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Fourth Edition. 3s. 6c?. SOLUTIONS OF THE EXAMPLES IN THE PLANE TRIGONOMETRY. Crown Svo. 10s. 6^^. THE ELEMENTS OF HYDROSTATICS, being a Companion Volume to the Elements of Statics and Dynamics. Ex. Fcp. Svo. Fourth Edition. 4s. Qd. SOLUTIONS OF THE EXAMPLES IN THE ELE- MENTS OF HYDROSTATICS. Ex. Fcp. Svo. 5s. CAMBRIDGE UNIVERSITY PRESS Eontion: FETTEK LANE, E.G. C. F. CLAY, Managek ©Hinburfllj: lOc, PRINCES STREET AN ELEMENTARY TREATISE ON THE DYNAMICS OF A PARTICLE AND OF RIGID BODIES CAMBRIDGE UNIVERSITY PRESS Eoncon: FETTER LANE, E.G. 0. F. CLAY, Manager Cfifntuu!?)-. 100, PRINCES STREET Berlin: A. ASHER AND CO. Ecip.ns: F. A. BROCKHAUS i^eir gorti: G. P. PUTNAM'S SONS ISombag antJ Calcutta: MACJIILLAN AND CO., Ltd. All rights reserved v7 AN ELEMENTARY TREATISE ON THE DYNAMICS OF A PARTICLE AND OF RIGID BODIES . BY ^. L. LONEY, M.A. Professor of Mathematics at the Royal Holloway College (University of London), sometime Fellow of Sidney Sussex College, Cambridge Cambridge : at the University Press 1913 O } Published, November. 1909. Second Edition, 1913. PREFACE TN the following work I have tried to write an elementary -*- class-book on those parts of Dynamics of a Particle and Rigid D3'naraics which are usually read by Students attending a course of lectures in Applied Mathematics for a Science or Engineering Degi-ee, and by Junior Students for Mathematical Honours. Within the limits with which it professes to deal, I hope it will be found to be fairly complete. I assume that the Student has previously read some such course as is included in my Elementary Dynamics. I also assume that he possesses a fair working knowledge of Differential and Integral Calculus ; the Differential Equations, with which he will meet, are solved in the Text, and in an Appendix he will find a summary of the methods of solution of such equations. In Rigid Dynamics I have chiefly confined myself to two- dimensional motion, and I have omitted all reference to moving- axes. I have included in the book a large number of Examples, mostly collected from University and College Examination Papers; I have verified every question, and hope that there will not be found a large number of serious errors. For any corrections, or suggestions for improvement, I shall be grateful. S. L. LONEY. eoyal holloway college, Englefield Green, Surrey. October 23, 1909. CONTENTS DYNAMICS OF A PAETICLE CHAPTER PAGE I. Fundamental Definitions and Principles .... 1 II. Motion in a Straight Line 10 Simple Harmooic Motion 13 Motion under the Earth's Attraction 21 III. Uniplanar motion where the accelerations parallel to fixed axes are given , 33 Composition of Simple Harmonic Motions .... 36 IV. Uniplanar motion referred to Polar Coordinates . . 45 Revolving Axes ........ 48 Central Forces 54 Apses and Apsidal distances 59 Stability of Orbits 69 V. Uniplanar motion when the acceleration is towards a fixed centre and varies as the inverse square of the distance 75 Kepler's Laws ......... 79 Time of describing any arc of the path .... 84 Planetary Motion 89 Disturbed Orbits ........ 92 VI. Tangential and Normal Accelerations 97 Constrained Motion 100 Conservation of Energy 101 The Simple Pendulum 105 Motion on a rough curve Ill VII. Motion in a Resisting Medium 119 Motion when the Mass varies 130 VIII. Oscillatory Motion 137 Oscillations in a Resisting Medium 143 Oscillations when the Forces are Periodic . . . . 147 Motion of a pendulum in a Resisting Medium . . . 151 IX. Motion in Three Dimensions. Accelerations in terms of Polar Coordinates . 155 Accelerations in terms of Cartesian Coordinates . . . 164 X. The Hodograph 170 Motion on a Revolving Cm-ve 173 Impulsive tensions of Chains ISO viii Contents DYNAMICS OF A KIGID BODY CHAPTER PAGE XI. Moments and Products of Inertia 185 The Momental Ellipsoid 193 Equi-momental Systems 196 Principal Axes 199 XII. D'Alembert's Principle 204 The general equations of Motion ..... 205 Independence of the Motions of Translation and Rotation 208 Impulsive Forces . . . . . . . . . 211 XIII. Motion about a Fixed Axis 213 The Compound Pendulum . 219 Centre of Percussion 230 XIV. Motion in Two Dimensions. Finite Forces . , . 238 Kinetic Energy in Two Dimensions 240 Moment of Momentum in Two Dimensions . . . 242 Varying Mass . . 261 XV. Motion in Two Dimensions. Impulsive Forces . . 269 Impact of a Rotating Sphere on the ground . . . 274 XVI. Instantaneous Centre 282 Composition of Angular Velocities 289 Finite Rotations 295 Moment of Momentum and Kinetic Energy in Three Dimensions 298 General equations of motion of a body in Three Dimensions 301 Motion of a Billiard Ball 302 XVII. Conservation of Linear and Angular Momentum . . 306 Conservation of Energy 313 XVIII. Lagrange's Equations in Generalised Coordinates . . 330 Principal or Normal Coordinates 338 Lagrange's Equations for Blows 339 XIX. Small Oscillations 346 Initial Motions 351 Tendency to break 356 XX. Motion of a Top 360 Appendix on Ditierential Equations 369 CHAPTER I FUNDAMENTAL DEFINITIONS AND PRINCIPLES 1. The velocity of a point is the rate of its displacement, so that, if P be its position at time t and Q that at time t + Lt, PQ the limiting value of the quantity -ry , as At is made very small, is its velocity. Since a displacement has both magnitude and direction, the velocity possesses both also; the latter can therefore be represented in magnitude and direction by a straight line, and is hence called a vector quantity. 2. A point may have two velocities in different directions at the same instant; they may be compounded into one velocity by the following theorem known as the Parallelogram of Velocities ; If a moving point possess simultaneously velocities which are represented in magnitude and direction hy the tiuo sides of a parallelogram dratvn from a point, they are equivalent to a velocity which is represented in magnitude and direction hy the diagonal of the parallelogram passing through the point. Thus two component velocities AB, AC are equivalent to the resultant velocity AD, where AD is the diagonal of the parallelogram of which AB, AC are adjacent sides. If BAChesi right angle and BAD = 6, then AB = ADcosd, AC= AD sin 6, and a velocity v along AD is equivalent to the two component velocities v cos 6 along AB and v sin 6 along AG. Triangle of Velocities. If a point possess two velocities completely represented {i.e. represented in magnitude, direction and sense) by two straight lines AB and BC, their resultant is 2 Dynamics of a Particle completely represented by AC. For completing the parallelo- gram ABGD, the velocities AB, BG are equivalent to AB, AD whose resultant is A G. Parallelepiped of Velocities. If a point possess three velocities completely represented by three straight lines OA, OB, OG their resultant is, by successive applications of the parallelogram of velocities, completely represented by OD, the diagonal of the parallelopiped of which OA, OB, OG are con- terminous edges. Similarly OA, OB and OG are the component velocities of OD. If OA, OB, and OG are mutually at right angles and u, v, w are the velocities of the moving point along these directions, the resultant velocity is s/ ic^ + v^ + w'^ along a line whose direction cosines are proportional to w, v, w and are thus equal to u V , w — , and =r. V M"-' + v^ + W wu^ + v^ + w- V 1*2 + v^ + w- Similarly, if OD be a straight line whose direction cosines referred to three mutually perpendicular lines OA, OB, OG are I, m, n, then a velocity V along OD is equivalent to component velocities lV,mV,nV along OA, OB, and OG respectively. 3. Change of Velocity. Acceleration. If at any instant the velocity of a moving point be represented by OA, and at any subsequent instant by OB, and if the parallelogram OABG be completed whose diagonal is OB, then OG ov AB represents the velocity which must be compounded with OA to give OB, i.e. it is the change in the velocity of the moving point. Acceleration is the rate of change of velocity, i.e. if OA, OB represent the velocities at times t and t + M, then the AB limiting value of -r— {i.e. the limiting value of the ratio of the change in the velocity to the change in the time), as At becomes indefinitely small, is the acceleration of the moving point. As in the case of velocities, a moving point may possess simul- taneously accelerations in different directions, and they may be compounded into one by a theorem known as the Parallclogx'am of Accelerations similar to the Parallelogram of Velocities. As also in the case of velocities an acceleration may be resolved into two component accelerations. Fundamental Definitions and Principles 3 The results of Art. 2 are also true for accelerations as well as velocities. 4. Relative Velocity. When the distance between two moving points is altering, either in direction or in magnitude or in both, each point is said to have a velocity relative to the other. Suppose the velocities of two moving points A and B to be represented by the two lines AP and Bq (which are not necessarily in the same plane), so that in the unit of time the positions of the points would change from A and B to P and Q. Draw BR equal and parallel to A P. The velocity BQ is, by the Triangle of Velocities, equivalent to the velocities BR, RQ, i.e. the velocity of B is equivalent to the velocity of A together with a velocity RQ. The velocity of B relative to A is thus represented by RQ. Now the velocity RQ is equivalent to velocities RB and BQ (by the Triangle of Velocities), i.e. to velocities completely represented by BQ and PA. Hence the velocity of B relative to A is obtained by com- pounding the absolute velocity of B ivith a velocity equal and opposite to that of A. Conversely, since the velocity BQ is equivalent to the velocities BR and RQ, i.e. to the velocity of A together with the velocity of B relative to A, therefore the velocity of any point B is obtained by compounding together its velocity relative to any other point A and the velocity of A. The same results are true for accelerations, since they also are vector quantities and therefore follow the parallelogram law. 5. Angular velocity of a point whose motion is in one plane. If a point P be in motion in a plane, and if be a fixed point and Ox a fixed line in the plane, the rate of increase of 1—2 4 Dynamics of a Particle the angle xOP per unit of time is called the angular velocity of P about 0. Hence, if at time t the angle xOP be 6, the angular velocity about C/ IS -TT . at If Q be the position of the point P at time t+ t^t, where A* is small, and V the velocity of the point at time t, then PQ If then ZP0Q = A^, and OP = r, OQ = r + Ar, r (r + Ar) sin A^ = 2 APOQ = PQ . F, where OF is the perpendicular on PQ. Hence, dividing by Af, and proceeding to the limit when M is very small, we have where p is the perpendicular from upon the tangent at P to the path of the moving point. Hence, if to be the angular velocity, we have ?•-&) = v .p. The angular acceleration is the rate at which the angular velocity increases per unit of time, and d d fv.p\ = dt^''^ = di[l^)' Areal velocity. The areal velocity is, similarly, the rate at which the area XOP increases per unit of time, where X is the point in which the path of P meets Ox. It ^ ^ area POQ , , = Lt. ^^ =hr-.o>. 6. Mass and Force. Matter has been defined to be " that which can be perceived by the senses " or " that which can be acted upon by, or can exert, force." It is like time and space a primary conception, and hence it is practically im- Fundamental Definitions and Principles 5 possible to give it a precise definition, A body is a portion of matter bounded by surfaces. A particle is a portion of matter which is indefinitely small in all its dimensions. It is the physical correlative of a geometrical point. A body which is incapable of any rotation, or which moves without any rotation, may for the purposes of Dynamics, be often treated as a particle. The mass of a body is the quantity of matter it contains. A force is that which changes, or tends to change, the state of rest, or uniform motion, of a body. 7. If to the same mass we apply forces in succession, and they generate the same velocity in the same time, the forces are said to be equal. If the same force be applied to two different masses, and if it produce in them the same velocity in the same time, the masses are said to be equal. It is here assumed that it is possible to create forces of equal intensity on different occasions, e.g. that the force necessary to keep a given spiral spring stretched through the same distance is always the same when other conditions are unaltered. Hence by applying the same force in succession we can obtain a number of masses each equal to a standard unit of mass. 8. Practically, different units of mass are used under different conditions and in different countries. The British unit of mass is called the Imperial Pound, and consists of a lump of platinum deposited in the Exchequer Office. The French, or Scientific, unit of mass is called a gramme, and is the one-thousandth part of a certain quantity of platinum, called a Kilogramme, which is deposited in the Archives. One gramme = about 15'432 grains = -0022046 lb. One Pound = 45359 grammes. 9. The units of length employed are, in general, either a foot or a centimetre. 6 Dy^iamics of a Farticle A centimetre is the one-hundredth part of a metre which = 39-37 inches = 32809 ft. approximately. The unit of time is a second. 86400 seconds are equal to a mean solar day, which is the mean or average time taken by the Earth to revolve once on its axis with regard to the Sun. The system of units in which a centimetre, gramme, and second are respectively the units of length, mass, and time is called the C.G.S. system of units. 10. The density of a body, when uniform, is the mass of a unit volume of the body, so that, if m is the mass of a volume F of a body whose density is p, then m = Vp. When the density is variable, its value at any point of the body is equal to the limiting value of the ratio of the mass of a very small portion of the body surrounding the point to the volume of that portion, so that p = Lt. ^ , when V is taken to be indefinitely small. The weight of a body at any place is the force with which the earth attracts the body. The body is assumed to be of such finite size, compared with the Earth, that the weights of its component parts may be assumed to be in parallel directions. If m be the mass and v the velocity of a particle, its Momentum is mv and its Kinetic Energy is \mv-. The former is a vector quantity depending on the direction of the velocity. The latter does not depend on this direction and such a quantity is called a Scalar quantity. 11. Newton's Laws of Motion. Law I. Every body continues in its state of rest, or of uniform motion in a straight line, except in so far as it be compelled by impressed force to change that state. Law II. The rate of change of momentum is proportional to the impressed force, and takes place in the direction in which the force acts. Law III. To every action there is an equal and opposite reaction. These laws were first formally enunciated by Newton in his Frincipia which was published in the year 1686. Fimdamental Definitions mid Principles 7 12. If P be the force which in a particle of mass m produces an acceleration /, then Law II states that P = \-r. (mv), where \ is some constant, If the unit of force be so chosen that it shall in unit mass produce unit acceleration, this becomes P = mf. If the mass be not constant we must have, instead, The unit of force, for the Foot-Pound-Second system, is called a Poundal, and that for the C.G.S. system is called a Dyne. 13. The acceleration of a freely falling body at the Earth's surface is denoted by g, which has slightly different values at different points. In feet-second units the ^alue of g varies from 32-09 to 32-25 and in the c.G.s. system from 978-10 to 98311. For the latitude of London these values are 32*2 and 981 very approximately, and in numerical calculations these are the values generally assumed. If W be the weight of a mass of one pound, the previous article gives that W = 1 .g poundals, so that the weight of a lb. = 32'2 poundals approximately. So the weight of a gramme = 981 dynes nearly. A poundal and a dyne are absolute units, since their values are the same everywhere. 14. Since, by the Second Law, the change of motion pro- duced by a force is in the direction in which the force acts, we find the combined effect of a set of forces on the motion of a particle by finding the effect of each force just as if the other forces did not exist, and then compounding these effects. This is the principle known as that of the Physical Independence of Forces. From this principle, combined with the Parallelogram of Accelerations, we can easily deduce the Parallelogram of Forces. 8 Dynamics of a Pari ids 15. Impulse of a force. Suppose that at time t the value of a force, whose direction is constant, is P. Then the impulse of the force in time t is defined to be I P . dt. Jo From Art. 12 it follows that the impulse r ^^^^. r T = 1 m-y- at= \ mv Jo dt I Jo = the momentum generated in the direction of the force in time r. Sometimes, as in the case of blows and impacts, we have to deal with forces, which are very great and act for a very short time, and we cannot measure the magnitude of the forces. We measure the effect of such forces by the momentum each pro- duces, or by its impulse during the time of its action. 16. Work.' The work done by a force is equal to the product of the force and the distance through which the point of application is moved in the direction of the force, or, by what is the same thing, the product of the element of distance described by the point of application and the resolved part of the force in the direction of this element. It therefore = JPds, where ds is the element of the path of the point of application of the force during the description of w^iich the force in the direction of ds was P. If X, Y, Z be the components of the force parallel to the axes when its point of application is {x, y, z), so that X = P -r , etc. then Lxdx + Ydy + Zdz)=j(P^£dx +... + ...) -H^f-m-(l^h-h = the work done by the force P. The theoretical units of work are a Foot-Poundal and an Erg. The former is the work done by a poundal during a dis- placement of one foot in the direction of its action ; the latter is that done by a Dyne during a similar displacement of one cm. One Foot-Poundal = 421390 Ergs nearly. One Foot-Pound is the work done in raising one pound vertically through one foot. Fundamental Definitions and Principles 9 17. Power. The rate of work, or Power, of an agent is the work that would be done by it in a unit of time. The unit of power used by Engineers is a Horse-Power, An agent is said to be working with one Horse-Power, or h.p., when it would raise 33,000 pounds through one foot per minute. 18. The Potential Energy of a body due to a given system of forces is the work the system can do on the body as it passes from its present configuration to some standard configuration, usually called the zero position. For example, since the attraction of the Earth (considered as a uniform sphere of radius a and density p) is known to be 7 . — ~- . — at a distance x from the centre, the potential energy of a unit particle at a distance' y from the centre of the Earth, the surface of the earth being taken as the zero position. 19. From the definitions of the following phj^sical quantities in terms of the units of mass, length, and time, it is clear that their dimensions are as stated. Quantity Volume Density Velocity Acceleration Mass 1 Dimensions in length -3 Time -1 -2 Force -2 Momentum -1 Impulse -1 KJnetic Energy Power or Rate of Work 2 2 -2 -3 Angular Velocity -1 CHAPTER II MOTION IN A STRAIGHT LINE 20. Let the distance of a moving point P from a fixed point be ^ at any time t. Let its distance similarly at time f + A^ be a; + ^x, so that PQ = Ax. t- The velocity of P at time t PQ . . = Limit, when At = 0, of -~ At = Limit, when A^ = 0, of -r-^ = -r- . At dt Hence the velocity v= -j-. Let the velocity of the moving point at time t + At be V + Av. Then the acceleration of P at time t = limit, when A^ = 0, of -r- At _ dv ~di _d^x ~dt'' 21. Motion in a straight line with constant accelera- tion /. Let X be the distance of the moving point at time t from a fixed point in the straight line. Motion in a straight line 11 Tt™ §=/ (!>• Hence, on integration, v = -r=ft-]r A (2), where A is an arbitrary constant. Integrating again, we have x = \ft' + At-^B (3), where B is an arbitrary constant. Again, on multiplying (1) by 2-77, and integrating with respect to t, we have v^~=(J^) =2fa^ + C (4). where G is an arbitrary constant. These three equations contain the solution of all questions on motion in a straight line with constant acceleration. The arbitrary constants A, B, C are determined from the initial conditions. Suppose for example that the particle started at a distance a from a fixed point on the straight line with velocity w in a direction away from 0, and suppose that the time t is reckoned from the instant of projection. We then have that when ^ = 0, then v=u and a; = a. Hence the equations (2), (3), and (4) give u = A, a = B, and vJ'^C + ^fa. Hence we have v = u +ft, £c — a = ut+ ^fP, and v- = u'' + 2f{x - a), the three standard equations of Elementary Dynamics. 22. A particle moves in a straight line OA starting from rest at A and moving with an acceleration which is always directed towards and varies as the distance from 0; to find the motion. Let X be the distance OP of the particle from at any time t ; and let the acceleration at this distance be fix. The equation of motion is then d'x ... 5^ = -^" '^^- 12 Dynamics of a Particle [We have a negative sign on the right-hand side because -J— is the acceleration in the direction of x increasing, i.e. in the direction OP ; whilst fix is the acceleration towards 0, i.e. in the direction PO.] Multiplying by 2 -j- and integrating, we have di) =-M^+^- If OA be a, then -Tf = ^ when a; = a, so that = - fia^ + C, .-. | = -V;iVa?:r^' (2). [The negative sign is put on the right-hand side because the velocity is clearly negative so long as OP is positive and P is moving towards 0.] Hence, on integration, dx , X _ [ dx _ tA^/j, = — I = cos~^ - + C, , a where 0= cos~- - + G,, i.e. C, = 0, a if the time be measured from the instant when the particle was at A. .'. x = acos'\/fit (3), When the particle arrives at 0, x is zero ; and then, by (2), the velocity = — a V/*- The particle thus passes through and immediately the acceleration alters its direction and tends to diminish the velocity; also the velocity is destroyed on the left-hand side of as rapidly as it was produced on the right- hand side ; hence the particle comes to rest at a point A' such that OA and OA' are equal. It then retraces its path, passes through 0, and again is instantaneously at rest at A. The whole Simple Harmonic Motion 13 motion of the particle is thus an oscillation from A to A' and back, continually repeated over and over again. The time from ^ to is obtained by putting x equal to zero ,— TT in (3). This gives cos (v/ii) = 0, i.e. t = ttt ' The time from A to A' and back again, i.e. the time of a 27r complete oscillation, is four times this, and therefore = ~ . This result is independent of the distance a, i.e. is inde- pendent of the distance from the centre at which the particle started. It depends solely on the quantity ft which is equal to the acceleration at unit distance from the centre. 23. Motion of the kind investigated in the previous article is called Simple Harmonic Motion. 27r The time, — - , for a complete oscillation is called the Periodic Time of the motion, and the distance, OA or OA', to which the particle vibrates on either side of the centre of the motion is called the Amplitude of its motion. The Frequency is the number of complete oscillations that the particle makes in a second, and hence 1 _ VM; Periodic time ~ 27r * 24. The equation of motion when the particle is on the left-hand side of is d'^x -7- = acceleration in the direction P'A dv = fl . P'O = fJL(— X)= — fix. Hence the same equation that holds on the right hand of holds on the left hand also. As in Art. 22 it is easily seen that the most general solution of this equation is x = acoB [\//x^ + e] (1), which contains two arbitrary constants a and e. This gives -^' = — aVft sin (Vyu,i + e) (2). 27r (1) and (2) both repeat when t is increased by -— , since the sine 14 Dynamics of a Particle and cosine of an angle always have the same value when the angle is increased by 27r, Using the standard expression (1) for the displacement in a simple harmonic motion, the quantity e is called the Epoch, the angle V/i^ + e is called the Argument, whilst the Phase of the motion is the time that has elapsed since the particle was at its maximum distance in the positive direction. Clearly a; is a maximum at time to where V/a^o + e = 0. Hence the phase at time t -t-t -t\ ^ _^^t + e Motion of the kind considered in this article, in which the time of falling to a given point is the same whatever be the distance through which the particle falls, is called Tauto- chronous. 25. In Art. 22 if the particle, instead of being at rest initially, be projected from A with velocity V in the positive direction, we have Hence (^Y =V' + fi (a' - x") = fjL(b'-a;% where h'' = a'' + — (1), and t\//j, = — cos~^-r+Gi, where = — cos~' y- + (7i. .". t \/fJb = COS~^-r — COS~^j (2). From (1), the velocity vanishes when and then, from (2), ^ / ,<^ • ^ 1 * t J ix = cos ^ T , I.e. t = —1- cos b KJii / V The particle then retraces its path, and the motion is the same as in Art. 22 with h substituted for a. Simple Harmonic Motion 15 26. Compounding of two simple harmonic motions of the same period and in the same straight line. The most general displacements of this kind are given by a cos {nt + e) and h cos {nt + e'), so that x = a cos {nt + e) + 6 cos {nt + e') = cos nt {a cos e + 6 cos e') — sin nt {a sin e + 6 sin e'). Let a cos e + 6cos 6' = J. cos^l , and a sin 6 + 6 sin e' = J. sin £"1 ^ ^' so that ^=Va^ + 62+2a6cos(e-e'), and tan ^ = ^^i^^±^^ . a cos 6 + 6 cos 6 Then a? = J. cos {nt + E), so that the composition of the two given motions gives a similar motion of the same period whose amplitude and epoch are known. If we draw OA (= a) at an angle e to a fixed line, and OB {= b) at an angle e' and complete the parallelogram OA CB then by equations (1) we see that 00 represents A and that it is inclined at an angle E to the fixed line. The line repre- senting the resultant of the two given motions is therefore the geometrical resultant of the lines representing the two com- ponent motions. So with more than two such motions of the same period. 27. We cannot compound two simple harmonic motions of different periods. The case when the periods are nearly but not quite equal, is of some considerable importance. In this case we have x = a cos (ni + e) + 6 cos {n't + e'), where n' — n is small, = \ say. Then « = a cos {nt + e) + 6 cos [nt + e^'J, where ei' = \t + e. By the last article x=- Acos{nt + E) (1), where A'^ = a- + b^ + lab cos (e — e^) = (i' + ¥ + 2ab cos [e - e' - {u -n)i] (2), 16 Dynamics of a Particle J X T^ a sin 6 + 6 sine/ and tan E = 5 -. a cos e + 6 cos ej _ a sin e 4- 6 sin [e' + (w' — n) t] . a cos 6 + 6 cos [e' + {n — n)t] The quantities A and ^ are now not constant, but they vary slowly with the time, since n' — n is very small. The greatest value of A is when e — e' — {n' — n)t = any even multiple of tt and then its value is a + &. The least value of A is when e — e — {n — n) t = any odd multiple of tt and then its value is a — h. At any given time therefore the motion may be taken to be a simple harmonic motion of the same approximate period as either of the given component motions, but with its amplitude A and epoch E gradually changing from definite minimum to definite maximum values, the periodic times of these changes , . 27r bemg ~ . ° n —n [The Student who is acquainted with the theory of Sound may compare the phenomenon of Beats.] 28. Ex. 1. Shew that the resultant of two simple harmonic vibrations in the same direction and of equal periodic time, the amplitude of one being twice that of the other and its phase a quarter of a period in advance, is a simple harmonic vibration of amplitude J5 times that of the first and tan ~ ^ 2 ■whose phase is in advance of the first by -^ of a period. Ex. 2. A particle is oscillating in a straight line about a centre of force 0, towards which when at a distance r the force is m . nh', and a is the amplitude of the oscillation ; when at a distance ^ from 0, the particle receives a blow in the direction of motion which generates a velocity na. If this velocity be away from 0, shew that the new amplitude is a /J3. Ex. 3. A particle P, of mass m, moves in a straight line Ox under a force «i/x (distance) directed towards a point A which moves in the straight line Ox with constant acceleration a. Shew that the motion of P 2_. is simple harmonic, of period -t-- , about a moving centre which is always VM at a distance - behind A. Ex. 4. An elastic string without weight, of which the unstretched length is I and the modulus of elasticity is the weight of ?i ozs., is suspended by one end, and a mass of m ozs. is attached to the other ; shew that the time of a vertical oscillation is 2n -■ ng /ml V ng ' Simple Harmonic Motion 17 Ex. 5, One end of an elastic string, whose modulus of elasticity is X and whose unstretched length is a, is fixed to a point on a smooth horizontal table and the other end is tied to a particle of mass m which is lying on the table. The particle is pulled to a distance where the extension of the string is b and then let go ; shew that the time of a complete oscillation is 2 ( tt + -r- j » / — - . Ex. 6. An endless cord consists of two portions, of lengths U and 21' respectively, knotted together, their masses per unit of length being m and m'. It is placed in stable equilibrium over a small smooth peg and then slightly displaced. Shew that the time of a complete oscillation is A' {m-m)g Ex. 7. Assuming that the earth attracts points inside it with a force which varies as the distance from its centre, shew that, if a straight frictionless airless tunnel be made from one point of the earth's surface to any other point, a train would traverse the tunnel in slightly less than three-quarters of an hour. 29. Motion when the motion is in a straight line and the acceleration is proportional to the distance from a fixed point in the straight line and is always away from 0. Here the equation of motion is d'a; dF = ^^ (!)• Suppose the velocity of the particle to be zero at a distance a from at time zero. The integral of (1) is l-Tij =fiOc- + A, where = /jLa^ + A. .-. ^ = V/.(^^-aO (2), the positive sign being taken in the right-hand member since the velocity is positive in this case. r dx . •■• ^^^ = jvFz9 = ^°g[^ + ^^'-«']+^.- where = log [a] + B. . ^ . , x-Y^/'aF^^^ .'. t V/i = log . 18 Dynamics of a Particle .•. X + '^x- — a^ = ae*^". /. x—Nx^-ar — / , • , = ae~V^ x-\- w x^ — a- Hence, by addition, ^ = |,s/.-.. + |,-sV-. (3). As t increases, it follows from (3) that x continually increases, and then from (2) that the velocity continually increases also. Hence the particle would continually move along the positive direction of the axis of x and with continually increasing velocity. Equation (3) may be written in the form x = a cosh {\/ fxt), and then (2) gives v=a^fM sinh {"J fit). 30. In the previous article suppose that the particle were initially projected towards the origin with velocity V; then we should have -tt equal to — F when x=a ; and equation (2) would be more complicated. We may however take the most general solution of (1) in the form x = Ce'^^^ + De-^'^^ (4), ■where C and D are any constants. dx Since, when i = 0, we have x = a and — ^ = — F, this gives a=C+D, and - V = '^^G- V]iD. 1 / V\ 1 / V Hence G= ^{a - ^^) ^nd D ^ ^^{a + -^ •'• (4) gives = acosh(Vy[i«)--r-sinh(\/yL4 (6). In this case the particle will arrive at the origin when Motion in a straight line, 19 I.e. when gW^ « = _ i/i K — asjyi, , , 1 , F + a V/tA I.e. when t = ^r—r- log ^ j- . In the particular case when V=a V/w- this value of t is infinity. If therefore the particle were projected at distance a towards the origin with the velocity a\//ji, it would not arrive at the origin until after an infinite time. Also, putting V=a ^//x in (5), we have X = ae""^^'^ *, and v = -y- = ~ a V'/xe"'^'^ '. The particle would therefore always be travelling towards with a continually decreasing velocity, but would take an infinite time to get there. 31. A particle moves in a straight line OA ivith an accelera- tion luhich is always directed towards and varies inversely as the square of its distance from ; if initially the particle were at rest at A, find the motion. P' , p _ i + 1 ^ Let OP be x, and let the acceleration of the particle when at P be -^ in the direction PO. The equation of motion is therefore d^x a -5-^ = acceleration along OP — — -^ (1), Multiplying both sides by 2 -^ and integrating, we have at/ X where = — + 0, from the initial conditions. a 2—2 20 Dynamics of a Par-tide Subtracting, (^) =2/z(-- X a dx di --JTyJ'^^ (2). '^ V ax the negative sign being prefixed because the motion of P is towards 0, i.e. in the direction of x decreasing. Hence f. —.t = -\ K dx. V a j V a — X To integrate the right-hand side, put x = a cos^ Q, and we have — , . 2a cos 6 sin B dd sm d /2fi [CO = a [ (1 + cos 2^) de=^a(d+h sin 2^) + G = a cos~^ A / — ■ V a .(3). + \/ax - a,'2 + C, where = a cos-^ (1 ) + + C, i.e. C = 0. ••• ^= \/^ [V^^^^+acos-y^^] Equation (2) gives the velocity at any point P of the path, and (3) gives the time from the commencement of the motion. The velocity on arriving at the origin is found, by putting «; = in (2), to be infinite. Also the corresponding time, from (3), « r -1 AT "^ ^^ 2^[acos..0] = ^^. The equation of motion (1) will not hold after the particle has passed through 0; but it is clear that then the acceleration, being opposite to the direction of the velocity, will destroy the velocity, and the latter will be diminished at the same rate as it was produced on the positive side of 0. The particle will therefore, by symmetry, come to rest at a point A' such that AO and OA' are equal. It will then return, pass again through and come to rest at A. -■/, Motion under the EartKs Attraction 21 The total time of the oscillation = four times the time from A to = 27r — = . 32. By the consideration of Dimensions only we can shew that the I time X ^ . For the only quantities that can appear in the answer are a and fi. Let then the time be a^/x'. Since , ,. .^ r^ is an acceleration, whose dimensions are [Zl [TY^, (distance)^ the dimensions of /* are [X]^ {Ty^ ; hence the dimensions of a^^i are Since this is a time, we have p-\-Zq = and -2q = \. .-. q= -- and jo = -. Hence the required time x -j-. 33. As an illustration of Art. 31 let us consider the motion of a particle let fall towards the earth (assumed at rest) from a point outside it. It is shewn in treatises on Attractions that the attraction on a particle outside the earth (assumed to be a homogeneous sphere), varies inversely as the square of its distance from the centre. The acceleration of a particle outside the earth at distance x may therefore be taken to be — . If a be the radius of the earth this quantity at the earth's surface is equal to g, and hence — ^ = g, i.e. /j, = ga^. For a point P outside the earth the equation of motion is therefore (l^^_gar df a? ^ ^' X If the particle started from rest at a distance h from the centre of the earth, this gives (r:T--'(i-F) •••• (^). and hence the square of the velocity on reaching the surface of the earth = 2(/afl — -,) (3). 22 Dynamics of a Particle Let us now assume that there is a hole going down to the earth's centre just sufficient to admit of the passing of the particle. On a particle inside the earth the attr-action can be shewn to vary directly as the distance from the centre, so that the acceleration at distance x from its centre is fji,-^x, where /ija = its value at the earth's surface =^g. The equation of motion of the particle when inside the earth therefore is d'x a dv a /dx\^ a and therefore \-^A = -- x" ->r G.. \dtj a Now when x = a, the square of the velocity is given by (3), since there was no instantaneous change of velocity at the earth's surface. 2,.(l-|) = -|.a= + a, ^)=-^' + »a[3-T]- dx\^ _ g dtj a On reaching the centre of the earth the square of the b velocity is therefore ^a (3 34. Ex. 1. A particle falls towards the earth from infinity ; shew that its velocity on reaching the earth is the same as it would have acquired in falling with constant acceleration g through a distance equal to the earth's radius. Ex. 2. Shew that the velocity with which a body falling from infinity reaches the surface of the earth (assumed to be a homogeneous sphere of radius 4000 miles) is about 7 miles per second. In the case of the sun shew that it is about .360 miles per second, the radius of the sun being 440,000 miles and the distance of the earth from it 92,500,000 miles. Ex. 3. If the earth's attraction vary inversely as the square of the distance from its centre, and g be its magnitude at the surface, the time of falling from a height h above the surface to the surface is /a + hfa + h . , / A , //il where a is the radius of the earth and the resistance of the air is neglected. Motion in a straight line 23 If h be small compared with a, shew that this result is approximately 35. It is clear that equations (2) and (3) of Art. 31 cannot be true after the particle has passed 0; for on giving x negative values these equations give impossible values for v and t. When the particle is at P', to the left of 0, the acceleration is tJ^ , i.e. — „ , towards the right. Now ^— means the accelera- tion towards the positive direction of x. Hence, when P' is on the left of 0, the equation of motion is (Px fl 'dt^'^x-' giving a different solution from (2) and (S). The general case can be easily considered. Let the accelera- tion be /x (distance)" towards 0. The equation of motion when the particle is on the right hand of is clearly d-x When P' is on the left of 0, the equation is -y- = acceleration in direction OA df =fiiP'or=fi{-xr. These two equations are the same if - /j,.x''= /x{- xY, i.e. if ( - 1)" = - 1, i.e. if n be an odd integer, or if it be of the form ^ — - , where p and q are integers; in these cases the same equation holds on both sides of the origin ; otherwise it does not. 36. JEx. A small bead, of mass m, moves on a straight rough wire under the action of a force equal to mfi times the distance of the bead from a fixed point A outside the wire at a perpendicular distance a from it. Find the motion if the bead start from rest at a distance c from the foot, 0, of the per pendic^dar from A upon the toire. Let P be the position of the bead at any time t, where OP=x and AP=y. 24 Dynamics of a Particle Let R be the normal reaction of the wire and ;xi the coefficient of friction. Resolving forces jjerpendicular to the wire, we have R=m\i.y sin OP A =mfxa. Hence the friction fiiR=mfifiia. The resolved part of the force wi/iy along the wire = m/xy cos OP A = vifijs. Hence the total acceleration = /iptia-/n.r. The equation of motion is thus d'\v ■^^=liHia-iix=-fi{x-iiia) (1), so long as P is to the right of 0. [If P be to the left of and moving towards the left, the equation of motion is -TY= acceleration in the direction OC = fMfiia+fi{ — x:), as in the last article, and this is the same as (1) which therefore holds on both sides of 0.] Integrating, we have where 0=-fi{c- fiicc)'^ + G, ••• «''^ = (^^y = M[(^-Mi«)--(-^-Mi«n "'•.(^^. and therefore, as in Art. 22, J u.t = co%~'^ — ^ + C'i, where o=cos-i^^^^^ + e,, i.e. Ci=0. .'. Vm< = 003-1-:^:^^ (3), (2) and (3) give the velocity and time for any position. From (2) the velocity vanishes when x- fiia= ±(c- fiia), I.e. when x=c=OC\ and when x = - {c - 2ayi{), i.e. at the point C", where 0C' = c — 2afii, and then from (3) the corresponding time =-7- cos 1 ^-^ = -7- cos M — 1) = -7-. Motion in a straight line. Examples 25 The motion now reverses and the particle comes to rest at a point C" on the right of where OC" = OC' - 2fX]a=0C- A^na. Finally, when one of the positions of instantaneous rest is at a distance which is equal to or less than /nja from 0, the particle remains at rest. For at this point the force towards the centre is less than the limiting friction and therefore only just sufficient friction will be exerted to keep the particle at rest. o It will be noted that the periodic time -r- is not affected by the friction, but the amplitude of the motion is altered by it. 37. Ex. A particle, of mass m, rests in equilibrium at a point N, being attracted by two forces equal to m^i^ {distanceY and m^'" {distance)^ towards two fixed centres and 0'. If the particle be slightly displaced from iV", and if n be positive, shew that it oscillates, and find the time of a small oscillation. O N O' Let 0(y = a, ON=d and NO' = d', so that /x".C^» = /.'".C^''' (1), since there is equilibrium at N. .-. ^ = ^ = ^, (2). Let the particle be at a distance x from N towards 0', The equation of motion is then ^=-^^.OF>' + ,x\PO'^=-^»id+xy+lx'^'{d'-x)^ (3). If X is positive, the right-hand side is negative ; if j? is negative, it is positive ; the acceleration is towards iV in either case. Expanding by the Binomial Theorem, (3) gives = - nx [/ix"e/" ~ ^ + m' "<^'" ~ ^] + terms involving higher powers of ^ = -n.m"-' y^"^\ -i-...by(2). If X be so small that its squares and higher powers may be neglected, this gives df^- '\,.+^r-^'' ^^^- 26 Dynamics of a Particle Hence, as in Art. 22, the time of a small oscillation If n be negative, the right-hand member of (4) is positive and the motion is not one of oscillation. EXAMPLES ON CHAPTER II. 1. A particle moves towards a centre of attraction starting from rest at a distance a from the centre ; if its velocity when at any distance x from the centre vary as */ — ^^ , find the law of force. 2. A particle starts from rest at A and moves towards a centre of force at ; if the time to any position P varies inversely as the distance AP moved through, shew that the attraction towards varies as the cube of^P. 3. Prove that it is impossible for a particle to move from rest so that its velocity varies as the distance described from the commencement of the motion. If the velocity vary as (distance)", shew that n cannot be greater than - . 4. A point moves in a straight line towards a centre of force / M 1 |(distance)^J ' starting from rest at a distance a from the centre of force ; shew that the time of reaching a point distant h from the centre of force is , and that its velocity then is — ^ Ja^ — V^. 5. A particle falls from rest at a distance a from a centre of force, where the acceleration at distance x is nx~ "^ ; when it reaches the centre 2a^ shew that its velocity is infinite and that the time it has taken is -t^= . 6. A particle moves in a straight line under a force to a point in it varying as (distance) ~ s^ ; shew that the velocity in falling from rest at infinity to a distance a is equal to that acquired in falling from rest at a distance a to a distance -^ . Motion in a straight line. Examples 27 7. A particle, whose mass is w, is acted upon by a force toju ( x-\- -^ \ towards the origin ; if it start from rest at a distance a, shew that it will arrive at the origin in time — — - . 8. A particle moves in a straight line with an acceleration towards a fixed point in the straight line, which is equal to ^ — ;^ when the particle is at a distance x from the given point ; it starts from rest at a distance a ; shew that it oscillates between this distance and the distance — — - , and that its periodic time is 9. A particle moves with an acceleration which is always towards, and equal to fi divided by the distance from, a fixed point 0. If it start from rest at a distance a from 0, shew that it will arrive at in time a */ ^ . Assume that 1 e~=^ dx = ^-- 10. A particle is attracted by a force to a fixed point varying inversely as the nth. power of the distance ; if the velocity acquired by it in falling from an infinite distance to a distance a from the centre is equal to the velocity that would be acquired by it in falling from rest at a distance a to a distance j , shew that n = -. 11. A particle rests in equilibrium under the attraction of two centres of force which attract directly as the distance, their attractions per unit of mass at unit distance being fi and n' ; the particle is slightly displaced towards one of them ; shew that the time of a small oscillation is — , . 12. A uniform chain, of length 2a, is hung over a smooth peg so that the lengths of it on the two sides are a + b and a-b ; motion is allowed to ensue ; shew that the chain leaves the peg at the end of time v/; - log- g ° b 13. A particle moves in a straight line with an acceleration equal to /x-rthe nth power of the distance from a fixed point in the straight line. If it be projected towards 0, from a point at a distance a, with the velocity it would have acquired in falUng from infinity, shew that it will reach in time 1 \/ -?: — . a 2 , n + l V 2fx 28 Dynamics of a Fartide 14. In the previous question if the particle started from rest at distance a, shew that it would reach in time according as w is > or < unity. 15. A shot, whose mass is 50 lbs., is fired from a gun, 4 inches in diameter and 8 feet in length. The pressure of the powder-gas is inversely proportional to the volume behind the shot and changes from an initial value of 10 tons' weight per square inch to 1 ton wt. per sq. inch as the shot leaves the gun. Shew that the muzzle velocity of the shot is approximately 815 feet per second, having given log« 10 = 2-3026. 16. If the Moon and Earth were at rest, shew that the least velocity with which a particle could be projected from the Moon, in order to reach the Earth, is about 1^ miles per second, assuming their radii to be 1100 and 4000 miles respectively, the distance between their centres 240,000 miles, and the mass of the Moon to be -^ that of the Earth. oi 17. A small bead can slide on a smooth wire AB, being acted upon by a force per unit of mass equal to /ii-=-the square of its distance from a point which is outside AB. Shew that the time of a small oscillation about its position of equilibrium is -^b^ , where h is the perpendicular distance of from AB. 18. A solid attracting sphere, of radius a and mass J/, has a fine hole bored straight through its centre ; a particle starts from rest at a distance b from the centre of the sphere in the direction of the hole produced, and moves under the attraction of the sphere entering the hole and going through the sphere ; shew that the time of a complete oscillation is ^[V2a'^sin- x/g^.+Z'^cos- ^/^Waft (6 - a)] , ■where y is the constant of gravitation. 19. A circular wire of riidius a and density p attracts a particle according to the Newtonian law y ,. ^ — ^g '■> ^^ ^^^ particle be placed on the axis of the wire at a distance b from the centre, find its velocity when it is at any distance x. If it be placed on the axis at a small distance from the centre, shew that the time of a complete oscillation is a a/ — . Motion ill a straight line. Examples 29 20. In the preceding question if the wire repels instead of attracting, and the particle be placed in the plane of the wire at a small distance is 2a a/ - from its centre, shew that the time of an oscillation yp 21. A particle moves in a straight line with an acceleration directed towards, and equal to /x times the distance from, a point in the straight line, and with a constant acceleration / in a direction opposite to that of its initial motion ; shew that its time of oscillation is the same as it is when / does not exist, 22. A particle P moves in a straight line OOP being attracted by a force mft . PC always directed towards C, whilst C moves along DC with constant acceleration /. If initially C was at rest at the origin 0, and P was at a distance c from and moving with velocity V, prove that the distance of P from at anv time t is - +c\cos^at+-7= sin Vm t- - -k--^ t\ 23. Two bodies, of masses M and 31', are attached to the lower end of an elastic string whose upper end is fixed and hang at rest ; M' falls off; shew that the distance of Af from the upper end of the string at time t is Wi')' where a is the unstretched length of the string, and b and c the distances by which it would be stretched when supporting M and M' respectively. 24. A point is performing a simple harmonic motion. An additional acceleration is given to the point which is very small and varies as the cube of the distance from the origin. Shew that the increase in the amplitude of the vibration is proportional to the cube of the original amplitude if the velocity at the origin is the same in the two motions. 25. One end of a light extensible string is fastened to a fixed point and the other end carries a heavy particle ; the string is of unstretched length a and its modulus of elasticity is n times the weight of the particle. The particle is pulled down till it is at a depth b below the fixed point and then released. Shew that it will return to this position at the end of time 2 /v/ — ^ + cosec -^p + Jp- - 1 , where p = ('i + 1), provided ^ ng\_J, J a that jo is not >\/l+4?i. If jt)>\/i + 4», shew how to find the corresponding time. 26. An endless elastic string, whose modulus of elasticity is X and natural length is 2ttc, is placed in the form of a circle on a smooth horizontal plane and is acted upou by a force from the centre equal to 30 Dynamics of a Particle ft. times the distance per unit mass of the string. Shew that its radius will vary harmonically about a mean length ^ ^^ , where m is the mass of the string, assuming that 27rX>m^c. Examine the case when 27rX = mfj.c. 27. An elastic string of mass m and modulus of elasticity X rests imstretched in the form of a circle whose radius is a. It is now acted on by a rejDulsive force situated in its centre whose magnitude per unit mass of the string is (distance)^ ' Shew that when the circle next comes to rest its radius is a root of the quadratic equation ttA 28. A smooth block, of mass M, with its upper and lower faces horizontal planes, is free to move in a groove in a parallel plane, and a particle of mass m is attached to a fixed point in the upper face by an elastic string whose natural length is a and modulus E. If the system starts from rest with the particle on the upper face and the string stretched parallel to the groove to (n + 1) times its natural length, shew that the block will perform oscillations of amplitude ' ^ ^ ) ^^ j^ ^j^q ^ M+m periodic time 2 ( tt + - j */ a Mm ElM+mj' 29. A particle is attached to a point in a rough plane inclined at an angle a to the horizon ; originally the string was unstretched and lay along a line of greatest slope ; shew that the particle will oscillate only if the coefficient of friction is < ^ tan a. o 30. A mass of m lbs. moves initially with a velocity of u ft. per sec. A constant power equal to If horse-power is applied so as to increase its velocity ; shew that the time that elapses before the acceleration is reduced to -th of its original value is - _ -, ^^^ /,— . 31. Shew that the greatest velocity which can be given to a bullet of /2Y\V mass M fired from a smooth-bore gun is \/ *" ^ , (m log m + 1— m}, where changes of temperature are neglected, and the pressure H in front of the bullet is supposed constant, the volume V of the powder in the cartridge being assumed to turn at once, when fired, into gas of pressure will and of volume V, Motion m a straight line. Examples 31 32. Two masses, rtix and m-i^ are connected by a spring of such a strength that when nix is held fixed m.' and dl = ~^2/ (2). Composition of Simple HarmoniG Motions 35 The solutions of these equations are, as in Art. 22, a; = ^ cos [V/Ii + £] (3), and 2/ = (7 cos [\/yu,i + D] (4). The initial conditions are that when ^ = 0, then c = OA = a, ^ = 0, y = 0, and %=V. dt '^ ' dt Hence, from (3), a = ^ cos 5 and = - J. sin 5. These give B = and A = a. Also, from (4), similarly we have = cos D, and V= - C ^f/lsin D. .'. i)=?r, and C= 7- . 2 V/t .*. (3) and (4) give x = a cos (\/ /it) (5), and ?/ = — 7- cos V/i«+^ = — -sin(V/xO (6). • 1- i'— = 1 a^ V- The locus of P is therefore an ellipse, referred to OX and OF as a pair of conjugate diameters. V Also, if the ellipse meet OY in B, then OB = -r- , i.e. F = Vy"- >< semi-diameter conjugate to OA. Since any point on the path may be taken as the point of projection, this result will be always true, so that at any point the velocity = \/fi X semi-conjugate diameter. [This may be independently derived from (5) and (6). For (Velocity at P)^ = x^ + y^ + 2xy coso) = a^ix sin2 {Jixt) + V^ cos2 {Jfj.t) - 2a V^/x sin (Jfj.t) cos (V/"0 ^°^ " = ix\ a2 + a2 cos2 ( J /It) sin2 Jj^t - -^- sin {J]j.t) cos {Jjit} cos w | L ^ M V/* -J = fj.[a'^ + — -x^- y2-2xy cos J] =/x{a^ + — -0P2\ = iu X square of semi-diameter conjugate to OP.] 36 Dynamics of a Particle From equations (5) and (6) it is clear that the values of X and y are the same at time i + -^ as they are at time t. 2_ Hence the time of describing the ellipse is -r- . 41. If a particle possess two simple harmonic motions, in perpendicular directions and of the same period, it is easily seen that its path is an ellipse. If we measure the time from the time when the a;-vibration has its maximum value, we have x = a cos wi (1), and 2/ = 6 cos (n^ + e) (2), where a, h are constants. (2) gives y = cos nt cos e — sin nt sin e = - cos e — sin e a/ 1 — — , • , cos e = sin^ el pa/ V ci-, x^ Ixy y^ . „ ,_, I.e. — ^ cos e + TT, = sm^ e (3). a^ ah h' This always represents an ellipse whose principal axes do not, in general, coincide with the axes of coordinates, but which is always inscribed in the rectangle x = ±a, y = ±h. The figure drawn is an ellipse where e is equal to about -^ . If 6 = 0, equation (3) gives h.~^' *'"^" ^^® straight line AG. If e = TT, it gives - + r = 0, i.e. the straight line BD. In the particular case when e = ^ , i.e. when the phase of the ^/-vibration at zero time is one-quarter of the periodic time, equation (3) becomes i.e. the path is an ellipse whose principal axes are in the direction Composition of Siinjjle Hat^monic Motions 37 of the axes of x and y and equal to the amplitudes of the component vibrations in these directions. If in addition a = b, i.e. if the amplitudes of the component vibrations are the same, the path is a circle. B Y \ y^ X 42. If the period of the 2/-vibration is one-half that of the a;-vibration, the equations are a; = a cos nt, and y=b cos (2nt + e). Hence, by eliminating t, we have as the equation to the path y^, '2x'' ^1 . 2x L x" —7 - 1 - sm e . — A / 1 J On rationalization, this equation becomes one of the fourth degree. Y E A \ i / / V \ ' X / \ X 38 Dynamics of a Particle IT The dotted curve in the figure is the path when e , ..^. when the phase of the ^/-vibration at zero time is negative and equal to one-quarter of the period of the 2/-vibration. When 6 = TT, i.e. when the phase of the y- vibration at zero time is one-half of the y-period, the path becomes ^=-|(y-6). i.e. the parabola GEB. When e = Oj the path is similarly the parabola ^=g(j/+6). For any other value of e the path is more complicated. Curves, such as the preceding, obtained by compounding simple harmonic motions in two directions are known as Lissajous' figures. For other examples with different ratios of the periods, and for different values of the zero phases, the student may refer to any standard book on Physics. These curves may be drawn automatically by means of a pendulum, or they may be constructed geometrically. 43. Ex. 1. A point moves in a plane so that its projection on the axis of x performs a harmonic vibration of period one second with an amplitude of one foot; also its projection on the perpendicular axis of y performs a harmonic vibration of period two seconds with an amplitude of one foot. It being given that the origin is the centre of the vibrations, and that the point (1, 0) is on the path, find its equation and draw it. Ex. 2. A point moves in a path produced by the combination of two simple harmonic vibrations in two perpendicular directions, the periods of the components being as 2 : 3 ; find the paths described (1) if the two vibrations have zero phase at the same instant?, and (2) if the vibration of greater period be of phase one-quarter of its period when the other vibration is of zero phase. Trace the paths, and find their equations. 44. If in Art. 40 the acceleration be always from the fixed point arid varying as the distance from it, we have similarly X = a cosh V/A<, and y = -j- sinh V/ii. .'. — — p- = 1, so that the path is a hyperbola. Accelerations parallel to fixed axes 39 45. A 'particle describes a catenary under a force which acts parallel to its accis; find the law of the force and the velocity at any point of the path. Taking the directrix and axis of the catenary as the axes of cc and y, we have as the equation to the catenary c / £ _£\ y = 2\^" + 6 '') (1). Since there is no acceleration parallel to the directrix, .' — = • • df' • .*. -77 = const. = M (2). Differentiating equation (1) twice, we have dt lft£ _^\ dx 1 / ? --IN Also (velocity)' = (|y+(|J = «= + f (.!-.-!)' W" / '1 _i'\- IV so that the velocity — ~ y. Hence the velocity and acceleration at any point both vary as the distance from the directrix. 46. A particle moves in one plane ivith an acceleration which is always toivards and. perpendicidar to a fixed straight line in the plane and varies inversely as the cube of the distance from it; given the' circumstances of projection, find the path. Take the fixed straight line as the axis of x. Then the equations of motion are S=« w- -•J %-^f (^)- (1) gives x==At-\rB (3). 40 Dynamics of a Particle Multiplying (2) by -^ and integrating, we have ydy 1 Let the particle be projected from a point on the axis of y distant h from the origin with component velocities u and v parallel to the axes. Then when ^ = we have r^ i dx , dxi x = 0, v = o, -T- = u, and -77 = v. ' ^ ' dt dt .'. A=u, B = 0, G^v"--^^, and i) = - ,, f "" . 0^ b^v- — fj, .'. (3) and (4) give x = tit, and [t + j— =7TT-:, z. + jA . V b-v — /xj [b-v- — fi)- b^v — fi Eliminating t, we have as the equation to the path fx ¥v Y y"b'~ _ fj,¥ \u~ /M- bv) "*" yu, - b-'v' ^ ifM- b'u-y ' This is an ellipse or a hyperbola according as /j,< 6V. If /Lt = ¥v', then C = and equation (4) becomes Hence the path in this case is y'^ — b^ = 2^//^ - , i.e. a parabola. The path is thus an ellipse, parabola, or hyperbola according as v A /t"j. *-6- according as the initial velocity perpendicular to the given line is less, equal to, or greater than the velocity that would be acquired in falling from infinity to the given point with the given acceleration. For the square of the latter =— 1 2^t/2/=p^ =,^. Cor. If the particle describe an ellipse and meets the axis of X it will not then complete the rest of the ellipse since Accelerations parallel to fixed axes 41 the velocity parallel to the axis of x is always constant and in the same direction; it will proceed to describe a portion of another equal ellipse. 47. If the velocities and accelerations at any instant of particles m^, w^, m^ ... parallel to any straight line fixed in space he Vi,v^ ... dindf,f ...to find the velocity and acceleration of their centre of mass. If 0?!, x^, ... be the distances of the particles at any instant measured along this fixed line from a fixed point, we have _ TOja?! + m^X2 + . . . 7?li + ???2 + • • • Differentiating with respect to t, we have _ dx Wit'i + W2V2+ ... /iv 1} zzz —- =• V-l), dt m^ + mo + ... and /=g = -"'/■ + '"'.^+- (2), •^ dt^ nia + «'2 + • • • where v and /are the velocity and acceleration required. Consider any two particles, m-^ and m^, of the system and the mutual actions between them. These are, by Newton's Third Law, equal and opposite, and therefore their impulses resolved in any direction are equal and opposite. The changes in the momenta of the particles are thus, by Art. 15, equal and opposite, i.e. the sum of their momenta in any direction is thus unaltered by their mutual actions. Similarly for any other pair of particles of the system. Hence the sum of the momenta of the system parallel to any line, and hence by (1) the momentum of the centre of mass, is unaltered by the mutual actions of the system. If Pj, P2, ... be the external forces acting on the particles nil, ^2 ••• parallel to the fixed line, we have 'mifi + vi2f2 + ... = (Pi-h P2+ '■■) + (i^& sum of the com- ponents of the internal actions on the particles) = P, + P2+..., since the internal actions are in equilibrium taken by them- selves. 42 Dynamics of a Particle Hence equation (2) gives (mi + m. + . . . ) 7= Pi + Po + . . ., i.e. the motion of the centre of mass in any given direction is the same as if the whole of the particles of the system were collected at it, and all the external forces of the system applied at it parallel to the given direction. Hence also If the sum of the external forces acting on any given system of particles parallel to a given direction vanishes, the motion of the centre of gravity in that direction remains unaltered, and the total momentum of the system in that direction remains constant throughout the motion. This theorem is known as the Principle of the Conservation of Linear Momentum. As an example, if a heavy chain be falling freely the motion of its centre of mass is the same as that of a freely falling particle. EXAMPLES ON CHAPTER III 1. A particle describes an ellipse with an acceleration directed towards the centre ; shew that its angular velocity about a focus is inversely proportional to its distance from that focus. 2. A particle is describing an ellipse under a force to the centre ; if V, Vi and V2 are the velocities at the ends of the latus-rectum and major and minor axes respectively, prove that v^v^=Vi^{2v^^v-^). 3. The velocities of a point parallel to the axes of x and y are u + ay and v + ta'x respectively, where m, v, a> and a are constants j shew that its path is a conic section. 4. A particle moves in a plane under a constant force, the direction of which revolves with a uniform angular velocity ; find equations to give the coordinates of the particle at any time t. 5. A small ball is projected into the air ; shew that it appears to an observer standing at the point of projection to fall past a given vertical plane with constant velocity. 