Fundamental mechanics : newtonian mechanics for engineering

ENGINYERIES INDUSTRIALS Fundamental Mechanics. Newtonian Mechanics for Engineering Xavier Jaén Josep Salud Carina Serra Jaume Calaf Maria Khoury www.upc.edu/idp Fundamental Mechanics. Newtonian Mechanics for Engineering This book is addressed at a first course in engineering mechanics. Newtonian mechanics is studied, a non-relativistic classical mechanics and, therefore, applied to objects that are neither extremely small nor excessively fast. The newtonian mechanics affects a good part of the world around us. It is the mechanics of the everyday world. The book includes the basic concepts of mechanics with theoretical demonstrations. You will also find issues and problems, some resolved. The text includes graphs, figures, and diagrams for facilitate understanding. Mainly the foundations of mechanics are studied, but you will also find an introduction to wave phenomena and a small foray into analytical mechanics. The authors are professors from the Physics Department of the Universitat Politècnica de Catalunya, who teach physics, some with more than 20 years of experience. ENGINYERIES INDUSTRIALS UPCGRAU UPCGRAU Xavier Jaén, Josep Salud Carina Serra, Jaume Calaf Maria Khoury 9788419 184641 Fundamental Mechanics. Newtonian Mechanics for Engineering ENGINYERIES INDUSTRIALS UPCGRAU Fundamental Mechanics. Newtonian Mechanics for Engineering Xavier Jaén Josep Salud Carina Serra Jaume Calaf Maria Khoury The authors and collaborators, Manel Canales, Josep Sempau and Claudia Grossi, are part of the teaching staff in the UPC Physics Department Design and layout made by the authors, based on Tufte-Latex by Bil Kleb, Bill Wood and Kevin Godby (2007-2015) inspired by the works of Edward Tufte. First edition: january 2023 © The authors, 2023 © Iniciativa Digital Politècnica, 2023 Oficina de Publicacions Acadèmiques Digitals de la UPC Edifici K2M, Planta S1, Despatx S103-S104 Jordi Girona 1-3, 08034 Barcelona Tel.: 934 015 885 www.upc.edu/idp mail: [email protected] ISBN digital: 978-84-19184-65-8 ISBN paper: 978-84-19184-64-1 DL B. 3334-2023 the contents of this work are subject to the Creative Commons license: Attribution-NonCommercial-NoDerivs 3.0. Contents Prologue 11 1 Physico-mathematical foundations of mechanics 17 Introduction.............................................. 17 1.1 Space, time and reference frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 1.2 Euclidean Cartesian coordinates. Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 1.3 Scalarsandvectors ....................................... 18 1.4 Principleofsymmetry...................................... 24 1.5 Measurement and treatment of experimental data . . . . . . . . . . . . . . . . . . . . . . . 25 1.6 Newton’s first law. Inertial reference frames . . . . . . . . . . . . . . . . . . . . . . . . . . 33 1.7 Point kinematics: Position, trajectory, velocity and acceleration . . . . . . . . . . . . . . . . 36 2 Dynamics of a particle 43 Introduction.............................................. 43 2.1 Newton’s first and second laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 2.2 Forceandmomentum...................................... 49 2.3 Torque and angular momentum of a particle . . . . . . . . . . . . . . . . . . . . . . . . . . 51 2.4 Work, kinetic energy and potential energy. Power . . . . . . . . . . . . . . . . . . . . . . . 54 2.5 Someforces........................................... 63 3 Dynamics of Nparticles 75 Introduction.............................................. 75 3.1 Forces between particles. Newton’s second and third laws . . . . . . . . . . . . . . . . . . 75 3.2 Netforceandcentreofmass .................................. 76 3.3 Momentum ........................................... 81 3.4 Angularmomentum....................................... 82 3.5 Work, kinetic energy and potential energy . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 3.6 Collisions............................................ 89 3.7 Gravitational and electromagnetic interaction . . . . . . . . . . . . . . . . . . . . . . . . . 92 3.8 Constraints and reactions. Possible displacements and virtual displacements... . . . . . . . . 94 Contents 3.9 The general equation of dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 3.10 Conservative system with constraints. Energy conservation . . . . . . . . . . . . . . . . . . 99 3.11Rigidbody ........................................... 