the acceleration is opposite to that of s, we have a = k 2 s, or d z s/df + k 2 s = 0. If this be integrated as a linear equation with constant coefficients (see also the exercise below), we have for the solution s = c l cos kt+ c 2 sin kt. It is shown in Analytic Geometry that two numbers c x and c 2 are always proportional to the sine and cosine of some angle e, say, c l = A sin e, c 2 = A cos e. Then s becomes s = A sin (kt + e). The student should verify by direct * Examples of forces of this sort are elastic forces, for small displacements, due to springs and elastic bands; gravity in mines and beneath the surface of the earth; and, approximately, the forces involved in pendulum motion and the motion of mag- netic needles. SPECIAL KECTILIKEAR MOTIONS 49 substitution that this satisfies the differential equation. This motion the student has studied in problem 9 of the previous par- agraph. The motion repeats itself at intervals such that kt n+l +e = kt n +e + 27r, or t n+l - t n = T = ^-- K This quantity T is called the period of the simple harmonic motion. Ex. 3. Integrate the above differential equation by the method of case (c) in 16 and reconcile the results. Ex. 4. Discuss dampened harmonic motion in which there is a resistance due to air or friction proportional to the velocity, so that a = k 2 s bv. Compare with problem 13 of 16, which is a special case. (c) Fall of a body from a great height. The force of gravity acting upon a body is sensibly constant if the distance through which the body moves is small. Careful measurements, however, do reveal a variation, and, according to Newton's law of universal gravitation, the force upon a body outside the earth's surface varies with the inverse square of the distance from the earth's center. Let us take for origin the point from which the body falls, say, at a height h from the earth's center, and let the direction toward the center be taken as positive. Then h s is the distance of the body from the center, and / = ma = mc/(h s) 2 . If R denote the radius of the earth, we have, since at the surface a = g, g = c/R 2 , or c = gR 2 . Thus our equation becomes gR 2 d 2 s _ gR 2 ~(k-s) 2 ' d f~(h~s) 2 ' which comes under case (c) of the last paragraph. We have, then, dv 1 ds v 2 gR 2 v = gR - > whence = + - - -f c.. dt (h-s) 2 dt 2 h-s As the body falls from rest, v = when s = 0, so that Q=ff ~h + v and therefore v 2 = 2 gR 2 (-^ -- 1\ = 2 gR 2 / \h 8 h] \ h(h 50 DYNAMICS OF A PARTICLE ds 2 q s Whence V = = R \\-f- \ T dt " h vi s This admits of further integration by the ordinary methods, but the integral is more useful when obtained in the form of a series. We have ,- dt = -\ i'^^Yds. For the motion which interests us < s < h, so that we develop the radical as a series in s/li. h -/^Yri-VV Ll/V_iiiJ/Y -(s)[ -nsri^vy i^i w fV i 1/fV-i - -/- A/ 2*4\A/ 2 "I" 6^ Using this series in the differential equation and integrating, we have _ 1 lA/o^^---- 1 - ---- - - - - z ~E^2a( (IS >* 3k* 2'4'5P 2*4'6'77i ,, 113 2 * _ , _ 05 I V _ . _ _ . _ _ _ / _ 1 _ _ , B _ _ _ " 2 4 5 2 4 6 l /s\ 3 I _ 1 _ I \h) J where we have put c 2 = on the supposition that s = when t = ; that is, that we count the time from the instant the body starts to fall. If s is small compared with h, this series converges and a few terms give a good approximation to the true motion. Ex. 5. Write out the nth term of the above series. Ex. 6. Taking simply the first term of the series above and putting h = R, show that we have the ordinary law for falling bodies at the earth's surface. .Ex. 7. The terms of the series after the first are all negative, so that the actual motion differs from that considered in the preceding exercise in a way easy to state. Make the statement. (d) Body falling through a resisting medium. Raindrops and me- teorites are examples of this case ; similar forces act upon ships SPECIAL RECTILINEAR MOTIONS 51 and trains moving with small velocities under propulsion of their own engines. The resistance we shall assume to be proportional to the square of the velocity, and take our s-axis downward. We have, then, where g is the acceleration due to gravity and g/k 2 a convenient pro- portionality factor. The differential equation is, writing a = dv/dt, dv g , sinh (g/k) t + k cosh (g/k) f\ where - sinh u = , and cosh u = Ex. 9. Study the motion of a body on which no constant force, but the resistance alone, acts, there being an initial velocity v . * This equation holds only during downward motion, for if the velocity were upward, i.e. negative, the resistance is added to the acceleration due to gravity. For upward motion the law is a = g + (jv^/ld 2 -. The student should bear in mind that any simple law of resistance like the above is merely an approximation. See Osgood, Differential and Integral Calculus, p. 216. 52 DYNAMICS OF A PAETICLE Vm. PROBLEMS ON BODIES MOVING IN STRAIGHT LINES UNDER THE ACTION OF GIVEN FORCES* 1. A body falls from a height 100 m. After falling for 2 sec. a second body is projected vertically upward from the earth toward the first with a velocity 40 m. per sec. Find the time and height at which they meet. Solution. Let us count the time from the instant the first body begins to fall, and the distance vertically upward from the earth. If we use sub- scripts to distinguish the two bodies, we then have the conditions a^ y, with v t = 0, s l = 10,000 when t = for the first body, the units being centi- luctcrs and seconds ; and a 2 = y, with v z = 4000 and s 2 = when t = 2. These give s l = - gt 2 /2 + 10,000, * a = -m 1 , show that the * Except when otherwise stated, the force of gravity is to be considered constant and resistances are to be neglected. 53 tension T on the string is 2m l m 2 /(m l + m 2 ) and determine the motion of the system. (Note that if s l and s 2 are the distances of the particles below the pulley, s l + s 2 = constant and hence v l + v 2 = a l + a 2 = 0.) 9. Show that during the motion in the above problem the pressure on the axle is less than the sum of the weights of the two bodies. (Be careful to express all forces in the same units. The theorem may be proven by the fact that a perfect square, (m 2 n^) 2 , is positive.) 10. Over a pulley passes a string to one end of which is attached a weight 10 lb., and to the other a weight 8 Ib. with a rider 4 Ib. After being in motion 5 sec. the rider is removed without checking the velocity. I low much farther will the system move? 11. With what velocity must a particle be projected downward in order to overtake in 10 sec. a body that has already fallen 100 ft. from rest at the same point ? 12. Given a point and a vertical line a distance d from it, find the inclination of the straight line which would guide a particle acted upon by gravity only, from the point to the line in the briefest time. 13. Consider the same problem when the line is not vertical. 14. What is the weight by a spring balance of a man of 160 lb. descend- ing in an elevator with an acceleration 2 ft. per sec. per sec. ? 1"). A high jumper in jumping raises his center of mass 3 ft. How high could he jump on the surface of the moon, where he weighs one sixth as much? How long is he off the ground in both cases? (Assume he leaves the ground with the same velocity in both cases.) 10. If a motor car with speed 40 mi. per hr. can be stopped by its brakes in a hundred yards, find the inclination of a hill on which the brakes would just hold it. 17. A train with velocity 30 mi. per hr. runs with steam shut off against a resistance of 10 lb. per T. How far will it go? 18. A train running at 15 mi. per hr. with steam shut off strikes an up grade of 1 in 300. The resistance due to friction and air averages 3 lb. per T. Find how far the train will run before stopping. 19. A train of weight 200 T. descends a grade of inclination = arcsin ^g-. It its velocity is initially 40 mi. per hr., what frictional resistance will stop it in half a mile? 20. The attraction of the earth for a particle of mass m beneath the surface of the earth varies as the distance from the earth's center, and it is my at the surface. Find how soon the body would reach the center if dropped into a hole through the center; also its maximum speed, and at what point this speed is attained. 21. An elastic band is stretched between two points on a smooth table. A weight fixed at the mid-point of the band moves in the line of the band, 54 DYNAMICS OF A PARTICLE the elastic force being proportional to the displacement. Discuss the motion from its equations. 22. To one end of an elastic cord of natural length 2 ft. is attached a weight 2 lb., which when hanging in equilibrium extends the cord to a length 2^ ft. Assuming Hooke's law, that the elastic force exerted by the cord is proportional to its increase in length, determine how far the weight would fall if allowed to drop from the point at which the suspending cord has its natural length 2 ft. 23. A cylinder of radius 5 cm. is weighted at one end so as to float vertically in water. If forced downward into the water a distance 5 cm. below its position of equilibrium and then released, it is found to oscillate vertically with a period 2 sec. Determine the cylinder's weight. 24. Find the velocity with which a body reaches the earth's surface in falling from a height equal to the earth's radius, resistance of the atmos- phere being neglected. Find the time occupied by the fall. (Use about five terms of the series.) 25. As we increase the distance through which a body falls to the earth, show that the velocity with which it reaches the surface approaches an upper limit V2 gll. If a body were shot upward with this velocity, what would happen ? 26. A ship steaming at its maximum rate of 12 mi. per hr. is stopped by reversed engines and resistance in 6 sec. What is the value of k, and what is its velocity after 2 sec. ? (See footnote, p. 51.) 27. In the motion studied under (d) of the present section, character- ized by the equation ma = m(g g/k*v 2 ), the coefficient of resistance c = mg/k 2 depends only upon the size and shape of the body, and not upon its mass. Using this fact, compare the ultimate speeds k l and & 2 acquired by a falling raindrop and a falling bullet of the same size, the specific gravity of lead being 11.3. 28. In problem 21 suppose the body experience a resistance equal to 2 kv as well as the elastic force. Show that the motion becomes a dampened harmonic one of the type s = ^4e- A ''sin (/x/ + .B), A and B being constants of integration. 29. Study the motion of a particle repelled from a center by a force proportional to the distance of the particle from the center. Such a force might arise in studying electrified bodies. 