title: Outlines of physics; a concise discussion of the laws and relationships of physics and their mathematical expression.
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"o o' 1 • . ^ *- v' i' ' \ 0O X. - ^ V. 1 V y V c» s sX'“ S // ' \^ <X * *%^' **'' J * A .- ^ £\V//C ^ A X ^ -|%c ^ "Vf« "bo^ *' ^ "A ' ' ^ y * -f. \ ■ » I 0*0 ■ Ujf 01 i*- t OUTLINES OF PHYSICS A CONCISE DISCUSSION OF THE LAWS AND RELATIONSHIPS OF PHYSICS AND THEIR MATHEMATICAL EXPRESSION. BY CHARLES C. BIDWELL, Ph D. Vi Assistant Professor of Physics Cornell University Ithaca, N. Y. THE COMSTOCK PUBLISHING CO. 1922 QC 2.1 ,35 COPYRIGHT 1922 BY THE COMSTOCK PUBLISHING COMPANY ITHACA, N. Y. Press of W. F. Humphrey Geneva, N. Y. NOV 27 ’22 © Cl A 6 92068 PREFACE This text is a development from notes used for several years in mimeograph form by students of engineering and chemistry at Cornell University as a basis of a second course in physics. It is designed for those students who h^ve completed an intro¬ ductory course in college physics, and have had a first course in differential and integral calculus. A great deal of the descriptive material properly found in introductory texts is omitted. It is the purpose of the text to study the fundamental laws of physics, and to show how these laws are expressed in mathematical form. It is hoped thereby to bridge the gap between the introductory descriptive texts, and the more advanced mathematical treatises on special topics. It is regretted that through lack of time certain material on sound, which is in course of preparation, has had to be deferred for a future edition. October, 1921. C. C. B. CONTENTS MECHANICS PAGE Fundamental Concepts of Position and Motion. Displacement and Vector Quantities. Vector Operations. Velocity and Acceleration. Uniformly Accelerated Motion. Thi Motion of a Projectile. Uniform Circular Motion. Problems. 1 Newton’s Laws of Motion. The First Law. The Second Law. Momentum. Impulse of a Force. The Third Law. 6 Systems of Units. The c. g. s. System. The English Gravitational System. Dimensions. Problems. 8 Simple Harmonic Motion. Definition. Solution of the Equation of S. H. M. Superposition of Simple Harmonic Motions. Expressions for Acceleration, Period, Force. Problems. 12 Work and Energy. Mathematical Expression for Work. Units. Power. Graphi¬ cal Representation of Work. The Stretched Spring. Energy. Problems. 17 Angular Motion. General Equations. Vector Representation of Angular Quanta ties. Moment of Force. Problems. 23 Center of Mass. Definition. Velocity and Acceleration of Center of Mass. Action of a Couple. Motion of a Rigid Body under Action of a Single Force not through the center of Mass. Center of Gravity. Problems. 25 Moment of Inertia. Definition. Angular Momentum. Impulse of a Torque. Work and Energy in the Case of Angular Motion. Moments of Inertia about Parallel Axes. Motion of a Rolling Body. Equation of Motion in the Case of Rigid Bodies. Problems... 29 Gyroscopic Motion. 35 Elasticity. Definitions. Types of Strain. Limitations of Hooke’s Law. The Twisted Shaft or Wire. Problems. 37 HEAT Thermometry. Temperature Scales. Expansion. Problems. 41 The Fundamental Gas Laws and the Equation of State. Boyle's Law. Charles’ Law. The Equation of State. The Absolute Zero. Work on the (p. v) diagram. Problems. The Kinetic Theory of Gases. Assumptions of the Kinetic Theory. Pressure and Temperature in terms of the Kinetic Theory. Avagadro’s Law. Brownian Movements. Deviations from Boyle’s Law. Van der Waal’s Equation. The Porous Plug Experiment. Mechanical Refrig¬ eration. Heat as Energy. Quantity of Heat. The First Law of Thermodynamics. Problems......*.. t. Change of State. Vaporization. Fusion. Problems. Properties of Gases. Application of the First Law to Gases. The Specific Heats of a Gas. Isothermal Changes. Adiabatic Changes. Problems. Thermodynamics. Cyclic Processes; Reversible and Irreversible Processes. The Carnot Cycle. Efficiency of a Carnot Engine. The Second Law of Thermodynamics. Carnot’s Theorem. The Thermodynamic Temperature Scale. Entropy. The Entropy- Temperature Diagram. Comparison of the Perfect Gas Scale with the Thermodynamic Scale. Deviations of Gas Thermom¬ eters from the Thermodynamic Scale. The use of Thermo¬ dynamic Equations. Maximum Efficiency of a Heat Engine. Problems. The Transfer of Heat The Radiation Laws and the Measurement of High Tem¬ peratures. Newton’s Law of Cooling. Kirchhoff’s Law. Uniform Temperature. Enclosures. Stefan’s Law. Wien’s Displacement Law. Wien’s Energy Distribution Law. Opti¬ cal Pyrometry. LIGHT The Nature of Light. Light as Wave Motion. Huyghen's Principle; Rectilinear Propagation. Color. Reflection. Reflection at a Plane Sin-face. Image in Plane Mirror. Re¬ flection at Spherical Surfaces. The Sagitta Theorem. Con¬ jugate Focal Relations for Concave Spherical Mirror. Convex Mirror. The Parabolic Mirror. Spherical Aberration. Refraction. Snell’s Law. Critical Angle. Refraction through a Plate with plane parallel Faces. Refraction through a Prism. The Spectrometer... PAGE Dispersion. Dispersion and Dispersive Power. Dispersion without Devia¬ tion; Direct Vision Spectroscope. Deviation without Disper¬ sion. Problems. 