PROBLEMS ON MECHANICS 1 INTRODUCTION 2 FIRST LAWS THEORETICAL BASIS

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1 PROBLES ON ECHANICS Jaan Kalda Version: 1.2β 1, 7th June 2018 Partially translated by: S. Ainsaar, T. Pungas, S. Zavjalov 1 INTRODUCTION This booklet is a sequel to a siilar collection of probles on kineatics and has two ain parts: Section 3 Statics and Section 4 Dynaics; Section 5 contains revision probles. The ain ai of this collection of probles is to present the ost iportant solving ideas; using these, one can solve ost (> 95%) of olypiad probles on echanics. Usually a proble is stated first, and is followed by soe relevant ideas and suggestions (letter K in front of the nuber of an idea refers to the correspondingly nubered idea in the kineatics booklet; cross-linking works if the kineatics booklet is stored in the sae folder as the echanics one). The answers to the probles are listed at the end of the booklet (Section 7). They are preceded by quite detailed hints (Section 6), but it is recoended that you use the hints only as a last resort, after your very best efforts at tackling a proble fail (still, once you have solved a proble successfully by yourself, it is useful to check if your approach was the sae as suggested by the hints). The guiding principle of this booklet argues that alost all olypiad probles are variations on a specific set of topics the solutions follow fro corresponding solution ideas. Usually it is not very hard to recognize the right idea for a given proble, having studied enough solution ideas. Discovering all the necessary ideas during the actual solving would certainly show uch ore creativity and offer a greater joy, but the skill of conceiving ideas is unfortunately difficult (or even ipracticable) to learn or teach. oreover, it ay take a long tie to reach a new idea, and those relying on trying it during an olypiad would be in disadvantage in coparison to those who have astered the ideas. In science as a whole, solution ideas play a siilar role as in olypiads: ost scientific papers apply and cobine known ideas for solving new (or worse, old) probles, at best developing and generalising the ideas. Genuinely new good ideas occur extreely rarely and any of the are later known as asterpieces of science. However, as the whole repertoire of scientific ideas encopasses iensely ore than ere echanics, it is not so easy to reeber and utilise the in right places. The respective skill is highly valued; an especial achieveent would be eploying a well-known idea in an unconventional (unexpected, novel) situation. In addition to ideas, the booklet also presents facts and ethods. The distinction is largely arbitrary, soe ideas could have been called ethods or facts and vice versa; attept has been ade to pursue the following categorization. Facts are fundaental or particular findings, the knowledge of which can be useful or necessary for proble solving, but which are not forulated as ready recipes. While in theory, all probles can be solved starting fro the first principles (the fundaental facts ), but typically such a brute force approach leads to long and soeties unrealistically coplex calculations; the ideas are recipes of how to solve probles 1. INTRODUCTION ore easily. The ethods are powerful ideas of particularly wide applicability. Several sources have been used for the probles: Estonian olypiads regional and national rounds, Estonian-Finnish Olypiads, International Physics Olypiads, journal Kvant, Russian and Soviet Union s olypiads; soe probles have been odified (either easier or tougher), soe are folklore (origins unknown). Siilarly to the kineatics booklet, probles are classified as being siple, noral, and difficult : the proble nubers are coloured according to this colour code (keep in ind that difficulty is a subjective category!). Finally, don t despair if there are soe things (or soe sections) which you are not able to understand for the tie being: just forward to the next topic or next proble; you can return to those parts which you didn t understand later. 2 FIRST LAWS THEORETICAL BASIS Those who are failiar with the basic laws of echanics can skip this Section (though, you can still read it, you ay get soe new insight), and turn to Section 3. In fact, it is expected that ajority of readers can do this because alost all the physics courses start with echanics, and it is unlikely that soeone is drawn to such a booklet aiing to develop advanced proble-solving skills without any prior experience in physics. However, attepts have been ade to keep this series of study guides self-contained; this explains the inclusion of the current chapter. Still, the presentation in this Section is highly copressed and in soe places involves such atheatical foralis which ay see intiidating for beginners (e.g. usage of the suation sybol and differentials), therefore it is not an easy reading. If you find this section to be too difficult to start with, take a high-school echanics textbook and turn here to the Section Statics. 2.1 Postulates of classical echanics Classical echanics, the topic of this booklet, is a science based entirely on the three Newton s laws 2, forulated here as facts. fact 1: (Newton s 1 st law.) While the otion of bodies depends on the reference frae (e.g. a body which oves with a constant velocity in one frae oves with an acceleration in another frae if the relative acceleration of the fraes is nonzero), there exist so-called inertial reference fraes where the facts 2 5 are valid for all the bodies. fact 2: (Newton s 2 nd law.) In an inertial frae of reference, a non-zero acceleration a of a body is always caused by an external influence; each body can be characterized by an inertial ass (in what follows the adjective inertial will be dropped), and each influence can be characterized by a vectorial quantity F, henceforth referred to as the force, so that equality F = a is valid for any influence-body pair. 1 As copared with v. 1.0, introductory theory sections are added 2 I. Newton which will also serve as the ass unit 4 or a cylinder of height and diaeter, ade of platinu-iridiu alloy Pt-10Ir (the official SI definition). page 1

2 Please note that once we establish an etalon for the ass 3, e.g. define 1 kg as the ass of one cubic decieter of water 4, the fact 2 serves us also as the definitions of the ass of a body, and of the agnitude of a force. Indeed, if we have a fixed reference force which is (a) guaranteed to have always the sae agnitude, and (b) can be applied to an arbitrary body (e.g. a spring defored by a given aount) then we can define the ass of any other body in kilogras nuerically equal to the ratio of its acceleration to the acceleration of the etalon when the both bodies are subject to the reference force. Newton s 2 nd law is valid if this is a self-consistent definition, i.e. if the obtained ass is independent of what reference force was used. Siilarly, the agnitude of any force in Newtons (denoted as N kg /s 2 ) can be defined to be equal to the product of the ass and the acceleration of a body subject to that force; Newton s 2 nd law is valid and this definition is self-consistent if the result is independent of which test body was used. To su up: the Newton s 2 nd law F = a serves us both as the definition of the ass of a body (assuing that we have chosen a ass etalon), and the force of an interaction; the law ensures that these are self-consistent definitions: the ass of a body and the agnitude of a force are independent of the easureent procedure. fact 3: Forces are additive as vector quantities: if there are any forces F i (i = 1... n) acting on a body of ass then the fact 2 reains valid with F = i F i. The vector su i F i can be calculated using either the triangle/parallelogra rule, or coponent-wise arithetic addition: F x = i F ix, where an index x denotes the x-coponent (projection onto the x-axis) of a vector; siilar expressions can be written for the y- and z-axis. fact 4: asses are additive as scalar quantities: if a body is ade up of saller parts of asses j (j = 1... ) then the total ass of the copound body equals to the su of the asses of its coponents, = j j. fact 5: (Newton s 3 rd law.) If a body A exerts a force F on a body B then the body B exerts siultaneously the body A with an equal in odulus and antiparallel force F. 2.2 Basic rules derived fro the postulates The facts 1 5 can be considered to be the postulates of classical (Newtonian) echanics, confired by experients. All the subsequent facts, theores, etc. can be derived atheatically using these postulates. Thus far we have used a vague concept of the acceleration of a body. Everything is fine as long as a body oves translationally, i.e. so that all its points have the sae acceleration vector. However, if a body has a considerable size and rotates then different points have different accelerations, so that we need to clarify, the acceleration of