104 3.12 Topics in rigid body kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 3.13 Equations of motion for a rigid body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 3.14Torque.............................................. 113 4 The statics of rigid bodies 117 Introduction.............................................. 117 4.1 The statics of a body: Equilibrium conditions . . . . . . . . . . . . . . . . . . . . . . . . . 117 4.2 Weight and centre of gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 4.3 Forces on bodies due to gravitating fluids: Archimedes Principle . . . . . . . . . . . . . . . 120 4.4 Constraints and reaction forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 4.5 The statics of Nrigidbodies.................................. 125 4.6 Virtualworkprinciple...................................... 127 4.7 Equilibrium and stability in conservative systems . . . . . . . . . . . . . . . . . . . . . . . 130 5 Dynamics of a rigid body in a plane 135 Introduction.............................................. 135 5.1 Translationequationin3D ................................... 136 5.2 Rotation equation for (2D) plane motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 5.3 Kinetic energy of rotation and translation. Energy conservation . . . . . . . . . . . . . . . . 144 6 Small oscillations 153 Introduction.............................................. 153 6.1 Small oscillations around a stable equilibrium position . . . . . . . . . . . . . . . . . . . . 153 6.2 Simple harmonic motion (SHM) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 6.3 Damped harmonic motion (DHM) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 6.4 Forced harmonic motion (FHM) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 7 Mechanical waves 167 Introduction.............................................. 167 7.1 Waves.............................................. 167 7.2 Plane waves and the wave equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 7.3 From Newton’s laws to the wave equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 7.4 Fourier analysis and synthesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 8 Wave phenomena 189 Introduction.............................................. 189 8.1 The power and intensity of plane waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 8.2 The power and intensity of spherical waves . . . . . . . . . . . . . . . . . . . . . . . . . . 194 6 Contents 8.3 Transmission and reflection of a wave at a change in medium . . . . . . . . . . . . . . . . . 196 8.4 Interferenceandbeats...................................... 198 8.5 Standingwaves ......................................... 202 8.6 Doppler effect and shock waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 8.7 Diffraction............................................ 211 9 Lagrange equations 215 Introduction.............................................. 215 9.1 Lagrange equations of the second kind . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 9.2 Lagrangeequations....................................... 217 1 Problems and questions 223 1.5 Measurement and treatment of experimental data . . . . . . . . . . . . . . . . . . . . . . . . 224 1.7 Point kinematics: position, trajectory, velocity and acceleration . . . . . . . . . . . . . . . . . 224 2 Problems and questions 227 2.1 Newton’s first and second laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 2.2Forceandmomentum....................................... 228 2.3 Torque and angular momentum of a particle . . . . . . . . . . . . . . . . . . . . . . . . . . . 230 2.4 Work, kinetic energy and potential energy. Power . . . . . . . . . . . . . . . . . . . . . . . . 231 2.5Someforces............................................ 236 3 Problems and questions 243 3.2Netforceandcentreofmass ................................... 243 3.3Momentum............................................ 247 3.4Angularmomentum........................................ 248 3.5 Work, kinetic energy and potential energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 3.6Collisions ............................................. 250 3.7 Gravitational and electromagnetic interaction . . . . . . . . . . . . . . . . . . . . . . . . . . 252 3.10 Conservative system with constraints. Energy conservation . . . . . . . . . . . . . . . . . . 253 3.12 Topics of rigid body kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 3.13 Equations of motion of the rigid body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 4 Problems and questions 257 4.2Weigthandcentreofgravity ................................... 