zotero/storage/NVUFBC5F/.zotero-ft-cache

THE COLLECTED PAPERS OF ALBERT EINSTEIN
Volume 3
The Swiss Years:
Writings, 1909-1911
Anna Beck, Translator
Don Howard, Consultant
Princeton University Press
Princeton, New Jersey


Copyright © 1993 by the Hebrew University of Jerusalem
Published by Princeton University Press, 41 William Street, Princeton, New Jersey 08540 In the United Kingdom: Princeton University Press, Chichester, West Sussex
All rights reserved
ISBN: 0-691-10250-3
Publication of this translation has been aided by a grant from the National Science Foundation
Princeton University Press books are printed on acid-free paper and meet the guidelines for permanence and durability of the Committee on Production Guidelines for Book Longevity of the Council on Library Resources
Printed in the United States of America
1098765432


Contents
Publisher’s Foreword
Preface
List of Texts
1. Lecture Notes for Introductory Course on Mechanics at the University of Zurich, Winter Semester 1909/1910
The Principle of Relativity and Its Consequences in Modern Physics (Le principe de relativité et ses conséquences dans la physique moderne), Archives des sciences physiques et naturelles 29 (1910)
Response to Manuscript of Planck 1910a (Antwort auf Planks Manuskript), Archives des sciences physiques et naturelles 29 (1910)
Lecture Notes for Course on the Kinetic Theory of Heat at the University of Zurich, Summer Semester 1910
On the Theory of Light Quanta and the Question of the Localization of Electromagnetic Energy (Sur la théorie des quantités lumineuses et al question de la localisation de l’énergie électromagnetique), Archives des sciences physiques et naturelles 29 (1910)
On the Ponderomotive Forces Acting on Ferromagnetic Conductors Carrying a Current in a Magnetic Field (Sur les forces pondéromotrices qui agissent sur des conducteurs ferromagnétiques disposés dans un champ magnétique et parcourus par un courant), Archives des sciences physiques et naturelles 30 (1910)
On a Theorem of the Probability Calculus and Its Application in the Theory of Radiation (Uber einen Satz der Wahrescheinlichkeitsrechnung und seine Anwendung in der Strahlungstheorie), with Ludwig Hopf, Annalen der Physik 33 (1910)
Statistical Investigation of a Resonator’s Motion in a Radiation Field (Statistische Untersuchung der Bewegung eines Resonators in einem Strahlungsfeld), with Ludwig Hopf, Annalen der Physik 33 (1910)
The Theory of the Opalescence of Homogeneous Fluids and Liquid Mixtures near the Critical State (Theorie der Opaleszenz von homogenen Flissigkeiten und Flissigkeitsgemischen in der Nahe des kritischen Zustandes), Annalen der Physik 33, 1910
xi
143
144
207
209
211
220
231


10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
Comments on P. Hertz’s Papers: “On the Mechanical Foundations of Thermodynamics” (Bemerkungen zu den P. Hertzschen Arbeiten: ,,Uber die mechanischen Grundlagen der Thermodynamik“), Annalen der Physik 34 (1911)
Lecture Notes for Course on Electricity and Magnetism at the University of Zurich, Winter Semester 1910/11 (Einfiihrung in die Theorie der Elektrizitat und des Magnetismus)
Comment on E6tvés’s Law (Bemerkung zu dem Gesetz von Eétvés), Annalen der Physik 34 (1911)
A Relationship between Elastic Behavior and Specific Heat in Solids with a Monatomic Molecule (Eine Beziehung zwischen dem elastischen Verhalten und der spezifischen Warme bei festen K6rpern mit einatomigem Molekil), Annalen der Physik 34 (1911)
Correction to My Paper: “A New Determination of Molecular Dimensions” (Berichtigung zu meiner Arbeit: ,,Eine neue Bestimmung der Molekildimensionen“), Annalen der Physik 34 (1911)
Comment on My Paper: “A Relationship between Elastic Behavior. . . ” (Bemerkung zu meiner Arbeit: ,,Eine Beziehung zwischen dem elastischen Verhalten . . . “), Annalen der Physik 34 (1911)
Comment on a Fundamental Difficulty in Theoretical Physics (Bemerkung Uber Eine fundamentale Schwierigkeit in der Theoretischen Physik) (1911)
The Theory of Relativity (Die Relativitats-Theorie), Naturforschende Gesellschaft in Ziirich. Vierteljahrsschnift 56 (1911)
“Discussion” Following Lecture Version of “The Theory of Relativity,” Naturforschende Gesellschaft in Ziirich. Sitzungsberichte (1911)
Notes for a Lecture on Fluctuations (10 February 1911)
Statement on the Light Quantum Hypothesis, Naturforschende Gesellschaft in Ziirich. Sitzungsberichte (1911)
Elementary Observations on Thermal Molecular Motion in Solids (Elementare Betrachtungen tber die thermische Molekularbewegung in festen Korpern), Annalen der Physik 35 (1911)
vi
250
251
328
332
336
338
339
340
35]
359
364
365


22.
24.
25.
27.
On the Ehrenfest Paradox. Comment on V. Varitak’s Paper (Zum Ehrenfestschen Paradoxon. Bemerkung zu V. Varitaks Aufsatz), Physikalische Zeitschrift 12 (1911)
On the Influence of Gravitation on the Propagation of Light (Uber den Einflu8 der Schwerkraft auf die Ausbreitung des Lichtes), Annalen der Physik 35 (1911)
Excerpts of Discussions Following LecturesDelivered at 83rd Meeting of the Gesellschaft Deutscher Naturforscher und Arzte, 25 and 27 September 1911, Physikalische Zeitschrift 12 (1911)
Discussion Remarks Following Lectures Delivered at First Solvay Congress (1911)
On the Present State of the Problem of Specific Heats (Zum gegenwartigen Stande des Problems der spezifischen Warme), in Eucken, Arnold, ed., Die Theorie der Strahlung und der Quanten. Verhandlungen auf einer von E. Solvay einberufenen Zusammenkunft (30. Oktober bis 3. November 1911), mit einem Anhange tiber die Entwicklung der Quantentheorie vorn Herbst 1911 bis Sommer 1913. Halle a. S.: Knapp, 1914. (Abhandlungen der Deutschen Bunsen Gesellschaft fir angewandte physikalische Chemie, vol. 3, no. 7)
“Discussion” Following Lecture, “The Present State of the Problem of Specific Heats” (Doc. 26), 3 November 1911
Vil
378
379
388
391
402
426




Publisher’s Foreword
We are pleased to be publishing this translation of Volume III of The Collected Papers of Albert Einstein. While every effort has been made to ensure the scientific accuracy of this translation, it is not a literary translation and is not intended for use without the documentary edition. The documentary edition provides the extensive editorial commentary necessary for a full understanding of the source documents.
Recent advances in desktop typesetting are responsible for the enhanced quality of the typography in this and future translation volumes.
We are grateful to Dr. Anna Beck and Professor Don Howard for their efforts and dedication to this project. All translations in this volume were prepared by Dr. Beck in consultation with Professor Howard; all translations in the documentary edition were prepared separately by the editors of that volume.
We wish to acknowledge gratefully the continuing grant from the National Science Foundation which has made this publication possible.
Princeton University Press August 1993




Preface
This volume presents new English translations of all of the documents in Volume 3 of The Collected Papers of Albert Einstein, with the exception of the “Scratch Notebook, 1909-1914,” which is published as Appendix A in the documentary edition. The documentary edition presents twenty-four of these documents in German versions, the remaining three appearing in their published French versions. The translation volume does not reproduce the annotations or editorial apparatus of the documentary edition, which the reader should consult. We have, however, included in this volume the editorial footnote numbers that correspond to the footnotes in the documentary edition; they are placed within square brackets. Bracketed numbers in the margins that are preceded by a “p.” refer to pages in Einstein’s notebooks. Angle brackets indicate crossed-out material. For the most part, misprints and errors in the original documents have not been corrected, except for the occasional correction of misspelled names. The purpose of the translation project, in accordance with the agreement between Princeton University Press and the National Science Foundation, is to provide “a careful, accurate translation that is as close to the German original as possible while still producing readable English.” Therefore, our aim has not been to produce a “literary translation,” so style has been sacrificed to literalness in some places to enable readers who are not fluent in German to make a scholarly evaluation of the content of the documents. We hope, nevertheless, that the quality of the original German prose shines through. Some of the technical vocabulary found in the original documents is peculiar to the time and place of their composition. We have tried, whenever possible, to provide not modern translations but English equivalents commonly employed in the contemporary physics literature; otherwise we supply literal translations. Perhaps the most significant exception to this rule is our translation of “Spannung” as “voltage” or as “potential difference” (depending on the context), there being no one standard English equivalent in common use in the first two decades of this century. Similarly, we have reproduced all notations and equations in a form as close as possible to the original. Three documents in this volume presented a special challenge—the three sets of lecture notes (Docs. 1, 4, and 11). As might be expected, the style of these notes is often fragmentary and telegraphic. To the greatest extent possible, we have sought to reproduce Einstein’s abbreviations, repetitions, and errors of grammar and spelling, so as to preserve the feel of the original notes, except in the few cases where such literalness would have produced an impossibly unclear translation.
We would like to thank the staff at the Einstein offices in Boston for their help at various stages in the preparation of this translation. We owe a debt to Walter Lippincott, director of Princeton University Press, for his support and encouragement of this project. Alice Calaprice, senior editor at Princeton University Press, has provided invaluable assistance, for which we thank her. Our thanks go as well to Michael Perlman for his technical assistance in the preparation of the final camera-ready copy of the volume, and to Charles Creesy and Michael Volk for seeing to our computer needs.
ANNA BECK, TRANSLATOR DON HOWARD, CONSULTANT




Texts




DOC. 1 MECHANICS LECTURE NOTES 1
Doc. 1
Lecture Notes for Introductory Course on Mechanics at the University of Zurich, Winter Semester 1909/1910
[18 October 1909-5 March 1910]"!
Mechanics is the science of motion of ponderable matter. It establishes the conditions {[p. 1]
under which the motion of matter ceases (statics). It seeks to reduce the manifold phenomena of motion to the smallest possible number of elementary laws of the simplest possible form, from which it seeks to reconstruct the more complicated phenomena.
I. Mechanics of the Material Point
We shall first discuss the motion of a body whose dimensions are of no importance in the motions we will discuss, that is, can be regarded as © small. While in motion, such a body will, in general, carry out rotations and change its shape. But we disregard these
circumstances, that is, treat it as if it were pointlike; we designate it as a “material point.”
Before we investigate the motion of a m. p."! as a function of the motive causes, we must discuss the means and the auxiliary quantities that we use in order to describe the motion of an m. point.
A. Kinematics of the M. P.
One cannot speak of the motion of a body (and hence also of am. p.) in and for itself, [p. 2] but only of a relative motion of bodies with respect to each other. If we wish to describe the motion of an m. p., we must describe its motion with respect to a second body. For the latter we choose a system of 3 mutually perpendicular rigid rods. (Coordinate system). We conceive of times as being measured by an arbitrary clock, in that we assume that means are available for ascertaining the readings of the clock that are simultaneous with particular individual positions that the m. p. assumes during its motion. Obviously, the motion of the <body> m. p. is given if the coordinates x, y, z are given with respect to the c.s.!" as a function of time. Equations of the following type obtain here: ax
x=o() y= z=x) .
Ma uM!
Rectilinear motion —* >
4x W
(a) uniform