6. A man starts from a point and walks, or runs, with a constant velocity u along a straight road, taken as the axis of x. His dog starts at a distance a from 0, his starting point being on the axis of y which is perpendicular to Ox, and runs with constant velocity r in a direction which Accelerations parallel to fixed axes. Examples 43 is always towards his master. Shew that the equation to his path is If X = 1, shew that the path is the curve 2f«4--j = ^-alog-. [The tangent at any point P of the path of the dog meets Ox at the point where the man then is, so that -y^ = -^ — . Also -n = 7 • ax ut — X at K dx ^ . .'. —y-^ = ut — x = \s-x, ^ dy d r dx'X ^ds dx . . d^x . /, , (dx\^ -, 7. A particle is fastened to one end, B, of a light thread and rests on a horizontal plane ; the other end, A, of the thread is made to move on the plane with a given constant velocity in a given straight line ; shew that the path of the particle in space is a trochoid. [Shew that AB turns round A with a constant angular velocity.] 8. Two boats each move with a velocity v relative to the water and both cross a river of breadth a running with uniform velocity V. They start together, one boat crossing by the shortest path and the other in the shortest time. Shew that the difference between the times of arrival is either -1 according as F or t> is the greater. [The angle that v makes with V being B, the length of the path V'F^jhJHTTjTcos^ , ,, J- X- • « TV,^ IS a. — 4--i and the corresponding time is . . . Ine V sin 6 '■ ° V sin o condition for a minimum path gives (y cos ^ + F) ( Fcos ^ + v) = 0.] 9. A particle moves in one plane with an acceleration which is always perpendicular to a given line and is equal to /x-j- (distance from the line)^. Find its path for different velocities of projection. If it be projected from a point distant 2a from the given line with a velocity a/- parallel to the given line, shew that its path is a cycloid. 10. If a particle travel with horizontal velocity u and rise to such a height that the variation in gravity must be taken account of as far as small quantities of the first order, shew that the path is given by the equation _ a-y = {a-k) cosh |^>/| -^J> where 2a is the radius of the earth ; the axes of x and y being horizontal and vertical, and h, k being the coordinates of the vertex of the path. 44 Dynamics of a Particle 11. A particle moves in a plane with an acceleration which is parallel to the axis of y and varies as the distance from the axis of x ; shew that the equation to its path is of the iovm.y = Aa'+ Ba~^, when the acceleration is a repulsion. If the acceleration is attractive, then the equation is of the form y = A cos \ax + B\ 12. A particle moves under the action of a repulsive force perpen- dicular to a fixed plane and proportional to the distance from it. Find its path, and shew that, if its initial velocity be parallel to the plane and equal to that which it would have acquired in moving from rest on the plane to the point of projection, the path is a catenary. 13. A particle describes a rectangular hyj^erbola, the acceleration being directed from the centre ; shew that the angle 6 described about the centre in time t after leaving the vertex is given by the equation tan ^ = tanh(\//x<), where /i is acceleration at distance unity. 14. A particle moves freely in a semicircle under a force perpendicular to the bounding diameter ; shew that the force varies inversely as the cube of the ordinate to the diameter. 15. Shew that a rectangular hyperbola can be described by a particle under a force parallel to an asymptote which varies inversely as the cube of its distance from the other asymptote. 16. A particle is moving under the influence of an attractive force m —^ towards the axis of x. Shew that, if it be projected from the point (0, k) with component velocities U and V parallel to the axes of x and y, it will not strike the axis of x unless \i> V^k^, and that in this case the distance of the point of impact from the origin is -^ . 17. A plane has two smooth grooves at right angles cut in it, and two equal particles attracting one another according to the law of the inverse square are constrained to move one in each groove. Shew that the centre of mass of the two particles moves as if attracted to a centre of force placed at the intersection of the grooves and attracting as the inverse square of the distance. CHAPTER IV UNIPLANAR MOTION REFERRED TO POLAR COORDINATES CENTRAL FORCES 48. In the present chapter we shall consider cases of motion which are most readily solved by the use of polar coordinates. We must first obtain the velocities and accelera- tions of a moving point along and perpendicular to the radius vector drawn from a fixed pole. 49. Velocities and accelerations of a particle along and perpendicidar to the radius vector to it from a fixed origin 0. Let P be the position of the particle at time t, and Q its position at time t + At. Let XOP = e, XOQ = e + AO, OF = r, OQ = r + Ar, where OX is a fixed line. Draw QM perpendi- cular to OP. Let u, V be the velocities of the moving point along and perpendicular to OF. Then / = Lt = Lt At=Q Distance of particle measured along the line OP at time ^ + A^ — the similar distance at time t OM-OP At _dr 'It At J {r + Ar) cos, A6 - r - Lit T— A^=0 At small quantities above the first order being neglected, .....(1). 46 Dynamics of a Farticle Also Distance of particle measured perpendicular to the line OF at time ^ + A^ —the similar distance at time t = Lt QM-0 = Lt At=0 At (r + Ar) sin A^ A^ M=o Ai = Lt -T~^ , on neglecting small quantities of the second order, = r* -^- , in the limit (2). The velocities along and perpendicular to OP being u and V, the velocities along and perpendicular to OQ are u + Au and v + Av. Let the perpendicular to OQ at Q be produced to meet OF at L. Then the acceleration of the moving point along OF u+Au = Lt its velocity along OF at time t + At — its similar velocity at time t A< = Ol A^ (u + Au) cos Ad — (v + Av) sin A6 — u A^ + Au) . 1 - (y + Ay) . A ^ - w' ^^ \{u + Au).l-{v+Av).Ae-u'] = lt\. A^ J ' on neglecting squares and higher powers of A0, ^^Au-vAd du cW . ^, ,. ., = Lt r =—1 v^r. , 'i^n the limit, by (1) and (2). ■(3), Polar Coordinates 47 Also the acceleration of the moving point perpendicular to OP in the direction of 6 increasing its velocity perpendicular to OP at time t+ ^t — its similar velocity at time t Lt Af _ J Uu + A^) sin /^9 ->r(v-\- Ay) cos A^ - ^ "1 ~ A^=o L A^ J , r(tt + A?0 . A^ + (y + At') . 1 - v1 ^ifoL ^t J ' on neglecting squares and higher powers of A^, dd dv . ,, ,. .^ dr^dd d ( dd\ , ,,. , ,^, d* cfi+* rf^^~rd«L dt\ ^*^- Cor. If r = a, a constant quantity, so that the particle is describing a circle of centre and radius a, the quantity (3) = — a62 and (4) = a^, so that the accelerations of P along the tangent PQ and the radius PO are a^ and a6^. 50- The results of the previous article may also be obtained by resolving the velocities and accelerations along the axes of x and y in the directions of the radius vector and perpendicular to it. For since x = r cos d and y = rsva. 9, dx dr ^ ■ ndO\ ••• Tt=dt'''''-'''^'di\ ^^^ .(2). and | = |sin^ + rcos4^j Also d'^x d^r ■ ^ ^drde . ^ n /dd\^ . „d^0 ■M=dfi''''-^dtdt''^'-''''\Tt)-'''^'^^ ( , d2y d'h . „ ^drdd „ . „ fd9\^ „d^e The component velocity along OP ^f'^^'^ + if^^^^l' ^yW- and perpendicular to OP in the direction of 6 increasing it = |oos.-|.in,=,.^i,b,,X,. 48 Dynamics of a Particle The component acceleration along OP •(S)^^^(^). and perpendicular to OP it rf2.r . ^ ^drde d^ ^ ,„, rdtL dtj 51. By the use of Arts. 4 and 49 we can obtain the accelerations of a moving point referred to rectangular axes Ox and Oy, which are not fixed in space, but which revolve in any manner about the origin in their own plane. Let OA be a line fixed in space, and, at time t, let 6 be the inclination of Ox to OA . Let P be the moving point ; draw PM and PN perpendicular to Ox and Oy. By Art. 49 the velocities of the point M are -^ along OM and X -J- along MP, and the velocities of N are ~ along ON at ctt J a and y -j- along PN produced. Hence the velocity of P parallel to Ox = the velocity of N parallel to Ox + the velocity of P relative to N. = vel. of N parallel to Ox 4- the vel. of M along OM dO dx = ~'^di^di .(1). Revolving Axes 49 So the velocity of P parallel to Oy = vel. of M parallel to Oy 4- the vel. of P relative to M = vel. of M parallel to Oy + the vel. of N along ON -4-t ^; (^)-^^ Again, the accelerations of M are, by Art. 49, -^ ~ ^ ( 7/7 ) along MP, and the accelerations X dt\ dt of N are g-2,(^y along ON, and^|(2/^|) along PN jDroduced. Hence the acceleration of P parallel to Ox = acceleration of N parallel to Ox + acceleration of P relative to N = acceleration of N parallel to Ox + acceleration of M along OM 1 d ( ^de\ ^ d'x (de\^ .„. --y-dt{yw^dt^-''\jt) (^>- Also the acceleration of P parallel to Oy = acceleration of M parallel to Oy + acceleration of P relative to M = acceleration of M parallel to Oy + acceleration of N along ON 1 d ( ,de\^d'y fddy = ^dt[^dt)+-^-y[di) ^^>- Cor. In the particular case when the axes are revolving dd with a constant angular velocity oo, so that -^ = o), these component velocities become ^ - 2/(y along Ox, and -^ + xfjo along Oy ; 50 Dynamics of a Particle also the component accelerations are J - ^-0,^-20, 1 along Oo., and ^ - 2/0)" + 2ft, J" along Oy. 52. -Eo;. 1. 5/(ejt) «/ia< t/je ipath of a point P which possesses Uoo constant velocities u and v, the first of lohich is in a fixed direction and the second of which is perpendicular to the radius OP drawn from a fixed point 0, is a conic whose focus is and ivhose eccentricity is -. With the first figure of Art. 49, let u be the constant velocity along OX and V the constant velocity perpendicular to OP. Then we have dr n J ''^^ ■ ^ I dr u cos 6 ;j- =M cos 6, and -^ = v - it sin d. .: - — = -. — ;, . at at r ad v — us\nd .'. logr= -log fi;-M sin ^) + const., i.e. r{v-usin6)=const. = lv, if the path cut the axis of a; at a distance I. Therefore the path is I 1 — sin ^ V . u i.e. a conic section whose eccentricity is - . V Ex. 2. A smooth straight thin tube revolves with uniform angular velocity w in a vertical plane about one extremity which is fixed ; if at zero time the tube be horizontal, and a particle inside it be at a distance a from the fixed end, and be moving with velocity V along the tube, shew that its distance at time t is a cosh (wt) + ( ~j sinh {«() + . -g sin wt. At any time t let the tube have revolved round its fixed end through an angle wt from the horizontal line OX in an upward direction; let P, where OP=r, be the position of the particle then. By Art. 49, d'^r ^-^i - rw^ z= acceleration of P in the direction OP dt- = - ^ sin wt, since the tube is smooth. The solution of this equation is : ^{-g&mut) 2w' where A and B, and so L and BI, are arbitrary constants. Motion referred to polar coordinates 51 The initial conditions are that r = a and f = F" when « = 0. .-. a = L, and V=Mw + S- • 2(1} .: r = a cosh wt+ \ -^ sinh (ut)+ ~ sin ut. If R be the normal reaction of the tube, then gicos £ot=the acceleration perpendicular to OP = ^~ (r2w), by Art. 49, =2rw = 2au' sinh (wt) + (2Vu- g) cosh {ij}t)+g cos tat. EXAMPLES 1. A vessel steams at a constant speed v along a straight line whilst another vessel, steaming at a constant speed F, keeps the first always exactly abeam. Shew that the path of either vessel relatively to the other is a conic section of eccentricity y.. 2. A boat, which is rowed with constant velocity «, starts from a point A on the bank of a river which flows with a constant velocity nu ; it points always towards a point B on the other bank exactly opposite to A ; find the equation to the path of the boat. If n be unity, shew that the path is a parabola whose focus is B. 3. An insect crawls at a constant rate u along the spoke of a cartwheel, of radius a, the cart moving with velocity v. Find the acceleration along and perpendicular to the spoke. 4. The velocities of a particle along and perpendicular to the radius from a fixed origin are Xr and fi6 ; find the path and shew that the accelerations, along and perpendicular to the radius vector, are XV-^ and /x^ fx + ' 5. A point starts from the oi'igin in the direction of the initial line with velocity — and moves with constant angular velocity w about the (I) origin and with constant negative radial acceleration -/. Shew that the rate of growth of the radial velocity is never positive, but tends to the limit zero, and prove that the equation of the path is a)V=/(l — e^e). 6. A point P describes a curve with constant velocity and its angular velocity about a given fixed point varies inversely as the distance from ; shew that the curve is an equiangular spiral whose pole is 0, and that the acceleration of the point is along the normal at P and varies inversely as OP. 4—2 52 Dynamics of a Particle 7. A jJoint P describes an equiangular spiral with constant angular velocity about the pole ; shew that its acceleration varies as OF and is in a direction making with the tangent at P the same constant angle that OP makes. 8. A point moves in a given straight line on a plane with constant velocity T', and the plane moves with constant angular velocity t. 12. A particle falls from rest within a straight smooth tube which is revolving with uniform angular velocity w about a point in its length, being acted on by a force equal to nifi (distance) towards 0. Shew that the equation to its path in space is r = acosh| . / —^ - ^ or 7-=a cos |^^^— ^ (9L according as fita^. If /I =0)2, shew that the path is a circle. 13. A particle is placed at rest in a rough tube at a distance a trom Motion referred to polar coordinates 53 one end, and the tube starts rotating with a uniform angular velocity w about this end. Shew that the distance of the particle at time t is ae~"^- **'^* [cosh [mt . sec e) + sin e siuh {oit sec e)], where tan e is the coefficient of friction. 14. One end ^ of a rod is made to revolve with uniform angular velocity w in the circumference of a circle of radius a, whilst the rod itself revolves in the opposite direction about that end with the same angular velocity. Initially the rod coincides with a diameter and a smooth ring capable of sliding freely along the rod is placed at the centre of the circle. Shew that the distance of the ring from A at time t is - [4 cosh (o)^) + cos 1a>t\. [If be the centre of the circle and P, where AP=r, is the position of the ring at time t when both OA and AP have revolved through an angle ^, ( = &)0, in opposite directions, the acceleration of A is aw^ along AG and the acceleration of P relative to is r - rff^, by Art. 49, i.e. r — rofi. Hence the total acceleration of P along AP is r - rw^ + aco^ cos 20^, and this is zero since the ring is smooth.] 15. PQ is a tangent at § to a circle of radius a; PQ\s equal to p and makes an angle 6 with a fixed tangent to the circle ; shew that the accelerations of P along and perpendicular to QP are respectively Id P< [The accelerations of Q along and perpendicular to QP are ad and aQ'^ ; the accelerations of P relative to Q in these same directions are p-pe^ and ^^yh] 16. Two particles, of masses m and m', connected by an elastic string of natural length a, are placed in a smooth tube of small bore which is made to rotate about a fixed point in its length with angular velocity «. The coefficient of elasticity of the string is 'imm'aa)^^{m-\-m'). Shew that, if the particles are initially just at rest relative to the tube and the string is just taut, their distance apart at time t is 2a — a cos mt. 17. An elastic string is just stretched against a rough wheel of radius a, which is set in rotation with uniform angular velocity co. Shew that the string will leave the wheel and expand to a maximum radius r given by rHr-a)^Ma?^ r + a ~ 2n\ ' where M and X are the mass and modulus of elasticity of the string. 18. A uniform chain AB is placed in a straight tube GAB which re- volves in a horizontal plane, about the fixed point 0, with uniform angular p-pd^ + ad, and -^(p^^h 54 Dynamics of a Particle velocity o). Shew that the motion of the middle point of the chain is the same as would be the motion of a particle placed at this middle point, and that the tension of the chain at any point P is ^mai'.AP. PB, where m is the mass of a unit length of the chain. 53. A particle moves in a plane with an acceleration which is aliuays directed to a fixed point in the plane ; to obtain the differential equation of its path. Referred to as origin and a fixed straight line OX through as initial line, let the polar coordinates of P be (r, 6). If P be the acceleration of the particle directed towards 0, we have, by Art. 49, s-Kir=-^ «■ Also, since there is no acceleration perpendicular to OP, we have, by the same article, rdt\ dt) " '^-''• (2) gives r^ -^ = const. = /i (say) (3). .*. -— = — = hu-, if u be equal to - . dt r- ^ r rvu dr _ d /1\ _ _ 1 die __1 du dd _ _.du ^^"^ Tt~dt[uJ ^'It' u'Td'dt" dd' J d^r d / J du\ , d (du\ dO , „ , d'^w ^^^ w^ = dt[-^mr-^'dd[ddydt=-^'^'dd^' Central Forces 55 Hence equation (1) becomes _,,.^;_U,.=_P, ...(4). Again, if p be the perpendicular from the origin the tangent at P, we have upon 1 11 fdrV , (du\^ Hence, differentiating with respect to 6, we have 2 dp c. du du dhi ~p'dti~'^''dd'^'^ddd6-'' 1 dp r dhil du ( d-u\ ( 1 \ 1 dp dhi ' ' uYdr~'^'''^de'' Hence (4) gives P = — --^ ...(5). Equation (4) gives the path in terms of r and 6, and (5) gives the (p, r) equation of the path. 54. In every central orbit, the sectorial area traced out hy the radius vector to the centre of force increases uniformly per unit of time, and the linear velocity varies inversely as the perpendicular from the centre upon the tangent to the path. Let Q be the position of the moving particle at time t + A*, so that Z POQ = M and OQ = r+ Ar. The area POQ =\OP.OQ sin POQ = \r{r+ Ar) sin A^. Hence the rate of description of sectorial area sinA^ A6'"l -r ,\r (r + Ar) sin A^ ^ i Jjt T-T = lit ir(r+Ar).-^^.^ ^ ?•- j^ , in the limit, .de dt = the constant ^A by equation (3) of the last article. 56 Dynamics of a Particle The constant h is thus equal to twice the sectorial area described per unit of time. Again, the sectorial area POQ = in the limit h ■ PQ x per- pendicular from on PQ, and the rate of its description 1 As = lA ^. -r-x perpendicular from on PQ. Now, in the limit when Q is very close to P, -r- = the velocity v, and the perpendicular from on PQ = the perpendicular from on the tangent at P =^. .*. fi = v.p, I.e. v = -. P Hence, when a particle moves under a force to a fixed centre, its velocity at any point P of its path varies inversely as the perpendicular from the centre upon the tangent to the path at P. Since v = -, and in any curve P 55. A 'particle moves in an ellipse under a force which is always directed towards its focus; to find the law of force, and the velocity at any point of its path. The equation to an ellipse referred to its focus is I.e. u=j + jCosd (1). 1 + e cos ^ I I d'u e a Hence equation (4) of Art. 53 gives P = .v[^ + .]4«' (2). The acceleration therefore varies inversely as the square of Central Forces 57 the distance of the moving particle from the focus and, if it be h = N iJbl = w/j, X semi-latus-rectum (3). Also = f [1 + 2. cos ^ + ..] = ;. [2 1+^ - 1^] =4'-9'^>-«' W' where 2a is the major axis of the ellipse. It follows, since (4) depends only on the distance r, that the velocity at any point of the path depends only on the distance from the focus and that it is independent of the direction of the motion. It also follows that the velocity V of projection from any point whose distance from the focus is ro, must be less than 2a — , and that the a of the corresponding ellipse is given by Vn aj Periodic time. Since h is equal to twice the area described in a unit time, it follows, that if T be the time the particle takes to describe the whole arc of the ellipse, then ^h X T = area of the ellipse = -rrab. Also h = ^I fjL X semi-latus-rectum = a / fi — . Hence T = — 7 — = -^a- . fi V/* 56. Ex. Find the laio of force toicards the pole under lohich the curve r^=a"cosnd can be described. Here u"a"cos?i^ = l. Hence, taking the logarithmic differential, we obtain du y- = uta,nnd. da dhi du .'. jg2 = Jg tan nd + nu sec2 nd = ii [tan2 jid + n sec^ nd]. d?u •' ^ "^ " ^ " (" + ^) ^®*^^ ne = {n^-\) a2''a2'H-i, 58 Dynamics of a Particle Hence equation (4) of Art. 53 gives P={n + l);i2a2''!(2«+3, i.e. the curve can be described under a force to the pole varying inversely as tho (2n + 3)rd power of the distance. Particular Gases. I. Let n= -~, so that the equation to the curve is a 2a ^e 1 + cos ^ ' cos2- i.e. the curve is a parabola referred to its focus as pole. Here Pxi. II. Let w = - , so that the equation is r = - (1 + cos 6), which is a cardioid. Here ^ ^ Zi' III. Let n=l, so that the equation to the curve is r=a cos ^, i.e. a circle with a point on its circumference as pole. Here Pen —.. ,.8 IV. Let n=2, so that the curve is r2 = a2 cos 25, i.e. a lemniscate of 1 rT V. Let n= - 2, so that the curve is the rectangular hyperbola a'^ = r'^ cos 26, the centre being pole, and P oc - r, since in this case (n + 1) is negative. The force is therefore repulsive from the centre. EXAMPLES A particle describes the following curves under a force P to the pole, shew that the force is as stated : 1. Equiangular spiral ; P x 2. Lemniscate of Bernouilli ; P x-.. ,.7 3. Circle, pole on its circumference ; Pa:—.. ,.3- 5. r^cosnd = a^; Pccr^^'^. 1 4. - = e"«, n6, cosh n6, or sin n6 ; 5. r"cos«^ = a''; 6. r'^ = A cos, n6 -\- B a\n nO ; Px y.2» + 3 • 2»2a2 n2 ._ 1 7. r=asmnd: P~ , , 8. aM = tanh f-^) or cothf-^j; P oc W'2^ Central Forces 59 cosh ^-2 coshg + 2 _ „ 1^ "" cosh ^ + 1 cosh 6 — \' r* * „ „ cosh2^-l cosh 2(9 + 1 _ 1 lO- "'"'= cosh2^ + 2 " %osh2^-2 ' ^^,-T- 11. Find the law of force to an internal point under which a body will describe a circle. Shew that the hodograph of such motion is an ellipse. [Use formula (5) of Art. 53. The hodograph of the path of a moving point P is obtained thus : From a fixed point draw a straight line OQ parallel to, and proportional to, the velocity of P ; the locus of the point Q, for the diflerent positions of P, is the hodograph of the path of P.] 12. A particle of unit mass describes an equiangular spiral, of angle a, under a force which is always in a direction perpendicular to the straight line joining the particle to the pole of the spiral ; shew that the force is /ir^*®°"*~^, and that the rate of description of sectorial area about the pole is - Vm sin a . cos a . r^^^"^": 13. In an orbit described under a force to a centre the velocity at any point is inversely proportional to the distance of the point from the centre of force ; shew that the path is an equiangular spiral. 14. The velocity at any point of a central orbit is - th of what it would be for a circular orbit at the same distance ; shew that the central force varies as — -, — and that the equation to the orbit is X-i=a''*-icos{(?i2-i)^}. 57. Apses. An apse is a point in a central orbit at which the radius vector drawn from the centre of force to the moving particle has a maximum or minimum value. By the principles of the Differential Calculus u is a maximum or a minimum if -j^ is zero, and if the first differential coefficient ciu of u that does not vanish is of an even order. If p be the perpendicular from the centre of force upon the tangent to the path at any point whose distance is r from the origin, then 1 , fdiC\ When -7^ is zero, — = w^ = — , so that the perpendicular in the case of the apse is equal to the radius vector. Hence at an apse the particle is moving at right angles to the radius vector. 60 Dynamics of a Particle 58. When the central acceleration is a single-valued function of the distance {i.e. when the acceleration is a function of the distance only and is always the same at the same distance), every o.pse-line divides the orbit into two equal and similar portions and thus there can only he tivo apse-distances. Let ABC be a portion of the path having three consecutive apses A, B, and C and let be the centre of force. c Let V be the velocity of the / /"^---.^^ V particle at B. Then, if the velocity / ./ ^n^ of the particle were reversed at B, / / JXP it would describe the path BPA. / / ^'^ \ For, as the acceleration depends /,/' .^^ \ -^ on the distance from only, the /^^^'"'^ velocity, by equations (1) and (3) O "" — ■""--— ~^^ / of Art. 53, would depend only on ~~"^ — 7 A the distance from and not on the / direction of the motion. Again the original particle starting from B and the reversed particle, starting from B with equal velocity F, must describe similar paths. For the equations (1) and (3) of Art. 53, which do not depend on the direction of motion, shew that the value of r and d at any time t for the first particle {i.e. OP' and Z BOP') are equal to the same quantities at the same time t for the second particle {i.e. OP and Z BOP). Hence the curves BP'C and BPA are exactly the same ; either, by being rotated about the line OB, would give the other. Hence, since A and G are the points where the radius vector is perpendicular to the tangent, we have OA = 00. Similarly, if D were the next apse after C, we should have OB and OD equal, and so on. Thus there are only two different apse-distances. The angle between any two consecutive apsidal distances is called the apsidal angle. 59. When the central acceleration varies as some integral power of the distance, say /i^i'^ it is easily seen analytically that there are at most two apsidal distances. A])ses and Apsidal Distances 61 For the equation of motion is The particle is at an apse when ;to=^> ^^d then this equation gives n — 1 h^ „ ^ ^ 2 fi Whatever be the values of n or C this equation cannot have more than two changes of sign, and hence, by Descartes' Rule, it cannot have more than two positive roots. 60. A particle moves luith a central acceleration , ,. ^ — - ; •^ {distancef to find the path and to distinguish the cases. The equation (4) of Art. 53 becomes d^ , Ai • d^ii f/jb .\ ,_. ^ + « = ^,«, ,.e. ^-g, = (p-lj« (1). Case I. Let A* < fi, so that |^ — 1 is positive and equal to n^, say. d-u The equation (1) is -^^ = «^it, the general solution of which is, as in Art. 29, u = Ae-"^ + 5e-"« = L cosh nd + M sinh nO, where ^, 5, or X, M are arbitrary constants. This is a spiral curve with an infinite number of convolutions about the pole. In the particular case when A on B vanishes, it is an equiangular spiral. Case II. Let A^ = /x, so that the equation (1) becomes dO"- "• .*. u=Ae^B = A{d-a\ where A and a are arbitrary constants. 62 Dynamics of a Particle This represents a reciprocal spiral in general. In the particular case when A is zero, it is a circle. Case III. Let /i^ > ^, so that |^ - 1 is negative and equal to — n-, say, nnti'nn CW is f.ViprpfnrA dd The equation (1) is therefore ■j^^ = — n^u, the solution of which is u = A cos {rtO + B) = A cos n {d — a), where A and a are arbitrary constants. The apse is given hj 6 = a, u = A. 61. The equations (4) or (5) of Art. 53 will give the path when P is given and also the initial conditions of projection. Ex. 1. A particle moves with a central acceleration which varies inversely as the cube of the distance ; if it be projected from an apse at a distance a from the origin with a velocity which is sJ2 times the velocity for a circle of radius a, shew that the equation to its path is rcos-^ = a. Let the acceleration be yuw^. If Fi be the velocity in a circle of radius a with the same acceleration, then Fi2 a — i- = normal acceleration = — „ . a a^ •• ^^ -a2- Hence, if V be the velocity of projection in the required path, ^ a The differential equation of the path is, from equation (4) of Art. 53, dhi fiu^ ij. dT2+"=Pir2 = p"- Hence, multiplying by -j^ and integrating, we have da r'=i"'[(sy"']=f"'-^ -'»• The initial conditions give that when m = - , then -r- = 0, and v — ^_:^. a dd a Hence(l)gives l'i = l^^'^[_^^ = £. + 0. h^ = 2ij. and C-- 2a2' from equation (1) we have 1 m^" 2- _ ,, . 2a2* Central Forces. Examples 63 ■■■%-^lif^) (^>- 6 adn . , If be measured from the initial radius vector, then ^ = when u = -, and therefore 7=-sin-i(l): 2 ■ .Fir e-\ cos x/2j v/2 Hence the path is the curve r cos -^ = a. If we take the negative sign on the right hand side of (2), we obtain the same result. Ex. 2. A particle, subject to a force producing an acceleration /j. — g— towards the origin, is projected from the point {a, 0) with a velocity equal to the velocity from infinity at an angle cot~^ 2 with the initial line. Shew that the equation to the path is r = a(l + 2sin 6), and find the apsidal angle and distances. The " velocity from infinity " means the velocity that would be acquired by the particle in falling with the given acceleration from infinity to the point under consideration. Hence if this velocity be V we have, as in Art. 22, SO that F2= 1^ (1). The equation of motion of the particle is d^u n •••r'=2rH^^)j='^L3+2«"'j+^ (2)- If Pq be the perpendicular from the origin upon the initial direction of pro- jection, we havepo = * sin a, where cot a = 2, i.e. Pq=—-. Hence, initially, we have „ (duy 1 5 '"nw -^7==^ '''• Hence (2) gives, initially, from (1) and (3) 5^ ;i2 5 r 1 1 "1 ^ so that C = and h^ = ~. da From (2) we then have i.e. (^) =w2[2au + 3a2u2-l] = M2[au+l][3a«-l]. 64 Dynamics of a Fartide On putting u = - , this equation gives ^^)=(a + .)(3a-r), and hence 0= ' ^ ){Ba-r) Putting r = a + y, we have 6= I , ^ - =pin-i -^- J sj{a+r I: '2a ' 2a If we measure 6 from the initial radius vector, then ^ = when r = a, and hence 7 = 0. Therefore the path is r = a{l + 2 sin 6). dr Clearly 37^ = 0, i.e. we have an apse, when at/ . TT Sir 5ir , ^ = 2' T' T' ^*''- Hence the apsidal angle is w and the apsidal distances are equal to 3a and a, and the apses are both on the positive directions of the axis of y at distances 3a and a from the origin. The path is a lima9on and can be easily traced from its equation. EXAMPLES 1. A particle moves under a central repulsive force ] = "^^ — t , and is projected from an apse at a distance a with velocity V. Shew that the equation to the path is rcos^^ = a, and that the angle 6 described in time t is -tan-iK^ ;; , where ^0^ = ^ . 2. A particle moves with a central acceleration, = -rTT—^^^ r-j, and is projected from an apse at a distance a with a velocity equal to n times that which would be acquired in falling from infinity ; shew that the other apsidal distance is . . ^ si if- -I If 72 = 1, and the particle be projected in any direction, shew that the path is a circle passing through the centre of force. 1^ ;o 3. A particle, moving with a central acceleration (distance)' projected from an apse at a distance a with a velocity V ; shew that the path is rcosh — ^- — ^ — 6 \ = a, or t-cos j^ — - 6 =a, according as F is $ the velocity from infinity. Central Forces, Examples (55 4. A particle moving under a constant force from a centre is projected in a direction perpendicular to the radius vector with the velocity- acquired in falling to the point of projection from the centre. Shew that its path is 3 , ($)' :C0S-2. and that the particle will ultimately move in a straight line through the origin in the same way as if its path had always been this line. If the velocity of projection be double that in the previous case, shew that the path is ^ , _i /r-a 1 , _, /r-a / ■2a'^\ 5. A particle moves with a central acceleration /^ ( '/* + —_.- ) , being projected from an apse at a distance a with twice the velocity for a circle at that distance ; find the other apsidal distance, and shew that the equation to the path is -^ = tan-i(W3)-^tan-(x/g^). where ^^ = 3^, 6. A particle moves with a central acceleration /x ( ?' + -3 j being projected from an apse at distance a with a velocity 2'\/na; shew that it describes the curve r^ [2 + cos >^3d] = 3a^. 7. A particle moves with a central acceleration iji{r'-c*r), being projected from an apse at distance c with a velocity ./ -^ c^ ', shew that its path is the curve A-* +2/* = c*. 8. A particle moves under a central force /nX I3a^u*+8au^] ; it is projected from an apse at a distance a from the centre of force with velocity ^^lOA ; shew that the second apsidal distance is half the first, and that the equation to the path is 2r=a\ l + sech-y^ . 9. A particle describes an orbit with a central acceleration fiu^—Xu^, being projected from an apse at distance a with a velocity equal to that from infinity ; shew that its path is r = a cosh - , where ^i^ + 1 = Tlr^^ _ n X Prove also that it will be at distance r at the end of time 6G Dynamics of a Particle 10. lu a central orbit the force is fxv? {Z + ^i'^h'^); if the particle be projected at a distance a with a velocity /—^ in a direction making tan"^ - with the radius, shew that the equation to the path is r = a tan 6. 11. A particle moves under a force m^ {3«?t* - 2 (a^-b^) u^}, a>b, and is projected from an apse at a distance a + b with velocity '\/fi-7-{a + b) ; shew that its orbit is r = a + b cosd. 12. A particle moves with a central acceleration \^ (8aii^ + ahc^) ; it is projected with velocity 9X from an apse at a distance f from the origin ; shew that the equation to its path is 1 /au + 5 J3\/ au-3 '^°^J6' 13. A particle, subject to a central force per unit of mass equal to fi{2{a^ + b'^)ii^-3a^'^vJ], is projected at the distance a with a velocity — - in a direction at right angles to the initial distance ; shew that the path is the curve 7*2 =a2 cos^ d+b"^ sin^ 6. 14. A particle moves with a central acceleration fi (u^--—ii~\ ; /25 it is projected at a distance a with a velocity */ -rr times the velocity for 4 a circle at that distance and at an inclination tan~' ^ to the radius vector. Shew that its path is the curve 15. A particle is acted on by a central repulsive force which varies as the nth power of the distance ; if the velocity at any point of the path be equal to that which would be acquired in falling from the centre to the point, shew that the equation to the path is of the form 71+3 , „ r ^ cos — ^ 6 = const. 16. An elastic string, of natural length I, is tied to a particle at one end and is fixed at its other end to a point in a smooth horizontal table. The particle can move on the table and initially is at rest with the string straight and unstretched. A blow (which, if directed along the string would make the particle oscillate to a maximum distance 2l from the fixed end) is given to the particle in a direction inclined at an angle a to the string. Prove that the maximum length of the string during the ensuing motion is given by the greatest root of the equation Central Forces. Examples 67 17. A particle of mass m is attached to a fixed point by an elastic string of natural length a, the coefficient of elasticity being nmg ; it is projected from an apse at a distance a with velocity ^2pgh ; shew that the other apsidal distance is given by the equation nr'^ (r—a) — 2pha {r + a) = 0. 18. A particle acted on by a repulsive central force /xr -=- (r^ - Qc^)- is projected from an apse at a distance c with velocity \/^ J shew that it will describe a three-cusped hypocycloid and that the time to the cusp [Use equation (5) of Art. 53, and we have Sp'-=Qc^-r'^. Also hdt=p .ds=pdr . , giving ht= I -^-a/ -^ 5-. Vr^ - p2 7 c d V r- - c- To integrate, put r^ = c- + 8c^cos^(p.] 19. Find the path described about a fixed centre of force by a particle, when the acceleration toward the centre is of the form ^ + ^, in terms of J.2 ^3' the velocity V at an apse whose distance is a from the centre of force. 20. Shew that the only law for a central attraction, for which the velocity in a circle at any distance is equal to the velocity acquired in falling from infinity to that distance, is that of the inverse cube. 21. A particle moves in a curve under a central attraction so that its velocity at any point is equal to that in a circle at the same distance and under the same attraction ; shew that the law of force is that of the inverse cube, and that the path is an equiangular spiral. 22. A particle moves under a central force m/x-r (distance)" (where n>l but not = 3). If it be projected at a distance ^ in a direction making an angle ^ with the initial radius vector with a velocity equal to that due to a fall from infinity, shew that the equation to the path is w-3 n-3 / _.) X If n>3 shew that the maximum distance from the centre is R cosed"' ~^, 3, and if ?i<3 then the particle goes to infinity. 23. A particle moves with central acceleration fiu^ + vu^ and the velocity of projection at distance R is V; shew that the particle will ultimately go ofi" to infinity if ^^> -/} + ^i- 5—2 68 Dynamics of a Particle 24. A particle is projected from an apse at a distance a with a velocity —^ and moves with a central attraction equal to ^(?i — l)a"-3r-" + Xr~3j where ?i>3, per unit of mass ; shew that it will arrive at the centre of force in time 25. In a central orbit if P=^iu'^{cu + Q,os,d)~'^, shew that the path is one of the conies (cm + cos 6Y = a-\-b cos {±6 + a), 26. A particle, of mass m, moves under an attractive force to the pole equal to -~ sin^ 6. It is projected with velocity a/ -^ from an apse at a distance a. Shew that the equation to the orbit is r(l-4-cos''6) = 2a, and that the time of a complete revolution is (3a) - x —^- . 27. If a particle move with a central acceleration ^ (1 + F sin^ 6) ~ ^, find the orbit and interpret the result geometrically. [Multiplying the equation of motion, h^ {u + u)=iji{l+k^sm^ 6)~ "^f by cos 6 and sin d in succession and integrating, we have h^ {u cosd + u sin 6) =fi sin (9 (1 +F siu^ 6) ' ^+A, and h^ {u sin ^- mcos ^)= -ja cos ^ (1 +F sin2 e)~^^i\ +k'^) + B. EUminating u, we have A2tt=/i(H-Fsin2^)2^(l + P) + ^sin^-5cos^.] 28. A particle moves in a field of force whose potential is fxr ~ - cos 6 and it is projected at distance a perpendicular to the initial line with 2 velocity - -Jy. ; shew that the orbit described is = a sec J 2 log tan -— — . 29. A particle is describing a circle of radius a under the action of a constant force X to the centre when suddenly the force is altered to \+fiii\n7it, where /x is small compared with X and t is reckoned from the instant of change. Shew that at any subsequent time t the distance of the particle from the centre of force is (^Vi:^^"(V?)-""^' 3X - an^ I What is the character of the motion if 2>\ = an'^ \ Stability of orbits 69 [Use equations (1) and (2) of Art. 53 ; the second gives r^ =■ sj\a?, and the first then becomes .. Xa3 . . ^ r — ^ = - A — /I sin 7it. Put r = a + ^ where ^ is small, and neglect squares of ^.] 62. A particle describes a path which is nearly a circle about a centre of force (= fxu^) ; find the condition that this may be a stable motion. The equation of motion is d^u /"- „ „ /-IN This will give a circle of radius - if h'^ = fic^''^ (2). Suppose the particle to be slightly displaced from the circular path in such a way that h remains unaltered (for example, suppose it is given a small additional velocity in a direction away from the centre of force by means of a blow, the perpendicular velocity being unaltered). In (1) put u = c + X, where x is small ; then it gives d^x , , _(c + ^)"-- Neglecting squares and higher powers of x, i.e. assuming that X is always small, we have d'^x ._ . __ = _(3-n)^. If n be < 3, so that 3 — n is positive, this gives a? = ^ cos [\/3 -nd -Y B]. If n be > 3, so that ?i — 3 is positive, the solution is x = A leV'^^s 6 + Bye--J~>^^ ^ so that X continually increases as 6 increases ; hence x is not always small and the orbit does not continue to be nearly circular. If n < 3, the approximation to the path is u = c + Acosl'^S-nd + B] (4). 70 Dynamics of a Particle c^n^tCl^■.^ c\r\ The apsidal distances are given by the equation ;77i = 0, i.e. by = sin[\/3-?i^ + 5]. The solutions of this equation are a series of angles, the difference between their successive values beinef , . This is therefore the apsidal angle of the path. If w = 3, this apsidal angle is infinite. In this case it would be found that the motion is unstable, the particle departing from the circular path altogether and describing a spiral curve. The maximum and minimum values of it, in the case n < 3, are c+A and c — A, so that the motion is included between these values. 63. The general case may be considered in the same manner. Let the central acceleration be ^ (u). The equations (1) and (2) then become and h'c' = fx,i> (o) (6). Also (3) is now de'^^'^''''J{cy {c±xy = c-1x + X -^tM , neglecting squares of x. (p{c) •• dd' { 4>o and the motion is stable only if In this case the apsidal angle is Central and transversal accelerations 1 64 If, in addition to the central acceleration P, we have an acceleration T perpendicular to F, the equations of motion are dt' and dt dd = -P Let r dt h. In this Then (2) gives T dh = ''dt dr drdd • • dt'- -dddt rdt\ dt In this case h is not a constant. dh dd , ,dh dd dt dd 1 du 'ic-'de hu- du dd and d'^r _ cZ / , du\ dd _ i ^[i ^"'^ ^^ ^1 di^~~dd\ddjTt'~~^"' Y'M'^dddd] = - hhC- d'-a dd" Tdji udd' by equation (3). Therefore (1) gives ,, „dhi Tdu ,„ 3 p dd^ u dd p_Tclu d-u _ u dd This may also be written in the form P_Tdu u- t<^ dd dhi d dd _ dd from equation (3) + u d ,,„, ^, dh d-d^^'^ = ^^'dd IT .(2> .(3). .(4). EXAMPLES 1. One end of an elastic string, of unstretched length a, is tied to a point on the top of a smooth table, and a particle attached to the other end can move freely on the table. If the path be nearly a circle of radius b, shew that its apsidal angle is approximately tt / , _- . 72 Dynamics of a Particle 2. If the nearly circular orbit of a particle be ^''^ (a'" ""*-?■"»- 2) = Jni^ shew that the apsidal angle is -j- nearly. [Using equation (5) of Art. 53 we see that P varies as r'""^; the result then follows from Art. 62.] 3. A particle moves with a central acceleration ,-(^ - -jj ; shew that the apsidal angle is 7r-=- */ 1 +T2> where - is the constant areal velocity. 4. Find the apsidal angle in a nearly circular orbit under the central force ar»' + 6>'". 5. Assuming that the moon is acted on by a force . ,. J^ r-j to the earth and that the effect of the sun's disturbing force is to cause a force m? X distance from the earth to the moon, shew that, the orbit being nearly circular, the apsidal angle is iri\+^~A nearly, where — is a mean lunar month, and cubes of m are neglected. 6. A particle is moving in an approximately circular orbit under the action of a central force ^ and a small constant tangential retardation/; 3 f shew that, if the mean distance be a, then 6 = nt + ^ - t\ squares of / being neglected. 7. T\uo particles of masses M and in are attached to the ends of an inextensible string which passes through a smooth fixed ring, the whole resting on a horizontal table. The particle m being projected at right angles to the string, shew that its path is \_VnVVJj^\- The tension of the string being T, the equations of motion are ^_ fdd\-^_T df ^\dt) m ^^^' -rdtV'^tr^ ^^)' and ^^^^-^'^^-J ^•^- Central Forces. Examples 73 (2) gives r^e=h (4), and then (1) and (3) give fH \r = -^. r^-d^ since r is zero initially, when r=^a. This equation and (4) give / m f adr . a ~ • '• ^^ \ rr= — /-i a = C03-1 - + (7, and C vanishes if 6 be measured from the initial radius vector. .-. a = rcos[^^^^^]isthepath. 8. Two masses i/, m are connected by a string which passes through a hole in a smooth horizontal plane, the mass m hanging vertically. Shew that M describes on the plane a curve whose diflferential equation is / m\ d'^u _ mg 1 \-^Jl)'d&^^^~~M A2^- Prove also that the tension of the string is 9. In the previous question if m=M, and the latter be projected on the plane with velocity a/ — ^ from an apse at a distance a, shew that the former will rise through a distance a. 10. Two particles, of masses M and m, are connected by a light string ; the string passes through a small hole in the table, in hangs vertically, and M describes a curve on the table which is very nearly a circle whose centre is the hole ; shew that the apsidal angle of the orbit of M is it * / ,.- ■ . 11. A particle of mass m can move on a smooth horizontal table. It is attached to a string which passes through a smooth hole in the table, goes under a small smooth pulley of mass M and is attached to a point in the under side of the table so that the parts of the string hang vertically. If the motion be slightly disturbed, when the mass m is describing a circle uniformly, so that the angular momentum is un- changed, shew that the apsidal angle is it a/ -t^ — • 74 Dynamics of a Particle 12. Two particles on a smooth horizontal table are attached by an elastic string, of natural length a, and are initially at rest at a distance a apart. One particle is projected at right angles to the string. Shew that if the greatest length of the string during the subsequent motion be 2a, then the velocity of projection is */ -^— , where m is the harmonic mean between the masses of the particles and X is the modulus of elasticity of the string. [Let the two particles be A and B of masses M and M', of which B is the one that is projected. When the connecting string is of length r and therefore of tension T, such that T=\ , the acceleration of .4 a T . T is Yf along AB, and that of B is — , along BA. To get the relative motion we give to both B and A an acceleration equal and opposite to that of A. The latter is then "reduced to rest" and the acceleration of B relative to A is along BA and _T T^_2 r-a _ 2X l-ait ~ M M' m a ~ ma u The equation to the relative path of B is now (Pu _ 2\ \ — au Integrate and introduce the conditions that the particle is projected trom an apse at a distance a with velocity V, The fact that there is another apse at a distance 2a determines F.] CHAPTER V UNIPLANAR MOTION WHEN THE ACCELERATION IS CENTRAL AND VARYING AS THE INVERSE SQUARE OF THE DISTANCE 65. In the present chapter we shall consider the motion when the central acceleration follows the Newtonian Law of Attrac- tion. This law may be expressed as follows ; between every two particles, of masses wij and m^ placed at a distance r apart, the mutual attraction is 7 — ~ units of force, where 7 is a constant, depending on the units of mass and length employed, and known as the constant of gravitation. If the masses be measured in pounds, and the length in feet, the value of 7 is 1*05 x 10~' approximately, and the attraction is expressed in poundals. If the masses be measured in grammes, and the length in centimetres, the value of 7 is 6'66 x 10~^ approximately, and the attraction is expressed in dynes. 66. A particle moves in a -path so that its acceleration is always directed to a fixed point and is equal to jj^. x; ; to sheiu that its path is a conic section and to distinguish between the three cases that arise. 76 Dynamics of a Particle When P= -, the equation (5) of Art. 53 becomes p'dr r' ^ ^' Integrating we have, by Art. 54, .^ = ^ = 2^ + (7 :....(2). jf- r Now the (p, r) equation of an ellipse and hyperbola, re- ferred to a focus, are respectively -= — -1, and - = — +1 (3), p- r p> r where 2a and 26 are the transverse and conjugate axes. Hence, when G is negative, (2) is an ellipse; when G is positive, it is a hyperbola. Also when (7=0, (2) becomes — = constant, and this is the r (p, r) equation of a parabola referred to its focus. Hence (2) always represents a conic section, whose focus is at the centre of force, and which is an ellipse "I negative] parabola > according as G is zero or hyperbola] or positive i.e. according as v^ | -^ , i.e. according as the square of the 2.-1 velocity at any point P is = ^ , where S is the focus. Again, comparing equations (2) and (3), we have, in the case of the ellipse, h^^/^^G_ b' a - 1 * .'. h = Kf /J, - = "J/xx semi-latus-rectum, and G = . Hence, in the case of the ellipse, v^ = fi ( ~^ — -] (4). \v aj (2 \\ So, for the hyperbola, v- = yu, ( - + - 1 , 2a and, for the parabola, v- = - . Central Forces. Law of the Inverse Square 11 It will be noted that in each case the velocity at any point does net depend on the direction of the velocity. Since h is twice the area described in the unit of time (Art. 54), therefore, if T be the time of describing the ellipse, we have ^ area of the ellipse irah Itr % so that the square of the periodic time varies as the cube of the major axis. Cor. 1. If a particle be projected at a distance R with velocity V in any direction the path is an ellipse, parabola or hyperbola, according as F^ < = > ^ . Now the square of the velocity that would be acquired in falling from infinity to the distance R, by Art. 81. -/:(-r^)-=HM- Hence the path is an ellipse, parabola or hyperbola accord- ing as the velocity at any point is < = > that acquired in falling from infinity to the point. Cor. 2. The velocity Vi for the description of a circle of radius R is given by -^ = normal acceleration = -^, , so that Vi^ = ^ . li JrC' K _. • V — '^^'^ocity from infinity 67. In the previous article the branch of the hyperbola described is the one nearest the centre of force. If the central acceleration be from the centre and if it vary as the inverse square of the distance, the further branch is described. For in this case the equation of motion is ^^£ = -ii. .-X^-^ + C (1). 78 Dynamics of a Particle Now the (p, r) equation of the further branch of a hyper- bola is • ^ - 1 _ ^^ f~ T' h- a and this always agrees with (1) provided that — = - = (7, so h'~ /I that h = v/i X semi-latus-rectum, and v' P 68. Construction of the orbit given the point of projection and the direction and magnitude of the velocity of projection. Let 8 be the centre of attraction, P the point of projection, TPT the direction of projection, and V the velocity of pro- jection. Case I. Let F^ < o^ ; then, by Art. QQ, the path is an ellipse whose major axis 2a is given by the equation V^ = ,j,(%-^, where R = SP, so that 2a = ■ ^^'^ M a) ' ' 2ix - V'U ■ Draw PS', so that PS' and PS are on the same side of TPT', making Z T'PS' = Z TPS, and take PS' =2a-SP = 2a-R = 2fi-V'R' Then S' is the second focus and the elliptic path is therefore known. Case II. Let V^ -■ ^^ , so that the path is a parabola. Draw the direction PS' as in Case I; in this case this is the direction of the axis of the parabola. Draw SU parallel to PS' to meet TPT in U; draw >SF perpendicular to TPT and YA perpendicular to SU. Then A is the vertex of the required parabola whose focus is S, and the curve can be constructed. SY- 2«o" The semi-latus-rectum = 2SA = 2 . -^^^ = -^ , where «o is the perpendicular from S on the direction of projection. Case III. Let V^ > ^^ , so that the path is a hyperbola of transverse axis 2a given by the equation and hence 2a = -,r., » — s~ • V K — Zfjb (l + a)' Kepler's Laivs 79 In this case PS' lies on the opposite side of TPT' from PS, such that Z TPS = z TPS', and S'P - SP = 2a, so that The path can then be constructed, since S' is the second focus. 69. Kepler's Laws. The astronomer Kepler, after many years of patient labour, discovered three laws connecting the motions of the various planets about the sun. They are : 1. Each planet describes an ellipse having the sun in one of its foci. 2. The areas described by the radii draum from the planet to the sun are, in the same orbit, proportional to the times of describing them. 3. The squares of the periodic times of the various planets are pt'^'opoi'tional to the cubes of the major axes of their orbits. 70. From the second law we conclude, by Art. 54, that the acceleration of each planet, and therefore the force on it, is directed towards the Sun. From the first law it follows, by Art. 55 or Art. 66, that the acceleration of each planet varies inversely as the square of its distance from the Sun. From the third law it follows, since from Art. 66 we have T■^ = ^-^.a^ H- that the absolute acceleration /x {i.e. the acceleration at unit distance from the Sun) is the same for all planets. Laws similar to those of Kepler have been found to hold for the planets and their satellites. It follows from the foregoing considerations that we may assume Newton's Law of Gravitation to be true throughout the Solar System. 71. Kepler's Laws were obtained by him, by a process of continually trying hypotheses until he found one that was suitable ; he started with the observations made and recorded for many years by Tycho Brahe, a Dane, Avho lived from A.P. 1546 to 1601. 80 Dynamics of a Particle The first and second laws were enunciated by Kepler in 1609 in his book on the motion of the planet Mars. The third law was announced ten years later in a book entitled On the Harmonies of the World. The explanation of these laws was given by Newton in his Principia published in the year 1687. 72. Kepler's third law, in the form given in Art. 69, is only true on the supposition that the Sun is fixed, or that the mass of the planet is neglected in comparison with that of the Sun. A more accurate form is obtained in the following manner. Let S be the mass of the Sun, P that of any of its planets, and 7 the constant of gravitation. The force of attraction between the two is thus 7 . — V- , where r is the distance r- between the Sun and planet at any instant. The acceleration of the planet is then a ( = ^ j towards the Sun, and that of the Sun is /3 ( = -^ j towards the planet. To obtain the acceleration of the planet relative to the Sun we must give to both an acceleration ^ along the line PS. The acceleration of the Sun is then zero and that of the planet is a + /3 along PS. If, in addition, we give to each a velocity equal and opposite to that of the Sun we have the motion of P relative to the Sun supposed to be at rest. The relative acceleration of the planet with respect to the Sun then=a + ^ = :yl^\ Hence the /x. of Art. 66 is 7 {S + P), and, as in that article we then have T JliS + P) Kepler's Laws 81 If 7\ be the time of revolution and a^ the semi-major axis of the relative path of another planet Pj, we have similarly rp ^ 27r I >S+P T^_^ JjiS + Pr)"^' "S + P^■T,^ a,^' T- . a? Since Kepler's Law, that 777- varies as ^, is very ^ 1\^ a/ ■^ Si 4- P approximately true, it follows that -^ — ^ is very nearly unity, and hence that P and Pj are either very nearly equal or very small compared with S. But it is known that the masses of the planets are very different ; hence they must be very small compared with that of the Sun. 73. The corrected formula of the last article may be used to give an approximate value to the ratio of the mass of a planet to that of the Sun in the case where the planet has a small satellite, whose periodic time and mean distance from the planet are known. In the case of the satellite the attraction of the planet is the force which for all practical purposes determines its path. If P be the mass of the planet and D its mean distance from the Sun, then, as in the previous article, T= , ^"^ D\ ^y{S + P) Similarly, if p be the mass of the satellite, d its mean distance from the planet, and t its periodic time, then 27r 3 ^ S + P T'_D' ~\/j(P + p) ' " P+p f- ~d'- The quantities T, t, D and d being known, this gives a value for ^ . P -^p As a numerical example take the case of the Earth E and the Moon m. Now T = 365i days, « = 27^ days, D= 93,000,000 miles, and d = 240,000 miles, all the values being approximate. L. D. 6 82 Dynamics of a Particle Therefore .5 + 7? = 325900 times the sum of the masses of the Earth and Moon. Also m=^E nearly. .-. 5f = 330000 E nearly. This is a fairly close approximation to the accurate result. If the Sun be assumed to be a sphere of radius 440,000 miles and mean density n times that of the Earth, assumed to be a sphere of radius 4000 miles, this gives n X (440000)3 = 330000 x (4000)3. 330000 330 ,1 .'. 71= = = about - . 1103 1331 ''"""" 4* Hence the mean density of the Sun = - that of the Earth =- x 5 . 527 = about 1*4 grammes per cub. cm., so that the mean density of the Sun is nearly half as much again as that of water. 74. It is not necessary to know the mean distance and periodic time of the planet P in order to determine its mass, or rather the sum of its mass and that of its satelHte. For if E and m be the masses of the Earth and Moon, R the distance of the Earth from the Sun, r that of the Moon from the Earth, if Y denote a year and y the mean lunar month, then we have F= ,-^ ^ R^ (1), 9r 3 y= ,^ - - .r^ (2). ^ ^ry (E+ 711) Also, as in the last article, t= , ^„ - -d' (3). From (1) and (3), From (2) and (3), (P+^)| = (^+m).^ (5). Equation (4) gives the ratio oi P + p to S + E. Equation (5) gives the ratio of P + _p to ^ + m. Law of the Inverse Square. Examples 83 EXAMPLES 1. Shew that the velocity of a particle moving in an ellipse about a centre of force in the focus is compounded of two constant velocities, ^ perpendicular to the radius and ^ perpendicular to the major axis. 2. A particle describes an ellipse about a centre of force at the focus ; shew that, at any point of its path, the angular velocity about the other focus varies inversely as the square of the normal at the point. 3. A particle moves with a central acceleration = ,. f" — ^ ; it is projected with velocity V at a distance R. Shew that its path is a rectangular hyperbola if the angle of projection is . _i fi VR (--1/ 4. A particle describes an ellipse under a force ,. /^ -^ towards the focus ; if it was projected with velocity Ffrom a point distant r from the centre of force, shew that its periodic time is 27rr2_ r2-|-f 5. If the velocity of the Earth at any point of its orbit, assumed to be circular, were increased by about one-half, prove that it would describe a parabola about the Sun as focus. Shew also that, if a body were projected from the Earth with a velocity exceeding 7 miles per second, it will not return to the Earth and may even leave the Solar System. 6. A particle is projected from the Earth's surface with velocity v ; shew that, if the diminution of gravity be taken into account, but the resistance of the air neglected, the path is an ellipse of major axis —-^ — ?, where a is the Earth's radius. "zga — v'' 7. Shew that an unresisted particle falling to the Earth's surface from a great distance would acquire a velocity '^'iga, where a is the Earth's radius. Prove that the velocity acquired by a particle similarly falling into the Sun is to the Earth's velocity in the square root of the ratio of the diameter of the Earth's orbit to the radius of the Sun. 6—2 84 Dynamics of a F article 8. If a planet were suddenly stopped in its orbit, supposed circular, shew that it would fall into the Sun in a time which is ^ times the 8 period of the planet's revolution. 9. The eccentricity of the Earth's orbit round the Sun is — ; shew 60 that the Earth's distance from the Sun exceeds the length of the semi- major axis of the orbit during about 2 days more than half the year. 10. The mean distance of Mars from the Sun being 1-524 times that of the Earth, find the time of revolution of Mars about the Sun. 11. The time of revolution of Mars about the Sun is 687 days and his mean distance 141| millions of miles ; the distance of the Satellite Deimos from Mars is 14,600 miles and his time of revolution 30 hrs. 18 mins. ; shew that the mass of the Sun is a little more than three million times that of Mars. 12. The time of revolution of Jupiter about the Sun is 11-86 years and his mean distance 483 millions of miles ; the distance of his first satellite is 261,000 miles, and his time of revolution 1 day 18^ hrs. ; shew that the mass of Jupiter is a little less than one-thousandth of that of the Sun. 13. The outer satellite of Jvipiter revolves in 16| days approximately, and its distance from the planet's centre is 26j radii of the latter. The last discovered satellite revolves in 12 hours nearly ; find its distance from the planet's centre. Find also the approximate ratio of Jupiter's mean density to that of the Earth, assuming that the Moon's distance is 60 times the Earth's radius and that her siderial period is 27J days nearly. [Use equations (2) and (3) of Art. 74, and neglect m in comparison with E, and p in comparison with P.] 14. A planet is describing an ellipse about the Sun as focus ; shew that its velocity away from the Sun is greatest when the radius vector to the planet is at right angles to the major axis of the path, and that it then is , , where 2a is the major axis, e the eccentricity, and T the periodic time. 75. To find the time of description of a given arc of an elliptic orhit starting from the nearer end of the major aocis. The equation r^-j- =h of Art. 53 gives ht= j\"-dd = \\.^ — a^^.de (1) Jo Jo(l +ecos^)2 ^ ' Time of describing any arc If \ be > 1, we have 85 dO e \{\ +tan2-j + (l-tan2 I) /X + 1 , 6/ , . = , ^ tan-^ Itan ^ . /^^^ — ^ I . Hence, by differentiation with respect to the constant \, we have 2\ j(A, + cos^)2 2 dd = - tan" (V-l)S tan- ['""IVx-Tl] A. -}- i 2! 2>" . , r . ^ A -11 sin d 1 = — ■ ^ tan~^ tan ^ a / ^ r + ;^ ?i x r^; — r (\--l)* L 2V X + lJ \ + cos^ X^-l In this equation, putting A. = - , we have j(l + ec dd cos ey 2 ^ T. ^ /l^^l « ^ = 5 tan~^ tan tz\/ q — z , :; — (1 - e^) Hence substituting in equation (1) we have. sin ^ e cos ^ since Y = -r= = -;= = }. a^(l-e^)^ L(i-e^)2 = — = I 2 tan~^ I tan fvrr e sin B ~|^ 1 — e-1 + ecos ^Jo vr 1 + e cos f^ ■ 86 Dynamics of a Particle [An alternative method of obtaining this result will be given in Art. 82.] 76. To find the time similarly for a hyperbolic orbit. Here e is > 1, so that we put e = - where X < 1. Hence sec^ 2 ^^ l+\-{l-\) tan"- 1 setf 2 ^^ 1 + X , J 1 /i-x, Vi-x + tan /1 + A, tan V 1 + X + V 1 - A. tan ; =-. log ^1-^' "vrTx-Vl-Xtan| Differentiating with respect to X, we have, after some simplification, _[_dO__ (1-v)^ Vl + X + \/l - X tan ; Vl + X - Vl - X tan iin^ 1 - X^ ■ X + cos I Replacing X by - , we have •'(1 + d0 ; cos ey {e'' - 1)^ V e + 1 + V e - 1 tan t; 01 Ve + 1 - Ve - 1 tan + — sin 6 1 ■ 1 + e cos ^ ' Time of describing aiiy arc 87 Hence, since in this case ,a h 'Jim ^fj, * the equation (1) of the last article gives '\/fi . ^ Ve + 1 + A/e - 1 tan _ TZl ^15i_ _ log. ? 1+ecos^ ° . . ve + 1 — ve— 1 tan- 77. In the case of a paraholic orbit to find the corresponding time. The equation to the parabola is r = = -^ , where 2d is ^ ^ 1 + cos ^ ' the latus-rectum and is measured from the axis. Hence the equation (3) of Art. 53 gives h.t=(rW=[j^-^ J J (1 + coi '(l+cos6')^^^" . ht [6 de ir J j.a But h^^^^ d? d' d« if a be the apsidal distance. 78. Motion of a projectile, vacations of gravity being taken into consideration but the resistance of the air being neglected. The attraction of the Earth at a point outside it at a distance r irom the centre is •^. Hence the path of a projectile in vacuo is one of the cases of Art. QQ, one of the foci of the path described being at the centre of the Earth. 