which point needs to be used. In order to overcoe this difficulty and keep our set of postulates 2.2 Basic rules derived fro the postulates 1 5 as siple as possible, let us assue that the fact 2 is valid for so-called point asses, i.e. for very sall bodies the diensions of which are uch saller than the characteristic travel distances; then, the position of a point ass is described by a single point which has unabiguously defined velocity and acceleration. We can generalize the fact 2 to real finite-sizedbodies by dividing these fictitiously into tiny pieces, each of which can be treated as a point ass. To begin with, one can derive (see appendix 1) the generic forulation of the Newton s 2 nd law. fact 6: (oentu conservation law/generalized Newton s 2 nd law.) For the net oentu P = i i v i of a syste of point asses 5, d P dt = F, (1) where F is the net force (the su of external forces) acting on the syste. In particular, the net oentu is conserved ( P =const) if F = 0. i i r i By substituting P = i i v i = i i d ri dt = d dt (where r i denotes the position vector of the i-th point ass), we can rewrite Eq (1) as d2 r C dt 2 = F, where r C = i i r i i i is called the centre of ass. This result clarifies: in the case of acroscopic bodies, the fact 2 reains valid if we use the acceleration of the centre of ass. According to the Newton s 2 nd law, once we know how the interaction forces between bodies depend on the inter-body distances and on the velocities, we can (in theory) calculate how the syste will evolve in tie (such systes are refereed to as deterinstic systes). Indeed, we know the accelerations of all the bodies, and hence, we can deterine the velocities and positions after a sall tie increent: if the tie increent t is sall enough, the changes in accelerations a can be neglected, which eans that the new velocity for the i-th body will be v i = v i + a i t, and the new position vector r i = r i + v i t; the whole teporal dependences v i (t) and r i (t) (with i = 1... N where N is the nuber of bodies) can be obtained by advancing in tie step-by-step. In atheatical ters, this is a nuerical integration of a syste of ordinary differential equations: the second tie derivatives of coordinates ẍ i, ÿ i, and z i are expressed in ters of the coordinates x i, y i, z i, and the first derivatives ẋ i, ẏ i, ż i 6. While in principle, these calculations can be always ade, at least nuerically and assuing that we have enough coputational power, in practice the atheatical task can be very difficult 7. Apart fro the facts 1 5, the Newtonian echanics is a collection of recipes for easier solution of these differential equations. Aong such recipes, finding and applying conservation laws has a central role. This is because according to what has been said above, the evolution of echanical systes is described by a syste of differential equations, and each conservation 5 such as rigid bodies in which case the point asses are bound by inter-olecular forces together into a acroscopic body 6 Here we assue that the force depends only on the coordinates and velocities of the bodies; with the exception of the Abraha-Lorentz force (accounting for the cyclotron radiation), this is always satisfied. 7 Also, there is the issue of possibly chaotic behavour when in any-body-systes, sall differences in initial conditions lead to exponentially growing differences in the sae way as it is ipossible to put a sharp pencil vertically standig on its tip onto a flat surface. page 2 (2)

3 law reduces the order of that syste by one; this akes the atheatical task uch sipler. The conservation laws can be derived atheatically fro the Newton s laws; while it is definitely useful to know how it is done, ajority of echanics probles can be solved without being failiar with this procedure. Because of that, the conservation laws are derived in Appendices 1,2, and 3; here we just provide the forulations. We have already dealt with the oentu conservation law (see fact 6), so we can proceed to the next one. fact 7: (Angular oentu conservation law.) For the net angular oentu L = i i r i v i of a syste of bodies, d L dt = T, (3) where T = r i F i i is the net torque acting on the syste; here F i stands for the net force acting on the i-th point ass. In particular, the net angular oentu of the syste is conserved if T = 0. Eq. (3) is derived in appendix 2, and can be considered to be the Newton s 2 nd law for rotational otion of bodies. In three-diensional geoetry, calculating the vector products to deterine the net torque and angular oentu ay be quite difficult. Luckily, ost of the Olypiad probles involve two-diensional geoetry: velocities, oenta, and radius vectors lie in the x y-plane, and vector products (torques and angular oenta) are parallel to the z-axis, i.e. we can consider L = r p and T = r F as scalars, characterized by their projection to the z-axis (in what follows denoted as L and T, respectively). According to the definition of the vector product, the sign of such torque is positive if the rotation fro the vector r to the vector F corresponds to a clockwise otion, and negative otherwise. Thus we can write T = r F sin, where is the angle between the radius vector and the force and can be either positive (rotation fro r to F is clockwise) or negative. We can introduce the ar of the force h = r sin (see figure), in which case T = F h; siilarly we can use the tangential coponent of the force F t = F sin and obtain T = r F t. Siilar procedure can be applied to the angular oenta: L = r p sin = h p = r p t. O origin (axis) h area = torque F t r F force application point The discipline of statics studies equilibria of bodies, i.e. conditions when there is an inertial frae of reference where a body reains otionless. It is clear that both the oentu and angular oentu of a body at equilibriu needs to be 2.2 Basic rules derived fro the postulates constant, hence the su of all the forces, as well as the su of all the torques acting on a body ust be zero; this applies also to any fictitious part of a body. While there are statics probles which study deforable bodies (which change shape when forces are applied to it), there is an iportant idealization of rigid body: a body which preserves its shape under any (not-too-large) forces. While for the Newton s 2 nd law [Eq. (1)], and for the static force balance condition, it doesn t atter where the force is applied to, in the case of angular oentu [Eq. (3)] and for the static torque balance condition, it becoes iportant. In classical echanics, the forces are divided into contact forces which are applied at the contact point of two bodies (elasticity forces in its various fors, such as noral and friction forces, see below), and body forces which are applied to every point of the solid body (such as gravity and electrostatic forces). The application point of contact forces is obviously the contact point; in the case of body forces, the torque can be calculated by dividing the entire body (syste of bodies) into so sall parts (point asses) and by integrating the torques applied to each of these. It is easy to see that with the total body force (i.e. the su of all the body forces applied to different parts of the body) F and total torque T applied to a body, one can always find such a radius vector r that T = r F, i.e. although the body forces are applied to each point of the body, the net effect is as if the net force F were to applied to a certain effective application centre; in soe cases, there are siple rules for finding such effective application centres, e.g. in the case of an hoogeneous gravity field, it appears to be the centre of ass. 8. At the icroscopic level of quantu echanics, such a division of forces becoes eaningless, because on the one hand, fields which ediate body forces are also aterial things and in this sense, all the forces are contact forces. On the other hand, classical contact forces are also ediated at the icroscopic level via fields so that in a certain sense, all the forces are body forces. Regardless, at the acroscopic level such a division still reains helpful. fact 8: (Energy conservation law; for ore details, see appendix 3.) If we define the kinetic energy for a syste of point asses (or translationally oving rigid bodies) as K = 1 i v i 2, 2 and the total work done by all the forces during infinitely sall displaceents d r i of the point asses as dw = F i d r i i then the change of the kinetic energy equals to the total work done by all the forces, i dk = dw ; here, F i denotes the total force acting on the i-th point ass. The work done by so-called conservative forces depends only on the initial and final states of the syste (i.e. on the positions of the point asses), and not along which trajectories the point asses oved. This eans that the work done by conservative 8 Siilarly, if two bodies ake contact over a finite-sized area (rather than at just few contact points), we would need to find the total torque by integrating over the contact area, and one can always find the effective application point of these forces 9 dπ Π( r 1 + d r 1, r 2 + d r 1,...) Π( r 1, r 2,...) page 3