2 DOC. 1 MECHANICS LECTURE NOTES
x=a+bt
x + Ac=a + b(+tAt)
Ax =bAt —=6
(b) non-uniform
(p. 3] Arbitrary Curvilinear Motion
(Definition of mean and instantaneous velocities)
6)
ym M' W
/—\ tea?
Velocity is a vector (structure defined by magnitude, direction, and orientation). Graphically represented by an arrow of a given direction & magnitude. Usually denoted by German letters (e.g., A). Components &, 4, &..
Two vectors & = (1, A, 4)
and ® = (B®, B,, B,)
One speaks of the sum @ + B of these vectors. By this one means the vector
(+B, A+B, +B)
Geometrically The commutative law applies. In the case of several vectors, the associative and distributive


DOC. 1 MECHANICS LECTURE NOTES 3
laws apply."
Acceleration [p. 4]
If one divides these by At by wt
and passes to the limit, L&E
one obtains _ gt se
2
dx dy d&z )
d? dt dt A‘, wh?
‘a ot
ST Aether
“ i y) seach hevantpm
a!
begs 1 TES
& 4+
Hodograph”TM eet [mean acceleration passage to the limit 4 acceleration at the time]
0 +20


4 DOC. 1 MECHANICS LECTURE NOTES
[p. 5] Tangential and Normal Acceleration
There exists a particularly noteworthy way of resolving the acceleration vector into components, namely, the resolution into a tangential and into a normal component.
10 4+470
v e ”
a a
i
e e
d t
A N
7 ”
\
P p
Direction cos of N: a B y Direction cos of tangt a’ B’ y 4
(3)
d*x dv v?
epee ee ‘ds
dt? dt ew -- UHay
7 0< & aa ------- sy ih
+a av e
[p. 6] B. Dynamics of the Material P."!
1. <Galileo’s> principle of inertia a m.p. that is <present alone in space> not acted upon by other bodies! moves without acceleration. (a) In a certain sense, this law is an empirical law; (billiard ball, railway car). Strictly speaking, however, it has the character of a definition. For we say that other bodies do not act on a body if this body moves uniformly in a straight line. But for all practical purposes the law can be designated as empirical, because experience happens to be such that the law can be carried out or maintained without any artificial-looking assumptions."4)


DOC. 1 MECHANICS LECTURE NOTES 5
(b) But this law does not hold for an arbitrary state of motion of the c.s. But it holds to
acertain degree of approximation for systems at rest relative to the earth, and to an even
closer approximation for a system whose origin is at rest relative to the center of gravity
of the solar system, and whose axes are directed permanently toward 3 fixed stars.
er bodies act upon a material point, the acceleration vector Ee dy z is
d? dP df generally different from zero. The acceleration of a m.p. is called a force.
We have a certain direct representation of force, and thi T ing of exertion or
pressure that we experience when, for example, we use our hand to set into mot
body that was originally at rest.
2. The accelerations imparted by A to B, & and by B to A are directed along the connecting line and are oppositely oriented.
[p.7]
73.
3. The ratio of accelerations of two mat. p. defines the ratio of masses. Explanation of the empirical laws involved.
One mass can be chosen arbitrarily. The rest of the masses can be derived from it by
experiment. 4. The addition theorem for accelerations.
If one introduces a vector (x, y, z) that is equal to the acceleration of the point
multiplied by m, that is, if one sets
then, for two masses interacting with each other, this vector has the property of being
equal for the two, and of opposite direction. We call this vector the force acting on the
mass point. Thus, this force always fulfills the condition of equivalence of action and [p. 8] reaction.


[p.9}
{p. 10]
6 DOC. 1 MECHANICS LECTURE NOTES
The equations of motion given above have the character of definitional equations for the force, thus they can be neither confirmed nor refuted by experience. Nonetheless, we could find ourselves compelled by experience to abandon them, this would happen
if the description of facts by means of the equations m S = X -- would lead to our
t
having to assume expressions for the force components X - - - in a very complicated manner. One would then reject the equations of motion as unsuitable. Example: identical springs, stretched in the same way, act in the same direction upon
a free body. If the acceleration were not proportional to the number of springs acting,
then it would follow from the equations that the force would also not be proportional to the number of springs. This does not represent a logical contradiction, but it would
result in our presuming that we could arrive at a simpler, ie., preferable theory of motion, if we based ourselves on other equations of motion.
General Remarks on the Motion of the Material Point
For our equations of motion to be useful, the expressions for the force components Xx etc. may not contain higher than first-order time derivatives of the coordinates. Because the second derivative can be eliminated by solving the equations. However, the occurrence of higher derivatives would make a solution for the second derivative seem unjustified. Hence, for a general theory we have to consider X etc. as functions of
xy ... and? alone. We have then 3 simultaneous equations of the second order.
The general integrals of these eq. contain 6 arbitrary constants. For the motion is
completely determined only if, for a time t,,x-- and xyz are given. If X -- are
unique functions, then the solution is thereby uniquely det. For we can write
— = — X(x..%...0) ee dx=X( )dt dx=xdt
m at
Thus, if x --% -- are given for a time t, they can be calculated for the time 1 + dt etc. i es it proves possible to integrate the equations of motion once (first integrals), so that one arrives at firs i (1) The eq. of mot. can be written


DOC. 1 MECHANICS LECTURE NOTES 7
If the right-hand side can be directly integrated with respect to time, if X<YZ> vanishes or <are> is at least independent of x --x -- etc. Example. The force is everywhere parallel to a given direction. We choose the one parallel to the direction Z. Th
From this, x =a’t+c, y=b't+c, The motion takes place on a plane, because a dy - bdx = 0, ay - bx ® Remark: the above equation contains the vector (mx, my, mz), the v multiplied by mass. It plays a role in many derivations. We call it b = (6, We have
db, = Xdt_—b, = [Xat
Fhe momentum is equal to the time integral of the force acting on the body (material [p. 11]


[p. 12]
[p. 13}
8 DOC. 1 MECHANICS LECTURE NOTES
Analogous equations hold for the other axes.
<Free Fall. Force of Gravity> Practical and CGS-Unit of Mass
We measure time in <average> seconds, of the aver. solar day, and
1
24 -60 -60 lengths in cm. 1 cm is the hundredth part of the distance between two marks of a
specific meter-stick kept in Sévres near Paris.
Besides the quantities that depend only on length and time Ee | , our equation
t
of motion contains two additional quantities, namely m and X-- It suffices to establish
a unit for one of these quantities, because the equations of motion make it possible then
to establish the second one.
For if we have defined a unit for m, then we can define
mB =K
as the unit of force that force which imparts to the unit of mass the acceleration 1.
Conversely, if we have established a unit for the force, than the unit mass is that
mass to which the force 1 imparts the acceleration 1.
From the theoretical point of view it does not matter for which of the quantities we
will establish a unit, but from a pragmatic point of view it does.
<Earlier (before Gauss) one could>"! A unit for the force can be defined in the
following way in Paris the earth exerts a quite specific force on 1 cc water at 4°. I call
this unit of force 1 gram. (In addition, the kg is also used.)
This definition suffers (for precise investigations) from the following drawback. If
people who are not in Paris wish to measure a force accurately, they must compare the
force to be measured with the force that terrestrial gravitation exerts on 1 cc water in
Paris. To this extent the application of the definition is cumbersome.
But the situation is different if the unit of mass (also called “gram”) as the mass of
one cc water at 4°. In this definition, <which now in physical investigations> no
particular location on the earth plays a role. The mass 1 gr can be realized at once at


DOC. 1 MECHANICS LECTURE NOTES 9
any location at which a cm-measuring rod and water are available. For this reason, this system of measures seems now generally to be used in physics. The advantage over the other system is only a formal one.
x Free Fall At a place close to the earth’s surface we imagine a coord. sys. whose Z-axis is directed vertically upwards. We inquire into the motion of a material point with respect to this system. In order to solve this problem, we must know the ~y Magnitude of the force exerted by the earth on the <materi| al> body. One would expect a priori that this force
1) is proportional to the <mass of the> m. p. 2) depends on the physical quality of the point 2 3) The force could also depend on the velocity.!"“!
<For reasons of symmetry> From the choice of the position of the coordinate system it follows that [p- 14]
X=0 Y=0
Further, one arrives at a correct description of the phenomena if one assumes that gravity does not depend either on the quality or the velocity of the m.p. In this way one obtains
fl 0
df
dy 9
d?
aazvi
From the two first equations one obtains


10 DOC. 1 MECHANICS LECTURE NOTES
a" dy
dnyn? dx
ady —bdx =0
ay—bx=c
The motion takes place in the vertical plane. We choose this to be the x-Z-plane. In that case, we have permanently y = 0, and we obtain by direct integration of our equations
x= ct + Cc, dx ot Cc,
dat
z - 3¢ +o,t +e, é =c¢, + gt
[p. 15] We shall now assume that fort = 0, x =z =0 and a. e = 0; in that case all
c = 0, and we obtain
The familiar formula for free fall. By calculating the constants c from the conditions of the problem, we can solve each problem concerned with free fall. Example: On a hill of height A is a cannon whose elev. is «. The initial velocity of the shell is vo. Where will it strike? The most favorable elevation angle? (Air resistance neglected).