88 Dynamics of a Particle If R be the radius of the Earth, then ~ R" the value of gravity at the surface of the Earth = g, so that yu, = gR-. The path of a projectile which starts from a point on the Earth's surface is therefore an ellipse, parabola, or hyperbola according as V^ = ~ , i.e. according as V^ = 2gR. 79. The maximum range of a particle starting from the Earth's surface with a given velocity may be obtained as follows : Let S be the centre of the Earth and P the point of projection. Let K be the point vertically above P to which the velocity, V, of projection is due, so that, by Art. 31, we have where R is the radius of the Earth and PK is h. (1), Hence, by equation (4) of Art. 66, By comparing this with equation (1) we have PH=h, so that the locus of the second focus is, for a constant velocity of projection, a circle whose centre is P and radius h. It follows that the major axis of the path is SP + PH or SK. Planetary Motion 89 The ellipse, whose foci are S and H, meets a plane LPM, passing through the point of projection, in a point Q, such that SQ + QH—SK. Hence, if SQ meet in T the circle whose centre is S and radius SK, we have QT=QH. Since there is, in general, another point, H', on the circle of foci equidistant with H from Q, we have, in general, two paths for a given range. The greatest range on the plane LPM is clearly Pq where qt equals qO. Hence Sq + qP=Sq + qO + OP = Sq + qt + PK^SK+ PK. Therefore q lies on an ellipse, whose foci are the centre of the Earth and the point of projection, and which passes through K. Hence we obtain the maximiun range. 80. Suppose that the path described by a planet P about the Sun S is the ellipse of the figure. Draw PN perpendicular to the major axis and produce it to meet the auxiliary circle in Q. Let G be the centre. The points A and A' are called respectively the Perihelion and Aphelion of the path of the planet. The angle ASP is called the True Anomaly and the angle ACQ the Eccentric Anomaly. In the case of any of the planets the eccentricity of the path is small, being never as large as ■1 except in the case of Mercury when it is "2 ; the foci of the path are therefore very near C, the ellipse differs little in actual shape from the auxiliary circle, and hence the difference between the ,True and Eccentric Anomaly is a small quantity. 27r If — be the time of a complete revolution of the planet, so that n is its mean angular velocity, then nt is defined to be the Mean Anomaly and n is the Mean Motion. It is clear 90 Dynamics of a Particle therefore that nt would be the Anomaly of an imaginary planet which moved so that its angular velocity was equal to the mean angular velocity of P. c- 27r 27r „ V/i Since —=—^ (^^.,. ggx . ,^ ^ _ . Let be the True Anomaly ASP, and ^ the Eccentric Anomaly ACQ. If h be twice the area described in a unit of time, then -^t = Sectorial area ASP = Curvilinear area ANP + triangle SNP = - X Curvilinear area ANQ + triangle SNP = - (Sector ACQ- triangle GNQ) + ^SN.NP — ~ {2'^' 4* ~ i^^ ^^^ 4* co^ ^) + i (^ cos (f) — ae) . b sin ^ = ^ ( - e sin (^). .*. n^ = -T- (^ — esin (f>) = (f) — e sin ^ (1). By the polar equation to a Conic Section, we have SP = ^ ^ Q^ (1 - e") 1 + e cos ^ 1 + e cos ^ ' and SP = a — e. ON = a{l- e cos ^). /. (1 - e cos <^) (1 + e cos ^) = 1 - e^ , ^ cos (f) — e ... and .*. cos^ = - (2). 1 — e cos 9 81. If e be small, a first approximation from (1) to the value of ^ is nt, and a second approximation is nt + e sin nt From (2), a first approximation to the value of 6 is (/>, and a second approximation is + A, where ^ • , cos c^ — e cos — X sm = , ^- -, ^ 1 — e cos J ^ e sin rf) . , ana .'. X = = '—, = e sin 6 approx. 1 — e cos ^ ^^ Planetary Motion 91 Hence, as far as the first power of e, ^ = ^ -)- e sin ^ = w^ + e sin ??^ + e sin {nt + e sin nt) = nt + 2e sin ?ii. Also 8P = ''^-^"^'^^ = a (1 - e cos ^), 1 + e cos ^ ^ ^ to the same approximation, = a — ae cos (n< + 2e sin ?2i) = a — ae cos nt. If an approximation be made as far as squares of e, the results are found to be g2 ) _ l+g (^ ^^"^ 2~l + cos^""(l-e)(H-cos<^)"l-6'^'' 2' so that = 2tan-MA/tj tan- , and VfT 2 tan ^ -/I/ -, --" ^ . /, 2 V 1 + e 2 /- ^ sm 6 sm«^= -= 77 = vl i,,o0 1,1-6, ,,6' 1 + ecos^' l+tan-| 1+^-p^tan--^ Hence, from equation (1) of the same article, remembering that n = ^ , we have t = -r 2tan-M «/ — — - tan^^ -eVl -e^-— . Va*. |_ ( V 1 + e 2j 1 + e cos ^ J This is the result of Art. 75 and gives the time of describing any arc of the ellipse, starting from perihelion. 83. When a particle is describing an elliptic orbit, it may- happen that at some point of the path it receives an impulse so that it describes another path ; or the strength of the centre of force may be altered so that the path is altered. To obtain the 92 Dynamics of a Particle new orbit we shall want to know how the major axis has been altered in magnitude and position, what is the new eccentricity, etc. 84. Tangential disturbing force. Let APA' be the path of a particle moving about a centre of force at S, and let H be the other focus. When the particle has arrived at P let its velocity be changed to v + u, the direction being unaltered ; let 2a' be the new major axis. Then we have Hence, by subtraction, we have — . -^ a Since the direction of motion is unaltered at P, the new focus lies on PH ; and, if H' be its position, we have EH' = {H'P + SP) - (HP + SP) = 2a' - 2a. If the change of velocity u be small and equal to 8v, say, then by diffei'entiating the first of equations (1) we have 2v8v = —Ja, a^ [For SP is constant as far as these instantaneous changes are concerned.] Hence Sa, the increase in the semi-major axis, ^^EA'^' (2) Again, since HH is now small, we have HH sin H 2Ba . sin H tan HSH 2ae + HH'cosH Disturbed Orbits 93 Hence S\^, the angle through which the major axis moves, = HSH = = — .smH.Sv (3). ae e/j, Since the direction of motion at P is unaltered by the blow, the value of h is altered in the ratio , so that 8h = — h. V V But h'' = fia(l-e'). .'. 2hdh = fiSa{l-e')-fia.2eBe. .-. fia.2e8e = 2vBv.a'(l-e')-2~h\ so that ce= —.- -. — (4). V e fM ^ ^ This gives the increase in the value of the eccentricity. Since the periodic time T= — a^. BT 3 Sa S vaSv y-2¥— ^ <5). 85. If the disturbing force is not tangential, the velocity ib produces must be compounded with the velocity in the orbit to give the new velocity and tangent at the point P. The equations (1) or (2) of the last article now give the magnitude, 2a', of the new major axis. Also since the moment of the velocity of the point P about the focus 8 is equal to V/u. X semi-latus-rectum, i.e. to /x \/a' (l — e'% we obtain the new eccentricity. Finally by drawing a line making with the new tangent at P an angle equal to that made by SP, and taking on it a point H', such that SP + H'P is equal to the new major axis, we obtain the new second focus and hence the new position of the major axis of the orbit. 86. Effect on the orbit of an instantaneous change in the value of the absolute acceleration /*. When the particle is at a distance r from the centre of force, let the value of [x be instantaneously changed to //,', and let the new values of the major axis and eccentricity be 2a' and e. 94 Dynamics of a Particle Since the velocity is instantaneously unaltered in magnitude, we have K^-D=--'a-?) w- an equation to give a'. The moment of the velocity about S being unaltered, h remains the same, so that ^/fia {l-e') = h = \/fi.'a' (1 - e'') (2), giving e'. The direction of the velocity at distance r being unaltered, we obtain the new positions of the second focus and of the new major axis as in the previous article. If the change B/u, in /j, be very small the change 8a in a is obtained by differentiating the first equation in (1), where v and r are treated as constants, and we have 8a - Sa a~ fjb- So, from (2), we have, on taking logarithmic differentials, 8/x 8a _ 2e8e _ fj, a 1 — e"~ ' 2e8e 8/j, v^a ^ B/j, f v'^a\ Again, since the periodic time T= 7— ^^ EXAMPLES 1. If the period of a j)lanet be 365 days and the eccentricity e is — , shew that the times of describing the two halves of the orbit, bounded by the latus rectum passing through the centre of force, are 2 L ISttJ very nearly. 2. The perihelion distance of a comet describing a parabolic path is - of the radius of the Earth's j)ath supposed circular ; shew that the time that the comet will remain within the Earth's orbit is 2 71 + 2 n-\ ^ — . . ^/ — — ot a year. Law of the Inverse Square. Examples 95 [If S be the Sun, a the radius of the Earth's path, A the perihelion of the comet's path, and P the intersection of the paths of the earth and comet, then a = SP—. -, so that cos^ = -- 1, and therefore 2ir s Now use the formula of Art. 77, remembering that -^a'2' = one year.] 3. The Earth's path about the Sun being assumed to be a circle, shew that the longest time that a comet, which describes a parabolic path, can 2 remain within the Earth's orbit is 5— of a year. OTT 4. A planet, of mass 3f and periodic time T, when at its greatest distance from the Sun comes into collision with a meteor of mass m, moving in the same orbit in the opposite direction with velocity v ; if jj be small, shew that the major axis of the planet's path is reduced by 4m im vT / l-i 5. When a periodic comet is at its greatest distance from the Sim its velocity v is increased by a small quantity dv. Shew that the comet's least distance from the Sun is increased by the quantity 48v. \ —j- (l . 6. A small meteor, of mass m, falls into the Sun when the Earth is at the end of the minor axis of its orbit ; if Jlf be the mass of the Sun, shew that the major axis of the Earth's orbit is lessened by 2a. j^, that the periodic time is lessened by -j^ of a year, and that the major axis of its orbit is turned throusrh an angle — . t>. ° ae M 7. The Earth's present orbit being taken to be circular, find what its path would be if the Sun's mass were suddenly reduced to - of what it is now. 8. A comet is moving in a parabola about the Sun as focus ; when at the end of its latus-rectum its velocity suddenly becomes altered in the ratio n:\, where n<\ ; shew that the comet will describe an ellipse whose eccentricity is Jl-'ln^ + 'in'^, and whose major axis is = ^j where 21 was the latus-rectum of the parabolic path. 9. A body is moving in an ellipse about a centre of force in the focus ; when it arrives at P the direction of motion is turned through a right 96 Dynamics of a Particle angle, the speed being unaltered; shew that the body will describe an ellipse whose eccentricity varies as the distance of P from the centre. 10. Two pai'ticles, of masses m^ and wig, moving in co-planar parabolas round the Sun, collide at right angles and coalesce when their common distance from the Sun is R. Shew that the subsequent path of the combined particles is an ellipse of major axis ^^i + "'2-> ^ 11. A particle is describing an ellipse under the action of a force to one of its foci. When the particle is at one extremity of the miuor axis a blow is given to it and the subsequent orbit is a circle ; find the magnitude and direction of the blow. 12. A particle m is describing an ellipse about the focus with angular momentum mA, and when at the end of the minor axis receives a small impulse mu along the radius vector to the focus. Shew that the major axis of the path is diminished by —y — , that the eccentricity is increased by -7- (1 — e^)'!'^ and that the major axis is turned through the angle — —f -^ where a, h are the semi-axes and e the eccentricity of the ellipse. 13. A particle is describing a parabolic orbit (latus-rectum 4a) about a centre of force (/n) in the focus, and on its arriving at a distance r frona the focus moving towards the vertex the centre of force ceases to act for a certain time r. When the force begins again to ojserate prove that the new orbit will be an ellipse, parabola or hyperbola according as Ir—a 14. Shew that the maximum range of a projectile on a horizontal plane through the point of projection is 2A ,-p — -^ , where R is the radius of the Earth, and h is the greatest height to which the projectile can be fired. [Use the result of Art. 79.] 15. When variations of gravity and the spherical shape of the Earth are taken into account, shew that the maximum range attainable by a gun placed at the sea level is 2i?sin-M „ j , and that the necessary angle of elevation is - cos~^ ( "n ) > where R is the Earth's radius and h is the greatest height above the surface to which the gun can send the ball. 16. Shew that the least velocity with which a body must be projected from the Equator of the Earth so as to hit the surface again at the North Pole is about 4^ miles per second, and that the corresponding direction of projection makes an angle of 67^° with the vertical at the point of projection. CHAPTER YI TANGENTIAL AND NORMAL ACCELERATIONS. UNIPLANAR CONSTRAINED MOTION 87. In the present chapter will be considered questions which chiefly involve motions where the particle is constrained to move in definite curves. In these cases the accelerations are often best measured along the tangent and normal to the curve. We must therefore first determine the tangential and normal accelerations in the case of any plane curve. 88. To shew that the accelerations along the tangent and d?s ( clv\ v^ normal to the path of a particle are j^ ( = ^ ;7- ) '^''^d - , where p is the radius of curvature of the curve a.t the point considered. Let V be the velocity at time t along the tangent at any point P, whose arcual dis- tance from a fixed point C on the path is s, and let v+ Av be the velocity at time t + At along the tan- gent at Q, where PQ = As. Let and 4> + ^4> ^^ the angles that the tan- gents at P and Q make with a fixed line Ow, so that A(f> is the angle between the tangents at P and Q. 98 Dynamics of a Particle Then, by definition, the acceleration along the tangent at P velocity along the tangent at time t + Ai ^ L — the same at time t = Lit . , ■ — ^ (y + Aw) cos A0 — V = -Lt 7— At=o A^ ^ i> + Av — w on neglecting small quantities of the second order, _dv _ d^s ^dt~dt^' . dv_dvds_ dv dt ~~ ds dt ds ' Again the acceleration along the normal at P I velocity along the normal at time t + At~\ At -r L — the same at time t = Lit T.(v + Av) sin A — -Lit T— A<=o Ai T , . . , sin A0 A<^ As ^ 1 v" = Lt (v + Av) . ~-r-~ .-r-^.'.^ = V.l.-.V = -. M=o^ ^ A(f> As At p p Cor. In the case of a circle we have p = a, s = ad, v = aO and the accelerations are a6 and aQ'^. 89. The tangential and normal accelerations may also be directly obtained from the accelerations parallel to the axes. dx _dx ds ' di~d^'dt' d^_^ /dsy dxd^ ^° d^~ dp \dt) '^ ds dfi' But, by Differential Calculus, '■X d^ 1 df _ ds2 p~ dy ~ dx' ds ds and d^x dy 1 df2 - ~ ds'y^ (dsy dxd^s sin^ \dt) +ds dfi- p + d^y dx 1 /ds dt^ ~ds'p'\di \2 dy (/2,s- cos (A „ (i-N :)+d'-dt2- /•^^+dt2 . COS 0, Tangential and Normal Accelerations. Examples 99 Therefore the acceleration along the tangent ' cm ^^~^ ^„„ u. , '^^il „:„ ^ _ <^^s dv _dvds _ dv and the acceleration along the normal = - --7 sin + — | cos = — , cli- dt- p 90. -Ej;. a curve is described by a particle having a constant acceleration in a direction inclined at a constant angle to the tangent; shew that the curve is an equiangular spiral. Here -3-=/ cos o and — =/sin o, where /and a are constants, 1 ds .'. jr 3-, =s cot a +^, where 4 is a constant. 2 d^ .: log (s cot tt + .4) = 2^ cot a + const. .-. s= -^tano + jBe^'/'Cota which is the intrinsic equation of an equiangular spiral. EXAMPLES 1. Find the intrinsic equation to a curve such that, when a point moves on it with constant tangential acceleration, the magnitudes of the tangential velocity and the normal acceleration are in a constant ratio. 2. A point moves along the arc of a cycloid in such a manner that the tangent at it rotates with constant angular velocity ; shew that the acceleration of the moving point is constant in magnitude. 3. A point moves in a curve so that its tangential and normal accelerations are equal and the tangent rotates with constant angular velocity ; find the path. 4. If the relation between the velocity of a particle and the arc it has described be find the tangential force acting on the particle and the time that must elapse from the beginning of the motion till the velocity has the value V. 5. Shew that a cycloid can be a free path for a particle acted on at each point by a constant force parallel to the corresponding radius of the generating circle, this circle being placed at the vertex. 6. An insect crawls at a constant rate u along the spoke of a cart- wheel, of radius a, whose centre is moving in a straight line with velocity V. Find its accelerations along and perpendicular to the spoke. 7. A circle rolls on a straight line, the velocity of its centre at any instant being v and its acceleration /; find the tangential and normal accelerations of a point on the edge of the circle who&e angi.dar distance from the point of contact is 6. 7—2 100 Dynamics of a Particle arcual distance from a Q 91. A particle is compelled to move on a given smooth plane curve under the action of given forces in the plane; to fnd the motion. Let P be a point of the curve ^v fixed point C is s, and let v be the velocity at P. Let X, Y be the components parallel to two rectangular axes Ow, Oy of the forces acting on the particle when at P ; since the curve is smooth the only reaction will be a force R along the normal at P. Resolving along the tangent and normal, we have vdv -^ = force along TP = X cos (j>+ Fsin cfi = xf-+Y'^^ ds ds ■in and = — X sin(f)+ Ycos (f) + R = -X^+Y — + R ds ds .(2). When V is known, equation (2) gives R at any point. Equation (1) gives ^mv'- = j{Xdx+Yd:/) (3). Suppose that Xdw + Ydy is the complete differential of some function d) (x, y), so that X = -^ and Y= -f^ . ^^ dx dy Then 1 ., [ d (x, y), it follows ft-om (5) that The change in the Kinetic Energy of the particle = the Work done by the External Forces. Forces of this kind are called Conservative Forces. The quantity <^ {x, y) is known as the Work-Function of the system of forces. From the ordinary definition of a Potential Function, it is clear that (f){x,y) is equal to the Potential of the given system of forces added to some constant. If the motion be in three dimensions we have, similarly, that the forces are Conservative when j{Xdx + Ydy + Zdz) is a perfect differential, and an equation similar to (5) will also be true. [See Art. 131.] 92. The Potential Energy of the particle, due to the given system of forces, when it is in the position P = the work done by the forces as the particle moves to some standard position. Let the latter position be the point (xi, y^). Then the potential energy of the particle at P r(x, , Vd fix, , y,) /(JU. dfk ■ r -|(^,,2/,) = <^ («» 3/) = <^ ('-^'i >yi)-4> (''> y)- L J(«.2/) Hence, from equation (4) of the last article, (Kinetic Energy -f Potential Energy) of the particle when at P = (fi (x, y) + C + cfi {x„ y,) - 4> {X, y) = G -\- (f)(xi, 2/i) = a constant. 102 Dynamics of a Pat'ticle Hence, when a particle moves under the action of a Con- servative System of Forces, the sum of its Kinetic and Potential Energies is constant throughout the motion. 93. In the particular case when gravity is the only forde acting we have, if the axis of y be vertical, X = and F= — mg. Equation (3) then gives ^mv'^ = - mr/j/ + C. Hence, if Q be a point of the path, this gives kinetic energy at P - kinetic energy at Q = mgx difference of the ordinates at P and Q = the work done by gravity as the particle passes from Q to P. This result is important ; from it, given the kinetic energy at any known point of the curve, we have the kinetic energy at any other point of the path, if the curve be smooth. 94. If the only forces acting on a particle be perpendicular to its direction of motion (as in the case of a particle tethered by an inextensible string, or moving on a smooth surface) its velocity is constant ; for the work done by the string or reaction is zero. 95. All forces tvldch are one-valued functions of distances from fixed points are Conservative Forces. Let a force acting on a particle at the point (cc, y) be a function -y^ (r) of the distance r from a fixed point (a, h) so that r^={^x-ay + {y-h)\ Also let the force act towards the point {n, b). Then r ^ =(oc — a\ and da; ^ dr 'Wy^y-''- The component X of this force parallel to the axis of x = - a/t (r) X X -a r ' if the force be an attraction, and ih le component Y parallel to y = -^(r)x y-b r Hence Xdx + Ydy = - yfr (r) > (x - a)dx + (y r - f (,/■) dr. -h)dy Conservation of Energy 103 Hence, if ^(r) be such that -^ i?' (r) = - -v^ (?•) (1), we have \{Xdx + Ydy) = f^F (r) dr = F (r) + const. Such a force therefore satisfies the condition of being a Conservative Force. If the force be a central one and follow the law of the inverse square, so that ^ (r) = ^ , then F (r) = - I \p (r) dr-'^ and hence / (Xdx + Ydy) = ^ + constant. 93. The work done in stretching an elastic string is equal to the extension produced inultiplied by the mean of the initial and final tensions. Let a be the unstretched length of the string, and \ its modulus of elasticity, so that, when its length is x, its tension = A, . , by Hooke's Law. a -^ The work done in stretching it from a length 6 to a length c = 2^J(„_„)..-(6-a).-]=(o-i)[x^+x"-^«]xi = (c — 6) X mean of the initial and final tensions. Ex. AandBare two points in the same horizontal plane at a distance 2a apart ; AB is an elastic string whose unstretched length is 2a. To O, the middle point of AB, is attached a particle of mass m lohich is allowed to fall under gravity ; find its velocity -when it has fallen a distance x and the greatest vertical distance through which it moves. When the particle is at P, where OF = x, let its velocity be v, so that its kinetic energy then is imv^. The work done by gravity = TOr; . x. The work done against the tension of the string = 2 X (£P - £0) X J \ :^^^J=i^ = ^ (L'P - a)2 = ^ [v/^;;r^2 _ a]2. Hence, by the Principle of Energy, ^mj;2 = mgx - '- [sjx^ + a'^ - ap. The particle comes to rest when i; = 0, and then x is gi\'en by the equation ingxa — X [\x^ + a^ - a]^. 104 Dynamics of a Farticle EXAMPLES 1. If an elastic string, whose natural length is that of a uniform rod, be attached to the rod at both ends and suspended by the middle point, shew by means of the Principle of Energy, that the rod will sink until the strings are inclined to the horizon at an angle 6 given by the equation cot3--cot- = 2«, given that the modulus of elasticity of the string is 7i times the weight of the rod. 2. A heavy ring, of mass m, slides on a smooth vertical rod and is attached to a light string which passes over a small pulley distant a from the rod and has a mass M (> m) fastened to its other end. Shew that, if the ring be dropped from a point in the rod in the same horizontal plane as the pulley, it will descend a distance j^ — ^ before coming to rest. Find the velocity of m when it has fallen through any distance x. 3. A shell of mass M is moving with velocity V. An internal explosion generates an amount of energy E and breaks the shell into two portions whose masses are in the latio m^ : m^. The fragments continue to move in the original line of motion of the shell. Shew that their , ... „ /^m^E , „ f^miE velocities are I + * / rp and k — * / r? • V 'iHiM A' Hijjil/ 4. An endless elastic string, of natural length 'Ina, lies on a .smooth horizontal table in a circle of radius a. The string is suddenly set in motion about its centre with angular velocity w. Shew that if left to itself the string will expand and that, when its radius is r, its angular velocity i& -^a, and the square of its radial velocity from the centre is — ^ (r^ — a?-) ^ , where m is the mass and X the modulus of r^ ^ ' ma elasticity of the string. 5. Four equal particles are connected by strings, which form the sides of a square, and repel one another with a force equal to /x x distance ; if one string be cut, shew that, when either string makes an angle Q with .,..,.,.., , 1 •* • /4/xsiu(9(2 + sin^) its original position, its angular velocity is a / -^-—- — -^^r^. • ▼ Jf^" Sin u [As in Art 47 the centre of mass of the whole system remains at rest ; also the repulsion, by the well-known property, on each particle is the same as if the whole of the four particles were collected at the centre and = 4^1 X distance from the fixed centre of mass. Equate the total kinetic energy to the total work done by the repulsion.] The Simple Pendulum 105 6. A uniform string, of mass M and length 2a, is placed symmetrically over a smooth peg and has particles of masses m and ra' attached to its extremities ; shew that when the string runs off the peg its velocity is J J/+ 2 (m - m') Jf+m+m' *^' 7. A heavy uniform chain, of length 21, hangs over a small smooth fixed pulley, the length l + c being at one side and ^ — c at the other ; if the end of the shorter portion he held, and then let go, shew that the chain will slip off the pulley in time f - j log . 8. A uniform chain, of length I and weight W, is placed on a line of greatest slope of a smooth plane, whose inclination to the horizontal is a, and just reaches the bottom of the piano where there is a small smooth pulley over which it can ruu. Shew that, wheu a length x has run oftj the tension at the bottom of the plane is TF(1 — sina) X (I - x) 9. Over a small smooth pulley is placed a uniform flexible cord ; the latter is initially at rest and lengths I — a and l + a hang down on the two sides. The pulley is now made to move with constant vertical accelera- tion /. Shew that the string will leave the pulley after a time V: f+g cosh~i - . 97. Oscillations of a Simple Pendulum. A particle m is attached hy a light string, of length I, to a fixed paint and oscillates under gravity through a small angle; to find the period of its motion. When the string makes an angle 6 with the vertical, the equation of motion is m-j^ = — mg sin But s = W. (j = —^ ain = — J 6,to a, first approxima- tion. If the pendulum swings through a small angle a on each side of the vertical, so that 6 = a and ^ = when t = 0, this equation gives _ ^ = a cos U /? t\ , lVT 106 JJynamics of a Fartide so that the motion is simple harmonic and the time, Ti, of a very small oscillation = 27r a/- , as in Art. 22. For a higher approximation we have, from equation (1), W' = 2g (cos 6 - cos a) (2), since 6 is zero when = (x. [This equation follows at once from the Principle of Energy.] ... J^.t-T _ ^^ V I Jo Vcos ^ — cos a' where t is the time of a quarter-swing. jo V°'2 •Put sin 2 = sin 2- sin <^. / „ cos K • SI sm-2 cos (^f?^ 1/ . a ^ . sm ^ cos (f> ^ Jo fl-sin-^^sin2<^)- (3) ""^Wo L 2'"'"2-'^^"'^+2":4'''' 2'^^^ + -/'^ + (,2-^6) ^^^^2 + -J <^>- Hence a second approximation to the required period, T2, =r,[i+.|.sin=g=r.[i+f;], if powers of a higher than the second are neglected. Even if a be not very small, the second term in the bracket of (4) is usually a sufficient approximation. For example, The Simple Pendulum 107 suppose a. = 30°, so that the pendulum swings through an angle of 60"^ ; then sin^ -^ = sin- 15° = '067, and (4) gives t = I ^- [1 + -017 + -00063 + ...]. [The student who is acquainted with Elliptic Functions will see that (3) gives sin ^ = sn U a/?) , [mod. sin ^ J , so that sin - = sin ^ sn (^ a/i ) , (mod. sin-j . ^/: The time of a complete oscillation is also, by (3), equal to - multiplied by the real period of the elliptic function with modulus sin ^ .] 98. The equations (1) and (2) of the previous article give the motion in a circle in any case, when a is not necessarily small. If CO be the angular velocity of the particle when passing through the lowest point A, we have le-' = 2g cos 6 + const. = Im^ - 2^ (1 - cos ^) (5). This equation cannot in general be integrated without the use of Elliptic Functions, which are beyond the scope of this book. If T he the tension of the string, we have T — mg cos 6 = force along the normal PO = mie^ = mlw'' - ^vig (1 - cos 6\ .-. T = m{la>^-g{2-Scose)} (6). Hence T vanishes and becomes negative, and hence circular motion ceases, when cos = ~^^ . % Particular Case. Let the augular velocity at A be that due tu a fall from the highest point A', so that i2w2 = 2(7.2?, i.e. ^ - ^ Then (5) gives 108 ^/l' Dynamics of a FarticLe , . l^l-[ ^^^ 1 /"JL n- giving the time t of describing an angle d from the lowest point. Also in this case T = in {Ag -2g + Bg cos e}=mg[2 + 3 cos 0], tan H=v/^' Therefore the time during which the circular motion lasts Vi Ioge(v/5 + V6). 99. Ex. 1. Shew that a pendulum, which beats eeconds when it swings through 3° on each side of the vertical, will lose about 12 sees, per day if the angle be 4° and about 27 sees, per day if the angle be 5°. Ex. 2. A heavy bead slides on a smooth fixed vertical circular wire of radius a; if it be projected from the lowest point with velocity just sufbcient to carry it to the highest point, shew that the radius to the bead is at time t inclined to the vertical at an angle 2 tau-i sinh ^/-t , and that the bead will be an infinite time in airriving at the highest point. 100. Motion on a smootk cycloid vertex lowest. whose axis is vertical and Motion on a smooth Cycloid 109 Let AQD be the generating circle of the cycloid GPAC, P being any point on it; let PT be the tangent at P and PQN perpendicular to the axis meeting the generating circle in Q. The two principal properties of the cycloid are that the tangent TP is parallel to AQ, and that the arc ^P is equal to twice the line A Q. Hence, if PTx be 0, we have e = zQAx = AI)Q, and s = arc^P = 2.^Q = 4asin^ (1), if a be the radius of the generating circle. If R be the reaction of the curve along the normal, and m the particle at P, the equations of motion are then m -j-„ = force along PT = — mg sin ^ (2), and ni . ~ = force along the normal = R — mg cos 6 .. ,(3). From (1) and (2), we then have ^=-£/ w- so that the motion is simple harmonic, and hence, as in Art. 22, the time to the lowest point TT _ 2 /a V 4a and is therefore always the same whatever be the point of the curve at which the particle started from rest. Integrating equation (4), we have = 4a^ (sin^ 0^ - sin^ Q), if the particle started from rest at the point where = 0^. [This equation can be written down at once by the Principle of Energy.] Also P ~ rf^~ ^^ ^^^ ^' 110 Dynamics of a Particle Therefore (3) gives „ ^ . sin'' ^0 - sin" ^ cos2^+sin2< K = mg cos v + mg — = mg cos 6 ' giving the reaction of the curve at any point of the path. On passing the lowest point the particle ascends the other side until it is at the height from which it started, and thus it oscillates backwards and forwards. 101. The property proved in the previous article will be still true if, instead of the material curve, we substitute a string tied to the particle in such a way that the particle describes a cycloid and the string is always normal to the curve. This will be the case if the string unwraps and wraps itself on the evolute of the cycloid. It can be easily shewn that the evolute of a cycloid is two halves of an equal cycloid. For, since p = 4a cos 6, the points on the evolute corre- sponding to A and C are A', where AD = DA', and C itself. Let the normal PO meet this evolute ' in P', and let the arc CP' be o". By the property of the evolute o- = arc P'G = P'P, the radius of curvature at P, = 4(* cos (9 = 4a sin P'GD. Hence, by (1) of the last article, the curve is a similar cycloid whose vertex is at G and whose axis is vertical. This holds for the arc CA. The evolute for the arc C'A is the similar semi-oycloid C'A'. Hence if a string, or flexible wire, of length equal to the arc GA', i.e. 4a, be attached at A' and allowed to wind and unwind itself upon fixed metal cheeks in the form of the curve GA'G', a particle P attached to its other end will describe the cycloid GAG', and the string will always be normal to the curve GAG'; the times of oscillation will therefore be always isochronous, whatever be the angle through which the string oscillates. In actual practice, a pendulum is only required to SAving through a small angle, so that only small portions of the two arcs near A' are required. This arrangement is often adopted in the case of the pendulum of a small clock, the upper end of the supporting wire consisting of a thin flat spring which coils and uncoils itself from the two metal cheeks at A', Motion on a rough Curve TU 102. Motion on a rough curve under gravity. Whatever be the curve described under gravity with friction, we have, if ^ be the angle measured from the horizontal made by the tangent, and if s increases with 4), and dv = ^ sin - p~ gcos (p- R m IX - = g (sin (j) — [Ji cos (p). 'ifxr'^ = 2gp (sin ^ — ^ cos , this sfives v\ and hence ^^- I ( -^^ L Hence -^ is known, and dt gives .-,an.inence^^j(-^^^. therefore theoretically t in terms of (p. 103. // the cycloid of Art 100 he rough with a coefficient of friction fi, to find the motion, the particle sliding downwards. In this case the friction, fiR, acts in the direction TP produced, so that the equations of motion are dv d ■^^=fiR-mg sine (1), and m. — =R — mg cos 6 dv V- . . ^, V ■^- fi — = g i^fx cos o — am o). .(2). ^5 112 Dynamics of a Particle /. ^'-2y^i-^ = 2r/(^cos^-sin^)^ = Sag (fico2 — sin 6) . cos = ^ag {/Jb+ fi cos 20 — sin 20). To integrate this equation, multiply by e'^^e^ and we have ^•2g 2^6 ^ 4a^je-2,x9 (^ + ^ cos 26 - sin 29) = - 2age-^>'' + ^^- e'^'^^ [2yu, sin 26* + (1 - fi'^) cos 2^] + ^, i.e. v'=A e''^' - 2ag + =^^, [2/z sin 2^ + (1 - /ti*) cos 2(9]. The constant A is determined from the initial conditions. The equation cannot be integrated further. EXAMPLES ON CHAPTER VI, 1. A particle slides down the smooth curve y = a sich - , the axis of x being horizontal, starting from rest at the point where the tangent is inclined at a to the horizon ; shew that it will leave the curve when it has fallen through a vertical distance a sec a. 2. A particle descends a smooth curve under the action of gravity, describing equal vertical distances in equal times, and starting in a vertical direction. Shew that the curve is a semi-cubical parabola, the tangent at the cusp of which is vertical. 3. A particle is projected with velocity V from the cusp of a smooth inverted cycloid down the arc ; shew that the time of reaching the vertex is 2 ^-tan-M '—^ . 4. A particle slides down the arc of a smooth cycloid whose axis is vertical and vertex lowest ; prove that the time occupied in falling down the first half of the vertical height is equal to the time of falling down the second half. 5. A particle is placed very close to the vertex of a smooth cycloid whose axis is vertical and vertex upwards, and is allowed to run down the curve. Shew that it leaves the curve when it is moving in a direction making with the horizontal an angle of 45°. 6. A ring is strung on a smooth closed wire which is in the shape of two equal cycloids joined cusp to cusp, in the same plane and sym- metrically situated with respect to the line of cusps. The plane of the wire is vertical, the line of cusps horizontal, and the radius of the generating circle is a. The ring starts from the highest point with Constrained motion. Examples 113 velocity v. Prove that the times from the upper vertex to the cusp, and from the cusp to the lower vertex are respectively 7. A particle moves in a smooth tube in the form of a catenary, being attracted to the directrix by a force proportional to the distance from it Shew that the motion is simple harmonic. 8. A particle, of mass m, moves in a smooth circular tube, of radius a, under the action of a force, equal to »i/i x distance, to a point inside the tube at a distance c from its centre ; if the particle be placed very nearly at its greatest distance from the centre of force, shew that it will desci'ibe the quadrant ending at its least distance in time 7: logU/2 + 1). 9. A bead is constrained to move on a smooth wire in the form of an equiangular spiral. It is attracted to the pole of the spiral by a force, = m/i (distance)- 2, and starts from rest at a distance h from the pole. Shew that, if the equation to the spiral be r = ae^'^°*", the time of arriving at the pole is ^ */ „- . sec a. Find also the reaction of the curve at any instant. 10. A smooth parabolic tube is placed, vertex downwards, in a vertical plane ; a particle slides down the tube from rest under the influence of gravity ; prove that in any position the reaction of the tube is Iw , where w is the weight of the particle, p the radius of curvature, 4a the latus rectum, and h the original vei'tical height of the particle above the vertex. 11. From the lowest point of a smooth hollow cylinder whose cross- section is an ellipse, of major axis 2a and minor axis 26, and whose minor axis is vertical, a particle is projected from the lowest point in a vertical plane perpendicular to the axis of the cylinder ; shew that it will leave the cylinder if the velocity of projection lie between V2(/a and 12. A small bead, of mass wi, moves on a smooth circular wire, being acted upon by a central attraction ,_,. f''^ — r-„ to a point within the circle (distance)^ ^ situated at a distance h from its centre. Shew that, in order that the bead may move completely round the circle, its velocity at the point of the wire nearest the centre of force must not be less than * / „~-„ . 114 Dynamics of a Particle 13. A small bead moves on a thin elliptic wire under a force to the focus equal to ^ + -3 . It is projected from a point on the wire distant R from the focus with the velocity which would cause it to describe the ellipse freely under a force -^ . Shew that the reaction of the wire is p yj^ or A -J where p is the radius of curvature. 14. If a particle is made to describe a curve in the form of the four-cusped hypocycloid x^+y'^ = a^ under the action of an attraction perpendicular to the axis and varying as the cube root of the distance from it, shew that the time of descent from any point to the axis of x is the same, i.e. that the curve is a Tautochrone for this law of force. 15. A small bead moves on a smooth wire in the form of an epi- cycloid, being acted upon by a force, varying as the distance, toward the centre of the epicycloid ; shew that its oscillations are always isochronous. Shew that the same is true if the curve be a hypocycloid and the force always from, instead of towards, the centre. 16. A curve in a vertical plane is such that the time of describing any arc, measured from a fixed point 0, is equal to the time of sliding down the chord of the arc ; shew that the curve is a lemniscate of Bernouilh, whose node is at and whose axis is inclined at 45° to the vertical, 17. A particle is projected along the inner surface of a rough sphere and is acted on by no forces ; shew that it will return to the point of projection at the end of time — j^(e^'"^- 1), where a is the radius of the sphere, V is the velocity of projection and /x is the coefficient of friction. 18. A bead slides down a rough circular wire, which is in a vertical plane, starting from re.st at the end of a horizontal diameter. When it has described an angle 6 about the centre, shew that the square of its angular velocity is y [(1 - 2/^2) sin 6 + 3^ (cos 6 - e'^^^ where /i is the coefficient of friction and a tlie radius of the rod. 19. A particle falls from a position of limiting equilibrium near the top of a nearly smooth glass sphere. Shew that it will leave the sphere at the point whose radius is inclined to the vertical at an angle where cos a = |, and /x is the small coefficient of friction. 20. A particle is projected horizontally from the lowest point of Constrained motion. Examples 115 a rough sphere of radius a. After describing an arc less than a quadrant it returns and comes to rest at the lowest point. Shew that the initial velocity must be sin a */ 2ga _^ ^ , where ft is the coefficient of friction and aa is the arc through which the particle moves. 21. The base of a rough cycloidal arc is horizontal and its vertex downwards ; a bead slides along it starting from rest at the cusp and coming to rest at the vertex. Shew that /xV^=l. 22. A particle slides in a vertical plane down a rough cycloidal arc whose axis is vertical and vertex downwards, starting from a point where the tangent makes an angle 6 with the horizon and coming to rest at the vertex. Shew that fxe*^^ = sin ^ — /i cos 9. 23. A rough cycloid has its plane vertical and the line joining its cusps horizontal. A heavy particle slides down the curve from rest at a cusp and comes to rest again at the point on the other side of the vertex where the tangent is inclined at 45° to the vertical. Shew that the coefficient of friction satisfies the equation 3;x7r + 4log,(l+;x) = 2log,2. 24. A bead moves along a rough cvu-ved wire which is such that it changes its direction of motion with constant angular velocity. Shew that the wire is in the form of an equiangular spiral. 25. A particle is held at the lowest point of a catenary^ whose axis is vertical, and is attached to a string which lies along the catenary hut is free to unwind from it. If the particle be released, sheio that the time that elapses before it is moviaj at an angle to the vertical is and that its velocity then is 2 i\Jgc sin ^ , where c is the parameter of the catenary. Find also the tension of the string in terms of (f). At time t, let the string PQ be inclined at an angle (p to the horizontal, where P is the particle and Q the point where the string touches the catenary. A being the lowest point, let s=arc AQ = line PQ. The velocity of P along QP=ve\. of Q along the tangent + the vel. of P relative to Q = (-s)+s==0 (1). The velocity of P perpendicular to QP similarly = ^•0 (2). The acceleration of P along QP (by Arts. 4 and 49) = acc. of Q along the tangent QP+the ace. of P relative to Q = -s + (i--4^)=-5<^2 (3). 8—2 116 Dynamics of a Particle The acceleration of P perpendicular to QP = acc. of Q in this direction + ace. of P relative to Q =-s(j) + s^ (4). These are the component velocities and accelerations for any curve, whether a catenary or not. The equation of energy gives for the catenary \m. (c tan<^0)2 = wi^ (c— ccos<^) (5). Resolving along the line PQ, we have Mictan(^.02_2T_^^gjjj ^ (^Q^_ (5) and (6) give the results required. 26. A particle is attached to the end of a light string vprapped round a vertical circular hoop and is initially at rest on the outside of the hoop at its lowest point. When a length aO of the string has become unwound, shew that the velocity v of the particle then is \/2ag {6 sin ^+cos 6 - 1), and that the tension of the string is (3sin^H -j times the weight of the particle. 27. A particle is attached to the end of a fine thread which just winds round the circumference of a circle from the centre of which acts 2 a repulsive force nifi (distance) ; shew that the time of unwinding is -p , and that the tension of the thread at any time t is 2/u- .a.t, where a is the radius of the circle. 28. A particle is suspended by a light string from the circumference of a cylinder, of radius a, whose axis is horizontal, the string being tangential to the cylinder and its unwound length being a/S. The particle is projected horizontally in a plane perpendicular to the axis of the cylinder so as to pass ixnder it; shew that the least velocity it can have so that the string may wind itself completely up is \'2(/a(/3-sin/3). 29. A particle P moves in a plane under a force acting in the direction of the tangent from /• to a fixed circle and inversely propor- tional to the length of that tangent. Shew how to solve the equations of motion, and shew that in one particular case the particle moves with constant velocity. 30. If a particle can describe a certain plane curve freely under one set of forces and can also describe it freely under a second set, then it can describe it freely when both sets act, provided that the initial kinetic energy in the last case is €<^ual to the sum of the initial kinetic eneryies in the first two cases. Constrained 'motion. Examples 117 Let the arc s be measured from the point of projection, and let the initial velocities of projection in the first two cases be U\_ and Ui. Let the tangential and normal forces in the first case be T^ and N^, when an arc s has been described, and T^ and N^ similarly in the second case ; let the velocities at this point be Vi and ^2. Then as p dr Vo"- mvo -y- = To : and m -^ = JV, . as ^' p '■ and ^wva^ = / T^ds + \m U^. .: hn {vi^ + v^^) = P T^ds + r T2ds + ^m U^^ + ^m U<^ (1 ), and m'^-^^^^N^^-N^ (2). P If the same curve be described freely when both sets of forces are acting, and the velocity be v at arcual distance s, and U be the initial velocity, we must have similarly i97iv^=j\T,+ T2)ds + lmU^ (3), and m-=JVi + iyo^ (4). P Provided that hyiU^^hnUi^ + hnU^^ equations (1) and (3) give and then (4) is the same as (2), which is true. Hence the conditions of motion are satisfied for the last case, if the initial kinetic energy for it is equal to the sum of the kinetic energies in the first two cases. The same proof would clearly hold for more than two sets of forces. Cor. The theorem may be extended as follows. If particles of masses mi, m^, ms... all describe one path under forces Fi, F2, F^...; then the same path can be described by a particle of mass M under all the forces acting simultaneously, provided its kinetic energy at the point of projection is equal to the sum of the kinetic energies of the particles ??ij, wjgj *'J3--- at the same point of projection. 31. A particle moves under the influence of two forces ^ to one point and ^ to another point ; shew that it is possible for the particle to /5 describe a circle, and find the circle, 118 Dynamics of a Particle 32. Shew that a particle can be made to describe an ellipse freely under the action of forces Ar+^, X/4-^ directed towards its foci. 33. A circle, of radius a, is described by a particle under a force .■,. , Tg to a point on its circumference. If, in addition, there be a constant normal repulsive force — ., shew that the circle will still be described freely if the i^article start from rest at a point where V 2^- 2/ 34. Shew that a particle can describe a circle under two forces ^ and 1^ directed to two centres of force, which are inverse points for the circle at distances / and /' from the centre, and that the velocity at any point is f (- f )• 35. A ring, of mass m, is strung on a smooth circular wire, of mass M and radius a ; if the system rests on a smooth table, and the ring be started with velocity v in the direction of the tangent to the wire, shew Mm v''- that the reaction of the wire is always -^ . M + m a 36. 0, A and B are three collinear points on a smooth table, such that OA=a and AB = b. A string is laid along AB and to B is attached a particle. If the end A be made to describe a circle, whose centre is 0, with uniform velocity v, shew that the motion of the string relative to the revolving radius OA is the same as that of a pendulum of length ^-5- , and further that the string will not remain taut unless a > 46. CHAPTER YII MOTION IN A EESISTING MEDIUM. MOTION OF PARTICLES OF VARYING MASS 104. When a body moves in a medium like air, it ex- periences a resistance to its motion which increases as its velocity increases, and which may therefore be assumed to be equal to some function of the velocity, such as kpf{v), where p is the density of the medium and k is some constant depending on the shape of the body. Many efforts have been made to discover the law of resistance, but without much success. It appears, however, that for projectiles moving with velocities under about 800 feet per second the resistance approximately varies as the square of the velocity, that for velocities between this value and about 1350 feet per second the resistance varies as the cube, or even a higher power, of the velocity, whilst for higher velocities the resistance seems to again follow the law of the square of the velocity. For other motions it is found that other assumptions of the law for the resistance are more suitable. Thus in the case of the motion of an ordinary pendulum the assumption that the resistance varies as the velocity is the best approximation. In any case the law assumed is more or less empiric, and its truth can only be tested by enquiring how far the results, which are theoretically obtained by its use, fit with the actually observed facts of the motion. Whatever be the law of resistance, the forces are non- conservative, and the Principle of Conservation of Energy cannot be applied. 120 Dynamics of a Particle 105 In the case of a particle falling under gravity in a resisting medium the velocity will never exceed some definite quantity. For suppose the lavi^ of resistance to be kv"^ . m. Then the downward acceleration \b g — kv'\ and this vanishes when hv'^ = g, i.e. when the velocity = (yjn . This therefore will be the maximum velocity possible, and it is called the limiting or terminal velocity. It follows from this that we cannot tell the height from which drops of rain fall by observing their velocity on reaching the ground. For soon after they have started they will have approximately reached their terminal velocity, and will then continue to move with a velocity which is sensibly constant and very little differing from the terminal velocity. In the case of a ship which is under steam there is a full speed beyond which it cannot travel. This full speed will depend on the dimensions of the ship and the size and power of its engines, etc. But whatever the latter may be, there will be some velocity at which the work that must be done in overcoming the resistance of the water, which varies as some function of the velocity, will be just equivalent to the maximum amount of work that can be done by the engines of the ship, and then further increase of the speed of the ship is impossible. 106. A 'particle falls under gravity {supposed constant) in a resisting medium whose resistance varies as the square of the velocity ; to find the motion if the particle starts from rest. Let V be the velocity when the particle has fallen a distance X in time t from rest. The equation of motion is d'x Let ;. = |, so that ^ = ,(1-1) (1). From (1) it follows that if v equalled k, the acceleration would be zero; the motion would then be unresisted and the velocity of the particle would continue to be k. For this reason k is called the " terminal velocity." Motion in a Resisting Medium 121 dv f v^ From(l), ^dx^^V'!^ so that -^x=\.rr—^.= -^o^(k''-v'') + A f 2vdv , ,,, ,. Since v and x are both zero initially, .. A = log A;^ j2gx (2). It follows that a; = 00 when v=k. Hence the particle would not actually acquire the " terminal velocity " until it had fallen an infinite distance. Again (1) can be written dv /, v"^ , gt _ f dv _ 1 , k+v j^ ' ' k"^ J k'-v'~2k^^k-v^ • Since v and t were zero initially, .-. B = 0. k + v -'- Hence i = e * . k — V \v = kK::^=k\,^n\. (l) (3). e^- +1 From (2) and (3), we have k^ A; , , at cosh- ~ k so that c''^'=cosh^, and a;=— logcosh^ (4). 107. If the 'particle were projected upwards instead of doiumuards, to find the motion. Let V be the velocity of projection. The equation of motion now is _ = _^_^,> = -^(^l + _j (5), where x is measured upwards. 122 Dynamics of a Fartiele Hence .^ = -5,(1+^;). where = - log ( F- + A--) + A. 2gx , V^ + k' '-'-W^^'^Wk^ ; «5). Again (5) gives dv ( v"-\ gt f dv 1, ,v , „ 1 V where ^ ~ T^ tan"^ -y + B. .••f = tan-^-tan-| (7). Equation (6) gives the velocity when the particle has described any distance, and (7) gives the velocity at the end of any time. 108. Ex. A person falls by means of a parachute from a height of 800 yards in 2^ minutes. Assuming the resistance to vary as the square of the velocity, sheio that in a second and a half his velocity diners by less than one per cent, from its value when he reaches the ground and find an approximate value for the limiting velocity. When the parachute has fallen a space x in time (, we have, by Art. 106, if (1 V^^k^[l-e~ k^J (1), t;=^tanh^|*^ (2), and «= -log cosh (■^) (3). \ «/' Here 2400 ^, = log cosh fl^\ . 150g _ l.-flgr . 2400f, e k +c 'k ••• e k^ = 2 (^)- The second term on the right hand is very small, since k is positive. 2400| '-^ Hence (4) is approximately equivalent to e =le .-. 2400.% i^- log 2= ^^, nearly. K- It K Motion in a Resisting Medium. Examples 123 Hence fc = 16 is a first approximation. Putting ft=16 (1 + 2/), (4) gives, for a second approximation, 800(1-2,) / «0(l-a/) + ,-300a-.) e^OOg-^) ^auuu -y) _ _ _ _ _ yery approx. Therefore a second approx. is A; = 16 (1 + -0023), giving the terminal velocity. Also the velocity i>i , when the particle reaches the ground, is, by (1), given by 2.32.2400-1 ri2=/,2|_i = fc2, for aU practical purposes. When V is 99 "/o of the terminal velocity, (2) gives :. e k =199 = 6" , from the Tables. k 16 .*. t = p— X 5'3 = --- X 5"3 = 1'325 approx., 2g bi i.e. t is less than l^secs. EXAMPLES 1. A particle, of mass m, is falling under the influence of gravity- through a medium whose resistance equals fx times the velocity. If the particle be released from rest, shew that the distance fallen 1 ( tL f\ through in time t is a -Ae '»-l + ^L 2. A particle, of mass m, is projected vertically under gravity, the resistance of the air being mk times the velocity ; shew that the greatest y-i height attained by the particle is — [\-log(l+X)], where F is the terminal velocity of the particle and X F is its initial vertical velocity. 3. A heavy particle is projected vertically upwards with velocity ic in a medium, the resistance of which is gw'^tB.n'^a times the square of the velocity, a being a constant. Shew that the particle will return to the point of projection with velocity u cos a, after a time «^-icota(a + log , '"'"^" )■ ■^ \ ° 1 - sui a J 4. A particle falls from rest under gravity through a distance a; in a medium whose resistance varies as the square of the velocity ; if y be the velocity actually acquired by it, v^ the velocity it would have acquired had there been no resisting medium, and F the terminal velocity, shew that Vi? 2 K^"^2.3 V^ 2.3.4 V^'^'" 124 Dynamics of a Particle 5. A particle is projected with velocity V along a smooth horizontal plane in a medium whose resistance per unit of mass is /* times the cube of the velocity. Shew that the distance it has described in time t is and that its velocity then is ■ , — . 6. A heavy particle is projected vertically upwards with a velocity « in a medium the resistance of which varies as the cube of the particle's velocity. Determine the height to which the particle will ascend. 7. If the resistance vary as the fourth power of the velocity, the energy of m lbs. at a depth x below the highest point when moving in a vertical line under gravity Nvill be E tan — ^ when rising, and ^tanh^" when falling, where E is the terminal energy in the medium. 8. A particle is projected in a resisting medium whose resistance varies as (velocity)", and it comes to rest after describing a distance s in time t. Find the values of s and t and shew that s is finite if n < 2, but infinite if «= or > 2, whilst t is finite if n < 1, but infinite if n= or > 1. 9. In the previous question if the resistance be k (velocity) and the V initial velocity be V, shew that v= Fe"** and s= t-(1 -e"*^'). 10. A heavy particle is projected vertically upwards in a medium the resistance of which varies as the square of the velocity. It has a kinetic energy K in its upward path at a given point ; when it passes the same point on the way down, shew that its loss of energy is =r-— -=^ , where A'' is the limit to which its energy approaches in its downward course. 11. If the resistance to the motion of a railway train vary as its mass and the square of its velocity, and the engine work at constant h. p., shew that full speed will never be attained, and that the distance traversed from rest when half-speed is attained is — logg ^ , where fj. is the resistance per unit mass per unit velocity. Find also the time of describing this distance. 12. A ship, with engines stopped, is gradually brought to rest by the resistance of the water. At one instant the velocity is 10 ft. per sec. and one minute later the speed has fallen to 6 ft. per sec. For speeds below 2 ft. per sec. the resistance may be taken to vary as the speed, and for higher speeds to vary as the square of the speed. Shew that, before coming to rest, the ship will move through 900[l-|-loge 5] feet, from the point when the first velocity was observed. Motion in a Resisting Medium. Examples 125 13. A particle moves from rest at a distance a from a fixed point under the action of a force to equal to fx times the distance per imit of mass ; if the resistance of the medium in which it moves be k times the square of the velocity per unit of mass, shew that the square of the velocity when it is at a distance x from is ^' - ^ e^'''~°> + 2X2 [1 - e^'"^""*]- Shew also that when it first comes to rest it will be at a distance h given by {\-2l-b)e'^''^ = {l-\-2l-a)e-^''K 14. A particle falls from rest at a distance a from the centre of the Earth towards the Earth, the motion meeting with a small resistance proportional to the square of the velocity v and the retardation being /x for unit velocity ; shew that the kinetic energy at distance x from the centre is mgr^ l-2/i( 1 — ) - 2/i loge -I , the square of /x being neglected, and r being the radius of the Earth. 15. An attracting force, varying as the distance, acts on a particle initially at rest at a distance a. Shew that, if V be the velocity when the particle is at a distance x, and V the velocity of the same particle when the resistance of the air is taken into account, then "=''[i-|*'-^^^^r^'] nearly, the resistance of the air being given to be k times the square of the velocity per unit of mass, where k is very small. 109. A 23cirt{cle is projected xmder gravity and a resistance equal to mk {velocity) luith a velocity u at an angle a to the horizon; to find the motion. Let the axes of x and y be respectively horizontal and vertical, and the origin at the point of projection. Then the equations of motion are .. _ jds dx _ jdx ~ dt' ds di' . .. , ds dy J dy Integrating, we have log X = — kt + const. = — kt + log (u cos a), and log {ky + g) = — kt + const. = — kt + log (kit sin a + g); .'. a; = I/, cos ae~^* (1), and k;^ + g = {ku sin a + g) e-''^ (2). 126 Dynamics of a Particle ... ^^.'i^ ,-M + ,o„st. = ^" (1 - .-«) (3). , , , ku. sin a + 7 , , and ky -\- gt = r e~*^ + const. Ju^ma^g ^^_^_,,^ ^^^_ tC Eliminating t, we have y ^log(l--^)+_^(.sina + f) ...(5), k^ ^\ ucosaJ MCOsaV kj ^ ^ which is the equation to the path. The greatest height is attained when ^ = 0, i.e. when i.e. at time t logr 1 + / ku sin a\ kus,\noL-\- g' ' ' k ° \ g g , ( ^ ku sin a and then 2/ = —^ P^'^V'^ g It is clear from equations (3) and (4) that when t= oo , X = — y — and 2/ = — 00 , Hence the path has a vertical Vj cos ol asymptote at a horizontal distance — y — from the point of wn projection. Also, then, x = Q and ^ = — '|, i.e. the particle then have just attained the limiting velocity. Cor. If the right-hand side of (5) be expanded in powers of k, it becomes _g Y kx 1 k'^x' 1 k'af _ ] ^^.|_ wcosa 2w"cos^a 3w*cos^a ""J ■ u sm cf + ' %cosa gx^ 1 gkx? 1 gk\\ I.e. y = xiaxia. ^ ^ . ^ „ „ . , Sw^cos^a dM^cos-'a 4M*cos-»a On putting k equal to zero, we have the ordinary equation to the trajectory for unresisted motion. 110. A particle is moving under gravity in a medium luliose resistance = mfju {velocity)- ; to find the motion. When the particle has described a distance s, let its tangent make an angle (p with the upward drawn vertical, and lot v be its velocity. Motion in a Resisting Medium 127 The equations of motion are then '»j^ = -gcos(f>-fMv'- (1), and - = ^sin^ (2). (1) gives -j^ -^ = -2gcos(f>- Ifiv, i.e., from (2), - -7-7 (p sin <^) = — 2 cos ^ — 2/Ap sin <^. Idp p d(f) sin ^ + 3 cos ^ = - '2/xp 1 A. (}.\ 1 _ ?^£1^ 1 _ 2/A ^p/ ' sin^ (^ sin"* ^ ' p sin* <^ * p sin^ J sur 9 ^ cos (i , 1 + cos (6 , . .„ ^ = - /ti -^-^ - /i log — ^-7-^ + ^ ...(3 . ^sm-<^ ^ * sin0 ^ (2) then gives „ r . cos , 1 + cos 01 n v^ A-iJi ^^^ - /x log ^-r^ = -^Vt • |_ ^ sm' '^ o sm J sin^ Equation (3) gives the intrinsic equation of the path, but cannot be integrated further. 111. A bead moves on a smooth wi7'e in a vertical plane under a resistance {= k {velocityy} ; to find the motion. When the bead has described an arcual distance s, let the velocity be v at an angle to the horizon (Fig., Art. 102 ), and let the reaction of the wire be R. The equations of motion are — T- =g sin — kv^ (1), and - = gcos(b—R (2). p ^ T Let the curve be s =/"(0). 128 Dynamics of a F article Then (1) gives a linear equation to give v". Particular case. Let the curve be a circle so that s = a(p, if s and ^ be measured from the highest point. (1) then gives — (v^) + 2akv^=2ag sin d>. a(p 'a vh'^^^'l' = 2agjsm