257 4.3 Forces on bodies due to gravitating fluids. Archimedes Principle . . . . . . . . . . . . . . . . 257 4.5 Statics of Nrigidbodies ..................................... 260 4.7 Equilibrium and stability in conservative systems . . . . . . . . . . . . . . . . . . . . . . . . 269 5 Problems and questions 277 5.2 Rotation equation for (2D) plane motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 5.3 Kinetic energy of rotation and translation. Conservation of energy . . . . . . . . . . . . . . . 284 7 Contents 6 Problems and questions 291 6.2 Simple harmonic motion (SHM) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291 6.3 Damped harmonic motion (DHM) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298 6.4 Forced harmonic motion (FHM) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301 7 Problems and questions 311 7.2 Plane waves and wave equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311 7.3 From Newton’s laws to the wave equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313 7.4 Fourier analysis and synthesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314 8 Problems and questions 317 8.1 Power and intensity of plane waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 8.2 Power and intensity of spherical waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 8.3 Transmission and reflection of a wave at a change in medium . . . . . . . . . . . . . . . . . . 318 8.4Interferenceandbeats....................................... 320 8.5Standingwaves .......................................... 323 8.6 Doppler effect and shock waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327 Solutions to the questions 331 Tables 335 8 1 Physico-mathematical foundations of mechanics Introduction This chapter combines physics and mathematics because it is a very fine line that separates the most basic concepts of physics from mathematics. 1.1 Space, time and reference frames Fig. 1.1: Observer Physics in general, and mechanics in particular, study the behaviour of objects in space time. Although some current physical theories debate the reality of this spacetime, we will not do so here. It is enough to view spacetime as a very useful intellectual construction for systematising everything that happens, as we all agree on how to organize our perceptions of events by placing them at different instants of time in a three-dimensional space. An observer (see Figure 1.1) is a recording system (human or automatic) of the position of objects at each instant in space. He or she can assign a triad of numbers to each point Awhich will apply a unique label to it: the coordinates of point A= (xA, yA, zA). He or she will be able to assign to each point Aa time tA, which is synchronized with the time of clock tOusing the maximum available speed c: tA−tO=dAO/c(1.1) where dAO is the distance between Aand the point where the observer’s clock is located. Later, we will see that using any speed other than the maximum gives rise to inconsistencies. The set of objects and labels used by the observer constitute the reference frame. By international agreement, all observers define time and space patterns in the same way. The BIPM (Bureau International des Poids et Mesures) is the institution that oversees the SI (International System of Units). The final definitions are: 1 Physico-mathematical foundations of mechanics Unit of time. The second, s:a time interval of 1second is equal to 9192631770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium 133 atom (Adopted in 1967). Unit of distance. The meter, m:a spatial interval of 1meter is equal to the distance travelled by light in vacuum during a time interval of 1/299792458 s (Adopted in 1983). 1.2 Euclidean Cartesian coordinates. Distance Fig. 1.2: Euclidean Cartesian coordinates of the Apoint A remarkable fact about physical space is that it can be labelled using Euclidean Cartesian coordinates (simply Cartesian, if there is no doubt). An important property of these coordinates is that the distance, or shortest path between two points, A= (xA, yA, zA)and B= (xB, yB, zB)(see Figures 1.2 and 1.3) can be expressed as d(A, B) = √(xA−xB)2+ (yA−yB)2+ (zA−zB)2(1.2) This is not possible for the space points on the surface of a sphere. Fig. 1.3: Distance from point Ato B Point (0,0,0) is the origin of coordinates; and the straight lines (x, 0,0),(0, y, 0) and (0,0, z)are the coordinate axes. 1.3 Scalars and vectors Physical quantities have some characteristics in relation to the observer measuring them. According to this criterion, we