DOC. 1 MECHANICS LECTURE NOTES 11
Example 2: Water pipe Equation of the curve.
<For> The fact that the gravitational force is independent of the material <we have no explanation. > shows a close relationship between inertial mass, on the one hand, and the effect of gravitation, on the other hand.!*! Let us now find the law of interaction between <masses> the sun & the planets [p. 16] through <gravity> gravitation, the way Newton found it from Kepler’s laws. These laws of Kepler are as follows: (1) The radius vector sun-planet sweeps out equal areas in equal times. es (2) The planet travels in an ellipse, in Be which the sun occupies one focus. . (3) The squares of the planets’ peri- “ t ods of orbit vary as the third power of the Mer major axes of the ellipses. The nature of the present problem re a makes it seem expedient to use polar coordinates for the description of planetary motion. In order to apply these, we will seek to express the acceleration vector in polar coordinates. First, we have
xX =rcos »


12 DOC. 1 MECHANICS LECTURE NOTES
eps dy _ «dr , do
y=rsin @ a sin oF, r cos FF
. d?x dr . arde do\?
17 —sin 0: — = —— maripueae aaa
[p. 17] 9g COs ~ dt = SP a2 2sing a. dt reoso(“?)
2
—rsing Ge
. d*y | dr dr dp do\?
co: +> = —- — — —fsi ae
Sp sing a SIN 9 2 + 2008 97 dt rsing (22)
2
+ reosp
d? d’y .
B, = 2 O89 + “7sing
d*x . d?
B, = — Fring + G3 c0s@
d?r do 2 Ap ner ed r dt :
drdg _ dp 1d(,dg Fi
B, = == —- = 205 \
dt dt" dt? rat\" 4)
a
From (1) it follows that r-rdp=cdt & rae =¢ x
From this it follows first that B, = 0. Thus, the acceleration vector of planetary motion lies in the direction sun-planet. We now calculate B, by means of (1) and (2) Because of (2) we have


DOC. 1 MECHANICS LECTURE NOTES 13
_ P dr p cee AP 2. do 1—ecose dt (1—ecosg) esin Oa > nar’
ad csin
dt ?
Gat ccos
dt? a)
_r—p | d’r_c?r—p4
80 er | dep
dp _ c do ? iG
tor? Na) ~B
ec 1
fr ep re
const. [p. 18]
The acceleration imparted to a planet = There is still the question whether
r
this constant has the same value for all planets. To find this out, we must introduce the period of orbit. We have
The area of the ellipse = hef = abn cel
But since us =p
a


[p. 19]
[p. 20]
14 DOC. 1 MECHANICS LECTURE NOTES
3
But since according to Kepler’s 3d law @ “has the same value for all planets, we can
“3
set the acceleration] caused by the sun = 5 Ad Force with which the sun attracts
a planet = mass-accel. = uf , where f a factor that is independent of the planet. For
Pr
reasons of symmetry, the numerator must depend on the mass of the sun M just as it depends on the mass of the planet, hence f = Mx, where «x does not depend either on the sun or on the planet. Hence we have
Mm _M
force = K:__ or from accel. = k
P Pr
There is still the question: What is the value of constant x? In order to find it, we must know for one case both masses, the force, and the distance. In order to determine x, we must know the magnitude of the acting mass. This is only possible for relatively small masses. (Example of this. Earth as gravitating center mass of the earth)
Tt has been found that « = 6.70 - 10°°.
More about the methods later on."®
4 20°
B=67-10°*--7
10°85 555 13.6
7 6.7-1078-8- 10! - 13.6- 3.14 3
_ :i
=7,6-104¢ 
® m weniger als i000 TMTM
. 10-7 der Schwere
‘
6 - 1 9 *
2 , 5 - 2 0
Q O
" 2 0
6 - 1 0 8
RdRdg- Rsin p<dy)2n
p cos y/!191


DOC. 1 MECHANICS LECTURE NOTES 15
. 2nR? si
Anz[iehung] a cos
bd i d
{ = Kmp2nR? (*s P cosy
au
u? = R? +r? + 2Rrcosg?TM
udu = Rrsingdg
d
sing dg = “a
r? + uy? — R?
cosy = .an
1 [du
ae (Se a R? + u’)
__! J2_ pry | a!
= ape Re) [Sa dub = oe
11
{ (r? — R?)
7u
— —(p2 — R2 _fr ryt r—R
= —(2R) + 2R
+ut konst}
- > ba@ Rr Re


{p. 21]
16 DOC. 1 MECHANICS LECTURE NOTES
Integrating the equations of motion between two time limits 1, and 4, one obtains
etc.
determined change of m Lind "
dt?
components of the momentum (mv) of the m. p. Conversely, we see that the total effect that a force lasting for a certain time has on the state of motion of the m. p. is
General Remarks on the Motion of the M. P.
1) mi =X
dt®
mo? =Y
dt?
moe =Z
dt
Thus, the time integral of the force acting on a m. p.
determined only by [Xdt etc. (impulse. 2) The momentum <motion in a plane>
dx
mi =X =
dt 4
mit? - yt
dt’ dt
Fee! =Y | tx
dt
But since x—d=y ye d dy _ yds
de “de dt\dt ~ dt} ’
causes a completely
of the material point. These quantities are called the
=xY -yX .


DOC. 1 MECHANICS LECTURE NOTES 17
one obtains
d dy dx
= — -y—|b =x¥ - p¥
dt { E dt yo calli
d ge po) =yZ -zY
dt dt dt
d m Pci ~x% = 2X -xZ
dt dt dt
d dy
al — -~y—b =xY - yX
dt {m ie "a I aire
If the right-hand side of one of these equations vanishes, i.e., if - = ,ie., if the
force intersects the Z-axis, one obtains an integral.
We have then xf - ye = const.
dt dt
foe , alia dx)
2 xt+dx ytdy 2 u 2
ds 1f dy dx
dt = 5(x2 ond yo) = konst.
The areal velocity of the rad. vect. of the x-y projection is const. The reverse also holds.
ya
If central force, then 3 integrals, because all three right sides are then = 0. Then
y& _, =A
dt dt
Phas agt =B
[p. 22]


Ip. 23)
{p. 24]
18 DOC. 1 MECHANICS LECTURE NOTES
Mult by x y z and add. Ax + By + Cz = 0. Equation of a plane passing through [sic] through the O-point.
Geometrical Interpretation of the Law of Areas
Given are two vectors @ and ®, with the components &, @, @, and $B, B,, B, In addition to these two vectors we can construct a third vector » in the following way:
1. v is perpendicular to the plane laid through @ and B
2. The quantity, or (as it is called) the tensor”?! of the vector is twice the area of the triangle that is to be constructed from A & B 3. The direction of » is such that a rotary motion of @ toward B, together with a translational motion in i the direction of the arrow of » leads to a right-hand coil
This vector v is called the vector product of @ and B. Components of the vector product. The plane (43) is 1 to t. Hence the angle between b and the Z-axis is equal to the angle between the plane @® and the plane XY
jo] = A
, = Acos b, = A,,, where A,, denotes the area of the proj[ection] of A on thexy plane.
b, a A,B, 7 AB. Ne f
Analogously for the other two components of. ’[EA = We consider the special case where one of the vectors is the radius vector drawn from O to the point of application of the other vector. The vector product of the radius vector and the given vector B is called the moment of the vector % with respect to the point oO. It is perpendicular to the plane OB, and its tensor is equal to twice A, and thus to the product of the magnitude [%| of the vector and the latter’s distance d from O.


DOC. 1 MECHANICS LECTURE NOTES 19
According to the aforesaid, the components are
yB, - zB,
In the equation developed above, components of moments appear on both sides. On the left, the moment of the momentum, on the right, the moment of the forces acting on the mp. Ina central motion, the moment of momentum, and thus also that of the velocity, is a spatially and temporally constant vector.
The Law of Kinetic Energy
d2x iS where dx is the projection of the element of the me a ant = dx, trajectory.
dy dy
d?z dz
dx d?x dyd*y dzd?z
o-(2) +)
dt dt dt
dxd*x | ; a ae tt
dv?) = 24
From this A | = Xdx + Ydy + Zdz
mTM is called kinetic energy. The right-hand side is the product, resultant force 
path element - cos of the angle between the two. Because one can write it as
[p. 25}


20 DOC. 1 MECHANICS LECTURE NOTES
R-ds ER LID, Zk = Rds cos @ ,
Rds Rds R ds
where the bracketed fractions are direction cos. of R & ds.
ote
This is the work that the force X Y Z transmits to the mat. p. during time dt. The law can also be derived directly, by resolving the acceleration into a tangential & 1 component, B, and B..
dv
. 26 B=
[p. 26] 7
F ds = mv at -a{m
‘ dt
Integrating the equation obtained, one gets
2
mv —MV— = [' Xd+xYdy + Zdz .
road, 7
In the special case where X¥ Y Z depend only on x y z, it is possible to calculate the integral on the right-hand side if the trajectory is given. But in an even more special case, namely if X Y Z are of the form
+U wy
ox
X=


DOC. 1 MECHANICS LECTURE NOTES 21
Yontaedey
Zu +U |
oz
it is not even necessary to know the trajectory in order to carry out that integration. <Such forces we call forces derivable from a potential> For in this case we have
[Xax + Yay + Zaz = (Fe + yWyy . a +fau
+(U - U,)
2
My,
In this case [rm + aE> + v] = const., where U is, thus, a function of the
2
coordinates alone. If the m. p. turns up twice at the same spatial point, then P, and thus also v, will have the same value if P is single-valued.! The temporally constant [p. 27]
quantity that we found here for the case when a m. p. is under the influence of temp. const. forces that are derivable from a <potential> force function, we will call the “energy” of the system considered. —P is called the potential energy. The law we found can then be formulated thus: “The sum... . remains constant.” As the condition for the work [ to be independent of the integration path, we found the equations
This condition can also be expressed in another form. If one differentiates the third d. eq. with respect to y, and the second with respect to z, one obtains
eZ _ a _9 ro a 4
analog. ax _ 92 | 0


22 DOC. 1 MECHANICS LECTURE NOTES
An example of multivalued UU = arctg ¥x
17% #
xd-yydx r+y
xX + yY = 0, thus the force is perpendicular to the radius. Its
Y= = magnitude is Z .
Pr r
“TA
{K NV >’
[p. 28] Equilibrium of the Material Point
If force function present
aU _ aU _ wu |
a dy a
X=Y=Z=
The law of kinetic energy reveals a case in which it is certain that the equilibrium is stable.
.
v + P =
o Y
+ P ,
— _
E q u i l i b r i u m
a t
P , ( X %
y o
2 0 )


DOC. 1 MECHANICS LECTURE NOTES 23
=P, +e
my +P-P=e
2
P-P,se
Central Forces That Depend Only on the Distance
We have seen that motion takes place in a plane. 1. Law of areas
dy dx |
x2 -y— =e
dt dt
do
or also P= =c 1
; (1)
2. The energy law. ; 7
om] g [ae + Ydy + <a]
r
=Fdr (2)
These two equations determine completely r & @ as funct of time. We have
y? a dr + rd¢
dt
From this we obtain by means of the law of areas
v= (Z] BS 3)
Ip. 29]


24 DOC. | MECHANICS LECTURE NOTES
1¥
vec | +3 4)
By substituting 2 IF dr for v*, one obtains dt and dq as funct of r.
m
Now we write (2) in the form
ldmv? _ ,dr
2 dt a
d|m|\ (dry ~COl pt
dt\2\\at) PF dt
Differentiating,
migdr _ Cd _p
2| ¢@t dt P gt t
& mit Clap
d’ Pf
& dr =T ee oo (5)?!
dt Pr
2
[p. 30] From ldmv* pdr
2 do do
one obtains, by inserting v’ from 4


DOC. 1 MECHANICS LECTURE NOTES 25
a!
d imC Hr 1V\$ _ par
—— |} ] + fe =F
dp | 2 \lde) (7 dp
al
mC) 2 7 _ 297 _2 Pig ag reo )
ee 1
pe-m@) rile 8 6
P log r
We now determiner& as funct. ofr. We set 2 fF dr = p(r) +h =v’ according
m to eq. 2 Then eq. 3 becomes
e dr
Wo =o) -Fth]es
dt = dr
tyr)
Suppose we know the sign of a for t =f,
The sign of the square root is thereby determined up to the moment when =a
f
becomes 0 again. Then a usu. changes sign (at r,). This is generally easy to detect
in the special case under discussion. It can be determined unambiguously from the sign
2
of met apg ME
dt’
F r o m
t h i s
i t
i s
e a s y
t o
d e r i v e
d q ,
s i n c e ,
a c c o r d i n g
t o
t h e
l a w
o f
a r e a s ,
[ p .
3 1 ]