90 Spectra. Cont inuous Spectra. Wave-Length Units; The Spectral Range. Bright Line Spectra. Absorption Spectra. The Color of Bodies. Luminescence. 91 Lenses. Refraction through Lenses; Types of Lenses. Path of Ray through a Lens; Construction of Image. Conjugate Focal Relations; Huyghen’s Construction for Converging Lens. Rule of Signs. The Diverging Lens. Spherical and Chromatic Aberrations. Combinations of Lenses. Achromatic Com¬ bination . 94 Optical Instruments. Lens as Magnifier. Simple Compound Microscope. Astron¬ omical Telescope. Galileo’s Telescope. 99 Interference; Diffraction. The Zone Plate. Diffraction passed a Straight Edge. The Diffraction Grating. 102 Polarization. Double Refraction and Polarization. Polarization by Re¬ flection. The Nicol Prism; the Polariscope. Rotation ofthe Plane of Polarization. 107 ELECTRICITY Electrostatics. Conlomb’s Law. Electric Field and Electric Induction. Gausses Theorem. Potential. Capacity. Condensers. Dielectric Constant. Units. Capacity of a Sphere. Capaci¬ ty of a Plate Condenser. Capacity of a Spherical Condenser. Capacity of a Cylindrical Condenser. Capacity of Condensers in Parallel and in Series. Energy of a Charged Condenser. 112 Magnetism. Conlomb's Law. Magnetic Field and Magnetic Induction. Gausses Theorem. Induction and Field in a Transverse Gap in an Iron Ring. Induction and Field in a Transverse Gap in a Straight Iron Bar. Case of Bunched Windings on an Iron Ring. Case of Field due to Current in a Wire. Perme¬ ability. Torque experienced by a magnet. Magnetic Poten¬ tial. Problems. 120 The Electric Current and its Effects. Production and Detection of Electric Current. Side Push on a Wire carrying Current. Magnetic Field due to Electric PAGE Current. Magnetic Field at the Center of a Circular Loop. Units. The Field on the Axis of a Circular Coil. The Field near a Long Straight Wire. Field at Center of a Solenoid. Work to Carry Unit Pole around a Wire Carrying Current. Quantity of Electricity. 127 Measurement of Current. The Tangent Galvanometer. Torque on a Coil Suspended in a Magnetic Field. The D’Arsonval Galvanometer. 132 Ohms's Law. Joules Law; Resistance. Ohm's Law for a Portion of a Circuit. Ohm's Law for a Complete Circuit. General Expression for Ohm's Law. Kirchhoff’s Laws. Application of Ohm's Law to a Simple Circuit. 134 Measurement of Resistance and E. M. F. The Wheatstone Bridge. The Potentiometer. Voltmeters and Ammeters. Resistances in Parallel. Effect of Tem¬ perature on Resistance. Power and Energy in the Electric Circuit. 138 Induced Electromotive Forces. Lenz’s Law. Alternating Current Dynamo. Direct Current Dynamo. The Direct Cm rent Motor. Efficiency of a Motor. The Induction Motor. 140 Self and Mutual Induction. Self Induction. Mutual Induction. Ohm’s Law for a Circuit Containing Inductance. 147 The Ballistic Galvanometer and Measurement of Capacity. The Ballistic Galvanometer. Logarithmic Decrement. 149 Magnetic Induction in Iron. Effect of Poles. Hysteresis. Experimental Methods of Obtaining the Hysteresis Loop. Work to Magnetize a Sample of Iron. Relation between Flux of Induction and Ampere turns. The Magnetic Circuit. Method of Solving Problems on the Magnetic Circuit. 153 Electrolysis. Conduction of Electricity. Faradays Laws of Electrolysis. Energy Equation of an Electrolytic Cell. The Voltaic Cell. Theory of the Chemical Action of a Simple Voltaic Cell. Energy equation of a Voltaic Cell. Source of Electrical Energy in a Daniell Bell. Secondary Cells. 160 Thermo-Electricity. 165 Alternating Currents. The Fundamental Equations of the Alternating Current Circuit. Circuit with Inductance and Resistance—The Equation for Current. Vector Representation of Voltages in an A. C. Circuit. Circuit with Inductance, Resistance and PAGE Capacity. Vector Diagram for Circuit with Capacity. Average current and E. M. F. Virtual Current and E. M. F. Power in an A. C. Circuit; Power Factor. The Wattmeter. Application of Ohm’s Law to A. C. Circuits. Addition of Electromotive Forces out of Phase. Addition of Currents. Impedances in Series. Impedances in Parallel. Inductance and Capacity in Series with Resistance negligible. Oscillating Circuits. Coupled Circuits; the Transformer. 166 Electromagnetic Radiation. Electric Displacement. Displacement Current. Electric Oscillations and Radiation. The Detection of Electromag¬ netic waves. The Three-Electrode Vacuum Tube. The Vacuum Tube as a Detector. The Vacuum Tube as an Amplifier. The Vacuum Tube as a Generator. Heterodyne Reception by Vacuum Tubes. Radio-Telephony; Voice Modulation. 