30. A chain of length 5 ft. rests upon a smooth table with 1 ft. of its length hanging over the edge. Determine the motion of the chain as it slides off ; also the time when it leaves the table and its velocity at this moment. Hint. Treat the problem as if the chain were falling vertically, but only the weight of the part of the chain below the level of the table were effect- ive in producing motion. PLANE MOTION 55 18. Motion of a particle in a plane. In taking up motion in the plane it is essential to recall that velocity, acceleration, and force are no longer mere numbers, but directed magnitudes, that is, vectors. We shall, as a rule, think of them as fixed by their two projections on the axes. We consider first a complete description of the motion (cf. p. 42). This will evidently be attained if we know the coor- dinates of the point at any given time ; in other words, the complete description will consist in two equations :* X-point motion FIG. 19 It is thus seen that the complete description of a plane motion consists in the description of two straight-line motions, namely, those of a point on the #-axis directly beneath f (or above) the point in the ^ plane, and of a point on the y-axis on a level with it. This dependence of the mo- tion of a point in the plane upon two straight-line mo- tions is fundamental, as our whole analytic treatment of plane motion depends upon it, and a similar statement holds for space. The coordinates x and y of the moving point may be considered the projections of a vector whose beginning is at the origin of coordinates and whose end is at the moving point; we shall call it the position vector. We now define the velocity vector as the derivative with respect to the time of the position vector, and simi- larly we define the acceleration vector as the derivative with respect * Another form of description sometimes of value consists in the equation of the path f(x, y) = 0, or x =f(s), y g(s), together with a relation showing how the path is describee! , say s=4>((), where s represents the length of the path measured from some convenient point on it. The question of passing from one kind of description to the other is left for the student to consider. 56 DYNAMICS OF A PARTICLE to the time of the velocity vector; * or, referring to 5 and 6, the velocity vector V and the acceleration vector A are defined as fol- lows by means of their pro- jections, denoted in the usual manner : '=!=/'> O dx dt F IG .20 dt df and if f x and f y are the projections of the force vector F, we have expressing analytically the fundamental relation between accelera- tion and force vectors. For the magnitudes of these vectors we have 4 at- The magnitude of the ve- locity is called the speed. The essential distinction between speed, which is a positive number, and veloc- ity, which is a vector, is ^ one frequently overlooked, and serious errors result. Finally, we have for the direction of the velocity vector * Xote the close analogy of these definitions with the definitions of velocity and acceleration in a straight line (p. 42). The only difference is that vectors replace numbers. (a, is negative) FIG. 21 dx dt- PLANE MOTION 57 cos (x, v) = sin (y, v) = tan (x, v) = > cos (y, v) = + sin (x, v) = and similarly for the other vectors. 19. Some geometric properties of plane motion. In considering rectilinear motion our first concern was to fix a positive sense along the line. So in the plane curve, which is now to be our path, we fix a convenient point from which to measure the arc s, and fix a positive direction along the curve. We shall then denote by t the direction of the tangent to the curve at the point considered, the positive sense of its tangent agreeing with the positive sense of the curve. We shall denote by n the normal obtained by rotat- ing the tangent through an angle + 7r/2. The terms tangential velocity, normal velocity, tangential acceleration, and normal accel- eration will then readily be understood as the projections of the vectors named upon the directions named. We are now concerned with these vectors and with these projections. (a) The velocity vector is tangent to the path (though it may be pointed in the negative sense of the tangent). For, from the cal- culus, r being the angle between the tangent to the curve and the a-axis, tan T = (dy/dt) -r- (dx/dt). But this is v y /v x = tan (x, v), so that tan T = tan (x, v) and the theorem is true. (b) Tangential and normal velocities. From the above it is at once evident that the normal velocity is zero. For the tangential velocity we have, referring to IV, 5, v t = v x cos (x, t} + v y cos (y, t). But in the calculus it is shown that dx dx ds COS(X, t) = COS T = = H- , ds dt dt <1< I ds cos (?/, {) = --%- > dt dt ds\ 2 /dx\ 2 . /dy\ 2 ~r~ I and that =^7 + , \dt/ \dt / \dt Using these values with r, = ilr/dt and v y = dy/dt, we obtain 58 DYNAMICS OF A PARTICLE v t = ds/dt; that is,, the tangential velocity is the derivative of the arc with respect to the time.* (c) The acceleration vector is not, in general, tangent to the path. For this would mean a y /a x = v y /v x , or (dv y /dt) -5- (dv x /dt) = v v /v x ; i.e. dv y /v y = dv x /v x , and integrating, log ?^ = log 7^+ log ?^, whence v y = mv x or dy/dt = m (dx/dt], and integrating again, y = mx + c. We therefore have the theorem : If the acceleration vector is continually tangent to the path, the path is straight. As a corollary we see that where the path is curved the acceleration is not tangent to the path. O (a a is negative) (d) The tangential and FIG. 22 normal accelerations. These, like the tangential velocity, we obtain by application of IV, 5. First a t = a x cos (x, ) + a y cos (y, t) dx dy _ / dx dv\ ds ( ^ ' dt* + dt ' dt 2 dt 2dt\dt \dt ds It 2 dt\dt/ I dt that is, the tangential acceleration is the second derivative of the arc with respect to the time. For the normal acceleration a n = a x cos (x, n) + a y cos (y, n) = a x cos (y, t) + a y cos (x, t) dy . dx I dx dy\ dt ds * Notice that the tangential velocity may be positive or negative according as the point is moving in the positive or negative direction along the curve. The speed is the (positive) numerical value of the tangential velocity. PLANE MOTION 59 To obtain an interpretation for this we refer again to the calculus, where we find for the radius of curvature the expression * 2 ~ * dx d 2 x whence a y v x - a x v y = l \+. \ ai / /ds\ 2 v 2 We have, therefore, a n = I 1 -^- R = or, the normal acceleration is the square of the speed divided by the radius of curvature. The usual formula for the radius of curvature is , , . dy dy dx and the values = -. -- dx dt dt , d~y _^d/dy\T\ dx_/dxd' 2 t/ dyd 2 x\ dx* ~ Idi (dx) \ dt \dtdP dt d?) * /dx\ s di) reduce this to the form given. In the numerator of R occurs a square foot, and the reduction made in deriving the expression for a n amounts to giving to this square root the sign of ds/dt. The result is that R must be considered positive or negative according as the center of curvature lies in the positive direction of the normal to the curve as above fixed, or in the negative direction. The formulas found in (b) and (d) are important, and are therefore gathered together here : ds n d 2 s ir * = *' y " = ' *=**' n = R' The two projections of the acceleration are interesting in connection with their effect. The formulas show that the effect of the tangential accelera- tion is exclusively to change the tangential velocity ; the effect of the normal acceleration is exclusively to change the curvature of the path. It is easily seen from physical considerations that the normal component of the acceleration points from the curve toward the center of curvature. IX. PROBLEMS ON PLANE MOTION In the following problems derive, when possible, the equation of the path by eliminating t from the parameter equations ; find also expressions for v.0 v y and a^., a y . Plot the path, and draw in different-colored inks for various values of t the velocity vector (red, say) and the acceleration vector 60 DYNAMICS OF A PARTICLE (blue, say), and the tangential and normal components of the latter (blue dotted, say). Add a description of any salient characteristics of the motion. A different scale may be used for position vector, velocity vector, and acceleration vector in case any become too large or too small to yield a clear figure, but in this case the scales should be clearly designated. 1 Solution. The path is seen to be y 2 = X s , the " semicubical parabola." By differentiation we find v x = 2 t, v y = 3 t 2 , and x = 2, a y = 6 t. We next proceed to draw the path and the components parallel with the axes of velocity and the total velocity, first making a table (see Fig. 23). We may always compute the magnitude of the velocity, or the speed: v = Vrj 2 + c 2 = \/4 t 2 + 9 t* = \i\- (The vertical bars about t here denote that the positive numerical value of t is to be taken.) We see from this that the speed vanishes for t = and increases with \t\. As i-y = 3 1 2 is always positive except for t = 0, it follows that the motion is always upward. We next turn to a consideration of the acceleration. The components may be drawn as before for the velocity, being previously tabulated (see Fig. 24). The resultant should then be drawn. and a tangent and normal. The tangent should be drawn by means of its slope, and not merely by t a ruler laid along the curve. The tangential and normal accelerations are the projections on tangent and normal of the acceleration vector. Remark. Analytic expressions for tangential and normal acceleration may be calculated by the formulas given on p. 59. The only difficulty con- sists in determining the sign of the radical in ds/dt = v t = V (dx/df) 2 + (dy/dt) 2 . In our present problem if we measure s, say, from the origin, and ahvays increasing with y, then as y always increases with t, so also does s, and ds/dt is positive or zero. Hence (3 i' 2 )' 2 = T ^/' T ' ^'j/'' * ' I A *-* L and a t = = == > and a n = 2. x = t 2 3. x = t 4. x = l + cost "1 5. x-t Take for t in problem 4 a series of values 0, Tr/3, 27r/o. TT, . 6. x = t _ \ 8. x = cost "1 y = 2\ / l-t 2 j 7/ = 2sin. These notions have also an important application to the study of the rotation of a rigid body about an axis (see p. 91). As an illustration of the general method for obtaining projections upon given lines of the velocity and acceleration vectors, we shall derive these results directly, obtaining incidentally an interesting verification of out- general theory. Using IV of 5, we have v t = v x cos(x, t) + v y cos(y, t). But as (x, t) = + 7T/2 and (jj, I) = 6, cos (x, t) = sin 6 and cos (y, /) = cos 6 ; moreover \' and . a < i'= r sin dt dt '/'/ a (1 v,. = = r cos o * dt dt Using these values, we have r, = r(sin20 + cos 2 0) = r . Again, by differentiating r z and /. we have FIG. 25 a.. = r sin r/0\ 2 z ) r cos 6 dt) Forming now a, after the manner of r ( , we have a t - r ('1-6 /dt 2 ) ; and forming , keeping in mind that (x, n) = + ir, (y, n) = + tr/'2, so that cos(x, n) = cos 6, cos(?