26 DOC. 1 MECHANICS LECTURE NOTES
Insert (should come before “Central Forces.”)TM!
Let us discuss the case of two mat. points acting on each other by central forces that depend only on the distance.
x, ~ %,
dx.
m, = FO)
v
-- aot (ee ryt
These equations do not change their form with the introduction of a uniformly moving coordinate system
Second derivat. & differ[ence] x, - x,
‘ ' ' ' i ! ' '
i!] 4 1
1
1 =t _e
1%, =X +art+B, x, =x,’ ta +r B,
‘ ' ' \ ‘
; do not change under transformation.
\
“
Neel ere
‘de * dP
gam + m2X2
7 m, + m2
/ mx, +m x, = at +B n=
é=
In order to interpret, we define


DOC. 1 MECHANICS LECTURE NOTES
the so-called center of gravity of both masses
— =(m, + m,) (at + B) moves uniformly
New coordin. syst. that mov. unif., hence is at rest relative to the cen. of grav. O placed at the center of gravity. Then
m,x, + m,x, = 0
or mx, = —m,xX, My ty 2y My Vy + 2
my, = “My, sq. & add
pas mr = Mr,
m1
r= —r
21
my
m, +m,
rerntrne= r, m,
This is the same equation as the equation of motion with a fixed center of force.
Application to the Solar System. Sun & Planet
Force F(r) =« = K—r_
mm, mM
re


28 DOC. 1 MECHANICS LECTURE NOTES
3rd Kepler’s law not strictly valid. Neither is the second law.
Example with regard to central forces Force law between two identical gas molecules
F = 3 Collision law"!
B
a1 Tila) = G55 = F
Problem reduced to a central-force problem.
ee ee ee
gr) +h = faa ap atk v


DOC. 1 MECHANICS LECTURE NOTES 29
If v, denotes the velocity at distance, then fh = v,”
wn =e +h -f = -28.1 fy [p33]
Pr m5 fA Pf
dt = f dr
tv Here c = +bv,
dp = ef dr
Pv Wr)
dr
@ = 2a = 2c? “_ Py Wr)
Example. Mass penetrates into the solar system, branch of hyperbola. What direction does it have afterwards?!
‘Ss |4
From the last equation (6) it is very easy to derive the force law from Kepler’s 2nd law.
gel
me4 or 1
aa
2 |dg@ r
1_l-ecsg
rP
1 wl P
ft Secale
o¢ P


30 DOC. 1 MECHANICS LECTURE NOTES
[p. 34] The Motion of a Point That Must Stay on a <Plane> Surface
mo* =X +X,
dt’ a f
m =e¥ +Y,
dp * Ff
mez =Z +Z
df? = Z
surface exerts on the point a the components (X; : Y; : Z)) v:
nterpressure that is normal to the former. In that case as the direction cosines of the normal to the surface.
Se, 28: 28 , hence x,-a2e
@ (x, y, Zz, t) =
Together, these equations determine the four variables x, y, z\and A. If surface at rest & forces derivable from a potential, then the\gonservation of energy holds. Examples: Simple Pendulum Z-axis downwards Equation of the surface
{p. 35] 89, 99. 99 _


DOC. 1 MECHANICS LECTURE NOTES 31
ence equations
& substituting 2a instead of A
m
oie A string tension 2A’ = mal
ae gta.
We need two relations for a complete solution. 1) Energy principle. Because xdx + ydy + zdz = 0
(8 (a -m
& ade + dy’ + dz = (Qgz + hd?
xdy - yd=xedt
Because the distance from the coordinate origin is const., it is advantageo' polar coordinates.
to introduce
x =lsin0 cosw dx = I{cos? coswd?d - sind sinw dy y=lsindOsinw dy = l{cosd sinwdd + sind cosw du} z =I cosd dz = l{-sind-dd}


Ip. 36]
MECHANICS LECTURE NOTES
32 poc. 1
de +dy +de=Ft
xdy - ydx = Psin’d dod
Inserting this into our equation, we obtain
Pid? + sin’ddei} = (2gl cos® + hyd’
Psin? dds” = cdt.
Motion of a Point Along a Given Fixed Curve
§ 1.
Thus far we have been addressing the problem of
finding the force when the motion was given, OF
finding the motion when the force was given. But
there are problems in which conditions for the motion
are given. Imagine, for example, a small perforated
body pulled along a rigid wire and acted upon by given
external forces. Besides the given external force, 4
reactive force of the wire, to be viewed as unknown
for the time being, also acts upon the point. Aji that
we assume for the time being about this reactive force
& is that it is perpendicular to the tangent on the wire,
so that we will have
R dx + Rdy + dz =
%..
vw
\F
0.
This implies that the reactive force does not perform any work.
To find the mot. of the p., we can replace the wire by the re
it. Formally, this reduces the case of the point pulled
freely moving point. We can therefore set
ax
mor =X + &,
dt
mo =Y+
dt® y
a&z
mo =Z + &,
dt’
along the wire
:
active force exerted by
to the case of the


DOC. 1 MECHANICS LECTURE NOTES 33
We can evaluate the condition assumed for & by multiplying these equations by dx dy dz [p. 37] & summing, & we obtain
2
d me| = Xd+xYdy +Zdz (1)
Thus, the equation for the kinetic energy is here valid and is sufficient for the solution of any problem of motion, as can be seen from the following. A single variable (q)""! suffices for the descrip. of the motion of the point. x, y, & z are to be considered the
2.
given functions of this single variable s. First, we have v* = (Z]
a , f ds?
ey ae
Further, we have
dx dy dz ds
Xx 1, Z, —, —, —, F], th ae : rt bd
x2 EE us al 5,¥—y2". (s =]
Further, we have to set (as known,
x=@s) y=xs) z= Ks)
Further,
ds?
v= (g!2 + "2 2 as"
(pe +x? ey’) 7
X dep.on xyz, & . . andt, Hence, since one has to set x = »(q).. (known func.
of s), “ =q’ @. Thus, X a known func. of g and da Thus, the above
dt
equation yield diff. eq. for s. We can write above equation as
2
aye? + x? +? (Z) = (XQ + Vy’ + ZW }ds = Qds. (1’)


34 DOC. 1 MECHANICS LECTURE NOTES
where X, Y, Z are to be thought of as expressed in the new variables. If X, Y, Z depend only on the coordinates, then { } of the right-hand side (Q) depends only on s. The equation can then be integrated right away. <If we set [Q ds = f> one obtains
Mya yh = Joao or (solved for v* = Sf, (Z] = fg) (1"))
<From this, f - 1, = f dq >
+ fl?
[p. 38] §2 Geometric Derivation of the Fundamental Equation
We resolved the acceleration into normal and a tang. components.TM! We resolve the total force R acting on the point in an analogous way.
_ a = K<@>
B,- = R, = Kr _
analogously, total force R.,,, = Ke aN
B-zkt We
a R,=0=K,+N'.,.
| — From the tangential components
| mB, = K, ....08 (2a)
K, is gen. known as a func of sf & t. Mult both sides by ds = vdt, we obtain
deve 1) Ki
ne (ds
& d (5 = Kds_ integrable if K, dep[{ends] only on s.
The total force K is composed of the external force K” and the reactive force of the curve of magnitude N this is perpendicular to the curve & just like K, is taken as positive with respect to the center of the curvature. After the solution of the motion problem, the normal components yield the reaction of the curve. We have


DOc. 1 MECHANICS LECTURE NOTES 35
jue = Ke@? +N (2b)
Pp
& N= ve - K® (2b) If’ is found as a funct of s, then this eq. yields norm.
p reaction.
§ 3 [p. 39] There Exists a Single-valued Force Function. Physical Meaning.
We return to equation (1)
a5" = Xdx + Ydy + Zdz
We have already seen that this eq. is integrable if X, Y, Z depend only on s. We now further assume that there exists a force function for X, Y, Z that only depends on x y z, so that
<Since v= é , therefore> The solution of the problem is obtained from eq. (3) by
a single integration. Let us add here a general remark. Suppose the force X Y Z derives from a system of bodies that does not experience any spatial or other kind of change during the motion of them p. <If <force U> dep. only on the position of the m. p., then> What does the existence of a single-valued force function mean in this case? Suppose the point moves, perchance, along an endless wire without changing the sign me o of its vel. Then ” is always of the same magnitude at the < same location otherwise mechanism for the construction 6 of perpetuum mobile Forces exerted by unchang. sys., which must depend only on the position
eo
—_


Ip. 40]
[p. 41]
36 DOC. 1 MECHANICS LECTURE NOTES
<5> Gravitation z-axis upwards
a
Then Xdx + Ydy + Zdz = - mg dz U = - mgz v2 v2
I = -mg +h' a = gz +h’
If we write A =a
v’ = 2g (a - z) Vv,” = 2g (a - z,) thus a can be made arbitrarily large.
4%
Suppose we lay on a plane z =a, then this plane either cuts the curve or lies above it.
1) @ -z cannot become negative, hence, in the first case, the curve cannot cross the plane z =a. The material point turns around at z = a, but at no other point, because at no other point can we have v = 0. 2) If z= a lies above the curve, then a - z is positive for all points of the curve. Then the point travels without turning around. Ist Case thus shuttling to & fro between A and A’, with the velocity being the same at each point of the curve. We calculate the time the mobile needs from M, to P.
vw _ ds = 2g(a - 2z) [32} dt?
1 fP ds
2B i, ty a-z
In the integralz & s are related by the equation of the curve.
dt =


DOC. 1 MECHANICS LECTURE NOTES
§ (5). Example. Circle in a Vertical Plane (Simple Pendulum)"
v = 2g (a - z)
Constant from velocity at the lowest point
vo =2ga+D a=-l+
Ist Case. z =a intersects the circle
2
isos! v < 2ylg
We set z=-lcos? a=-lcosea
ya _ 1
dt ot (1 — cosa) — (1 — cos 9)
2
P (4) = 2gl(cos 9 — cos a)
dg\? o A)
or patel = in? _ — ein2
(22) 4g {sin 37 sin 4
(5) 9 a2
ie : if | :
dt = 7t=
| sin?S — eee Pd fine® — sin?9
22 22
. . a _ oO 3
sin, = usin, a/ =sing/1l—u


38 DOC. 1 MECHANICS LECTURE NOTES
. oO . oO . a
3 sin; : sin du sin 5 du 5=
_a where * =
ff = { . du if K infinitely small
Po Jo S(t — u?)(1 — xu?)
ae = arcsinu
sin 
[p. 421 7-2 stitn( 41)
sin = sin’stita( /2e)
cos = ./1—x’s[i]n( ) = au( |e)
The duration of a simple oscillation


DOC. 1 MECHANICS LECTURE NOTES 39
[ez du -K 4 ova — wa - ee)
We develop T as a funct of x.
1 _ 2,,2\-1/2 _ 1 22
ae =i+3ku
1:3----2n—1
2 ptyFt a ce eee 2n,,2n
+5 are + 240-9, *
i 1? 4a (1-3\? 4a
=m [lr +() sin “+ (73) sin 2
2
2nd approx[imation] 1 + F1i6ns
2nd Case
v
==> I
2g
2
2 =) = 29(a + Icos9) = 2a(a +1-— 2isin*?)
= a. 48
= 2g(a+ n(1 ra jain 4
XY


DOC. 1 MECHANICS LECTURE NOTES
a.
a +1 _ __iel _
du At = , Ja u?)(1 — x?u?)
ir=2["=xfi+(5)e
(3
ye
(p.43] 3rd Case. Limiting case
d§\? $
2 — — — 2=
l ( i) 2g(!+ lcos3) = 4glcos 5
3
2g
d in>d =
7 7 One <sin)> 5
I cos 3, I 3
g,_ © 2x
fie togto($ +7)
S t r i n g
t e n s i o n
R ,
=
K ,
+ N
mM
N= TT + mg cos $ = "92a — 3z}
we 2g(a—z) ~ to discuss!