182 MECHANICS FUNDAMENTAL CONCEPTS OF POSITION AND MOTION 1. Displacement and Vector Quantities. The position of a particle with reference to some arbitrary origin is usually desig¬ nated either by means of rectangular coordinates x, y and z, or by means of vectors. The line drawn from the origin to the particle is called the position vector , s. When the position of a particle is changed, the displacement is given by specifying the changes Ax, Ay, Az or by As, the change in the position vector. All physical quantities may be classified as either scalar quantities or vectors , the former having magnitude only, the latter magnitude and direction. A vector quantity may be represented by a line whose length and direction indicate re¬ spectively the magnitude and direction of the quantity. Thus in addition to the position vector, we have the displacement vector which is the line drawn from the point representing the initial position of a particle to the point representing the final position; we have velocity vectors, lines specifying direction and magnitude of velocities; acceleration vectors, force vectors and various others. 2. Vector Operations. If a particle is subjected to two or more displacements the resultant displacement may be determined graphically by laying off the corresponding vectors end to end and drawing the line from the initial to the final position as the resultant displacement. Thus the sum of displacements oa and ab (Fig. 1) is the displace¬ ment ob. The resultant displacement may also be computed by trigonometry since the vectors form the two sides of a triangle with lengths and directions known. The above method of addition applies in general to any like vectors. If several like vectors when placed end to end form a closed polygon the resultant vector, that is, the sum of all the vectors, is zero. (Fig. 2). This is a test which may be applied in the case of balanced forces; the sum of all the forces vectorially added must give a closed figure. 2 MECHANICS A single vector may be resolved into any number of component vectors the only condition being that the sum of the components vectorially added equal the original vector. The vector ob (Fig. 1) may be resolved into vectors oa and ab and is equivalent to these two. We are often concerned in physics with the multiplication and division of vectors. The product or quotient of parallel vectors is a scalar. Thus the product of force and distance traveled in the direction of the force is the scalar quantity work. Force divided by area gives the scalar quantity, pressure. The product or quotient of two mutually perpendicular vectors is a third vector at right angles to the two. Thus the product of two sides of a rectangle is the vector quantity area, the direction of an area being given by the normal to it. The product of force and lever arm is moment of force or torque, indicated by a vector along the axis of rotation. An angular displacement measured in radians may be considered as a linear distance along an arc divided by a radius, thus giving radians and indicated by a vector along the axis of rotation. 3. Velocity and Acceleration. If, s, represents the displace¬ ment of a particle from some origin (in such a case the displace¬ ment and position vectors are the same), then ds/dt, the rate of change of position with time at some particular instant is called the velocity, v, at that instant. Magnitude, direction and sense must be expressed in specifying a velocity. (By sense is meant the sign, away from the origin usually being positive). If, v, represents the velocity of a particle then dv /dt, the rate of change of velocity with time at a particular instant, is called the acceleration, a, at that instant. Magnitude, direction and sense must here also be specified. In mathematical language we may express these definitions as v = ds /dt 1 a = dv/dt = d 2 s/dt 2 2 Equation 1 may be written ds = vdt f rom which by integration « e FUNDAMENTAL CONCEPT 3 s = f vdt 3 Likewise for 2 v = f adt 4 Equations 1 «rrd 4 are general equations fitting any type of motion. 