/. n) = - sin ^, we have a n r (dd/tatfd : The tangential velocity is r times the angular velocity; the tangential acceleration /x / PROJECTILES 65 times the angular acceleration; and the normal acceleration is r times the square of the angular velocity. Ex. 1. If to is constant, 6 = < is the projection of the moving point. Its motion is simple harmonic motion (see p. 48). (b) Projectiles. If a. body is thrown into the air and left to the action of gravity alone, resistance of the air and the effect of the rotation of the earth being neglected, the differential equations become m = dt 2 ^= m -: = dt* = mg or dt *=~9 If we take for origin of coordinates the point from which the body was thrown, and if the body was thrown in a direction making an V sin FIG. 26 angle T O with the horizontal with an initial speed v , the auxiliary conditions are x = 0, y = ; v x = v cos T O ,V V = v sin T O for t = 0. A first integration of the differential equations gives d<-/dt, or v x =c^ dy/dt, or v y = gt + c a . When t = these reduce to v cos T O and v sin T O respectively. Hence c 1 = v cos T O and c 2 = v sin T O . Hence 66 DYNAMICS OF A PARTICLE dx/dt = v cos T O and dy / dt = gt + v siiiT . Integrating again, we have x = (v cos T O ) t + c 3 , y=\ (j? + (%sin T O ) t + c 4 . As a? = and y = where = 0, c 3 = and c 4 = 0. Hence we have the complete description X = OoCOST )f y = I gt 2 + Oo sin T O ) t ^1 j Some important results in the theory of projectiles are now left for the student to derive in the ensuing exercises and in some later problems. Ex. 2. Verify, by eliminating <, that the equation of the path of the projectile is gx ( tan T O) * , v 3 2 (VQ cos r ) 2 What is the nature of this curve? Draw it for T O = 45, v = 40 ft. per sec., using g = 32. Draw the velocity vectors for t = 0, t = ^, t = 1, t = ^. Ex. 3. Range. By the range, R, of a projectile is meant the horizontal distance covered before it reaches the ground again. Show that z^ K 9 What elevation gives the greatest range? Show that any two values of TO differing by equal amounts on either side of this elevation give the same range. Ex. 4. Find the time of flight. The initial speed V Q being given, what elevation gives the greatest time of flight? Ex. 5. Suppose the initial speed v be given. Then for various values of T we have a system of parabolas with vertical axes and all passing through the same point. Any point in the plane which is covered by this system is iritl/in r(i,,cje, that is, can be hit with the given initial velocity. In other words, all points within the envelope of the system are within range. Find the envelope, writing the system in the form y <7 2 (1 + m 2 )/2 t' 2 + mx. The result should be a parabola whose highest point is that of the highest path of the system and whose breadth at the level of the initial point is twice the maximum range. Show that your result agrees with these statements. Ex. 6. Refer the path of the projectile of Ex. 2 to its vertex by a shift of the axes (see Analytic Geometry). What are the coordinates of the vertex? What is the latus rectum? Show that all the trajectories corre- sponding to a given initial speed have the same directrix. (c) Pendulum. We are dealing with a particle moving in a circular arc under gravity. Knowing the path, we proceed as PENDULUM 67 indicated in 20, except that instead of s it will be convenient to use the angle between the pendulum and a line vertically down- ward. Then s = Id, I denoting the length of the pendulum; and as the tangential compo- nent of the force is mg sin 6, we have d 2 s d 2 or d g . . -7 = 7 sm 6. dt 2 I To avoid difficulties in integra- tion it is usual to substitute for sin 6 on the right the quantity 6, which approximation * is in error by less than # 3 /6, where 6 is measured in radians, and which is quite satisfactory when 6 does not exceed a twentieth of a radian. The resulting equa- tion, \0 FIG. 27 (j sin 6 my = ^ 6 df 2 I ' is easily integrated (see (b), 16), and yields t] -f- B cos 6 = A sin * If this approximation is not used, the method is as follows: Multiplying by dd/dt (see (c), 10), the equation becomes d IfddV q d fd9V 2g ~J- t n TT; = 7 ' T f cos whence -77 = r cos e + c i- dt2\dt/ I dt \, by the relation sin 6/2 = sin (0Q/2) si n0, we obtain (ft X FIG. 28 J V 1 - 81 "*^ If t when 6=0, and hence when = 0, we have, since .FIO, sin -)= 0, / / 6 \ When 6 = , = ir/2, so that the whole time of an oscillation T, being four times the time consumed as goes from to , is 4 (l/g) F[(ir/2), sin (0 /2)]. As an illustra- tion the student may show that the period of a pendulum swinging through an angle 60 to either side of the vertical is 6.744 ~Vl/g. A further study of the properties of elliptic integrals is recommended at this point to students who find the subject inter- esting (see Byerly, Integral Calculus). 69 Vre 2 + if, the force center being taken as origin, and cos (x, F) = x/r, cos (y, F} = y/r, so that f x = mk*x and / = - rntfy. The differential equations of motion therefore take the form dt 2 df 2 ~ Referring to (b), 17, we see that the projections of the motion on the axes are both simple harmonic motion. To find the path, we endeavor to eliminate t from the solution x = c l cos kt + c 2 sin kt v = c, cos kt + c. sin kt U tj 4 This is best done by solving for sin kt and cos kt and squaring and adding. We find * c.x c,?/ . , , c.x + c.ii cos kt = -^ > sin kt = i j whence (c 4 .r c 2 y) 2 + ( c 3 + c^) 2 = (c x c 4 c/ 3 ) 2 . This is a conic section referred to its center, for it is of the second degree and the linear terms are absent. That it is an ellipse is easily verified by the test given in Analytic Geometry, namely that B- 4 A C < where A, B, and C are the coefficients of ar 2 , xy, and y 2 respectively. The motion is therefore called elliptic har- monic motion. Ex. 7. Show that the force producing the motion x = a cos kt, y = b sin fct, is toward the origin and varies with the distance from the origin. Returning to the general problem of central motion, we shall have the same direction cosines for the force, but the magnitude / will be some function of r, say,/= ^(r). The equations become - df 2 * Unless 0^4 c 2 c 3 = 0. The motion in this case is simple harmonic motion in a straight line, as the student may show as an exercise. 70 DYNAMICS OF A PARTICLE These equations present the difficulty mentioned in both equations contain both variables, since r = 20, that Two 2 + y l . integrals may, however, be obtained, first, the integral of areas. Multiplying the first equation by y and the second by + x and adding, we have * that is, d dt .4 dt . = or dx dt The meaning of this is clearer upon using polar coordinates, when it becomes a dr . r cos v I sin a + r cos dt dt \ (dr dO\ ] r sin 6 1 cos a r sin 6 - I = c., ] \dt dt) v or rd6 dt Now let A represent the area swept out by the moving radius vector. We have, by the calculus, r*d0 = \ C or dA = l f*d&, so that dt 2dA dt and the equation becomes dA = -* > so that ^ = t=3t = -3 axis, we have / 7T\ sin(0 c 4 ) = sin [ 6 + \ = cos 0, \ */ and the equation may be written PROBLEMS ON PLANE MOTION 73 Comparing this with the equation (see Analytic Geometry) l/r = 1 e cos 6, we see that the path is a conic section referred to its focus as origin. This fact was discovered by observation by Kepler, and bears the name Kepler's first law. The eccentricity is 6= A 1 + fc 8 and is |1, according as 2mc 1 s c,/A; = 0, that is, according as c 3 =0. But, referring to the" energy integral, we find an interpretation for c y For if we denote by V Q and r the speed of the planet and its distance from the sun when t = 0, we have, since the energy equation holds for every t, c s = 1 mv (&/r ). Hence e 1 1 accord- ing as c s 1 0, that is, according as v* = 2 Jc/mr . Thus the path will be a parabola if the initial speed is v 2 k/mr , and this no matter what the initial direction of the planet. .For larger initial speeds the paths are hyperbolas, for smaller ones ellipses. Planets move only in ellipses. Comets have orbits of any of the types. The earth's orbit is very nearly a circle, its eccentricity is about ^. The motion of the body in its path is determined by the inte- gral of areas. Equal areas being swept out in equal times, it follows that the planets move fastest when nearest the sun. This result may also be read from the energy integral X. PROBLEMS ON BODIES MOVING IN A PLANE UNDER THE ACTION OF FORCES 1. A snowball fight is arranged between two boys who can throw with speeds of 78 ft. per sec. and 83 ft. per sec. respectively. In order to equal- ize the contest the boy with the less speed is given a station G| ft. higher than that of his comrade. Assuming that the effect of the snowball varies with the square of the speed, does this arrangement equalize the fight? Solution. We must first find the expression for the speed of a projectile. Taking the equations of motion, finding v x and , and squaring and adding, we have v 2 = v$ 2 gt (?> sin T O ) + g 2 t 2 . We wish to know the value of this expression for a given value of y. To do this it is only necessary to find t from the equation for y, y = gt 2 /2 + (c sin T O )<, and substitute it in the equation for v 2 , or, in other words, to eliminate t. This is easily done by multiplying the second equation by 2 g and adding. The result is r 2 + 2 gy = v 2 , or v 2 = v 2 2 yy. Now let v' be the speed with which the stronger boy 74 DYNAMICS OF A P ARTICLE hits the weaker at a height 61 ft. above him ; then v' 2 = (S3) 2 - and if v" be the speed with which the weaker boy hits the stronger at a height 6{ ft. below him, then v" 2 = (78) 2 + 2 2 = (2 g/T) cos + c determine c by the data w = o> when = 0. Then find what the initial velocity w must be in order that the pendulum (considered as rigid) rise to the vertical and continue revolving, rather than oscillate. 19. An automobile weighing 1000 Ib. turns a corner on the arc of a circle of radius 50 ft. at a speed of 6 mi. per hr. Find the " skid "-producing force. In what ratio would this force be reduced if the speed were halved? 20. A centripetal force of 131 oz. holds a body in a circle of radius 100 ft. with a uniform velocity of one revolution per hr. Find the weight of the body. 21. Formerly railroad curves were built in the form of a circle touched by two straight lines. A later method consisted in using hyperbolic arcs. Explain how this was an improvement. 22. At what angle must the track be raised at a point where its radius of curvature is 400 ft. in order that for trains with a speed of 10 mi. per hr. gravity may exactly balance the pressure on the outer rail due to the normal acceleration? 76 DYNAMICS OF A PARTICLE 23. What proportion of its weight does a body lose at the equator because of the earth's rotation ? How many times as fast would the earth have to rotate in order that bodies should have no weight at the equator ? 24. A hammer thrower whirls a hammer, of weight 16 lb., in a circle of radius 4 ft., and the pull on his arms just before letting go is 180 lb. Find the speed as he lets go, and the length of his throw, assuming the hammer to rise initially at an angle of 45. 25. A body moves with uniform speed in the parabola y i = 4 ax. Deter- mine the normal component of the force. Show that it is greatest at the vertex. 26. Show that if the velocity vector always has the same direction, the motion will be rectilinear. 27. In the elliptic harmonic motion of the present section determine the constants of integration by the following data : when t = the body is moving directly upward from the point (a, 0) with a speed kb. Find the equation of the path. 28. On buoyant bodies the upward pressure of the air might be taken into account by assuming for the force f x =0, f 9 = mg + mk(h y), where h is a h'eight where the upward pressure practically ceases. Study the motion. Let the body rise from rest. 29. Study the motion of a point repelled from a fixed center by a force proportional to its distance from the center. Show that the path is a hyperbola. Compare with elliptic harmonic motion, and also, as in prob- lem 27 of this section, work out a special case with simple initial conditions. 