, IS given by — ^ ---^' =log ( l+Xseca], where X is the ratio of the ^ '' cosa+X ° "■ ■' velocity of projection to the terminal velocity. 3. A particle acted on by gravity is projected in a medium of which the resistance varies as the velocity. Shew that its acceleration retains a fixed direction and diminishes without limit to zero. 4. Shew that in the motion of a heavy particle in a medium, the resistance of which varies as the velocity, the greatest heiglit above the level of the point of projection is reached in less than half the total time of the flight above that level. 5. If a particle be moving in a medium whose resistance varies as the velocity of the particle, shew that the equation of the trajectory can, by a proper choice of axes, be put into the form Motion in a Eesisting Medium. Examples 129 6. If the resistance of the air to a particle's motion be n times its weight, and the particle be projected horizontally with velocity F, shew that the velocity of the particle, when it is moving at an inclination d> w-l _n+\ to the horizontal, is F(l-sin(^) 2 (l+sin0) 2 . 7. A heavy bead, of mass m, slides on a smooth wire in the shape of a cycloid, whose axis is vertical and vertex upwards, in a medium whose resistance is m — and the distance of the starting point from the vertex is c ; shew that the time of descent to the cusp is \/ ^^ Q^-cj V gc where 2a is the length of the axis of the cycloid. 8. A heavy bead slides down a smooth wire in the form of a cycloid, whose axis is vertical and vertex downwards, from rest at a cusp, and is acted on besides its weight by a tangential resistance proportional to the square of the velocity. Determine the velocity after a fall through the height x. 9. If a point travel on an equiangular spiral towards the pole with uniform angular velocity about the pole, shew that the projection of the point on a straight line represents a resisted simple vibration. 10. A particle, moving in a resisting medium, is acted on by a central force -^ ; if the path be an equiangular spiral of angle a, whose pole is at the centre of force, shew that the resistance is ^^^^ ^^'^^^ . 2 7^ 11. A particle, of mass m, is projected in a medium whose resist- ance is mJc (velocity), and is acted on by a force to a fixed point (=7«. /I. distance). Find the equation to the path, and, in the case when 2/^-2 = 9//, shew that it is a parabola and that the particle would ultimately come to rest at the origin, but that the time taken would be infinite. 12. If a high throw is made with a diabolo spool the vertical resistance may be neglected, but the spin and the vertical motion together account for a horizontal drifting force which may be taken as proportional to the vertical velocity. Shew that if the spool is thrown so as to rise to the height h and return to the point of projection, the spool is at its greatest distance c from the vertical through that point when it is at a height -5- ; and shew that the equation to the trajectory is 3 of the form 4^3.^2 = 21 cY' {h-y). 13. If a body move under a central force in a medium which exerts a resistance equal to k times the velocity per unit of mass, prove that -T7^+u=To—5-e-^''i where h is twice the initial moment of momentum about the centre of force. L. D. 9 130 Dy7iamics of a Particle 14. A particle moves with a central acceleration P in a medium of ■which the resistance is k . (velocity)^ ; shew that the equation to its path is -TTi, +«= r^— 5 e^**, where s is the length of the arc described. 15. A particle moves in a resisting medium with a given central acceleration P ; the path of the particle being given, shew that the 1 d ( ,dr ^\ resistance is -■^^J■^ \r ^^ 112. Motion where the mass moving varies. The equation P = mf is only true when the mass m is constant. Newton's second law in its more fundamental form is P = JtO^'^) (!)• Suppose that a particle gains in time Bt an increment Bm of mass and that this increment Bm was moving with a velocity u. Then in time Bt the increment in the momentum of the particle = m .Bv + B7n (v + Bv — u), and the impulse of the force in this time is PBt. Equating these we have, on proceeding to the limit, dv dm dm „ dt dt dt J^(^^^> = ^ + ^^-^ (2>- When u is zero we have the result (1). 113. Ex. 1. A spherical raindrop, falling freely, receives in each instant an increase of volume equal to \ times its surface at that instant; find the velocity at the end of time t, and the distance fallen through in that time. When the raindrop has fallen through a distance .x in time t, let its radius be /• and its mass M. Then T.["'i>'" "»• 4 dr dM Kow 5/=- 7r/»-3, so that Awr^p — = — = p . iXirr-, by the questiou. .*. -r =^, aud r = a + Xt, dt where a is the initial radius. Hence (1) gives ^ |^(a + Xfj3 '-^^=^{a + Mi^g, Motion when the mass varies 131 dt 4\ since the velocity was zero to start with. dx 9 r , since x and t vanish together. .-. x = ^J(a + \t)'^-2a'i + + xd ~ 8 La + XtJ ■ Ex. 2. J mass in the. form of a solid njllnder, the area of ic^ose cross- section is A, moves parallel to its axis, being acted on by a constant force F, through a uniform cloud of fine dust of volume density p which is moving in a direction opposite to that of the cylinder ivith constant velocity V. If all the dust that meets the cylinder clings to it, find the velocity and distance described in any time t, the cylinder being originally at rest, and its initial mass m. Let M be the mass at time t and v the velocity. Then ill. 5v + 5M {v + bv + F)=increase in the momentum in time dt = FSt. ^^dv dM ^^dM „ •• ^^5-t+^dF + ^T.=^ (1) in the limit. Also ^=Ap(v+V) (2). (1) gives Mv + MV=Ft + coQst.=Ft + mV. M^=Ap(Ft + mV). M^^Ap(Ft^ + 2mVt) + m^. Therefore (2) gives M^=Ap (Ft + mV). Therefore (2) gives .= -F+^t;^^=-F+_=£i±^!£= (3)- ^ Jm^ + 2mApVt + AFpt^ Also if the hinder end of the cylinder has described a distance x from rest, '^'^ ■' Vt+ -?- Jnfi + 2mApVt + AFpt^ -~. Ap^ Ap From (3) we have that the acceleration dv _ nfi(F-ApV^) ^* (}ifi + 2mApVt + AFpt'i)^* so that the motion is always in the direction of the force, or opposite, according &S, F^ApV^. Ex. 3. A uniform chain is coiled up on a horizontal plane and one end passes over a small light pulley at a height a above the plane; initially a length b, >a, hangs freely on the other side; find the motion. When the length b has increased to x, let v be the velocity ; then in the time dt next ensuing the momentum of the part (x + a) has increased by 9—2 132 Dynamics of a Particle m(a: + a)5?;, where m is the mass per unit lenp;th. Also a lenc^th m^x has been jerked into motion, and given a velocity v + 5v. Hence TO (a; + a) 5v + mdx (v + Sv) = claange in the momentum = impulse of the acting force = hi.^/ [x-a) . 8t. Hence, dividing by ot and proceeding to the limit, we have .'. V ^-^ . {x + a) +v^^ (x - a) g. .-. f2 (x + ar' = jy (.r2 - «2) g = 2 j^^ - a2 (.r - ^)| g, 2_ % {x-b){x^ + bx + b^-3a^) so that This equation cannot be integrated further. 2a In the particular case when b = 2a, this gives v'^= ~-(x-b), so that the end descends with constant acceleration |, The tension T of the chain is clearly given by T8t = mdx . v, so that T=zmv^. EXAMPLES 1. A spherical raindrop of radius a cms. falls from rest through a vertical height h, receiving throughout the motion an accumulation of condensed vapour at the rate of k grammes per square cm. per second, no vertical force but gravity acting ; shew that when it reaches the ground its radius will be k \/ —\ ^ + \/ ^ + ^2 • 2. A mass in the form of a solid cylinder, of radius c, acted upon by no forces, moves parallel to its axis through a uniform cloud of fine dust, of volume density p, which is at rest. If the particles of dust which meet the mass adhere to it, and if 31 and m be the mass and velocity at the beginning of the motion, prove that the distance x traversed in time t is given by the equation {31+ pird^x)^ = M^ + 2f}7ruc^ML 3. A particle of mass 3f is at rest and begins to move under the action of a constant force F in a. fixed direction. It encounters the resistance of a stream of fine dust moving in the opposite direction with velocity F, which deposits matter on it at a constant rate p. Shew that it« mass will be m when it has travelled a distance -Jm-31 |l+log ^n where k = F-pV 4. A spherical raindrop, whose radius is -04 inches, begins to fall from a height of 64U0 feet, and during the fall its radius grows, by precipitation of moisture, at the rate of 10"* inches per second. If its motion be unresisted, shew that its radius when it reaches the ground is '0420 inches and that it will have taken about 20 seconds to fall. Motion when the mass varies. Examples 133 '' \/o 5. Snow slides off a roof clearing away a part of uniform breadth ; shew that, if it all slide at once, the time in which the roof will be cleared J^^ wf) ' ^"* *^**' ^f ^^^ *0P ™ove first and gradually set the rest in motion, the acceleration is ^g sin a and the time will be / ^^ \' g&in a ' where a is the inclination of the roof and a the length originally covered with snow. 6. A ball, of mass m, is moving under gravity in a medium which deposits matter on the ball at a uniform rate /x. Shew that the equation to the trajectory, referred to horizontal and vertical axes through a point on itself, may be written in the form k-uy =kx{g-it-kv)+gu{\-e^)^ where ^f, v are the horizontal and vertical velocities at the origin and mk = 2fj.. 7. A falling raindrop has its radius uniformly increased by access of moisture. If it have given to it a horizontal velocity, shew that it will then describe a hyperbola, one of whose asymptotes is vertical. 8. If a rocket, originally of mass M, throw off every unit of time a mass eM with relative velocity F, and if M' be the mass of the case etc., shew that it cannot rise at once unless eV>g, nor at all unless ^-jp->g. If it rises vertically at once, shew that its greatest velocity is and that the greatest height it reaches is 9. A heavy chain, of length I, is held by its upper end so that its lower end is at a height I above a horizontal plane ; if the upper end is let go, shew that at the instant when half the chain is coiled up on the plane the pressure on the plane is to the weight of the chain in the ratio of 7 : 2. 10. A chain, of great length a, is suspended from the top of a tower so that its lower end touches the Earth ; if it be then let fall, shew that the square of its velocity, when its upper end has fallen a distance .^•, is 2gr log , where r is the radius of the Earth. 11. A chain, of length I, is coiled at the edge of a table. One end is fastened to a particle, whose mass is equal to that of the whole chain, and the other end is put over the edge. Shew that, immediately after leaving the table, the particle is moving with velocity - * > 1 /5gl^ 134 Dynamics of a Particle 12. A uniform string, whose length is I and whose weight is W, rests over a small smooth pulley with its end just reaching to a horizontal plane ; if the string be slightly displaced, shew that when a length x has been deposited on the plane the pressure on it ia •^[^'o^z^-f]- l — 2x and that the resultant pressure on the i>ulley is W -= , 13. A mass M is attached to one end of a chain whose mass per unit of length is m. The whole is placed with the chain coiled up on a smooth table and Mis, projected horizontally with velocity V. When a length x MV of the chain has become straight, shew that the velocity of M is -,> , M+mx and that its motion is the same as if there were no chain and it were acted on by a force varying inversely as the cube of its distance from a point in its line of motion. Shew also that the rate at which kinetic energy is dissipated is at any instant proportional to the cube of the velocity of the mass. 14. A weightless string passes over a smooth pulley. One end is attached to a coil of chain lying on a horizontal table, and the other to a length I of the same chain hanging vertically with its lower end just touching the table. Shew that after motion ensues the system will first be at rest when a length x of chain has been lifted from the table, 2x such that {l—x)e ' =1. Why cannot the Principle of Energy be directly applied to find the motion of such a system ? 15. A ship's cable passes through a hole in the deck at a height a above the coil in which the cable is heaped, then passes along the deck for a distance b, and out at a hole in the side of the ship, immediately outside of which it is attached to the anchor. If the latter be loosed find the resulting motion, and, if the anchor be of weight equal to 2a+\b of the cable, shew that it descends with uniform acceleration \g. 16. A mass M is fastened to a chain of mass m per unit length coiled up on a rough horizontal plane (coefficient of friction = /i). The mass is projected from the coil with velocity V ; shew that it will be brought to rest in a distance — -^ 1 + — ^ — - 1 V . m\\ 2M,xgJ J 17. A uniform chain, of mass M and length I, is coiled up at the top of a rough plane inclined at an angle a to the horizon and has a mass M fastened to one end. This mass is projected down the plane with velocity F. If the system comes to rest when the whole of the chain is just straight, shew that 7^= -^ sec e sin(e — a), where e is the angle o of friction. Motion ichen the mass varies. Examples 185 18. A uniform chain, of length I and mass ml, is coiled on the floor, and a mass mc is attached to one end and pi'ojected vertically upwards with velocity \llgh. Shew that, according as the chain does or does not completely leave the floor, the velocity of the mass on finally reaching the floor again is the velocity due to a fall through a height where a?—c'''(c-\r 8k). 19. A uniform chain is partly coiled on a table, one end of it being just carried over a smooth pulley at a height h immediately above the coil and attached there to a weight equal to that of a length 2h of the chain. Shew that until the weight strikes the table, the chain uncoils with uniform acceleration ^cf, and that, after it strikes the table, the x-h velocity at any moment is sj^ghe 2A ^ where x is the length of the chain uncoiled. 20. A string, of length Z, hangs over a smooth peg so as to be at rest. One end is ignited and burns away at a uniform rate v. Shew that the other end will at time ^ be at a depth x below the peg, where x is given f d^x \ dx by the equation {l-vt) ( -j-, -Vg] - '^ 77 — 2gx=0. [At time t let x be the longer, and y the shorter part of the string, so that x+9/=l — vi. Also let V, (=x), be the velocity of the string then. On equating the change of momentum in the ensuing time dt to the impulse of the acting force, we have {x+y-v 8t) ( V+ 8 V) - {x+2/) V= (x - y) ght, dV giving {x-\-y)-^-vV= {x -y)g={2x-l + vt) g, etc.] 21. A chain, of mass m and length 2^, hangs in equilibrium over a smooth pulley when an insect of mass M alights gently at one end and begins crawling up with uniform velocity V relative to the chain ; shew that the velocity with which the chain leaves the pulley will be [Let To be the velocity with which the chain starts, so that V- Fq is the velocity with which the insect starts. Then ir(F- Fo) = the initial impulsive action between the insect and chain = ?nFo, so that At any subsequent time t let x be the longer, and y the shorter part of the chain, 2 the depth of the insect below the pulley, and P the force exerted by the insect on the chain. We then have 136 Byiimuics of a Far tide Also x + i/ = 2l. These equations give ( J/+ m) i'2 = 2 {M- m) gx + ^-^ x"- + A. Also, when x=l^ x= Vq, etc.] 22. A uniform cord, of length I, hangs over a smooth pulley and a monkey, whose weight is that of the length k of the cord, clings to one end and the system remains in equilibi'ium. If he start suddenly, and continue to climb with uniform relative velocity along the curd, shew that he will cease to ascend in space at the end of time (T)^^--(-l)' CHAPTER VIII OSCILLATORY MOTION AND SMALL OSCILLATIONS 114. In the previous chapters we have had several examples of oscillatory motion. We have seen that wherever the equation of motion can be reduced to the form x = — n^x, or Q = —v?Q, the motion is simple harmonic with a period of oscillation equal 27r to — . We shall give in this chaj)ter a few examples of a more difficult character. 115. Small oscillations. The general method of finding the small oscillations about a position of equilibrium is to write down the general equations of motion of the body. If there is only one variable, x say, find the value of x which makes x, X ... etc. zero, i.e. which gives the position of equilibrium. Let this value be a. In the equation of motion put ic = a + f , where | is small. For a small oscillation ^ will be small so that we may neglect its square. The equation of motion then generally reduces to the form | = — X^, in which case the time of a small oscillation . 27r For example, suppose the general equation of motion is <^^* Cr X (dx\" „ . , For the position of equilibrium we have F (x) = 0, giving x = a. Fat x=a+ ^ and neglect ^\ 188 Dynamics of a Particle The equation becomes ^, = F{a + ^) = F{a) + ^F'{a)+..., by Taylor's theorem. Since F{a) = this gives -^= ? • F'{a). If F'{a) be negative, we have a small oscillation and the position of equilibrium given by a; = a is stable. If F'{a) be positive, the corresponding motion is not oscillatory and the position of equilibrium is unstable. 116. Ex. 1. A uniform rod, of length 2a, is supported in a horizontal position by tivo strings attached to its ends luhose other extremities are tied to a fixed point; if the unstretched length of each string be I and the modulus of elasticity be n times the toeight of the rod, sliew that in the position of equilibrium the strings are inclined to the vertical at an angle a such that acot a - I cos a — ^r-. In and that the time of a small oscillation about the position of equilibrium is V^iT , „ cot a 2?i cos^ a * When the rod is at depth x below the fixed point, let d be the incliuatiou of each string to the vertical, so that x = a cot Q and the tension sin 6 nmg a - isin 6 "^ I I sin^ The equation of motion is then „ nmq a-lsinO mx = mq - 2 . — r^ . -. — - — cos d, a ■■ 2a cos 6 ., ^ng a-lsin I.e. — ^^^g+ ■ „ ^ e'=g-2~ — ^— r— cos(?, sm^d sm3^ '' I sine i.e. 6i-2cot^^2=:_£sin2 6'+^sin^cos tf (a-Zsin(9) (1). a al ^ ^ In the position of equilibrium when 6 = a, we have ^ = and i) = 0, and .". a cot a- I cos « = ;,— (2)- For a small oscillation put 6 = a + ip, where i// is small, and .•. sin ^ = sin a + ^cosa, and cos 5 = cos a- i/- sin a. In this case 6'^ is the square of a small quantity and is negligible, and (1) gives Small Oscillations . 139 ^= -- (sin a + -^008 0)2 -\ ^ (sin a + 1/- cos a) (cos a -^.t sin a) [a - I (sin o + i^ cos o)] = — (siu2 a + 2^ sin a cos a) H j- [sin a cos a + !/■ (cos^ a - sin^ a)] ( — tan a - /•■/' cos a 1 bj' equation (2j Hence the required time=27r\/-- — =- , V gr 1 + 2« cos3 a Making use of the principle of the last article, if the right-hand side of (1) be/(^), the equation for small oscillations is and /' (a) = — - sin a cos a H f (cos2 a - sin2 a) (a - I sin a) sin a cos2 a = etc., as before. Ex. 2. ^ heavy particle is placed at the centre of a smooth circular table; n strings are attached to it and, after passing over small pulleys symmetrically arranged at the circumference of the table, each is attached to a mass equal to that of the particle on the table. If the particle he slightly displaced, shew that the time of an oscillation is ^wk/ - ( 1 + - Let be the centre of the board, Ai, A^, ..., A^ the pulleys, and let the particle be displaced along a line OA lying between OA^ and OAi. When its distance OP-x, let PA^=yy. and lPOAj.=aj.. Also, let a be the radius of the table and I the length of a siring. Then y^- sja^ + a;^ - 2a a; cos a^ = a ( 1 - - cos a,, j , since x is very small. Let Ty be the tension of the string PA^. Then mg -T^ = m—^[l-yj)— mx cos o^. ,'. Tj. = m (g - i' cos a^). , a cos a,. -as / , x \ = vi (a -xeosar) 1 + - cos o- I " a V a ; = -2 (g - X COS Ur) [_a^ COS a^- ax + ax cos^ aj.]. Nowif PO^i = a, then 2cosa^=cosa-)-cos ( a+ —j + ... ton terms = 0, Scos2a,. = - l-l-cos2a + l + cos( 2a + — j + ... =ni and 2cos3oy=2S[3cosa^ + cos3ar] = 0. 140 . Dynamics of a F article Therefore the equation of motion of P is vix = S2V cos APAr = "2 - (19^ ■ " + O^^"^ 9 ~ "^^' n « a (2 + 71) ' the time of a complete oscillation = 27r . / — — . It can easily be shewn that the sum of the resolved parts of the tensions perpendicular to OP vanishes if squares of x be neglected. Ex. 3. Two particles, of masses m and m', are connected by an elastic string of natural length a and modulus of elasticity X; m is on a smooth table and describes a circle of radius c with uniform angular velocity ; the string passes through a hole in the table at the centre of the circle and m' hangs at rest at a distance c' below the table. Shew that, if m be slightly disturbed, the periods — of small oscillations about this state of steady motion are given by the equation ahmm'pi - {mc + (4c + 3c' - Ba) m' } a\p^ + B(c + c' -a)\^ = 0. At any time during the motion let x and y be the distances of m and m' from the hole and T the tension, so that the equations of motion are m(x-rf')=-T=-xiit^ (1), ilfi'-'^'" <^'- and m'y-m'g-T = m'g-\ '- (3). (2) gives x^d = const. = h, so that (1) gives x = -^ " ^ ^'^ "*" '-^ " "^ ^^^' When x = c, y — c' we have equilibrium, so that x = y — then, and hence from (3) and (4) 7(2 \(c + c'-a) ,. vi'g^m-^= (5). fience (4) and (3) give, on putting a; = c + ^ and j/ = c' + r? where f and ^ are small, .. '7,2 / 3t\ X , . , X r4c + 3c'-3a^ , "1 ^ c3 V c / 7na ^ ^ " am\_ c J and ^■=-A,(.+ ,). To solve these equations, put ^ = ^cos (pt + ^) and 7? = Bcos (pf + j3). On substituting we have A and A [_ ^ am C J am A, + z7r_^2 + A1.o. am' L «'»J Small Oscillations. Examples 141 Equating the two values of -j- thus obtained, we have, on reduction, a^cmm'p* - {mc + m' (4c + 3c' - 3a) } a\p^ + 8 (c + c' - a) \2 = 0. This equation gives two vaUies, p{^ and ps^, for p^, both vaUies being positive. The solution is thus of the form ^= Ai cos (pit + ^i) + A2C0S (pot + ^2) with a similar expression for 77. Hence the oscillations are compounded of two simple harmonic motions . - 27r , 27r whose periods are — and — . Pi P2 EXAMPLES 1. Two equal centres of repulsive force are at a distance 2a, and the law of force is 4 + ^ ; find the time of the small oscillation of a particle on the line joining the centres. If the centres be attractive, instead of repulsive, find the corresponding time for a small oscillation on a straight line perpendicular to it. 2. A heavy particle is attached by two equal light extensible strings to two fixed points in the same horizontal line distant 2a apart ; the length of each string when unstretched was b and the modulus of elasticity is X. The particle is at rest when the strings are inclined at an angle a to the vertical, and is then slightly displaced in a vertical direction ; shew that ,, .,, ,. . ^ /a cot a a-6sino the time of a complete small oscillation is Stt a / . 7-^—0— • 3. Two heavy particles are fastened to the ends of a weightless rod, of length 2c, and osciUate in a vertical plane in a smooth sphere of radius a ; shew that the time of the oscillation is the same as that of a simple pendulum of length — . 4. A heavy rectangular board is symmetrically suspended in a horizontal position by four light elastic strings attached to the corners of the board and to a fixed point vertically above its centre. Shew that the period of the small vertical oscillations is 27r (- + j^j?) where c is the equilibrium-distance of the board below the fixed point, a is the length of a semi-diagonal, k = >\la?-^c' and X is the modidus. 5. A rod of mass m hangs in a horizontal position supported by two equal vertical elastic strings, each of modulus X and natural length a. Shew that, if the rod receives a small displacement parallel to itself, the period of a horizontal oscillation is ^tt kI " ( ~ + h^ ) 142 Dy7iamics of a Particle 6. A light string has one end attached to a fixed point A, and, after passing over a smooth peg B at the same height as A and distant 2a from A, carries a mass F at the other end. A ring, of mass M, can slide on the portion of the string between A and B. Shew that the time of its small oscillation about its position of equilibrium is 477 [aMP iM+ P)-^g {4P^ - M^)^^, assuming that 2P> M. 7. A particle, of mass m, is attached to a fixed point on a smooth horizontal table by a fine elastic string, of natural length a and modulus of elasticity X, and revolves uniformly on the table, the string being stretched to a length b ; shew that the time of a small oscillation for a small additional extension of the string is 27r \/ >. MA_q v 8. Two particles, of masses OTj and »i2, are connected by a string, of length ai + a2, passing through a smooth ring on a horizontal table, and the particles are describing circles of radii oti and 02 with angular velocities ©i and (02 respectively. Shew that miaj(oi'^ = m2a2a)2'^, and that the small oscillation about this state takes place in the time y. mi + TO2 9. A particle, of mass m, on a smooth horizontal table is attached by a fine string through a hole in the table to a particle of mass m' which hangs freely. Find the condition that the particle m may describe a circle uniformly, and shew that, if m' be slightly disturbed in a vertical direction, the period of the resulting oscillation is 27r a/ . . '■ > where a is the radius of the circle. 10. On a wire in the form of a parabola, whose latus-rectum is 4a and whose axis is vertical and vertex downwards, is a bead attached to the focus by an elastic string of natural length - , whose modulus is equal to the weight of the bead. Shew that the time of a small oscillation is Vff' 11, At the corners of a square whose diagonal is 2a, are the centres of four equal attractive forces equal to any function m.f{x) of the distance x of the attracted particle m ; the particle is placed in one of the diagonals very near the centre ; shew that the time of a small oscillation is ^v^2 {!/(«) +/'(a)} . 12. Three particles, of equal mass m, are connected by equal elastic strings and repel one another with a force n times the distance. In Small Oscillations. Examples 143 equilibrium each string is double its natural length ; shew that if the particles are symmetrically displaced (so that the three strings always form an equilateral triangle) they will oscillate in period 2: 13. Every point of a fine uniform circular ring repels a particle with a force which varies inversely as the square of the distance ; shew that the time of a small oscillation of the particle about its position of equilibrium at the centre of the ring varies as the radius of the ring. 14. A uniform straight rod, of length 2a, moves in a smooth fixed tube under the attraction of a fixed particle, of mass m, which is at a distance c from the tube. Shew that the time of a small oscillation is ^.^r^t^ 15. A uniform straight rod is perpendicular to the plane of a fixed uniform circular ring and passes through its centre ; e\ery particle of the ring attracts e\^ery particle of the rod with a force varying inversely as the square of the distance ; find the time of a small oscillation about the position of equilibrium, the motion being perpendicular to the plane of the ring. 16. A particle, of mass M, hangs at the end of a vertical string, of length I, from a fixed point 0, and attached to it is a second string which passes over a small pulley, in the same horizontal plane as and distant I from 0, and is attached at its other end to a mass «i, which is small compared with M. When m is allowed to drop, shew that the system oscillates about a mean position with a period 2n 1 + 5^(2+^2) \/- approximately, and find the mean position. 17. A heavy particle hangs in equilibrium suspended by an elastic string whose modulus of elasticity is three times the weight of the particle. It is then slightly displaced ; sliew that its path is a small arc of a parabola. If the displacement be in a direction making an angle cot~i 4 with the horizon, shew that the arc is the portion of a parabola cut ofi" by the latus-rectum. 117. A particle of mass m moves in a straight line under a force mn- {distance) towards a fixed point in the straight line and under a small resistance to its motion equal to w,, jj, {velocity); to find the motion. The equation of motion is d'x „ dx d'x dx 144 Dynamics of a Particle [This is clearly the equation of motion if the particle is moving so that x is increasing. If as in the second figure the particle is moving so that X decreases, i.e. towards the left, the frictional resistance is towards the right, and equals m . fjLV. But in this case -^ is negative, so that the value of -y is — -i- ; the frictional resistance is thus m[i ( — 777 ) "^- The equation of motion is then d?x , / dx^ which again becomes (1). in/xv. Hence (1) gives the motion for all positions of P to the right of 0, irrespective of the direction in which P is moving. Similarly it can be shewn to be the equation of motion for positions of P to the left of 0, whatever be the direction in which P is moving.] To solve (1), put x = LeP\ and we have jf + fip + w^ = 0, giving p = -^±i^n'-fl, ie. x = Ae~^i*cos \^n^-l^t + B] (2), where A and B are arbitrary constants. If /i be small, then Ae~'i* is a slowly varying quantity, so that (2) approximately represents a simple harmonic motion of period 27r-rA/ ^i* — X' whose amplitude, Ae~^* , is a slowly Oscillations. Resisted motion 145 decreasing quantity. Such a motion is called a oscillation and /a measures the damping. This period depends on the square of fi, so that, to the first order of approximation, this small frictional resistance has no effect on the period of the motion. Its effect is chiefly seen in the decreasing amplitude of the motion, which = A f 1 —^t\ when squares of //, are neglected, and therefore depends on the first power of //-. Such a vibration as the above is called a free vibration. It is the vibration of a particle which moves under the action of no external periodic force. If fjb be not small compared with n, the motion cannot be so simply represented, but for all values of fx, < 2n, the equation (2) gives the motion. From (2) we have, on differentiating, that ^ = when tan [y^»'-^( + i;] =-^=^, = tano.(say)...(3), giving solutions of the form /■ n^ -^ t + B = a, TT + a, 27r + a, ... . Hence x is zero, that is the velocity vanishes, at the ends of periods of time differing by tt 4- a / ''i'^—^ • The times of oscillation thus still remain constant, though they are greater than when there is no frictional resistance. If the successive values of t obtained from (3) are t^, ti, tg, ... then the corresponding values of (2) are --t --t -^f Ae 2 ' cos a, — Ae 2'cosof, Ae 2^»cosa, ... so that the amplitudes of the oscillations form a decreasing g.p. whose common ratio = e~2^*''~*'^ = e 2'^^**°~T. If /i > 2n, the form of the solution changes ; for now p=-lW' L. D. 10 146 Dynamics of a Particle and the general solution is = e~f ^1 cosh \J{^ - n^ . ^ + A ] • In this case the motion is no longer oscillatory. If /i = %i, we have by the rules of Differential Equations a; = Z6>-"«+Ltilfe- <«+■)"« y = = Le-""^ + Lt il/e-"« (1 - 7^ + squares) Y = = Xie-«« + lUe-^ = e-"« (Z, + M^t). Ex. The time of oscillatiou of a particle when there is no frictional resistance is 1^ sees. ; if there be a frictional resistance equal to J x m x velocity, find the consequent alteration in the period and the factor which gives the ratio of successive maximum amplitudes. 118. The motion of the last article may be represented graphically ; let time t be represented by distances measured along the horizontal axis and the displacement x of the particle by the vertical ordinates. Then any displacement such as that of the last article will be represented as in the figure. The dotted curve on which all the ends of the maximum -^< ordinates lie is a? =± J. e ^ cos a. The times ^1^12,^2^3; ^3^4,-. • of successive periods are equal, whilst the corresponding maxi- mum ordinates A^B^, A^B^, ... form a decreasing geometrical progression whose ratio A cosae"2-^^ '_ _^.t J.cosae~2-^^' where t is the time of an oscillation. Oscillations. Periodic Forces 147 If we have a particle moving with a damped vibration of this character, and we make it automatically draw its own dis- placement curve as in the above figure, we can from the curve determine the forces acting on it. For measuring the successive distances G-JJ^, C.^G^, ... , etc, and taking their mean, we have the periodic time r which we found in the last article to be 'iTT-^A/n'-^, so that ^ = n^-^. V 4 -r 4 Again, measuring the maximum ordinates AjB^, A^B^, A B A B AzB^, ..., finding the values of -r-jf y -r^ •••> and taking their —It mean, X, we have the value of the quantity e 2'', so that We thus have the values of 'n? and fx, giving the restorative force and the frictional resistance of the motion. 119. A point is moving in a straight line with an accelera- tion fxx towards a fixed centre in the straight line and with an additional acceleration L cospt; to find the motion. The equation of motion is d'^x -^ = — fix + L cos pt. The solution of this is x = A cos(^/ fit + B) + L jy^ cospt = Aco8['^Jit + B] + _ ^ cos pt (1). If the particle starts from rest at a distance a at zero time, we have B = and A = a „ . .'. x=\ a cos ^iit -\ ,cos pt (2). The motion of the point is thus compounded of two simple harmonic motions whose periods are -7- and — . From the right-hand side of (2) it follows that, if p be 10—2 148 Dynamics of a Particle nearly equal to \//*> tlie coefficient ^ becomes very e^reat ; in other words, the effect of the disturbing acceleration L cos pt becomes very important. It follows that the ultimate effect of a periodic disturbing force depends not only on its magnitude L, but also on its period, and that, if the period be nearly that of the free motion, its effect may be very large even though its absolute magnitude L be comparatively small. If p= V/^, the terms in (2) become infinite. In this case the solution no longer holds, and the second term in (1) = L J- cos [V;Ltf] = Z Lt J- cos Wfi + 7) t = Z Lt T7^7^^'> '^'^^ t^^ + 'y^ ^ 7 = ytt - (V/i + 7)- = — Z Q-T- [something infinite — t sin V/u-i]. Hence, by the ordinary theory of Differential Equations, the solution is x = Ai cos [V/A^ + B^ + 27- t sin Vyu.i. If, as before, x = a and x = when i = 0, this gives x = a cos '^ fit + ^r-— t sin \Jiit, and hence x = ( ^-^ a \/fi j sin ^f fxt + -^^ cos V/u.^. It follows that the amplitude of the motion, and also the velocity, become very great as t gets large. 120. If, instead of a linear motion of the character of the previous article, we have an angular motion, as in the case of a simple pendulum, the equation of motion is ^, = -jd + Lcospt, and the solution is similar to that of the last article. In this case, if Z be large compared with j or if p be very nearly equal to a/j, the free time of vibration, is no longer Oscillations. Periodic Forces 149 small throughout the motion and the equation of motion must be replaced by the more accurate equation ^'^ 9 • a , T ^- = — T sin d -\- L cos pt. at- I ^ 121. As an example of the accumulative effect of a periodic force whose period coincides with the free period of the system, consider the case of a person in a swing to whom a small impulse is applied when he is at the highest point of his swing. This impulse is of the nature of a periodic force whose period is just equal to that of the swing and the effect of such an impulse is to make the swing to move through a continually increasing angle. If however the period of the impulse is not the same as that of the swing, its effect is sometimes to help, and sometimes to oppose, the motion. If its period is very nearly, but not quite, that of the swing its effect is for many successive applications to increase the motion, and then for many further applications to decrease the motion. In this case a great amplitude of motion is at first produced, which is then gradually destroyed, and then produced again, and so on. 122. A particle, of mass m, is moving in a straight line under a force mn^ (distance) towards a fivced point in the straight line, and under a frictional r^esistance equal to ni. /j, {velocity) and a periodic force mL cos pt ; to find the motion. The equation of motion is d-x , dx J — =-n^a^-f.-^^+Lcospt, d^x , dx _ . J- . I.e. -J- + /J, -J- + n^x = L cos pt. The complementary function is Ae--2* cos r^w«-^« + £l (1). assuming fj, < 2n, and the particular integral 1 I- ,,, ,,, _ r (n'-p') COS pt + fip sinpt - n^ -p^ + f,D^ ""^^ P^~ \n' - p'f + ^lY = Q,os(pt-e) (2), fip 150 Dynamics of a Particle where tan e = „^~,. n^ — p^ The motion is thus compounded of two oscillations ; the first is called the free vibration and the second the forced vibration. 27r Particular case. Let the period — of the disturbing force 27r be equal to — , the free period. The solution is then, for the forced vibration, a? = — sin nt fin If, as is usually the case, /* is also small, this gives a vibra- tion whose maximum amplitude is very large. Hence we see that a small periodic force may, if its period is nearly equal to that of the free motion of the body, produce effects out of all proportion to its magnitude. Hence we see why there may be danger to bridges from the accumulative effect of soldiers marching over them in step, why ships roll so heavily when the waves are of the proper period, and why a railway-carriage may oscillate considerably in a vertical direction when it is travelling at such a rate that the time it takes to go the length of a rail is equal to a period of vibration of the springs on which it rests. Many other phenomena, of a more complicated character, are explainable on similar principles to those of the above simple case. 123. There is a very important difference between the free vibration given by (1) and the forced vibration given by (2). Suppose for instance that the particle was initially at rest at a given finite distance from the origin. The arbitrary constants A and B are then easily determined and are found to be finite. The factor e~^* in (1), which gradually diminishes as time goes on, causes the expression (1) to continually decrease and ultimately to vanish. Hence the free vibration gradually dies out. The forced vibration (2) has no such diminishing factor but Motion of a pendulum in a resisting medium 151 is a continually repeating periodic function. Hence finally it is the only motion of any importance. 124. Small oscillations of a simple pendulum lender gravity, where the resistance = /n {velocityf and /x is small. The equation of motion is I'd = - g9 + fjLp6- (1). [If the pendulum start from rest at an inclination a to the vertical, the same equation is found to hold until it comes to rest on the other side of the vertical.] For a first approximation, neglect the small term /xl-d^, and we have = Acosly/^^t + B]. For a second approximation, put this value of 6 in the small terms on the right-hand side of (1), and it becomes • d + ^^e = fxl.^^A'sm'\^'^^t + B] = ^^[l-cos(Vf....)]. .-. e = Acos[^^t + By^ + ^cos[2^it+2B\ (2), where a = AcosB + -— ^ -\ — ^^ cos 2B, 2 6 and = - ^ sin 5 - ^-^ sin 2B. o .'. B = 0, and A = a — ^ ol^/jlI, squares of /j, being neglected. Hence (2) gives »-|.w)cos(yf<) + ^^foos(Vf<) (3), :• (4). , 6 is zero when sin a/ y f =^ 0, i.e. when t= ir \/ -. and hence 152 Dynamics of a Particle The time of a swing from rest to rest is therefore unaltered by the resistance, provided the square of fju be neglected. Again, when 1 = 17 \/ - , Hence the amplitude of the swing is diminished by f a^^- Let the pendulum be passing through the lowest point of its path at time J- (^ + 2^) . where T is small. Then (3) gives = (a - I aV^) (- sin T) + ^^ - ^^^os 2T, and .-. r=«|-^-. Hence the time of swinging to the lowest point ^/: I fir (xixl ^V2+^ and of swinging up to rest again EXAMPLES 1. Investigate the rectiliuear motion given by the equation and shew that it is compounded of two harmonic oscillations if the equation At/^ + Bi/ + C=0 has real negative roots. 2. A particle is executing simple harmonic oscillations of amplitude a, under an attraction ^ . If a small disturbing force ^ be introduced (the amplitude being imchanged) shew that the period is, to a first approximation, decreased in the ratio 1 — 5- : 1. 8/i Oscillatory Motion. Examples 153 3. Two heavy particles, of masses m and m', are fixed to two points, A and B, of an elastic string GAB. The end is attached to a fixed point and the system hangs freely. A small vertical disturbance being given to it, find the times of the resultant oscillations. 4. A particle hangs at rest at the end of an elastic string whose unstretched length is a. In the position of equilibrium the length of the string is h, and — is the time of an oscillation about this position. At time zero, when the particle is in equilibrium, the point of suspension begins to move so that its downward displacement at time t is c&mpt. Shew that the length of the string at time t is , cnp . ^ , cf^ ■ — .-, ^ „ sni )it H — sin pt. ji- —JO'' n' —p'' ' lip=n, shew that the length of the string at time t is h+-s,in'nt- -^cQBnt, 5. A helical spring supports a weight of 20 lbs. attached to its lower end ; the natural length of the spring is 12 inches and the load caases it to extend to a length of 13| inches. The upper end of the spring is then given a vertical simple harmonic motion, the full extent of the displace- ment being 2 inches and 100 complete vibrations occurring in one minute. Neglecting air resistance and the inertia of the spring, investigate the motion of the suspended mass after the motion has become steady, and shew that the amplitude of the motion set up is about 3^ inches. 6. If a pendulum oscillates in a medium the resistance of which varies as the velocity, shew that the oscillations are isochronous. 7. Shew that the time of oscillation of a simple pendulum in a medium whose resistance varies as the cube of the velocity bears a constant ratio to that in a vacuum. 8. A pendulum performs small oscillations in a medium of which the resistance varies as the square of the velocity ; given the number of oscillations in which the arc of oscillation is reduced one-half, compare the original resistance with the weight of the pendulum. 9. The point of suspension of a simple pendulum of length I has a horizontal motion given by x=acos,mt. Find the effect on the motion of the particle. Consider in particular t£e* motion when m^ is equal, or nearly equal, to J. In the latter case if the pendulum be passing through its vertical position with angular velocity a at zero time, shew that, so long as it is small, the inchnation to the vertical at time t J][^-^>^^^^h 154 Dynamics of a Particle [If Cy be the position of the point of suspension at time t its accelera- tion is X. Hence the accelerations of P, the bob of the pendulum, are l() perpendicular to O'P, W along PO, and x parallel to OCf, Hence resolving perpendicular to O'P^ W+a'cos5= -g^\\i6= -(jO, I.e. 0=-j-$+ —— cos mt, suice 6 is small. No-w solve as in Art. 119.] 10. The point of support of a simple pendulum, of weight to and length /, is attached to a massless spring which moves backwards and forwards in a horizontal line; shew that the time of vibration = 27r a/ - I l + "n/)> 'where W is the weight required to stretch the spring a distance I. 11. Two simple pendulums, each of length a, are hung from two points in the same horizontal plane at a distance b apart ; the bob of each is of mass m and the mutual attraction is , ,. , ,, , where X is small compared (dist.)2 ' '■ with g ; shew that, if the pendulums be started so that they are always moving in opposite directions, the time of oscillation of each is 27r k/ - ( H — T§— ) nearly, about a mean position inclined at -j^ radians nearly to the vertical. 12. A pendulum is suspended in a ship so that it can swing in a plane at right angles to the length of the ship, its excursions being read off on a scale fixed to the ship. The free period of oscillation of the pendulum is one second and its point of suspension is 10 feet above the centre of gravity of the ship. Shew that when the ship is rolling through a small angle on each side of the vertical with a period of 8 sees., the apparent angular movement of the pendulum will be approximately 20 per cent. greater than that of the ship. 13. The point of suspension of a simple pendulum of length I moves in a horizontal circle of radius a with constant angular velocity o> ; when the motion has become steady, shew that the inclination a to the vertical of the thread of the pendulum is given by the equation co^ {a + 1 sin a) — g ta,n a = 0. 14. A pendulum consists of a light elastic string with a particle at one end and fastened at the other. In the position of equilibrium the string is stretched to | of its natural length I. If the particle is slightly displaced from the position of equilibrium and is then let go, trace its subsequent path and find the times of its component oscillations. CHAPTER IX MOTION IN THREE DIMENSIONS 125. To find the accelerations of a particle in terms of polar coordinates. Let the coordinates of any point P be r, 6, and (p, where r is the distance of P from a fixed origin 0, 6 is the angle that OP makes with a fixed axis Oz, and <^ is the angle that the plane zOP makes with a fixed plane zOx. Draw PN perpendicular to the plane xOy and let ON = p. Then the accelerations of „ d'x d^y , d^'z , P are -rnr , -tit , and -j- , where df dt- dt- X, y and z are the coordinates of P. Since the polar coordinates of N, which is always in the plane xOy, are p and increasing, we have /de\^ . afdcpy [dt)-p''''^[-i) =s-Kfr— K^y ^^>' and ^ = lil/A) 1 ^f^sin'^^) (3). ^ pdtV dt) r sm e dt\ dt ) ^ ^ 126. Cylindrical coordinates. It is sometimes convenient to refer the motion of P to the coordinates z, p, and ^, which are called cylindrical coordinates. As in the previous article the accelerations are then Motion in three dimensions Vol and -T- f|0^ -^ j perpendicular to the plane zPK, -7- parallel to Oz. 127. A particle is attached to one end of a string, of length I, the other end of which is tied to a fixed point 0. When the string is inclined at an acute angle a to the downward-drawn vertical the particle is projected liorizontally and perpendicular to the string with a velocity V ; to find the resulting motion. In the expressions (1), (2) and (3) of Art. 125 for the accelerations we here have r=l. The equations of motion are thus W-l sin- 64)'' = + q cos 9. Id -I cos 6 sin e4> 1 sin 6 dt The last equation gives sin^ d^} = constant = sin^ a [ cos- a + '2n^ cos a. sm a 4 cos a i.e. „ „ V^ ^ Ig sin a tan a. The tension of the string at any instant is now given by equation (1). In the foregoing it is assumed that T does not vanish during the motion. The square of the velocity at any instant = (W)^ + (l sin e(j>y = ^2 (6^ + 4>' sin'' 6). Hence the Principle of Energy gives ^l' (6^ + (j)^ sin^ (9) = iw F2 - mgl (cos a - cos 0). [On substituting for from (4) we have equation (6).] (1) then gives T o (vel.)' a .^^- %^ (cos a - cos d) = -V- + 5r (3 cos ^ — 2 cos a). 128. In the previous example d is zero when 6 = a., i.e. the particle revolves at a constant depth below the centre as in sin^ CL the ordinary conical pendulum, if F^ =gl . Suppose the particle to have been projected with this velocity, and when it is revolving steadily let it receive a small displacement in the plane NOP, so that the value of ^ was not Motion in three dimensio7is 159 instantaneously altered. Putting 6 = a-\-'f, where -v/r is small, the equation (6) of the last article gives V a sin* a cos (a + ilr) g . ^ Zcosa sin*(a + '«|r) I lin a r 1 — yfr tan a -v/r cot a)* (1 + \lf cot a) , neglecting squares of yjr, gsin — —J — •^ (tan a + 4 cot a) g 1 + 3 cos^ g I cos a so that the time of a small oscillation about the position of , . •,•! • ■ -. /l cos a relative equilibrium is ^ir ^ -z. — 7^ z- . ^ V ^ 1 + 3 cos^ a Again, from (4), on putting ^ = a + -v/r, we have so that during the oscillation there is a small change in the value of (p whose period is the same as that of -\|r. 129- ^ particle moves on the inner surface of a smooth cone, of vertical angle 2a, being acted on by a force towards the vertex of the cone, and its direction of motion always cuts the generators at a constant angle /3; find the motion and the law of force. Let F.m he the force, where m is the mass of the particle, and B the reaction of the cone. Then in the accelerations of Ai't. 125 we have 6 = a and therefore ^ = 0. Hence the equations of motion are and Also, since the direction of motion always cuts OP at an angle /3, r sin ac Oh- . , /rf0\2 ....(1), rsinacosa(^~j =-- . ....(2), r dt\ dt } (3). (3) gives nnd therefore, from (4), = tanS . r "^ r^ -r- = constant = A (If di . A — =:8inacot a . — dt ^ r 160 Dynamics of a Particle Substituting in (1), we have - f'= - 8in2 a cot2 /3 . -3- - sin'^ a . ^ , ^2sin2a 1^_^ sin2|3 V~^ i.e. F='^-^^.=,^r (7). „ /dr\2 „ . „ /d0\2 ^2sin2a so that ''=^- , , . 12 42 gin d cos a _ 8in2 j8 cos a Again, (2) gives — = 5 =•* ; 6 ' ' / f' m 1-3 sin a From (4), the path is given by r = ro.e«'""°°^^-"''. EXAMPLES 1. A heavy particle moves in a smooth sphere j shew that, if the velocity be that due to the level of the centre, the reaction of the sui-face will vary as the depth below the centre. 2. A particle is projected horizontally along the interior surface of a smooth hemisphere whose axis is vertical and whose vertex is downwards ; the point of projection being at an angular distance /3 from the lowest point, shew that the initial velocity so that the particle may just ascend to the rim of the hemisphere is ^/2agsec^. 3. A heavy particle is projected horizontally along the inner surface of a smooth spherical shell of radius -j- with velocity \/ ~^ at a depth — below the centre. Shew that it will rise to a height - above the centre, and that the pressure on the sphere just vanishes at the highest point of the path. 4. A particle moves on a smooth sphere under no forces except the pressure of the surface ; shew that its path is given by the equation cot ^ = cot /3 cos (^, where 6 and (f> are its angular coordinates. 5. A heavy particle is projected with velocity V from the end of a horizontal diameter of a sphere of radius a along the inner surface, the direction of projection making an angle /3 with the equator. If the particle never leaves the surface, prove that 3siu2/3<2 + ( — 1 . 6. A particle constrained to move on a smooth spherical surface is projected horizontally from a point at the level of the centre so that its angular velocity relative to the centre is a. If a>^a be very great compared with g, shew that its depth z below the level of the centre at time t is -|sin2^ approximately. Motion in three dimensions. Examples 161 7. A thin straight hollow smooth tube is always inclined at an angle a to the upward drawn vertical, and revolves with uniform velocity o> about a vertical axis which intersects it. A heavy particle is projected from the stationary point of the tube with velocity ^ cot a : shew that in time t it has described a distance , . ° n _g-^= — , ^ . „ .—. and COS^asm'^a r* r= — ■ --1 ) gi^ng »'^= — ., ^ ■ ^ I ^, — 5 ) , where o? is a constant, cos'^asm^a r* cos- a sin'' a V" *"/ Hence I -T7 1 =r^. ,^ . Hence (^='y-sm"i- . . •. -;- = sm(7-<^)=cos^, if the initial plane for <^ be properly chosen. This is the plane ^=c?sin a, which is a plane parallel to the axis of the cone. The locus is thus a hyperbolic section of the cone, the parallel section of which through the vertex consists of two straight lines inclined at 2a. Hence, etc.] 23. If a particle move on the inner surface of a right circular cone under the action of a force from the vertex, the law of repulsion being rrifA — -g 9^ ' '"'here 2a is the vertical angle of the cone, and if it be projected from an apse at distance ' a with velocity /v/ - sin a, shew that the path will be a parabola. [Show that the plane of the motion is parallel to a generator of the cone.] 24. A particle is constrained to move on a smooth conical surface of vertical angle 2a, and describes a plane curve under the action of an attraction to the vertex, the plane of the orbit cutting the axis of the cone at a distance a from the vertex. Shew that the attractive force 1 a cos a must vary as 5 — . 25. A particle moves on a rough circular cylinder under the action of no external forces. Initially the particle has a velocity F in a direction making an angle a with the transverse plane of the cylinder ; shew that the space described in time t is ^log 1 + ^^ " t . /I [_ « J [Use the equations of Art. 126.] 11—2 1G4 Difnamics of a Particle 130. A point is moving along any curve in three dimensions; to find its accelerations along (1) the tangent to the curve, (2) the principal normal, and (3) the binomial. If (x, y, z) be the coordinates of the point at time t, the accelerations parallel to the axes of coordinates are x, y, and z. dx _ dx ds dt ds dt ' Now So and d^'x dx dh ~d^dt' d^y dt^' dydh ~dsdf d'z dt' '' _dzd\ ds dt o + d?x /ds\ d? [Jtj d^y /ds\ dtj d^z fds ds- [dt The direction cosines of the tangent are , Hence the acceleration along it dx d-x dy d^y dz d-z Tslt'^'^'dsW'^dsW- dx dy ds and •(I). •(2), .(3). dz ds ' _ d?s T/^Y (dyV- fdz\f\ fclsV p ~ dt' [[ds] ■*" [dsj "^ [dsj J "^ [dtj dx d'x dy d^y ds ds" ds ds^ dzdH ds ds' _d^ ~dt' since and therefore •(4), ^hm-m-^' dx d'x dy d'y dz d-z _ _ ds ds^ ds ds' ds d^ d'X d'u The direction cosines of the principal normal are p ^^, p -r^ d'z and p -1—, where p is the radius of curvature. '^ ds' '^ Hence the acceleration along it _ d'x d'x d-y d'y d'z d-z ~^'d£-'dl''^^d^' IX''^ ^di'dt' _ d's r^ d'x dy d'y , dz d,'z' ~Pdt' d^dT''^ d'y dzd^l ds' ds ds'] Motion in three dimensions 165 dsV 1 1 /ds The direction cosines of the binormal are proportional to dy d"z dz d^y dz^ d'^x _ dx^ d'^z , dx d^y dy d'^x ds ds^ ds ds^ ' ds o?s^ ds ds- ds ds^ ~ ds ds^ ' On multiplying (1), (2) and (3) in succession by these and adding, the result is zero, i.e. the acceleration in the direction of the binormal vanishes. The foregoing results might have been seen at once from equations (1), (2), (3). For if (l^, Wi, %), (Zj, m^, n^) and (^3, iris, W3) are the direction cosines of the tangent, the principal normal, and the binormal, these equations may be written d?x_d?s fl (dsY\ dt^-^' dt^ '^'" \p[dij]' d^y d^s fl /dsV) , d^z d^'s (1 fds\-\ These equations shew that the accelerations along the axes are the components of d^s an acceleration -^-j along the tangent, an acceleration - [-t.) along the principal normal, and nothing in the direction of the binormal. We therefore see that, as in the case of a particle describing a plane curve, the accelerations are -r^ , or v -7- , along the tan- gent and — along the principal normal, which lies in the osculating plane of the curve. 131. A particle moves in a curve, there being no friction, under forces such as occur in nature. Shew that the change in its kinetic energy as it passes from one position to the other is 166 Dynamics of a Farticle independent of the path pursued and depends only on its initial and final positions. Let X, Y, Z be the components of the forces. By the last article, resolving along the tangent to the path, we have m^ = X—+ Y-^ + Z—. dt^ ds ds ds' I • *"" iltf " /^^^"^ "^ ^^^ "^ '^'^^^• Now, by Art. 95, since the forces are such as occur in nature, i.e. are one-valued functions of distances from jfixed points, the quantity Xdx + Ydy + Zdz is the differential of some function ^ {x, y, z), so that 1 , 1 (ds^ , , , ^ where \mv^ = <|) («o, 2/o . ■^o) + G, (xq, 2/o, Zo) being the starting point and Vq the initial velocity. Hence ^mv- — ^mv^ = (f> (x, y, z) — (j) (x^, y^, Zo). The right-hand member of this equation depends only on the position of the initial point and on that of the point of the path under consideration, and is quite independent of the path pursued. The reaction R of the curve in the direction of the principal normal is given by the equation p where p is the radius of curvature of the curve. 132. Motion on any surface. If the particle move on a surface whose equation is f(x, y, z) = 0, let the direction cosines at any point (x, y, z) of its path be (^i, m-^, n-^, so that ^1 mj Wi 1 "5"= ^= 1= , /JWuW^\'' dx dy dz V \dxj "^ \dyj "^ \dz) Then, if R be the normal reaction, we have d'^x „ „- d'^y -.^ „ , d-z „ „ m -J- = Z -t- Rl, m -^, = Y+ Rm^, and m ^- = Z + Rn^, dt- dt- dt- where A', Y, Z are the components of the impressed forces. Motion in three dimensions 167 Multiplying these equations by -^ , ~, -^ and adding, we have 1 d \(dx\^^(dyV r'dtVKdtj'^Kdt)^ for the coefficient of R , dx dy dz C?2 dt dt dt dx Ts dy dz\ ds ''^ ds ^ ds) dt = J- x the cosine of the angle between u tangent line to the surface and the normal = 0. Hence, on integration, ^mv' =J(Xdx + Ydy + Zdz), as in the last article. Also, on eliminating R, the path on the surface is given by d'-x "^d^-^ d'y V d-z „ dt^ ^ ^rf^-^ ^1 m^ Hi ' giving two equations from which, by eliminating t, we should get a second surface cutting the first in the required path. 133. Motion under gravity of a particle on a smooth surface of revolution whose axis is vertical. 168 Dynamics of a Particle Use the coordinates z, p and of Art. 126, the ^•-axis being the axis of revolution of the surface. The second equation of that article gives ~ j: (p^ -?? ) = 0, i.e. />' -^ = constant = /i (1). Also, if s be the arc AP measured from any fixed point A, ds the velocities of P are -j along the tangent at P to the gene- rating curve, and p ~ perpendicular to the plane zAP. Hence the Principle of Energy gives H(S)'-^K§)}=— ^^ <^>- Equations (1) and (2) give the motion. Equation (1) states that the moment of the momentum of the particle about the axis of z is constant. , equation (2) easily gives which gives the differential equation of the projection of the motion on a horizontal plane. EXAMPLES 1. A smooth helix is placed with its axis vertical and a small bead slides down it under gravity ; shew that it makes its first revolution from rest in time 2 ^ / — ^-^ , where a is the angle of the helix. V g sm a cos a 2. A particle, without weight, slides on a smooth helix of angle a and radius a under a force to a fixed point on the axis equal to Motion in three dimensions. Examx^les 169 TO/A (distance). Shew that the reaction of the curve cannot vanish unless the greatest velocity of the particle is a \j \i sec a. 3. A smooth paraboloid is placed with its axis vertical and vertex downwards, the latus-rectum of the generating parabola being 4a. A heavy particle is projected horizontally with velocity F at a height h above the lowest point ; shew that the particle is again moving hori- F2 zontally when its height is -^ . Shew also that the reaction of the paraboloid at any point is inversely proportional to the corresponding radius of curvature of the generating parabola. 4. A particle is describing steadily a circle, of radius h, on the inner sui'face of a smooth paraboloid of revolution whose axis is vertical and vertex downwards, and is slightly disturbed by an impulse in a plane through the axis ; shew that its period of oscillation about the steady hi. ^ yi motion is tt */ — y~ ■> wtiere I is the semi-latus-rectum of the paraboloid. 5. A particle moving on a paraboloid of revolution under a force parallel to the axis crosses the meridians at a constant angle ; shew that the force varies inversely as the fourth power of the distance from the axis. 6. A particle moves on a smooth paraboloid of revolution under the action of a force directed to the axis which varies inversely as the cube of the distance from the axis ; shew that the equation of the projection of the path on the tangent plane at the vertex of the paraboloid may, under certain conditions of projection, be written V4a^+r'' + a log -V= =k.d. 'sj4.a? + r^ + 2a where 4a is the latus-rectum of the generating parabola. CHAPTER X MISCELLANEOUS THE HODOGRAPH. MOTION ON REVOLVING CURVES. IMPULSIVE TENSIONS OF STRINGS 134. The Hodograph. If from any fixed point we draw a straight line OQ which is parallel to, and proportional to, the velocity of any moving point P, the locus of Q is called a hodograph of the motion of P. If P' be a consecutive point on the path and OQ' be parallel and proportional to the velocity at P', then the change of the velocity in passing from P to P' is, by Art. 3, represented by QQ'. If T be the time of describing the arc PF\ then the accele- ration of P = Limit — -^ = velocity of Q in the hodograph. T=0 T Hence the velocity of Q in the hodograph represents, both in magnitude and direction, the acceleration of P in its path. The Hodograph 171 It follows that the velocity, or coordinate, of Q in any direction is proportional to the acceleration, or velocity, of P in the same direction. The same argument holds if the motion of P is not coplanar. If at any moment x and y be the coordinates of the moving- point P, and ^ and tj those of the corresponding point Q of the hodograph, we have where A, is a constant. The values of -^ and -^ being then known in terms of t, we eliminate t between these equations and have the locus of (^, ??), i.e. the hodograph. So for three-dimensional motion. 135. The hodograph of a central orbit is a reciprocal of the orhit with respect to the centre of force S turned through a right angle about S. Let SY be the perpendicular to the tangent at any point P of the orbit. Produce SY to P' so that SY. SP' = ^, a constant; the locus of P' is therefore the reciprocal of the path with respect to >S'. By Art. 54, the velocity t; of P = ^= ^ . ^P'. Hence SP' is perpendicular to, and proportional to, the velocity of P. The locus of P' turned through a right angle about S is therefore a hodograph of the motion. The velocity of P' in its path is therefore perpendicular to and equal to y times the acceleration of P, i.e. it = -y- X the central acceleration of P. h EXAMPLES 1. A particle describes a parabola under gravity ; shew that the hodograph of its motion is a straight line parallel to the axis of the parabola and described with uniform velocity. 172 Dynamics of a Particle 2. A particle describes a conic section under a force to its focus ; shew that the hodograph is a circle which passes through the centre of force when the path is a parabola. 3. If the path be an ellipse described under a force to its centre, shew that the hodograph is a similar ellipse. 4. A bead moves on the arc of a smooth vertical circle starting from rest at the highest point. Shew that the equation to the hodograph IS / = Xsin -. 5. Shew that the hodograph of a circle described under a force to a point on the circumference is a parabola. 6. The hodograph of an orbit is a parabola whose ordinate increases uniformly. Shew that the orbit is a semi-cubical parabola. 7. A particle slides down in a thin cycloidal tube, whose axis is vertical and vertex the highest point ; shew that the equation to the hodograph is of the form r^ = '2g\a + hcQS.'i6\ the particle starting from any point of the cycloid. If it start from the highest point, shew that the hodograph is a circle. 8. A particle describes an equiangular spiral about a centre of force at the pole ; shew that its hodograph is also an equiangular spiral. 9. If a particle describe a lemniscate under a force to its pole, shew TT — ^d that the equation to the hodograph is r^ = a^ sec^ — - — . o 10. If the hodograph be a circle described with constant angular velocity about a point on its circumference, shew that the path is a cycloid. 11. Shew that the only central orbits whose hodographs can also be described as central orbits are those where the central acceleration varies as the distance from the centre. [In Art. 135 if SP meet the tangent at P' in T, then S7' is perpen- dicular to Y'P' and = ^^ . The hodograph is described with central acceleration to S if the velocity of P' x SY' is constant, i.e. if the central acceleration of Px ^ is constant. Hence the result.] 