4. Uniformly Accelerated Motion. Motion in which the acceleration is constant is said to be uniformly accelerated motion. In this case the integration of equation 4 gives v = at + c where c is the constant of integration. The physical meaning of c is found by introducing the boundary conditions. Thus it is evident that c = v 0 , i. e., the velocity when t = 0 or in other words, the initial velocity. Therefore we have v = at + v 0 If this value of v be substituted in equation 3 we have s = a f tdt + v 0 f dt 6 Upon integration this gives s = i at + v 0 t + c' 7 c' is the constant of integration. Interpreting as before, we find c' = s 0 . Therefore * 1 2 s = - at + v 0 t + s 0 QUESTIONS 1. Following the method employed in the case of uniformly accelerated motion discuss the case where a = 0, (uniform motion) finding v and s. 2. Draw curves representing (a) velocity and (b) distance covered from starting point as a function of time, (body initially at rest) for (a) acceleration = 0 (b) acceleration = const. ( = a) (c) acceleration = const, x time (= ct) (d) acceleration = — A sin pt (where p and A are constants). (Suggestion: In each case find the expression first for velocity and then for displacement by integrating equations 4 and 3 4 MECHANICvS respectively. Assume convenient numerical values for the constants and plot curves with time as abscissa. In each of the above cases as in uniformly accelerated motion, the process of integration is merely that of anti-differentiation, that is, the integral of a given quantity is the function which when differ¬ entiated gives the quantity. There may be several specific functions which will satisfy this condition. It is necessary to find the most general function. For instance, the integral of equation 4 may be v = at, but this is the special case of starting from rest. The addition of the constant of integration makes the solution general. In substituting values in the equation for question (d) above, namely a = — A sin pt, and in the corresponding expressions for v and s if values of pt are found as multiples of t /6 the sin and cosine computations will be much simplified. Thus let pt = 0. 7r/6, 27r/6, 3x/6, 47r/6, 57 t/ 6 etc. carrying this at least up to 24 tt/6). 5. The Motion of a Projectile. Suppose a body is projected into the air with an initial velocity, v 0 , at an angle, 0, with the horizontal. Required to find expression for (a), the range , (the horizontal distance tra¬ veled when the body reaches the ground; (b), the time of flight; (c), the maximum height and (d) ,the equation of the path. It is convenient to resolve the motion into its horizontal and vertical components. The horizontal component is subjected to no acceleration (ig- noringair friction), the vertical component has an acceleration, —g, since upward is taken as positive. We may then find hori¬ zontal velocities, v x , and displacements ,s x , by applying the expressions for uniform motion; and vertical velocities, v y , and displacements, s y , by applying the expressions for uniformly accelerated motion. Thus, taking the origin at the starting point, we have for the horizontal component v x = v 0 cos 0 8 $ > v 0 cos 0 * t 9 For the vertical component v y = (-g)t + v 0 sin 0 10 s y = -*(-g)t 2 + v 0 sin 0 • t. 11 FUNDAMENTAL CONCEPTS 5 The range, that is, the horizontal displacement when the pro¬ jectile has again reached the ground, may be found by substi¬ tution s y = 0 in equation 11, [why?], solving for t which gives the time of flight and substituting this value of t in equa¬ tion 9. At the instant of maximum height the vertical velocity is zero. The time for the body to attain its maximum height may therefore be found by substituting v y = 0 in equation 10. This value substituted in equation 11 gives the value of maximum height. The equation of the path may be obtained by eliminating t between equations 9 and 11. The equation thus obtained may be shown to be a parabola. The student should obtain for himself the above expressions. PROBLEM In the case of a projectile show, (a), that the time of ascent and descent are the same; (b), that the initial and final vertical velocities are the same. 6. Uniform Circular Motion. This is the motion of a particle traveling in a circle with constant speed. Suppose the particle, traveling around the circle with center at o, Fig. 4, is at a given instant, in position (1) and after the lapse of a small time, At, in position (2) , having traveled along the arc a distance, As. Let Vi and V 2 represent the two velocities equal in magnitude but differing in direction. The velocity vectors may be placed in juxta-position as shown, the vector Av being the change in velocity in the time At. Since the points (1) and (2) are very close together, the arc As may be considered as sensibly straight. The triangles r, r, As and Vi, v 2 , Av are similar (homologous sides being perpendicular). Hence Av/v = As/r 12 Dividing each” side by At we have A v _ As 13 v A t r A t Fig. 4 A v _ dv At" dt a and =4 t = v, therefore At at In the limit 6 MECHANICS v 2 14 r The vector diagrams indicate that the direction of the accelera¬ tion is perpendicular to that of the velocity at any instant, the direction of the velocity being along the tangent, and that of the acceleration being toward the center. PROBLEMS 1. A train gains a speed of 40 miles per hour in 6 minutes. Find the average acceleration in miles per hour per second, and in miles per second, per second. 