highlight and use two categories of physical magnitudes: scalars and vectors. Scalar field is a map that associates points and instants with values that are independent of the observer’s orientation (or rotation), f(x, y, z, t). It is also called ascalar. Vector field is a map that associates points and instants with vectors: a positive scalar (modulus or magnitude) and a direction (line of action and sign),  V(x, y, z, t). It is also called a vector. Examples of scalar magnitudes are temperature, volume and density, among others. Examples of vector magnitudes are the velocity of a particle associated with the points and instant it passes through, the gravitational field associated with all points at each instant, and the force exerted by an electrical charge wherever one may be. The idea that a vector exists at each point in space is why we use the name of field. Let us imagine a wheat field in which each ear of wheat is the vector of a point in space. The set of ears of wheat, that is, the entire field of wheat, is what we call the vector field or simply vector. We can also speak of a scalar field such us a 18 1.3 Scalars and vectors field of densities or temperatures, although not in the sense of a point’s density or temperature in spacetime, but of a function that assigns at each point and at each instant a density or temperature that may be different. Fig. 1.4: At each point in space we have the three basis vectors {ˆı, ˆȷ, ˆ k} The Cartesian basis of vectors is formed by the unit vectors {ˆı, ˆȷ, ˆ k}that run parallel to the coordinate axes and are defined at all points in space (see Figure 1.4). Any vector  Vcan be written as  V=Vxˆı+Vyˆȷ+Vzˆ k= (Vx, Vy, Vz)(1.3) where Vx,Vyand Vz, are the Cartesian components of the vector, which can be functions of (x, y, z, t).1 1If any of the components is numeric and we are using a comma as a separator, we can use a semicolon (;) instead of a comma (,). For example, (2; 5,603; 7.2). In this book we will always use a comma The basis vectors {ˆı, ˆȷ, ˆ k}have no units and are not associated with any magnitude. They represent directions of space, which in this case are the three directions of the coordinate axes. Note that a component of a vector is not a scalar. Algebraic operations with vectors If we have two vectors  Aand  B, it can be shown that (Ax+Bx, Ay+By, Az+Bz) is a vector. The sum of two vectors is defined according to this property. Sum: Given two vectors  Aand  B, their sum  A+ Bis the vector which, in Cartesian coordinates, is expressed as  A+ B= (Ax+Bx, Ay+By, Az+Bz)(1.4) If Uis a scalar and  Ais a vector, it can be shown that (UAx, UAy, UAz)is a vector. According to this, the external product or the product of a scalar and a vector can be defined. External product: Given a scalar Uand a vector  A, the external product is the vector U A, which in Cartesian coordinates can be expressed as U A= (UAx, UAy, UAz)(1.5) If we have two vectors  Aand  B, it can be shown that AxBx+AyBy+AzBzis a scalar. According to this, we define the scalar (or dot) product. Scalar product: Given two vectors  Aand  B, the scalar product is the scalar  A· Bwhich in Cartesian coordinates can be expressed as  A· B=AxBx+AyBy+AzBz(1.6) 19 1 Physico-mathematical foundations of mechanics The modulus of a vector  Ais the scalar: A=√ A· A(1.7) The cosine of the angle 0≤θ≤π(see Figure 1.5) between two vectors  Aand  Bcan be expressed as Fig. 1.5: Angle θbetween two vectors cos θ= A· B AB (1.8) The unit vector ˆ Aof a vector  Ais the unit module vector ˆ A= A A(1.9) Given two vectors  Aand  B, it can be proved that (AyBz−AzBy, AzBx−AxBz, AxBy−AyBx) is a vector. Accordingly, we have defined the vector product. Vector product: Given two vectors  Aand  B, the vector product is the vector  A× B(see Figure 1.6), which in Cartesian coordinates can be expressed as  A× B≡ ˆıˆȷˆ k AxAyAz BxByBz  = (AyBz−AzBy, AzBx−AxBz, AxBy−AyBx) (1.10) The sine of the angle 0≤θ≤πbetween two vectors  Aand  Bcan be expressed as Fig. 1.6: Vector product  A× B between two vectors. Corkscrew rule. sinθ= A× B AB (1.11) Corkscrew rule:The vector product  A× Bcan be expressed as Fig. 1.7: Reference frame {x, y, z}of positive orientation  A× B=B A sinθˆu(1.12) where ˆuis a unit vector normal to  Aand  Bthe direction given by an advancing dextro-rotatory screw rotating from  Ato  Bon the shortest path (see Figure 1.6). It is especially important to use {x, y, z}reference frames of positive orientation, that is, frames that can be juxtaposed onto the frame in Figure 1.7 only by translation and rotation. 