30. Study the motions obtained from x a cos mt, ?/ = 1> sin nt, by giving m and n different integral values. The paths are called Lissajou's curves, and are the result of the composition of two harmonic motions with . due to gravity, and of 400 Ib. due to resistance, in all S400 Ib. This is hauled at 20 mi. per hr., that is, at 17GO ft. per min. But each foot means Moo ft. Ib. of work. Tims ve have S400 x 1700= ll.7si.ooo ft. Ib. per min., or 448 horse power, about. 2. Calculate the work done in raising a weight of 2 T. from the ground to a point a yard above the ground. Suppose inclined planes of inclinations 15, :>o'. 10. be used, what will be the work done in each case, the final height above the --round being always 1 yd.? 78 FOKCE FIELDS 79 3. A pump discharges water at the rate of 20 cu. ft. per min. from an artesian well 1000 ft. deep. What is its power? Note. 1 cu. ft. of water weighs 62 Ib. 4. A body of weight m Ib. is to be raised from the ground to a height of h ft. by use of an inclined plane making an angle i with the horizontal. If the coefficient of friction is /*, show that the work done will be Inn (1 + /x.cott) ft. Ib. 5. One end of a spring whose other end is clamped is pulled out a distance 18 in., the average force exerted by the spring being 3 poundals. What is the work done? 6. The stroke of a piston whose head has an area of 100 sq. in. is 13 in. long. The average pressure is 80 Ib. per sq. in. What is the work done? If the fly wheel makes 10 revolutions per min., what is the estimated horse power ? 7. Show that the work done in moving a body from the bottom to the top of a smooth inclined plane is the same as that done in raising the body vertically from the level of the bottom to the top. 23. Force fields ; work on curved paths. If a particle mov- ing in space is subject at each point to some definite force which is either constant like gravity or varies from point to point con- tinuously, like the force of a magnet on a piece of iron, the particle is said to move in afield of force. Work is done against the field when a body is moved against its forces. Moreover, the path may be curved and the force may vary. In order to extend our defini- tion of work given in the last section, we imagine a polygon inscribed in the curved path and the body to be moved along the sides of this polygon. We shall suppose the force to retain along each side the value it had at the beginning of that side. This will give us an approximation to reality which is closer and closer as the sides of the polygon are decreased in length and increased in number. For each side the work will be / g (s f )As f , and the defini- tion of the amount of work done in moving the body from the point Sj of the curve to a point s 2 is accordingly Ex. 1. Show that the work done in raising a particle against gravity from a height h l to a height h 2 is independent of the path along which the body is raised, and has therefore the value (Ji z h^g. 80 WORK AND ENERGY Ex. 2. Show that the work done against gravity on a solid body by any motion whatever is equal to (h 2 hj) g where h l and h z are the heights at the beginning and the end of the motion, respectively, of the center of mass of the body. Hint. Assume the whole work is the sum of the work done on the separate parts. Note that the results of these exercises are valid not only for gravity, but for any force field where the force on the unit mass is constant in magnitude and direction. XI. PROBLEMS ON WORK AND POWER (Continued) 8. Show that the work done in raising a body from the lowest to the highest point of a vertical circle is the same whether a semicircle or the diameter be used as path. Do this as an independent verification of Ex. 1, not as an application. 9. Find the horse power necessary to haul a train at a speed of 30 mi. per hr., the frictioual resistance being equivalent to the weight of 10,000 Ib. 10. Let A be the modulus of elasticity of an elastic string (i.e. the tension required to double its length, it being assumed that increase in the length is proportional to the tension) and x its length. Then the ten- sion is given by T = (x /) A//, being measured in pounds. Find the work done in stretching the string from a length x = a to a length x = b. 11. An automobile ascends a hill, whose inclination is i = arctan (1/30), a distance of 200 yd. per min. If the automobile weighs IT., find the horse power required. 24. Conservative fields ; potential energy. If work be done upon a body so as to lift it against gravity, the body, in turn, by means of pulleys, may be made to do work in lifting other bodies. If the force field is such that exactly as much work will be done by the body in returning by any path to its original position as was done against it in moving it from this to its final position, the field is called conservative. Otherwise it is called nonconservative ; an example of a nonconservative force is friction. Conservative fields have the property, which the student has proved in the case of gravity, that the work done in moving a body from a point J[ to a point 1% depends only upon the positions of these points and in no sense upon the path between them. If a body moves in a conservative field from a point P^ to a point P^ against the forces of the field, it acquires a certain ability to do POTENTIAL ENERGY 81 ivorJc. This ability to do work is called energy, and as it depends upon the position of the point it is called potential energy, that is, possible energy, ready to manifest itself as soon as the body is allowed to move. It will readily be seen that this energy is so far a relative thing. The excess of energy at 7^ over that at P is denned as equal to the amount of work that can be done by the body in passing from P 1 to P^, or It is, however, customary to speak of the energy at a point P 1} by which we mean this same difference 7F 01 , the point P Q being con- sidered the standard point of zero energy. There will, in general, be points of negative potential energy, they being of course points such that in moving the body from them to P , work must be done against the field. To change the standard point merely means to add a constant to the energy, and this in no wise inconveniences us in our use of the idea. We therefore write t/S as the expression for the potential energy.* Ex. What are the equipotential surfaces of a field of central forces? (See p. 68.) The projections on the axes of the force at any point of a conserv- ative force field are the negatives of the derivatives of the potential energy with respect to the corresponding coordinates, that is, dW , _ dW "" 5J' fy ~ ~~dj" For, taking two points on a parallel with the a?-axis, with abscissas x v and 1 + A./.' 1 , we have * If all along the path used/,, = 0, that is, if tlie path is at all points perpendicular to the direction of the force, we see that no work is done over the path. All such paths radiating from a point form, in the case of a conservative field, a surface whose nor- mals are the force vectors. Such a surface is called an equipotential surface, for the potential energy is the same at all of its points. Thus if the force of gravity be con- sidered constant, the equipotential surfaces are level planes. From this fact equipo- tential surfaces for other force fields are frequently called level surfaces. 82 WOKK AND ENEEGY XX! + Ax t -ox t nx + Ax f x dx+ I f x dx I f x dx, C/Xo l/Xj or, applying the law of the mean for integrals, /wfc+Aa* where < 6 < 1. Dividing by Aa^ and taking the limit, cW dW =f x , and similarly =/,, From this we derive an interesting corollary : A particle is in equi- librium when its potential energy is at a maximum or minimum. The converse, that if the body is in equilibrium, the potential energy is at a maximum or minimum, is not always true, though frequently so stated. For example, if W = x 2 y i ->f x = f y = at the origin, but the surface z = x 2 y 2 is saddle shaped here, and has neither a maximum nor a minimum. XII. PROBLEMS ON POTENTIAL ENERGY 1. The potential energy of a solid body or a system is a maximum or minimum when the height of the center of mass is a maximum or minimum. Two rods of length Z hinged together like an inverted V lie over a cylinder of radius a. Let 2 be the angle between them in a position of equilibrium. Show that satisfies the equation 2 a cos I sin 3 0. 2. A right cylinder with elliptic cross section is so weighted that its center of mass is in a plane halfway between its parallel bases and one quarter of the way along its maximum diameter. Find the possible posi- tions of equilibrium of the cylinder lying horizontally on a smooth plane, and show that there are four or two according or not as the eccentricity is greater than 1/V2. Hint. This requires finding the distance of the point (fl/2, 0) from the point (x, ?/) on the ellipse x = a cos 0, y = b sin where b 2 = a 2 (1 e 2 ). 25. Kinetic energy. A body may have energy because of motion. Thus if a body is thrown upward against gravity it con- tinually gams potential energy until it reaches its highest point. The energy which its motion gives it is called its kinetic energy, and is measured by the amount of work the body does against the forces of the field before coming to rest. For this we find (confining ourselves to the plane) KINETIC ENERGY 83 /0 /A where s determines the position of rest. But by IV of 5 this is / a o [fx cos (*, s ) + / cos (y, )] ds, d# V or, since cos (a?, s) = > cos (y, s) = > as efc ! V_ CV and since f x = m f, f= m f at dt J t \dt x dt r'd /I A m 2 m 2 = m I ( - v }dt = v* -\ v , J t dt\2 ) 2 2 or, since the position characterized by t is one of rest, V Q = 0, and we have k = | mv 26. Conservation of energy. Suppose a body move in a force field from a point I* to a point P z ; the gain in kinetic energy is |mv 2 2 \mvl. But this is equal to the work done by the forces of the field, or r ^ j f s ds=-W(s. 2 ) + W( Sl ), so that, equating and transposing, \ mvl + W(s z ) = 1 mr? + W( Sl ), that is, the sum of the kinetic and potential energies of a body is always the same during motion in a conservative force field. This result is known as the principle of the conservation of energy. As the object of a perpetual-motion machine is to do work (at least against the friction of its own parts) perpetually, always returning to some standard position with the same kinetic energy, we see that a perpetual-motion machine is impossible, provided the forces on which it depends are conservative. 84 WORK AND ENERGY Ex. Suppose a body moves on a curved path under the influence of gravity, like a bead upon a smooth wire. By the principle that the energy is constant, show that the speed of the body is the same at all points where it has the same heights (cf. problem 10 of 21). XHI. PROBLEMS ON ENERGY [See note under Suggestions and Answers.] 1. A hammer weighing 1 Ib. strikes a nail with a velocity 8 ft. per sec., driving it a quarter of an inch into a plank. Find the average force exerted upon the nail during its motion. Solution. The kinetic energy of the hammer is transferred to the nail, except a small amount which is dissipated in heat, which we neglect here. Thus 1 mv* = \ (1/32.2) S 2 = .99 ft. Ib. work is done against resistance by the nail. If the force is constant, we have for the work f- s =f- ^ J^, and as this is equal to the kinetic energy, we have/= 48 .99 = 48.3 Ib. 2. How much work is accumulated in a body weighing 300 Ib. and moving 64 ft. per sec.? 3. A fly wheel 12 ft. in diameter, whose rim weighs 12 T., makes 50 revolutions per min. What is its kinetic energy ? 