4 forces can be expressed as the decrease of a certain function of state Π( r 1, r 2...) which is referred to as the potential energy; for infinitesial displaceents we can write dw cons = dπ 9. Therefore, if we define the full echanical energy as E = K +Π then de = dw, where dw stands for the work done by the non-conservative forces; if there are no such forces then de = 0, hence E = Π + K = const. (4) Here few coents are needed. First, while the oentu of a body is the oentu of its centre of ass, the kinetic energy of a coposite body is not just the kinetic energy of its centre of ass: the kinetic energy in the frae of the centre of ass ( C-frae ) needs to be added, as well, K = 1 2 v2 C + 1 i u 2 i, 2 where where u i = v i v C is the velocity of the i-th point ass in the C-frae 10. How to calculate the kinetic energy in the frae of the centre of ass for rotating rigid bodies will be discussed soewhat later. Finally, let us notice that forces depending on the velocities (e.g. friction forces) and/or on tie (e.g. noral force exerted by a oving wall), cannot be conservative because the work done by such forces along a path depends clearly on how fast the bodies ove. An exception is provided by those velocitydependent forces which are always perpendicular to the velocity (e.g. for the Lorentz force and noral force) and for which dw = F d r = F vdt Basic forces Gravity. Now, let us consider the case of a gravity field in ore details; it can be described by the free fall acceleration vector g. Fro the Kineatics we know that then all the bodies ove with the acceleration g; then, according to the Newton s 2 nd law, this should be caused by a force F = g. i 2.3 Basic forces Fro the Newton s 3 rd law we know that each force is caused by soe other body: a gravity force acting on a body A needs to be caused by a body B. We also know that the gravity force is caused by and is proportional to the ass of a body, and apparently this applies both to the body A, and body B. Hence, the force needs to be proportional to the product of asses, F = c A B, where the coefficient of proportionality c can be a function of distance. It appears that c is inversely proportional to the squared distance, c = G/r 2 ; let us consider this as an experiental finding. Here G kg 1 s 2 is called the gravitational constant. Now it is easy to guess that this force ust be parallel to the only preferred direction for a syste of two point asses, the line connecting the two points. This is indeed the case; furtherore, the gravitational force appears to be an attractive force. fact 9: The gravitational force acting on the i-th point ass due to the j-th point ass can be expressed as G i j F i = ˆr ij rij 2, (5) where ˆr ij = ( r j r i )/ r j r i stands for the unit vector pointing fro the i-th body to the j-th body and r ij = r j r i 11. The presence of a third body does affect the validity of this law, i.e. the superposition principle holds: total gravitational force can be found by suing up the contributions fro all the gravitating bodies according to Eq. (5). 12 Eq. (5) reains also valid when the gravitationally interacting bodies have spherically syetric ass distribution in that case, r i and r j point to the respective centres of syetry (which coincide with the centres of ass) 13 NB! in the case of arbitrarily shaped bodies, using the centres of ass would be incorrect; the force needs to be calculated by dividing the bodies into point asses and taking integral. For the gravitational pull of the Earth, we can typically approxiate r ij with the radius of the Earth R E, so that F = g, g = ẑ G E R 2 ẑ 9.81 /s 2, (6) E This is called the gravity force. The fact that in a given gravity where ẑ stands for a downwards pointing unit vector and E field, the gravity force is proportional to the ass of a body is denotes the Earth s ass. to be considered as an experiental finding. Let us recall that the ass is introduced via the Newton s 2 nd Note that the force due to a hoogeneous gravity field g is law and describes applied effectively to a centre of ass of a body, regardless of the inertia of a body, i.e. the capability of a body to retain its its shape. Indeed, the torque of the gravity force is calculated velocity; because of this we can call it also the inertial ass. then as Here, however, the ass enters a copletely different law: the gravity force is proportional to the ass. It is easy to iagine T = ( ) r i g i = r i i g = r C g, that the gravity force is defined by another characteristic of i i a body, let us call it the gravitational ass, unrelated to the where = i i. inertial ass. Experients show that the gravitational ass is always equal to the inertial ass and thus we can drop the adjectives gravitational and inertial. As a atter of fact, Gravity force is a conservative force because for any pair of point asses, the force is directed along the line connecting these point asses and depends only on the distance between the equivalence of inertial and gravitational ass has a great these (cf. appendix 3). The work done by a gravity force significance in physics and represents the ain postulate and F = g due to a hoogeneous gravity field can be expressed cornerstone of the general theory of relativity. as da = g d r, hence Π = g d r; upon integration we 10 Indeed, we divide the body into point asses and write K = 1 2 i i( v i v C + v C ) 2 = 1 2 i i( u i + v C ) 2 ; now we can open the braces and 1 factorize v C : 2 i ( u i + v C ) 2 = 1 2 i i u 2 i + v C i i u i v2 C i i. Here, i i u i is the total oentu in the C-frae, which is zero according to the definition of the centre of ass. 11 I. Newton The superposition principle corresponds to the linearity of the non-relativistic equations of the gravity field and can be treated as an experientally verified postulate. 13 In the booklet of electroagnetis, this property will be derived fro Eq. (5) using the superposition principle. page 4

5 2.3 Basic forces obtain Π = g ( r r 0 ), where r 0 is the vector pointing to an arbitrarily chosen reference point. Now, keeping in ind that the energy is additive, we can write an expression for the gravitational potential energy of N bodies: Π = g i ( r i r 0 ) = g ( r C r 0 ) i. (7) i Gravitational energy of two point asses can be calculated siilarly by integration; for two point asses, it is usually convenient to take the reference configuration (for which the potential energy is zero) such that the distance between the two bodies is infinite. For sall displaceents, the work done by the gravity forces acting on the both bodies dw = G 1 2 ˆr 12 (d r 1 d r 2 ) = G 1 2 r 2 12 r 2 12 dr 12 [here we took into account that d r 1 d r 2 = d( r 1 r 2 ), and ˆr 12 d( r 1 r 2 ) = dr 12 ]. So, r12 Π = G 1 2 r 2 dr = G 1 2. r 12 If there are ore than two interacting bodies, we can use the superposition principle to find Π = G i j ; (8) r i>j ij note that the interaction energy of any pair of bodies needs to be counted only once, hence we su over pairs with i > j. fact 10: Potential energy of two gravitationally interacting spherically syetric bodies is given by Eq. (8); in the case of hoogeneous downwards-directed gravity field of strength g, the change of potential energy of a body of ass is Π = g h, where h is the change of height. a ; however, then the stiffness (the proportionality coefficient k) depends also on â, and force is not necessarily antiparallel to the displaceent 16. Elasticity forces are conservative, potential energy can be found by siple integration Π = F d a = kada = 1 2 ka2. fact 12: Under the assuptions of fact 9, the potential energy of an elastically deforable body is given by Π = 1 2 ka2. (10) It is quite clear that if we take, for instance, a rubber band of length l and stiffness k, aking it twice longer will reduce its stiffness by a factor of two. Indeed, one can divide the longer band fictitiously into two halves, each of length l, which eans that if we apply now the sae force F to the endpoints of the long band, both halves will be defored by x = F/k and hence, the entire band is defored by x = x + x = 2F/k so that the stiffness k = F/x = k/2. Siilarly, aking the band twice thicker will increase the stiffness by a factor of two because we can consider the thicker band as being ade of two parallel thinner bands. NB! this does not apply to the springs and bending deforation: while aking a spring twice longer will, of course, still decrease k by a factor of two, aking it fro a twice thicker wire will increase the stiffness ore than two-fold: by the sae bending angle the thicker wire will be defored ore than a thinner wire. This paragraph can be suarized as the following fact. fact 13: For a band of elastic aterial of length L and crosssectional area