12. If the path be a helix whose axis is vertical, described under gravity, shew that the hodograph is a curve described on a right circular cone whose semi-vertical angle is the complement of the angle of the helix. Revolving Curves 173 136. Motion on a Revolving Curve. A given curve turns in its own plane about a given fixed point with constant angular velocity a> ; a small bead P moves on the curve under the action of given forces wJiose components along and perpen- dicular to OP are X and Y ; to find the motion. Let OA be a line fixed in the plane of the curve, and OB a line fixed with respect to the curve, whichat zero time coincided with 0^, ^ / so that, at time ^, Z ^ Oi? = ft)t N\ -/^ At time t let the bead be at P, '^^^•'^--^^p where OP = r, and let cf) be the angle between OP and the tangent to the curve. Then, by Art. 49, the equations of motion are jo^ d'r (de y X R . ^ °^'' ' ^ r dt [ \dt J] m m ^ d'r fdev ^ ^ de X R . ^ Ihese ffive t:: —t\-j-] = rco^ + Zrco -77 + sin 0, ° dt-" \dtj dt m m ^ 1 d ( ,dd\ „ dr Y R and ~ ji I ^ "77 = ~ 2a) -7- H h — cos rf>. r dt\ dtj dt m m ^ Let V be the velocity of the bead relative to the wire, so that . dr , . , dd V cos C) = -TT and i; sin = r -7- . ^ dt ^ dt The equations of motion are then d'r fdey ^ X R' . ^ ,^, 1 d f ,dd\ Y R' /?' 7? where — = 2a)V (3). mm, ^ ' These equations give the motion of the particle relative to the curve. 174 Dynamics of a Particle Now suppose that the curve, instead of rotating, were at rest, and that the bead moved on it under the action of the same forces X and Y as in the first case together with an additional force mwl^r along OP, and let 8 be the new normal reaction. The equations of motion are now dP \dtj mm ^ ^ ^ , \dfdd\ Y 8 ^ ,., and - J. k^ :7T = — + - cos (^ . . .(o). rdt\ dtj mm ^ ^ ^ These equations are the same as (1) and (2) with S substi- tuted for R'. The motion of the bead relative to the curve in the first case is therefore given by the same equations as the absolute motion in the second case. The relative motion in the case of a revolving curve may thus be obtained as follows. Treat the curve as fixed, and put on an additional force on the bead, away from the centre of rotation and equal to mar.r; then find the motion of the bead ; this will be the relative motion when the curve is rotating. The reaction of the curve, so found, will not be the actual reaction of the moving curve. To get the latter we must by (3) add to the reaction, found by the foregoing process, the quantity 2m(ov, where v is the velocity of the bead relative to the curve. The above process is known as that of "reducing the moving curve to rest." When the moving curve has been reduced to rest, the best accelerations to use are, in general, the tangential and normal ones of Art. 88. 137. If the angular velocity about the centre of the tube equal to 4 tan-i ( tanh ~ j . Shew also that the reaction between the tube and the particle is then equal to 2wiaw2cos| ( 3 cos | -2 ) . P being the position of the particle at time t, and 0= / ACP, we may treat the tube as at rest if we assume an additional force along OP equal to mw^ . OP, i.e. mu^. 2a cos ^. R' being the normal reaction in this case, we have, on taking tangential and normal accelerations, a^= -w2. 2acos^sin ^ (1), and a^2=__„2,2acos2| (2). (1) gives ^2 = 2w2cos0 + 4 (3). Now, if the tube revolves J , then, since the particle was initially at rest, its initial relative velocity was w . OA, i.e. w . 2a, in the opposite direction. Hence ^ = 2w initially. Therefore (3) gives (^2 = 2c>j2 (1 + cos = Q), d?x . R . xoi- H — sm 136 be revolving with uniform dt"" S Z - + - m m and d?y df' R . Y ■■ cos U -\ m m .(1), .(2), •(3), where X, F, and Z are the components of the impressed forces parallel to x and y, and perpendicular to the plane of the curve, and d is the inclination of the normal to the axis of y. Equations (1) and (3), which give the motion of the bead relative to the wire, are the same equations as we should have if tve assumed the wire to be at rest and put on an additional Revolving Curves 177 force 77? ft)' X distance perpendicular to and away from the axis of rotation. Hence, applying this additional force, we may treat the wire as at rest, and use whatever equations are most convenient. 140. As a numerical example, let the curve be a smooth circular wire revolving about its vertical diameter, G being its centre and a its radius. Let the bead start from rest at a point indefinitely close to the highest point of the wire. Treat the circle as at rest, and put on an additional force 7710)2 . NP{ = mt, the particle and being initially at a distance a apart ; shew also that the reaction between the tube and the particle then is maaP' [2 cosh o)^ - 1]. 2. A circular tube, of radius a, revolves uniformly about a vertical diameter with angular velocity \/ ~ , and a particle is projected from its lowest point with velocity just sufficient to carry it to the highest point ; shew that the time of describing the first quadrant is 3. A particle P moves in a smooth circular tube, of radius a, which turns with uniform angular velocity w about a vertical diameter ; if the angular distance of the particle at any time t from the lowest point is d, and if it be at rest relative to the tube when 6 = a, where cos^ = - a /- , 2 w V a then, at any subsequent time t, cot - = cot - cosh { at sin - ) . L. D. 12 y; 178 Dy7iamics of a Particle 4. A thin circular wire is made to revolve about a vertical diameter with constant angular velocity. A smooth ring slides on the wire, being attached to its highest point by an elastic string whose natural length is equal to the radius of the wire. If the ring be slightly displaced from the lowest point, find the motion, and shew that it will reach the highest point if the modulus of elasticity is four times the weight of the ring. 5. A smooth circular wire rotates with uniform angular velocity « about its tangent line at a point A. A bead, without weight, slides on the wire from a position of rest at a point of the wire very near A. Shew that the angular distance on the wire traversed in a time t after passing the point opposite J. is 2 tan"^ a)t. 6. A small bead slides on a circular arc, of radius a, which revolves with constant angular velocity &> about its vertical diameter. Find its position of stable equilibrium according as ©^ < - , and shew that the time of a small oscillation about its position of equilibrium is, for the two ,. , , . 27r(Ba J 27r cases, respectively equal to - ana . - . V a"" 7. A parabolic wire, whose axis is vertical and vertex downwards, rotates about its axis with uniform angular velocity w. A ring is placed at any point of it in relative rest ; shew that it will move upwards or downwards according as ©^ < ^, and will remain at rest if cos a ,v/ y • 12. A particle, of mass m, is placed on a horizontal table which is lubricated with oil so that the force on the particle due to viscous friction is mku, where tt is the velocity of the particle relative to the table. The table is made to revolve with uniform angular velocity w about a vei-tical axis. Shew that, by properly adjusting the circumstances of projection, the equations to the path of the particle on the table will be /^2 where a + z/3 = */ — + ia^k. [With the notation of Art. 51, the equations of motion are {D'^-\-kD-a>^)x-2o>Dy = Q and {D^+kD-(o^)y+2a>Dx = 0. Hence \^D^ +kD — ay^ + 'iiu}D']{x + iy) = 0. Now solve in the usual manner.] 13. A smooth horizontal plane revolves with angular velocity ds~ d\p' c ' d27'_d2r cos^i^ 2 sin i/> cos3 )/ > dr ds^'df^ ~c2~ ^ d^ ' Hence equation (4) gives d^T , . ,dT ^ , . , dT t.e. — 2Cos^-sm\^^ = Tcosnji - 2Enl - 2Flm, 1 _ M.IO "^ OQ' OQ' ' where M is the mass of the body and K is some linear factor. If (x, y, z) are the coordinates of Q referred to the axes OX, OF, 0^ this gives Ax' + Bif + Gz' - 2Byz -2Ezx- 2Fxy = MK' . . .(1). The locus of the point Q is thus an ellipsoid, which is called the momental ellipsoid of the body at the point 0. Since the position of Q is obtained by a physical definition, which is independent of any particular axes of coordinates, we arrive at the same ellipsoid whatever be the axes OX, Y, OZ with which we start. It is proved in books on Solid Geometry that for every ellipsoid there can be found three perpendicular diameters such that, if they be taken as axes of coordinates, the resulting equation of the ellipsoid has no terms involving yz, zx or xy, and the new axes of coordinates are then called the Principal Axes of the Ellipsoid. Let the momental ellipsoid (1) when referred to its principal axes have as equation A'x'-vB\f-\-G'z'=MK' (2). L. D. 13 194 Dynamics of a Rkjid Body The products of inertia with respect to these new axes must be zero; for if any one of them, say D', existed, then there would, as in equation (1), be a term — ^D'yz in (2). Hence we have the following important proposition : For any body ivhatever there exists at each point a set of three perpendicular axes {which are the three principal diameters of the mo/nental ellipsoid at 0) such that the products of inertia of the body about them, taken two at a time, all vanish. These three axes are called the principal axes of the body at the point ; also a plane through any two of these axes is called a principal plane of the body. 152. It is also shewn in Solid Geometry that of the three principal axes of an ellipsoid, one is the maximum radius vector of the ellipsoid, and another is the minimum. Since the square of the radius vector of the momental ellipsoid is inversely proportional to the corresponding moment of inertia of the body, it follows that of the three principal axes, one has the minimum moment of inertia and another has the maximum. If the three principal moments of inertia at are equal, the ellipsoid of inertia becomes a sphere, all radii of which are equal ; in this case all moments of inertia about lines through are equal. Thus, in the case of a cube of side 2a, the principal moments of inertia at its centre are equal, and hence the moment of inertia about any line through its centre is the same and equal to if. -7-. o If the body be a lamina, the section of the momental ellipsoid at any point of the lamina, which is made by the plane of the lamina, is called the momental ellipse at the point. If the tAVO principal moments in this case are the same, the momental ellipse becomes a circle, and the moments of inertia of the lamina about all lines through are the same. 153. To shew that the moments and products of inertia of a uniform triangle about any lines are the same as the moments and products of inertia, about the same lines, of three particles placed at the middle points of the sides, each equal to one-third tlte mass of the triangle. Moments of Inertia of a Triangle 195 Divide up the triangle ABC into narrow slips by a large number of straight lines parallel ^ to its base. Let x, = AP, be the distance of one of them from A. Then B'C' = ra, where AD=h, and h the mass M of the triangle = ^ ah . p, where p is the density. The moment of inertia about a line AK parallel to BG ^r[7-H-=i ^ 1;.. M (!)• The moment of inertia about AD, by Art. 147, where E is the middle point of BG, = lpah\^^ + DE^'^=^[(bcosC + ccosBy + 3 (6 cos C - c cos By] M = -^ [¥ cos' G + c' cos"" B- be cos B cos G] (2). The product of inertia about AK, AD = r^dx.a;.PE', by Art. 147, = ^pa^.D£:.dx = '^^.^¥.DE=^[bcosG-ccosB]...(S). M If there be put three particles, each of mass -^ , at ^, F, and G, the middle points of the sides, their moment of inertia about ^A' fi-Hir- D] M ¥ Their moment of inertia about A D M[{a M = zr^ [(b cos G - c COS Bf + ¥ cos^ G + c' cos- B] 21 [b- cos^ G + c'"' cos- B — be cos B cos 6*]. 13—2 196 Dynamics of a Rigid Body Also their product of inertia about AK, AD f[ ad.de+\ad.^dg-\ad.^bd^\ Mh \' p.Tp & cos C — c cos B ' ¥h r . = — [6 cos C — c cos B]. The moments and products of inertia of the three particles about AK, AD are thus the same as those of the triangle. Hence, by Art. 149, the moments of inertia about any line through A are the same ; and also, by the same article, the products of inertia about any two perpendicular lines through A are the same. Also it is easily seen that the centre of inertia of the three particles coincides with the centre of inertia of the triangle. Hence, by Art. 147, it follows that the moments and products of inertia about any lines through the common centre of gravity are the same for the two systems, and therefore also, by the same article, the moments and products about any two other perpendicular lines in the plane of the triangle are the same. Finally, the moment of inertia about a perpendicular to the plane of the triangle through any point P is equal to the sum of the moments about any two perpendicular lines through P in the plane of the triangle, and is thus the same for the two systems, 154. Two mechanical systems, such as the triangle and the three particles of the preceding article, which are such that their moments of inei'tia about all lines are the same, are said to be equi-momental, or kinetically equivalent. If two systems have the same centre of inertia, the same mass, and the same principal axes and the same principal moments at their centre of inertia, it follows, by Arts. 147 and 148, that their moments of inertia about any straight line are the same, and hence that the systems are equi-momental. Momental ellipsoids. Examples 197 EXAMPLES 1. The momental ellipsoid at the centre of an elliptic plate is ^+f-!+22n + n=const. 2. The momental ellipsoid at the centre of a solid ellipsoid is ( 62 + c2) a;2 + (c2 + a2) 2^2 + (^2 + 62) ^2 ^ gongt. 3. The equation of the momental ellipsoid at the corner of a cube, of side 2a, referred to its principal axes is 2;!;2+ll (?/2 +22) = const. 4. The momental ellipsoid at a point on the rim of a hemisphere is 2^2 + 7 (y2 ^ g2) _ lAxz = const. 5. The momental ellipsoid at a point on the circular base of a solid cone is {da^+2h^)x'^ + [23a^ + 2h^)f + 26ah^-10akxz= const, where h is the height and a the radius of the base. 6. Find the principal axes of a right circular cone at a point on the circumference of the base ; and shew that one of them will pass through its centre of gravity if the vertical angle of the cone is 2tan-i|. 7. Shew that a uniform rod, of mass in, is kinetically equivalent to three particles, rigidly connected and situated one at each end of the rod and one at its middle point, the masses of the particles being im, ^m and |;n. 8. A BCD is a uniform parallelogram, of mass M; at the middle points of the four sides are placed particles each equal to — , and at the intersection of the diagonals a particle, of mass — ; shew that these five particles and the parallelogram are equi-mo mental systems. 9. Shew that any lamina is dynamically equivalent to three particles, each one-third of the mass of the lamina, placed at the corners of a maximum triangle inscribed in the ellipse, whose equation referred to the principal axes at the centre of inertia is ^ +^ = 2, where viA and mB are the principal moments of inertia about Ox and Oy and m is the 10. Shew that there is a momental ellipse at an angular point of a triangular area which touches the opposite side at its middle point and bisects the adjacent sides. [Use Art. 153.] 11. Shew that there is a momental ellipse at the centre of inertia of a uniform triangle which touches the sides of the triangle at the middle points. 198 Dynamics of a Rigid Body our *\p 12. Shew that a uniform tetrahedron is hineticalhj equivalent to fi particles, each of mass — , at the vertices of the tetrahedron, and a fifth 4 ¥ particle, of mass -^ , placed at its centre of inertia. Let OABG be the tetrahedron and through the vertex draw any three rectangular axes OX, OY, OZ. Let the coordinates oi A, B and C referred to these axes be (^i, yi, Zi), (^2, ?/2, ^2) and {x^, y^, 23), so that the middle point of BG is (^^^ ^-^'^^ ^') • Take any section PQ,R parallel to J5C at a perpendicular distance ^ from ; its area is A^^„, where Aq is the area of ^5Cand p is the perpendicular from P on ABC. By Art. 153 the moment of inertia of a thin slice, of thickness d^, about Ox 1 ^^ = that of three particles, each ^.A^p.^^.d^ placed at the middle points of QR, RP and PQ = - A(,o.-d^\(^. y^^^\ + two similar terms 3 ■ "'^ p'^ ^ \_\p 2 / ( • ^ n ) + ^^^ similar terms . Hence, on integrating with respect to ^ from to p, the moment of inertia about OX of the whole tetrahedron = — . ^oP- -[{(3/2+^3)^ + (^2 + 23)^} + two similar expressions] _ifr ^12+^2^^3^+^2^3+3/3^1+^13/2! /jN 10 L+ V + ■^2^ + %^ + ^2^3 + 23^1 + ^l22j Now the moment of inertia about OXoi four particles, each of mass — , 4i/ at the vertices of the tetrahedron, and of -— at its centre of inertia 5 and this, on reduction, equals (1). Similarly for the moments about the axes OF and OZ. In a similar manner the product of inertia of the tetrahedron about 07, OZ = ^[(_y2+y3) (22 +23) + two similar expressions] (2), and that of the five particles M, , , T 4il/(yi+.y2+y3)(fi+f2+f3) = 20 l>2^2+^3^3+3/i2i] + -5- 4 ■ 4 ' and this is easily seen to be equal to (2). Principal Axes 199 Also it follows at once that the centre of inertia of the tetrahedron coincides with the centre of inertia of the five particles. The two systems are therefore equal momental, for, by Arts. 147 and 148, their moments of inertia about any straight line is the same. 13. Shew that a tetrahedron is kinetically equivalent to six particles at the middle points of its edges, each Jjjth of the mass of the tetrahedron and one at the centroid §th of the mass of the tetrahedron. 155. To find whether a given straight line is, at any point of its length, a principal axis of a given material system, and, if so, to find the other two principal axes. Take the given straight line as the axis of z, and also any origin on it and any two perpen- dicular lines OX, OF as the other two axes. Assume that OZ is a principal axis at a point G of its length and let GX', GY' be the other two principal axes where GX' is inclined at an angle d to a line parallel to OX. Let 0(7 be h. Let X, y, z be the coordinates of any particle m of the material system referred to OX, OY, OZ and x, y', z' its coordinates referred to GX', GY', GZ. Then z= z' -^tK x = x' cos 6 — y' sin 6, and y = x' sind + y' cos 6, so that x' = x cos 6 -^-y sin 6, y' = —x sin 6 + y cos 6, and z' = z — h. .'. Imy'z' = Xm (— xz sin 9 + yz cos + hx sin 6 — hy cos 6) = Dcos0-Esm6 + Mh (xsind-ycosd) (1), with the notation of Art. 148, ^mz'x = l.m [xz cos 6 + yz sin 6 - hx cos — hy sin 6] = Dsind + Ecosd- Mh (x cosO + y sin 6) . . .(2), and Xmx'y' = 2m [- aP sin 6cos0 + xy (cos^ - sin^ 0) + y"^ sin 9 cos 6] =^ sin 29 {A -B) + F cos 29 (3). If GX', GY', GZ are principal axes, the quantities (1), (2), (3) must vanish. '? V The latter gives tan 29 = ^ — j (4). 200 Dynamics of a Rigid Body Mh. From (1) and (2) we have Esm0-Dcose _ D sin^+^ cos^ X sin — y cos 6 x cos 6 + ysin6 These give - = — (5), and h = ^~ = -jYz (6). ihy Mx ^ ' (5) is the condition that must hold so that the line OZ may be a principal axis at some point of its length, and then, if it be satisfied, (6) and (4) give the position of the point and the directions of the other principal axes at it. 156. If an axis be a principal axis at a point of its length it is not, in general, a principal axis at any other point. For if it be a principal axis at then D, E and F are all zero ; equation (6) of the previous article then gives h = 0, i.e. there is no other such point as G, except when x = and y=0. In this latter case the axis of z passes through the centre of gravity and the value of h is indeterminate, i.e. the axis of z is a principal axis at any point of its length. Hence if an axis passes tltroagh the centre of gravity of a body, and is a principal axis at any point of its length, it is a principal axis at all points of its length. 157. If the body be a lamina, as in the figure of Art. 149, the principal axes at a point are a normal OZ to its plane, and two lines OX', OY' inclined at an angle 6 to OX and OY. In this case, since z is zero for every point of the lamina, both D and E vanish. Hence equation (6) of Art. 155 gives ^ == 0, and 6 is given by 2F tan 2^ = ^-^ , . B-A As a numerical illustration, take the case of the triangle of Art. 158. Here A=M^; B=—\ , „ ^ 2 6 I — 6c cos 5 cos (7 and F — -2- (b cos G — c cos B). Principal Axes 201 The inclination 6 of one of the principal axes to AK is then given by the above formula. 158. The principal axes at any point P of a lamina may be constructed as follows. The plane of the lamina being the plane of the paper, let G be its centre of inertia and GX, GY the principal axes at G, the moments of inertia about which are A and B, A being greater than B. On GX take points 8 and H, such that G8=GH = ^^^. Then, by Art. 147, the moment of inertia about SY', parallel to GY, =B + M.GS' = A, so that the moments of inertia about SX and SY' are both equal to A. Also the product of in- ertia about SX, SY' H G G' s x = ItVi (x — GS) . y = Xmxy — GS . %iiiij = 0, since GX and G Y are the principal axes at G, and since G is the centre of inertia. Hence /S is a point such that SX and SY' are the principal axes, and the moments about each are equal to A. Hence, by Art. 149 or 152, any line through S in the plane of the paper is a principal axis at S, and the moment of inertia about it is ^. Similarly for any line through IT. Hence the moments of inertia about SP and HP are each equal to A. Also the normal at P to the lamina is clearly one of the principal axes at P, so that the other two lie in the plane of the lamina. If then we construct the momental ellipse at P, its radii-vectores in the directions PS and PU must be equal, since we have shewn that the moments of inertia about rS and PH are the same. Also in any ellipse equal radii- vectores are equally inclined to its principal axes, so that the latter bisect the angles between equal radii-vectores. Hence the principal axes of the momental ellipse at P, 202 Dynamics of a Rigid Body i.e. the principal axes of the lamina at P in its plane, bisect the angles between PS and PH. If then, with S and H as foci, we describe an ellipse to pass through any point P of the lamina, the principal axes of the lamina are the tangent and normal to this ellipse at P. The points >S^ and H are hence known as the Foci of Inertia, 159. The proposition of the preceding article may be extended to any body, if G be the centre of inertia, OX, 07, and GZ its principal axes at and P be any point in the plane of ZF. EXAMPLES 1. If A and B be the moments of inertia of a uniform lamina about two perpendicular axes, OA'and OV, lying in its plane, and Fhe the product of inertia of the lamina about these lines, shew that the principal moments at are equal to^[A+B± \/(A-By'+4F^. 2. The lengths AB and AD of the sides of a rectangle A BCD are 2a and 26 ; shew that the inclination to ^ J? of one of the principal axes at A . 1 _j 3a6 '^2^^'' 2(a2-62)* 3. A wire is in the form of a semi-circle of radius a ; shew that at an end of its diameter the principal axes in its plane are inclined to the diameter at angles - tan-i - and jr + „ tan"* - . 2 TT 2 2 IT 4. Shew that at the centre of a quadrant of an ellipse the principal axes in its plane are incHned at an angle - tan"^ ( 5 — 5-= ) to the axes. 2 \7r a^ — b^J 5. Find the principal axes of an elliptic area at any point of its bounding arc. 6. At the vertex C of a triangle A BG, which is right-angled at C, the principal axes are a perpendicular to the plane and two others inclined to the sides at an angle - tan~i 2_h'i ' 7. ABC is a triangular area and AD \s perpendicular to BC ; E is the middle point of BC and the middle point of DE ; shew that BC is a principal axis of the triangle at 0. [Use the property of Art. 153.] 8. A uniform square lamina is bounded by the axes of x and y and the lines a;=2c, y=2c, and a corner is cut oflf by the line -+| = 2. Shew that the principal axes at the centre of the square are inclined , , ^„ a6-2(a + 6)c-f3c'- to the axis of x at angles given by tan 2^= (a-6)(a + 6-2c) ' Principal Axes. Examples 203 9. A uniform lamina is bounded by a parabolic arc, of latus-rectum 4a, and a double ordinate at a distance b from the vertex. If &=- (7 + 4^/7), 3 shew that two of the principal axes at the end of a latus-rectum are the tangent and normal there. 10. Shew that the principal axes at the node of a half-loop of the lemniscate r'^=a'^ cos, ^6 are inclined to the initial line at angles gtan-i-and 2+2tan-i2- 11. The principal axes at a corner of a cube are the line joining to the centre of the cube and any two perpendicular lines. 12. If the vertical angle of a cone is 90°, the point at which a generator is a principal axis divides the generator in the ratio 3 : 7. [Use Art. 159.] 13. Three rods AB, 5(7 and CD, each of mass m and length Sa, are such that each is pei'pendicular to the other two. Shew that the principal moments of inertia at the centre of mass are ma^t ^J-tna^ and ima^. CHAPTER XII D'ALEMBERT'S PRINCIPLE THE GENERAL EQUATIONS OF MOTION 160. We have already found that, if x, y, z be the co- ordinates of a particle m at time t, its motion is found by d?x equating w -^ to the force parallel to the axis of x, and similarly for the motion parallel to the axes of y and z. If m be a portion of a rigid body its motion is similarly given, but in this case we must include under the forces parallel to the axes not only the external forces acting on the particle (such as its weight), but also the forces acting on the particle which are due to the actions of the rest of the body on it. d?x The quantity m -,— is called the effective force acting on the particle parallel to the axis of x. [It is also sometimes called the kinetic reaction of the particle.] Thus we may say that the ^--component of the effective force is equivalent to the ^-component of the external forces together with the ^r-component of the internal forces, or again that the ^r-component of the reversed effective forces together with the a;-components of the external and internal forces form a system in equilibrium. So for the components parallel to the axes of y and z. Hence the reversed effective force, the external force, and the internal force acting on any particle m of a body are in equilibrium. General equations of motion 205 So for all the other particles on the body. Hence the reversed effective forces acting on each particle of the body, the external forces, and the internal forces of the body are in equilibrium. Now the internal forces of the body are in equilibrium amongst themselves ; for by Newton's third law there is to every action an equal and opposite reaction. Hence the reversed effective forces acting on each particle of the body and the external forces of the system are in equilibriuin. This is D'Alembert's principle. It was enunciated by him in his Traite de Dynamique published in the year 1743. It will be noted however that it is only a deduction from Newton's Third Law of Motion. 161. Let X, Y, Z be the components parallel to the axes of the external forces acting on a particle m whose coordinates are x, y, z at the time t. Then the principle of the preceding article says that forces whose components are ^ d^x „ d'u ^ d-z X-m-j-, Y-m-j^, Z-m-f- dt- dt- df acting at the point {w, y, z), together with similar forces acting at each other such point of the body, form a system in equi- librium. Hence, from the ordinary conditions of equilibrium proved in Statics, we have 2(Z-.§) = 0, and 4(^-§)-.(x-»>g)]=o, 206 Dynmaics of a Rigid Body These give 2m^ = 2X (1), 2»S=2r (2). ^"'i-s^ (3). Hy%-^%'^^y'-'^) W' ^™(^§-4')=^(^^--^) (5)- and 2m(.;g-2,^) = 2(<»7-yX) (6). These are the equations of motion of any rigid body. Equations (1), (2), and (3) state that the sums of the components, parallel to the axes of coordinates, of the effective forces are respectively equal to the sums of the components parallel to the same axes of the external impressed forces. Equations (4), (5), (6) state that the sum of the moments about the axes of coordinates of the effective forces are re- spectively equal to the sums of the moments about the same axes of the external impressed forces. 162. Motion of the centre of inertia, avid motion relative to the centre of inertia. Let {x, y, z) be the coordinates of the centre of inertia, and M the mass of the body. Then Mx = Xma: throughout the motion, and therefore T,,d^x _, (Px Hence equation (1) of the last article gives *S=^^ w- S» *S=2^ C2). and -^2=^^ <*>• But these are the equations of motion of a particle, of mass M, placed at the centre of inertia of the body, and acted General equations of motion 207 on by forces parallel to, and equal to, the external forces acting on the different particles of the body. Hence the centre of inertia of a body moves as if all the mass of the body were collected at it, and as if all the external forces were acting at it in directions parallel to those in which they act. Next, let {x, y , z') be the coordinates, relative to the centre of inertia, G, of a particle of the body whose coordinates referred to the original axes were {x, y, z). Then x = x-\-x', y=iy-\-y' and z = z-\-z' throughout the motion. ^_d?x_ d?x_ d^_d^ dy d'^z _ d^'z d^z' •'• dt'~ dt''^ dt" dF'~d^'^W'^^^dt'~dF''^d¥' Hence the equation (4) of the last article gives d^z -d^y\ ^ ( ,d''z' ,d''y' dt' df dt^ df df" — y X[{y+y')Z-{z+z)Y]...{^). Now -r~^ = the ■y-coordinate of the centre of inertia referred to G as origin = 0, and therefore Xniy' = and Swi -^ = ; d^z' so ^mz' = and Sm ^- = 0. Hence (4) gives ^A-d''z dry~\ ^ ( ,d?z' ,d^y'\ = X[yZ-zV + y'Z-/7]...i5). But equations (2) and (3) give .*. (5) gives 208 Dynamics of a Rigid Body But this equation is of the same form as equation (4) of the last article, and is thus the same equation as we should have obtained if we had regarded the centre of inertia as a fixed point. Hence the motion of a body about its centre of inertia is the same as it would be if the centre of inertia were fixed and the same forces acted on the bodi/. 163. The two results proved in the previous article shew us that the motion of translation of the body can be considered independently of the motion of rotation. By the first result we see that the motion of the centre of inertia is to be found by the methods of Dynamics of a Particle. By the second result we see that the motion of rotation is reduced to finding that of a body about a fixed point. As a simple example, consider the case of a uniform stick thrown into the air in such a way that at the start its centre is moving in a given direction and at the same time it is rotating with given angular velocity about its centre. [Neglect the resistance of the air and suppose gravity to be constant.] By the first result the motion of the centre of inertia is the same as if there were applied at it all the external forces acting on the body in directions parallel to that in which they act. In this case these external forces are the weights of the various elements of the body; when applied at the centre of inertia they are equivalent to the total weight of the body. Hence the centre of the stick moves as if it were a particle of mass M acted on by a vertical force Mg, i.e. it moves just as a particle would under gravity if it were projected with the same velocity as the centre of the stick. Hence the path of the centre of the stick would be a parabola. In a subsequent chapter it will be seen that the angular velocity of the stick will remain unaltered. Hence the centre of the stick will describe a parabola and the stick revolve uniformly about it. As another example consider a shell which is in motion in the air and suppose that it bursts into fragments. The internal forces exerted by the explosion balance one another, and do not exert any influence on the motion of the centre of inertia of D'Alemhert's Principle. Examples 209 the shell. The centre of inertia therefore continues to describe the same parabola in which it was moving before the explosion. [The motion is supposed to be in vacuo and gravity to be constant.] 164. Equation (1) of Art. 161 may be written in the form dx' It [-^^ dt S(X), i.e. -T [Total momentum parallel to the axis of x] = Sum of the impressed forces parallel to OX. So for the other two axes. Also (4) can be written ^^.„ ^_.^^1=' [Total moment of momentum about the axis of x] liiHyt-ty^^y'-^^^- dt Sum of the moments of the impressed forces about OX. 165- -A-s an example of the application of D'Alembert's principle let us consider the following question. A uniform rod OA, of length 2a, free to turn about its end 0, revolves with uniform angular velocity w about the vertical OZ through 0, and is inclined at a constant angle a to OZ ; find the value of a. Consider an element PQ of the rod, such that OP = i and PQ = d^. Draw FN perpendicular to OZ. By Elementary Dynamics, the acceleration of P is w- . FN along PN. Hence the reversed effective force is — -r — . I sin a — . ?» . 0)2 . 1 sin o as marked. All the reversed effective forces acting at different points of the rod, together with the external force, i.e. the weight mg, and the reactions at 0, form a system of forces in statical equilibrium. Taking moments about to avoid the reactions, we therefore have vig . a sm ( moment about of all the several effective forces 2 — . ?» . w^^ sm a X I cos [la mw'sm a cos I ??iw'^siu a cos a . 4a2 14 210 Dynamics of a Rigid Body Hence either a = 0, or cos a = -~- , U 3q-> iw-a, i.e. it u^-c-^, the second equation gives an impossible value for a, and the only solution in this case is a = 0, i.e. the rod hangs vertically. If 3g^<^ , and then 6 is zero. yCta-J a 7. A thin heavy disc can turn freely about an axis in its own plane, and this axis revolves horizontally with a uniform angular velocity w about a fixed point on itself. Shew that the inchuation 6 of the plane of the disc to the vertical is cos "^7^,, where A is the distance of the K-'W Impulsive Forces 211 centre of inertia of the disc from the axis and k is the radius of gyration of the disc about the axis. If 0)2 < Y^j , then the plane of the disc is vertical. 8. Two uniform spheres, each of mass M and radius a, are firmly fixed to the ends of two uniform thin rods, each of mass m and length I, and the other ends of the rods are freely hinged to a point 0. The whole system revolves, as in the Governor of a Steam- Engine, about a vertical line through with angular velocity w. Shew that, when the motion is steady, the rods are inclined to the vertical at an angle 6 given by the equation M{l + a) + m- cos^=4 i. Impulsive Forces 166. When the forces acting on a body are very great and act for a very short time, we measure their effects by their impulses. If the short time during which an impulsive force rT X acts be T, its impulse is Xdt. -'o In the case of impulsive forces the equations (1) to (6) of Art. 161 take a different form. Integrating equation (1), we have Xm ^ = r^X .dt = X I^Xdt If u and u' be the velocities of the particle m before and after the action of the impulsive forces, this gives 2m(«' — w) = 2 A' , where X' is the impulse of the force on m parallel to the axis of*'. This can be written Xmu' — Xmu = IX' (1), i.e. the total change in the momentum parallel to the axis of x is equal to the total impulse of the external forces parallel to this direction. Hence the change in the momentum parallel to Ox of the ivhole mass M, supposed collected at the centime of inertia and moving with it, is equal to the impulse of tJie external forces parallel to Ox. 14—2 212 Dynamics of a Rigid Body So for the change in the motion parallel to the axes of y and z, the equations being ^mv -tmv = tY' (2), and 2ww' — Xmiu = 2-^' (3). Again, on integrating equation (4), we have i.e. Xm [y (w' — iv) — z {v — v)] = 2 [yZ' - z Y'\ Hence 2m \yv} — zv'^ — Sm \yw — zv\ = 2 (yZ' — zY') . . .(4). Hence the change in the moment of momentum about the cuvis of X is equal to the moment about the axis of x of the impulses of the external forces. So for the other two axes, the equations being 2m {zu — xw') — 2m (zu — xw) = 2 {zX' — xZ') . . .(5), and 2m {xv — yu) — 2m {xv — yu) = 2 {coY' — yX') . . .(6). 167. The equations of Arts. 161 and 166 are the general equations of motion of a rigid body under finite and impulsive forces respectively, and always give the motion. They are not however in a form which can be easily applied to any given problem. Different forms are found to be desirable, and will be obtained in the following chapterSj for different classes of Problems. CHAPTER XIII MOTION ABOUT A FIXED AXIS 168. Let the fixed axis of rotation be a perpendicular OZ at to the plane of the paper, and let a fixed plane through OZ cut the paper in OA. Let a plane ZOG, through OZ and fixed in the body, make an angle 6 with the fixed plane, so that /.AOG=e. Let a plane through OZ and any point P of the body make an angle ^ with ZOA and cut the plane of the paper in OQ, so that zAOQ = (f). As the body rotates about OZ the angle QOG remains the same always, so that the rate of change of 6 is the same as that of along PM and r -^ perpendicular to PM. Hence its effective forces in these directions are and mr ^d^V- dt) Hence the moment Oz is dt' ' ' of its .e. mr ( -j- I and mr effective forces dt^' about the axis df" i.e. mr"^ d^e dt' 214 Dynamics of a Rigid Body Hence the moment of the effective forces of the whole body about OZ is since -^ is the same for all particles of the body. Now 2«ir^ is the moment of inertia, Mk-, of the body about the axis. Hence the required moment of the effective forces is Mk" . -T-^ , where 6 is the angle any plane through the axis which is fixed in the body makes with any plane through the axis which is fixed in space. 169. Kinetic energy of the body. The velocity of the particle mis r~- , i.e. r -,- . Its energy is therefore h''^ (^777) • Hence the total kinetic energy of the body ~^2"" \Tt) -2\dtJ ^^"" -2^""- KW 170. Moment of momentum of the body about the fixed ^1 fdOV l/dOY K« . l.r/.^^^V The velocity of the particle m is r -^ in a direction perpen- dicular to the line, of length r, drawn from m perpendicular to the axis. Hence the moment about the axis of the momentum of m is mr x r -v- , i.e. mr^ -rr • Hence the moment of momentum dt dt of the body ^ de de ^ ^ ,,,, d0 dt dt dt 171. To find the motion about the axis of rotation. Art. 161 tells us that in any motion the moment of the effective forces about the axis is equal to the moment of the impressed forces. Hence, if L be the moment of the impressed forces about the axis of rotation, in the sense which would cause 6 to increase, we have Motion about a fixed axis 215 do This equation on being integrated twice will give and -j- in terms of the time t. The arbitrary constants which appear in the integration will be known if we are given the position of the plane ZOG, which is fixed in the body, and its angular velocity at any time. 172. Ex. 1. A uniform rod, of mass m and length 2a, can turn freely about one end which is fixed; it is started with angular velocity u from the position in ichich it hangs vertically ; Jind the motion. The only external force is the weight Mg whose moment L about the fixed axis is Mg .a Bind, when the rod has revolved ^ through an angle 6, and this moment tends to lessen 6. Hence the equation of motion is dt^ - Mga sin $, or, since ft^ ; Aa^ ^ 3 ' d^e dt^' 3/7 . , Integrating, we have mr^ 37 „ ^ = 7^C0S^+C 4a where 5 w^ ^ 3.7 4a + C. fdd \dt y... -|(l-cos^) (1), giving the angular velocity at any instant. In '• general the equation (1) cannot be integrated further, so that t cannot be found in terms of 6. The angular velocity — gets less and less as 6 gets bigger, and just vanishes .e. when the rod is in its highest position, if w = a/— . This is the least value of the angular velocity of the rod, when in its lowest position, so that the rod may just make complete revolutions. With this particular angular velocity the equation (1) gives when I =!!<'- = -^ . C0s2 - =21ogtan(| + - tan i)! giving the time of describing any angle 6 in this particular case. Energy and Work. The equation (1) may be written in the form 'dey 1 ,, 4a2 ^dt) -2^-X i.e., by Art. Ifi9, the change in the kinetic energy of the body is equal to the work done against the weight of the body. 1 ,r 4«2 c^= - Mga {1- cose). 216 Dynamics of a Rigid Body Ex. 2. A fine string has two masses, M and M', tied to its ends and passes over a rough pulley, of mass m, whose centre is fixed ; if the string does not slip over the pulley, shew that M will descend loith acceleration ,-, . g, M+M' + m~ a^ where a is the radius and k the radius of gyration of the pulley. If the pulley be not sufficiently rough to prevent sliding, and M he the descending mass, shew that its acceleration is M + M'e'^'' 2MM'pa{e'^''-l) g, and that the pulley will now spin with an anaular acceleration equal to ^ ^ mk^iM + M'ef^n Let T and T' be the tensions of the string when the pulley has turned through an angle 6; and let the depths of M and M' below the centre of the pulley be x and y. Then, by Art. 171, the equation of motion of the pulley is mk^-d^(T-T')a (1). Also the equations of motion of the weights are Mx = Mg-T and M'y = M'g-T' ...(2). Again x + y is constant throughout the motion, so that y=-x (3). First, let the pulley be rough enough to prevent any sliding of the string, so that the string and pulley at A are always moving with the same velocity. Then x = ad always, and therefore x = ad {i). 31 -M' Equations (1) to (4) give x-a9 = M + M' + m^^ g, giving the constant acceleration with which M descends. If the pulley be a uniform disc, ^2; 2 ' M-M' M + M' + If it be a thin ring, k- = a^, and the acceleration is and this acceleration is M-M' M+M'+m-" Secondly, let the pulley be not rough enough to prevent all sliding of the string. In this case equation (4) does not hold ; instead, if /* be the coethcient of friction, we have, as is proved in books on Statics, r=T'.e'^'^ (5). Solving (2), (3) and (5), we have T'ei^'"=T= 2mrge^ M+M'ei^''' Iga (£>*"■ - 1) and x = M+M'ei^"' MM' and then (1) gives „ — , „ . The result of the first case might have been easily obtained by assuming the Principle of Work and Energy ; in the second case it does not apply. Motion about a fixed axis. Examples 217 EXAMPLES 1. A cord, 10 feet long, is wrapped round the axle, whose diameter is 4 inches, of a wheel, and is pulled with a constant force equal to 50 lbs. weight, until all the cord is unwound. If the wheel is then rotating 100 times per minute, shew that its moment of inertia is ^ ft.-lb. units. 2. A uniform wheel, of weight 100 lbs. and whose radius of gyration about its centre is one foot, is acted upon by a couple equal to 10 ft.-lb. units for one minute ; find the angular velocity produced. Find also the constant couple which would in half-a-minute stop the wheel if it be rotating at the rate of 15 revolutions per second. Find also how many revolutions tlie wheel would make before stopping. 3. A wheel consists of a disc, of 3 ft. diameter and of mass 50 lbs., loaded with a mass of 10 lbs. attached to it at a point distant one foot from its centre ; it is turning freely about its axis which is horizontal. If in the course of a single revolution its least angular velocity is at the rate of 200 revolutions per minute, shew that its maximum angular velocity is at the rate of about 204-4 revolutions per minute. 4. Two unequal masses, J/ and J/', rest on two rough planes inclined at angles a and /3 to the horizon ; they are connected by a fine string passing over a small pulley, of mass wi and radius a, which is placed at the common vertex of the two planes ; shew that the acceleration of either mass is ^[i/(sina — /icosa) — i/''(sin/3 + /a'cos^)]^ Jf+J/' + m^ , where \i, fi' are the coefficients of friction, k is the radius of gyration of the pulley about its axis, and M is the mass which moves downwards. 5. A uniform rod aB is freely movable on a rough inclined plane, whose inclination to the horizon is i and whose coefficient of friction is fi, about a smooth pin fixed through the end A ; the bar is held in the horizontal position in the plane and allowed to fall from this position. If 6 be the angle through which it falls from rest, shew that sin ^ , . -^-=/xcot?.. 6. A uniform vertical circular plate, of radius a, is capable of revolving about a smooth horizontal axis through its centre ; a rough perfectly flexible chain, whose mass is equal to that of the plate and whose length is equal to its circumference, hangs over its rim in equilibrium ; if one end be slightly displaced, shew that the velocity of the chain when the other end reaches the plate is » / — ^ [Use the Principle of Energy and Work.] 218 Djjnamics of a Rigid Body 7. A uuiform chain, of length 20 feet and mass 40 lbs., hangs in equal lengths over a solid circular pulley, of mass 10 lbs. and small radius, the axis of the pulley being horizontal. Masses of 40 and 35 lbs. are attached to the ends of the chain and motion takes place. Shew that the time taken by the smaller mass to reach the pulley is /15 ^— logj (9 + 4^/6) sees. 4 8. A heavy fly-wheel, rotating about a symmetrical axis, is slowing down under the friction of its bearings. D%iring a certain minute its angular velocity drops to 90 °/„ of its value at the beginning of the minute. What will be the angular velocity at the end of the next mimtte on the assumption that the frictional moment is (1) constant, (2) proportional to the angular velocity, (3) proportional to the square of the angular velocity? Let / be the moment of inertia of the body about its axis, a its angular velocity at any time t, and Q its initial angular velocity. Let xQ, be the angular velocity at the end of the second minute. (1) If F be the constant frictional moment, the equation of Art. 171 is .-. Ico=-Ft+C=-Ft+lQ, where I. — .a=-F.eO+lQ, and I.xQ= - F . 120 + lQ. 80 (2) If the frictional moment is Xw, the equation of motion is ,-. /log 0)= -X;+ const. ^- t -^- t .-. (o = Ce I' =Qe 1' , 90 „ -ieo , „ „ -fi20 where —=Qe^ , and xQ.=ae -' _/9V__81^ • • "^"Vio/ "100- (3) Let the frictional moment be /xw^, so that the equation of motion is ^dt = --^''- ... /.i=^, + (7 = ^^+|, where /.9^=M.60 + |, and I .^ = ^^-^^^^i' 81 A The Compound Pendulum 219 With the three suppositions the angular velocity at the end of two minutes is therefore 80, 81, and 81^j°/„ of the initial angular velocity. 9. A fly-wheel, weighing 100 lbs. and having a radius of gyration of 3 ft., has a fan attached to its spindle. It is rotating at 120 revolutions per minute when the fan is suddenly immersed in water. If the resistance of the water be proportional to the square of the speed, and if the angular velocity of the fly-wheel be halved in three minutes, shew that the initial retarding couple is 20it ft.-pouudals. 10. A fly-wheel, whose moment of inertia is /, is acted on by a variable couple G cos pt ; find the amplitude of the fluctuations in the angular velocity. THE COMPOUND PENDULUM 173. If a rigid body swing, under gravity, from a fixed horizontal axis, to shew that the time of a complete small oscilla- tion is 277 a/ t— , ivhere k is its radius of gyration about the fixed axis, and h is the distance between the fi^ed axis and the centre of inertia of the body. Let the plane of the paper be the plane through the centre of inertia G perpendicular to the fixed axis; let it meet the axis in and let 6 be the angle between the vertical OA and the line OG, so that 6 is the angle a plane fixed in the body makes with a plane fixed in space. The moment L about the horizontal axis of rotation OZ of the impressed forces = the sum of the moments of the weights of the component particles of the body = the moment of the weight Mg acting at G = Mgh sin 6, where OG = h, and it acts so as to diminish d. Hence the equation of Art. 171 becomes T,.j^d^O n^ 1 ■ a ■ d-9 gh . If 6 be so small that its cubes and higher powers may be neglected, this equation becomes d'0_ gh W--T^^ ^'>- 220 Dynamics of a Rigid Body The motion is now simple harmonic and the time of a complete oscillation is 27r . ^ IT' V A;^ By Art. 97 the time of oscillation is therefore the same as that of a simple pendulum of length -y-. This length is that of the simple equivalent pendulum. Even if the oscillation of the compound pendulum be not small, it will oscillate in the same time as a simple pendulum of length y- . For the equation of motion of the latter is, by Art. 97, d?e - g sin 6 gh sin^ •(3), which is the same equation as (1). Hence the motion given by (1) and (3) will always be the same if the initial conditions of the two motions are the same, e.g. if the two pendulums are instantaneously at rest when the value of 6 is equal to the same value a. in each case, or again if the angular velocities of the two pendulums are the same when each is passing through its position of stable equilibrium. 174. If from we measure off, along OG, a distance 00^, equal to the length of the simple equivalent pendulum -y- , the point Oi is called the centre of oscillation. We can easily shew that the centres of suspension and oscillation, and 0^, are con- vertible, i.e. that if we suspend the body from Oi instead of from 0, then the body will swing in the same time as a simple pendulum of length OiO. -I-Oi 4 0s For we have 00. '' ^^^^^^ OG OG The Compoivnd Pendulum 221 where K is the radius of gyration about an axis through (r parallel to the axis of rotation. Hence K^ = OG .00,-0G'= OG . GO, (1). When the body swings about a parallel axis through Oi, let O2 be the centre of oscillation. We then have, similarly, K'=0,G.GO, (2). Comparing (1) and (2) we see that O2 and are the same point. Hence when 0, is the centre of suspension, is the centre of oscillation, so that the two points are convertible. This property was used by Captain Kater in determining the value of g. His pendulum has two knife-edges, about either of which the pendulum can swing. It also has a movable mass, or masses, which can be adjusted so that the times of oscillation about the two knife-edges are the same. We then know that the distance, I, between the knife-edges is the length of the simple equivalent pendulum which would swing in the observed time of oscillation, T, of the compound pendulum. Hence g is obtained from the formula T=1'k sj - . For details of the experiment the Student is referred to practical books on Physics. 175. Mwimum time of oscillation of a compound pendulum. If K be the radius of gyration of the body about a line through the centre of inertia parallel to the axis of rotation, then k' = K"' + h\ Hence the length of the simple equivalent pendulum The simple equivalent pendulum is of minimum length, and therefore its time of oscillation least, when ^(^^ + nr)=^> i.e. when 1 — ;— = 0, i.e. when h = K, ic- and then the length of the simple equivalent pendulum is 2K. If /i = or infinity, i.e. if the axis of suspension either passes through the centre of inertia or be at infinity, the corresponding simple equivalent pendulum is of infinite length and the time of oscillation infinite. 222 Dynamics of a Rigid Body The above gives only the minimum time of oscillation for axes of suspension which are drawn in a given direction. But we know, from Art. 152, that of all axes drawn through the centre of inertia G there is one such that the moment of inertia about it is a maximum, and another such that the moment of inertia about it is a minimum. If the latter axis be found and if the moment of inertia about it be K^, then the axis about which the time is an absolute minimum will be parallel to it and at a distance Ki. 176. -Ea;. Find the time of oscillation of a compound pendulum, consisting of a rod, of mass m and length a, carrying at one end a sphere, of mass mi and diameter 2h, the other end of the rod being fixed. Here (m + mi) k-~m .^ +mi\ {a + 6)2 + ^ , and {m + m-i) h = m.- + mi{a + b). Heiice the length of the required simple pendulum a2 r, ,,„ 2Z;2-| /( 0. , , , m- + mi(a + fc) 177. Isochronism of Torsional Vibrations. Suppose that a heavy uniform circular disc (or cylinder) is suspended by a fairly long thin wire, attached at one end to the centre C of the disc, and with its other end firmly fixed to a point 0. Let the disc be twisted through an angle a about 00, so that its plane is still horizontal, and let it be then left to oscillate. We shall assume that the torsion-couple of the wire, i.e. the couple tending to twist the disc back towards its position of equi- librium, is proportional to the angle through which the disc has been twisted, so that the couple is \6 when the disc is twisted through an angle 0. Let M be the mass of the disc, and k its radius of gyration about the axis of rota- tion 00. Torsional Vibrations 223 By Art. 171 the equation of motion is The motion is therefore simple harmonic, and the time of oscillation = 27r^y^-^=27r^^^.. (1). This time is independent of a, the amplitude of the oscilla- tion. We can hence test practically the truth of the assumption that the torsion-couple is \d. Twist the disc through any angle a and, by taking the mean of a number of oscillations, find the corresponding time of oscillation. Repeat the experi- ment for different values of a, considerably differing from one another, and find the corresponding times of oscillation. These times are found in any given case to be approximately the same. Hence, from (1), the quantity \ is a constant quantity. 178. Experimental determination of moments of inertia. The moment of inertia of a body about an axis of symmetry may be determined experimentally by the use of the preceding article. If the disc be weighed, and its diameter determined, then its Mk^ is known. Let it be /. Its time of oscillation is then T, where T=2'rr^^ (1). Let the body, whose moment of inertia /' about an axis of symmetry is to be found, be placed on the disc with this axis of symmetry coinciding with GO, and the time of oscillation T' determined for the compound body as in the previous article. '^'=^W-¥ "■'>■ (1) and (2) give giving i ' in terms of known quantities. Then 224: Dynamics of a Eiyld Body EXAMPLES Find the lengths of the simple equivalent pendulums in the following cases, the axis being horizontal : 1. Circular wire ; axis (1) a tangent, (2) a perpendicular to the plane of the wire at any point of its arc. 2. Circular disc ; axis a tangent to it. 3. Elliptic lamina ; axis a latus-rectum. 4. Hemisphere ; axis a diameter of the base. Resxilt. — a. 5. Cube of side 2a ; axis (1) an edge, (2) a diagonal of one of its faces. [Results. (l)|v/2a; {2)^ . 6. Triangular lamina ABC; axis (1) the side BC, (2) a perpendicular to the lamina through the point A. [W«. (l)46si„e;(2>l.|i±J|^.] 7. Cone; axis a diameter of the base. Result. .h. 8. Three equal particles are attached to a rod at equal distances a apart. The system is suspended from, and is free to tm-n about, a point of the rod distant x from the middle point. Find the time of a small oscillation, and shew that it is least when d7='82a nearly. 9. A bent lever, whose arras are of lengths a and b, the angle between them being a, makes small oscillations in its own plane about the fulcrum ; shew that the length of the corresponding simple pendulum is 2 a? + h^ 3 \/a* + 2a^6'^cosa+6'*' 10. A solid homogeneous cone, of height h and vertical angle 2a, oscillates about a horizontal axis through its vertex ; shew that the length of the simple equivalent pendulum is -(4 + tan2a). 11. A sphere, of radius a, is suspended by a fine wire from a fixed point at a distance I from its centre ; shew that the time of a small oscillation is given by '^ y -^-^|^Tl +iSin2|J, ^here a represents the amplitude of the vibration. 12. A weightless straight rod A BC, of length 2a, is movable about the end A which is fixed and carrieo two particles of the same mass. itself. The Compound Pendulum. Examples 225 one fastened to the middle point B and the other to the end C of the rod. If the rod be held in a horizontal position and be then let go, shew that its angular velocity when vertical is sj -S- ■> ^^^ ^tiat -^ is the length of the simple equivalent pendulum. 13. For a compound pendulum shew that there are three other axes of support, parallel to the original axis and intersecting the line from the centre of inertia perpendicular to the original axis, for which the time of oscillation is the same as about the original axis. What is the practical application of this result ? 14. Find the law of graduation of the stem of the common metronome. 15. A simple circular pendulum is formed of a mass M suspended from a fixed point by a weightless wire of length ^ ; if a mass m, very small compared with i/, be knotted on to the wire at a point distant a from the point of suspension, shew that the time of a small vibration of the pendulum is approximately diminished by -^prf%\ 1 -7 ) of 16. A given compound pendulum has attached to it a particle of small mass ; shew that the greatest alteration in the time of the pendulum is made when it is placed at the middle point of the line bisecting the distance joining the centi'es of oscillation and of suspension ; shew also that a small error in the point of attachment will not, to a first ajiproximation, alter the weight of the particle to be added to make a given difference in the time of oscillation. 17. A uniform heavy sphere, whose mass is 1 lb. and whose radius is 3 inches, is susjjended by a wire from a fixed point, and the torsion- couple of the wire is proportional to the angle through which the sphere is turned from the position of equilibrium. If the period of an oscillation be 2 sees., find the couple that will hold the sphere in equilibrium in the position in which it is turned through four right angles from the equi- librium-position. 18. A fly-wheel is hung up with its axis vertical by two long ropes parallel to and equidistant from the axis so that it can perform torsional vibrations. It is found that a static-couple of 60 ft. -lbs. will hold it when it is turned through ^igth of a radian, and that if it be turned through any small angle and let go it will make a complete oscillation in 5 sees. Shew that when this fly-wheel is revolving at the rate of 200 revolutions per minute the energy stored up in it will be about 31 ft. -tons. 179. Reactions of the axis of rotation. Let us first consider the simple case in which both the forces and the body are symmetrical ^\\ki respect to the plane through the centre of gravity perpendicular to the fixed axis, t.e. with respect to L. D. 15 226 Dynamics of a Rigid Body the plane of the paper, and let gravity be the only external force. By symmetry, the actions of the axis on the body must reduce to a single force acting at in the plane of the paper; let the components of this single force be P and Q, along and perpendicular to 00. By Art. 162 the motion of the centre of gravity is the same as it would be if it were a particle of mass M acted on by all the external forces applied to it parallel to their original directions. Now describes a circle round as centre, so that its accelerations along and perpendicular to GO are A ( ^7 ) and h j^ I and ,. , . Hence its equations of motion are M.h and M . Also, as in Art. 171, Mk' dtl we have 'dt^~ d'd Mg cos 6 2-Mgsme 3Igh sin 6 .. •(1). .(2). .(3). Q is given by eliminating -r-^ between (2) and (3). If (3) be integrated and the resulting constant determined from the initial conditions, we then, by (1), obtain P. As a particular case let the body be a uniform rod, of length 2a, turning about its end 0, and let it start from the position in which it was vertically above 0. In this case h = a, k^=a^+--- =-^, Hence equation (3) becomes sin 9 3.<7 4a ■const. (l + cos< since (5 is zero wheu d- Reactions of the axis of rotation 227 (1) and (5) give F=Mg 3 + 5 cos 5 2 (2) and (4) give Q = ^Mg sin d. Hence the resulting reaction of the fixed axis. When 9 is zero, i.e. when the rod is in its lowest position, this reaction is four times the weight. The vertical reaction for any position of the rod ^ . „ /I + 3 cos 61 \2 = Pcose + Qsmd = Mg I ^ j , and therefore vanishes when ^ = cos~i {-^)- The horizontal reaction = P sin ^ - Q cos ^ = ^Mg sin (9 (2 + 3 cos 6). 180. In the general case when either the external forces acting on the body, or the body itself, is not symmetrical about the axis of rotation we may proceed as follows. Let the axis of rotation be taken as the axis of y, and let the body be attached to it at two points distant h^ and h^ from the origin. Let the component actions of the axis at these points parallel to the axes be X^, Fj, Z^ and X^, T^, Z.^, respectively. 2' P / "^irct) /r.^. X -/ ■ y '<, Y M / _ Let P be any point {x, y, z) of the body, Avhose perpendicular distance PM from OF is of length r and makes an angle 6 with a line parallel to OZ. Then during the motion P describes a circle about M as centre, so that r is constant throughout the motion and hence r is zero. Now a; = r sin ^ ; z — r cos 6. .'. cb^r cos 66, and z= — rsm6d. .'. x = -7^s'm 6 6^ + rcos 66; 2= -r cos66^ -rsinOe. 15—2 228 Dynamics of a Rigid Body Hence, if 6 be denoted by w, x=z — xw^ + zod; y = 0; z = —zai crw. [These results may also be obtained by resolving parallel to the axes the accelerations of P, viz. raP' along PM and rm perpendicular to MP^^ The equations of motion of Art. 161 now become, if Z, F, Z are the components parallel to the axes of the external force acting at any point {x, y, z) of the body, S X + Xi + X^ = '^uix = Ini [— xm^ + zw] = -Mx.(o^+Mz.w (1); 2F+ Y,+ Y,= l.my=0 (2); tZ + Z,+ Zo_ = Imz = tm (- z + xzttP' + x^io) = ay . Mk^ (5), where k is the radius of gyration about OY; and ^(xY-yX)-XA-XA — Xm (xy — yx) = — Smy (— xo)- + zw) = oi^Xmxy - bj^myz (6). On integrating (5) we have the values of w and m, and then, by substitution, the right-hand members of equations (1) to (4) and (6) are given. (1) and (6) determine X^ and X^. (3) and (4) determine Z^ and Z^. Fi and Y^ are indeterminate but (2) gives their sum. It is clear that the right-hand members of (4) and (6) would be both zero if the axis of rotation were a principal axis at the origin 0; for then the quantities "Xmxy and Xmyz would be zero. In a problem of this kind the origin should therefore be always taken at the point, if there be one, where the axis of rotation is a principal axis. Reactions of the axis of rotation. Examples 229 EXAMPLES 1. A thin uniform rod has one end attached to a smooth hinge and is allowed to fall from a horizontal position ; shew that the horizonal strain on the hinge is greatest when the rod is inclined at an angle of 45° to the vertical, and that the vertical strain is then ^ times the weight of the rod. 2. A heavy homogeneous cube, of weight W, can swing about an edge which is horizontal ; it starts from rest being displaced from its unstable position of equilibrium ; when the perpendicular from the centre of gravity upon the edge has turned through an angle 6, shew that the components of the action at the hinge along, and at right angles to, this W W perpendicular are -^ (3 - 5 cos 6) and -— sin 6. 