4. An automobile running 30 mi. per hr. comes to the foot of a hill. To what height will it ascend without power, friction and other resistance being neglected ? 5. Compare the kinetic energy of a mass of 20 Ib. falling from rest at the end of the fifth second with the energy at the end of the sixth second. 6. A mason's helper throws bricks up to him through a vertical distance of 14 ft., so that when the mason catches them they have a vertical velocity of 6 ft. per sec. By what proportion would the helper reduce his work by throwing the brick so as to reach the mason with no speed remaining? 7. If a pendulum hanging at rest is given an initial velocity i - < how high will it rise ? 8. A gun carriage of 2 T. recoils horizontally with a velocity of 12 ft. per sec. Find the constant force which will take up the recoil within a distance of 2 ft. 9. The 500-lb. hammer of a pile driver falls 10 ft. on to the head of a pile, which is forced thereby an inch into the ground. Find the average force exerted by the hammer. 10. A river has a cross section of area 16,000 sq. ft. and flows with a mean speed of 4 mi. per hr. Find the horse power that would be developed if all the energy of the river could be utilized. Note. The weight of 1 cu. ft. of water is 62^ Ib. 11. A particle slides from the highest point on the outer surface of a smooth sphere of radius a. Find the point where the particle leaves the sphere. ANALYSIS 85 ANALYSIS OF CHAPTER IV 1. Definitions and units of work and power. 2. Definition of work for a curved path and varying force. 3. Definitions of conservative fields and potential energy. 4. The force components given by the negative of the derivatives of the potential energy ; equilibrium in terms of potential energy. 5. Definition of and expression for kinetic energy. 6. The principle of the conservation of energy. CHAPTER V MECHANICS OF RIGID BODIES 27. Instantaneous motion of a rigid body. When a body moves from one position to another, and we consider only the positions, and leave out of account the way in which the body moved from one to the other, we speak of the displacement of the body. The displacement of any point of the body may be represented by a vector joining its initial and final positions. "We next consider two special motions of a body, in which the intermediate positions are considered : namely, translation, in which all points move in con- gruent curves, for which we may usually take straight lines, and all lines in the body remain parallel to themselves ; and rotation, in which there is a line, or axis, in the body, all of whose points are fixed during the motion. The following theorem is impor- tant : Any displacement of a rigid body may be brought about by a translation and a rotation. To see this we remark that the motion of a body may be studied by fixing a set of coordinate axes in the body and considering their various positions relative to a set of axes fixed in space. Let us denote by O^X V O^Y^, 1 Z 1 the initial position of the axes fixed in the body, and by O z X a , 2 Y 2 , 2 Z 2 their final position. The translation desired is, then, the one which carries 1 over into 2 . Let us suppose it carries O l X l , 0-Y V O l Z l into the parallel set Z X[, 0. 2 Y[, 2 Z[. We have now to show that a rotation may be found which carries 0. 2 X[, 2 Y{, 2 Z{ into 2 X 2 , 2 F 2 , 2 Z 2 respectively. Let A'/, r/, Z[, A' 2 , Y.,, Z. 2 , denote the points where the corresponding axes pierce a unit sphere about the point 2 (see Fig. 31). Then we need merely show that we can find a diameter such that, if the sphere be rotated about it through the proper angle, X[ will pass into A" 2 and Y[ into F 2 , for then Z[ will DISPLACEMENTS 87 also become Z z because each Z-point is a quadrant's distance from the corresponding X- and F-poiuts and on the same side of the AT-plane in both cases. Now in this rotation which we seek, all points of the sphere move on parallel circles with the required diameter as axis. Hence they remain unchanged in their distance from this pole ; the pole is equidistant from X[ and X a , and is therefore in the plane which perpendicularly bisects the line join- ing them. Similarly, it lies hi the plane which perpendicularly bisects Y[ Y a . The axis is thus determined, unless the planes coin- cide. It is left to the student to consider the meaning of this case, and further to prove that an dmjle can always be found which will serve to bring loth X[ into X 2 and I'/ into I", at the same time. It is evident that this resolution of a displacement into transla- tion and rotation can be effected in many ways, for instead of 88 MECHANICS OF RIGID BODIES choosing the particular point O l as origin of our axis in the body we might have chosen any other point. It is important to notice that the translation and rotation are by no means the actual motion of the body. They simply give a way in which it may be brought from its initial to its final position. The real value of this resolu- tion consists in its application as follows : In a short interval of time, A, the body suffers a small displacement, which may be resolved as indicated. The displacement of an}' given one of its points is thus also resolved into one due to the translation and one due to the rotation. The smaller A the more accurately do these two displacements actually represent the real motion of the point. Considering the displacements as vectors, and dividing by At and taking the limit, we see that at any instant the velocity of any given point may be con- sidered as decomposed into a rdocity of translation and a cdocity of rotation; or, as we frequently express it, the in- stantaneous motion consists in an instantaneous transla- tion and an instantaneous ro- tation. The important thing, however, is that the velocity of translation and the angular velocity of the rotation are the same for all points of the rigid body. Thus instead of having an infinite number of velocities to deal with corresponding to the infinitely many points, as we should have in a nonrigid body, we have merely two. Similar results hold for the acceleration. The actual finite motion of the body is generated by a translation with a velocity that, in general, is continually changing, and by a rota- tion with varying velocity about a varying axis. The foregoing remarks make clear the importance of studying two particular cases of motion of a rigid body, namely pure translation FIG. 31 TRANSLATION 89 and pure rotation. Before doing so, however, two general remarks should be made. The first concerns the internal forces and reactions of the body. We shall make the assumption embodied in Newton's third law of motion, that the forces exerted by two particles upon each other lie in the line joining them, and are equal in magnitude and opposite in direction. This assumption will enable us to leave out of con- sideration entirely the effect of internal forces in the study of rigid bodies, as we shall see later on. The second concerns the energy of a system of a number of par- ticles. If two weights are raised against gravity, the amount of work to be obtained by letting them fall is the sum of the amounts to be obtained from each singly. Similarly, for any number of particles. We are thus led to the more general statement : The potential and kinetic energies of any system of masses are the sums of the potential and kinetic energies, respectively, of the- individual masses, although we shall not take the space to consider the com- plete establishment of tin's principle. It would be a mistake, however, to assume that if a velocity is resolved into two component velocities, the kinetic energy is the sum of the energies due to the component velocities, as is shown by the simple case in which we consider rest as the resultant of two equal but oppositely directed veloc- ities. If the speed is c. we should have for the sum of the kinetic ener- gies lmr- + i tnr-. whereas the- total energy is obviously 0. We shall see later what statement can be made in this respect concerning a rigid body ( 31). 28. Pure translation. Tlir in,, -I.- /<' mass were concentrated at the center of mass. To see this we imagine the bod}- split up, after the fashion of the Integral Calculus, into a set of masses which are approximately particles. Then all the elements of mass Aw move along congruent curves with the same acceleration, and the same is true of the center of mass. If a s is the projection of this com- mon acceleration upon the common direction of the paths, the projection of the force acting upon the whole mass M as if con- centrated at the center of mass upon the direction of its path is 90 MECHANICS OF RIGID BODIES a s M, and hence the work done on this hypothesis is 2 a.Mds. As a t is also the acceleration of the mass Am, the total force act- ing upon it is oAw, though it must not be inferred that this force comes alone from the external field, for the internal stresses of the body will contribute a part. These forces contribute no work, how- ever, because they occur always in pairs, and the distances between their points of application remain unchanged. Hence the work for the mass A?/i is ,, W. = | a \m..ds\ ) t/S! and summing and taking the limit, we have for the total work r - = I ^ s li J*l which agrees with the above result. In replacing the summations by the integrals which are their limits, we eliminate any error arising from the fact that the masses Am f are not particles, and thus the theorem is established. Ex. Prove similarly that the kinetic energy also of a rigid body whose motion is a translation is the .same as if tin; mass were concentrated at the center of mass. 29. Pure rotation. "We ask first for the work done by a rota- tion, in which we consider one force F as acting, applied at a point ( ', //, z). Let us take the 2-axis along the axis of rotation. The work done against the force is, by definition, X-'2 . Let r denote the length of the perpendicular upon the axis from (x, y, z), (r =Var J + y 2 ), and let 6 be the angle this line makes with the .Kz-plane. Then if s is measured from the point where the circular path of (./-, y, z) pierces the ?