A, the stiffness k = AY/L, (11) Elasticity. Siilarly to gravity forces, elasticity forces can where Y denotes the so-called Young odulus of the aterial 17. be et literally at each our step. While icroscopically, all elasticity forces can be explained (at least in principle) in ters This equality akes it possible to give an alternative forulation of the Hooke s law. To that end, let us introduce the of electrostatic interactions using quantu echanics, acroscopically it can take different fors. First of all, there is the concepts of the strain, which is defined as the relative deforation ε = a/l, and stress which is defined as the elasticity Hooke s law which describes elasticity forces for deforable bodies (e.g. a rubber band or a spring); there are also noral force per unit area, σ = F/A. Then, the Hooke s law can be force and dry friction force which seeingly have nothing to do rewritten as with elasticity, but in reality, both noral force and dry friction force have icroscopically the sae origin as the Hooke s One can also introduce the concept of energy density of a σ = εy. law. defored aterial, the ratio of the potential energy and the volue, w = Π/(LA) = 1 fact 11: (Hooke s law 14 2 ) If the deforation of an elastically Y ε2. Now we can also address the question, which deforations deforable body is not too large, the defored body exerts a can be considered as sall enough so that the Hooke s law force which (a) is antiparallel to the deforation vector a and reains valid. One ight think that we need to have ε 1, (b) by odulus is proportional to the deforation, i.e. but typically this is a too loose requireent: ajority of elastic F = k a. (9) aterials break far before deforations ε 1 are reached. Typically, the Hooke s law starts failing when deforations are so This law is valid for sall deforations of all elastic aterials including rubber bands, springs, etc, as long as the deforation is not too large, and the body deforation includes deforations, the aterial is no longer fully elastic: when the large that the aterial is close to breaking. Also, for such large only stretching (or copression), and does not involve bending or shear 15. If bending and shear deforations are involved, tial shape and soe residual deforation will reain; such force is reoved, the aterial does not fully restore its ini- with a fixed deforation direction (described by the unit vector â a/ a ), the force odulus reains to be proportional to also aterials (which can be referred to as plastic aterials deforations are referred to as plastic deforations. There are or 14 R. Hooke Deforation of springs, in fact, does involve bending, but the Hooke s law reains in that case nevertheless valid. 16 A proper description of the elasticity forces when bending and shearing are involved requires tensorial description and is beyond the scope of this booklet 17 L. Euler 1727, G. Riccati 1782 page 5

6 plastically deforable aterials) which defor plastically over a very wide range of strain values before breaking into pieces. In such cases, the Hooke s law reains valid only for extreely sall strains by which the deforation still reains elastic. Note that there are hyperelastic aterials for which the deforation can be very large, ε > 1; then, indeed, the condition ε 1 is required for the applicability of the Hooke s law. ajority of the aterials have rather large values of Y which eans that unless we have really long and thin threads or wires, oderate forces will cause only a really inute and un-noticeable deforations. This is typically the case for wires, ropes, rods, and solid surfaces. In those cases, while the geoetrical effect of the deforations can be neglected, such a deforation will be fored that the elasticity force copensates any applied external force. If we are dealing with a solid surface, such an elasticity force is called the noral force; in the case of rods, wires and ropes, we call it the tension force. Unless otherwise ephasized, it is assued that the tension force is parallel to the rod, wire or rope. In the case of a rope or a thin wire, this is essentially always the case: there is no possibility of having an elastic shear or bending because the rope is typically very soft with respect to such deforations: if we try to create a perpendicular elasticity force by applying a peperpendicular external force, the rope will be bended without giving rise to any noticeable force. In the case of a rod, this is not true: if we apply an external perpendicular force, the rod resists elastically to the attepts of bending and creates a perpendicular coponent of the tension force. Still, if all the external forces applied to a rod are parallel to it, according to the Newton s 3 rd law, the tension force will be also parallel to the rod. In the case of a stretched string (rod, rope, etc), we can divide it fictitiously into two parts. Then, at the division point P, the two pieces attract each other with a certain elasticity force. The direction of this force depends on which part of the string is considered, but due to the Newton s 3 rd law, the odulus of the force reains the sae. The force with which the two fictitious parts of the rope interact with each other at the point P describes the state of the rope at that point, and is referred to as the tension. So, we ll distinguish the force F which is applied to the endpoint of a rope, and the tension T, which is defined for any point of the rope and describes the state of it; note that when external forces are applied only to the endpoints of a rope at equilibriu, F = T (this follows fro the force balance condition for any fictitious part of the rope). fact 14: Tension is an elasticity force in a linear construction eleent such as string (rod, wire, etc) 18. For a non-stretchable string 19, if it is being pulled (or pushed, which can happen in the case of a rod) the tension adjusts itself to the externally applied force to prevent stretching. If the ass of a thread or rope can be neglected, the tension is constant along it. In a freely bending rope, the tension force at a point P is parallel to the tangent drawn to the rope at point P. 3. STATICS fact 15: Noral force is the perpendicular to the surface coponent of an elasticity force at the surface of a rigid (nondeforable) body with which the rigid body acts on a contacting body; it adjusts itself to the externally applied force preventing thereby the rigid body fro being defored. Note that if the externally applied force is not perpendicular to the rigid body surface then due to the Newton s 3 rd law, the elasticity force will have both perpendicular to the surface and tangential (parallel to the surface) coponents. The latter anifests itself at the contact points of two solid bodies as the friction force. ore accurately, the friction force is the force at the contact point of two bodies due to the interaction of the olecules of one body with the olecules of the other body when the bodies try to slip one over the other; the surface olecules are kept at their place due to the elasticity forces inside each of the bodies; these elasticity forces are caused by the (typically unnoticeably sall) shear deforation of the bodies. In the case of solid bodies, if the external tangential forces are not too large as copared with the noral forces, the bodies will not slip, and the friction force adjusts itself so as to copensate the external tangential forces; this is called the static friction force. fact 16: (Aontons law.) The axial static friction force at the contact area of two bodies F ax = µ s N, where N is the noral force at that contact area and µ s is a constant dependent on the two contacting aterials, referred to as the static coefficient of friction; it ay also depend on the teperature, huidity, etc. Thus, F ax is independent of the contact area. fact 17: (Coulob s friction law.) When two bodies ove with respect to each other, the friction force at the contact area of these bodies F = µ k N, where N is the noral force at that contact area and µ k is a constant dependent on the two contacting aterials, referred to as the kinetic coefficient of friction; it does depend slightly on the slipping speed, but this dependence is weak and typically ignored. In the case of Olypiad probles, ost often it is assued that µ s = µ k, but soeties these are taken to be different, with µ s > µ k. While the friction laws are very siple and have been forulated a long tie ago 20, deriving it fro the icroscopic (olecular) dynaics turns out to be a very difficult task (there are still papers being published on that topic in research papers, c.f..h. üser et al., Phys. Rev. Lett. 86, 1295 (2000), and O.. Braun et al., Phys. Rev. Lett. 110, STATICS When solving probles on statics, one can always use standard brute-force-approach: equations (1) and (3) tells us that for each body at equilibriu, F = 0 and T = 0. So, for each solid body, we have a force balance condition, and a torque balance condition. According to the standard procedure, these equations are to be projected onto x-, y- and z-axis yielding us 18 In the bulk of three-diensional elastic bodies, instead of tension, the concept of stress is used; the respective description is atheatically ore coplicated, e.g. the stress is a tensor quantity. 19 ore precisely, a string ade of a aterial with a very large value of Young odulus Y. 20 The friction laws were developed by L. da Vinci 1493, G. Aontons 1699, and C.A. Coulob page 6