2. A column with circular base is under construction. The height is increasing at the rate of 8 feet per day. How fast is the volume increasing? 3. A weight is dropped from an aeroplane traveling northward at a speed of 40 miles per hour. The aeroplane is at an altitude of 1,000 feet. How long before the weight reaches the ground and what velocity does it acquire? What is the equation of the path and what sort of a curve does this represent? (Ignore air 4. A gun which produces a projectile velocity of 200 ft. per second is mounted aboard a car with its barrel at right angles to the di¬ rection of motion of the car, as shown in figure. The car is traveling 50 ft. per second. The sight would have to be arranged at an angle 6 to the gun barrel as shown. Find the value of 6 so that the ball may hit any object which, at the instant of firing is in the line of sights. (From Nichols and Frank- / lin, Elements of Physics) 1 5. A stone weighing 200 Fi &- 5 . grams is whirled vertically in a sling, the cord of which is 100 cm. long. What is the minimum linear velocity that will prevent the stone from falling out? NEWTON’S LAWS OF MOTION 7. The First Law. If a body is at rest it will continue at rest or if in motion it will continue in motion in the same straight line resistance m this problem.) /it dr rO / / c - ^ NEWTON’S LAWS OF MOTION 7 unless acted upon by some agency or influence external to the body which is said to cause the change. This agency, which must be acting when any change in the motion of a body occurs, is called a force. This law is based upon experimental evidence. It gives a qualitative definition of the term force, as the cause of any change in the motion of a body. •8. The Second Law. Experiment shows that for a given force the acceleration produced depends upon the mass upon which the force acts—the greater the mass the smaller the ac¬ celeration and vice versa. For example, consider a locomotive capable of exerting a definite constant pull.—With twice as long a train (twice the mass) it will take twice as long to attain a given speed (neglecting friction considerations). The acceleration thus varies inversely with the mass. For a given train two engines will develop the given speed in half the time—or the acceleration is twice as great—thus acceleration is proportional to force. For a given force experimental evidence thus shows that in¬ creasing the mass acted upon results in decreasing the acceler¬ ation in direct proportion and vice versa. In other words the product of mass times acceleration is constant and is a measure of the force acting. Hence the defining equa¬ tion F = k ma 15 (k) is a proportionality constant. Its value depends upon the units. This law describes the behavior of a body when acted upon by an unbalanced force. It may be thus stated—“when acted upon by a force a body gains velocity in the direction of the force at a rate which is proportional to the force and inversely proportional to the mass.” Equation 15 is the fundamental equation of mechanics. 9. Momentum. The product, mv, is called momentum. The rate of change of momentum of a particle, m, is a measure of the force acting upon it. That is, F = k ^(mv) 15 a This is simply another expression of equation 15. 10. Impulse of a Force. The product of force times time is called the impulse of a force. In general the impulse of a force is J\Fdt. For a constant force impulse is Ft. The impulse of a force is proportional to the momentum acquired under the action of the force. This follows directly from equation 15. Multiply- 8 MECHANICS ing both sides by t, assuming the body originally at rest, Ft = kmat = kmv 16 Or, if the force is a variable, multiplying equation 15 by dt and integrating, we have f Fdt = km f adt by definition adt = dv, hence Impulse = / Fdt = km /'dv = kmv 17 If the body has an initial velocity v 0 the integration is between the values v and v 0 or Impulse = f Fdt = km f dv = k (mv-mvo) 18 Or the impulse is proportional to the gain in momentum. 11. The Third Law. A change of motion, or the effect of a force, can only be obtained through the mutual interaction of one body upon another. That is, the action of a force requires two bodies and the effect of a force is the same upon each but oppositely directed. In other words to every action on one body there is an equal and opposite reaction on some other body* PROBLEM Show from these laws that when two bodies collide the total momentum is unchanged. SYSTEMS OF UNITS 12. The C. G. S. System. In this system three fundamental units are selected, namely, the centimeter, the gram and the second. We may have as many units as we have kinds of quanti¬ ties but these are called “ derived units ” since they may all be expressed in