-plane, s = rO and ds = rdd ; ROTATION 91 then, as cos (x, s) = cos [0 + (7r/2)] = sin0 = y/r, cos(y, s) = cos0 = x/r, and cos (z, s)=0,f s = f x cos (x, s) + f y cos (y, s) + / 2 cos (2, s) = ( x f y yf*)/ r > and w e have f+i W=- I (a/,- %/! yf x }dd. The quantity xf y yf x is the moment of the force about the 2-axis (see 11, II, also Ex. 2). Thus the work done against a force ly a rotation is the negative of the integral of the moment of the force over the angle rotated through. If we have a number of forces applied at different points, Here again the interior forces may be left out of consideration, for their moments are equal and opposite in pairs. The kinetic energy of the mass m rotating about the z-axis is ^ mv*. But as s = rd, v = ds/dt = r(d6/dt] = rw, this expression becomes JL m /'-'or, and for a number of masses If our masses form a rigid continuous body, we find, by the process of the Integral Calculus, K = liin ^ % 8 (#., y., z.) rf A^Aj/.Az, o> 2 = T) IIJ S (* y> z ) r*dxdydz 2 in the above expressions for the kinetic energy, are called the moments of inertia of the particle, system of particles, or the continuous body respectively. It will be noticed that if we consider angular velocity in rotation as corresponding to speed in translation, they occupy exactly the same place with respect to the energy of rotation that the mass does with respect to the energy of translation ; they are the coefficients of half the angular speed ; they measure the inertia or resistance of the body 92 MECHANICS OF RIGID BODIES to turning forces. This fact makes them of great importance both in the theory of rotating bodies and also in other problems of engineering, where similar expressions find application to the theory of strength of materials. We shall therefore devote a separate paragraph to them. 30. Moments and products of inertia. Radius of gyration. In the integral / = |M & (x, y, z) r 2 dxdydz the limits are determined just as in the volume problems of the Integral Calculus, by the boundaries of the body considered. The expression holds for any axis, where r denotes the distance of the point x, y, z from that axis. In particular A = /Yjf 8 (x, y, z) (y 2 + z^ dxdydz, B = CCC& (x, y, z) (z* + x^ dxdydz, C = jfJTa (x, y, z} (x 2 + z/ 2 ) dxdydz, are the moments of inertia about the three coordinate axes. If we have any other axis a, the distance r of the point (x, y, z) from this axis is one side of a right tri- angle, of which the hypot- enuse OP is the distance Vie 8 +f+ of (x, y, z) from the origin, and one side ON is the projection * a) + y cos (y, a) + z cos (z, a) of this distance upon the axis have, therefore, FIG. 32 X COS (x, a. We * This is merely an extension to space of IV of 5. OP is a vector with projec- tions x, y, z on the axes. MOMENTS OF INERTIA 93 r*=OP 2 -ON 2 = [x 2 + y 2 + z 2 ] [x cos (x, a) + y cos (y, a) + z cos (z, a)] 2 = x 2 [1 cos 2 (x,a)] + if [1 cos 2 (y, a)] + z 2 [1 cos 2 (z, a)] 2 yz cos (y, a) cos (z, a)2,zx cos (z, a) cos (a?, a) Ixy cos (#, a) cos (y, a] ; or, remembering that cos 2 (, a) + cos 2 (y, a)+ cos 2 (2, a) = 1, r 2 = cos 2 (a, a) [y/ 2 + s 2 ] + cos 2 (y, a] [z 2 + x 2 ] + cos 2 (z, a) [x 2 + f] 2 cos (y, a) cos (2, a) 2 cos (z, a) cos (a?, a) 20; 2 cos (x, a) cos (y, a) yx. Multiplying by 8 (x, y, z) and integrating, and remembering that cos(x,a), cos(y, ), cos (2, a), being the cosines of the angles be- tween the axis a and the coordinate axes are therefore constant, we find / = I I I &(x, y, z) r^dxdydz = A cos 2 (x, a) + B cos 2 (y, a) + C cos 2 (z, a) 2 D cos (y, a) cos (z, a) 2 cos (z, a) cos (x, a) 2 F cos (JT, a) cos (y, a), (1) where D = | I I &(x, y, z)yzdxdydz, E III 8 (x, y, z) zxdxdydz, F = III S(x, y, z)xy dxdydz, are the so-called products of inertia. Thus having calculated the six coefficients of inertia, A, B, C, and D, E, F, we can find from them the moment of inertia about any axis through the origin by means of the formula (1) above. The coefficients are visually calculated for coordinate axes with origin at the center of mass. The follow- ing theorem then enables vis to find the moment of inertia about any axis, even if not through the center of mass, w ill i out further integrations, unless we need one to find the mass: The moment of inertia about any given axis is equal to the moment about a parallel 94 MECHANICS OF RIGID BODIES axis through the center of mass increased by the moment of inertia about the given axis of the body considered concentrated at its center of mass. In symbols, if /' denote the required moment of inertia and / the moment about a parallel through the center of mass, and a the distance between the parallel axes, I' = I + Ma i . (2) To prove this, we take our coordinate axes through the center of mass, so that the z-axis is parallel to the given axis, and so that the 2!-plane passes through it. The equations of the given axis will then be x = a, y = 0. If, then, r' denote the distance of the point (x, y, z) from the given axis, we have r' 2 = (a; - a) 2 + if = x 2 + f-2ax + a 2 = r 2 + a 2 - 2 ax. Hence /' = III B(x, y, z)r' 2 dxdydz = I I I $(', y, z)r*dxdydz + a 2 I I | 8 (x, y, z) dxdydz 2 a I I I 8 (x, y, z}x dxdydz. The first integral on the right is /, the second M, and the third is xM (see 13). But as the origin was taken at the center of mass, x = 0, so that this term drops out and equation (2) is thus established. PROBLEMS ON MOMENTS OF INERTIA 95 If the whole body were concentrated on an axis, the moment of inertia about that axis would vanish. We ask, How far from the axis should the mass of the body be concentrated in order that its moment of inertia with respect to that axis be the same as when the body has its git-en form ? The distance sought is called the radius of gyration and is denoted by R. From its definition MR 2 = I. Hence R*=-Jji. XTV. PROBLEMS ON MOMENTS AND PRODUCTS OF INERTIA AND RADII OF GYRATION Unless otherwise specified, take the density (r, ?/, z) equal to 1. Choose the coordinate axes as symmetrically as possible. Calculate A, B, C, D, E, F and the radii of gyration about the coordinate axes. Straight wires should be laid along the a;-axis, and plates in the z^-plane. The student should observe that sometimes we can tell in advance of a definite integral that it vanishes, namely, in cases where the field of inte- gration is symmetric with respect to a point, line, or plane, and the inte- grand has values which are equal in absolute value but opposite in sign at pairs of symmetric points. Thus /+! />+a nit I xdx = 0, I x 5 dx = 0, I cos xdx = 0, J-\ J-a /0 6 / /+a /> a v pa. ~b ~c / x y ~\ xydydx = 0, I sin irl - + - + -} dzdydx = 0. / b . J-aJ-bJ-c \Cl b C/ a a Va The truth of this statement is obvious when we recall the fact that the integral is the limit of a sum. Two bodies with the same coefficients of inertia are sometimes "called ili/niniticitlli/ cffiiirfi/'-iif. 1. Calculate the coefficients of inertia of an equilateral triangular plate of side a, and of unit surface density ; also the moment of inertia about a side, about an axi.s perpendicular to the plate and through one corner, and finally about an axis through one corner and making an angle of 45 with the plate and lying in the plane that perpendicularly bisects the opposite edge of the plate. Give also the radii of gyration. ution. The center of mass is at the intersection of the medians, which we take for origin of coordinates, taking one median along the y-axis. \Ve find the equations of the sides to be 3 x + V$ y a = 0, 3 z VJ3 y + a = 0, and 2 V% y + a = 0. Then 96 MECHANICS OF RIGID BODIES and since we have a plate, 2 = 0, and with unit density dm = dxdy, so that we have _?_ n-Vsj /** n ^ t> y s\ s\ I Vs f* 3 / V3 /~ A = ( I trdxdy I y^dxdy = I w 2 f(a Vi JJ ' J g_ / g-Vsy J_^_ 2 Vi 3 2 Vi Similarly, x 3 2 Vi 2 l/a-V5yV VI V3 4\ 3 C = rf(a; 2 + y*) 2 = 1. Calculate its moments of inertia about axes through its focus parallel with its minor axis and perpendicular to its plane. 98 MECHANICS OF EIGID BODIES 13.* Calculate the coefficients of inertia of the homogeneous ellipsoid x 2 /a 2 + y*/b 2 + z*/c 2 = 1. 14. Show that if a homogeneous body is symmetric with respect to the a:?/-plane, D = 0. What further similar statements can you make regarding bodies with one or more planes of symmetry ? 15.* Find the moment of inertia of the cardioid plate bounded by p = a (1 cos 0) about an axis through the origin and perpendicular to its plane ; also about the axis of symmetry in the plane. 1(5. Determine the coefficients of inertia of a cuboid of dimensions a, b, and c ; also moments of inertia about edges and about a diagonal. 17. Find the moments of inertia of the above cuboid about the diagonals of the rectangles forming its faces. 18.* Find the moment of inertia of the elliptic plate l/r 1 ccos$ about a perpendicular through the focus (which is the origin as the equa- tion is written). Compare the result with problem 12, noting that e 2 = (a 2 -i 2 )/a 2 and l = b' 2 /a. 19. Find the moment of inertia about its axis of a right circular cone of altitude a and radius of base r. 20. Find the coefficients of inertia of a right circular cylinder of height a and radius of base r. 21. Calculate the coefficients of inertia of a flat angle iron as indicated in Fig. 17, p. 39, the origin being the center of mass. 22. Do the same for the plates of Figs. 14 and 15, taking simple numer- ical values for the constants. 23. What is the effect on the moments of inertia and the radius of gyration of a homogeneous body if the density is multiplied by a constant k, that is, increased in the ratio 1 : k 'i 24. In problem 5 calculate the moment of inertia of the wire about an axis through its mid-point and making an angle with it. How does the moment of inertia change as 6 varies from to ?r/2 ? Can you explain this from mechanical considerations? 25. Given a circular plate of radius r from which have been removed four circular pieces of radius r' with centers at the mid-points of four equally spaced radii of the plate. What is the value of r' if the radius of gv ration of the )>lat<> about an axis perpendicular to its plane through its center is ^ the radius of the plate ? 26. Suppose a body expand in such a way that it always remains similar to itself in form and that its mass is constant in amount. How are its coefficients of inertia and its radii of gyration affected ? 31. Work and energy in the case of a rigid body. Let us con- sider first a system of particles m f with coordinates (x { , y { , z.) with * The reckonings in the starred problems are rather protracted. Integration tables should be used, e.g. Peirce (formulas 308 and 300 for problem 18). 