7 six equations (assuing 3-diensional geoetry); if there are N interacting bodies, the overall nuber of equations is 6N. For a correctly posed proble, we need to have also 6N unknown paraeters so that we could solve this set of algebraic equations. The description of the procedure sounds siple, but solving so any equations ight be fairly difficult. In the case of 2-diensional geoetry, the nuber of equations is reduced twice (while the nuber of force balance equations coes down to two, all torques will be perpendicular to the plane, so there is only one torque equation), but even with only two bodies, we have still 6 equations. Luckily, there are tricks which can help us reducing the nuber of equations! Usually the ain ingenuity lies in idea 1: choose optial axes to zero as any projections of forces as possible. It is especially good to zero the projections of the forces we do not know and are not interested in, for instance, the reaction force between two bodies or the tensile force in a string (or a rod). To zero as any forces as possible it is worthwhile to note that a) the axes ay not be perpendicular; b) if the syste consists of several bodies, then a different set of axes ay be chosen for each body. idea 2: for the torques equation it is wise to choose such a pivot point that zeroes as any oent ars as possible. Again it is especially beneficial to zero the torques of uninteresting forces. For exaple, if we choose the pivot to be at the contact point of two bodies, then the oent ars of the friction force between the bodies and of their reaction force are both zero. As entioned above, for a two-diensional syste, we can write two equations per body for the forces (x- and y- coponents) and one equation (per body) for the torques. We could increase the nuber of equations either by using ore than two projections for force balance equations, or ore than one pivot point ( axis of rotation) for the torque balance. However, fact 18: the axiu nuber of linearly independent equations (describing force and torque balance) equals the nuber of degrees of freedo of the body (three in the two-diensional case, as the body can rotate in a plane and shift along the x- and y-axis, and six in the three-diensional case). So, if we write down two force balance conditions with two torque balance conditions then one of the four equations would always be a redundant consequence of the three others. Equations (1) and (3) see to tell us that for 2-diensional geoetry, we should use one torques equation and two force equations; however, each force equation can be traded for one torques equation. So, apart for the canonical set of two force equations and one torque equation, we can use one force equation with two torque equations (with two different pivot points), and we can also use three torques equations with three different pivot points which ust not lie on a single line. Indeed, let us have two torque balance conditions, i OP i F i = 0 and i O P i F i = 0 where P i is the application point of the force F i. Once we subtract one equation fro the other, we obtain OO F i = OO i F i = 0, which is the projection of the i 21 unless there are parallel forces page 7 3. STATICS force balance condition to the perpendicular of OO. It should be ephasized that at least one torques equation needs to be left into your set of equations: the trading of force equations for torques equations works because a rotation around O by a sall angle dφ followed by a rotation around O by dφ results in a translational otion by OO dφ, but there is no such sequence of translational otions which could result in a rotational otion. idea 3: Using torque balance conditions is usually ore efficient than using force balance conditions because for any force balance condition, we can eliinate only one force 21 by projecting the condition to the perpendicular of that force; eanwhile, if we choose the pivot point as the intersection point of the two lines along which two unparallel forces are applied, both forces disappear fro the equation. pr 1. An end of a light wire rod is bent into a hoop of radius r. The straight part of the rod has length l; a ball of ass is attached to the other end of the rod. The pendulu thus fored is hung by the hoop onto a revolving shaft. The coefficient of friction between the shaft and the hoop is µ. Find the equilibriu angle between the rod and the vertical. µ ω r l This proble is classified as a difficult one because ost people who try to solve it have difficulties in drawing a qualitatively correct sketch. What really helps aking a correct sketch is relying on the idea 2. atheatical siplifications are further offered by fact 19: on an inclined surface, slipping will start when the slope angle fulfills tan = µ. pr 2. On an incline with slope angle there lies a cylinder with ass, its axis being horizontal. A sall block with ass is placed inside it. The coefficient of friction between the block and the cylinder is µ; the incline is nonslippery. What is the axiu slope angle for the cylinder to stay at rest? The block is uch saller than the radius of the cylinder. Here we can again use fact 19 and idea 2 if we add idea 4: soeties it is useful to consider a syste of two (or ore) bodies as one whole and write the equations for the forces and/or the torques for the whole syste.

8 Then, the net force (or torque) acting on the copond body is the su of external forces (torques) acting on the constituents. Our calculations are siplified because the internal forces (torques) between the different parts of the copound body can be ignored (due to Newton s 3 rd they cancel each other out). For the last proble, it is useful to asseble a copound body fro the cylinder and the block. pr 3. Three identical rods are connected by hinges to each other, the outost ones are hinged to a ceiling at points A and B. The distance between these points is twice the length of a rod. A weight of ass is hanged onto hinge C. At least how strong a force onto hinge D is necessary to keep the syste stationary with the rod CD horizontal? A C Again we can use idea 2. The work is also aided by fact 20: if forces are applied only to two endpoints of a rod and the fixture(s) of the rod at its endpoint(s) is (are) not rigid (the rod rests freely on its supports or is attached to a string or a hinge), then the tension force in the rod is directed along the rod. Indeed, at either endpoints, the applied net external force F ust point along the rod, as its torque with respect to the other endpoint ust be zero. Further, according to the Newton s 3 rd law, the external force F ust be et by an equal and opposite force exerted by the rod, which is the tension force T, so F = T. Soe ideas are very universal, especially the atheatical ones. idea K-2 soe extrea are easier to nd without using derivatives, for exaple, the shortest path fro a point to a plane is perpendicular to it. pr 4. What is the iniu force needed to dislodge a block of ass resting on an inclined plane of slope angle, if the coefficient of friction is µ? Investigate the cases when a) = 0; b) 0 < < arctan µ. idea 5: force balance can soeties be resolved vectorially without projecting anything onto axes. Fact 19, or rather its following generalisation, turns out to be of use: idea 6: if a body is on the verge of slipping (or already slipping), then the su of the friction force and the reaction force is angled by arctan µ fro the surface noral. This idea can be used fairly often, for instance in the next proble. F D F B 3. STATICS pr 5. A block rests on an inclined surface with slope angle. The surface oves with a horizontal acceleration a which lies in the sae vertical plane as a noral vector to the surface. Deterine the values of the coefficient of friction µ that allow the block to reain still. Here we are helped by the very universal µ idea 7: any probles becoe very easy in a non-inertial translationally oving reference frae. To clarify: in a translationally oving reference frae we can re-establish Newton s laws by iagining that each body is additionally acted on by an inertial force a where a is the acceleration of the frae of reference and is the ass of a given body. Indeed, we have learned in kineatics that for translational otion of a reference frae, the accelerations can be added, cf. idea K-19; so, in a oving frae, all the bodies obtain additional acceleration a, as if there was an additional