20 1.3 Scalars and vectors Algebraic relations with vectors These are equalities (which we will give without proof) involving the above defined operations:  A·( B× C) = (  A× B)· C(1.13)  A×( B× C) =  B( A· C)− C( A· B)(1.14) ( A× B)·( C× D) = (  A· C)(  B· D)−( A· D)(  B· C)(1.15) Differential operations with vectors Given a function f(x, y, z, t), we may want to derive it with respect to some of the variables. For example, we will indicate its xderivative with the operator ∂f ∂x and will say that we make the partial derivative with respect to x. If we want to derive it only with respect to t, we will indicate it with ∂f ∂t and will say that we make the partial derivative with respect to t. It must be remembered that the chain rule for functions states that if F(λ) = f(g(λ), λ), then d dλ F=∂F ∂g ∂g ∂λ +∂F ∂λ . When we derive vectors, we can use the usual rules of derivation, especially Leibniz’s rule for the product. In general, scalars and vector components are functions of the type f(x, y, z, t). In addition, it can happen that the coordinates {x, y, z} are part of a trajectory and, therefore, may depend on tor on some other λparameter: {x(t), y(t), z(t)}or {x(λ), y(λ), z(λ)}. The derivation operator Dmay be: D=∂ ∂x ,D=∂ ∂y ,D=∂ ∂z ,D=∂ ∂t ,D=d dt ,D=∂ ∂λ ,D=d dλ . All of them will comply with the usual rules of derivation and especially with the Leibniz rule. The Leibniz rule for functions fand gwith the usual product is D(fg) = (Df)g+ f(Dg). When the expressions involve scalars Uand vectors  Aand  Bwith the scalar ·and vector ×products, we have: D(U A) = (DU) A+U(D A)(1.16) D( A· B) = (D A)· B+ A·(D B)(1.17) D( A× B) = (D A)× B+ A×(D B)(1.18) An important property of the Cartesian basis (associated with the {x, y, z}coordinates) {ˆı, ˆȷ, ˆ k}is that Dˆı=Dˆȷ=Dˆ k= 0 (1.19) 21 1 Physico-mathematical foundations of mechanics This is so because at each space point the vectors of the fields {ˆı, ˆȷ, ˆ k}remain parallel (see Figure 1.8).This does not happen if, for example, a spherical basis is used. At each point in space, the radial unit field vectors ˆrdo not remain parallel. Nor do the other two fields completing the spherical basis (see Figure 1.9). Fig. 1.8: The ˆ kbasis vectors are parallel. So are ˆıand ˆȷ Property (1.19) is very useful. By deriving a vector on a Cartesian basis D V=D(Vxˆı) + D(Vyˆȷ) + D(Vzˆ k) = D(Vx)ˆı+D(Vy)ˆȷ+D(Vz)ˆ k it is enough to derive its components: D V= (DVx, DVy, DVz) In the case of derivation with respect to t, it is interesting to note that the variables x,yand zmay also depend on tbecause they are part of a trajectory. In other words, they are in fact functions of t, and the function fis a function only of t, f(x(t), y(t), z(t), t), although we do not make it explicit. In this case, the derivative with respect to tis called the total derivative and is written as ˙ f=df dt . To calculate this, we will use the chain rule for each coordinate as well as the time partial derivative, i.e.: ˙ f=∂f ∂x ˙x+∂f ∂y ˙y+∂f ∂z ˙z+∂f ∂t (1.20) Fig. 1.9: The ˆrbasis vectors are not parallel. Nor are ˆ θand ˆ ϕ Concerning total derivatives, we have the following two important results. Given a scalar f, the total derivative df dt is a scalar. Given a vector v, the total derivative dv dt is a vector. Other basic differential operations between scalars and vectors are the following. Given a scalar U, the gradient  ∇Uis the vector that in Cartesian coordinates can be written as  ∇U=∂U ∂r ≡(∂U ∂x ,∂U ∂y ,∂U ∂z )(1.21) The differential of U(x, y, z)is the infinitesimal scalar expression dU, which in Cartesian coordinates can be written as dU =∂U ∂x dx +∂U ∂y dy +∂U ∂z dz (1.22) We can also express as dU = ∇U·dr =∂U ∂r ·dr where dr ≡(dx, dy, dz). In physics, we can identify dx with an arbitrary increase in the magnitude x, with the condition that it is of the same order or smaller than the error εx, with which we measure x:dx .εx. The same can be stated for dy,dz and dt. We will also have dU .εu(see the subsection Error propagation in Section 1.5). 