99 respect to a fixed set of axes. As we are about to study how the work and energy of the body depends upon the work and energy due to translation and rotation separately, we shall introduce also a system of axes moving parallel with the old, with origin always at the center of mass x, y, z. If (x!, y[, z/) are the coordinates of a point with respect to these axes, we have, by Analytic Geometry, (i) The work done against the force field in moving m t will be ; P/.A /% S*8f = - / '[/ cos (a;., s.) + /. cos (y, , s { ) + /. cos (z i} s t )]d Si , s 'i dx,. diL dz,- or, as cos (x., s,) = - > cos (y {) SA = -^ > cos (,, s,) = ! > _ dx fr v 'vi> v 'zi> are the velocities relative to the center of mass. The considerations apply in particular to rigid bodies. In either case the theorem is true : The work done in moving a rigid body is equal to the work done in moci/tg the center of mass computed as if all the forces were applied there, plus the whole work done on the body computed as if its center of mass were at rest. In a conservative force field, inasmuch as the work done is for each particle the same, no matter what path is taken, so for a rigid body the work depends only upon the initial and final positions. As the change may be brought about by a translation and a rota- tion about the center of mass, we may say that in moving a rigid body against a conservative force field, the work done is the sum of the work done in translating its center of mass, the forces being regarded as concentrated there, phis the work done in rotating the body about its center of mass. Similar results hold for the kinetic energy. We have for the particle m f , - ^ , + i vf + m. [v-v' xi + v-v'^ + v-v(:\. When we add for all particles, the terms in the bracket lead to three sums like ^-\ / x~^ I ^ 'i X~^ ' 2, " l Mrf* = r- 2 ,<- = r- x 2 m f ~ = *' 5 jj m i i i i But 2?tt v /:. is nothing else than M times the x of the center of PROBLEMS ON EIGID BODIES 101 mass referred to the same axes as x\, y\, z(, that is, to axes through the center of mass. It is therefore 0, and we have IT/ 2 ' l i v i 5 and extending to rigid bodies, we may say that the kinetic energy of a rigid body is the kinetic energy computed as if the body were con- centrated at its center of mass, plus the kinetic energy of the motion relative to the center of mass computed as if the point were fixed. We have seen (p. 88) that the instantaneous motion of a body may be resolved into the instantaneous translation of one of its points, for which we here take the center of mass, and a corre- sponding instantaneous rotation. If &> is the instantaneous angular velocity and R the radius of gyration about the instantaneous axis, we have for the kinetic energy, k = i my 2 + | mR 2 a) 2 , where, in general, v, J?, to, and the axis of rotation are continually varying. XV. PROBLEMS ON WORK AND ENERGY IN THE CASE OF RIGID BODIES The following problems are intended mainly as applications of the principle of the conservation of energy. 1. Study the motion of a homogeneous sphere of mass M and radius a rolling down an inclined plane of height h and inclination i. Show that at all points the speed of its center is less than if it slid down, in the ratio of v5:\/7. (Note that for a sphere of radius r rolling so that its center describes a straight line, ~v = rw.) 2. Show that the time of rolling down the incline is Vl-1 h/5 g cosec i. 3. Show that for a homogeneous sphere the times of rolling down the chords of a vertical circle joining the highest point of the circle to other points of the circumference are all the same. 4. A man has a hollow iron ball and a solid aluminum ball of the same radius and weight. They are painted so that they appear the same. Explain how he can distinguish them by means of an inclined plane, telling w r hich is which. 5. A pendulum is formed by a homogeneous ball rolling on a circular track, the plane of the circle being vertical. Show that the motion is that of the ordinary simple pendulum whose length is 7/5 times the radius of the path of the center of the ball. 102 MECHANICS OF KIGID BODIES 6. Show that if a homogeneous right circular cylinder roll down any incline, the ratio of the kinetic energy of rotation to the kinetic energy of translation is constant. What is the constant ratio? Show that a similar statement holds for a homogeneous sphere rolling on any track. 7. In problem 11 of 26 a particle sliding down a sphere was considered. Change this by asking at what point a rolling sphere of radius r leaves the surface of the given sphere of radius a. 8. The moon rotates about the earth, keeping the same side always toward the earth. Suppose that it always faced the same direction in space, would it have more or less kinetic energy? 32. Compound pendulum ; experimental determination of mo- ments of inertia. In the simple pendulum we have, theoretically, a particle moving in a circular path. We now consider any heavy body free to turn under the influence of gravity about a horizon- tal axis. Such a body is called a compound pendulum. If 6 is the angle between the vertical plane and the plane containing the cen- ter of gravity, both planes passing through the axes, we have for the kinetic energy, I /(W\ 2 1 . 1 - L - j while the potential energy, or work done, is Mgh, li being the height through which the center of mass is raised. Let d be the distance of the center of mass from the axis. Then h = d d cos 6, and as the sum of the energies is constant, 1 MRW + Mcjd(l - cos 0) = c. Let us suppose the pendulum falls from an angle # . At this point &) = 0, so we have Hence, eliminating c, J,- .l/A'-or = Mfjd (cos 6 cos ), 2 /^ or o>": Comparing this with the formula for the simple pendulum, (c) of 21, we see that the motions are the same, provided 1gd/R 2 = 2 g/l, or provided I = R 2 /d ; that is, the compound pendulum COMPOUND PENDULUM 103 behaves just like a simple pendulum of length R*/d. For small oscillations the period is T 9 J. = 2 .Point of suspension Center of mass As the position of the center of mass of a body can usually be simply determined, particularly if the body have symmetry, and hence d may be found, and as T can be readily observed, the above formula determines the radius of gyration R, and hence also the moment of inertia about the axis s. of suspension. The point in a line perpendicu- lar to the axis of suspension and through the center of mass, and distant I from the axis, is called the center of oscillation, and is the point at which the weight of the equivalent simple pendulum would be concentrated. Let its distance from the center of mass loed'. Let us take an axis through it parallel with the axis of sus- pension, and call the radius of gyration about this axis R'. If R c be the radius of gyration of the F IG . 34 body about a parallel axis through the center of mass, we have seen ( 30, (2), remembering that I=MR 2 ) R 2 = R*+d* and Jt'*= li*+ d'*, so that, subtracting, R*jl ia = d*d'*. Suppose now^the body be suspended by the parallel axis through the center of oscillation, and let I' be the length of the equivalent simple pendulum. As I = R*/d, R 2 = Id, and, similarly, R n = I'd', the above equation gives But d + d' = 1. Hence Id - I'd' =l(d- d'), or I'd' = Id', and since d'^Q, 1 = I', that is, the body suspended from either axis is 104 MECHANICS OF RIGID BODIES equivalent to the same simple pendulum. In other words, point of suspension and center of oscillation may be interchanged, it being understood that the axes used are parallel. 33. General equations of motion of a rigid body. Any set of magnitudes which determine the position of a body are called the coordinates of the body. The motion of the body is determined when these coordinates are determined as functions of the time. If the motion is given through the forces that act, we shall have differ- ential equations from which to determine these functions, in general one differential equation for each coordinate. Let us ask how many coordinates (and hence how many differential equations) are neces- sary to fix the position (or the motion) of a rigid body. To fix one point demands three coordinates. The body is then free to turn about the point. To fix the direction of a line through the point demands two more magnitudes, for instance, two of the three direc- tion cosines (the third may be found from the fact that the sum of their squares is unity). The body is then free to turn about an axis, and if the angle through which it may be supposed to have turned is fixed, the body is fixed also. In all, then, there are six coordi- nates, which, however, might have been selected in a multitude of other ways. We must therefore establish six differential equations, and we shall briefly indicate how this may be done. If we have a set of particles m { , the coordinates of each satisfy the equations ax rili lie (i) where the quantities /.., f yi , f zi embody all forces, not only external ones but the reactions of the particles on each other. The reactions we endeavor to eliminate. Our first step is to write down the first equation (1) for each body and add them all; similarly, for the other two equations. In a rigid body, thought of as a set of EQUATIONS OF MOTION 105 particles, the reactions between two particles are equal and oppo- site in sign, and hence drop out in the sum. So if X, Y, Z, represent the projections on the axes of the resultant of all the external forces, we have or, as 2mfXf=Mx t where M is the total mass, and ~x the x of the center of mass, (2) that is, the center of mass moves as if the whole mass ivere con- centrated there and all the external forces acted there. This gives us three of our differential equations of motion. Another way to eliminate the effects of internal forces is suggested by the definition of the moment of a force about an axis as the prod- uct of its magnitude by the perpendicular upon its line from the axis (see 9). As the internal forces pair off into equal and oppo- site forces acting along the same lines, their total moments about any axis must vanish. Hence we form the moments, first of the forces acting upon m i and about the z-axis. Calling it N t , we have / . \ ,, i ~dt~ yi ~ai = V "" " '^ ''"' / = ''' "' C S V * C S where s { denotes the direction perpendicular to the radius r t . This is called the moment of the velocity about the z-axis, and 106 MECHANICS OF RIGID BODIES the mass times the moment of the velocity is called the moment of momentum. We may therefore interpret the equation d as follows : the time rate of change of the moment of momentum of a particle about an axis is equal to the moment about the axis of the forces acting upon it. Compare this equation and its interpretation with the equation m^d-Xf/dt 2 ) =fxi, "which may be written d(tUfV xt ')/dt=f x i. The quantity ?n,-r,- is called the momentum of the particle m ( ; the equation states that the time rate of change of the momentum in any direction is equal to the total force acting in that direction. If we add the equations like the above for all masses, the inter- nal forces disappear as above explained, and writing the analogous equations for the other axes, we have d { dz, fw v; . f* ~ z ; -'] = L d_ dt l \ l dt ' dt d I dy i dx dt \ ' dt ' dt (3) where L, M, and N are the sums of the moments about the axes of all the external forces acting upon the body. These, with equa- tions (2), form the required six differential equations. The last three (3) have the disadvantage, however, that they do not contain a limited number of coordinates which simply fix the position of the body in case it is rigid. This cannot be satisfactorily done without introducing moving coordinate systems, which would lead us too far at present, although it would furnish a basis for con- sidering most interesting motions like those of rotating planets, or of gyroscopes and "tops," the latter being the name applied in mathematical physics to any rigid body with one point fixed. It is the problem which naturally follows the compound pendulum, which has two points, or an axis, fixed. ANALYSIS 107 We close by pointing out the results which obtain where no external forces are present. Equations (2) and (3) admit of an integration at once, the first set giving the result valid for a rigid body or for a system of particles : the center of mass moves in a straight line with constant velocity; and the second, the sum of the moments of momentum about any fixed axis is constant. These results find interesting application in our solar system. The student who is interested in a further study of the motion of rigid bodies is referred to more extended works on dynamics.