terms of the three fundamental units of length, mass and time. The gram mass is 1 /1000 of the mass of a certain block of platinum kept at the International Bureau of Weights and Measures at Sevres, near Paris, and known as the Kilogram prototype. The centimeter is 1 /100 of the distance between two marks on a platinum bar preserved also at Sevres. The second is 1 /86400 of the mean solar day. In this system the unit of force, called the dyne, is defined as that force which will accelerate a gram mass at the rate of one *Care should be taken to distinguish between (a) action and reaction and (b) cases of balanced forces. Thus, a book lies at rest on the table. The earth pulls on the book giving it its weight, the book also pulls with an equal force on the earth. This is action and reaction. Note that the action is on one body, the reaction on the other. The weight of the book is balanced by the upthrust of the table. Here are two balanced forces. They act on the same body. Each of these balanced forces has its action and reaction. vSYSTEMS OF UNITS 9 cm. per second, every second. We see that this is a derived unit since it is expressed in terms of the fundamental units of length and time. With this choice of units k in equation 15 becomes unity, and usually is not expressed. A system based upon the funda¬ mental units of length, mass and time in which an absolute force unit rather than a gravitational force unit is chosen is called an absolute system. Force is sometimes expressed in gravitational units, namely, in terms of the weight of the gram mass. We speak of a force of so many grams meaning so many times the weight of the gram mass. Whenever such a unit is used it should be noted that k in equation 15 is not unity but has the value, 1 /g. This follows from the fact that the weight of a gram mass is the pull of gravity upon it and that gravity accelerates a gram mass at the rate of g cm. per sec., per sec. Therefore, allowing the weight of one gram, as unit force, to act upon one gram mass we find ex¬ perimentally that a in equation 15 is g cm. per sec., per sec. and that therefore k = 1 /g. Such a system of units is called a gravi¬ tational system. Formulas in physics are usually derived on the basis of the absolute c.g.s. system in which the fundamental equation is F = ma. This fact should be remembered and substitutions in such formulas made accordingly, that is, forces should be con¬ verted to dynes. Throughout the work on mechanics in this text the k will not be dropped. Its presence will serve two purposes (1) to show the fundamental dependence of these formu¬ las on Newton’s Second Law and (2) to enable one to use any system of units he pleases, the value of k being selected to correspond. 13. The English Gravitational System (Engineering System). In this system the weight at London of a certain mass of platinum (the pound mass) kept at the Exchequer in London is taken as the unit of force. The foot is the unit of length in physical measurements although legally the yard is the unit and is de» fined as the distance between two lines on a bronze bar kept at the Exchequer in London. In engineering computations the pound mass is usually not used as the unit, the term w/g being arbitrarily substituted for mass. If w/g is taken to mean mass we have unit mass when w/g = 1. This says unit mass weighs (g) pounds. In this system the fundamental equation is written This is simply a statement of the proportionality between forces and accelerations. 10 MECHANICS In this system the mass of a man weighing 160 pounds is Jf, 1 since mass = w/g (assuming g = 32). This unit of mass is sometimes called the “slug.” The engineering system is not an absolute one since mass does not enter as a fundamental unit. It should be noted that in the absolute c.g.s. system it is also true that w/g = m, (w) being of course expressed in absolute units (dynes). In fact, in any system in which k = 1, w/g = m. (An English absolute system, which has long been urged but not generally adopted in practice as yet, selects the three funda¬ mental units of length, mass and time, as in the c.g.s. absolute system, namely, the foot, the pound mass, and the second to¬ gether with an absolute unit of force, the latter being that force which will accelerate a pound mass at the rate of one foot /sec. /sec. This unit of force is called the poundal. k in this system is unity as it would be in any absolute system). 