force F = a acting on a body of ass. Note that that due to the equivalence of the inertial and heavy ass (cf. Gravity, Section 2.3) the inertial foce is totally analogous to the gravitational force 22. Because of that, we can use idea 8: the net of the inertial and gravitational forces is usable as an effective gravitational force. pr 6. A cylinder with radius R spins around its axis with an angular speed ω. On its inner surface there lies a sall block; the coefficient of friction between the block and the inner surface of the cylinder is µ. Find the values of ω for which the block does not slip (stays still with respect to the cylinder). Consider the cases where (a) the axis of the cylinder is horizontal; (b) the axis is inclined by angle with respect to the horizon. ω idea 9: a rotating frae of reference ay be used by adding a centrifugal force ω 2 R (with ω being the angular speed of the frae and R being a vector drawn fro the axis of rotation to the point in question) and Coriolis force. The latter is uniportant (a) for a body standing still or oving in parallel to the axis of rotation in a rotating frae of reference (in this case the Coriolis force is zero); (b) for energy conservation (in this case the Coriolis force is perpendicular to the velocity and, thus, does not change the energy). Warning: in this idea, the axis of rotation ust be actual, not instantaneous. Expressions for the centrifugal force and Coriolis force are derived in appendix 4. For the proble 6, recall also idea K-2b and idea 6; for part (b), add 22 Their equivalence is the cornerstone of the theory of general relativity (ore specifically, it assues the inertial and gravitational forces to be indistinguishable in any local easureent). page 8 a

9 idea K-11: in case of three-diensional geoetry, consider twodiensional sections. It is especially good if all interesting objects (for exaple, force vectors) lie on one section. The orientation and location of the sections ay change in tie. pr 7. A cart has two cylindrical wheels connected by a weightless horizontal rod using weightless spokes and frictionless axis as shown in the figure. Each of the wheels is ade of a hoogeneous disc of radius R, and has a cylidrical hole of radius R/2 drilled coaxially at the distance R/3 fro the centre of the wheel. The wheels are turned so that the holes point towards each other, and the cart is put into otion on a horizontal floor. What is the critical speed v by which the wheels start juping? This proble is soewhat siilar to the previous one, and we would be able to solve it using those ideas which we have already studied. Indeed, if we consider the process in a frae co-oving with the cart, we can just apply Newton s 2 nd law to the centripetal acceleration of the wheel s centre of ass. However, let us solve it using few ore ideas. idea 10: Gravity force (or a fictitious force which is proportional to the ass of a body) can be considered to be applied to the centre of ass of a body only in the following cases: (a) the effective gravity field is hoogeneous; (b) the body has a spherically syetric ass distribution; (c) the effective gravity field is proportional to the radius vector, e.g. centrifugal force field if the otion is constrained to the plane perpendicular to the frae s axis of rotation. In all the other cases, it ay happen by coincidence that the gravity force is still applied to the centre of ass, but typically it is not. For instance, the Coriolis force can be considered to be applied to the centre of ass only if the body is not rotating (as seen fro the rotating frae). The part (a) of this clai has been otivated in the paragraph following the idea 9; parts (b) and (c) will be otivated in the booklet of electroagnetis (electrostatic and non-relativistic gravitational fields obey siilar laws). In order to prove that the part (d) is valid, we need to show that total force and total torque exerted by the gravitational have the sae agnitudes what would be obtained if the body were a point ass at the position of the centre of ass. So, using the attraction(repulsion) centre as the origin and assuing g = k r, let us express the total torque as T = r i ki r i r i 0; the sae result would be obtained for a centre of ass as the gravity force would have a zero ar. Further, let us express the total force as F = k i r i = k( i r i /) = k r C ; here = i is the total ass of the body. There are two ore ideas which can be used here, idea 11: In order to achieve a ore syetric configuration or to ake the situation sipler in soe other way, it is soeties useful to represent a region with zero value of soe page 9 3. STATICS quantity as a superposition of two regions with opposite signs of the sae quantity. This quantity can be ass density (like in this case), charge or current density, soe force field etc. Often this trick can be cobined with idea 12: ake the proble as syetric as possible. This goal can be reached by applying idea 11, but also by using appropriate reference fraes, dividing the process of solving into several phases (where soe phases use syetric geoetry), etc. pr 8. A hollow cylinder with ass and radius R stands on a horizontal surface with its sooth flat end in contact the surface everywhere. A thread has been wound around it and its free end is pulled with velocity v in parallel to the thread. Find the speed of the cylinder. Consider two cases: (a) the coefficient of friction between the surface and the cylinder is zero everywhere except for a thin straight band (uch thinner than the radius of the cylinder) with a coefficient of friction of µ, the band is parallel to the thread and its distance to the thread a < 2R (the figure shows a top-down view); (b) the coefficient of friction is µ everywhere. Hint: any planar otion of a rigid body can be viewed as rotation around an instant centre of rotation, i.e. the velocity vector of any point of the body is the sae as if the instant centre were the real axis of rotation. v a This is quite a hard proble. It is useful to note idea 13: if a body has to ove with a constant velocity, then the proble is about statics. Also reeber ideas 1 and 2. The latter can be replaced with its consequence, idea 14: if a body in equilibriu is acted on by three forces at three separate points, then their lines of action intersect at one point (note that the intersection point can be infinitely far lines intersecting at infinity eans that the lines are parallel to each other). If there are only two points of action, then the corresponding lines coincide. This very useful idea follows directly fro the torque balance condition if the intersection point of two lines of action is taken for the pivot point (with two ars and the total torque being equal to zero, the third ar ust be also equal to zero). Another useful fact is fact 21: the friction force acting on a given point is always antiparallel to the velocity of the point in the frae of reference of the body causing the friction. Fro tie to tie soe atheatical tricks are also of use; here it is the property of inscribed angles, and ore specifically the particular case of the Thales theore (aong geoet- µ

10 h rical theores, this is probably the ost useful one for solving physics probles), fact 22: a right angle is subtended by a seicircle (in general: an inscribed angle in radians equals half of the ratio between its arc-length and radius). The property of inscribed angles is also useful in the next proble, if we add (soewhat trivial) idea 15: in stable equilibriu the potential energy of a body is iniu. pr 9. A light wire is bent into a right angle and a heavy ball is attached to the bend. The wire is placed onto supports with height difference h and horizontal distance a. Find the position of the wire in its equilibriu. Express the position as the angle between the bisector of the right angle and the vertical. Neglect any friction between the wire and the supports; the supports have little grooves keeping all otion in the plane of the wire and the figure. pr 10. A rod with length l is hinged to a ceiling with height h < l. Underneath, a board is being dragged on the floor. The rod is eant to block the oveent the board in one direction while allowing it ove in the opposite direction. What condition should be fulfilled for it to do its job? The coefficient of friction is µ 1 between the board and the rod, and µ 2 between the board and the floor. µ 2 µ 1 Let s reeber fact 6: if the relative sliding between two bodies has a known direction then the direction of the su of the friction and reaction force vectors is always uniquely deterined by the coefficient of friction. If a force akes one of the bodies ove in such a way that the reaction force grows then they ja: the larger the forces we try to drag the bodies with, the larger friction and reaction forces restrain the. idea 16: Friction can block oveent. In such a case, all forces becoe negligible except for the friction