3. A circular area can turn freely about a horizontal axis which passes through a point of its circumference and is perpendicular to its plane. If motion commences when the diameter through is vertically above C, shew that, when the diameter has turned through an angle 6, the components of the strain at along, and perpendicular to, W W this diameter are respectively -^(7 cos ^ — 4) and -5- sin 6. 4. A uniform semi-circular arc, of mass m and radius a, is fixed at its ends to two points in the same vertical line, and is rotating with constant angular velocity u>. Shew that the horizontal thrust on the , . g + ar'a upper end is m. 5. A right cone, of angle 2a, can turn freely about an axis passing through the centre of its base and perpendicular to its axis ; if the cone starts from rest with its axis horizontal, shew that, when the axis is vertical, the thrust on the fixed axis is to the weight of the cone as 1 + ^ cos2 a to 1 - ^ cos- a. 6. A regular tetrahedron, of mass M, swings about one edge which is horizontal. In the initial position the perpendicular from the centre of mass upon this edge is horizontal. Shew that, when this line makes an angle 6 with the vertical, the vertical component of thrust is ^(2sin2^-fl7cos2^). 181. Motion about a fixed axis. Impulsive Forces. By Ai-t. 166 we have that the change in the moment of momentum about the fixed axis is equal to the moment L of the impulsive forces about this axis. 230 Dynamics of a Rigid Body But, as in Art. 170, the moment of momentum of the body about the axis is Mk"^ . ft, where O is the angular velocity and Mk- the moment of inertia about the axis. Hence, if co and to' be the angular velocities about the axis just before and just after the action of the impulsive forces, this change is Mk- (&>' — &>), and we have Mk- (co' — ai)= L. Ex. A uniform rod OA, of jnass M and length la, rests on a smooth table and is free to turn about a smooth pivot at its end O ; in contact with it at a distance b from is an inelastic particle of mass m; a horizontal blow, of impulse P, is given to the rod at a distance x from in a direction perpendicular to the rod; find the resulting instantaneous angular velocity of the rod ajid the impulsive actions at and on the particle. If w be the angular velocity required and S the impulse of the action between the rod and particle, then, by the last article, we have M~u, = P.x-S.b (1). Also the impulse S communicates a velocity bio to the mass m, so that m.bu = S (2). 4a2 (1) and (2) give w = Px /(m ■mb^ Again, let X be the action at O on the rod. Then, since the change in the motion of the centre of gravity of the rod is the same as if all the impulsive forces were applied there, .-. 3I.au = P-S-X. (Ma +7nb) x' .: X=P-{Ma + mb)u,= P mPbx M — + mb- Also (2) gives 5 = M^ + mb^ 182. Centre of percussion. When the fixed axis of rotation is given and the body can be so struck that there is no impulsive action on the axis, any point on the line of action of the blow is called a centre of percussion. As a simple case consider a thin uniform rod OA (= 2a) suspended freely from one end and struck by a horizontal blow at a point G, where OC is x and P is the impulse of the blow. Centre of Percussion 231 Let ft)' be the instantaneous angular velocity communicated to the rod, and X the impulsive action upon the rod of the axis about which it rotates. The velocity of the centre of gravity Q immediately after the blow is rm. Hence the result (1) of Art. 166 gives Mato' = P + X (1). Also the moment of momentum of the rod about immediately after the P blow is Mk-w, where k is the radius of gyration of the rod about 0, i.e. k- = -^ . o Hence the result (4) of Art. 166 gives Mk"(o' = P .X (2). when Hence X Hence X = Maw - M - w = Maw . (3). X X ^ ^ 3ro, i.e. there is no impulsive action at 0, , and then 0C=^ the length of the simple equivalent pendulum (Art. 173). In this case G, the required point, coincides with the centre of oscillation, i.e. the centre of percussion with regard to the fixed axis coincides with the centre of oscillation with regard to the same axis. If X be not equal to - , then, by (3), X is positive or negative according as x is greater or less than — , i.e. the impulsive stress at on the body is in the same, or opposite, direction as the blow, according as the blow is applied at a point below or above the centre of percussion. 183. For the general case of the motion of a body free to move about an axis, and acted on by impulsive forces, we must use the fundamental equations of Art. 166. With the notation and figure of Art. 180, let {X, Y, Z) be the components of the impulsive forces at any point {x, y, z) and (Xj, Fi, Z^ and (Xg, Fj, Z.^ the components of the corre- sponding impulsive actions at B^ and Bo. 232 Dynamics of a Rigid Body Then, as in Art. 180, u = X = zo) ; v= y = ; w — z = — xoa ; u' = zoj'; ?;' = ; and w' = — xw, where cd' is the angular velocity about OF after the blows. The equations (1) to (6) of Art. 166 then become XX + Xi + X2 = Xmzo)' — Xmzw = Mz .{a - co) .. .(1) ; 2F+F, + F,= (2); X Z + Zi + Z2==^ Xm (— ccco') — 'S.in (— xo)) = -i]Ix.((o'-a}) (.3); X(yZ-zY) + ZJ), + ZJ), = tm [- xy') — tm (— yzu)) = — (&)' — «). tmyz (6). The rest of the solution is as in Art. 180. 184. Centre of Percussion. Take the fixed axis as the axis of y; let the plane of xy pass through the instantaneous position of the centre of inertia Q ; and let the plane through the point of application, Q, of the blow perpendicular to the fixed axis be the plane of xz, so that G is the point (x, y, 0) and Q is the point (^, 0, ^). Let the components of the blow parallel to the axes be X, Fand Z, and let us assume that there is no action on the axis of rotation. The equations of the previous article then become ^-0 (1), F=0 (2), Z = - Mx {(o' - (o) ...(3), ^y •(4), ^Y={(0'-(0)tl ^X-^Z = {a>'-a>)Mk'...{5), and ^Y = -(q}' -o})tmyz (6). Equations (1) and (2) shew that the blow must have no components parallel to the axes of x and y, i.e. it must be Centre of Percussion 233 perpendicular to the plane through the fixed axis and the instantaneous position of the centre of inertia. (4) and (6) then give %mxy=0, and ^myz = 0, so that the fixed axis must be a principal axis of the body at the origin, i.e. at the point where the plane through the line of action of the blow perpendicular to the fixed axis cuts it. This is the essential condition for the existence of the centre of percussion. Hence, if the fixed axis is not a principal axis at some point of its length, there is no centre of percussion. If it be a principal axis at only one point of its length, then the blow must act in the plane through this point perpendicular to the axis of rotation. k- Finally, (3) and (5) give ^= = . It follows, therefore, from Art, 173, that when a centre of percussion does exist, its distance from the fixed axis is the same as that of the centre of oscillation for the case when the body oscillates freely about the fixed axis taken as a horizontal axis of suspension. Corollary. In the particular case when y = and the centre of inertia G lies on Ox, the line of percussion passes through the centre of oscillation. This is the case when the plane through the centre of inertia perpendicular to the axis of rotation cuts the latter at the point at which it is a principal axis, and therefore, by Art. 147, the axis of rotation is parallel to a principal axis at the centre of inertia. The investigation of the three preceding Articles refers to impulsive stresses, i.e. stresses due to the blow, only ; after the rotation has commenced there will be on the axis the ordinary- finite stresses due to the motion. 185. A rough example of the foregoing article is found in a cricket-bat. This is not strictly movable about a single axis, but the hands of the batsman occupy only a small portion of the handle of the bat, so that we have an approximation to a single axis. If the bat hits the ball at the proper place, there is very little jar on the batsman's hands. Another example is the ordinary hammer with a wooden handle ; the principal part of the mass is collected in the iron hammer-head ; the centre of percussion is situated in, or close 234 Dynamics of a Rigid Body to, the hammer-head, so that the blow acts at a point very near the centre of percussion, and the action on the axis of rotation, i.e. on the hand of the workman, is very slight accordingly. If the handle of the hammer were made of the same material as its head, the effect would be different. 186. Ex. A triangle ABC is free to move about its side BC ; find the centre of percussion. Draw AD perpendicular to BC, and let E be the middle point of BC and F the middle point of DE. Then, as in Ex. 7, page 202, F is the point at which BC is a principal axis. li AD=p, then, by Art. 153, the moment of inertia about BC ■ 6* Also "[©'Ki)1-°'^-'=' In the triangle draw FF' perpendicular to BC to meet AE in F', so that FF' = ^.AD--^. oscillation for a rotation about BC as a horizontal axis of suspension. The points E' and F' coincide only when the sides AB, AC of the triangle are eqiial. EXAMPLES Find the position of the centre of percussion in the following cases : 1. A uniform rod with one end fixed. 2. A uniform circular plate ; axis a horizontal tangent. 3. A sector of a circle ; axis in the plane of the sector, perpendicular to its symmetrical radius, and passing through the centre of the circle. 4. A uniform circular lamina rests on a smooth horizontal plane, shew that it will commence to turn about a point on its circumference if it be struck a horizontal blow whose line of action is perpendicular to the diameter through and at a distance from equal to three-quarters of the diameter of the lamina. 5. A pendulum is constructed of a solid sphere, of mass M and radius a, which is attached to the end of a rod, of mass m and length b. Shew that there will be no strain on the axis if the pendulum be struck at a distance {M[f^d'^ + {a-\-b)^] + ^mb'}-i-[M{a + b) + -^mb] from the axis. 6. Find how an equilateral triangular lamina must be struck that it may commence to rotate about a side. Motion about a fixed axis. Examples 235 7. A uniform beam AB can turn about its end A and is in equi- librium ; find the points of its length where a blow must be applied to it so that the impulses at A may be in each case -th of that of the blow. 8. A uniform bar AB, of length 6 feet and mass 20 lbs., hangs vertically from a smooth horizontal axis &i A ; it is struck normally at a point 5 feet below J. by a blow which would give a mass of 2 lbs. a velocity of 30 feet per second ; find the impulse received by the axis and the angle through which the bar rises. 9. A rod, of mass m and length 2a, which is capable of free motion about one end A, falls from a vertical position, and when it is horizontal strikes a fixed inelastic obstacle at a distance b from the end A. Shew that the impulse of the blow is r?t. ^. */ -^ , and that the impulse of the reaction at ^ is TO a/ -|^ 1 - g| vertically upwards. 10. A rod, of mass nM, is lying on a horizontal table and has one end fixed ; a particle, of mass M, is in contact with it. The rod receives a horizontal blow at its free end ; find the position of the particle so that it may start moving with the maximum velocity. In this case shew that the kinetic energies communicated to the rod and mass are equal. 11. A uniform inelastic beam can revolve about its centre of gravity in a vertical plane and is at rest inclined at an angle a to the vertical. A particle of given mass is let fall from a given height above the centre and hits the beam in a given point P ; find the position of P so that the resulting angular velocity may be a maximum. 12. A rod, of mass M and length 2a, is rotating in a vertical plane with angular velocity a about its centre which is fixed. When the rod is horizontal its ascending end is struck by a ball of mass m which is falling with velocity u, and when it is next horizontal the same end is struck by a similar ball falling with the same velocity « ; the coefficient of restitution being unity, find the subsequent motion of the rod and balls. 13. A uniform beam, of mass m, and length 2?, is horizontal and can turn freely about its centre which is fixed. A particle, of mass m' and moving with vertical velocity u, hits the beam at one end. If the coefiicient of restitution for the impact be e, shew that the angular velocity of the beam immediately after the impact is Zm' {\ + e)uj{m-\-'im')l, and that the vertical velocity of the ball is then u (em - 3m')/{m -\-37n'). 236 Dynamics of a Rigid Body 14. Two wheels on spindles in fixed bearings suddenly engage so that their angular velocities become inversely proportional to their radii and in opposite directions. One wheel, of radius a and moment of inertia /j, has angular velocity o) initially ; the other, of radius h and moment of inertia I^, is initially at rest. Shew that their new angular velocities are 7,62 , I^ab J- T, , T — .-, a> and -f-yr, — F — :> (i>- Jib^+l/x- Jib^ + La- 15. A rectangular parallelepiped, of edges 2a, 2b, 2c, and weight W, is supported by hinges at the upper and lower ends of a vertical edge 2a, and is rotating with uniform angular velocity dx dx _dx ^ dx' _ dt' dt dt' dt ~ ' Hence the last two terms of (1) vanish, and the kinetic energy Hif)<^)'HH(^h(^ dt. = the kinetic energy of a particle of mass M placed at the centre of inertia and moving with it + the kinetic energy of the body relative to the centre of inertia. Now the velocity of the particle m relative to G d(b dS — f L — - v> _ — dt dt' and therefore the kinetic energy of the body relative to G Hence the required kinetic energy where v is the velocity of the centre of inertia G, 6 is the angle that any line fixed in the body makes with a line fixed in space, and k is the radius of gyration of the body about a line through G perpendicular to the plane of motion. L. D. 16 242 Dynamics of a Rigid Body 191. Moment of momentum about the origin O of a body moving in two dimensions. With the notation of the last article, the moment of momentum of the body about the origin ^ r dii dx~\ -:, f- dy , dy _ dx , dx\ , , + -'»L''i*"i-'''i«-^aJ ^'>- But, as in the last article, 1,mx = and 2m -j- = 0. at ... Smo^'f = f 2W=0, dt dt and l.my^^=ylm-^^=0. So also for the corresponding y terms. Hence, from (l),the moment of momentum about = moment of momentum about of a particle of mass M placed at the centre of inertia G and moving v/ith it + the moment of momentum of the body relative to Q. Now the velocity of the particle m relative to G _ d4 _ dd ~^^ dt~'^ dt' and its moment of momentum about G dO JO = ''''^Tt=' dt' Therefore the moment of momentum of the body relative to G dt dt Hence the total moment of momentum = Mvp + Mm (2), where p is the perpendicular from upon the direction of the velocity v of the centre of inertia. L dt y dt\ Motion in two dimensions 243 Or again, if the polar coordinates of the centre of inertia G referred to the fixed point as origin be {R, yjr), this expression may be written MI,f^M,^f^ (3>- 192. The origin being a fixed point, the rate of change of the moment of momentum about an axis through it perpendicular to the plane of rotation (for brevity called the moment of momentum about 0) is, by equation (3) of Art. 187, equal to the moment of the impressed forces about 0. For the moment of momentum we may take either of the expressions (1), (2), or (3). Thus taking (1) we have the moment of the forces about 0. Hence M\x^-y'^] + Mt9 - L. Similarly, if we took moments about the point (xo, y^), the equation is M r _ d-y ._ . d-x + Mm = the moment of the impressed forces about (a^o, ^o)- The use of the expressions of this article often simplifies the solution of a problem ; but the beginner is very liable to make mistakes, and, to begin with, at any rate, he would do well to confine himself to the formulae of Art, 187. 193. Instead of the equations (1) and (2) of Art. 187 may be used any other equations which give the motion of a particle, e.g. we may use the expressions for the accelerations given in Art. 49 or in Art. 88. The remainder of this Chapter will consist of examples illustrative of the foregoing principles. 16—2 244 Dynamics of a Rigid Body 19-4. A uniform sphere rolls donm an inclined plane, rough cnoiigh to prevent any sliding; to find the motion. Let be the point of contact initially when the sphere was at rest At time t, when the centre of the sphere has described a distance x, let A be the position ^ — ^/^ of the point of the sphere which was originally / pjV at 0, so that CA is a line fixed in the body. I CW; I ^^^ Let I KCA, being an angle that a line fixed vJJ-r'^^'^^^ in the body makes with a line fixed in space, be 0. Let R and F be the normal reaction and the friction. Then the equations of motion of Art. 187 are M^^ = Jfg^ma-F (1), = Mgcoiia-R (2), and 3fk^^,=F.a (3). Since there is no sliding the arc KA=\ine KG, so that, throughout the motion, x=-ad (4). d^x k^ d^Q (1) and (3) give ^"^ + a "S^^ =^ «"^ «' *Xby(4), (l+^J) =g sm a. rf% a2 Hence the centre of the sphere moves with constant acceleration -^-^ a sin a, and therefore its velocity v = ^^j., a sm a . t and 1 a2 . „ -?? = 7; o , ,o <7sma. t'. ^^TT^S'sma, 2(^2 5 In a sphere ^■^=-^ ; hence the acceleration =-^ sin a. [If the body were a hollow sphere, k^ would =-o-. ^^^ t^ie acceleration be f (7 sin a. If it were a uniform solid disc, k"^ would =— , and the acceleration be ^g sin a. If it were a uniform thin ring, k^ would =a^, and the acceleration be ^g sin a.] From (1 ) we have F= Mg sin a - ^Mg sin a = f % sin o and (2) gives R = % cos a. Motion in two dimensions. Examples 245 F 2 Hence, since -^ must be < the coefficient of friction /x, therefore - tan a must be < /i, in order that there may he no shding in the case of a soHd sphere. Equation of Energy. On integrating equation (5), we have the constant vanishing since the body started from rest. Hence the kinetic energy at time t, by Art. 190, = My .X sin a = the work done by gravity. Ex. 1. A uniform solid cylinder is placed with its axis horizontal on a plane, whose inclination to the horizon is a. Shew that the least coeflBcient of friction between it and the plane, so that it ra&j roll and not slide, is ^ tan a. If the cylinder be hollow, and of small thickness, the least value is \ tan a. Ex. 2. A hollow cylinder rolls down a perfectly rough inclined plane in one minute ; shew that a solid cylinder will roll down the same distance in 52 seconds nearly, a hollow sphere in 55 seconds and a solid sphere in 50 seconds nearly. Ex. 3. A uniform circular disc, 10 inches in diameter and weighing 5 lbs., is supported on a spindle, \ inch in diameter, which rolls down an inclined railway with a slope of 1 vertical in 30 horizontal. Find (1) the time it takes, starting from rest, to roll 4 feet, and (2) the linear and angular velocities at the end of that time. Ex. 4, A cylinder roUs down a smooth plane whose inclination to the horizon is a, unwrapping, as it goes, a fine string fixed to the highest point of the plane ; find its acceleration and the tension of the string. Ex. 5. One end of a thread, which is wound on to a reel, is fixed, and the reel falls in a vertical line, its axis being horizontal and the unwound part of the thread being vertical. If the reel be a solid cylinder of radius a and weight W, shew that the acceleration of the centre of the reel is f (/ and the tension of the thread ^W. Ex. 6. Two equal cylinders, each of mass m, are bomid together by an elastic string, whose tension is T, and roll with their axes horizontal down a rough plane of inclination a. Shew that their acceleration is 3 i.. mg sm aj where /x is the coefficient of friction between the cylinders. Ex. 7. A circular cylinder, whose centre of inertia is at a distance e from its axis, rolls on a horizontal plane. If it be just started from a position of unstable equilibrium, shew that the normal reaction of the plane when the 246 Dynamics of a Rigid Body 4,c2 centre of mass is in its lowest position is 1 + , ^o (a-c)2- where k is the radius of gyration about an axis through the centre of mass. 195. A uniform rod is held in a vertical position with one end resting upon a perfectly rough table, and when released rotates about the end in contact with the table. Find the motion. Let R and F be the normal reaction and the friction when the rod is inclined at an angle 6 to the vertical ; let x and y be the coordinates of its centre, so that x = asm6 and y=aGosd. i / The equations of Art. 187 are then '" j. / fiir ■■ ■ -yfa F=M^^=M[acos66-afim6e^] ...(1). 2/ a/ L R-Mg = M^^.=M[-aamed-acoad.6^]...{2\ and J/. — .^ = ^asin ^ — i^acos^ = Mga sin 6 - Ma^6, by (1) ana (2), Affi .. so that M—e = Mgas,\n6 (.3). [This latter equation could have been written down at once by Art. 171, since the rod is rotating about J. as a fixed point.] (3) gives, on integration, 6^ = ~ (1 -cos^), since 6 is zero when ^ = 0. Hence (1) and (2) give i^=i/.^sin^(3eos^-2) and i? = ^(l -3cos ^)2. It will be noted that R vanishes, but does not change its sign, when cos ^=-g-. The end A does not therefore leave the plane. The friction F changes its sign as 6 passes through the value cos ~ ^ § ; hence its direction is then reversed. F The ratio „ be infinitely rough there must be sliding then. In any practical case the end A of the rod will begin to slip for some value of 6 less than cos~^|-, and it will slip backwards or forwards according as the slipping occurs before or after the inclination of the rod is cos~if. 196. A uniform straight rod slides doxon in a vertical plane, its ends being in contact with two smooth planes, one horizontal and the other vertical. If it started from rest at an angle a with the horizontal, find the motion. Motion in two dimensions. Examples 247 Let R and S be the reactions of the two planes when the rod ia inclined at 6 to the horizon. Let x and y be the coordinates of the centre of gravity G. Then the equations of Art. 187 give ^S=^ M^=S-Mg dfi and ifF d^ df^'' ■ R.a sin 6 — Sa cos 6 . ■(1), •(2), .(3). Since k'^= — , these give o -^sm^ -^-cos^.^'-cos^ ^ dfi df^ .(4). and Now a;=a cos ^ and y=a sin 6, so that d-\v = —aQ.o^Qt- — asm.6 .6^ i %- = - a sin ^ . ^2 4- d cos 66. Hence (4) gives 4 d'^6 ^(^-:m=-9^^^G Hence, on integration, - (1)'=^^-^-'^.-- 0=-||sina+C. • •• \dt) =2^(«in«-«i"^) ( From (1), ^2-asin^^' (5). (6). 3.^ ^ (sin a - sin ^) cos (9 + ^ sin ^ cos (9, from (5) and (6), .(7). = ^cos^(3sin(9-2sina) From (2), j^ =5r - a sin 66"^ + a cos 66 = |[l-6sinasin(9 + 9sin2^] = |[(3sin^-sina)2 + cos2a]...(8). From (7) it follows that R is zero when sin 5 = f sin a, and, for a smaller value of 6, R becomes negative. Hence the end A leaves the wal l whe n sin ^ = |sina, and its angular velocity is then, by (6), equal to la sin a . , , ,, . . \/ 2a~' instant the horizontal velocity of G dx = —a sm d6 1 ^ ■Yf~'^ V Sag' sin^ a. 248 Dynamics of a Rigid Body The equations of motion now take a different form. They become ^w^-^ (^')' J/g='^.-% (2'), and Jf.|'.^=-*Si.acos0...(3'). Also y=asin0, so that -^= -asin (/). <^2^.c(cos0. '^=R-Mg cos cj) (2), and M(a-b)'(i) = F-Mgsm(f) (3), where R is the normal reaction, and F is the friction at B as marked. Also for the motion relative to the centre of inertia, we have i/^-^ = moment of the forces about G= -F.b, M. M Ifi a — b (f>=-Fb .(4). These equations are sufficient to determine the motion. Eliminating F between (3) and (4), we have .(5). ^ S(a-b) ^ Integrating this equation, we have where Q is the value of when the cylinder is in its lowest position. This equation cannot in general be integrated further. (2) and (6) give -^= (a - 6) Q^ + 1 [7 cos 0-4] .(7). In order that the cylinder may just make complete revolutions, R must be just zero at the highest point, where cp — ir. 252 Dynamics of a Rigid Body In this case (a-b)Q,^-—^, and hence the velocity of projection = (a-6)J2 = V-y-gf(a-6). If Q be less than this value, E will be zero, and hence the inner cylinder will leave the outer, when cos^ = - 4 ^^~ ^ — . (4) and (6) give F=^sin(j) (8). The friction is therefore zero when the cylinder is in its lowest position, and for any other position i^is positive, and therefore acts in the direction marked in the figure, Equation of Energy. The equation (6) may be deduced at once by assuming that the change in the kinetic energy is equal to the work done. When the centre is at C the energy (by Art. 190) Hence the loss in the kinetic energy as the cylinder moves from its lowest position = fi/ (a -5)2 (122 _02)_ Tj^jg equated to the work done against gravity, viz. Mg (a - b) (I - eos (f)), gives equation (6). Small oscillations. Suppose the cylinder to make small oscillations about the lowest position so that is always small. Equation (5) then 2(7 gives ^= — , •_ , r about the centre of the cylinder in time x/^--^''«s -(,-!)■ E.V. 2. A solid homogeneous sphere is rolling on the inside of a fixed hollow sphere, the two centres being always in the same vertical plane. Shew that the smaller sphere will make complete revolutions if, when it is in its lowest position, the pressure on it is greater than -'j- times its own weight. Ex. 3. A circular plate rolls down the inner circumference of a rough circle under the action of gravity, the planes of both the plate and circle being vertical. When the line joining their centres is inclined at an angle 9 the plate. Motion in two dimetisiotis. Examples 253 Ex. 4. A cylinder, of radius a, lies within a rough fixed cylindrical cavity of radius 2a. The centre of gravity of the cylinder is at a distance c from its axis, and the initial state is that of stable equilibrium at the lowest point of the cavity. Shew that the smallest angular velocity with which the cylinder must be started that it may roll right round the cavity is given by where k is the radius of gyration about the centre of gravity. Find also the normal reaction between the cylinders in any position. 199. An imperfectly rough sphere moves from rest doxmi a plane inclined at an angle a to the horizon; to determine the motion. Let the centre G have described a distance x in time t, and the sphere have rolled through an angle 6 ; so that 6 is the angle between the normal CB to the plane at time t and the radius GA v?hich was normal at zero time. Let us assume that the friction was not enough to produce pure rolling, and hence that the sphere slides as well as turns ; in this case the friction will be the maximum that the plane can exert, viz. [x/i, where /x is the coefficient of friction. Since the sphere remains in contact with the plane, its centre is always at the same distance a from the plane, so that y and i/ are both zero. Hence the equations of motions are M^=Mgsma-f.R (1), 0=R — Mgcost and M.k^.'^^^.R.a. •(2), .(3). Since k^^"-^ , (2) and (3) give -r^ = -^ cos a. dd baa 2a2 6 dt 2a •(4), and , hu.q t^ l = -p.cosa.- 2a 2 •(5), the constants of integration vanishing since 6 and 6 are both zero initially. d^x So (1) and (2) give ^ =5- (sin a- /i cos a). dx =g (sin a — /i cos a) t ' dt and x=g(sma—ncoaa)~. the constants vanishing as before. •(6), •(7). 254 Dynamics of a Rigid Body The velocity of the point B down the plane = the velocity of C+the velocity of B relative to C= ^ - a ^ =^ ( sin a - ^ m cos a j ?. Firsts suppose sinu — f /icosa to be positive, i.e. /Lt f tan a. In this case the velocity of B appears to be negative which is impos- sible ; for friction only acts with force sufficient at the most to reduce tlie point on which it acts to rest ; and then is only sufficient to keep this point at rest. In this case then pure rolling takes place from the start, and the maximum friction jxE is not always exerted. The equations (1), (2), (3) should then be replaced by ify = %sina-i^ (8), = R-Mgcosa (9) and MB'^^^F.a (10). Also, since the point of contact is at rest, we have dx d6 -dt-^'^r'' ™ , , , „^, . d^x 2ad'^6 (8) and (10) now give -^ + — -^=^ sui a Therefore, by (11), x=a6=^g sin a. .-. x = ad = jgsma.t (12), 5 t- and x=ae = -gsma. ^ (13), the constants of integration vanishing as before. Equation of Energy. The work done by gravity when the centre has described a distance x=Mg . x sin a, and the kinetic energy then In the first case the energy, by (4) and (6), =^if^2^2[-(sin„_^cosa)2+ t/x2cos2a] (14), and the work done by gravity, by (7), = il/<7. a;sina=2i/^V'^sin a(sina-/xco3ay (15). Motion in two dimensions. Examples 255 It is easily seen that (14) is less than (15) so long as /u < f tan a, i.e. so long as there is any sliding. In this case then there is work lost on account of the friction, and the equation of work and energy does not hold. In the third case the kinetic energy, by (12) and (13), and the work done, by (13), 5 fi \ b =Mff .X sin a = Mg sin a. ^grsina- = ~ 31 . = g'^ sin^ aA In this case, and similarly in the second case, the kinetic energy acquired is equal to the work done and the equation of work and energy holds. This is a simple example of a general principle, viz. that where there is no friction, i.e. where there is pure sliding, or where there is pure rolling, there is no loss of kinetic energy ; but where there is not pure rolling, but sliding and rolling combined, energy is lost. Ex. 1. A homogeneous sphere, of radius a, rotating with angular velocity w about a horizontal diameter, is gently placed on a table whose coefficient of friction is /i. Shew that there will be slipping at the point of contact for a time ^ — , and that then the sphere will roll with angular velocity -^. Ex. 2. A solid circular cylinder rotating about its axis is placed gently with its axis horizontal on a rough plane, whose inclination to the horizon is a. Initially the friction acts up the plane and the coefficient of friction is /x. Shew that the cylinder will move upwards if /n>tana, and find the time that elapses before rolling takes place. Ex. 3. A sphere is projected with an underhand twist down a rough inclined plane ; shew that it will turn back in the course of its motion if 2aw {/J, - tan o) > 5upL, where u, w are the initial linear and angular velocities of the sphere, n is the coefficient of friction, a the inclination of the plane and /i>f tano. Ex. 4. A sphere, of radius a, is projected up an inclined plane with velocity V and angular velocity i2 in the sense which would cause it to roll up ; if Y>aU, and the coefficient of friction >| tano, shew that the sphere will cease to ascend at the end of a time - — : , where o is the inclination of bg &in a the plane. Ex. 5. If a sphere be projected up an inclined plane, for which ix = j tan a, with velocity V and an initial angular velocity i2 (m the direction in which it would roll up), and if r>afi, shew that friction acts downwards at first, and upwards afterwards, and prove that the whole time during which the . 17r+4afi sphere rises is - "^ I8g sm a 256 Dynamics of a Rigid Body Ex. 6. A hoop is projected with velocity Fdown a plane of incliuation a, the coefficient of friction being fi ( > tan o). It has initially such a backward spin O that after a time tj it starts moving uphill and continues to do so for a time t2 after which it once more descends. The motion being in a vertical plane at right angles to the given inclined plane, shew that («i + «9)5fsina = (ii2- V. Ex. 7. A uniform sphere, of radius a, is rotating about a horizontal diameter with angular velocity and is gently placed on a rough plane which is incUned at an angle a to the horizontal, the sense of the rotation being such as to tend to cause the sphere to move up the plane along the line of greatest slope. Shew that, if the coefficient of friction be tan a, the centre of the sphere will remain at rest for a time •;; — . — , and will then move downwards with o^sina acceleration y^r sin a. If the body be a thin circular hoop instead of a sphere, shew that the time is —. — and the acceleration - g sin o. ^ siu a .i 200. A spliere, of radius a, whose centre of gravity G is at a distance c from its centre G is placed on a rough plane so that CO is horizontal; shew that it will begin to roll or slide according as ^ is > ^ , ichere k is the radius of gyration about a horizontal axis through G. If ft, is equal to this value, what happens ? When CG is inclined at an angle 6 to the horizontal let A, the point ol contact, have moved through a horizontal distance X from its initial position 0, and let OA = x. Assume that the sphere rolls so that the friction is F; since the point of contact A is at rest, .-. x=ae (1). The equations of motion of Art. 187 are d^ F=M-^\_x+c cos 6] = M{{a-c&m6)e d^ ■c cos 66'^']... {2), R- Mg = M-Y7^{a -csmffl^M [-C cos 66 + esmee'^] dt and ^ccos 6- F{a-csm6) = Mk-'6 We only want the initial motion when 6=0, and then i is not zero. The equations (2), (3), (4) then give F=Ma'e, ] R = Mg-Mcd, I for the MWiia^ values. and Rc-Fa=MIc%\ 92. •(3), (4). _ is zero but 6 Hence we have F + a^ + c^' P + rt2 , F_ gae Motion in two dimensions 257 In order that the initial motion may be really one of rolling, we must have F< fiR, i.e. ,x > j^^^ . If /i be < this value, the sphere will not roll, since the friction is not sufficient. Critical Case. If ii = jj, 2 ^^ "^^^^ ^® necessary to consider whether, p when 6 is small but not absolutely zero, the value of ^ is a little greater or a little less than ft. The question must therefore be solved from the beginning, keeping in the work first powers of 6 and neglecting 6-, 6^, ... etc. (2), (3) and (4) then give, on eliminating F and R, S [k'^ + a^ + c^ -2ac sin e}-ac cos 06'^ =ge cos d (5). Hence, on integration, 6^ [t^ + a-^ + c^-2ac sin d] = 2gc sine (6). If Z'2=F + a2 + c^ these give, neglecting squares of 6, Hence, to the first power of 6, we have from (2) and (3), F _^ {a-c6)e-ce\ _ ac V ZB-a'^ ~] R g-c'e ^•'- + a2L ''a^lc' + a'f]' on substitution and simplification. lik^> — , then -5 is less than y^ „ , i.e. -^ is less than the coefficient of friction and the sphere rolls. If ^'^ < — , then ^ > the coefficient of friction and the sphere slides. Ex. 1. A homogeneous sphere, of mass M, is placed on an imperfectly rough table, and a particle, of mass m, is attached to the end of a horizontal diameter. Shew that the sphere will begin to roll or slide according as /i is greater or less than „ ,,,, — i— 7, . If w be equal to this value, shew that the sphere will begin to roll. Ex. 2. A homogeneous solid hemisphere, of mass M and radius a, rests with its vertex in contact with a rough horizontal pJane, and a particle, of mass m, is placed on its base, which is smooth, at a distance c from the centre. Shew that the hemisphere will commence to roll or slide according as the coefficient of friction g — ~ -^ — — — . 2b(il/ + m)a2 + 40mc-2 L. D. 17 258 Dynamics of a Rigid Body Ex, 3. A sphere, of radius a, whose centre of gravity G is not at its centre 0, is placed on a rough table so that OG is inclined at an augle a to the upward drawn vertical ; shew that it will commence to slide along the csina (a + ccoso) table if the coefiScient of friction be will roll. A;"+(a + ccosa)^ but that otherwise it Ex. 4. If a uniform semi-circular wire be placed in a vertical plane with one extremity on a rough horizontal plane, and the diameter through that extremity vertical, shew that the semi-circle will begin to roll or slide according as fi r2-2 If fx has this value, prove that the wire will slide. Ex, 5. A heavy uniform sphere, of mass M, is resting on a perfectly rough horizontal plane, and a particle, of mass m, is gently placed on it at an angular distance a from its highest point. Shew that the particle will at ,, , ., sin a ^7il/-(-5ni (1-f cos a)} , . ,, once slip on the sphere if /*< irirT— ' ? — A r,' , where u is the coefficient of friction between the sphere and the particle. 201. A uniform circular disc is projected, with its plane vertical, along a rough horizontal plane toith a velocity v of translation, and an angular velocity o about the centre. Find the motion. Case I. V -^, aa>. In this case the initial velocity of the point of contact is v — aoa in the direction -*• and the friction is fiAlg ~f-. When the centre has described a distance x, and the disc has turned through an angle 6, the equations of motion are Mx= - fiMg, and J/. -- . '6 = ixMga, ,',it=v-ngt and -0 = ~(o + iJigt ...(1). o~ « P Hence the velocity of the point of contact P =x-a6=v-aai -'iy-gt. Sliding therefore continues until t = — — ^ and pure rolling then begins. Also at this time the velocity of the centre = The equations of motion then become Mx=-F .(2). and J/| = i^.aJ l.w here F is the friction -*-. Also i=«(i), since the point of contact is now at rest ; . •. x=a<^. ■ These three equations give F=0, i.e. no friction is now required. Also a(^=i = constant = the velocity at the commencement of the rolling =—^—, by (2). Motion in two dimensions 259 The disc therefore continues to roll with a constant velocity which is less than its initial velocity. Case II. V ^- , « ") and v < aa. Here the initial velocity of the point of contact is -*- and hence the friction is fiJIg -*-. The equations of motion are then Mx = \iMg^ and M .—6= -fj. MgOy giving x=v + ngt and -d^^co-iigL Hence pure rolling begins when x = aB, i.e. when t= ^~ . Zfig The velocity, x, of the centre then = -^^ — , and, as in Case I, the disc rolls on with constant velocity which is greater than the initial velocity of the centre. Case III. v-^,io^. Initially the velocity of the point of contact is v+aa -^, so that the friction is fj,Mg -«-. The equations of motion are Mx= — fiMg, and M—d=^i.Mga, ,'. x=v-iigt and - 6 = fj.gt - - a. Pure rolling begins when x=^ad, i.e. when t= — — ^, and the velocity of the centre then = — - — . If 2i> > aw, this velocity is -»-, and the motion during pure rolling is ^- with constant velocity as before. If however 2y < aw, the velocity of the centre when pure rolling commences is -e- and the disc rolls back towards 0. In this particvdar case the velocity of the centre vanishes when t= — which is less than -jr , if 2v < ao) : hence the disc begins to move in the direction -^ Zfxg before pure rolling commences. [In this last case the motion is of the same kind as that in the well- known experiment of a napkin-ring projected along the table with a velocity v~^ and a sufficient angular velocity to in the direction '\.'\ Ex. A napkin-ring, of radius a, is propelled forward on a rough horizontal table with a linear velocity u and a backward spin w, which is > »/a. Find the motion and shew that the ring will return to the point of projection in time — L- — -^ , where a is the coefficient of friction. 4/t^ (aw - u) What happens if u^aul 17—2 260 Dynamics of a Rigid Body and b, and G their centre of 202- Two unequal smooth spheres are placed one on the top of the other in unstable equilibrium, the lower sphere resting on a smooth table. The system is slightly disturbed; shew that the spheres will separate ichen the line joining their centres makes an angle 6 with the vertical givenby the eqvMion -zrz. cos^^ — 3 COS ^ + 2 = 0, \chere M is the mass of the lower, and m of the M+m upper, sphere. Let the radii of the two spheres be < gravity, so that CG_^C;G^ a-\-b m M M+ m ' There being no friction at the table the resultant horizontal force on the system consisting of the two spheres is zero. Hence, by Art. 162, the horizontal velocity of the centre of gravity is constant, and equal to its value at the commence- ment of the motion, i.e. it is always zero. Hence the only velocity of G is vertical, and it therefore describes a vertical straight line GO, where was the initial position of the point of contact B, so that is a fixed point. For the horizontal motion of the lower sphere, we thus have AS'sin^ = J/^2[(7G^.sin^J [cos<9i9-sin^^-'] ...(1). Mm{a + b) M+m For the vertical motion of the upper sphere •,os 6 -mg=m -j-^[a + {a + b) cos e] = m {a + b)[- sin ^<9 - cos (9(92] ___(2), Eliminating S, we have 6 [M+ m sin^ d] + m sin 6 cos Hence, by integration, {3f+m)g a + b (3). ^2 [M+m sin2 61=— K {M+m) (1 -cos 6) since the motion started from rest at the highest point. By (1), S vanishes, i.e. the spheres separate, when .(4), cos (9(9 = sin (9.52^ .(5). (3) and (5) give 6^=^ j- at this instant and then (4) gives, on sub- stitution, mcos'5=(i/'-l-m)(3cos 6 — 2). There are no forces acting so as to turn either sphere about its centre, so that neither of them has any rotatory motion. Varying mass 261 Work and Energy. Equation (4) may be obtained thus, by assuming the principle of work. The horizontal velocity of the lower sphere = -^(CG sm 6) = — jj- -' cos 00, dt^ M+m ' so that its kinetic energy is i M -r~, — ^cos2 66^. ^ {M+mY The horizontal and vertical velocities of the upper sphere are ^ {CG ' sin &\ and ^ [a + (a + 6) cos 6], i.e. ^}^'^^h os6d and -{a + b)sm0d, so that its kinetic energy is \JM+mf J Equating the sum of these two energies to the work done, viz. rag (a + b){\- cos 6), we obtain equation (4). 203. Varying mass. In obtaining the equations of Art. 161 we assumed the mass of the body to remain constant. If the mass m of a particle is not constant, the component effective „ . d ( dx\ 1 , d'^x The equation (1) of Art. 161 is then ^ _ ^ d f dx\ d ^ dx d f T,^ dx\ Also the equation (6) of the same article is ^ d [ dy dx~\ d ^ [ f dy dx =i[-f as in Art. 187. Ex. A cylindrical mass of snow rolls doion an inclined plane covered with snow of uniform depth E, gathering up all the anow it rolls over and ahcays remaining circular; find the motion of the snow, and shew that it luill move with an acceleration -^g sin a, if initially, lohen its radius is a, it be started with velocity a W ^ , where a is the inclination of the plane. 262 Dynamics of a Rigid Body At time t from the start, let x be the distance described down the plane, and r be the radius, so that IT (r2 - a2) = the amount of snow picted up = E.x (1). If F be the friction up the plane, and the angle turned through by the snow-ball, we have ^[7r)-2p.i] = 7rr2r7psina-F (2), and ^[7rr2p.fc2^] = ir.r (3), where p is the density of the snow-ball. Also x-rd^O (4), since there is no sliding. Since lfi = -- , the equations (2) and (3) give i.e. 3i'-i-7-x = 2gsina, or, from (1), x + l --^J^ = \g,\na. On putting i^^xi and hence 2i; = — , this equation becomes linear, and its solution is i2 (TTcfi + Ex)l = ^gs,ina.^ (7rn2 + ExfT -)- C. O iUxi .„ 2a sin a , „ „ , C i. e. a;2= ±--- {^a^ + Ex) + . (Tra^ + Ex)S This equation cannot in general be integrated further. If, however, x = a\/ -— „-= — . when x — 0, we have G = 0, and then ' ' \ 5E 'sina ^ 27r^a2 sin a _ . 2 2/7 sin a 5 ■ 5E ■■ ■'«=0 ' 6 so that the acceleration is g sin a MISCELLANEOUS EXAMPLES ON CHAPTER XIV. 1. A uniform stick, of length 2a, hangs freely by one end, the other being close to the ground. An angular velocity « is then given to the stick, and when it has turned through a right angle the fixed end is let go. Shew that on first touching the ground it will be in an upright position if 2a 3 + -^-— , where p is any odd multiple of 2. A circular disc rolls in one plane upon a fixed plane and its centre describes a straight line with uniform acceleration /; find the magnitude and line of action of the impressed forces. Motion in two dimensions. Examples 263 3. A spindle of radius a carries a wheel of radius 6, the mass of the combination being M and the moment of inertia /; the spindle rolls down a fixed track at inclination a to the horizon, and a string, wound round the wheel and leaving it at its under side, passes over a light puUejr and has a mass m attached to the end which hangs vertically, the string ■between the wheel and pulley being parallel to the track. Shew that the acceleration of the weight is g(b-a) [Ma sin a + ?w (6 - a)} -^ [/+ Ma^ + m{b- af]. 4. Three imiform spheres, each of radius a and of mass m, attract one another according to the law of the inverse square of the distance. Initially they are placed on a perfectly rough horizontal plane with their centres forming a triangle whose sides are each of length 4a. Shew that the velocity of their centres when they collide is */ y y-^ , where y is the constant of gravitation. 5. A unifoiTH sphere, of mass m and radius a, rolls on a horizontal plane. If the resistance of the air be represented by a horizontal force acting at the centre of the sphere equal to — v~ and a couple about it equal to m^v% where v is the velocity of the sphere at any instant, and if V be the velocity at zero time, shew that the distance described by the centre in time ^ is — log Fl + -^ Vt\ where A = '- . J. ■■ ■' 7 a 6. A uniform sphere rolls in a straight line on a rough horizontal plane and is acted upon by a horizontal force X at its centre in a direction opposite to the motion of the centre. Shew that the centre of the sphere moves as it would if its mass were collected there and the force reduced to jX, and that the friction is equal to jX and is in a direction opposite to that of X. 7. A man walks on a rough sphere so as to make it roll straight up a plane inclined at an angle a to the horizon, always keeping himself at an angle /3 from the highest point of the sphere ; if the masses of the sphere and man be respectively If and ?«., shew that the acceleration of the sphere 5cf {m sin /3 - (M+m) sin a} '^ 7if+5m{l+cos(a+^)} • 8. A circular cylinder, of radius a and radius of gyration k, rolls inside a fixed horizontal cylinder of radius b. Shew that the plane through the axes will move like a simple circular pendulum of length (6-<.)(i4:). If the fixed cylinder be instead free to move about its axis, and have 264 Dynamics of a Rigid Body its centre of gravity in its axis, the correspouding pendulum will be of length (6 - a) (1 + ?i), where m and M are respectively the masses of the inner and outer cylinders, and K is the radius of gyration of the outer cylinder about its axis. [In the second case, if the outer cylinder has at time t turned through an angle -v//-, the equations of motion are, as in Art. 198, m{b-a)4''^=R-mffcos(f) ; m{h-a)'<^ = F~mgsm(P; mk^ =-F.a; and 3/K 2^ = - Fb. Also the geometrical equation is a {6 + (j)) = b{(})-\j/).] 9. A uniform circular hoop has a fine string wound round it. The hoop is jjlaced upright on a horizontal plane, and the string, leaving the hoop at its highest point, passes over a smooth pulley at a height above the plane equal to the diameter of the hoop and has a particle attached to its other end. Find the motion of the system, supposed to be all in one vertical plane ; and shew that whether the plane be smooth or rough the hoop will roll without slipping. 10. A disc rolls upon a straight line on a horizontal table, the flat surface of the disc being in contact with the plane. If v be the velocity of the centre of the disc at any instant, shew that it will be at rest after a time ■^, — , where u is the coefficient of friction between the disc and 64,xg table. 11. A perfectly rough cylindrical grindstone, of radius a, is rotating with uniform acceleration about its axis, which is horizontal. If a sphere in contact with its edge can remain with its centre at rest, shew that the angular acceleration of the grindstone must not exceed ^ . 12. A perfectly rough ball is at rest within a hollow cylindrical garden roller, and the roller is then drawn along a level path with uniform velocity V. If V'^ > ^-f- g {b - a), shew that the ball will roll completely round the inside of the roller, a and b being the radii of the ball and roller. 13. A solid uniform disc, of radius a, can turn freely about a horizontal axis through its centre, and an insect, of mass - that of the disc, starts from its lowest point and moves along the rim with constant velocity relative to the rim ; shew that it will never get to the highest point of the disc if this constant velocity is less than - j2ga{n + 2). Motion in two dimensions. Examples 265 14. Inside a rough hollow cylinder, of radius a and mass i/, which is free to turn about its horizontal axis, is placed an insect of mass m ; if the insect starts from the lowest generator and walks in a plane perpendicular to the axis of the cylinder at a uniform rate v relatively to the cylinder, shew that the plane containing it and the axis never makes with the r^ Mk'^ - | upward drawn vertical an angle < 2cos~^ — - / , „,„ rr , where Mk'^ is the moment of inertia of the cylinder about its axis 15. A rough lamina, of mass i/, can turn freely about a horizontal axis passing through its centre of gravity, the moment of inertia about this axis being Mk"-. Initially the lamina was horizontal and a particle of mass m was placed on it at a distance c from the axis and then motion was allowed to ensue. Shew that the particle will begin to slide on the lamina when the latter has turned through an angle tan" where /ix is the coefficient of friction. 16. A uniform rod, of mass 3f and length I, stands upright on perfectly rough ground ; on the top of it, which is flat, rests a weight of mass m, the coefficient of friction between the beam and weight being fi. If the beam is allowed to fall to the ground, its inclination 6 to the vertical M /AM \ when the weight slips is given by -— sin ^ + ( ---+3m J cos 6 = M+2'm. 17. A rough cylinder, of mass M, is capable of motion about its axis, which is horizontal ; a particle of mass m is placed on it vertically above the axis and the system is slightly disturbed. Shew that the particle will slip on the cylinder when it has moved through an angle 6 given by fi {M-^ 6m) cos 6 — M sin 6 = A^iifx, where /a is the coefficient of friction. 18. A hemisphere rests with its base on a smooth horizontal plane ; a perfectly rough sphere is placed at rest on its highest point and is slightly displaced. Shew that in the subsequent motion the angular velocity of the line joining the centres, when its inclination to the vertical is 6, is T 5ng -f \\_c{7n-5cos^d)J ' hemisphere when 6 satisfies the equation 5 ^3-^") 0053^ + 20 cos2^ + 7(15-17%)cos^ + 70(ji-l)=0, where c is the sum of the radii and n the ratio of the sum of the masses of the sphere and hemisphere to that of the sphere. [Use the Principles of Linear Momentum and Energy.] 19. A thin hollow cylinder, of radius a and mass M, is free to turn about its axis, which is horizontal, and a similar cylinder, of radius b and mass m, rolls inside it without slipping, the axes of the two cylinders being 2 sin - -j^;- — / , ^, I , and shew also that the sphere will leave the 266 Dynamics of a Rigid Body parallel. Shew that, when the plane of the two axes is inclined at an angle 6 to the vertical, the angular velocity i2 of the larger is given by a^{M+m) (2J/+m) Q.^ = 2gm-{a-b) (cos 5- cos a), provided both cylinders are at rest when 6 = a. 20. A perfectly rough solid cylinder, of mass m and radius r, rests symmetrically on another solid cylinder, of mass M and radius R, which is free to turn about its axis which is horizontal. If 7n rolls down, shew that at any time during the contact the angle (j) which the line joining the centres makes with the vertical is given by Find also the value of ^ when the cylinders separate. 21. A locomotive engine, of mass M, has two pairs of wheels, of radius a, the moment of inertia of each pair about its axis being l/X;^ ; and the engine exerts a couple Z on the forward axle. If both pairs of wheels commence to roll without sliding when the engine starts, shew that the friction between each of the front wheels and the line capable of being L k'^ + a? called into action must be not less than rs „ . 22. A rod, of mass to, is moving in the direction of its length on a smooth horizontal plane with velocity u. A second perfectly rough rod, of the same mass and length 2a, which is in the same vertical plane as the first rod, is gently placed with one end on the first rod; if the initial inclination of the second rod to the vertical be a, shew that it will just rise into a vertical position if Sw^ sin^ a = \ga (1 — sin a) (5+3 cos^ a). 23. A rough wedge, of mass M and inclination a, is free to move on a smooth horizontal plane ; on the inclined face is placed a uniform cylinder, of mass m j shew that the acceleration of the centre of the cylinder down , ... i •, • ^ • i/+?rtsin2a the face, and relative to it, is a^rsina.— y^^- -— — ^-^ . ' " 3if+?rt + 2?nsm2a 24. A uniform circular ring moves on a rough curve under the action of no forces, the curvature of the curve being everywhere less than that of the ring. If the ring be projected from a point A of the curve without rotation and V)egin to roll at B, then the angle between the normals at A AH- log 2 and B is — s— . 25. A uniform rod has one end fastened by a pivot to the centre of a wheel which rolls on a rough horizontal plane, the other extremity resting against a smooth vertical wall at right angles to the plane containing the rod and wheel ; shew that the inclination 6 of the rod to the vertical, when it leaves the wall, is given by the equation Qi/cos^ 6 + ^m cos 6 — 4m cos a — 0, where M and m are the masses of the wheel and rod and a is the initial inclination to the vertical when the system was at rest. Motion in two dimensions. Examples 267 26. Rope is coiled round a drum of a feet radius. Two wheels each of radius b are fitted to the ends of the drum, and the wheels and drum form a rigid body having a common axis. The system stands on level ground and a free end of the rope, after passing under the drum, is inclined at an angle of 60° to the horizon. If a force P be applied to the rope, shew that the drum starts to roll in the opposite direction, its centre having acceleration , " , ., — f^jr , whore M is the mass of the system and k its radius 2.1/ (6^ + F) of gyration about the axis. 27. A thin circular cylinder, of mass M and radius 6, rests on a perfectly rough horizontal plane and inside it is placed a perfectly rough sphere, of mass in and radius a. If the system be disturbed in a plane perpendicular to the generators of the cylinder, obtain the equations of finite motion and two first integrals of them; if the motion be small, shew that the length of the simple equivalent pendulum is .. . 28. A uniform sphere, of mass J/, rests on a rough plank of mass m which rests on a rough horizontal plane, and the plank is suddenly set in motion with velocity u in the direction of its length. Shew that the sphere will first slide, and then roll, on the plank, and that the whole system will come to rest in time — r^? r? where a is the coefiicient of friction at each of the points of contact. 29. A board, of mass M, whose upper surface is rough and under surface smooth, rests on a smooth horizontal plane. A sphere of mass m is placed on the board and the board is suddenly given a velocity V in the direction of its length. Shew that the sphere will begin to roll after a time (^5) MS' 30. On a smooth table there is placed a board, of mass M, whose upper surface is rough and whose lower surface is smooth. Along the upper surface of the board is projected a uniform sphere, of mass m, so that the vertical plane through the direction of projection passes through the centre of inertia of the board. If the velocity of projection be u and the initial angular velocity of the sphere be « about a horizontal axis perpendicular to the initial direction of projection, shew that the motion will become uniform at the end of time zrr? — t^ > and that the tM + 'zra fig velocity of the board will then be ,. {u — aca). 31. A perfectly rough plane turns with uniform angular velocity a> about a horizontal axis lying in its plane ; initially when the plane was horizontal a homogeneous sphere was in contact with it, and at rest 268 Dynamics of a Rigid Body relative to it at a distance a from the axis of rotation ; shew that at time t the distance of the point of contact from the axis of rotation was a cosh (y^ ^^ + _0 sinh [y^^ ««] - ^, sin ^t. Find also when the sphere leaves the plane. [For the motion of the centre of gravity use revolving axes, as in Art. 51.] 32. In the previous question the plane turns about an axis parallel to itself and at a distance c from it ; when the plane is horizontal and above the axis the sphere, of radius 6, is gently placed on the plane so that its centre is vertically over the axis ; shew that in time t the centre of the sphere moves through a distance CHAPTEH XV MOTION IN TWO DIMENSIONS. IMPULSIVE FORCES 204. In the case of impulsive forces the equations of Art. 187 can be easily transformed. For if T be the time during which the impulsive forces act we have, on inte- where X' is the impulse of the force acting at any point {x, y). Let u and v be the velocities of the centre of inertia parallel to the axes just before the impulsive forces act, and u' and v' the corresponding velocities just after their action. Then this equation gives M{u'-u) = ^X' (1). So M{v'-v) = tY' (2). These equations state that the change in the momentum of the mass M, supposed collected at the centre of inertia, in any direction is equal to the sum of the impulses in that direction. So, on integrating equation (4), we have i.e. if ft) and on' be the angular velocities of the body before and after the action of the impulsive forces, we have Mk' {co' - ft)) = S {x Y' - y'X'). 270 Dynamics of a Rigid Body Hence the change produced in the momentum about the centre of inertia is equal to the moment about the centre of inertia of the impulses of the forces, 205- -Ex. 1. A uniforvi rod AB, of length 2a, is lying on a smooth horizontal plane and is struck by a horizojital blow, of impulse P, in a direction perpendicular to the rod at a point distant b from its centre; to find the motion. Let u' be the velocity of the centre of inertia perpendicular to the rod after the blow, and w' the corresponding angular velocity about the centre. Then the equations of the last article give Mu' = P, and M^w'=P.b. Hence we have u' and w'. Ex. 2. A uniform rod at rest is struck by a blow at right angles to its length at a distance x from its centre. Fiiid the point about which it will begin to turn. Let be the required centre of motion, GO = y, where G is the centre of inertia, and GA = GB = a. Let tlie impulse of the blow be P, and the resulting angular velocity about be w. The velocity acquired by G is yui. Hence, from Art. 204, we have Myu, = P (1), and M^f- + "^'joj^P{y + x) (2). Solving, w = ' ^ and y = w-,, giving the resulting angular velocity and the p position of O. The velocity of the centre of inertia G = yw=—- The kinetic energy acquired, by Art. 190, If the end A were fixed, the resulting angular velocity wj would be given by the equation M \ a^+ - \wi = P (a + x), so that wi = jr^. -^ , and the kinetic energy generated would 1 -. 4a2 3P2(a+x)2 The ratio of the energies given by (iJJ and U)^- \ . ' ^ ' 3 (a + a;)2 Irtipidsive motion in two dimensions 271 The least value of this ratio is easily seen to be unity, when x = — . Hence the kinetic energy generated when the rod is free is always greater than that when the end A is fixed, except when x=-, in which case A is the centre of rotation. Ex. 3. Two uniform rods AB, BC are freely jointed at B and laid on a horizontal table ; AB is struck by a horizontal blow of impuUe P in a direction ■perpendicular to AB at a distance c from its centre; the lengths of AB, BC being 2a and '2b and their masses M and M' , find the motion immediately after the blow. Q Let Ml and wj be the velocity of the centre of inertia of AB and its angular velocity just after the blow ; uj and W2 similar quantities for BC. There will be an impulsive action between the two rods at B when the blow is struck ; let its impulse be Q, in opposite directions on the two rods. Then for the rod AB, since it was at rest before the blow, we have Mui = P-Q (1), and M^ .wi = P.c-Q.a (2). So, fov BC, we have M'u2 = Q (3), and 3I'--.W2=-Q.b (4). Also, since the rods are connected at B, the motion of B, as deduced from each rod, must be the same. .-. Ui + aui=^U2-bcj2 (5). These five simple equations give wj, wi, M2, <>>2 and Q. On solving them, we obtain 3Prc 1 M' /, 3c\-| IP/ 3c\ , 3 P /, 3c\ Ex. 4. Three equal uniform rods AB, BC, CD are hinged freely at their ends, B and C, so as to form three sides of a square and are laid on a smooth table ; the end A is struck by a horizontal bloiv P at right angles to AB. Shew that the initial velocity of A is nineteen times that of D, and that the impulsive 5P P actions at B and C are respectively — and — . 272 Dynamics of a Rigid Body B The initial motion of the point B must be perpendicular to AB, so that the action at B must be along BC ; similarly the action at C must be along CB. Let them be Xi and X^ as marked. Let the velocities and angular velocities of the rods be «i and wi, W2> ^^^ "3 ^^^ "a as in the figure. For the motion of AB we have MlWi = P + Zi (1), Gi a2 .(2), G3 where m is the mass and 2a the length of each rod. For BC, we have muo = Xi - X2 For CD, we have mus^Xz .. P A and .(3). .(•5). Also the motion of the point B of the rod AB is the same as that of the same point B of the rod BC. .: Ui-au}i= -U2 (6). So, for the point G, Us + au3 = i(2 (7). On substituting from (1)...(5) in (6) and (7), we obtain 5Zi-Z2 = 2P and Xi^oX^, giving Hence we have 17P Xi^^ and Z2 ^ 12 12 7P P P '^<^i = xz:' «2=^; "3= 4/« 3m 12/u velocity of the point A _ wi + awx velocity of the point D au3 - M3 and (1(^3- 19. P^ 4hi* EXAMPLES 1. AB, BC are two equal similar rods freely hinged at B and lie in a straight line on a smooth table. The end A is struck by a blow perpen- dicular to AB; shew that the resulting velocity of A is 3^ times that of B. 2. Two uniform rods, AB and BO, are smoothly jointed at B and placed in a horizontal line ; the rod BC is struck at 6* by a blow at right angles to it ; find the position of O so that the angular velocities of AB and BC may be equal in magnitude. 3. Two equal uniform rods, AB and AC, are freely hinged at A and rest in a straight line on a smooth table. A blow is struck at B perpen- dicular to the rods ; shew that the kinetic energy generated is -j times what it would be if the rods were rigidly fastened together at A. Impulsive motion hi two dimensions. Examples 273 4. Two equal uniform rods, AB and BG, are freely jointed at B and ■ turn about a smooth joint at A. When the rods are in a straight line, o) being the angular velocity oi AB and u the velocity of the centre of mass oi BG, jBC impinges on a fixed inelastic obstacle at a point J); shew that the rods are instantaneously brought to rest if BI) = 2a ~ — , where 2a •' ° 3u + 2aa) ' is the length of either rod. 5. Two rods, AB and BG, of lengths 2a and 2b and of masses propor- tional to their lengths, are freely jointed at B and are lying in a straight line. A blow is communicated to the end A ; shew that the resulting kinetic energy when the system is free is to the energy when G is fixed as (4a + 36) (3a + 46) : 12 (a + 6)2. 