14. Dimensions. In any absolute system all derived units may be expressed in terms of the three fundamental units of length, mass and time. The dimensions of a unit are the powers of the fundamental units to which it is proportional. For in¬ stance a velocity unit must express a length divided by time or its dimensions are (+1) in length, and (—1) in time. Acceleration is expressed in centimeters per sec. per sec. and its dimensions are (+1) i n length, (—2) in time. The dimensional formula for velocity is [L] [T] -1 , for acceleration [L] [T] ~ 2 . From equation 15 we see that the dimensional formula for force is [M] [L] [T] ~ 2 , k being merely a number and having no dimensions. Dimensions are convenient in specifying a derived unit which has no name, thus both velocity and acceleration are expressed in terms of their dimensions, vel. as cms./sec., acceleration as cms. /sec 2 etc. Dimensions afford a valuable method of checking the accuracy of an equation. Both sides, to stand for the same thing, must be of the same dimensions. Thus suppose we have a case of potential energy converted to kinetic and we equate potential energy lost to kinetic energy acquired, obtaining such an equation as mgh=^ mv Expressed dimensionally this is [M] [Lf[T] " 2 = [M] [L] 2 [T] - 2. The equation is therefore dimensionally correct. Every correct equation must satisfy such a relation. We sometimes are enabled to determine physical relations by means of this property of dimensional formulae. For example suppose we know from experiment that the velocity of sound in an elastic medium depends only upon two factors, the density (5) and elasticity of the medium. Density is mass per unit volume; the modulus of SYSTEMS OF UNITS 11 elasticity (. M) concerned is the stretch modulus, which is force per unit area divided by elongation per unit length. The equation is Velocity = k 5 X M y where k is a proportionality constant and x and y are unknown powers of the density and elastic modulus respectively. Dimensionally this is [L] [T] -i ,= [ [M] [L] ‘ * * 3 4 5 6 ] ' ' [ [M] [L]-> [T] A 7 From observation we see that the only values of x and y which will satisfy this equation are x = —y = +-. Hence Z — the preceding equation becomes v = k\/ M/8 PROBLEMS 1. A mass is clamped near the outer end of one of the spokes ot a wagon wheel. Describe the change in direction and magnitude of the force acting on the mass during one complete revolution (a) when the wagon is speeding up (assume uniform acceleration) (b) when it is going with uniform velocity, (c) when slowing down. Assume no friction 2. M = 10 kg., m = 100 g. Find acceleration of the “system”, (a) when 0 = 90°, (b) when 0 = 120°. Three seconds after starting the string is cut just above m. How far will M go in the next 5 seconds ? 3. A mass of 2 kgm is acted upon by a force of 5,000 dynes for 3 minutes. No other force is acting. How far will the body go in that time, and what velocity will it acquire ? 4. An elevator reaches full speed of 3 meters per sec. 2 sec. after starting. With what force in kg. wt. will a man “weighing” 75 kg. push down on the floor, (a) when the elevator is starting upward, (b) when it is starting downward ? 5. A cord is hung over a pulley. At one end is a mass of 10 kg. At the other end a mass of 11 kg. Find the acceleration of each mass and the tension in the cord. Find the tension in the cord in problem 2. 6. Two buckets, each weighing 500 lbs. are suspended from the ends of a rope passing over a windlass. A gallon (10 lbs.) of EJ Fig. 12 MECHANICS water is poured into one of the buckets. Find how far it will descend in 10 seconds, neglecting friction. 7. a) A train has a mass of 350,000 kg. Starting from rest it reaches a speed of 2 yi kilometers per hour in 5 minutes. If the acceleration is uniform, what is the average pull of the locomotive in kg. wt.? (Assume that friction is constant and equals 4 kg. force per 1,000 kg. mass). b) The above train, moving at a speed of 50 km. per hr. is brought to a standstill in 16 seconds. Find the average pull of the locomotive in kg. wt. 8. Solve problem 7 for a train moving up a ^ percent grade. 9. A train starts from rest on a level line and moves through 1,200 feet in the first minute. It then begins to ascend a uni¬ form incline, up which it is found to run with uniform velocity. Find the inclination of this portion of the line on the supposition that the engine exerts a constant pull. 10. A locomotive is traveling on a curve. If the radius of the curve is r and the velocity (v) the mass (m), find the angle (6) of inclination such that no outward force is exerted against the rails by the flanges of the wheels. 11. An 80 ton locomotive goes round a railway curve of which the radius is 500 ft. at a velocity of 50 ft. per sec. With what force in pounds weight do the flanges of the wheels of the loco¬ motive push against the outer rail, when the outer rail is not elevated? What is the proper elevation to give the outer rail, the width of track being 4 ft. 8 ^ inches. 