22 1.3 Scalars and vectors Integral operations with vectors There are three basic integral operations relating scalars and vectors. The temporal integral of a scalar U(r, ˙ r, t)defined on a trajectory r(t)is the scalar ∫Udt =∫U(r(t),˙ r(t), t)dt (1.23) The temporal integral of a vector  Vdefined on a trajectory r(t)is the vector ∫ V dt =(∫Vxdt, ∫Vydt, ∫Vzdt)(1.24) Fig. 1.10: Circulation of  Falong a path C The integral of a space defined vector  Fon a path C, which is also called circulation of  Falong C, if the path is given parametrically by r(λ) = (x(λ), y(λ), z(λ)) from λ1to λ2(see Figure 1.10), is the scalar ∫ C  F·dr = λ2 ∫ λ1  F·dr dλdλ = λ2 ∫ λ1(Fx dx dλ +Fy dy dλ +Fz dz dλ)dλ (1.25) Problem 1.3.1. Given the vectors  A= 5ˆı+ 4ˆȷ+ 3ˆ kand  B=−2ˆȷ+ˆ k, calculate: a) The modules. b) The scalar and vector products c) The angle between them d) Find a unit vector perpendicular to  Aand  B Solution a) A=√ A· A=√AxAx+AyAy+AzAz=√52+ 42+ 32= 5√2 B=√ B· B=√BxBx+ByBy+BzBz=√02+ (−2)2+ 12=√5 b)  A· B=AxBx+AyBy+AzBz= 5 ·0+4·(−2) + 3 ·1 = −5  A× B= ˆıˆȷˆ k AxAyAz BxByBz  = 10ˆı−5ˆȷ−10ˆ k c) From the scalar product, we get the angle between the vectors  A· B=AB cos θ⇒cos θ= A· B AB =−1 √10 ⇒θ= 108.4◦ 23 1 Physico-mathematical foundations of mechanics d) The vector product gives the expressions of a vector  C= A× Bperpendicular to  Aand  B ˆ C= C C= A× B  A× B =10ˆı−5ˆȷ−10ˆ k √102+ (−5)2+ (−10)2=2 3ˆı−1 3ˆȷ−2 3ˆ k  Problem 1.3.2. Let  A(t)be a vector that is a function of time. Prove that  A·d A dt =AdA dt and that if  Ahas a constant modulus, then it is perpendicular to d A dt . Solution A2=AA = A· Ad(A·A) dt = 2AdA dt d( A· A) dt = 2  A·d A dt ⇒ A·d A dt =AdA dt if  Ahas a constant modulus, dA dt = 0,⇒ A·d A dt = 0, and therefore these vectors are perpendicular.  1.4 Principle of symmetry Fig. 1.11: Pierre Curie (1859-1906) was a French physicist The idea of symmetry is not exclusive to mechanics, nor even to physics. It has been used in many fields of science. Reasons of symmetry or arguments of symmetry are given to justify considerations that greatly simplify some problems, but it was not until relatively recently that any attempt has been made to specify what we mean by “reason of symmetry”. Pierre Curie was the first to state a principle of symmetry, in 1894: An effect cannot lack a symmetry if this symmetry is not lacking in the cause. Fig. 1.12: Were it not for the vertices being labelled, the two triangles would be indistinguishable That is, if the cause has a symmetry, it must be present in the effect. We will give a more detailed version of this idea, first by specifying what symmetry is. With an object Sand one transformation Tof any kind (see Figure 1.12), we say that Tis a symmetry transformation of Sor that Shas the Tsymmetry if Sremains unchanged when applying T T(S) = S(1.26) 24 1.5 Measurement and treatment of experimental data Principle of symmetry: If the causes have symmetries, then the effects have at least the same symmetries. Note that we need to make sure that we control all the causes, because it could happen that some of the causes are symmetrical and we have no information about the rest. In this case, the effects would not have to be identical! We use this principle many times without being aware of it, because it is very intuitive. Other times it does not seem like much to us, but it can also be applicable. Problem 1.4.1. A uniform spherical mass distribution creates a gravitational field. Arguing exclusively with the symmetry principle, what can we say about the gravitational field? Solution The cause, that is, the uniform spherical mass distribution, has spherical symmetry: If we make any rotation around its centre, it remains unchanged. The effect, in this case the gravitational field, must therefore also have spherical symmetry. The gravitational field must be radial and depends only on the distance to the centre of the distribution.  1.5 Measurement and treatment of experimental data We can never give a result in the form of a real number. This is so for many reasons. For example, suppose we want to measure the length of a rod like the one in Figure 1.13. Fig. 1.13: The error in the measurement may be inherent to the measured object 1) The assumption that the rod has a well-defined length is an idealization. Real rods are not a sharply cut prism. 2) The measuring device has some limitations due to construction. No matter how well built the ruler is, it has divisions with a certain thickness that we can see. These divisions are separated in such a way that they coincide with the end of the rod by chance. 