force, reaction force and the externally applied force that tries to ake the syste ove, because gravitational (and such) forces are fixed, but the said forces becoe the larger the harder we push or pull. pr 11. Four long and four half as long rods are hinged to each other foring three identical rhobi. One end of the contraption is hinged to a ceiling, the other one is attached to a weight of ass. The hinge next to the weight is connected to the hinge above by a string. Find the tension force in the string. a page STATICS This proble is the easiest to solve using the ethod of virtual displaceent. ethod 1: Iagine that we are able to change the length of the string or rod the tension in which is searched for by an in- nitesial aount x. Equating the work T x by the change Π of the potential energy, we get T = Π/ x. Generalisation: if soe additional external forces F i (i = 1, 2,...) act on the syste with the displaceents of their points of action being δ x i, while the interesting string or rod undergoes a virtual lengthening of x, then T = ( Π δ x i F i )/ x. i The ethod can also be used for finding soe other forces than tension (for exaple, in probles about pulleys): by iaginarily shifting the point of action of the unknown force one can find the projection of this force onto the direction of the virtual displaceent. pr 12. A rope with ass is hung fro the ceiling by its both ends and a weight with ass is attached to its centre. The tangent to the rope at its either end fors angle with the ceiling. What is the angle β between the tangents to the rope at the weight? β Let us recall the fact 14: The tension in a freely hanging string is directed along the tangent to the string. In addition, we can eploy idea 17: for hanging ropes, ebranes etc. it is usefult to consider a piece of rope separately and think about the coponentwise balance of forces acting onto it. In fact, here we do not need the idea as a whole, but, rather, its consequence, fact 23: the horizontal coponent of the tension in a assive rope is constant. Using the idea 17 and fact 23, it is relatively easy to show that the following approxiation is valid. idea 18: If the weight of a hanging part of a rope is uch less than its tension then the curvature of the rope is sall and its horizontal ass distribution can quite accurately be regarded as constant. This allows us to write down the condition of torque balance for the hanging portion of the rope (as we know the horizontal coordinate of its centre of ass). The next proble illustrates that approach. pr 13. A boy is dragging a rope with length L = 50 along a horizontal ground with a coefficient of friction of µ = 0.6,

11 l l l holding an end of the rope at height H = 1 fro the ground. What is the length l of the part of the rope not touching the ground? pr 14. A light rod with length l is hinged in such a way that the hinge folds in one plane only. The hinge is spun with angular speed ω around a vertical axis. A sall ball is fixed to the other end of the rod. (a) Find the angular speeds for which the vertical orientation is stable. (b) The ball is now attached to another hinge and, in turn, to another identical rod (see the figure below); the upper hinge is spun in the sae way. What is now the condition of stability for the vertical orientation? ω ω a) b) Here the following idea is to be used. idea 19: For analysing stability of an equilibriu, there are two options. First, presue that the syste deviates a little fro the equilibriu, either by a sall displaceent x or by a sall angle φ, and find the direction of the appearing force or torque whether it is towards the equilibriu or away fro it. Second, express the change of total potential energy in ters of the sall displaceent to see if it has a iniu or axiu (for a syste at equilibriu, its potential energy ust have an extreu); iniu corresponds to stability, and axiu to instability (for a otivation and generalization of this ethod, see appendix 5). NB! copute approxiately: when working with forces (torques), it is alost always enough to keep only those ters which are linear in the deviation; when working with potential energy, quadratic approxiation is to be used. It is extreely iportant in physics to be able to apply linear, quadratic, and soeties also higher order approxiations, which is based on idea 20: Taylor series: f(x + x) f(x) + f (x) x + f (x) x , for instace: sin φ tan φ φ; cos φ 1 x2 2 ; ex 1 + x + x2 2, (1 + x) a 1 + ax + a(a 1) 2 x 2, ln 1 + x 1 + x x2 2. Analogous approach can be use for ultivariable expressions, e.g. (x + x)(y + y) xy + x y + y x. Consider using such approxiations wherever initial data suggest soe paraeter to be sall. The case (b) is substantially ore difficult as the syste has two degrees of freedo (for exaple, the deviation angles φ 1 and φ 2 of the rods). Although idea 19 is generalisable for ore than one degrees of freedo, apparently it is easier to start fro idea 15. idea 21: The equilibriu x = y = 0 of a syste having two degrees of freedo is stable if (and only if 23 ) the potential 3. STATICS 23 We assue that apart fro the energy, there are no other conserved quantities for this syste. page 11 energy Π(x, y) as a function of two variables has a local iniu at x = y = 0, i.e. for any pair of values x, y within a sall neigbourhood of the equilibriu point (0, 0), inequality Π(x, y) > Π(0, 0) ust hold. pr 15. If a bea with square cross-section and very low density is placed in water, it will turn one pair of its long opposite faces horizontal. This orientation, however, becoes unstable as we increase its density. Find the critical density when this transition occurs. The density of water is ρ w = 1000 kg/ 3. idea 22: The torque acting on a body placed into a liquid is equal to torque fro buoyancy, if we take the latter force to be acting on the centre of the ass of the displaced liquid. The validity of the idea 22 can be seen if we iagine that the displaced volue is, again, filled with the liquid, and the body itself is reoved. Then, of course, the re-filled volue is at equilibriu (as it is a part of the resting liquid). This eans that the torque of the buoyancy force ust be balancing out the torque due to the weight of the re-filled volue; the weight of the re-filled volue is applied to its centre of ass, and according to the idea 14, the buoyancy force ust be therefore also acting along the line drawn through the centre of ass. Apart fro the idea 22, solution of the proble 15 can be siplified by using the ideas 11 and 12. pr 16. A heispherical container is placed upside down on a sooth horizontal surface. Through a sall hole at the botto of the container, water is then poured in. Exactly when the container gets full, water starts leaking fro between the table and the edge of the container. Find the ass of the container if water has density ρ and radius of the heisphere is R. idea 23: If water starts flowing out fro under an upside down container, noral force ust have vanished between the table and the edge of the container. Therefore force acting on the syste container+liquid fro the table is equal solely to force fro hydrostatic pressure. The latter is given by ps, where p is pressure of the liquid near the tabletop and S is area of the container s open side. pr 17. A block is situated on a slope with angle, the coefficient of friction between the is µ > tan. The slope is rapidly driven back and forth in a way that its velocity vector u is parallel to both the slope and the horizontal and has constant odulus v; the direction of u reverses abruptly after each tie interval τ. What will be the average velocity w of the block s otion? Assue that gτ v. x y v u v u R τ z t