DOC. 1 MECHANICS LECTURE NOTES 41
§ 3. Approxmative Treatment of the Pendulum Problem
i me ALD, =) —regh0 7
iand_ = -gt
dt? 8 h
A
or, if we introduce the abscissa x, d/ = x 
MES
dt? l
Solution A sin fe + B cos fs , as differentiation shows at once This can be
r e f o r m u l a t e d
V A ’
+
B ’
a a
d +
3
c o s
p
=
W
s i n
| e

;
~ ~
”
T = aft 8g This is also the form of the general solution. Graphic representation
Rotating vector. X, Amplitude. |e +8
Phase angle. 8 = phase A for t = 0. This graphic representation corresponds to calculation with complex numbers."
[p. 44]


42 DOC. 1 MECHANICS LECTURE NOTES
If G(x) = 0 is a homogeneous linear diff. eq. in the variable ¢ with real, constant coefficients, which is solved by means of the complex numbers a(t) + jP(t), where a & 6 are real, then we can write symbolically
G(« +jP) =0
Now, the real funct. remains real when it is diff, and the im function remains likewise im when diff. It is therefore easy to prove that
G(a + jB) = G(a) + jG (B)
Thus, the equation G(a + jB) = 0 is equivalent to
G(a) + jG(B) =0 & to the two equations
G(a) = 0 and G(f) =0
Thus, if we found the complex function a +7 that satisfies the eq. G = 0, then its real component also satisfies the equation. Application to the prev examp
adtx?
We seek the solution of the form e” Inserted
I t1
,
Solution e? Real part cos {e
+ Sx = 0 is linear eq.
Since the starting point of ¢ is arbitrary, we thus arrive at the previous solution.
[p. 45) 2nd Example. Infinitely small pendular oscillation with friction
moe = més x Re
dt l dt
ax Ra& 8. _9


DOC. 1 MECHANICS LECTURE NOTES 43
2
eTM solution a? +
RR,
Re“) =e TMcos¥ t =e“ cos ae
« determines damping
me ley = fe fe -2Leeby-dtg)
vee]! b205]
The effect of friction on oscillation is of the second order.
Brachystochrone
ds\? a x
(4 = 2gz + (h) h=0
Ve)
B
./ 2g dt = | <, Minimum A/Z <
Motion of a M P. Relative to a Fixed or Movable Plane [p. 46]
fx y, 2, t) =0
Reaction 1 to the plane, hence proportion to


44 DOC. 1 MECHANICS LECTURE NOTES
FS of
ax’ ay’ az
f
- of a
Bi ee
These 4 eq. determine x, y, z & A completely.
Example Plane rotates with const. angular velocity w about the z-axis. How does a point move on it? sin wt COs wt
Eye g
f=xsinwt — ycoswt = 0
:2 1
: D4 :
cos : mz = Asinwt
: d? t
sin : mo 7 = —Acoswt
nome =0
z linear function of time. Assume special case where z = 0. dt’ J
Z cos wt + dy sinat = 0
dt? dt?
dx dp .
x = pcos¢e = pcosat A gee
. . dy dp.
y= psing = psinot sin + wp cos
at dt


DOC. 1 MECHANICS LECTURE NOTES 45
d*x dp dp
qe = yr 8 — 2o7, sin — w’pcos | cos | sin
d*y dp. dp 2acin | sj
qe az in + 7a es —w*psin | sin | —cos
Tt
dq*ep oP =0
e=e", Thene? =? a=+0 [p. 47] p = Ae+TMBe“ @ = wl. It remains for us to find A. From the two equations
dp a
A= m|sin af FE _ ccooss eosp2D = -208P SF = -208 {{4eA'e - Bee“ }
If B = 0 logarithmic spiral.
Special case: curve at rest. dx dy dz is then a line element on the plane, so that also
S x +.4+.=0
ox
ne a .
Multiplying the eq. by dx = a etc., & summing, one gets
4 ‘| = Xd+xYdy + Zdz


[p. 48]
46 DOC. 1 MECHANICS LECTURE NOTES
Further, in the special case when X Y Z are derivable from a potential, then integrable.
2
m_=U+h
2
ie., <there exists> one can give an integral equation (law of the kinetic energy.)
Example. Spherical Pendulum
Point remains on the sphere f = 1? - x? + y? +z? =0
mek = -2Axr >
dt
may =-2dy 5
dt?
mz = -2Az +mg 
or, if we introduce on =p,
m Px _
a BX
ay __
rT = —py
Parzy = “yz +8
e small oscillations z =7/ #2 =
dr
Thus, from the last equation yp = g
The other two equations become:
dx _ 8,
dt’ 1
dy _ 8
we oP
x & y components behave totally independently of each other. We obtain


poc. 1 MECHANICS LECTURE NOTES 47
x ~A cos | sin F _ sin + BY cos
y =A'cos + B’ sin = -av sin +BY cos
There will exist points of time at which velocity is perpendicular to radius vector. We count the time from such a point on, and choose the X-axis such that the ZX plane passes through this point.
Thfoernt=0 y =O and & =0 —___t4
A’=0 B=0. thus
x=A cos |
y=B wn |
22
Thus, (3) + (5) =1 Ellipse
Oscillation period = a |b &
Oscillations of Arbitrary Amplitudes [p. 49]
We have <U> = mgz | mx = mgz + const. , hence
Vi=Ige th (1)
ax uy i -y
drt* :
d’y :
—2 =-py i x
dv? % :
Since neither the external force nor the reaction have a momentum with respect to the Z-axis, the law of areas is valid with respect to the xy-plane. In fact; if one mult. the second eq by x and the first by -y, & adds them up, one obtains


48 DOC. | MECHANICS LECTURE NOTES
x) - yt* 0
dt dt
ga ae c_ (law of areas) or also
re =c; if one setsx? +y’? =7? & fis the A betwetehen x axis & r
We choose r, 0, and z as coordinates. v is to be expressed in them. We have
at
yp ed? _ dP + Pdt? + dz
2 2 , ee
dt dt a tPA
so that equation (1) becomes
Grr ate th —_ (’)
We have further 7 +z? = @--- -(3) By eliminating ® and r with the aid of (2) and (3), we obtain an equation betweenz and
ft.
rdr +zdz =0; dr= tind
12 — z
dt cdt
do =- = _ =
— r? [@-2z
[p. 50] Inserting this in (1‘), we obtain:
P dz? = [(2gz + h\(? — 22) — C?] dt?
ye)
Idz aaee (1")
+y Wz)
never changes its sign & never becomes zero. In contrast, 
dt =
Because of (2), 
becomes zero if p(z) = 0. Only values between -/ and +/ come under consideration.


poc. 1 MECHANICS LECTURE NOTES 49
Positive values of (z) are certain to exist between them, because otherwise (1”) could not be satisfied. Thus, there are at least two zeropoints between them, and no more than two because is of the third order. We call them a and §. Thus, the m. p. moves always to and fro between two <points> planesz=a & z=. To traverse the space between two specific horizontal planes, it always requires the same time.
From 2) dd = ee ee
7 Pp -2Wu@)
The A descr[ibed] between two planes is thus also always the same.
6 & 3 for the case where «
« © --2a)m VuWzw) & B are positive
Laws of Motion Relative to the Earth!” (p. 51)
d*x .
—sin | cos | m 7 - =X, X, etc. shall be indep. of earth’s rotation
es a xX, * d’y, y,
cos 7
-2 OX m -_ Z;
4 dt
transferred to the comoving system We have here the equations
X, = X, COS wf +y, sin of X, = X; cos wf + Y, sin wt y, = -X, Sin wt + y, COs of --22 = 24


50 DOC. 1 MECHANICS LECTURE NOTES
This shows us the factors with which one must mult. the original eq. in order to obtain new ones.
X, = X2cos wt — y,sinwt Y, = X2Sin wt + y, cos wt
dx, dx, dy2. dy, dx dy
—! = —¢cos — —? —x, sin — t=? —=? si
FF de at sin + w(—x, sin — y, cos) 7 dt cos + dt sin
+ w(x, cos — y, sin)
d’x, dx d*y, . dx,. d
WE Fe OS + ae sin + 20 asin = 20s) + w?(—x, cos + y2sin) | cos | —sin
dy, : : .
aC cos cos —sin —sin — cos sin cos.
d’x, dy 2 ' ph
re 2 yz 2 . Ke Fuze,
dt? dt *26 = Xe
d*y, dx,
m} dt + 20 at SF wy, = Y,
dz,
mae 2 ,
d?x, ’ dy, : dx
d’y, 1 dx, : dy
dre 2 MO at
d*z, : de
m de = £2 +0 5 A
Interpret supplementary forces <> In the xY plane, thus 1 to w. 1 to velocit.


DOC. 1 MECHANICS LECTURE NOTES 51
—2m 0,8 0,8 oO,= —Wcos@ Ly Ferotnns ogh,
dz dy
O75, — Oxa, a, =0
2 @, = +wsin
at Ode OF y
d?x dy
pir = 0+ 2posing? x
d’y . dx dz
pid =0- 2p (sino + cos 0%) +ev. Y
pe = spa + Ypeocos 94? Z
Foucault’s pendulum. If motion takes place in the x-y plane, then a = 0. Then the
first two equat., in conjunction with the eq., set up earlier for x, etc., show that the system behaves as a system rotating with velocity w sin g. Thus, apparent rotation of the plane of the pendulum. The 3rd equation shows that, due to the rotation of the earth, reaction
dy dt We consider a free-falling m p.
force = -2mwucosp
daxe at wsiney =a + 2wsin g(d + ct)
© = b— gt + 2wcosy = b — gt + 2wcos g(d + ct)
Ip. 52]


52 DOC. 1 MECHANICS LECTURE NOTES
d’y .
az ~ 2w {asing + (b — gt)cos g}
dy .
7 a 2w{asing + bcos g}t + wg cos gt
.1
y =ct — w(asing + bcosg)t? + 32g cos ot?
72
x = (a+ 2wsin gd)t + coc sin p>
2
z =(b + 2wcosd)t —(g + eve 608 9) 5
Iffor ¢=0 #¥yz=0, a=0, b=0 c=0, and d=0
Then Reich’s experiments in Freiburg”!
x=0
1 1585m gy =ca.51°/
y= 329 cos pt? y = 27.5 mm Reich fand 28.4.4


DOC. 1 MECHANICS LECTURE NOTES 53
Be} (4) Elementary Consideration [p. 53]
op — p')t=y
p—p =zcos@
Factor ; missing.
Foucault's Pendulum
d?
Gi = 9x + 2osing? dx -y
d?y g . ax
—-y-—2 ~
2 jy — 2asing= dy x
d dy dx tes d 2
4(x2 - ye) = w'<sinp> ae )
G =) =—o'r? +C.
Set 8 + w’t = 6’. Then the ordinary pendulum law.
Dynamics of the Systems ay [p. 54]
For any of these points L yy,
m& = EY, + IX.
d? i a “S te,
Law of the momentum.