6. Three equal rods, AB, BG, GD, are freely jointed and placed in a straight line on a smooth table. The rod AB \s, struck at its end .4 by a blow which is perpendicular to its length ; find the resulting motion, and shew that the velocity of the centre of AB is 19 times that of GD, and its angular velocity 11 times that of CD. 7. Three equal uniform rods placed in a straight line are freely jointed and move with a velocity v perpendicular to their lengths. If the middle point of the middle rod be suddenly fixed, shew that the ends of the other two rods will meet in time —- — , where a is the length of each rod. 8. Two equal uniform rods, AB and AG, are freely jointed at A, and are placed on a smooth table so as to be at right angles. The rod ^4 C is struck by a blow at G in a direction perpendicular to itself ; shew that the resulting velocities of the middle points oi AB and AG are in the ratio 9. Two uniform rods, AB, AG, are freely jointed at A and laid on a smooth horizontal table so that the angle BAG is a right angle. The rod AB is struck by a blow P at B in a direction perpendicular to AB; shew 2P that the initial velocity of A is - — -. , where m and m' are the masses of •^ 41)1+7)1 AB, J. C respectively. 10. AB and GD are two equal and similar rods connected by a string BG; AB, BG, and GD form three sides of a square. The point A of the rod ^5 is struck a blow in a direction perpendicular to the rod; shew that the initial velocity of A is seven times that of D. 11. Three particles of equal mass are attached to the ends, A and G, and the middle point .B of a light rigid rod ABG, and the system is at rest on a smooth table. The particle G is struck a blow at right angles to the rod ; shew that the energy communicated to the system when A is fixed, is to the energy communicated when the system is free as 24 to 25. L. D. 18 274 Dynamics of a Rigid Body 12. A uniform straight rod, of length 2 ft. and mass 2 lbs., has at each end a mass of 1 lb., and at its middle point a mass of 4 lbs. One of the 1 lb. masses is struck a blow at right angles to the rod and this end starts off with a velocity of 5 ft. per second ; shew that the other end of the rod begins to move iu the opposite direction with a velocity of 2-5 ft. per sec. 206. A uniform sphere, rotating with an angular velocity to about an axis perpendicular to the plane of motion of its centre, impinges on a liorizontal plane ; find the resulting change in its motion. First, suppose the plane rough enough to prevent any sliding. Let u and v be the components of its velocity before impact as marked in the figure; u' and v' the components, and co' the angular velocity, just after the impact. Let R be the normal impulsive reaction and F the impulsive friction. Then the equations of Art. 204 give 3I{u'-u)=-F (1), M(v' + v) = R (2), and Mk- (co' - co) = Fa (3). Also since the point A is instantaneously reduced to rest, there being no sliding, u' — aco' = (4). Also, if e be the coefficient of restitution, (5). V = ev Solving (1), (3), and (4), we have , , 5?/. + 2aco u = aco = zz .(6), .(7). and F= M .f{ti-aco) Case I. u = aco. There is no friction called into play, and u and w are unaltered. Case II. u < aco. Then F acts — >- ; co' u. Hence when the point of contact A before impact is moving -* — , the angular velocity is decreased by the impact, the horizontal velocity is increased. Impact of a rotating sphere on the ground 275 and the direction of motion of the sphere after impact makes a smaller angle with the plane than it would if there were no friction. Case III. u> aw. Then F acts -« — ; w > a and u' < u. Hence when the point of contact A before impact is moving — >-, the angular velocity is increased, the horizontal velocity is diminished, and the direction of motion after impact makes a greater angle with the plane than it would if there were no friction. Case IV. Let the angular velocity before the impact be ^. We must now change the sign of &>, and have , , bu — 2aft) .„. u =a(o = s \p), and F=M.}{u+aco) (9). If w = —^r- then u' and &)' are both zero, and the sphere 5 rebounds from the plane vertically with no spin. If w < --^ , then u' is negative and the sphere after the impact rebounds towards the direction from which it came. [Compare the motion of a tennis ball on hitting the ground when it has been given sufficient " under-cut."] In each case the vertical velocity after the impact is ev and R = {l+e)v. In Cases I, II and III, in order that the point of contact F may be instantaneously brought to rest, we must have n fi(l+e)v, the friction is not sufficient to bring the point of contact A to instantaneous rest, equation (4^ will not hold and for equations (1), (2), (3), we must have M(u'-u) = -fMR (1'). M{v' + v) = R (2'), and Mk\/u,v(l + e) (5'), the friction is not sufficient, and we have equations similar to (!'), (2'), and (3'), but with the sign of &> changed. They will give u' = u — fiv {\ + e), v'=ev, and ft)' = 1^ V (1 + e) - ft). In this case it will, from (5'), be possible for u to be less than fxv {1 + e), if ft) be large enough; hence, if the ball has sufficiently large enough under-cut, u' can be negative, i.e. the ball can rebound backwards [Compare again the motion of a tennis ball.] 207. Ex. 1. A rod, of length 2a, is held in a position inclined at an angle a to the vertical, and is then let fall on to a smooth inelastic horizontal plane. Shew that the end which hits the plane will leave it immediately after the impact if the height through which the rod falls is greater than ^^a sec a cosec2 a (1 + 3 sin^ a)2. If u and w be the vertical and angular velocity just after the impact, V the vertical velocity before the impact and E the impulse of the reaction of the plane, then m(V-u)=R, mk^w = Ra Bin a, and m- aw sin a = vertical velocity of the end in contact with the plane = 0. T-T u SFsino Hence w = — ■. — = —„-, (1). asina a{l + 3sin2a) Assuming the end to remain in contact with the plane, and that S is the normal reaction when the rod is inclined at 6 to the vertical, we have S -mg = m -r- (a cos 6), and S. a sin d = )ii — d (2). Eliminating S, we have '(l + 3sin2^) + 3bin6'cos6i^2=— sin^ (3). Impulsive motion in two dimensions 277 Now S is negative when = a ii 6 is negative then, so that equation (3) then gives 3 sin a cos au)2 > -^ sin o, i. e. w^ a cos a Hence, from (1), r.^^ Ml + 3sin2a)2 ^, 9 sin2 a Hence the given answer ga(l + 3sin^a)a 9 cos a siu2 o Ex. 2. Four equal rods, each of mass m and length 2a, are freely jointed at their ends so as to form a rhombus. The rhombus falls icith a diagonal vertical, and is moving with velocity V when it hits a fixed horizontal inelastic plane. Find the motion of the rods immediately after the impact, and shew that their angular velocities are each equal to -^ „ ,, , o „:.,9. \ > where a is the angle each rod makes with the vertical. 2 a(l + 3sin2a) Sheto also that the impact destroys a fraction z — . „ of the kinetic i. -{- o sin Ot energy just before the impact. After the impact it is clear that AB is moving with some angular velocity (ci^ about A, and BC with an angular velocity W2 ^ about B. Since C is, by symmetry, moving vertically after the impact, its horizontal velocity is zero. .-. = horizontal velocity of B + horizontal velocity of G relative to B — 2a wx cos a + 2aoj2 cos a, ie- '^1= -'•'1 (!)• The horizontal velocity of G^, similarly = 2a(isx cos a + aw2 cos a = a wi cos a -^ , and its vertical velocity A = 2a wi sin a.-auii sin a = 3a wi sin o \ . If X be the horizontal impulse at G as marked (there being no vertical impulse there by symmetry) we have, as in Art. 192, on taking moments about A for the two rods AB, BC m -5- wi + m awi cos a . 3a cos o + 3awi sin a . a sin a + — W2 V 2X 2wi = — sin a H cos a . a ma 2mVa sin a = Z. 4a cos o, i.e. 2wi = — sin a + -::-^cosa (2). Similarly, taking moments about B for the rod BC, we have m awj cos a. a cos a-3«wi sin a. a 8ina + — - W2 -'"[- F]. a sin a = Z. 2a cos a, ^^ A ■ 9 \ V ■ 2A' .^. [■ e. Wi I TT -4sin'o ) = smaH cos a (3). ' \3 J a ma ^ ' Solving (2) and (3), we have wi and X, and the results given are obtained. 278 Dynamics of a Rigid Body The impulsive actions Zj -»- and Fj f at i? on the rod BG are clearly given by Xi->rX=m. horizontal velocity communicated to G2 = wi . awj cos a, and Yi — m x vertical velocity communicated to G^ = m(- 3awi sin a) - m [ - F] = 7« [ F - Sawj sin a]. Also the impulsive action A'2-*- at A on AB is given by X2 = Jn. horizontal velocity communicated to Gi = m. awj cos a. The total action Y^l^^t A = total change in the vertical momentum = Am V - Smaui sin a. On solving these equations, we have _3F sin a _ _j«Ftan a Scos^a- 1 '^^~2^'l + 3sin2a' ~ 2 1 + 3 sin2 « ' Y _^^ tana _7hF Scos^a- 1 _ '""2" l + 3sin2a' ^~ '¥ l + Ssin^a' _omFsinacosa j v _ ^^^ ^2--2" l + 3sin2a ' ^"'^ ^^'U^S^^^' Also the final kinetic energy = i . 2m . ^ wi2 + ^ . 2Hi ra2wi2 cos2 a + da^ui^ sin2 a + ~ u/] 2 3 sin2 , : original kinetic energy. l + 3sm2a It will be noted that, since we are considering only the change in the motion produced by the blow, the finite external forces (the weights of the rods in this case) do not come into our equations. For these finite forces produce no effect during the very short time that the blow lasts. Ex. 3. A body, whose mass is m, is acted upon at a given point P by a blow of impulse X. If V and V be the vilocities of P in the direction of X just before and just after the action of X, shew that the change in the kinetic energy of the body, i.e. the work done on it by the impulse, is 1(F+F')X Take the axis of x parallel to the direction of X. Let u and v be the velocities of the centre of inertia G parallel to Ox and Oy, and w the angular velocity round G just before the action of X. Let u', v', and w' be the same quantities just after the blow. The equations of Art. 204 then become m{u' -u)=X; m(v' -v}=0, and mk^{u' -a})= -y' .X (1), where (x', y') are the coordinates of P relative to G. By Art. 190, the change in the kinetic energy = bn (m'2 + 1;'2 + /(2a,'2) _ ^,« [ifi + v^ + k'^w^) = ^m (m'2 - u^) + |mA;2 (a;'2 - w2) = ^X {u' + u) -W . X{w' + w), by (l), = iX {(u'-y'w') + {u-y'^, \. Now F=the velocity of G parallel to Ox + the velocity of P relative to Q = u-. GP sin GPx = u- y'w, and similarly F' = u' - y'w. Hence the change in the kinetic energy = ^X(F'+ V). Impulsive motion in tivo dimensions. Examples 279 EXAMPLES ON CHAPTER XV 1. A uniform inelastic rod falls without rotation, being inclined at any angle to the horizon, and hits a smooth fixed peg at a distance from its upper end equal to one-third of its length. Shew that the lower end begins to descend vertically. 2. A light string is wound round the circumference of a uniform reel, of radius a and radius of gyration k about its axis. The free end of the string being tied to a fixed point, the reel is lifted up and let fall so that, at the moment when the string becomes tight, the velocity of the centre of the reel is u and the string is vertical. Find the change in the motion and k- show that the impulsive tension is mu . —. — t7, . 3. A square plate, of side 2a, is falling with velocity m, a diagonal being vertical, when an inelastic string attached to the middle point of an upper edge becomes tight in a vertical position. Shew that the impulsive tension of the string is jMu, where M is the mass of the plate. Verify the theorem of Art. 207, Ex. 3. 4. If a hollow lawn tennis ball of elasticity e has on striking the ground, supposed perfectly rough, a vertical velocity u and an angular velocity co about a horizontal axis, find its angular velocity after impact and prove that the range of the rebound will be - — eu. ^ 9 5. An imperfectly elastic sphere descending vertically comes in contact with a fixed rough point, the impact taking place at a point distant a from the lowest point, and the coefficient of elasticity being e. Find the motion, and shew that the sphere will start moving horizontally after the impact if V 5 ■ 6. A billiard ball is at rest on a horizontal table and is struck by a horizontal blow in a vertical plane passing through the centre of the ball ; if the initial motion is one of pure rolling, find the height of the point struck above the table. [There is no impulsive friction.] 7. A rough imperfectly elastic ball is dropped vertically, and, when its velocity is V, a man suddenly moves his racket forward in its own plane with velocity U, and thus subjects the ball to pure cut in a downward direction making an angle a with the horizon. Shew that, on striking the rough ground, the ball will not proceed beyond the point of impact, provided (£/■- Fsina)(l-cosa) > (1 + e) f 1 + ^-^j V sm a COS a. 280 Dynamics of a Rigid Body 8. An inelastic sphere, of radius a, rolls down a flight of perfectly rough steps; shew that if the velocity of the centre on the first step exceeds JgUi its velocity will be the same on every step, the steps being such that, in its flight, the sphere never impinges on an edge, [The sphere leaves each edge immediately.] 9. An equilateral triangle, formed of uniform rods freely hinged at their ends, is falling freely with one side horizontal and uppermost. If the middle point of this side be suddenly stopped, shew that the impulsive actions at the upper and lower hinges are in the ratio ,^13 : 1. 10. A lamina in the form of an equilateral triangle ABC lies on a smooth horizontal plane. Suddenly it receives a blow at ^ in a direction parallel to BC, which causes A to move with velocity V. Determine the instantaneous velocities of B and C and describe the subsequent motion of the lamina. 11. A rectangular lamina, whose sides are of length 2a and 26, is at rest when one corner is caught and suddenly made to move with pre- scribed speed V in the plane of the lamina. Shew that the greatest angular 37 velocity which can thus be imparted to the lamina is " — / „ ,g . 12. Four freely-jointed rods, of the same material and thickness, form a rectangle of sides 2a and 26 and of mass M'. When lying in this form on a horizontal plane an inelastic particle of mass M moving with velocity V in a direction perpendicular to the rod of length 2a impinges on it at a distance c from its centre. Shew that the kinetic energy lost in the impact IS 2 V^^\j^+jj, (1 + ^:^ -.)\- 13. Four equal uniform rods, AB, BC, CD, and DE, are freely jointed at B, C and D and lie on a smooth table in the form of a square. The rod AB is struck by a blow at A at right angles to AB from the inside of the square ; shew that the initial velocity of A is 79 times that of E. 14. A rectangle formed of four uniform rods ft-eely jointed at their ends is moving on a smooth horizontal plane with velocity Fin a direction along one of its diagonals which is perpendicular to a smooth inelastic vertical wall on which it impinges ; shew that the loss of energy due to the impact is „2 / f 1 3 cos^ " . 3 sin^ a 1 / \mi + vi^ 3^1 4-^2 ?Hi-f3???2l' where m^ and «i2 are the masses of the rods and a is the angle the above diagonal makes with the side of mass wij. 15. Of two inelastic circular discs with milled edges, each of mass m and radius a, one is rotating with angular velocity m round its centre which is fixed on a smooth plane, and the other is moving without spin in the plane with velocity v directed towards 0. Find the motion immediately afterwards, and shew that the energy lost by the impact is ^ m iv^+ -^ Impulsive motion in two dimensions. Examples 281 16. A uniform circular disc, of mass M and radius a, is rotating with uniform angular velocity <» on a smooth plane and impinges normally with any velocity ^l upon a rough rod, of mass m, resting on the plane. Find the resulting motion of the rod and disc, and shew that the angular velocity of the latter is immediately reduced to -^ — — - w. 17. An elHptic disc, of mass m, is dropped in a vertical plane with velocity V on a perfectly rough horizontal plane ; shew that the loss of 1 a;^ + p^ kinetic energy by the impact is -(1— e^)mF-. .^ ^^ , where r is the distance of the centre of the disc from the point of contact, p is the central perpendicular on the tangent, and e is the coeflBcient of elasticity. 18. Two similar ladders, of mass ?w and length 2a, smoothly hinged together at the top, are placed on a smooth floor and released from rest when their inclination to the horizontal is a. When their inclination to the horizontal is d they are brought to rest by the tightening of a string of length I which joins similarly situated rungs. Shew that the jerk in the string is V 3 ga (sin a - sin 6). 19. A sphere of mass m falls with velocity V on a perfectly rough inclined plane of mass M and angle a which rests on a smooth horizontal plane. Shew that the vertical velocity of the centre of the sphere immedi- , T ,, .1 • , . 5(M+m) Fsin^a ,, , j- i • ately after the impact is ---— — - — ^ -—k- , the bodies being all sup- 7i/+2m + 5msin2a' ° ^ posed perfectly inelastic. 20. A sphere, of mass m, is resting on a perfectly rough horizontal plane. A second sphere, of mass m', falling vertically with velocity V strikes the first ; both spheres are inelastic and perfectly rough and the common normal at the point of impact makes an angle y with the horizon. Shew that the vertical velocity of the falling sphere will be instantaneously reduced to F(m + to') -f ^ m sec^ y + to' + = m' tau^ (j + Tj , Shew also that the lower sphere will not be set in motion if siny=f, but that the upper sphere will be set spinning in any case. CHAPTER XVI INSTANTANEOUS CENTRE. ANGULAR VELOCITIES. MOTION IN THREE DIMENSIONS 208. To fix the position of a point in space we must know- its three coordinates ; this may be otherwise expressed by saying that it has three degrees of freedom. If one condition be given {e.g. a relation between its coordinates, so that it must lie on a fixed surface) it is said to have two degrees of freedom and one of constraint. If two conditions are given {e.g. two relations between its coordinates so that it must be on a line, straight or curved) it is said to have one degree of freedom and two of constraint. A rigid body, free to move, has six degrees of freedom. For its position is fully determined when three points of it are given. The nine coordinates of these three points are con- nected by three relations expressing the invariable lengths of the three lines joining them. Hence, in all, the body has 6 degrees of freedom. A rigid body with one point fixed has 6 — 3, i.e. three, degrees of freedom, and therefore three of constraint. A rigid body with two of its points fixed, i.e. free to move about an axis, has one degree of freedom. For the six co- ordinates of these two points are equivalent to five constraining conditions, since the distance between the two points is constant. 209. A rigid body has its position determined when we know the three coordinates of any given point G of it, and also the angles which any two lines, GA and GB, fixed in the body make with the axes of coordinates. Instantaneous Centre 283 [If G and GA only were given the body might revolve round GA.] Since there are the three relations (1) P + m'^-irn- = \, (2) r^ + m'^ + n''^ = 1 and (3) W + mm' + nn' = the cosine of the given angle AGB, between the direction cosines {I, m, n) and {V, m', n') of the two lines, it follows that, as before, six quantities, viz. three coordinates and three angles must be known to fix the position of the body. 210. Uniplanar motion. At any instant there is always an axis of pure rotation, i. a body can be moved from one position into any other by a rotation about some point with- out any translation. During any motion let three points A, B, G fixed in the body move into the positions A', B' , and G' respectively. Bisect A A', BB' at M and N and erect per- pendiculars to meet in 0, so that OA = OA' and OB = OB'. Then the triangles A OB, A' OB' are equal in all respects, so that Z. AOB = Z A' OB', and . aA0A' = zB0B' (1), and Z OB A = z OB' A'. But z CBA = z G'B'A'. .% by subtraction Z OBG=Z OB'C. Also OB=OB' and BC = B'C. Hence the triangles OBG, OB'C' are equal in all respects, and hence 0G= 00' (2), and Z COB = z C'OB', i.e. zCOC' = zBOB' = zAOA' (3). Hence the same rotation about 0, which brings A to A' and B to B', brings any point G to its new position, i.e. is the required centre of rotation. 284 Dynamics of a Rigid Body The point always exists unless A A' and BB' are parallel, in Avhich case the motion is one of simple translation and the corresponding point is at infinity. Since the proposition is true for all finite displacements, it is true for very small displacements. Hence a body, in uniplanar motion, may be moved into the successive positions it occupies by successive instantaneous rotations about some centre or centres. To obtain the position of the point at any instant let A and A' be successive positions of one point, and B and B' successive positions of another point, of the body. Erect perpendiculars to A A' and BB' ; these meet in 0. 211. The centre, or axis, of rotation may be either perynanent, as in the case of the axis of rotation of an ordinary pendulum, or instantaneous, as in the case of a wheel rolling in a straight line on the ground, where the point of contact of the wheel with the ground is, for the moment, the centre of rotation. The instantaneous centre has two loci according to whether we consider its position with regard to the body, or in space. Thus in the case of the cart-wheel the successive points of contact are the points on the edge of the wheel; their locus with regard to the body is the edge itself, i.e. a circle whose centre is that of the wheel. In space the points of contact are the successive points on the ground touched by the wheel, i.e. a straight line on the ground. These two loci are called the Body-Locus, or Body-Centrode, and the Space-Locus, or Space-Centrode. 212. The motion of the body is given by the rolling of the body-centrode, carrying the body with it, upon the space-centrode. Let C/, G2, Gi, Gl, ... be successive points of the body-centrode, and Cj, Cg, 63, 64 ... successive points of the space-cen- trode. At any instant let Gx and C/ coincide so that the body is for the instant moving Instantaneous Centre 285 about Oi as centre. When the body has turned through the small angle the point (7/ coincides with G^ and becomes the new centre of rotation ; a rotation about G^ through a small angle brings G^ to G^ and then a small rotation about Cg brings C/ to G^ and so on. In the case of the wheel the points 0/, C/ ... lie on the wheel and the points G-^, Cg . . . on the ground. Hx. 1. Bod sliding on a plane with its ends on two perpendicular straight lines CX and CY. At A and B draw perpendiculars to CX and GY and let them meet in 0, The motions of A and B are instantaneously along AX and BC, so that is the instantaneous centre of rotation. Since BOA is a right angle, the locus of with respect to the body is a circle on AB as diameter, and thus the body-centrode is a circle of radius 2 • ^B. Since CO = AB, the locus of in space is a circle of centre C and radius AB. Hence the motion is given by the rolling of the smaller ■ circle, carrying AB with it, upon the outer circle of double its size, the point of contact of the two circles being the instantaneous centre. Ex. 2. The end A of a given rod is compelled to move on a given straight line CY, whilst the rod itself always passes through a fixed point B. Draw BC { = a) perpendicular to CY. The instantaneous motion of along CY, so that the instantaneous centre lies on the perpendicular AO. The point B of the rod is for the moment moving in the direction AB, so that lies on the perpendicular OB to AB. Body-Centrode. By similar triangles OAB, ABC we have d^=JL. . A0= "■ AO AB' " cos^ OAB' so that with respect to the body the locus of is the curve A IS .(1). cos-^ Space-Centrode. If OM be perpendicular to CB, and C3I=x, MO = y, then x = a + y cotOBM=a + y tan, and y—CA = ata,n parallel to the axes, i.e. u — yw and v + xw. These are zero if a; = and y = ~ . CO ^ CO The coordinates of the centre of no acceleration are also easily found. For the accelerations of any point P relative to G are PG . u^ along PG and PG . w perpendicular to PG. Instantaneous Centre 287 Therefore the acceleration of P parallel to OX = u - PG . 0)2 . cos e - PG .usme = u- u'^x - uy, and its acceleration parallel to Oy = v-PG . 0)2, sin^ + PG .d}COse = v-(a'^y + (bx. These vanish at the point X ^ y 1 Mw2 - vw VW^ + nu W* + 0)2 ■ 214. The point P, whose coordinates referred to G are (x, y), being the instantaneous centre and L the moment of the forces about it, the equation of Art. 192 gives L = M [¥u) + yu- xv] , where Mk^ is the moment of inertia about G, Now, since P is the instantaneous centre, u^ + v^ = PG' . (i)\ ••• ^=£s[^''"'+^«="'J = £|f*>'J w. where ki is the radius of gyration about the instantaneous centre. (1) If the instantaneous centre be fixed in the body, so that /i-f is constant, this quantity = if Aji^w, (2) If PG (= r) be not constant, the quantity (1) Z(o at ii(o = Mki^co + Mrrw. Now if, as in the case of a small oscillation, the quantities r and 0) are such that their squares and products can be neglected, this quantity becomes Mk^w, so that in the case of a small oscillation the equation of moments of momentum about the instantaneous centre reduces to moment of momentum about the instantaneous . _ L centre / ~ Mki' ~ moment of inertia about / the squares of small quantities being neglected, i.e. as far as small oscillations are concerned we may treat the instantaneous centre as if it were fixed in space. 288 Dynamics of a Rigid Body 215. Motion in three dimensions. One point of a rigid body being fixed, to sheiu that the body may be transferred from one 2iOsitio7i into any other position by a rotation about a suitable axis. Let the radii from to any two given points a, /S of a body meet any spherical surface, of centre 0, in the points A and B, and when the body has been moved into a second position let A and B go to ^' and B' respectively. Bisect A A' and BB' in D and E and let great circles through D and E perpen- dicular to A A' and BB' meet in a Then GA = CA', CB = GB' and AB = A'B'. .'. ZAGB = ZA'GB'. .'. /.AGA' = /.BGB', so that the same rotation about OG which brings A to A' will bring B to B'. Now the position of any rigid body is given when three points of it are given, and as the three points 0, A, B have been brought into their second positions 0, A', B' by the same rotation about OG, it follows that any other point P will be brought into its second position by the same rotation. 216. Next, remove the restriction that is to be fixed, and take the most general motion of the body. Let 0' be the position of in the second position of the body. Give to the whole body the translation, without any rotation, which brings to 0'. 0' being now kept fixed, the same rotation about some axis OG, which brings A and B into their final positions, will bring any other point of the body into its final position. Hence, generally, every displacement of a rigid body is compounded of, and is equivalent to, (1) some motion of transla- tion whereby every particle has the same translation as any assumed point 0, and (2) some motion of rotation about some axis passing through 0. Composition of Angular Velocities 289 These motions are clearly independent, and can take place in either order or simultaneously. 217. Angular velocities of a body about more than one axis. Indefinitely small rotations. A body has an angular velocity w about an axis when every point of the body can be brought from its position at time t to its position at time t + ht\>y q> rotation round the axis through an angle wht. When a body is said to have three angular velocities Wi, w^, and 61)3 about three perpendicular axes Ox, Oy, and Oz it is meant that during three successive intervals of time ht the body is turned in succession through angles Wiht, w.^Zt and oic^U about these axes. [The angular velocity w^ is taken as positive when its effect is to turn the body in the direction from Oy to Oz ; so Wj and &)3 are positive when their effects are to turn the body from Oz to Ox, and Ox to Oy respectively. This is a convention always adopted.] Provided that Zt is so small that its square may be neglected it can be shev/n that it is immaterial in what order these rotations are performed, and hence that they can be considered to take place simultaneously. Let P be any point {x, y, z) of a body ; draw PM perpen- dicular to Ox and let PM be in- clined at an angle 6 to the plane p» xOy so that y = MP cos d,z = MP sin 6. Let a rotation asiht be made about Ox so that P goes to P' ^ \"f/W whose coordinates are X, y i- Sy, z + Sz. Then y+8y = MP cos {6 + (o,8t) = MP (cos e - sin e . co.Bt) = y- zoM, powers of ht above the first being neglected. So z-\-hz = MP sin {6 -f wM) = MP (^sin ^ + cos ^ . (a^U) = s + yw^Zt. L. D. 19 /' 290 Dynamics of a Rigid Body Hence a rotation Wiht about Ox moves the point {x, y, z) to the point (x, y — zwiht, z + ycoiSt) (1). So a rotation co^St about Oy would move the point (x, y, z) to the point {x-\-zco2ht, y, z — x(ti.M) (2). Also a rotation w^ht about Oz would move the point {x, y, z) to the point {x — yw^ht, y + xcosSt, z) (3). 218. Now perform the three rotations, about the perpen- dicular axes Ox, Oy, Oz, of magnitudes a)iht, w^U, oos^t respectively in succession. By (1) the rotation WiSt takes the point P (x, y, z) to the point Pi, viz. {x, y — zw^ht, z + yooiSt). By (2) the rotation cozBt takes Pj to the point Pg, viz. [a} + (z + yooiSt) w^ht, y — zcoiBt, z + yco^St — xw^St], i.e. [x + zco^St, y — ztOiBt, z + (yooi — xw^) Bt], on neglecting squares of Bt. Finally the rotation (OaBt about Oz takes P^ to the point Pg, viz. [x + zWiBt —{y — zcoiBt) oo^Bt, y — zcoiBt + (x + zw^Bt) WsBt, z + (2/&)i — xw^ Bi\, i.e. Pa is the point [sc + (zQ)2 — yws) Bt, y + (a^wg — zwi) Bt, z -f {ycoi — xw^) Bt], on again neglecting squares of Bt. The symmetry of the final result shews, that, if the squares of Bt be neglected, the rotations about the axes might have been made in any order. Hence when a body has three instantaneous angular velocities the rotations may he treated as taking place in any order and therefore as taking place simultaneously. If the rotations are of finite magnitude, this statement is not correct, as will be seen in Art. 225. Composition of Angular Velocities 291 219. If a body possesses two angular velocities coi and co^, about tivo given lilies which are represented in magnitude by distances OA and OB measured along these two lines, then the resultant angular velocity is about a line 00, where OACB is a parallelogram, and tvill be represented in magnitude by 00. Consider any point P lying on 00 and draw PM and PN perpendicular to OA and OB, The rotations Wiht and w^ht about OA and OB respectively N/ would move P through a small distance perpendicular to the /-^""^'^ ' /" plane of the paper which ' =.-PM.a>^U-vPN .^.U ^\[-PM.OA+PN.OB]k = 2\[-APOA+APOB]Bt = 0. Hence P, and similarly any point on 00, is at rest. Hence 00 must be the resultant axis of rotation ; for we know, by Art. 215, that there is always one definite axis of rotation for any motion. If (0 be the resultant angular velocity about 00, then the motion of any point, A say, will be the same whether we consider it due to the motion about 00, or about OA and OB together. Hence to x perpendicular from A on OC = twa x perpendicular from A on OB. .-. CO X OA sin AOG = i + Wi) about a line whose direction-cosines are ^,^^and-^ 220. A hodij has avgular velocities, coi and (O2, about two parallel axes ; to find the motion. Take the plane of the paper through any point P of the body perpendicular to the two axes, meeting them in 0^ and 0,. Then the velocities of P are ri&>i and r^w.i along PK^, and PK^ perpendicular to O^P and 0-iP respectively. Take N on 0^0., such that Wj . O^N = w^ . NO... The velocities of P are cdj . PO^ and cdj . PO2 perpendicular to POi and PO2 respectively. Hence by the ordinary rule their resultant is (wj + w.) PN perpendicular to PN, i.e. P moves as it would if it had an angular velocity (wj + Wo) about N. Hence two angular velocities Wj and w^ about two parallel axes Oi and O2 are equivalent to an angular velocity w^ + Wg about an axis which divides the distance O^O-z inversely in the ratio of (Wj to tOg. 221. If the angular velocities are unlike and w^ > a^ numeri- cally, then N divides O1O2 externally so that wi . OiN=a>. . OoN, and the resultant angular velocity is (o^ — ay^. Exceptional case. If the angular 'ti 6^ velocities are unlike and numerically equal, N is at infinity and the resultant angular velocity is P Composition of Angular' Velocities 293 The resultant motion is then a linear velocity. For, in this case, the velocities of P are perpendicular and proportional to OiP and PO2, and hence its resultant velocity is perpendicular and proportional to OiOn, i.e. it is N^-'-'^ ' ''^ Y^wa ft)i . O1O2 \ . o1^ Aliter. The velocity of P parallel to 0,0, = &)i . 0,P sin PO.Oo. - ft)i . OoP. sin PO^O, = 0, and its velocity perpendicular to OiO, = 0), . OiP cos PO1O2 + <0i . OoP. cos PO2O1 = ft)i . 0,0., j . 222. An angular velocity eo about an axis is equivalent to an angular velocity ca about a parallel axis distant a from the former together with a linear velocity to . a. Let the two axes meet the plane of the paper in Oi and O2 and be perpendicular to it. The velocity of any point P in the plane , ^^-^"^^ of the paper due to a rotation &> about 0, ^A"^'^''^ \ = a).OiP perpendicular to OiP, °^« * °' and this, by the triangle of velocities, is equivalent to velocities w . 0,0, and w . O2P perpendicular to 0,0, and 0,P in the same sense = (£) .a\ together with a velocity w . O2P perpendicular to O2P. Hence the velocity of any point P, given by an angular velocity « about 0,, is equivalent to that given by an equal angular velocity (o about 0,, together with a linear velocity o) . 0,0, perpendicular to 0,0,. 223. In practice the results of Arts. 220—222 are re- membered most easily by taking the point P on 0,0, \ Thus (1) the velocity of P = Wi . 0,P + (o, . 0,P = (O, (0,0, + 0,P) + CO, . O2P = {(o, + a>,) (0,P + -^^— . 0,0^ \ 0)1+0)2 / = (CO, + 0)2) . NP, where NO, = ""^ . 0,0^. 294 Dynamics of a Rigid Body (2) The velocity of P = ewj . O^P - w^ . O^P h(0 0, + 0,P) - CO, . O.P = (o), - CO,) \o,P + — ^^ . 0,0^ (&)i - f»2) . iV^P, where O^N-- CO, 0,0,. (3) The velocity of P = co . 0,P - co . 0,P Oj P = 0) . 0,0,= a constant velocity perpendicular to O1O2. (4) The velocity oi P = co . 0,P = a> . 0,0, + co . O^P, and is therefore equivalent to a Oi/(^ O2 P linear velocity a> . 0,0^ perpendicular to OiO, together with an angular velocity co about 0,. 224. To shew that the instantaneous motion of a body may be reduced to a twist, i.e. to a linear velocity along a certain line together with an angular velocity about the line. By Art. 216 the instantaneous motion of a rigid body is equivalent to a translational velocity of any point together with an angular velocity about a straight line passing through 0. vsin^ O' vsin Let OA be the direction of this linear velocity v, and Oz the axis of the angular velocity co. Instantaneous Motioti of a Body 295 In the plane zOA draw Ox at right angles to Oz and draw Oy at right angles to the plane zOx. Let Z zOA = 0. On Oy take 00' such that 00' . (o = v sin 6. Then, by Art. 222, the angular velocity m about Oz is equivalent to (o about a parallel axis OV to 0^ together with a linear velocity cd . 00', i.e. wsin^, through perpendicular to the plane zOO'. Also a linear velocity -y may be transferred to a parallel linear velocity through 0', and then resolved into two velocities V cos and v sin 6. We thus obtain the second figure. In it the two linear velocities v sin destroy one another, and we have left the motion consisting of a linear velocity vcos6 along OV and an angular velocity co about it. This construction is clearly similar to that for Poinsot's Central Axis in Statics; and properties similar to those for Poinsot's Central Axis follow. It will be noticed that, in the preceding constructions, an angular velocity corresponds to a force in Statics, and a linear velocity corresponds to a couple. 225. rinite Rotations. If the rotations are through finite angles it is easily seen that the order of the rotations about the axes is important. As a simple case suppose the body to be rotated through a right angle about each of two perpendicular axes Ox and Oy. The rotation through a right angle about Ox would bring any point P on Oz to a position on the negative axis of y, and a second rotation about Oy would not further alter its position. A rotation, first about the axis of y, would have brought P to a position on the axis of x, and then a second rotation about Ox would not have had any effect on its position. Thus in the case of finite rotations their order is clearly material. 226- 2'o Jind the effect of two finite rotations about axes OA and OB in succession. Let the rotations be through angles 2a and 2^ about OA and OB in the directions marked. On the geometrical sphere with as centre draw the arcs AC q and BC, such that ,-"-, /.BAG=a and /.ABC=^, the directions AG and BC being taken, one in the same direction as the rotation about OA, and the second in the opposite direction to the rotation about OB. Take C on the other side of AB, sym- metrical with C, so that iCAC' = 2o. and ACBC' = 2B. 296 Dynamics of a Rigid Body A rotation of the body through an angle 2a about OA would bring OC into the position OC, and a second rotation 2^3 about OB would bring OG back again into the position DC. Hence the effect of the two component rotations would be that the position of OC is unaltered, i.e. OC is the resultant axis of rotation. [If the rotations had been first about OB and secondly about OA it is clear, similarly, that OG would have been the resultant axis of rotation.] Magnitude of the resultant rotation. The point A is unaltered by a rotation about OA ; the rotation 2/3 about OB takes it to the point P, where lABP = 1^ and the arc BP=the arc BA, and therefore aBAP:^ iBPA. Hence the resultant rotation is through an angle AC'P{=x) about G', and CA = G'P. If BC meets AP in N, then N is the middle point of the arc AP and ACN=NCP=^ . If the axes OA and OB meet at an angle 7, then AB = y. Let AC be p. Then sin 7 sin j3= sin AN= sin p sin .(1). Also, from the triangle ABG', we have cos 7 cos a = sin7Cotp- which gives sin 7 ^ ^Ji + cot^p ^sin2 7+(cosacos7 + sinacotj3)2' Hence (1) gives . X I sin - = sin/3;^sin2 7+(cos 7 cos a + siuacotj8)2. Hence the position of the resultant axis OC, and the magnitude of the resultant rotation, are given for any case. Ex. 1. If a plane figure be rotated through 90° about a fixed point A, and then through 90° (in the same sense) about a fixed point B, the result is equivalent to a rotation of 180° about a certain fixed point G ; find the position ofC. Ex. 2. Find the resultant rotation when a body revolves through a right angle in succession about two axes which are inclined to one another at an angle of 60°. Ex. 3. When the rotations are each through two right angles, shew that the resultant axis of rotation is perpendicular to the plane through the two component axes, and that the resultant angle of rotation is equal to twice the angle between them. Motion in three dimensio7is 297 227. Velocity of any point of a body parallel to fixed axes in terms of the instantaneous angular velocities of the body about the axes. Let P be any point (w, y, z) of the body. Draw PM per- pendicular to the plane of xy, MN perpendicular to the axis of x, and PT perpendicular to NP in the plane NPM to meet NM in T. The angular velocity o)i about Ox gives to P a velocity along TP equal to twi . PN which is equivalent to a velocity — «Di . PN cos PTN, i.e. — coi . PN. sin PNT, i.e. -co^.z along NT, and a velocity ro,. PN sin PTN, i.e. Wi . PN cos PNM, i.e. toi . y along MP. Hence the Wi-rotation about Ox gives a component velocity — ©i.^ parallel to Oy and (o^.y parallel to Oz. So the rotation about Oy by symmetry gives component velocities — w^. x parallel to Oz and cog . z parallel to Ox. Finally the rotation about Oz gives — w-i.y parallel to Ox and (Wg . X parallel to Oy. Summing up, the component velocities are w^.z — w^.y parallel to Ox, Q)i.x — ci)i.z „ „ Oy, and (Oi.y — C0.2.X „ „ Oz. If be at rest, these are the component velocities of P parallel to the axes. If be in motion and u, v, w are the components of its velocity parallel to the fixed axes of coordinates, then the component velocities of P in space are u+ (ji^.z — oi^.y parallel to Ox, and V-{- W3.X — Wi.Z w + (Oi . y — (Oj . X Oy, Oz. 298 Dynamics of a Rigid Body 228. A rigid body is moving about a fixed point ; to find (1) the moments of momentum about any axes through fixed m space, and (2) the kinetic energy of the body. The moment of momentum of the body about the axis of x But, by the previous article, since is fixed, dy J dz -^ = (Oz.x— cox-z and -j- = w^.y - Wy.x where Wx, (Oy and cog are the angular velocities of the body about the axes. On substitution, the moment of momentum about Ox = Xm [(y^ + z^) cox — soywy — zxco^] = A .Wx — F.Wy-E.tOz' Similarly the moment of momentum about Oy = Bo3y — Dcoz — FtOx, and that about Oz = C0)z — Eq)x — DWy. (2) The kinetic energy = ^ Sm [{(Oy .z-a)z.yy-\-{(i)z.x-o3x. zf + (w^ . i/ - Wy . xY\ = iSmKH3/' + ^') + --+-"-2«2/«2. 2/^-. ••-•••] = \ {Aw^ + iiw/ + Gai - 2DcOyO)z - 2Ea)zCOx - 2FcoxO)y). ■ 229. In the previous article the axes are fixed in space, and therefore since the body moves with respect to them, the moments and products of inertia A, B, G ... are in general variable. Other formulae, more suitable for many cases, may be obtained as follows. Let Ox', Oy' and Oz' be three axes fixed in the body, and therefore not in general fixed in space, passing through and let G>i, 0)2, &)3 be the angular velocities of the body about them. The fixed axes Ox, Oy and Oz are any whatever, but let them be so chosen that at the instant under consideration the moving axes Ox', Oy' and Oz' coincide with them. Then <»a;=Wi, ft)y=CD2, &)2=Ci>3. The expressions for the moments of momentum of the last Motion in three dimeoisions 299 article are now Acoi— F(o.2— Ecos and two similar expressions, and the kinetic energy is ^ (A(o^^ + Bwi + Co)^^ - 2Dco.Ms - 2£'ft)3ft)i - 2F&)i&),), where A, B, G are now the moments of inertia and D, E, F the products of inertia about axes fixed in the body and moving with it. If these latter axes are the principal axes at 0, then D, E and F vanish and the expressions for the component moments of momentum are Aw^, Bco^ and Ccos, and that for the kinetic energy is ^ (Aq),' + Bco.,- + Cq)s% 230. General equations of motion of a body tvith one point fixed which is acted ^ipon by given bloius. The fixed point being the origin, let the axes be three rectangular axes through it. Let Wa, (Oy, (Oz be the angular velocities of the body about the axes just before the action of the blows, and oox, (Oy, &)/ the corresponding quantities just after. The moment of momentum of the body, just before, about the axis of x, is Aax — Fwy — Ewg and just after it is Acoas' — Fwy — Eq)z. Hence the change in the moment of momentum about the axis of a; is il ((Ox — ^x) — E((Oy — (Oy) — E (&>/ — <»z). But, by Art. 166, the change in the moment of momentum about any axis is equal to the moments of the blows about that axis. If then L, M, N are the moments of the blows about the axes of X, y and z, we have A {(Ox — (Ox) - F{(Oy' — (Oy) - E {(Oz — ft)z) = L, and similarly B {(Oy — (Oy) — D {(Oz —(Oz) — F {(Ox — cox) = M, and G {(Oz -(Oz) — E {(oj — (Ox)-D {(Oy — (Oy) = N. These three equations determine (Ox, (Oy and ; - (Oz) = N. 300 Dynamics of a Rigid Body 231. If the body start from rest, so that w^, (Oy and coj are zero, we have , L , M ^ , N «a; = J , «y = -g and wg =-^. Hence the direction cosines of the instantaneous axis are L M N' A' B^C) (1>- The direction cosines of the axis of the impulsive couple are {L, M, N) (2). In general it is therefore clear that (1) and (2) are not the same, i.e. in general the body does not start to rotate about a perpendicular to the plane of the impulsive couple. (1) and (2) coincide if A=B = G, in which case the momental ellipsoid at the fixed point becomes a sphere. Again if M=N = 0, i.e. if the axis of the impulsive couple coincides with the axis of w, one of the principal axes at the fixed point, then the direction cosines (1) become proportional to (1, 0, 0), and the instantaneous axis also coincides with the axis of ic, i.e. with the direction of the impulsive couple. Similarly if the axis of the impulsive couple coincides with either of the other two principal axes at the fixed point. In the general case the instantaneous axis may be found geometrically. For the plane of the impulsive couple is Lx + My +Nz = 0. Its conjugate diameter with respect to the momental ellipsoid Ax^ + By' + Gz^ = k is easily seen to be -r- = ~ = ^, A B G i.e. it is the instantaneous axis. Hence if an impulsive couple act on a body, fixed at a point and initially at rest, the body begins to turn about the diameter of the momental ellipsoid at ivhich is conjugate to the plane of the impulsive couple. Motion in three dimensions 301 232- -Ea?. 1. A lamina in the form of a quadrant of a circle OHO', whose centre is H, has one extremity of its arc fixed and is struck by a blow P at the other extremity 0' perpendicular to its plane ; find the resulting motion. Take OH as the axis of x, the tangent at as the axis of y, and a perpendicular to the plane at as the axis of z. Let G be the centre of gravity, GL perpendicular to OH, so that HL=LG = ^^. Then A=m'^; 4 B (by Art. U1) = M^- M . HL^ + 3I . OU = M (| - ^-^ a^; C = A+B; D = E = 0; : I i '^ rdedr (a - r C03 6) r Bin e = J\I.~a^. J J bir The equations of Art. 230 then give A ^^^ *^6 solution can be completed. If ^ be the inclination to Ox of the instantaneous axis, we have wj,' A-F IO-Stt Ex. 2. A uniform cube has its centre fixed and is free to turn about it ; it is struck by a blow along one of its edges ; find the instantaneous axis. Ex. 3. A uniform solid ellipsoid is fixed at its centre and is free to turn about it. It is struck at a given point of its surface by a blow whose direction is normal to the ellipsoid. Find the equation to its instantaneous axis. Ex. 4. A disc, in the form of a portion of a parabola bounded by its latus rectum and its axis, has its vertex A fixed, and is struck by a blow through the end of its latus rectum perpendicular to its plane. Shew that the disc starts revolving about a line through A inclined at tan"i^f to the axis. Ex. 5. A uniform triangular lamina ABG is free to turn in any way about A which is fixed. A blow is given to it at B perpendicular to its plane. Shew that the lamina begins to turn about AD, wliere Z) is a point on BG such that GD = lCB. 233. General equations of motion of a body in three dimen- sions, referred to axes whose directions ar^e fixed. If (x, y. z) be the coordinates of the centre of gravity of the d'^x body we have, by Art. 162, M-^ = sum of the components of the impressed forces parallel to Ox, and similar equations for the motion parallel to the other axes. 302 Dynavfiics of a Rigid Body If Q)x, coy, coz be the angular velocities at any instant about axes through the centre of inertia parallel to the axes of coordinates then, by Arts. 164 and 228, we have d dt [A (Ox — Fwy - Eoi^ = moment about a line parallel to Ox through O of the effective forces = L. d So dt [Bo)y - Dcoz - i^w J = ill and ^[Ca,,-^a,, DcOy] = N. [If the body be a uniform sphere, of mass M-^, then B = E = F 2tt2 = 0, and ^ = ^ = C 2a2 dcox -r M,. dt M . -^; these equations then become N.I 2a^^^ and if,.^^^" o dt o dt Impulsive forces. If u, v, w, m^, Wy, Wz be the component velocities of the centre of inertia and the component angular velocities about lines through G parallel to the fixed axes of coordinates just before the action of the impulsive forces, and u', v', w', Wx, (Oy, ft)/ similar quantities just after them, by Arts. 166 and 228 the dynamical equations are M {u' — u) = X^, etc. and A {w^ — a>x) — F {coy — coy) — E (ft)/ — oog) = the moment of the impulsive forces about Ox, and two similar equations-. 234. Em. A homogeneous billiard hall, spinning about any axis, moves on a billiard table which is not rough enough to always prevent sliding ; to shew that the path of the centre is at first an arc of a parabola and then a straight line. Take the origin as the ini- tial position of the point of contact, and the axis of x in the initial direction of its sliding. If u and V be the initial velocities of the centre parallel to the axes, General Motion of a Billiard Ball 303 angular velocities about the axes, then since the initial velocity of the point of contact parallel to the axis of y is zero, we have v + aD.^ = (1). At any time t let w-b, coy and w^ be the component angular velocities, and F^^, Fy the component frictions as marked. The equations of motion are Mx = -F^ \ My = -Fy (2), = R-Mgj and ,^ 2a- dwx r, M:^'^ = F^.a i (3). o at 5 at The resultant friction must be opposite to the instantaneous motion of A and equal to fiMg. Fy y + aoa; Hence -w = • (4), Fx x-acoy ^ '' and F^^ -¥ Fy^ = ix^Y (5). Equations (2) and (3) give Fy ^y ^ - (ba: ^ y + arji^ Fx ic (by id — awy ' Hence (4) gives y + ad>a; _x — ad}y if + aa>x X — aoDy ' .'. log {y + aoix) = log {x — acoy) + const. X — acoy u — ally •' Hence (4) and (5) give Fy=0 and F^ = /x3Ig. From (2) it follows that the centre moves under the action of a constant force parallel to the axis of x and hence it describes an arc of a parabola, whose axis lies along the negative direction of the axis of x. t = ^^—^ (u — aQy), 304 Di/ncwiics of a Rigid Body (1), (2), and (3) now give x = -ixgt^u\ 2/ = const. = w j and awx = const. = aD.^] 5 (7). ctwy =^figt + any At time t the velocity of the point of contact parallel to Ox 7 ~x — acoy = u — aVty — - figt, and parallel to Oy it = ^ + acoa; = v + aOa; = 0, by (1). The velocity of the point of contact vanishes and pure rolling begins when _2_ , , y V 7v and then -. = = = — —=,-?.- , X u — figt 0U+ zaily i.e. the direction of motion when pure rolling commences is inclined at tan ^ - — , _ „ to the original direction of motion ou + 2any ^ of the point of contact. On integrating (6), it is easily seen that pure rolling commences at the point whose coordinates are 2(u- aVty ) {Qu + gOy ) 2v (u - any) 4>9fig ^^ 7^ * It is easily seen that the motion continues to be one of pure rolling, and the motion of the centre is now in a straight line. EXAMPLES 1. If a homogeneous sphere roll on a fixed rough plane undei* the action of any forces, whose resultant passes through the centre of the sphere, shew that the motion is the same as if the plane were smooth and the forces reduced to five-sevenths of their given value. 2. A sphere is projected obliquely up a perfectly rough plane ; shew that the equation of the path of the point of contact of the sphere and plane is y=^tan/3 — :r- ~, — -^, where a is the inclination of the plane f ^ "^ 14 F'' cos^jS '■ to the horizon, and V is the initial velocity at an angle j3 to the horizontal. Motion in three dimensions. Examples 305 3. A homogeneous sphere is projected, so as to roll, in any direction along the surface of a rough plane inclined at a to the horizontal ; shew that the coefficient of friction must be > f tan a. 4i. A perfectly rough sphere, of mass 3/ and radius a, is rotating with angular velocity Q about an axis at right angles to the direction of motion of its centre. It impinges directly on another rough sphere of mass m which is at rest. Shew that after separation the component velocities of the two spheres at right angles to the original direction of motion of 2 m 2 J/ the first sphere are respectively = -j-. aQ. and = t? «Q. 5. A homogeneous sphere spinning about its vertical axis moves on a smooth horizontal table and impinges directly on a perfectly rough vertical cushion. Shew that the kinetic energy of the sphere is diminished by the impact in the ratio 26^ (5 + 7 tan^ 5) ; 10+49e2tan2^, where e is the coefficient of restitution of the ball and 6 is the angle of reflection. 6. A sphere, of radius a, rotating with angular velocity a about an axis inclined at an angle (3 to the vertical, and moving in the vertical plane containing that axis with velocity \t in a direction making an angle a with the horizon, strikes a perfectly rough horizontal j)lane. Find the resulting motion, and shew that the vertical plane containing the new direction of motion makes an angle tan~i with the original plane. 7. A ball, moving horizontally with velocity « and spinning about a vertical axis with angular velocity ^«>' Suppose the axis of x to be such that the sum of the resolved parts of the external forces parallel to it is zero throughout the motion, i.e. such that XmX = always. Equation (1) then gives d .^ dx r\ i.e. Sw^=constant (7), or M -^ = constant, dt where x is the a;-coordinate of the centre of gravity. Equation (7) states that in this case the total momentum of Conservation of Momentum 307 the body measured parallel to the axis of x remains constant throughout the motion. This is the Principle of the Conservation of Linear Momentum. Again suppose the external forces to be such that the sum of their moments about the axis of x is zero, i.e. such that Then, by equation (4), we have and .•. 2m I y -77 —0-7^1 = constant (8). Now y -, — ^ -r = the moment about the axis of x of the ^ dt dt velocity of the mass m, and hence equation (8) states that the total moment of momentum of the system about the axis of X is constant. Hence the Principle of the Conservation of the Moment of Momentum (or Angular Momentum), viz. If the sum of the moments of the external forces, acting on a rigid body, about a given line he zero throughout the motion, the moment of momentum of the body about that line remains un- altered throughout the motion. 236. The same theorems are true in the case of impulsive forces. For if the duration of the impulse be a small time t we have, as in Art. 166, on integrating equation (1), where Xi is the impulse of the forces parallel to the axis of x, i.e. the change in the total momentum parallel to the axis of x is equal to the sum of the impulses of the forces in that direction. If then the axis of x be such that the sum of the impulses parallel to it vanish, there is no change in the total momentum parallel to it, i.e. the total momentum parallel to the axis of x before the action of the impulsive forces = the total momentum in that direction after their action. 20— 2 308 Dynamics of a Rigid Body Again, integrating equation (4), we have i.e. the change in the angular momentum about the axis of x is equal to the sum of the moments of the impulses of the forces about that same direction. If then the axis of x be such that the sum of the moments of the impulsive forces about it vanishes, there is no change in the angular momentum about it, i.e. the angular momentum about it just before the action of the impulsive forces = the angular momentum about the same line just after their action. 237. -E.r. 1. A head, of mass vi, slides on a circular wire, of mass M and radius a, and the icire turns freely about a vertical diameter. If w and w' be the angular velocities of the wire when the head is respectively at the ends of 0}' M+2in a horizontal and vertical diameter, shew that — = — =-;— . (o M The moment of inertia of the wire about any diameter = it/— . Wherever the bead may be on the wire, the action of it on the wire is equal and opposite to that of the wire on it. Hence the only external forces acting on the system are (1) the action of the vertical axis AA', which has no moment about ^^', and (2) the weights of the bead and wire, neither of which has any moment with respect to the vertical axis AA'. Hence the moment of momentum of the system (wire and bead) about AA' is constant throughout the motion. Also the velocity of the bead along the wire has no moment about AA', since its direction intersects AA . When the bead is at ^, the moment of momentum about ^^' is ill— . w' ; at B, this moment is il/— w + ma^w. Equating these two, we also, when it M+2m have 31 Ex. 2. A rod, of length 2a, is moving on a smooth table with a velocity v perpendicular to its length and impinges on a small inelastic obstacle at a distance c from its centre. When the end leaves the obstacle, shew that the angular velocity of the rod is —^ . Both at the impact, and throughout the subsequent motion whilst the rod is in contact with the obstacle, the only action on the rod is at the obstacle itself. Hence there is no change in the moment of momentum about the obstacle. But before the impact this moment was Mcv. Also, if w be the Conservation of Momentum 309 angular velocity of the rod when its end is leaving the obstacle, its moment of momentum about the obstacle is, by Ai't. 191, M (--+a^\ u, i.e. il/ . -— w. Equating these two, we have '^=j-^- If w' were the angular velocity of the rod immediately after the impact, we have, similarly, Mcv=:M { — + c^\ u'. Ex. 3. A uniform circular plate is turning in its own plane about a point A on its circumference with unifoi'm angular velocity w ; suddenly A is released and another point B of the circumference is fixed ; shew that the angular velocity about B is —(1 + 2 cos a), where a is the angle that AB subtends at the centre. In this case the only impulsive force acting on the plate is at B and its moment about JB vanishes. Hence, by Art. 235, the moment of momentum about B is the same after the fixing as before. If w' be the required angular velocity, the moment of momentum after the fixing = lf (a2 + Z;2) w' = iU. — - &,'. The moment of momentum before the fixing = the moment of momentum of a mass M moving with the centre of gravity + the moment of momentum about the centre of gravity (Art. 191) = Mau . a cos a + Mk^u = iloja^ (cos a + 2) > since before the fixing the centre was moving at right angles to ^0 with velocity aw. „ ,r3n2 , f 1\ , 1 + 2 cos a Hence 3/-^ w' = il/wa2 ( cos tt + - j . .-. u=w . It is clear that w' is always less than w, so that the energy, ^m {k^ + a^) w'2, after the impact is always less than it was before. This is a simple case of the general principle that kinetic energy is always diminished whenever an impact, or anything in the nature of a jerk, takes place. If a = 120°, i.e. if the arc AB is one-third of the circumference, the disc is brought to rest. Ex. 4. A uniform square lamina, of mass M and side 2a, is moving freely about a diagonal with uniform angular velocity w when one of the corners not in and that T v/2 the impulse of the force on the fixed point is ~- . Mau. Let AC be the original axis of rotation. As in Art. 149, the moment of inertia about it is a2 il/ . — . Let the initial direction of rotation be such that B was moving upwards from the paper. Let D be the point that becomes fixed and w' the resulting angular velocity about DX, a hne parallel to AC. Since the impulsive force at the fixing acts at D its moment about DX vanishes. Hence the moment of momentum about DA' is unaltered by the fixing. 310 Dynamics of a Rigid Body After the fixing it = M7cV = M r^ + DO^I w' = M (^ +2a2) a,' = J/ . Z^ w'. Also before the fixing it, by Art. 191, = moment of momentum about ^C + the moment of momentum of a particle M at and moving with it = M . — w. Equating these two quantities, we have w' = -^ . Similarly, the moment of momentum about DB after the fixing = the moment of momentum before = zero. Hence after the fixing the square is moving about DA' with angular velocity - . Again, before the fixing the centre of gravity was at rest, and after the fixing it is moving with velocity DO . u', i.e. sj2a. ^, about D. The change in its momentum is therefore itf ^ "" , and this, by Art. 166, is equal to the impulse of the force required. EXAMPLES 1. If the Earth, supposed to be a unifonn sphere, had in a certain period contracted slightly so that its radius was less by - th than before, shew that the length of the day would have shortened by — hours. 2. A heavy circular disc is revolving in a horizontal plane about its centre which is fixed. An insect, of mass -th that of the disc, walks n from the centre along a radius and then flies away. Shew that the final angular velocity is --— ^ times the original angular velocity of the disc. 3. A uniform circular board, of mass M and radius a, is placed on a perfectly smooth horizontal plane and is free to rotate about a vertical axis through its centre ; a man, of mass M', walks round the edge of the board whose upper surface is rough enough to prevent his slipping ; when he has walked completely round the board to his starting point, shew that the board has turned through an angle „ „ , . 4 n. 4. A circular ring, of mass M and radius a, lies on a smooth horizontal plane, and an insect, of mass m, resting on it starts and walks round it with uniform velocity v relative to the ring. Shew that the centre of the ring describes a circle with angular velocity m V M+2m a' Conservation of Mo77ientum. Examples 311 5. If a merry-go-round be set in motion and left to itself, shew that in order that a man may (1) move with the greatest velocity, (2) be most likely to slip, he must place himself at a distance from the centre equal to (1) k^n, (2) k /^, k b&ing the radius of gyration of the machine about its axis and n the ratio of its weight to that of the man. 6. A uniform circular wire, of radius a, lies on a smooth horizontal table and is movable about a fixed point on its circumference. An insect, of mass equal to that of the wire, starts from the other end of the diameter through and crawls along the wire with a uniform velocity v relative to the wire. Shew that at the end of time t the wire has turned through an angle -p: tan-^ —p- tan — . [When the diameter OA has turned through an angle (\) from its vt initial position, let the insect be at P so that i AGP=6 = — , where C is the centre of the wire. Since the moment of momentum about is constant, r 6 ■ 6~\ .'. m (Ic^ -h a^) (/) + m 4a.- cos^ - -|- y . 2a cos^ - = constant = 0.] 7. A small insect moves along a uniform bar, of mass equal to itself and of length 2a, the ends of which are constrained to remain on the 2a circumference of a fixed circle, whose radius is -r-. If the insect start from the middle point of the bar and move along the bar with relative velocity V, shew that the bar in time t will turn through an angle -jr tan-1 — . 