* ANALYSIS OF CHAPTER V 1. Definitions of displacement, translation, and rotation. 2. The instantaneous motion of a rigid body consists in an instantane- ous translation and an instantaneous rotation. 3. Assumptions concerning internal forces. 4. The work and kinetic energy for a motion of translation. 5. The work and kinetic energy for a motion of rotation. 6. Definition of moments and products of inertia and the radius of gyration. 7. Two theorems enabling us to get the moment of inertia about any axis from the coefficients of inertia. 8. The work and kinetic energy for the general motion of a rigid body may be considered the sum of the work and kinetic energy, respectively, due to a translation plus the work and kinetic energy due to a rotation. 9. Compound pendulum. Experimental determination of moments of inertia. Intel-changeability of point of suspension and center of oscillation. 10. General equations of motion of a rigid body. The student should gain an idea of the nature of the problem as one of reducing the number of variables from an unlimited number to six. * For instance, Jeans, Theoretical Mechanics; Webster, Dynamics. SUGGESTIONS AXD ANSWERS NOTE. The numerical results following are, for the most part, computed with a 30-cm. slide rule. The student should therefore not expect agreement beyond i%. Page 4, Ex. 1. Use Fig. 1. Page 5, Ex. 3. Draw a diagram, indicating the sum vector. Next show that the order of any consecutive two of the given vectors may be changed without affecting the result. The theorem will then follow if it can be shown that any order may be made any other order by such interchanges of consecutive vectors. Pages 9-11. II. Problems on Vectors. 3. \/2 V2/= 1.85/. 4. v = V(7345 - 3408 V2) = 50.3 ft. per sec.; 47.5 E. of X. 8 (a). (16, 6, 6); (d) (2, 9.83, - .869). 9 (a). 2.38, (.728, - .485, -.485); (b) 0, (cosines indeterminate). 10. 22.9 ft. per sec.; 11.3 with track. 11. Against current at angle arccos(a/&) with bank. If a = b, im- possible. Shortest time, 90 with bank ; path then makes an angle arctan(/a) with bank. 12. 53 from the vertical. 13. 16.7. Page 13, Ex. jNIay be reduced in part to Ex. 3, p. 5. Pages 13-16. III. Problems on Equilibrium of Concurrent Forces. 2. Consider force polygon. 4. Use conditions (1) and Ex. 1, p. 4. 37.7, (- .647, - .727, - .262). 7. Let x, y, z be the coordinates of 0, and #!, ?/!, !, a- 2 , y 2 , z 2) , x n , y n , z n the coordinates of the other points. Express in terms of these the projections of the vectors OA, OB, , ON. 8. The components directly up the plane of the forces must be in equilibrium. 9. As a check note that the sum of the angles should be 360. 10. 52 Ib: 11. Lame's theorem may be used (problem 6) ; 84 and 112 Ib. 12. The chains are to be considered as of equal length. 70.7 Ib. 13. Solve a trigonometric equation. Get all roots. 15. .5 Ib. 17. The graph is a broken straight line. 18. The graph is partly curved and partly straight. 19. p. .3. The reac- tion of the plane is equal and opposite to the other forces acting upon the mass. 10.4 Ib. 16.7 with vertical. 21. i = arctau 3/t. 22. Height above bottom of bowl : /(! cos e) = r(l I/ Vl + /x' 2 ). 109 110 SUGGESTIONS AND ANSWERS Page 18, Ex. 4. Interpret the vanishing of the product f n -p = 0, supposing f= 0. Page 19, Ex. 1. This may be done by finding the moment with respect to any point (x, ?/) and showing that x and y do not enter the result. Take a simple position of the axes with respect to the forces. Ex. 2. Apply the preceding exercise and Ex. 2, p. 18. Page 22, Ex. 3. First show that if a set of forces is in equilib- rium, so also are their projections upon any plane ; do this by con- sidering the resultant and also the moment about any perpendicular to the plane. Then show that three forces in a plane which are in equilibrium must be concurrent or parallel ; do this by considering moments about the intersection of any two of the forces if any two intersect. Considering then the projections of the three given forces upon the three coordinate planes, show that the forces must be con- current or parallel, thus proving (b). To prove (a) apply the condi- tion for equilibrium of concurrent forces (p. 13), or, if they are parallel, consider moments about a line meeting and perpendicular to two of them. Considering then the resultant of parallel forces through the origin, (c) may be proved. Pages 23-26. IV. Problems on Moments and the Equilibrium of Forces. 2. The resultant of the parallel forces is 15 Ib. Consider- ing moments about the lighter end of the bar, the resultant is found to be applied at a distance 3| ft. from this end. 3. 2\ ^ ft. from the 5-lb. weight. 4. Call the lengths of the arms d and d' and the weight of the body w. Writing the two equations expressing equality of moments and solving for w, one finds 8.48 oz. 5 (a). G T. ; (b) 5:-J and GJ T. 6. T = 3 Ib., R = 2i Ib., i = 0. 7. 45. 8. The forces acting upon the gate are its weight and the reactions of the hinges, which may have various directions. 9. /= w(l rtan /)cot i/2 I ; the center is directly above the point of contact with the floor. 10. If the triangle of the rod and two strings be thought of as rigid, there will be seen to be two forces acting upon it, the upward pull of the hook and the downward pull of gravity upon the rod considered as acting at the mid-point. Hence the projections of a and b on a hor- izontal line may be shown to be equal. Then project upon a hori- zontal line the tensions of the strings and the weight of the bar. 11. T = 2cw/l; P = iv Vl - 2 c/l + (2 c/l}\ SUGGESTIONS AND ANSWEKS 111 Page 28. V. Problems on the Center of Mass. 1. 10 in. from the heavy end. Page 29, Ex. Consider whether the formulas of p. 27 hold for oblique axes ; then consider the relation of the product A#A?/A to the volume of the parallelepiped with sides parallel with the axes and of lengths A#, Ay, and A. Pages 30, 36-40. VI. Problems on Centers of Mass of Continuous Bodies. 2. On the line joining a vertex to the medial point of the opposite face triangle and three fourths the distance from the vertex. 3. 3 r/S from center. 4. Check by noting that when a = b = c = r the ellipsoid becomes the sphere of the preceding problem. 5. 2 h/3. 7 (a). 21/3] (e) check by considerations of symmetry; (f).636. 8. x = .425 a, y = .425 b. 10 (a). One third the way along the diagonal (the plates should, of course, be thought of as concentrated at their centers of mass) ; (b) one third the length of the side from the middle plate. 11.. Consider either as a conical surf ace or as a solid cone. The result may be obtained from the answer to problem 7 (c). 13. At cen- ter of axis. 14. See problem 7 (d). 16. y = .393. 19. r/2 from center. 20.2(a 2 + ab + b*)/3(a + b). 22.x = -4a/5. 23. z = (a 2 /mA)sin(A/ am), y = (a 2 /raA) [1 cos (A/am)], z = h/2. 24. x = C /M, y = S/M, where C= 2(V"" r -l)/(l + 4a 2 ), S = -(e 4a " -!)/(! + 4a 2 ), J|f=(e aair -l)/a. 25. 37i/8 from one end. 26. 3 (b + b 2 a + ba 2 + a s ) /8(b 2 + ba + a 2 ). 28. x = 2 r sin s a/3(a sin a cos a). Check by noting that as a = 0, x = r. 30. x = (3 a 2 + 16 ab + 3 b 2 )/5 (a + 6). 31. .052 a. 32. x =- ar 2 /2(ab - r 2 ). 34. y = 4 a/3. 36. x = 5a/6. 37. x = .555a. 38. x = .3SSa. 40. x = y = 2a/5. 41. T/TT from axis, A/4 from base. 42. Fig. 14. Distance from left-hand edge = (2 a 2 + be - 2 c 2 ) /2 (2 a + ft - 2 c) ; Fig. 15. Distance from bottom = (ac 2 + 2 6c + & 2 eZ) /2 (ac + id) ; Fig. 16. y = 5.41. 43. x = .538 a, y = 1.199 a. 44.5 = 8a/15, y = 152 a/525, i = 11 a/120. 45. x = 4m)(e 2 '" a 1). 46.418a. 47. Distance from edge = 3 Trr sin / 16 a. Check by putting a = 7r/2, and also a = 7r; cf. problem 14. 50. A = 4 7T 2 a, V = 2 TrWb. 51. x = 3 > (1 + cos a) /8. 52. In this problem the mass in question is understood to be outside the con- ical surface as opposed to problem 51. x = 3 r (1 cos a) / 8. 112 SUGGESTIONS AND ANSWERS 53. x = 3rsin 4 a/4(2 2 cos a sin 2 a cos a). 54. x = 3 Vr 2 -a 2 /8. s*2a 55. Use Ml) = I ya-ffa;. Apply to half the plate to the left of x = ira, Jo which half will have the same y as the whole plate, x = TTCI, 7/ = 5 a/6. 57. Use the integration tables, x = a [6 V2 - log e (3 + 2 V2)] / 8 [ V2 -f log e (l + V2)] = .305 a. 58. By " cone " is meant conical surface, and all matter is removed from the sphere which lies within the indefinitely prolonged conical surface, x = r sin 2 a. Pages 46-48. VII. Problems on Rectilinear Motion. 2 (a), t 1, ,s = G ; (b) points to left of s = 6 ; (c) forward to s = G, then back ; (d) part of line to left of s = 6, twice ; (e) to the left ; (f) for large negative t, the point is far to the left with large positive but dimin- ishing velocity ; for large positive t, the point is far to the left with large negative and numerically increasing velocity. 20, 21. The equations give more than one value of s for a given value of t. What would this mean ? 22. s = (kt? + 2)/2. 24. a = 2(e*' + k - 2)/A. Pages 52-54. VIII. Problems on Bodies moving in Straight Lines under the Action of Given Forces. 2. 403.6 ft. 5. Determine the height h of the body at its highest point and thus answer the problem. 6. 53 ft. 7. The weight of only one of the bodies is effective in pro- ducing motion, while the inertia of both bodies resists the motion. They are, of course, supposed to start from rest, (c) v = 242 cm. per sec. 8. Write an equation for each body in which its acceleration is equated to its mass times the resultant of the forces acting upon it, namely gravity and the tension of the cord. From these equations determine the unknown acceleration. 10. About 3 ft. 13. The path of the particle must bisect the angle between the vertical and the perpendicular to the line on to which it falls. 14. About 150 Ib. 15. Six times as high. 5.18 sec. 16. About 10.3. 17. 6012 ft. 18. 410 ft. 19. 5.7 T., or about 57 Ib. per T. 20. About 1268 sec., or 21.1 min. 25,950 ft. per sec. 22. 1 ft. 23. About 7.8 kgm. 24. 25,950 ft. per sec. 28.1 min. 26. k = 17.6 ; 10.15 ft. per sec. 27. Velocity of bullet is about 3.4 times that of raindrop. 30. Veloc- ity upon leaving table is V5/y log,, (5 -f- A/24) = .904 ft. per sec. Pages 73-76. X. Problems on Bodies moving in a Plane under the Action of Forces. 2. 106 ft. per sec. 4. 6.2. 6. 2.63 ft. 10.3 from vertical. 7. Solve the problem from the standpoint that the stream of water is sent from a fixed point, and consider the area above the SUGGESTIONS AND ANSWERS 113 level of that point only. 1100 sq. m. 9. Use equation of envelope. 16. By 2^ of its length. 19. The centrifugal force is mv 2 /p pounded*, if m is measured in pounds. 1.34 Ib. 20. 433 T. 22. 14.9. 23. About 1/291 and 17. 24. The data lead to a result 45 ft. As this is a very short throw, how should the data be changed ? 25. 4 a 2 mv 2 / (if + 4 a' 2 )! poundals. 28. y = ?/ + (kh g) /k (1 cos Vto). Pages 78, 79. XI. Problems on Work and Power. 3. 37.9 h. p. 5. .14ft. Ib. 6. 5i h. p. 9. 800 h. p. 10. (b + a - 2 l)(l> - a) A/2 I ft. Ib. 11. 1.21 h. p. Page 84. XIII. Problems on Energy. (NOTE. In these problems the student must be careful about units. In the derivation of the formula for k = \mv 2 , p. 83, the equations f x = m(dv x /dt), f y = m(dv y /dt~) hold if m is measured in pounds and/ x and/^ in poundals ; and also if m is measured in engineering units (weight -f-