12. Find the velocity at which the belt must travel to exert no pressure on the pulleys. Fig. 8 SIMPLE HARMONIC MOTION 15. Definition. Simple Harmonic Motion is motion in which the acceleration is proportional to the displacement and in the opposite direction. Forces brought into play by elastic dis¬ tortions produce motion of this type, the return force and there¬ fore the acceleration being proportional to the distortion and SIMPLE HARMONIC MOTION 13 in the opposite direction. Examples of such motions are the vibrations of stretched springs, twisted wires, etc. The defining equation of S. H. M. for the case of a particle (linear motion) is, a = —ks or d 2 s/dt 2 = —ks 20 Equation 20 is a differential equation involving two variables s and t. s can be expressed in terms of t and constants or s is said to be a function of t. The solution of the equation in this case is the particular function which, when differentiated twice, gives the original equation. 16. Solution of the Equation of S. H. M. Eq. 20 may be written as d 2 s + ks = 0 or a = —ks As general definitions we have ds 1 dv v ^and a From these we may write vdv = ads or for this case t- vdv = —ads k On integration this becomes 1 v 2 s 2 k2 = ~~2~ + C c is the constant of integration. We may write this v 2 = —ks 2 + kc'. c' must be the value of s 2 when v = 0. Let us call c' = s. o Then V = + -y/ k sj S 2 — S 2 0 ds If we substitute this in the equation v = we may write ds + V S 2 -! \/k dt The integration of the two equations here expressed gives _i g _i s _ sin -= x/T t + e and cos ~ = Vk t -f e' o o 14 MECHANICS or s = s 0 sin ( VIF t + e) 21 a and s = s 0 cos ( y/lTt + e') 21 b 17. Physical Meaning of these Solutions and Interpretation of the Constants. The solutions state that we can express the displacement as a sine or a cosine function of an angle which varies directly with the time. We may gain some light on this by considering the case of a particle on the rim of a wheel # rotating with constant angular velocity. Suppose co represents the angle swept over per second by a spoke (radius s 0 in Fig. 9). The angular dis¬ placement .(angle swept over) in time't, is cot. If a vertical line is dropped from the particle to the diameter, the displacement of the lower end of this line from the center of the circle, that is, the pro¬ jection of the motion of the particle on the diame¬ ter, is given by s = So sin (cot + e) or s = So cos (cot + e'). These equations are identi¬ cal in form with equations 21a and 21b. The pro¬ jection of a uniform circu¬ lar motion is therefore simple harmonic. For a simple harmonic motion we may therefore con¬ struct a “circle of reference” indicating the path of a particle for which the S. H. M. may be considered the projection. We use the sine function if we measure the angle from the mid¬ position where s = 0, the cosine function if we choose to measure the angle from the end position where s = s 0 . e or e' repre¬ sents the angular displacement in the circle of reference at the time t = 0. The angle (cot + e) or (cot -f- e') is called the “phase angle” and gives the phase of the motion at time, t. e or e' is the phase angle when t = 0, i. e., when we start to count time. When cot attains the value 2tt , the motion has passed through all its phases and is about to start a new cycle. The time re¬ quired for a complete cycle is called the period , T, of the motion. Thus when t = 2tt, t = T and co = 27r/T. The maximum dis¬ placement (called the amplitude of the motion) is obviously So. Equations 21 a and 21 b may thus be interpreted in terms SIMPLE HARMONIC MOTION 15 of a “circle of reference.” Vk may be written as 2tt T , for, at the time t = 0, s = Sp sin e, and at such time that Vk t = 2tt we have again s = So sin e, since sin (27r+e) = sine. That is, after such a time the displacement has again its initial value and the motion starts to repeat itself. This time is the period of the motion, T. Therefore \/k = We may now write the so¬ lution as s = s 0 sm ^ ( 2ivt 2irt \ T +V or s = s 0 cos /2irt f + O) 22 a 22 b In the latter case the reference position from which the phase angle is measured is 7r/2 radians behind the reference position for the first case. It should be understood that the circle of reference is merely an artifice of assistance in interpreting the equations. The phase angle has no physical significance as an angle in the circle of reference , for the circle of reference is purely imaginary. The equations simply state that the displacement varies with the time in the same way that a sine or cosine function of an angle changes when the angle varies directly with the time. The projection of a uniform circular motion on a diameter gives us a useful illustration of exactly this motion and enables us to visual¬ ize the phase angle. In strict terms the phase angle is merely the angle whose sine (or cosine) is s/s Q , i. e., the ratio of the dis¬ placement of any instant to the maximum displacement. 