54 DOC. 1 MECHANICS LECTURE NOTES
Law of the Center of Gravity
Dnt = EEX + LEX
dt’
for <m a we can then set
t
{ } = E¥ components of the momenta of all points of the system = momentum of the system. A different mode of expression We define the center of gravity Enf of a system of masses
ME=Um etc.
[p. 55] thus ome - Me
dt dt
Thus, ve = ,
dt etc.
The center of gravity of a syst moves like a mat p. of mass M that is acted upon by the resultant of all external forces of the system. Such a law is necessary because the dynamics of the mat. point can otherwise not be maintained. Examp. Heavy body falling freely in empty space. Examp. Masses attracted by the center proportionally to their distance & mass.
Force, = - mx , ii wa Resultant = - x Dmx = - MEx 5 4
LE i.
ae ES


DOC. 1 MECHANICS LECTURE NOTES 55
Identical with the equation of motion of a single mat. point. Central force, thus in the plane, motion in ellipse as with spherical pendulum with small amplitude. Examp. Recoil of firearms. MV + mv = 0
Examp. Vibration of the ground caused by the operation of a machine with reciprocating
parts.
Examp. Light pressure. Mass of energy.“!
Law of Moments of Momenta [p. 56]
Repetition of the law of areas for the mat point. Here
Im be -yG) = DDeY, -yX) + EEGY, - yX,)
xY, - yX,
Can also be conceived vectorially h
The sum of the moments of internal forces vanishes. What remains is.
d dy _ dr) _ 7
5m ie v5) ¥Y YY, - yX,.


[p. 57]
56 DOC. 1 MECHANICS LECTURE NOTES
If system is closed, then sum of areal velocities const with respect to every plane.
ym bs = v5] is then const. As regards rotation, analogous to the law of
t
mot. of the mom of momentum. Difference.
Law of Areas for Isolated Systems
<Graphic> Geometr. depiction.
If #4 = 0, then u is a constant. If the X - Y plane is chosen 1 to py, then only one
component of the resultant momentum, and this permanently. The moment of momentum of a solid body rotating about an axis.!!
2m ea -»9) - TGY, - yX) =0
dt dt at Iw + Mo’ = 0
1D + MRoE =0 R A
—- —~
Angle person traverses AIL — ) $ relative to the body is 0 - 0’ At the stat? =0 & 0’=0 At the end 0’ - 0 = 2x Now, Id + MO‘ = const = 0 (because for ¢ = 0)


Doc. 1 MECHANICS LECTURE NOTES 57
The story of the cat.“
. 58) Complete the discussion about reaction of machines on foundation.
The moment of the momentum must also be constant over time
ym bF - :2\ = const.
aot
extended over all moving masses of the machine. Otherwise torques & thus vibration of
the foundation.
The law of <areas> moments of mom. with respect to the center of gravity.
Derivation.
If the moments vanish.
Heavy rigid body in homog- gravitational field.
Moments vanish (mov), thus the law of areas.
Special case of rod.
de! ord I
xy =ra — =
r Ai ra
hee
dy’ :
tare Den
y= “i
dz!
zg =9e d=t r
Thus, the law of areas (ab - bay}, (mr) = ¢ etc.
Deformable bodies in gravitational field. The law of areas holds as regards the center
of gravity. The cat again.
>. 59)
The Law of Kinetic Energy”! x
SSS | g(r) eae
,r
Derivation .-.++°""
Work by internal forces
Sum over all combinations


[p. 60]
58 DOC. 1 MECHANICS LECTURE NOTES
on
FS x oy
k
<0>d®
- ae od hence X;=—,
= F, (x -x')d& - x’) = F.-dr or
;i
We assume that the F depend on the distance alone.
F iad ak T = Tdr —— d
= —= hen work (9,)
fi 0 tal Wor k d(X@,) ail
From this the energy principle holds in pure mechanics. A part of the external & internal forces of the system may consist in binding forces (threads stationary surfaces etc.), especially important the special force where binding forces do not perform work. In that case the energy principle holds without these bind. forces appearing in it. Systems with one degree of freedom can be solved compl. with the aid of the =o of kinetic energy. Example. Gliding chain.
z = @ (bend) ¢ = Work = -p da dz = -pgdd qi(o + A)do ee Integrated over A: -pgdo[g(o +1) - g(a - /)] Yee 4
kin. ener, pl doy ‘ 6
, gy a
d(kin. energy) = work. When is motion independent of /? Konig’s theorem. The law of kin energy with regard to the center of gravity.


DOC. 1 MECHANICS LECTURE NOTES 59
°
Work mgdz = - mgi sin 6 dt
mS id
4)
Simple Pendulum from the Energy Principle
z=lcost
a( 7? =) )
"OF dt TM“}
2.9147)
—78 in3 = a3 Pot Energ — mgz
d3
—mglcos 9 + Lemney = konst.
2
Scsos 9 + (3) = konst.


BOC. 1 MECHANICS LECTURE NOTES
60
Moments of Inertia!
1) Moment of inertia w. resp. to plane L2n8* DDrad 2) Moment of inertia w. resp. to axis Ln” Dan? + y’) 3) Moment of inertia w. resp. to point Un? + y? +27)
[p. 61] Example: moment of inertia of a sphere w. resp. to the center.
¥ mr first over shell of the sphere = 7? © m for the shell
4xdr-o = }>m_ over the shell
PYSm = Anp “dr
Integrated. over sphere yields amok
With respect to a plane arr
. A 8 2 |2
Witithh rersepspect" "axis —75 7PR S= r“RM k=R}2 r
Example: Homog ellipsoid > mz*
n
ao]
a
N,
W
me
Se
x=
RI
—_
Then boundaries unit sphere
di =do 2 =c%?
abc
3_1
hence I. = p abc? fao'z” = 4 pabe =Mc?
x 15 5
Example. Body of rotation bounded by two planes.
R
pdz f2nar P= aria
0


DOC. 1 MECHANICS LECTURE NOTES 61
For the whole body of rotation
I, = FJ Redz R = fz)
Examp cylinder. OR -mMe k= R
a ee
General Theorems on Moments of Inertia
1) Known 7 for axis through center of gravity. Sought for arbitrary axis.
I= Yim(x? + y?) = Ym{(x' + a)+?(y’ + 5} |;
<!
+x
= Di m(x’? + y?)+ Sahmaaxr! + Sombmy’
+ (a? + b?)}m
From that the theorem. [p. 62] analogous theorems for the other two kinds of moments of inertia. 2) Moment of inertia dependent on direction.
= 2 7) , 4
I=) m& = Y mr’ sin? 9 «fy yecap
=} m(r? — (rcos g)*) as
‘
rcos@ = ax + By + yz : wnat
r? — r* cos? e 7
= (x? + y? + 2?)(a? + B? + y?)
— (ax + By + yz)?
m|x?(B? + 7) +- +> — 2Byyz —-—


62 DOC. 1 MECHANICS LECTURE NOTES
In sum
a? {Di m(y? + 27)} + BED (2? + x?)} +»? mlx? + y?) — 2By yz +- +
I = Ac? + BB? + Cy? — 2DBy — 2Eya — 2F af.
aBy
Ait
a=X,/I etc.
1 = AX? + BY?....—2FXY
If the axes of this ellipsoid are chosen as coordinate axes, then D, E, F vanish. The axes are called major axes of inertia with respect to the point. Condition for the major axis of inertia z to be symmetrical to the xy plane, thus, for it not to change when the sign of y changes. So, D = E = 0. Each of the major axes of inertia with respect to the center of gravity is also the major axis of inertia with respect to any other of its points.
D=Syz=0 E=Yu=0 za +z
D =Y ye +2’) =0 Yay =-0
hence D = D’ = 0.


DOC. 1 MECHANICS LECTURE NOTES 63
Systems of Forces Acting on a Rigid Body
The position of a rigid body is determined by 6 variables (6 degrees of freedom). We need therefore 6 equations for the complete determination of its motions. These 6 necessary and also sufficient equations are furnished by the momentum law and the law of moments.
5 L(G} crx SF fOomps --ZI- Doz -%
Since these equations must completely suffice for the calculation of the motion of the motion of the rigid body, two systems of forces acting on a rigid body are equivalent if they have the same geometric sum of forces and moments. Such systems of forces can be substituted for one another. From this follows the elementary law of displaceability of the point of application of a force vector along the straight line in which the force vector lies.
X=X"=Ka
Proof also vectorially 2 vectors are called equal if they have equal components (XYZ) = (XYZ
[p. 63]
[p. 64]


64 DOC. 1 MECHANICS LECTURE NOTES
yZ -zY =L
zX -xZ =M (L MN) can be representebdy a vector that is constructed as follows
xY -yX =N
This construction shows immediately that the force can be displaced along its straight line without changing the moment. From this the law that two forces acting along the same line in opposite directions cancel each other.
We put (XX, =X ----PyZ,-2¥,= HNN.
The effect of the system of forces on rigid body is completely determined by X* Y*Z* -- - N*, X Y** Z* resultant force. L*M* N* resultant moment. In general, it is impossible to indicate one force that is equivalent to a system of forces. We seek simplest possible [p. 65] representation of a system of forces. To this end pairs of forces introduced. Two forces of equal magnitude and opposite direction. We investigate this system of forces and obtain


DOC. 1 MECHANICS LECTURE NOTES 65
* = @Y - yX) + @' (-y) - ¥(-®)
=@ -x)¥ -% -y)X
The force couple has no resultant force vector but only a moment of rotation. This is a vector product of the point-connecting vector and the force vector. 1) <Direction of ve> absolute position of the system without effect on the moment 2) Magnitude = 2 A = force - distance 3) Direction and sense of the vector 1 to the plane of force couple ({r ?], force, mom.) = right-hand system Force couple determined by vector with a totally arbitrary point of application. Each moment vector can be replaced by a force couple. From what has been said, it follows that 1) arbitrary system of forces acting on a rigid body can be replaced by the system X*Y*Z* L* M* N* thus also by x*y*z* 000 and 000 L*MN That is, by a force through the coordinate origin and force couple. [p. 66]


66 DOc. 1 MECHANICS LECTURE NOTES
Geometric derivation. K’ and M separately added geometrically and each combined to a resultant.
43/
x’ id Aa/A
Ain, Ph,
5 aad
XM =Y*=....... N* =0
Special case. All forces lie in a plane. Analytically
X* =X =0
y* = Yy=0
N* = Ex¥ -yX = 0
Graphically
Here replaceability by resultant force. [p. 67] Even more special, all forces [] X-axis
Then 2X =0 -pX =0 Example. Reaction of beam resting at 2 places"!