8. A circular disc is moAing with an angular velocity i2 about an axis through its centre perpendicular to its plane. An insect alights on its edge and crawls along a curve drawn on the disc in the form of a lemniscate with uniform relative angular velocity -gQ, the curve touching the edge of the disc. The mass of the insect being xgth of that of the disc, shew that the angle turned through by the disc when the insect gets to the centre is -j- tan ~ ^ — - - . V7 6 4 9. A rod OA can turn freely in a horizontal plane about tlie end and lies at rest. An insect, whose mass is one-third that of the rod, alights on the end A and commences crawling along the rod with uniform velocity V; at the same instant the rod is set in rotation about in such a way that the initial velocity of .4 is V ; when the insect reaches prove that the rod has rotated through a right angle, and that the angular velocity of the rod is then twice the initial angular velocity. 312 Dynamics of a Rigid Body 10. A particle, of mass m, moves within a rough circular tube, of mass M, lying on a smooth horizontal plane and initially the tube is at rest while the particle has an angular velocity round the tube. Shew that by the time the relative motion ceases the fraction ^r? — r— of the •' M+ 2iii initial kinetic energy has been dissipated by friction. [The linear momentum of the common centre of gravity, and the moment of momentum about it, are both constant throughout the motion.] 11. A rod, of length 2a, is moving about one end with uniform angular velocity upon a smooth horizontal plane. Suddenly this end is loosed and a point, distant b from this end, is fixed ; find the motion, considering the cases when &< = > — . 12. A circular plate rotates about an axis through its centre perpen- dicular to its plane with angular velocity a. This axis is set free and a point in the circumference of the plate fixed ; shew that the resulting angular velocity is - . 13. Three equal particles are attached to the corners of an equilateral triangular area ABC, whose mass is negligible, and the system is rotating in its own plane about A. A is released and the middle point of AB is suddenly fixed. Shew that the angular velocity is unaltered. 14. A uniform square plate ABCD, of mass M and side 2a, lies on a smooth horizontal plane ; it is struck at .4 by a particle of mass M' moving with velocity V in the direction AB, the particle remaining attached to the plate. Determine the subsequent motion of the system, and shew that its angular velocity is ,, ,,, . ^ . 15. A lamina in the form of an ellipse is rotating in its own plane about one of its foci with angular velocity w. This focus is set free and the other focus at the same instant is fixed ; shew that the ellipse now rotates 2 — be^ about it with angular velocity w . -g . 16. An elliptic area, of eccentricity e, is rotating with angular velocity w about one latus-rectum ; suddenly this latus-rectum is loosed and the other fixed. Shew that the new angular velocity is l-4e2 17. A uniform circular disc is spinning with angular velocity co about a diameter when a point P on its rim is suddenly fixed. If the radius vector to P make an angle a with this diameter, show that the angular velocities after the fixing about the tangent and normal at P are \ w sin a and o) cos a. Conservatio7i of Energy 313 18. A cube is rotating with angular velocity about a vertical axis through its centre; find its angular velocity when it has turned through any angle, and shew that it will rise through a distance -rr— . Prove also that the time of a small oscillation about the position of equilibrium T Let AB be the initial position of the rod with the strings CA and DB vertical, A'B' its position when it has risen through a vertical distance x and turned through an angle 6. Let the horizontal plane through A'B' cut GA and BD in K and L, and let j.A'GA=. The equation of energy then gives ^mifi + \mm^=\mk''-ur' ~ mgx...{l). Now, since the angle A'KC is a right angle, .-. x = AC-CK^l-lcosct> ...(2), where I is the length of a vertical string. Also lsm(j> = A'K=2a&m-^ .'. i=isin(/>^ = tan^ Hence equation (1) gives .(3). maH^ \^- 4a2 sin2 ^^ 1 a2 2"'- 3 4fl2 sin2 - .(4). This equation gives the angular velocity in any position. The rod comes to a2&j2 instantaneous rest when ^=0, i.e. when x = % For a small oscillation we have, on taking moments about 0', if T be the tension of either string, '= - 2rsin 4> X perpendicular from 0' on A'K = -T sin.(p .2a cos - = 2a2 Tsin Also 2T Gos (p-mg = mx = m d2 fi dt2 G*')' when is small. ma2 d2 (,,)^^^^.,^2^2^.^^, 21 dt^ ' ' 21 i.e. to the first order of small quantities, 2T-mg = 0. Therefore (5) gives m — i Hence the required time = 27r W ma- /T .(5). (6), 320 Dynamics of a Rigid Body Ex. 3. A uniform rod, of length 2a, is placed with one end in contact with a smooth horizontal table and is then alloiced to fall; if a be its initial inclination to the vertical, shew that its angular velocity when it is inclined at an angle 6 is (617 cos a - cos ^j h [a ' l + 3sin26' | * Find also the reaction of the table. There is no horizontal force acting on the rod ; hence its centre of inertia G has no horizontal velocity during the motion since it had none initially. Hence G describes a vertical straight line GO. When inclined at 6 to the vertical its kinetic energy 2 1,, a2. Equating this to the work done, viz. Mga (cos o - cos 6), we get ;„ 60 cosa- cos5 Differentiating, we have 6 = a l + 3sin2^ 3.9 sin ^ 4-6 cos a cos ^ + 3 cos2 d .(1). a (l + 3sin2^)2 Also, for the vertical motion of G, we have E-Mg = M -^ {a cos e) = M[-a sin 66 - a cos es"^]. On substitution, we have U = j 4 - 6 cos ^ cos o + 8 coB^ (l + 3sin2^2 • EXAMPLES 1. A uniform rod, of given length and mass, is hinged at one end to a fixed point and has a string fastened to its other end which, after passing over a light pulley in the same horizontal line with the fixed point, is attached to a particle of given weight. The rod is initially horizontal and is allowed to fall ; find how far the weight goes up. 2. A light elastic string of natural length 2a has one end, A, fixed and the other, B, attached to one end of a uniform rod BC of length 2a and mass m. This can turn freely in a vertical plane about its other end C, which is fixed at a distance 2a vertically below A. Initially the rod is vertical, and, on being slightly displaced, falls until it is horizontal, and then rises again. Shew that the modulus of elasticity is j»^(.3 + 2v'2). Conservatio7i of Energy. Examples 321 3. A uuiform rod moves in a vertical plane, its ends being in contact with the interior of a fixed smooth sphere ; when it is inclined at an angle 6 to the horizon, shew that the square of its angular velocity is ^2'Vr^i (^°^ ^ ~ ^^^ ")> "where a is the initial value of 6, 2a is the length of the rod, and c is the distance of its middle point from the centre of the sphere. 4. A hemisphere, of mass M and radius a, is placed with its plane base on a smooth table, and a heavy rod, of mass m, is constrained to move in a vertical line with one end P on the curved surface of the hemisphere ; if at any time t the radius to P makes an angle d with the vertical, shew that ad^ [M cos^ d + rn sin^ 6] = 2mg (cos a - cos 6). 5. A uTiiform rod, of length 2a, is held with one end on a smooth horizontal plane, this end being attached by a light inextensible string to a point in the plane ; the string is tight and in the same vertical plane as the rod and makes with it an acute angle a. If the rod be now allowed to fall under the action of gravity, find its inclination to the horizon when the string ceases to be tight, and shew that its angular velocity Q, just before it becomes horizontal is given by the equation 6aQ2=5' sin a (8 +cos2 a). 6. A uniform straight rod, of length 2a, has two small rings at its ends which can respectively slide on thin smooth horizontal and vertical wires Ox and Oy. The rod starts at an angle a to the horizon with an angular velocity */^(l-sina), and moves downwards. Shew that it will strike the horizontal wire at the end of time \/3^^^g{*^Kl cot - k 7. A straight uniform rod, of mass m, is placed at right angles to a smooth plane of inclination a with one end in contact with it ; the rod is then released. Shew that, when its inclination to the plane is 0, the . ,, , .„ , 3(1 -sin 0)2+1 reaction of the plane will be mg -^ , _, ,so cos a. (3 008^0+ 1)2 8. A hoop, of mass M, carrying a particle of mass vi fixed to a point of its circumference, rolls down a rough inclined plane ; find the motion. 9. Two like rods AB and BC, each of length 2a, are freely jointed at B; AB can turn round the end A and C can move freely on a vertical straight line through A. Initially the rods are held in a horizontal line, C being in coincidence with A, and they are then released. Shew that when the rods are inclined at an angle d to the horizontal, the angular velocity of either is x/ 3g sin 6 a" ■ l+3cos2^* 322 Dynamics of a Rigid Body 10. A sphere, of radius 6, rolls without slipping down the cycloid ^=a(^ + sin^), y=a(l-cos^). It starts from rest with its centre on the horizontal line y = 2a. Shew that the velocity Fof its centre when at its lowest point is given by P=J7V(2a-i). 11. A string, of length 2Z, is attached to two points in the same horizontal plane at a distance 26 and carries a particle m at its middle point ; a uniform rod, of length 2a and mass i/", has at each end a ring through which the string passes and is let fall from a symmetrical position in the straight line joining the ends of the string ; shew that the rod will not reach the particle if {l-Vh- 2a) . (i/+2TO) M< 2 (2a- 6) m^. If M=m and 6 = a, and the particle be given a small vertical dis- placement when it is in a position of equilibrium, shew that the time , n -11 .• • 27r /273a 01 a small oscillation is — w . 12. Two equal perfectly rough spheres are placed in unstable equi- librium one on the top of the other, the lower sphere resting on a smooth table. If the equilibrium be disturbed, shew that the spheres will continue to touch at the same point, and that when the line joining their centres is inclined at an angle 6 to the vertical its angular velocity a> is given by the equation a^u)^ (5 sin- ^ + 7) = lO^a (1 — cos 6), where a is the jadius of each sphere. 13. An inextensible uniform band, of small thickness r, is wound round a thin fixed axis so as to form a coil of radius h. The coil is unrolled until a length a hangs freely and then begins to unroll freely under the action of gravity, starting from rest. Shew that, if the small horizontal motion be neglected, the time which will elapse before the hanging part is of length x is approximately 6 ^ /Z flog ^_±^^I^ + _I_ V.r^^^l . 14. A roll of cloth, of small thickness e, lying at rest on a perfectly rough horizontal table is {jropelled with initial angular velocity Q so that the cloth unrolls. Apply the Principle of Energy to shew that the radius of the roll will diminish from a tor (so long as r is not small compared with a) in time — ./^y^^^-\/c^^^] , where 3Sl^a^ = A {c^ - a^) ff. 6 V oj Is the application of the principle correct ? 245. In many cases of motion the application of the principles of this Chapter will give two first integrals of the motion, and hence determine the motion, Ex. A perfectly rough inelastic sphere, of radius a, is rolling with velocity v on a horizontal plane tvhen it meets a fixed obstacle of height h. Find the Conservation of Momentum Mnd Energy 323 condition that the sphere will surmount the obstacle and, if it does, shew that it will continue rolling on the plane with velocity (l — s-) v. Let be the angular velocity immediately after the impact about the point of contact, K, with the obstacle. The velocity of the centre before the impact was v in a horizontal direction, and the angular velocity was - about the centre. Since the moment of momentum about K is unaltered, as the only impulsive force acts at K, we have m [k- + a'-) fl = mv {a - h) + vik^ - . ••• ^="^'Z''^ " (1). Let w be the angular velocity of the sphere about K when tbe radius to K is inchned at to the horizontal. The equation of Energy gives 1 7a2 -m.— (w2-02j= _mg (h + a sin 6 -a) (2). Also, if R be the normal reaction at this instant, we have, since the ac- celeration of the ceutre is aw^ towards K, mau- = mg sin e - R (3). (2) gives w2 = fi2__ .|^(/i + asin6>-a) (4), and (3) gives - = ^ [lOh - 10a + na sin d]-ail^ (5). In order that the sphere may surmount the obstacle without leaving it, (i) w must not vanish before the sphere gets to its highest point, i.e. w^ must be positive when d-dO°, and (ii) R must not be negative when it is least, I.e. when sm^= . a The first condition gives Q.'^>-=^ , and the second gives fi2<; ^(^~ \ ta^ a' Hence, from (1), JlOgh, and v < ;= — ^ Jg {a - h). la — 5h ' 7a — oh For both these conditions to be true it is clear that /«3>t^. If these conditions are satisfied so that the sphere surmounts, without leaving, the obstacle, its angular velocity when it hits the plane again is 0. If its angular velocity immediately after hitting the plane be wj, we have, by the Principle of the Conservation of Momentum, '''i^ ^ , ,, 2a2 wi = m . nQ . (a - h) + m ^- 0, since just before the impact the centre was moving with velocity aQ perpen- dicular to the radius to the obstacle. so that the sphere will continue to roll on the plane with velocity v (l ^ ) . 21—2 324 Dyiiwtmcs of a Rigid Body EXAMPLES 1. A smooth uniform rod is moving on a horizontal table about one end which is fixed ; it impinges on an inelastic particle whose distance from the fixed end was - th of the length of the rod ; find the ratio of the velocity of the particle when it leaves the rod to its initial velocity. 4(X^ 4(^2 4(^2 [For the impact we have M.~— co = J/ — - '. The Principles of Energy and Momentum then give 1 „ 4a2 .1 . 1 4^2 1 4q,2 2.¥.— e2 + _„,(^2 + ^2^2) = _J/._,2+_^__„'2 , ,, 4a2 . ,. ,,4«2 4Qr2 and M .-^r- e + mx^e = M -— » + m —k w . ] 2. A uniform rod, of mass i/, is moving on a smooth horizontal table about one end which is fixed ; it drives before it a particle, of mass nM, which initially was at rest close to the fixed end of the rod ; when the particle is at a distance - th of the length of the rod from the fixed end, shew that its direction of motion makes with the rod an angle 3. A uniform rod, of length 2a, lying on a smooth horizontal plane passes through a small ring on the plane which allows it to rotate freely. Initially the middle point of the rod is very near the ring, and an angular velocity of the rod when it is inclined at an angle 6 to the vertical is given by the equation w^ (1 + 3 cos^ 6) = j^g — ^ (^ ~ ®^" ^}« J\ Consei'vation of Momentum and Energy. Exs. 327 16. A hoop, of radius a, rolling on a horizontal road with velocity v comes into collision with a rough inelastic kerb of height h, which is perpendicular to the plane of the hoop. Shew that, if the hoop is to clear the kerb without jumping, v must be 17. An inelastic uniform sphere, of radius a, is moving without rotation on a smooth table when it impinges on a thin rough horizontal rod, at right angles to its direction of motion and at a height h from the plane ; shew that it will just roll over the rod if its velocity be a IWg a-bV 5 and 6 be < y= • 18. A sphere, of radius a, rolling on a rough table comes to a slit, of breadth b, perpendicular to its path ; if V be its velocity, shew that the condition it should cross the slit without jumping is 172^100 ._ , . „ 14- 10sin2a F2> -^ga (1 -cos a) sm^ « ^^-^^^^3 , where 6 = 2a sin a, and \lga cos a>lV^+lOga. 19. A sphere, of radius a, rests between two thin parallel perfectly roiigh rods A and B in the same horizontal plane at a distance apart equal to 2b ; the sphere is turned about A till its centre is very nearly vertically over A ; it is then allowed to fall back ; shew that it will rock between A and B if lQ)b'^-^ga. 24. A uniform cubical block stands on a railway truck, which is moving with velocity F, two of its faces being perpendicular to the direction of motion. If the lower edge of the front face of the block be hinged to the truck and the truck be suddenly stopped, shew that the block will turn over if V is greater than f \/^ga {^2 - 1), where 2a is the side of the block. 25. A string, of length b, with a particle of mass m attached to one end, is fastened to a point on the edge of a circular disc, of mass M and radius a, free to turn about its centre. The whole lies on a smooth table with the string along a radius produced, and the particle is set in motion. Shew that the string will never wrap round the disc if aM,...) (1), with similar expressions for y and z. These equations are not to contain 0, ^ ... or any other differential coefficients with regard to the time. As usual, let dots denote differential coefficients with regard Lagrange's Equations 331 to the time, and let 'Thy -tt--- denote partial differential coefficients. Then, differentiating (1), we have cLv dx i dx ^-dt+T0-' + d-^-^+ (2). ■(3). On differentiating (2) partially with regard to d, we have dec dx Te^dd Again, differentiating (2) with regard to 6, we have dx _ d^x d-x ^ d^x ■ dd ~ dMt ^d&'^^ ded4> • 'P + • • • ^^ rdxi dt Idd] .(4). If T be the kinetic energy of the system, then 2'=iSm[i;^+2/^+i^] (5). Now the reversed effective forces and the impressed forces form a system of forces in equilibrium, so that their equation of virtual work vanishes ; in other words the virtual work of the effective forces = the virtual work of the impressed forces. The first of these, for a variation of only, he by equations (3) and (4), :=^.4x^7n.^{x' + y'' + z')Sd-4-^lvi.^[x^- + f + z']Se dt dd dd If-S^^^^^'^"^''"'^^^* (^>' 332 Dynamics of a Rigid Body Again, if V be the Work, or Potential function, we have the virtual work of the impressed forces, for a variation of 6 alone, \_dx dd'^ dy dd "*" dz dd_ Equating (6) and (7), we have ^(dT\_dT_dV (dT\dT^dV dt\dd) de dd ^ '' Similarly, we have the equations dt\d^l dcfi d(}>' dt \dyjr) dy^ dyjr ' and so on, there being one equation corresponding to each independent coordinate of the system. These equations are known as Lagrange's equations in Generalised Coordinates. Cor. If K be the potential energy of the system, since F= a constant — K, equation (8) becomes d_(dT\_dT dK_Q dt\ddl dd^ dd~ ' If we put T — K = L, so that L is equal to the difference between the kinetic and potential energies then, since V does not contain 6, 4>, etc., this equation can be written in the form d_ dLr\_dL_ dt dd\ dd~ L is called the Lagrangian Function or Kinetic Potential. 248. When a system is such that the coordinates of any particle of it can be expressed in terms of independent coordinates by equations which do not contain differential coefficients with regard to the time, the system is said to be holonomous. 249. ^x- !• -^ homogeneous rod OA, of mass m^ and length 2a, is freely hinged at O to a fixed point; at its other end is freely attached another homogeneous rod AB, of mass m^ and length 2b; the syntein moves under gravity ; find equations to determine the motion. Lagrange's Equations 333 Let Gi and Gn be the centres of mass of the rods, and d and their inclina- tions to the vertical at time t. q. The kinetic energy of OA is 1 4a2 . G3 is turning round A with velocity 60, whilst ^ is turning round with velocity 2ad. Hence the square of the velocity of G^ = {2ad cos e + b

. NP, i.e. $sin e^, and it is moving perpendicular to OP in the plane VOA with velocity ^'d- Hence the kinetic energy of this element _1 rf£ ~2"2a Therefore the whole kinetic energy T Also the work function V = m .g . a cos Hence Lagrange's equations give d r4ma^ ;"1 2Hia' m[_^-&\\\-e^- + ^-0'-']. {(l)-s\\\^d + d-). '+0. dt L 3 and i.e. and ' - ^^^^ 0^ . 2 sin d cobO— - mga sm i e -qflam 6 cos^; %.,• sm d 4> sin^ d — constant = w sin- a (1) and (2) give, on the elimination of ^, u^ sin* a „ -'dg Ja cos5 = - sin d. •(1), .(2). .(3). Lagrange's Equations 335 Steady motion. The rod goes round at a constant inclination a to the vertical if d be zero when d = a, i.e. if u^-^^T^^ (4). 4a cos a ^ ' When w has not this particular value, equation (3) gives on integration from the initial conditions, ao 3;;k r, sin2 a"! 3ff , ^ 39 cos o-cos5^ „ „ „ ~ia — sin^"^ — [cos2^ + 2ncos^-l + 2ncosa]. Hence 6 is zero when d = a, i.e. initially, or when C0S2 ^ + 2«COS^-l + 2KCOStt = 0, I.e. when cos 6= -n+ Jl - 2n cos a + n^ (6). [The + sign must be taken ; for the - sign would give a value of cos d numerically greater than unity.] The motion is therefore included between the values d = a and 6 = 61 where cos 61 is equal to the right hand of (6). Now ^1 ^ a, i.e. the rod will rise higher than or fall below its initial position, according as cos 6^ < cos o, i.e. according as ^^1 - 2n cos a + n- > cosa + n, i.e. according as sin^ a ^ An cos a, siu2 a ^ aaj2 sin- a i.e. according as ofl $ -. , 4a cos a i.e. according as the initial angular velocity is greater or legs than that for steady motion at the inclination a. It is clear that equation (2) might have been obtained from the principle that the moment of momentum about the vertical OV is constant. Also the Principle of the Conservation of Energy gives — -— (02 sin2 e + e'') = mga (cos d - cos a) + -— - to^ sin^ a. o o On substituting for from (2) this gives equation (5). Small oscillations about the steady motion. then (3) gives e='/['^^^-sinel (7). 4a Lcos a sm'^ o J * Here put 6 = a + ^, where \f/ is small, and therefore sin ^ = sin a + ^ cos u, and cos ^= cos a- i/* sin a. 336 Dynamics of a Rigid Body Hence (7) gives • • _ 3^ sin a [(1 - \h tan a) (1 + 1// cot a)-3 - (1 + ^ cot a)] 3(7 sin o ,,, = - --J . ^[4 cot a + tau a] ^_ 3g(l + 3cos'a) on neglecting squares of \p. Hence the required time / 4a cos g "" V 35((l + 3cos2a) ' Ex. 8. Four equal rods, each of length 2a, are hinged at their ends so as to form a rhombus ABCD. The angles B and D are connected by an elastic string and the lowest end A rests on a horizontal plane whilst the end C slides on a smooth vertical tvire passing through A ; in the position of equilibrium the string is stretched to twice its natural length and the angle BAD is 2a. Sheio that the time of a small oscillation about this position is | 2a (l + 3sin2a)cosa ) ^ [ Sg cos 2a J When the rods are inclined at an angle 6 to the vertical, the component velocities of the centre of either of the upper rods are — [3a cos 6] and — (a sin 6), i.e. -3a&ind .6 and a cos d . 0. Hence T, the total kinetic energy, 1 r4a2 = ^■2 — ^2 + ( - 3a sin 69)"^ + {a cos 66}^ + o" ^M = 8ma26>2[l + sin2^]. Also the work function V f2asmB X - c = - mg . 2 . (a cos ^ + 3a cos ^) - 2\ I . dx J c c = - 8mga cos - - (2a sin - c)2, where 2c is the unstretched length of the string and \ the modulus of elasticity. Lagrange's equation is therefore J ri6a2m^ (^ + sin^yi - 16ma2^2 sin 6 cos = 8m(/asin a cos ^ (2a sin ^ -c) (1). Now we are given that d and '9 are zero when d=.a, and that c = a sin a. 2mgc Lagrange's Equations 337 In (1) putting 6 = a + \f/, where ^ is small, and neglecting squares and products of ip and ^, we have 16a2mi/;(| + sin2a) = %mga (sin a + vi- cos a) [cos a-ip sin a] [a sin o + 2iia cos al cos a r J L r J = - Qamg\p (cos a - sin a tan a), 3oco8 2a l.e. y}/— • 2a cos a (1 + 3 sin^ c Therefore the required time = 27r , /2« cos a (1 + 3 sin^ g) ^ V 3^ cos 2a. Ex. 4. Small oscillations about the stable position of equilibrium for the case of Ex. 1 ichen the masses and lengths of the rods are equal. When mi = m2 and a = b, the equations (3) and (4) of Ex. 1 become -^6 + 24^008 (9i-t?)-202sin {^, and putting 6 and for sin d and sin ^ Acqs [pit + ai), and 95 + (277"- 1)0 = 5 cos (2^2^ + 02). This method of solution has the advantage of only bringing in the four necessary arbitrary constants. 250. If in the last example we put 9^-(2V7+l)0 = Z, and 9^ + (2V7-l) and -^ in the case of a small oscillation about a position of equilibrium, so that T= Aj' + A^(j>''+A,,yjr^+2AJ4> + 2Ajyjr + 2A,4ylr ...(1), and V= G + a,6 + a^cj) + a^^jr + Ond^ + a^c^"" + a^r^y^r'' + . . .(2). If 6, (f>, yjr be expressed in terms of A^, Y, Z by linear equations of the form = \,X+\2Y+\sZ, = fl^X + /^2 F+ flsZ, and i^ = I'lX + 1^2!^+ Vs^, and \, ^-2) ^3, H-i) /^2> f^s, ^1, v^, Vg be so chosen that on substitution in (1) and (2) there are no terms in T containing YZ, ZX, X Y, and none in V containing Lagrange's Equations for Blows 339 YZ, ZX, XY, then X, Y, Z are called Principal or Norma Coordinates. For then T=A,^t^ + A^Y^ + A,:Z\ and F = Ci + a^X + ao!Y+ a^'Z + a^^X' + a^ Y^ + a,.^Z\ and the typical Lagrange equation is then 1A,^X = a/ + 2a„'X, i.e. an equation containing X only. On solving this and the two similar equations for F and Z, we have 6 given by sum of three simple harmonic motions. Similarly if the original equations contained more coordinates than three. 251. Lagrange's equations for Blows. Let i'o and i\ denote the values of x before and after the action of the blows. Since the virtual moment of the effective impulses Sm (^i — ^o). etc. is equal to the virtual moment of the impressed blows, we have, for a variation in 6 only, Let Tq and T-^ be the values of T just before and just after the blows. Then, from equations (3) and (5) of Art. 247, (dT \ ^ r . dx .dy . dz~\ \dd\ L d6 ^ de dd\^ . dy dz "Td^^'de z — de. K- y . dx and ( — r| =Sm \x — + y-^ + \de)r I de ^ dd Hence the left hand of (1) is ~dT-] rdT dd}^" Idd Also the right hand of (1) _\dVj,dx dV^dy dVj_dzl .. ^dV, ~\ dx dd^ dy dd^ dz dd\ de where 8 Fj is tiie virtual work of the blows. .(2), 340 Dynamics of a Rigid Body Hence if S V^ be expressed in the form the equation (1) can be written in the form \de)i \ddJo and similarly for the other equations. The equation (2) may be obtained by integrating equation (8) of Art. 247 between the limits and t, where t is the infinitesimal time during which the blows last. The integral of ^ (^^ is f^l^ i.e. ['^1 - f^l . ^'\deJ Ld^Jo' UJi UJo Since -j— is finite, its integral during the small time t is ultimately zero. do _,, . ^ . .dV . dVi The mtegral of -j^ is —t^ . ad do Hence equation (2). 252. We give two examples of the preceding Article. Many of the examples of pp. 272—274 and also Ex. 2 of p. 277 and Ex. 14 of p. 280 may well be solved by this method. Ex. 1. Three equal imiform rods AB, BG, CD are freely jointed at B and C and the ends A and D are fastened to smooth fixed pivots ivhose distance apart is equal to the length of either rod. The frame being at rest in the form of a square, a blow J is given perpendicularly to AB at its middle point and in Q T2 the plane of the square. Shew that the energy set up is — r— , lohere m is the mass of each rod. Find also the bloios at the joints B and C. When AB, or CD, has turned through an angle 6, the energy of either is 1 4a2 • 1 ^-m. -Q- ^^ ^^^ that of BC, which remains parallel to AD, is ^m{2ad)K ^ „ 1 4a^ •,, 1 , „ •„ 10;»((2 . .-. T = 2.-m. — 0-^ + ^mia^2= —^ . e\ 20ma2 (dT\ 20ma2. . fdT\ . ,-. ( — ) =-^-e, and — =0. Also 5\\ = J.a^d. u 2Qma- ■ ^ . ■ 3J Hence we have — - — 6 = J. a, i.e. 6= . 3 20ma . , 10ma2^2 3J2 ••• ^■^1""-^'^ energy = ^^— = ^^ . If Y and Y^ be the blows at the joints B and C then, by taking moments about A and D for the rods AB and DC, we have Tn.—d-J.a~Y.2a, and 7n . —e=Yi.2a. Lagrange's Equations. Examples 341 Ex. 2. Solve by the same method Ex. 12 of page 280. Let mi be the mass of the rod struck, and m^ that of an adjacent rod, so that mi _ OT2 _ 1 M' a " b ~2a + b' Let II be the velocity and w the angular velocity communicated to the rod that is struck. 1 / a2 \ 1 Then T= ^.2mi{u'^+-^-'A+~vi2l{u + au}Y + {ii-awf] = iw.|«v„=ttL' a). Also the blow X=M[V -u-cw] (2), and 5Fi = ilf [F-u-cco][5x + c56i] (3), where u = x and 01 = ^. Hence the equations of the last article give M'u= j^ = M{V-u-cu) (4), , M' „ rt + 36 dVi ^^ ^^^ ^""^ T'"''TTb = ^=^^'^^-''-"'^ (5)- Also, by Ex. 3, Art. 207, the total loss of kinetic energy = lA'[F+(M + cw)]-iZ[H + cw] = lA. F= 1 1/F [F-(w + ceo)] = etc. EXAMPLES 1. A bead, of mass M, slides on a smooth fixed wire, whose inclination to the vertical is a, and has hinged to it a rod, of mass m and length 21, which can move freely in the vertical plane through the wire. If the system starts from rest with the rod hanging vertically, shew that {AM+m (1 + 3 cos2 6)] W = Q (M+m) g sin a (sin ^ -sin a), where 6 is the angle between the rod and the lower part of the wire. 2. A solid uniform sphere has a light rod rigidly attached to it which passes through its centre. This rod is so jointed to a fixed vertical axis that the angle, 8, between the rod and the axis may alter but the rod must turn with the axis. If the vertical axis be forced to revolve constantly with uniform angular velocity, shew that the equation of motion is of the form d^ = n^ (cos 6- cos ^) {cos a -cos 6). Shew also that the total energy imparted to the sphere as 6 increases from 61 to d^ varies as cos^ 01 - cos"'' 62. 342 Dynamics of a Rigid Body 3. A uniform rod, of mass Zm and length 2Z, has its middle point fixed and a mass m attached at one extremity. The rod when in a horizontal position is set rotating about a vertical axis through its centre with an angular velocity equal to »/ -y^. Shew that the heavy end of the rod will fall till the inclination of the rod to the vertical is cos~^ [\/^^+ 1 — n\ and will then rise again. 4. A rod OA, whose weight may be neglected, is attached at to a fixed vertical rod OB, so that OA can freely rotate round OB in a horizontal plane. A rod XY, of length 2a, is attached by small smooth rings at X and Y to OA and OB respectively. Find an equation to give 5, the inclination of the rod XY to the vertical at time t, if the system be started initially with angular velocity fl about OB. Shew that the motion will be steady with the rod ZF inclined at a to the vertical, if Q.'^= — &eGa, 4a and that, if the rod be slightly disturbed from its position of steady motion, the time of a small oscillation is 47r ^ 3g{l+3cos^a) 5. If in the preceding question the rod OA be compelled to rotate with constant angular velocity to, shew that, if 4a>^a > 3g, the motion will be steady when cos a = ~j- , and that the time of a small oscillation will , 87raa> be Vl6a)*a2_9^2- [Reduce the system to rest by putting on the " centrifugal force " for each element of the rod XY, and apply the principle of Energy.] 6. Three equal uniform rods AB, BC, CD, each of mass m and length 2a, are at rest in a straight line smoothly jointed at B and 0. A blow / is given to the middle rod at a distance c from its centre in a direction 2/ perpendicular to it ; shew that the initial velocity of is ^ and that the 3 Ml initial angular velocities of the rods are {5a + 9c) I 6cl (5a - 9c) I lOma^ ' 57na-' lOma^ 7. Six equal uniform rods form a regular hexagon, loosely jointed at the angular points, and rest on a smooth table ; a blow is given perpen- dicularly to one of them at its middle point ; find the resulting motion and shew that the opposite rod begins to move with one-tenth of the velocity of the rod that is struck. 8. A framework in the form of a regular hexagon ^ .SCZ).£'i^ consists of uniform rods loosely jointed at the corners and rests on a smooth table ; a string tied to the middle point of AB is jerked in the direction of AB. Lagrange's Equations. Examples 343 Find the resulting initial motion and shew that the velocities along AB and DE of their middle points are in opposite directions and in the ratio of 59 : 4. [Let Ux and v^ be the resulting velocities of the middle point of AB along and perpendicular to AB and wi its angular velocity; and let ^2, ^2 and 0)2 gi^6 the motion of BC similarly, and so on. From the motion of the corners A, B, C, etc., we obtain Hence v, = ^ and a.,= ^-^%etc. Also h\\=J .bxi, where Ui=Xi. The equations written down by Art. 251 then completely determine the motion.] 9. A perfectly rough sphere lying inside a hollow cylinder, which rests on a perfectly rough plane, is slightly displaced from its position of equilibrium. Shew that the time of a small oscillation is J'- 14J/ g 10J/+7m' where a is the radius of the cylinder and b that of the sphere. 10. A perfectly rough sphere, of mass m and radius h, rests at the lowest point of a fixed spherical cavity of radius a. To the highest point of the movable sphere is attached a particle of mass m' and the system is disturbed. Shew that the oscillations are the same as those of a simple ^ ' , 7wi 47n+ — pendulum of length {a-h) m + m 2 11. A hollow cylindrical garden roller is fitted with a counterpoise which can turn on the axis of the cylinder ; the system is placed on a rough horizontal plane and oscillates under gravity ; if — be the time of a small oscillation, shew that p is given by jo2 [(2J/+ J/') y[;2 - i/'A2] = (2JI/+i/') gh where M and M' are the masses of the roller and counterpoise, k is the radius of gyration of M' about the axis of the cylinder and k is the distance of its centre of mass from the axis. 12. A thin circular ring, of radius a and mass M, lies on a smooth horizontal plane and two tight elastic strings are attached to it at opposite ends of a diameter, the other ends of the strings being fastened to fixed 344 Dynamics of a Rigid Body points in the diameter produced. Shew that for small oscillations in the plane of the ring the periods are the values of — given by —^ — 1=0 or - — 7 or - , where b is the natural length, I the equilibrium length, and T the equilibrium tension of each string. 13. A uniform rod AB, of length 2a, can turn freely about a point distant c from its centre, and is at rest at an angle a to the horizon when a particle is hung by a light string of length I from one end. If the particle be displaced slightly in the vertical plane of the rod, shew that it will oscillate in the same time as a simple pendulum of length - a2 + Zac cos2 a + 3c2 sin^ a a^ + Zac 14. A plank, of mass M, radius of gyration k and length 26, can swing like a see-saw across a perfectly rough cylinder of radius a. At its ends hang two particles, each of mass m, by strings of length I. Shew that, as the system swings, the lengths of its equivalent pendula are I and M-2 + 2m62 {M+ 2m) a ' 15. At the lowest point of a smooth circular tube, of mass M and radius a, is placed a particle of mass M' ; the tube hangs in a vertical plane from its highest point, which is fixed, and can turn freely in its own plane about this point. If the system be slightly displaced, shew that the periods of the two independent oscillations of the system are 27r \/ — , ~M ^ and 277 V. 16. A string AG is tied to a fixed point at A and has a particle attached at C, and another equal one at B the middle point of AG. The system makes small oscillations under gravity ; if at zero time ABG is vertical and the angular velocities of AB, BG are a and &>', shew that at time t the inclinations 6 and (f) to the vertical of AB, BG are given by the equations fh-c , where 2a = length of the rod, OC=c, and OB=k. Find also the time of a small oscillation and prove that it is not afifected by the elastic string. 11. A uniform rod AB can turn freely in a vertical plane about the end A which is fixed. B is connected by a light elastic string, of natural length ly to a fixed point which is vertically above A and at a distance h from it. If the rod is in equilibrium when inclined at an angle a to the vertical, and the length of the string then is k, shew that the time of a small oscillation about this position is the same as that of a simple pendulum of length - r-r^-9— • ^ ° 3 hi sm^ a 12. A rhombus, formed of four equal rods freely jointed, is placed over a fixed smooth sphere in a vertical plane so that only the upper pair are in contact with the sphere. Shew that the time of a symmetrical oscillation in the vertical plane is £77 * / —- — ^ ° ., -- , where 2a is the ^ V 3^(H-2cos^a)' length of each rod and a is the angle it makes with the vertical in a position of equilibrium. 350 Dynamics of a Rigid Body 13. A circular arc, of radiua a, is fixed in a vertical plane and a uniform circular disc, of mass i/and radius j, is placed inside so as to roll on the arc. When the disc is in a position of equilibrium, a particle of mass g- is fixed to it in the vertical diameter through the centre and at a distance ^ from the centre. Shew that the time of a small oscillation about the position of equilibrium is — » / — . 14. A uniform rod rests in equilibrium in contact with a rough sphere, under the influence of the attraction of the sphere only. Shew that if displaced it will always oscillate, and that the period of a small oscillation is Stt - ^^ r- , where v is the constant of gravitation, m the « (3ym)* mass and a the radius of the sphere, and 21 the length of the rod. 15. Two centres of force = J^ — -^ ^^e situated at two points S and H, where SH= 2b. At the middle point of Sff is fixed the centre of a uniform rod, of mass M and length 2a ; shew that the time of a small oscillation about the position of equilibrium is Stt {b'^-a^)-r-'\/G{jib. 16. A shop-sign consists of a rectangle A BCD which can turn freely about its side AB which is horizontal. The wind blows horizontally with a steady velocity v and the sign is at rest inclined at an angle a to the vertical ; assuming the wind-thrust on each element of the sign to be k times the relative normal velocity, find the value of a and shew that the time of a small oscillation about the position of equilibrium is 27r a/ - „— ; — ". „ , where BC= 2a. •)^ COS^ ( g 'i,v^ COS a-ga sin^ a ' 17. A heavy ring J, of mass nm, is free to move on a smooth horizontal wire ; a string has one end attached to the ring and, after passing through another small fixed ring at a depth h below the wii'e has its other end attached to a particle of mass m. Shew that the inclination 6 of the string OA to the vertical is given by the equation A (» + sin''' 6)6'^ = 2g cos* 6 (sec a - sec 6), where a is the initial value of 6. Hence shew that the time of a small oscillation about the position of equilibrium is the same as that of a simple pendulum of length nh. 18. A straight rod AB, of mass m, hangs vertically, being sujjported at its upper end A by an inextensible string of length a. A string attached to B passes through a small fixed ring at a depth b below B and supports a mass M at its extremity. Shew that the rod, if displaced to a neigh- Initial motions 351 bouring vertical position, will remain vertical during the subsequent lilh h oscillation if — i- = -r ; and that the equivalent pendulum is of length Find the times of the other principal oscillations. 19. A uniform heavy rod AB is in motion in a vertical plane with its upper end A sliding without friction on a fixed straight horizontal bar. If the inclination of the rod to the vertical is always very small, shew that the time of a small oscillation is half that in the arc of a similar motion in which A is fixed. 20. A uniform rod, of length 21, i-ests in a horizontal position on a fixed horizontal cylinder of radius a ; it is displaced in a vertical plane and rocks without any slipping ; if w be its angular velocity when inclined at an angle 6 to the horizontal, shew that (^ + a2^2 j ^2 ^ <2,ga (cos ^ + (9 sin <9) is constant. If the oscillation be small, shew that the time is 27r ^ / ^ — • V .iga 21. A smooth circular wire, of radius a, rotates with constant angular velocity oj about a vertical diameter, and a uniform rod, of length 26, can slide with its ends on the wire. Shew that the position of equilibrium in which the rod is horizontal and below the centre of the wire is stable if a)2< — — f^ — —^ , where c^^/a^ — b^, and that then the time of a small 3 (3a- — 46-) oscillation about the stable position is 27r * / ., ,_ ., — t-,,7 . ^ V 3gc - — I (sin a- 6 cos a) -^-a (1) , y = 1 cos, [a - 6) + a sin ^ = Z (cosa + ^ sina) + a^ (2), squares of d and

^ + m2 (2ae + b4>)2\ , and V=mig . ad + m2g (2a9 + b(p). MBq A" B Hence Lagrange's equations give (^+»"2) . iae + 2m2h^=-- g {mi + 2mi), 4b — „. If one end of the rods be fixed, shew that the initial radius of curvature of the other end is (aicoi + a2a>2 + • • • + o-n^^n)"^ Uico^^ + a2<02^ + ... + a^ton^' 17. A rod ABC, of length I, is constrained to pass through a fixed point B. A is attached to another rod OA, of length a, which can turn about a fixed point situated at a distance d from B. The system is arranged so that A, 0, B, C are all in a straight line in the order given. If A be given a small displacement, shew that the initial radius of cm'vature of the locus of C is --^ — r— — ^ . {a+d)^ — la 18. A uniform smooth circular lamina, of radius a and mass i/, movable about a horizontal diameter is initially horizontal, and on it is placed, at a distance c from the axis, a particle of mass m ; shew that the initial radius of curvature of the path of m is equal to 12 -tt^. [The distance of the particle from the axis being r when the inclination of the disc is a small angle d, the equations of motion are r-re- = gsin0 = ge (1), and -r [Mk-e + mr^d] = mrg cos e, dt' .e. [M j+7nr2 j e + 2mre = mrg (2). Now 6, and 6 are respectively of the order 0, 1, 2 in t, and hence, from (1), is of the order 2, and therefore r and r - c of the order 3 and 4 in t. Hence, from (2), on neglecting powers of t, ^ = j^^2 + lj i^2~^9' ^°-y- .-. d = Agt, and d:=iAgt2. Therefore (1) gives r=c . 42^2^2 + ^^^2,2=^^2(1 + 40) tK 12' M^- Hence 2p=Lt -r-^ =Lt ■ r cos a-c i ^■AY-t^ 4 '' = Lt : : etc.] 23—2 356 Dynamics of a Rigid Body 19. A uniform rod, of length 2a and mass M, can freely rotate about one end which is fixed ; it is held in a horizontal position and on it is placed a particle, of mass on, at a distance h from the fixed end and it is then let go. Shew that the initial radius of curvature of the path of the particle IS ^^-^^(^l + -j^J. Find also the initial reaction between the rod and the particle. 20. A homogeneous rod ACDB, of length 2a, is supported by two smooth pegs, C and D, each distant - from the end of the rod, and the peg D is suddenly destroyed ; shew that the initial radius of curvature of the path of the end £ is -^ , and that the reaction of the peg C is instantaneously increased in the ratio of 7 : 8. 21. In the previous question, if E be the middle point of AB, and the single rod ABhe replaced by two uniform rods AE, EB freely jointed at E and each of the same density, shew that the same results are true. 22. A solid cj'linder, of mass m, is placed on the top of another solid cylinder, of mass J/, on a horizontal plane and, being slightly displaced, starts moving from rest. Shew that the initial radius of curvtiture of the path of its centre is j „ A c, where c is the distance between its centres and all the surfaces are rough enough to prevent any sliding. Tendency to break 257. If we have a rod AB, of small section, which is in equilibrium under the action of any given forces, and if we consider separately the equilibrium of a portion PB, it is clear that the action of AP on PB at the section at P must balance the external forces acting on PB. Now we know, from Statics, that the action at the section at P consists of a tension T along the tangent at P, a shear *S' perpendicular to T, and a couple G called the stress-couple. The external forces acting on PB being known we therefore obtain T, S, and G by the ordinary processes of resolving and taking moments. If the rod be in motion we must, by D'Alembert's Principle, amongst the external forces include the reversed effective forces acting on the different elements of PB. Tendency to break 357 Now we know that in the case of a rod it is the couple G which breaks it, and we shall therefore take it as the measure of the tendency of the rod to break. Hence the measure of the tendency to break at P is the moment about P of all the forces, external and reversed impressed, on one side of P. The rod may be straight, or curved, but is supposed to be in one plane ; it is also supposed to be of very small section ; otherwise the problem is more complicated. If we had a string instead of a rod the couple G would vanish, and in this case it is the tension T which causes it to break. The following two examples will shew the method to be adopted in any particular case. 258. -Ea;. 1. A uniform rod, of length 2a, is moving in a vertical plane about one end luhich is fixed ; find the actions across the section of the rod at a point P, distant x from 0. Consider any element dy of the rod at a point Q distant y from P. Its weight is — . mg. The reversed effective forces are mdy •2a — {x + y)d and mdii , , •„ as marked. These three forces, together with similar forces on all the other elements of the body, and the external forces give a system of forces in equilibrium. The actions at P along and perpen- dicular to the rod and the stress couple at P, together with all such forces acting on the part PA, will be in equilibrium. Hence the stress couple at P in the direction ^ ^ r^"- mdy + 2/)^ 2a(^ + ^^'^|.m, 2a {x + y)e. mg sin 6 4a (2a e[li2o .t)2 + (2a - x} 1- But, by taking moments about the end of the rod, we have 3f) sin^ and therefore 0^ = :r^ (cos ^-cos a) 2a ^ where a. was the initial inclination of the rod to the vertical. : — - (cos ( 2a ^ ■(1), •(2). 358 Dynamics of a Rigid Body Hence the stress-couple This is easily seen to be a maximum when .r = — ; hence the rod will break, if it does, at a distance from equal to one-third of the length of the rod. The tension at P of the rod in the direction P0 = the sum of the forces acting on PA in the direction PA = 1^ (2a -x) [4a cos ^-1-3 {2a -(- x) (cos » - cos a)] (3), by equation (2). The shear at P perpendicular to OP and upwards = '^ (2a - a;) sin d +^^ (2a - x) (2a + x) mg sin ^ 16a2 (2«-^)(2«-33^)- Ex. 2. One end of a thin straight rod is held at rest, and the other is struck against an inelastic table until the rod breaks ; shew that the point of fracture /3 is at a distance from the fixed end equal to ^ times the total length of the rod. When the rod strikes the table let it be inclined at an angle a to the horizon ; let oj be the angular velocity just before the impact and B the blpw. Taking moments about the fixed end, we have 4rt2 ^ „ (1). where m is the mass and 2a the length of the rod. Let us obtain the stress-couple at P. The effective impulse on an element ^-m at Q, where PQ = y, is — ^ . (x + y) .w upwards. Hence the reversed effective impulse at Q is in the direction marked. Taking moments about P, the measure of the tendency to break _ f2a-x^y A ' = B{2a-x)cQsa- ^m(x + y)w.y ^ I (I '^"- m.r^ :B.(2a-a;)cosa-^ (2a -x)^ (\a + x) vi.^^{x + y). = Bcosa [(2. .- .) _ l?^iZ^i^^±£)] = ^ .. (4«- x^) . Tendency to break. Examples 359 This is a maximum when x= -— , and when B is big enough the rod will break here. EXAMPLES 1. A thin straight rod, of length 2a, can turn about one end which is fixed and is struck by a blow of given impulse at a distance h from the fixed end ; if 6>-^ , shew that it will be most likely to snap at a distance from the fixed end equal 4a If &< -„- > prove that it will snap at the point of impact. 2. A thin circular wire is cracked at a point A and is placed with the diameter AB through A vertical ; B is fixed and the wire is made to rotate with angular velocity w about AB. Find the tendency to break at auy point P. If it revolve with constant angular velocity in a horizontal plane about its centre, shew that the tendency to break at a point whose angular distance from the crack is a varies as sin- - . 3. A semi-circular wire, of radius a, lying on a smooth horizontal table, turns round one extremity A with a constant angular velocity w. If be the angle that any arc AP subtends at the centre, shew that the tendency to break at P is a maximum when tan (p = Tr-(p. If A be suddenly let go and the other end of the diameter through A fixed, the tendency to break is greatest at P where tan ~ = (p. 4. A cracked hoop rolls uniformly in a straight line on a perfectly rough horizontal plane. When the tendency to break at the point of the hoop opposite to the crack is greatest, shew that the diameter through the crack is inclined to the horizon at an angle tan"^ ( - ) • 5. A wire in the form of the portion of the curve r=a (l + cos(9) cut off by the initial line rotates about the origin with angular velocity w. Shew that the tendency to break at the point 6= ^ is measured by " '^ ' mu-a\ 6. Two of the angles of a heavy square lamina, a side of which is a, are connected with two points equally distant from the centre of a rod of length 2a, so that the square can rotate with the rod. The weight of the square is equal to that of the rod, and the rod when supported by its ends in a horizontal position is on the point of breaking. The rod is then held by its extremities in a vertical position and an angular velocity w given to the square. Shew that the rod will break if au^>3g. CHAPTER XX MOTION OF A TOP 259. A top, two of whose principal moments about the centre of inertia are equal, moves under the action of gravity about a fixed point in the axis of unequal moment; find the motion if the top be initially set spinning about ita axis which ivas initially at rest Let OGG be the axis of the top, G the centre of inertia, OZ the vertical, ZOX the plane in which the axis 00 was at zero time, OX and OY horizontal and at right angles. At time t let 00 be inclined at 6 to the vertical, and let the plane ZOO have turned through an angle i/r from its initial position ZOX. Let OA, OB be two perpendicular lines, each perpendicular to 00. Let A be the moment of inertia about OA or OB, and that about 00. Motion of a top 361 At time t let Wi, co^, and 6)3 be the angular velocities of the top about OA, OB, and 00. To obtain the relations between a)i, m^, wg and 6, <^, -v^ consider the motions of A and G. If 00 be unity, we have 6 = velocity of along the arc ZO = w^ sin (ft +CO2 cos (1), yjr .smd = ^|r X perpendiciilar from on OZ = velocity of perpendicular to the plane ZOO = — coi cos(j) + 0)2 sin (^ (2). Also C03 = velocity of A along AB = ^ +^ X perpendicular from N on OZ = 4>+^jrsin {90° -6) = (}>+^jr cos d (3). By Art. 229, the kinetic energy T=i[Aco^+Aai,' + Cco^] = ^ A (6' + yjr'sin'' d) + i 0(4, + ^jrcosey (4), by equations (1), (2) and (3). Also V= Mg(hcosi — hcosd) (5), where h = OG and i was the initial value of 6. Hence Lagrange's equations give j[A6]-Af'' sin ecosd + (j> + ^jr cos 6) ^jr sin = Mgh sin d (6), ^^[O(cp + ircose)] = (7), and j^[^^^sin2^ + Ccos6>(0+^cos^)]=O (8). Equation (7) gives (f) + '^coa6 = constant, i.e. ci)3 = Mghp, we have A sin^ 6.6^ = 2Mgh (cos i - cos 0) [sin- ^ - 2jj (cos i - cos ^)] = 2i\Igh (cos ^ - cos i) [(cos 6-pf- (p^ - 2p cos i + 1)] 2il/(//« (cos 6 — cos i) [cos 6 — p -\- Np^ — 2p cos t + 1] [cos ^-_p- \7)2-2p cost + 1] (11). Hence 6 vanishes when 6 = i or 6^ or d^, where cos 6i = p - Vjj- — 2p cos i + 1, and cos ^g = p + "V^^ ~ '^P ^^s i -f 1. [Clearly cos 6^ > unity and therefore 62 is imaginary.] Also ^1 > * since it is easily seen that cos d^ < cos i, since p — cos i < wp^ — 2p cos i + 1. Again, from (10), 6^ is negative if ^ < i, i.e. if cos 9 > cos i or again, from (11), if ^ > 6^, i.e. if cos ^ < p - V^2 - 2jj cos i + 1. Hence the top is never at a less inclination than i or at a greater inclination than 61, i.e. its motion is included between these limits. Now (9) gives A'^ sin- 6= Cn (cos i — cos 0) — a, positive quantity throughout the motion. Hence so long as the centre of inertia G is above the point 0, yjr is positive and the plane ZOO rotates in the same way as the hands of a watch when looked at from above. This is expressed often by saying that the precessional motion is direct. [If G be below the peg, this motion is found to be retro- grade.] It is clear from equations (9) and (11) that both 6 and -^ vanish when 6 = i. Also d^lA_ jl _± [ cos t- cos ^ "1 _ l-2cos^cost + cos''^ dd lOn "^j'dOl sin^ ]~ sin^ 6 which is always positive when 6> i. Motion of a top 363 Rence -^ continually increases, as 6 increases, for values of 6 between i and 6-^, and has its maximum value, which is easily seen to be —fT~ > when 6 = 6-y. The motion of the top may therefore be summed up thus ; its angular velocity about its axis of figure remains constant throughout the motion and equal to the initial value n; the axis drops from the vertical until it reaches a position defined by d = 9i, and at the same time this axis revolves round the vertical with a varying angular velocity which is zero when d = i and is a maximum when 6 = 6^. The motion of the axis due to a change in 6 only is called its " nutation." Ex. 1. If the top be started when its axis makes an angle of 60° with the . . A /SMcih upward-drawn vertical, so that the initial spin about its axis is -^ ^ / ■ , and the angular velocity of its axis in azimuth is 2 . / -~- , its angular velocity in the meridian plane being initially zero, shew that the inclination 6 of its axis to the vertical at any time t is given by the equation is/- so that the axis continually approaches to the vertical without ever reaching it. Ex. 2. Shew that the vertical pressure of the top on the point of support is equal to its weight when the inclination of its axis to the vertical is given by the least root of the equation ^AMgh cos2 5 - cos ^ [Chi^ + 2AMhgK] + C%^a - A Mgh = 0, where a and b are constants depending on the initial circumstances of the motion. 260. It can easily be seen from first principles that the axis of the top must have a precessional motion. Let OC be a length measured along the axis of the top to represent the angular velocity n at time t. In time dt the weight of the cone, if G be above 0, would tend to create an angular velocity which, with the usual convention as to sense, would be represented by a very small horizontal straight line OK perpendicular to OC. The resultant of the two angular velocities repre- sented by OK and 00 is represented by OD, and the motion of the axis is thus a direct precession. If the centre of inertia G be under 0, OK would be drawn in an opposite direction and the motion would be retrograde. sece = l-fsech-( ^ / ^^-|- i 364 Dynamics of a Rigid Body 261. Two particular cases. If, as is generally the case, n is very large, so that p is very large also, then /) r, /^ 2 . 1U1 . sin^z cos^,=^^l-(^l--cos* + -jJ=cos*-^^, on neglecting squares of - . Hence the motion is included between /I • 1 /I • sin i = 1 and 0=1 + ——, 2p . , ^ . 1 . 2AMqh sin i I.e. between ^ and i + — — ^^^ . Again if i = 0, then cos ^i = 1, so that 0^ is zero also and the axis remains vertical throughout the motion ; but, if the axis is slightly displaced, the motion of the top is not necessarily stable. 262. Steady motion of the top. In this case the axis of the top describes a cone round the vertical with constant rate of rotation. Hence all through the motion 0= a, 6 = 0, 6 = and -^ = const. = &>. The equation (6) of Art. 259 then gives ^6)2 cos a - Gnco + Mgh = (1). This equation gives two possible values of w ; in order that they may be real we must have G^n- >4in t V ^ / Cn \ A J , Mqh ^ AMqh . Cnt The first term increases uniformly with the time, and the second is periodic and smaller, containing — . Hence, to a first approximation, •>/r increases at a mean rate of -PT- per unit of time. On ^ Motion of a top 367 Thus, if a top be spun with very great angular velocity n, then, to start vs^ith, the axis makes small nutations of period Y? — , and it precesses with a mean angular velocity approxi- mately equal to -^ . At first these oscillations are hardly noticeable; as n diminishes through the resistance of the air and friction they are more apparent, and finally we come to the case of Art. 259. 264. A top is spinning with an angular velocity n about its axis which is vertical ; find the condition of stability, if the axis be given a slight nutation. The work of Art. 262 will not apply here because in it we assumed that sin a. was not small. We shall want the value of ^'when 6 is small; equation (6) of Art. 259 gives Ad = Aylr^s,m.ecosd- Cnyjr sin 6 +Mgh sin d ...(1). Also equation (9) gives A^ sin'' d = On (cos i - cos 6) = On (1 - cos 6) . . .(2), since the top was initially vertical. being small, (2) gives : Cn 1 On ^ • , • . ^ = X TT^^ ^ 21 + *"'^' mvolvmg 6^ etc. (1) then gives G'^n^ G^n^ A6=^ -g-T- . 6 - -x-j- . 6 + MghO + terms involving 6^ etc. 44 24 -\^-^-Mgh\e. ^44 Hence, if the top be given a small displacement from the vertical, the motion is stable, if 44 Also the time of a nutation > Mgh, I.e. if ?i > a/ — ^7.;^^ . = ^^V 0^ 44^ 4>AMgh 368 Dynamics of a Rigid Body Cor. If the body, instead of being a top, be a uniform sphere of radius a spinning about a vertical axis, and supported at its lowest point, then h = a, A=M.~ . and G = M~. 5 o Therefore n must be greater than a/ — . If a = one foot, the least number of rotations per second in order that the motion may be stable n 735732 , ^ ^, = ^^= o^ = about 51. iTT Ait Ex. A circular disc, of radius a, has a thin rod pushed through its centre perpendicular to its plane, the length of the rod being equal to the radius of the disc; shew that the system cannot spin with the rod vertical unless the angular velocity is greater than a/ — ~, APPENDIX ON THE SOLUTION OF SOME OF THE MORE COMMON FORMS OF DIFFERENTIAL EQUATIONS I- dx^^^^^' vvhere P and Q are functions of x. [Linear equation of the first order.] Multiply the equation by J^'^'', and it becomes Hence yeJ"^'^ = J<2e/^''^-+a constant. ■t,x. -f+y tan X = sec X Here ^/-Ptfa:_^Jtana;(to_ -logcosa;_ 1 COS X ' Hence the equation becomes 1 dit sin X 77 +y •-> =sec^.y. cos X ax "^ COS'' X .-. -^ = tiiux + C. cos.r ^^' d^^^K'ft^) ^^' ^^'^""^ ^ ^^^ ^ ^^^ functions of y. Onp„Ui„g(|;=r,„e.a™.|.g.f,.„t.atg = |f. The equation then becomes a linear equation between T and v, and is thus reduced to the form I. dx- III. :;4=-«V Multiplying by 2 ^^ and integrating, we have 24 370 Appendix. Differential Equations .'. y=C sm{n.v +!))== Lsm7ix + McosnJi!, where C, D, L, and M are arbitrary constants. We obtain, as in ITI, {J.\ = nY + a constant = n^ (ji/^ - C'-). .'. nx= \ ^ = cosh ~^^+ const. .'. 7/=Ccosh{nx + D)—Le'^''+Me-"'', where C, D, L, and M are arbitrary constants. Similarly, we have in this case J)='h^P''''^lf(>^''y- VI. Linear equation with constant coefficients, such as [The methods which follow are the same, whatever be the order of the equation.] Let T) be any solution of this equation, so that {D^ + aD' + bD + c)r]=f{x) (1). On putting y= T+rj, we then have {D^+aD'^ + bD + c)r=0 (2). To solve (2), put Y= e'*^, and we have p^ + ap'^ + bp + c = (3), an equation whose roots arepi^p^, and ^3. Hence Ae^''^, Be^^", Ce^^" (where A, B, and C are arbitrary constants) are solutions of (2), and hence Ae^'''+Be'^'^''+C'e''^'^ is a solution also. This solution, since it contains three arbitrary and independent constants, is the most general solution that an equation of the third order, such as (2), can have. Hence r=^e''''^ + 5e^2%Ce''^^ (4). This part of the solution is called the Complementary Function. Appendix. Differential Equations 371 If some of the roots of equation (3) are imaginary, the equation (4) takes another form. For let a+/3 \/-l, a-/3\/-l and p^ be the roots. = A e"^ [cos /3x + i sin ^x\ + 5e»^ [cos '^x - 1 sin /3.r] + 06"^' = e«=^ [J 1 cos ^x 4- -Si sin /3.i-] + Ce''^*, where ^i and Bi are new arbitrary constants. In some cases two of the quantities pu p^, pa are equal, and then the form (4) for the Complementary Function must be modified. Let p2=pi+y, where y is ultimately to be zero. Then the form (4) = Ai where A^, Bi are fresh arbitrary constants. If y be now made^ equal to zero, this becomes {Ai + Bix)e''^'' + Ce''^\ If three roots ^i, p2, ps are all equal, we have, similarly, as the form of the Complementary Function. The value of ?; given by (1) is called the Particular Integral. The method of obtaining rj depends on the form of f{x). The only forms we need consider are x'\ e^% ^^g Xx and e'^'' ^^g Xx. (i) f{x)=x\ Here, by the principles of operators, '^^Di + aD'^ + bD+c' '^'" on expanding the operator in powers of D. Every term is now known, and hence (ii) f{x) = e^'^. "We easily see that D''e^^ = X''e^. 372 Ap2>eiidix. Differential Equations = (.40 + ^1^+^2X2+.. O'^^'^ so that in ibis case r; is obtained by substituting X for D. (iii) /(a;) = sinXA'. "We know that D- sin X^= - X^ sin \x, and that Z>2'- sin \x={- \^y sin X^, and in general that i^ (i)2) sin X^ = Z' ( - X2) sin X^. = (i)3-aZ)2 + 6/)-c). _^,^^_^,^/_^_^^,^^^3 sinX.r 1 ( - X3 COS \x + aX^ sin X^' + b\ cos Xa; — c sin X.r) X2(X--6y^ + (aX2-c)2 (X3 - 6X) cos X.r — (aX^ - c) sin X.r X2(X2-6)2 + (aX2-c)2 (iv) /(.r) = e''^sinX.r. We easily obtain D (e*^^ sin X.r) = e''^ (D + /it) sin X.r, i)2 (e^st .sin X.r) = 6^^* {D+fif sin X.r, i)*" (e/^^ sin X.r) = ef"^ (-0 +/i)'' sin X.r, and, generally, i^ (/)) (ef^^^ sin X.r) = e*^^ F{D + ^x) sin X./;. Hence >? = ^ 3 + ,^2 + ;,z) + c ^^^^^°^^^ sinXjp, "'^ (i)+;i)3 + a(Z) + ;i)2 + 6(i> + /x) + C the value of which is obtained as in (iii). In some cases we have to adjust the form of the Particular lutegi-f Thus, in the equation {D-l){D-2){D-Z)y = e^-, Appendix. Differential Equations 373 the particular integral obtained as above becomes infinite ; to get the corrected form we may proceed as follows: {D-\){D-2){D-Z) _l 1 1 1 = _ Lt-e2«.ev« =._e2a:Ltiri+ya:+^'+...'l y=oyL 1-2 J = something infinite which may be included in the Complementary Function —xe^^. Hence the complete solution is y=Ae'+ Be^^ + Ce^^ - xe'^\ As another example take the equation (i)2 + 4)(Z)-3)y = cos2a\ The Complementary Function = ^i cos 2ar+5sin 2.r + C(e^. The Particular Integral as found by the rule of (iii) becomes infinite. But we may write 1 i? + 3 = - j^ Lt -p^ [3 cos (2 + y) .r - 2 sin (2+y) x\ - - YZ ]i, 4-(2+y)2 1^^ ^"' ^•'^ - ^ ^'" ^'^) '"' y-^ — (3 sin 2.r 4- 2 cos 'ix) sin y.r] = -^ U 347T,2[(3coB2.-2sin2..) (l-^~4-...) -(3sin2A' + 2cos2.r) fy^r-^+.-.j = something infinite included in the Complementary Function - — (3 sin 2a.' + 2 cos 'Ix) . x. 374 Appendix. Differential Equations VII. Linear equations with two independent variables, e.g. /i(^)y+/2(^)^=o ..(1), Fi{D)y+F^{D)z = (2), where D=~r . ax Perform the operation FiiD) on (1) and /a (2)) on (2) and subtract; we thus have {MD).F,{D)-MD)F,{D)]y = (\ a linear equation which is soluble as in VI. Substitute the solution for y thus obtained in (1), and we have a linear equation for z. i+^+«s=»l (U ^ + ^ + 2"^^-ol .(2), i.e. (Z)2+l)y+6Z>0=O,-| and i>y + (Z>2 + 2) 3=0.1 .-. [(i)2 + 2)(Z)2 + l)-2).6Z>]y = 0, i.e. (Z>2-i)(Z)2-2)?/ = 0. Hence (1) „dz ^^^+2.4e»: + 25e-'= + 3(7eV2a:4.3i>e-v2x=o, and hence we have the Vcilue of z. Cambritigt : PKINTED BY JOHN CLAY, JI.A AT THE UNIVERSITY PRESS,