12 h idea 24: If the syste changes at high frequency, then it is often pratical to use tie-averaged values X instead of detailed calculations. In ore coplicated situations a highfrequency coponent X ight have to be included (so that X = X + X). 3. STATICS pr 19. A horizontal platfor rotates around a vertical axis at angular velocity ω. A disk with radius r can freely rotate and ove up and down frictionlessly along a vertical axle which is fixed to a distance d > r fro the platfor s axis. The disk is pressed against the rotating platfor due to gravity, the coefficient of friction between the is µ. Find the angular velocity acquired by the disk. Assue that pressure is distributed evenly over the entire base of the disk. R d ω r ethod 2: (perturbation ethod) If the ipact of soe force on a body's otion can be assued to be sall, then solve the proble in two (or ore) phases: rst nd otion of the body in the absence of that force (so-called zeroth approxiation); then pretend that the body is oving just as found in the rst phase, but there is this sall force acting on it. Look what correction (so-called rst correction) has to be ade to the zeroth approxiation due to that force. In this particular case, the choice of zeroth approxiation needs soe explanation. The condition gτ v iplies that within one period, the block s velocity cannot change uch. Therefore if the block is initially slipping downwards at soe velocity w and we investigate a short enough tie interval, then we can take the block s velocity to be constant in zeroth approxiation, so that it is oving in a straight line. We can then ove on to phase two and find the average value of frictional force, based on the otion obtained in phase one. For proble 17, recall also a lesson fro kineatics, idea K-7 If friction aects the otion then usually the ost appropriate frae of reference is that of the environent causing the friction. pr 18. Let us investigate the extent to which an iron deposit can influence water level. Consider an iron deposit at the botto of the ocean at depth h = 2 k. To siplify our analysis, let us assue that it is a spherical volue with radius 1 k with density greater fro the surrounding rock by ρ = 1000 kg/ 3. Presue that this sphere touches the botto of the ocean with its top, i.e. that its centre is situated at depth r + h. By how uch is the water level directly above the iron deposit different fro the average water level? iron deposit idea 25: The surface of a liquid in equilibriu takes an equipotential shape, i.e. energies of its constituent particles are the sae at every point of the surface. If this was not the case, the potential energy of the liquid could be decreased by allowing soe particles on the surface to flow along the surface to where their potential energy is saller (cf. idea 15). Recall also the fact 10. r page 12 idea 26: If we transfor into a rotating frae of reference, then we can add angular velocities about instantaneous axes of rotation in the sae way as we usually add velocities. Thus ω 3 = ω 1 + ω 2, where ω 1 is angular velocity of the reference frae, ω 2 angular velocity of the body in the rotating frae of reference and ω 3 that in the stationary frae. In this question, we can use fact 21, ideas 2, 9, 13, K-7, and also idea K-33 Arbitrary otion of a rigid body can be considered as rotation about an instantaneous centre of rotation (in ters of velocity vectors of the body). ethod 3: (dierential calculus) Divide the object into in- nitesially sall bits or the process into innitesially short periods (if necessary, cobine this with idea 20). Within an infinitesial bit (period), quantities changing in space (tie) can be taken constant (in our case, that quantity is the direction of frictional force vector). If necessary (see the next question), these quantities ay be sued over all bits this is called integration. pr 20. A waxing achine consists of a heavy disk with ass densely covered with short bristles on one side, so that if it lies on the floor, then its weight is evenly distributed over a circular area with radius R. An electrical otor akes the disk rotate at angular velocity ω, the user copensates for the torque fro frictional forces by a long handle. The sae handle can be used to push the achine back and forth along the floor. With what force does the achine have to be pushed to ake it ove at velocity v? Assue that angular velocity of the disk is large, ωr v, and that the force needed to copensate for the torque can be neglected. The coefficient of friction between the bristles and the floor is µ. Here we need fact 21, ideas K-33, 11, and additionally idea 27: Try to deterine the region of space where forces (or torques etc) cancel at pairs of points. These pairs of points are often syetrically located. Idea 12 is relevant as well. pr 21. A hexagonal pencil lies on a slope with inclination angle ; the angle between the pencil s axis and the line of intersection of the slope and the horizontal is φ. Under what condition will the pencil not roll down? ϕ idea 28: When solving three-diensional probles, soeties calculating coordinates in appropriately chosen axes and

13 applying forulae of spatial rotations can be of use. For the rotation around z-axis by angle φ, x = x cos φ y sin φ and y = y cos φ + x sin φ. What (which vector) could be expressed in ters of its coponents in our case? The only proising option is the sall shift vector of centre of ass when its starts to ove; ultiately we are only interested in its vertical coponent. idea 29: Unfolding a part of the surface of a threediensional object and looking at the thereby flattened surface can assist in solving probles, aong other things it helps to find shortest distances. pr 23. A unifor bar with ass and length l hangs on four identical light wires. The wires have been attached to the bar at distances l 3 fro one another and are vertical, whereas the bar is horizontal. Initially, tensions are the sae in all wires, T 0 = g/4. Find tensions after one of the outerost wires has been cut. l/3 P l 4. DYNAICS they ust have the sae stiffness as well; the word wire hints at large stiffness, i.e. deforations (and the inclination angle of the bar) are sall. 4 DYNAICS A large proportion of dynaics probles consist of finding the acceleration of soe syste of bodies, or finding the forces acting upon the bodies. There are several possible approaches for solving such probles, here we consider three of the. pr 22. A slippery cylinder with radius R has been tilted to ake an angle between its axis and the horizontal. A string with length L has been attached to the highest point P of soe cross-section of the cylinder, the other end of it is tied to a weight with ass. The string takes its equilibriu position, how long (l) is the part not touching the cylinder? The the equation on x, y, and possibly z-axes). weight is shifted fro its equilibriu position in such a way that the shift vector is parallel to the vertical plane including the cylinder s axis; what is the period of sall oscillations? ethod 4: For each body, we nd all the forces acting on it, including noral forces and frictional forces 24, and write out Newton's 2 nd law in ters of coponents (i.e. by projecting NB! Select the directions of the axes carefully, cf. idea 1. In soe cases, it ay be possible (and ore convenient) to abstain fro using projections and work with vectorial equalities. Keep in ind that for a correctly posed proble, it should be possible to write as any linearly independent equations as there are unknowns (following idea 1 ay help to reduce that nuber). The guideline for figuring out how any equations can be found reains the sae as in the case of statics proble, see idea 18 (for tie being we consider probles where bodies do not rotate, so we need to count only the translational degrees of freedo). If the nuber of equations and the nuber of unknowns don t atch, it is either an ill-posed proble, or you need to ake additional physical assuptions (like in the case of proble 23). pr 24. A block with ass lies on a slippery horizontal surface. On top of it there is another block with ass which in turn is attached to an identical block by a string. The string has been pulled across a pulley situated at the corner of the big block and the second sall block is hanging vertically. Initially, the syste is held at rest. Find the acceleration of the big block iediately after the syste is released. You ay neglect friction, as well as asses of the string and the pulley. l idea 30: If ore fixing eleents (rods, strings, etc) than the necessary iniu have been used to keep a body in static equilibriu (i.e. ore than the nuber of degrees of freedo) and fixing eleents are absolutely rigid, then tensions in the eleents cannot be deterined. In order to ake it possible, the eleents have to be considered elastic (able to defor); recall the fact 13. idea 32: Let us note that this stateent is in accordance with fact 18 that gives the nuber of available equations (there can be no ore unknowns than equations). In this particular case, we are dealing with effectively one-diensional geoetry with no horizontal forces, but the body could rotate (in absence of the wires). Thus we have two degrees of freedo, corresponding to vertical and rotational otion. Since the wires are identical, This question can be successfully solved using ethod 4, but we need three ore ideas. idea 31: If a body is initially at rest, then its shift vector is parallel to the force acting on it (and its acceleration) right after the start of its otion. If bodies are connected by a rope or a rod or perhaps a pulley or one is supported by the other, then there is a linear 25 arithetic relation between the bodies shifts (and velocities, accelerations) that describes the fact that length of the string (rod, etc.) is constant. The relation for shifts is usually the easiest to find; if the otion is along a straight line, this relation can be differentiated 24 It is convenient to ake a sketch and draw all the force vectors fro their application points. 25 It is linear in shifts but ay contain coefficients which are expressed in ters of nonlinear functions, e.g. trigonoetric functions of angles. page 13