DOC. 1 MECHANICS LECTURE NOTES
A+B=ZP B=l=P8
If forces |, then treated graphically as follows
At {i--£F
ot
ee


68 DOC. 1 MECHANICS LECTURE NOTES
[p. 68] = th
c XN a;
‘yj “N -7
7
f Xa u fa a u
Ap Ae 4 1 Sy
’ aa l
aC
|G
e XJ
2. Special case, all points have the same direction, but points of application
X, =P a; Y,P.B; Z, = Py
Seeking resultant system
* = ay P, L* = ¥ 0,2, a z,Y,) = ydyP, ~ BY zP,
¥* = BYP, Me RR
Z*=yyP, N* =
Now we choose the coordinate origin such that Lx, P, = Ly, P, = Lz,P, = 0. Is always possible unless EP, = 0. Then moment vanishes for all «By. Thus, if one also changes directions, one can always replace the forces by resultants passing through the coordinate origin. Center of gravity of the force system Case of gravity special case. Here the magnitude of the force acting on the individual point is P, = m,g. Thus =P,x, [p. 69] =gXm,x, = 0. Le., resultant passes through the point that we have designated in the general dynamics of systems as the center of gravity of the system.


DOC. 1 MECHANICS LECTURE NOTES 69
Motion of a Rigid Body about an Axis
(40) = )(Xdx + Ydy + Zdz)
x =rcos¢ dx = —rsin 9d9 + cos9dr = —yodt
0
y=rsin8 dy = rco +ssi9n 9$drd= 9xwdt
z=2 dz=0
I2
d 50 = wdt)( —xYyX)
daa Y — yX
on yx y.
We now insert the reaction of the axis (X¥5Y4Z) & (X",Y4Z” & treat the body as free. We obtain®”
ax o.oo, d*x r
Lm ae XX + EX Em( Yvai “3)- LyZ —z¥)— hy
a?
ae
Lmaz=V+y'+yyY Em(24q2e - x4) - ¥ (2X — xZ) + hx”
Lng eZ +Z"4+PZ
dx _ d?x _ dy _ do ory — dw 2 
a ae at a vat Hey?
dy d’y dx dw A dw ae ae Oa tae OT ae ?
dz dz _o .
a ae? ~ ,


{p. 71]
70 DOC. 1 MECHANICS LECTURE NOTES
—w? ¥ mx — 2S my =X'+X"+ PX
— 0 Smy + PY mx = Y+Y'+yY
O=Z'4Z"4+FZ
2) —@}?mxz — “OS my = hX” + )\(2X — xZ)
dw is
1) +a?) myz — Fe as = —hY" + )(yZ — 2X)
dw
a = YY — yX)
We seek to reformulate equations in such a way that shall be the only variable occurring. To this end we introduce a co-rotating system
x =x’ cosy — y'sing
y=x'sing + y'cos@ /
zZ=2
Yi myz = sin> gx'z' + cose) y'z’
Y i
m x z
=
c o s e } .
x ' z '

s i n g
>
y ’ z ’
@
a
We consider the special case where the given external forces are not present. {?] In that case the sum on the right-hand side vanishes. fe When does P” experience no reaction? or when does one need no X”, Y’ to maintain the rotation about the axis? We must have Linx = Lnyz = > 0. The Z-axis, i.e., the axis of rotation, must be a major axis of inertia with respect to P". In addition to that, when does P’ experience no reaction? We must 4 have Ex = my = 0. Thus, the axis of rotation must pass through the
center of gravity, and one of the major axes of inertia must be through the center of gravity. So that none of the two bearings would experience a reaction. No reactions will then take place even if the body is accelerated by a torque N. Example of rigid bodies on elastic axis


DOC. 1 MECHANICS LECTURE NOTES 71
1) Disc perpendicular to the Z-axis. Symmetric with respect to the x - Y plane. Exz
= Dyz = 0. X" = Y" =0 The first two equations yield®"
-FTimx = (X' +X") - SME =X’ +X"
&
-3Ymy = (+ ¥") -GMn = y' +yY"
GMA = K, ws
Let us have an elastic axis
that is mounted somewhat “4. LA eccentrically. Rotation is
accompanied by a sag &'

A=A,+€ :
On the other hand, we will have
K, = E& The above reads accordingly es
wM(A,+ E) = Ek
Critical Angular Velocity for Vanishing Denominator [p. 72]
We have here considered the case in which the axis does not
pass through the center of gravity. Now let the axis pass through
the center of gravity but not be a major axis of inertia. Let the
center of gravity coincide with P’. No external forces may exist.
From equations 4 & 5 we then obtain
~ wo” Lym = hx"
~ wo? Dnyz = hY"
axis is the major axis of inertia, then the right-hand side
vanishes. But if the maj i inertia does not coincide with
the axis of rotation, but about Y-axis
The sign and the magnitude of the reaction are here
determined by the centrifugal moments.


72 DOC. 1 MECHANICS LECTURE NOTES
Physical Pendulum
{53]
128 = Say - yX = -M gl sind
& ifone sets J = Mk
ad = 8! sin 0
dt
[p. 73] This determines the motion. Synchronous with simple pendulum of length L,
2
#0 2-8 sind if L = ©.
df L l
We now introduce the radius of inertia (radius of gyration) for the center of gravity. We have
ke
1=1+M? eg 1-2-0
; al P
or Mk? = Mk? + MI? l=k Lain = 2k, & Rak? +P
Substituted in the above relation
ke
v,
We now imagine that the pendulum is suspended at O’ and that the L* of the accompanying seconds pendulum has been determined. For this we can apply the relation just obtained and insert the quantity J’ instead of /:
Kk kK
Lt =I’ + = or, because Ul’ = =
l i]
2
L* = 5 +1, hence = L :


DOC. 1 MECHANICS LECTURE NOTES 73
The relation between the points 0 and 0’ is thus reciprocal. Therefore, by observing the oscillation of any rigid pendulum one can find out the length of a mathematical pendulum of the same oscillation period. The above formula makes it possible to determine the minimum oscillation period on the axis. Further discussion of above formula. Minimal oscillation period that can be attained with the body about the axis of the minor moment of inertia.
Experimental Determination of Moments of Inertia & Torsional Forces by [p. 74] Means of Oscillations of a Suspended Rigid Body
d29
fiye-
a= ~ 9
. {oO 2n (2)
9= Asin [2 [9
r=2 [2
It is not possible to determine both quantites @ & J from such measurements, but this can be done using a second experiment in which the moment of inertia is increased by adding two cylinders.TM For each such cylinder
IT = m{k + 87} where k = a2
I= mfr + a
2
1, = miR® + 28°}
tot


[p. 75]
74 DOC. 1 MECHANICS LECTURE NOTES
1 = a9| Fm ~ 20
r)
The two equations yield R and 8. Modification of the method, in case torsion not independent of the added weights.
General Principles of Mechanics
Principle of virtual moments (statics). Equilibrium of the point
Equilibrium condition of a point
X=Y=Z-=0 5
We think of the point as infinit. displaced 8x dy 8z “Teby Se
Work of the force 8A = Xdx + Ydy + Z,z = 0. No joke. Happens only when a part of the forces is not given but determined by conditions (connections). These forces have the characteristic property that their work vanishes in its entirety. Let this always be assumed. Example Point is forced to stay on a stationary surface (f(xy,z,<!>) = 0
Force of the surface on the point af af af,
a ay a
Total force when, in addition, another force X, Y, Z acts on the point.
xeadax
yradoy
zeaoz


DOC. 1 MECHANICS LECTURE NOTES 75
Equilibrium condition. Vanishing of those J components. Can again be replaced by
X+aD lax + yraD ay + Z+aZ lee =0.
ax. oy oz
This holds for every arbitrary system 6x, dy, 5z. But for those displacements in which the point does not leave the surface, we have the relation
ater ee 4-=0.
ax
(Special case of the law that the connective forces do not perform work.) If we confine ourselves to the consideration of such displacements, then the connective forces make no contribution to the virtual work. Thus, for such displacements, which do not violate the conditions, we have the equation
X 8x + ¥ by + Z bz = 0
if 8x 8y & 8z are connected by the relation
Fax + Fay + Ff 52 =0
ox oy az
These equations are really sufficient for the calculation of the coordinates of the equilibrium position. Because if one eliminates 8x from the first equation by means of the second one, one obtains an equation of the form B ay + C 8z = 0. This is satisfied for an arbitrary choice of 5y & 8z only if one chooses B =0 and C =0. To these two equations is added as the third eq. f = 0. Generalization. Let there be a system of n material points P, P, ---P, We seek the general condition for the equilibrium of this system of points. Let each point be acted upon by connective forces X, Y, Z, and explicitly considered forces X Y Z --- Then we have for each point
X+X,=0 cece hence also (¥ + X,)éx +-+- =0
thus also © (X + X,)8x +: +-=0
This equation holds for any arbitrary displacement of the points (even for one incompatible with the given conditions). But if the displacements are chosen such that the
[p. 76]
[p. 77]


[p. 78]
76 DOC. 1 MECHANICS LECTURE NOTES
conditions are not violated by them, then the connective forces do not perform any work during displacement, i.e., we have (X,8x + - +-) = 0. Thus, we also have LY¥8x + - + - = 0. The sum of the virtual works vanishes for every virtual displacement compatible with the conditions of the system. Proof that sufficient restrictive eq. for the solution of the problem. The advantage of this principle consists in the fact that the connective forces do not have to be investigated & that the virtual work can often be calculated without using a Cartesian coordinate system.
Example. Epicyclic wheel. Consider infinitely small rotation of the outer wheel. Infinitely small rotation of the outer wheel a@,, the epicentric wheel a,, the epicentric arm ap,
There are two conditions between these displacements because 0-1 & 1-2 do not slide, i.e., they experience identical displacements with their places of contact. Hence, we must have


DOC. 1 MECHANICS LECTURE NOTES 77
(r, + 7,)a, — re, = 0 eliminate a, (Fy + FO — 8, = ye,
2(r, + F)O,= rye,
According to the principle of virtual moments
M,,¢,, + M,a, = 0
27, +7)
of
M,, + M, =0
Special case of the principle when the forces are derivable from a potential If ® is the potential energy, then we have for each point
ox, dy,
— a@ _
rinciple then takes the form >| 8x, + - + -| = 0
x,
or 8(®) = 0 for every displacement compatible with the conditions. Further, if only a part of the explicitly considered forces (e.g., all apart from the external [p. 79] forces) are derivable from a potential, and we call them residual (e.g., external) forces X YZ, then we can write
yy Xo + Ydy + Z8z - 8B=0
D’Alembert’s Principle
Considerations that are analogous to those regarding the eqilibrium of the material point. Equations of motion of the mat point (freely movable)


78 DOC. 1 MECHANICS LECTURE NOTES
-X=0] &
m#¥ -y =0 by [rite afr +-+ 20
- eee ee bz
Triviality Secondly, we assume that the point is subjected to two kinds of forces, namely explicitly considered forces and connective forces. Purely formally, as above, the following eq. will then be valid:
mi* ox -x lex + mf? -y ~y lay +-=0
dt’ dt?’
The quantities 8x 8y 8z determine for each moment a position infinitely close to the real position of the material point. We will now choose these infinitely close positions in such a way that at any moment the point could be displaced from x y z to x + 8x, y + dy, z + bz without violating the conditions of the system. [p. 80} If we are dealing, for example, with the motion of a m. p. on an arbitrarily moving surface, then let 8x Sy 8z for time ¢ be chosen such that f(x + 8x, :, 2) = 0 The law to the effect that the work of the connective forces vanishes holds for such displacements. Thus, for example, for the mat. p. on a surface, because connective force i to the plane, but displacement in the surface. We will have, therefore, for such a displacement X,8x + Y,8y + Z,8z = 0 Since the above equation is valid for every virtual displacement, and thus also for such ones that do not violate the conditions of the system, we also have for virtual displacements of the last-mentioned kind the equations
mo® -xlax+-+-=0
dt?
where 8x dy 8z are connected by those relations to which the connective forces are to be traced. Analogous argument for systems of material p. If we again introduce connective forces and other forces, we get


DOC. 1 MECHANICS LECTURE NOTES 79
es xox} essed
in
for every system of 8x Sy ’z. If &x - - are specially chosen such that the conditions of the system are not violated, then
WK sx +-+)=0,
len
so that subtraction yields [p. 81]
dx ax
p> "ap - } a0, tet =0
'q k
dx
Y | |m— - X Jax, +- +: [= 0
lton dt?
This is D’Alembert’s principle. Next we have to show that this equation gives the solution of every problem of motion. Let a virtual displacement that does not violate the conditions of the system be determined by k mutually independent quantities 5g, ---- 8g, (k degrees of freedom)
6x Beans; 4% 5 1 Ft 5g, dy, °° 6Z,
sq, 6q2 "> 4 ,
tf) Ox,
ox, = *" 5a, ~ — eee x Od, ee


[p. 82]
80 DOC. 1 MECHANICS LECTURE NOTES
ax ox a
If one sets Ym =-OQ EX m =R,
> ©’ «Og, > (Og,
dx, ox, ax
— —= Y—=R
re og, Q, Y "aq, 2 vv
then the above system of equations assumes the form
(Q, - R,)8q, + (Q, - R\)8q,-- °° - (Q, - R,)8q, = 9 Since all q, are independent of each other, we have
Q,=R, Q=R,--°°: Q =, R, These k equations are just sufficient for the solution of the problem.If the conditions between the 8x - - - - can be represented as equations between the x, ----z, & ¢, that is, in the form f(x, -- - -2z,f) = 0, then the system is said to be holonomic. The equations of motion for such a system can be found in the following fashion, which was first presented by Lagrange. We have
dx
fe ea as +++ | =O and f.(@, --z,,0 =0
me from 1 toh
For such displacements, which are compatible with these conditions, we have
of; a : A variations can be Ar » (# Lae ) =0 : expressed from the rest : 3n - h = k equations from the
X y 5x +:4:=0 : first
? 7 Ox, :
An Y phox, +-+-=0
Restrictive equations mult. by factors A & add.


DOC. 1 MECHANICS LECTURE NOTES 81
T K
a a F e
a
T h
a x
+ +
2 0
v
.
d r
o x ,
o x ,
a x ,
.
3n such {} are present. We can choose h of those = 0 by choosing A in the appropriate way. These terms of the sum will then vanish. The 8x dy 8z of the rest are then arbitrary, since of the 8 3n - h =k can be chosen arbitrarily. From this it follows that the remaining {} must also vanish. Thus, one also obtains the equations of motion of a system of points in the form
dx, =X at a Dt ga
” de v i 1 ax, ax,
4
a v from | ton
movey ea Dig
de v ax,
Equations of Motion of Lagrange
Of historical interest only."
<Principle of Least Action> Hamilton’s Principle. Lagrange’s Eq. of Motion
We start out from d’Alembert’s principle
dx
by [x 7 m, Ph, ca am = 0 v
for all virtual displacements compatible with the cond. of the syst. We can put
[p. 83]


82 DOC. 1 MECHANICS LECTURE NOTES
v dx dr d(&. dx. dx, | dx
djs, | ede)dtd Z| .
“a” ala | dod. ala’) a la
The second term can be written in another form. We shall show that it is equal to the variation of velocity
dt dt dt dt
dx, d Sdx, 1 dx?
ae? ° -3% ax} -o{} =}
d_ dx, m, (dx,\? _
EAI Tgens +pDOG (Gr) =0
\ , \ - } vu ~—’
6A 0 6L
Integr. over time limits. All 8 shall vanish at time limits.
J G, + dL)d = 0
b
[p. 84] We shall rewrite the work A, according to the follow. principle we choose indep. variables p, ----p,, Whose number is equal to the number of degrees of freedom. A, will then have the form P,3p, + P,8p, -----We now specialize the problem a little. Let the forces be derivable in part from a force function. Let II be the potential energy. The part of the virtual work A, derived
v
from themis -}* [2s tee | = -8II


DOC. 1 MECHANICS LECTURE NOTES 83
In addition, forces might be present that are either given or sought as functions of time. If X{ Y’ Z% are components of these forces for 1 point, then the corresponding term of A’ is
Y Max, ++ +)
ax, a
If we set 8x, = }> 5
P, rm
bp,
then we recognize that the part of work considered can be expressed in the form
> Pp, . If the P,, are to be viewed as functions of time only, then we can set
B
YP,dp, = 8 YP, P,
Substituting both terms for A, into the above formula, one obtains [p. 85]
5
3 fa -L-Y@, pat = 0
%
This is <the least action principle> Hamilton’s principle. If all forces can be derived from a potential (Il), the latter assumes the simple & familiar form
3 { r= Dat =0.
%
In this principle the Cartesian coordinates of mass points no longer occur. It is valid no matter what coordinates we choose to determine the position of the points of the system. Now we start out from the general form of Hamilton’s principle
4 (8L +A,)dt = 0
f:
Ly


84 DOC. 1 MECHANICS LECTURE NOTES
We use general coordinates, which determine completely the state of the system (P: --*"P,). Then we can set
= YP.op,
Further, we have to examine how L depends on the p,.
m,\ dey
We have L = pe = le x, = @.0,°" P,)
dx, — 09,, 4p,
a ba
d,
Thus, L, is afunctionof p, & - ae p.
{p. 86} For that reason we have to set
aL = r=8, +> oF ay). op.
Replacing A, and 8Z with their values, one obtains
fire, +p Hop! + DPapsa= 0
ap.
op, v
The factors of 8p, & 8p’, do not have to vanish individually! But we have
OF gp! = SE dap = 4) ay) - ap dob
ap! ap, at at| ap! at| ap’
But since the 8p, must vanish at the limits, it follows that


DOC. 1 MECHANICS LECTURE NOTES 85
Since the 8p, can be chosen totally arbitrarily, as long as they are continuously variable, one obtains
oL sed
— ee ~P =0
ap, dt v
SE) Pp -0.
ap,
“(= am -L
aS +
at} ap! ap,
These are the important equat. of motion of Lagrange. In the special case in which part of the forces can be derived from a potential, part of the
_ on
op,
forces has the form
Example. {p. 87]
Two identical rods are linked to each other at their ends by means of threads. One
of the rods is situated so as to pivot around its middle.
2M
L = 2M? + i707
Mee + 5 ie |
/ -M Zl? 4 Jl J
Mi takeg? + P04 aa
Il = - Mgloost
From this at once equations of motion


86 DOC. 1 MECHANICS LECTURE NOTES
dal | AM-L) _g at ag 39
dal , a-L) 9
dt at’ oo
The equations yield:
d 2a) = =
po ?) =0 q@ = const.
£ (MPP) +Mgl sin =O 0” =& sind.
2nd Example. y |
=Icos9 + ‘= —Is99i— n asingg’ l
&€ =Icos acos@ é 9e A ,
yn =Isin$ + asing n’ = Icos 99’ + acos gq’ A 4 2
= Oo
Tl = + Mg{i(i — cos 9) + a(1 — cos¢) | Ln 5
Calculation of the kin. energy
1) If the mass were concentrated
at the center of gravity x
(58)
Mee + nit) = Cero? + ag + 2al cos( ~ @)
2) Kin. energ. with respect to the center of gravity eMa'
M
’ L=
{p. 88] 5) lo? + sag? + 2ald’¢! cos(t - 9)


DOC. 1 MECHANICS LECTURE NOTES 87
For infinitely small oscillations only the smallest terms are retained
= Sue + ag)
L= Fae" _ sa + ald’)
The Lagrange eq. without P, @ ee + ig -L)=0 dt op’ op,
(MV’0" + Malo”) + Mgld =0 Pe’ +alg’ = -gld
4 2 nlf ft Ul 4 2 ff
aoe ” 6+ Mald" | + mgag = 0 alt” + aa = -gap
Linear homogeneous equations that can be solved trigonometrically. Set
0 = A, cos(wt + 8) (-w? + gla, - alw’A, = 0
@ = A, cos(wt + 8) -alw’ i,- sad + gad? =0
(wl - g) (Fe0" = s} -alw' =0
Biquadratic equation for frequency (w) From this wo, & @,
A
The equations also yield z
1
As the general solution, one finds
b= Aoi, cos(w,t + 8,) + aw5p, (COS wf + 8)
p = (g ~ Lwi)p, cos(wt + 3) +@ - 1y)u,(cos w,t + 8,)


88 DOC. 1 MECHANICS LECTURE NOTES
Superposition of two mutually completely independent oscillations of different periods.
[p. 89} Rigid Body Kinematics
Representation of rotation by vector w velocity of a mat. point at distance 1 from the axis of rotation. Can be represented by vector whose point of application is of no importance. Length w sense that rot with vector right-hand screw. We denote the projections of the rotation vector on the coordinate axes by p, g,r. We consider an arbitrary point of the body. Seeking vector V. It is perpendicular to vector (w), perpendicular to vector (X - Xp, ¥Y ~ Yo, Z ~ 2) = (r). Is equal to the product of the magnitudes of the two vectors mult by the sine of the enclosed angle. Sequence . . v, v, v, is the vector product of the vectors () and (r), hence
5x
aa < 47 Je, 2%-%
ws
v, = G(Z ~ 2) — ry - Yo)
v, = r(x - Xp) - pz - 2) P q r
¥, = PY ~ Yo) ~ F&-%) X-% YY Z-%
Important formulas.
[p. 90] Composition of Angular Velocities
Body with rotation (p qr) about a given point (coordinate origin) describes with its point xyz the path