3) Furthermore, our eyes have some limitations that may vary according to lighting, age, etc. Thus, the result of a measurement is not a real number; it is an interval. We can say, for example, that the length of the rod is within the range [27.30,27.40] cm (if the ruler has divisions up to mm). Otherwise expressed, we have 27.35 ±0.05 cm. This same thing can be expressed as 27.3cm, assuming that the number following the last one on the right can vary by one unit. In other words, we are working with an interval of [27.30,27.40] cm. 25 1 Physico-mathematical foundations of mechanics Data:R= 0.02 m, ρAl = 2700 kg/m3,ρOl = 960 kg/m3. The results of the table were obtained from measurements made between z= 0.25 m and z= 2.5m z(m) 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 t(s) 0.154 0.334 0.490 0.668 0.826 0.961 1.162 1.293 1.467 1.639 Table for Problem 1.5.1 a) Represent the points (t, z)according to the measurements made. b) Graphically find the straight line that best fits the points. c) Find by linear regression the straight line that best fits these points. Justify the goodness of this fit. d) Using the fitted straight line found in c) and the expressions (1, 2, 4), find vL,band ηOl. e) Find the slope error and the ordinate at the origin error of the fited straight line, and then calculate the propagation error in the viscosity ηOl. Solution a), b) and c). The represented points and the regression straight line can be seen in the figure. A spreadsheet has been used to do the linear regression. The analytic expression for the regression line gives: z=A t +B= 1.53091 t+ 0.00190, with a correlation coefficient of r= 0.9997 Solution to Problem 1.5.1 d) Relating the straight line and the fact that the movement is uniform according to z=vLt, we obtain vL= 1.5309 m/s . Taking into account (2), Pap = 0.5720 N, and according to (1), we obtain b= 0.3736 N s/m . By using (4), we have ηOl = 0.9911 N s/m2. e) To find the slope Aand the ordinate at origin Berrors, we use (1.39). With a number of measurements N= 10, we have σA= 0.01326 and, from Table 1.1, f= 0.7154. The obtained slope error is εA= 0.0095 m/s , which will also be the 32 1.6 Newton's first law. Inertial reference frames error of vL. Considering that, we can write ηOl =Pap 6πR 1 vL Given that the only magnitude subject to error on the right side is vL, and taking into account (1.34), we have εηOl =Pap 6πR 1 v2 L εvL where εvL=εA= 0.0095 m/s. We obtain εηOl = 0.0062 N s/m2, and the resulting ordinate at the origin Berror is εB= 0.0097 m . Observe that this value makes the Bvalue that we found by regression compatible with the value 0, which we expect according to the expression z=vLt. Finally, we can express the result of the experiment according to the value found for ηOl ηOl = (0.991 ±0.007) N s/m2 1.6 Newton's first law. Inertial reference frames Fig. 1.18: Isaac Newton (1643-1727) was an English physicist, mathematician and philosopher Newton's first law Newton’s first law was motivated by the need to clarify what happens when nothing happens. In other words, the spacetime conditions in which agents move: the interacting particles and their corresponding forces. Note that in those days it was commonly believed that things did not move by themselves; yet, Galileo had already conceived that bodies have inertia, a natural tendency to maintain their state of motion. Newton’s first law or the law of inertia: A body unaffected by any cause moves at a constant velocity (uniform rectilinear motion). Fig. 1.19: Galileo Galilei (1564-1642) was a Tuscan physicist, mathematician and philosopher We can view this as defining the absence of cause: A body observed to be moving at a constant velocity is being affected by nothing. We can therefore say that it has inertial motion. Inertial reference frames If an observer or reference frame moves inertially at a rectilinear and uniform speed, it is not affected by any external physical agent. How will he or she know how fast they are moving? Everything that is observed will seem to be a consequence of the same laws of physics that apply when not moving. This consideration invokes the definition of an inertial reference frame. Inertial reference frames: These are reference frames obtained by transforming a given reference frame in such a way that the laws of physics take the same form. The transformations from one to another are called inertial transformations (see Figure 1.20). 33 1 Physico-mathematical foundations of mechanics Fig. 1.20: Two inertial observers use the same laws of motion to explain their observations of the same event Inertial transformations Highly generalized spacetime hypotheses posit that two reference frames Sand S′ are inertial if they are related to each other by a composition of translation, rotation, change in the origin of time and a transformation of velocity. Both systems travel at a constant relative velocity  V. The expression of the latter depends on whether or not a maximum speed exists for synchronizing the different points in space. Galileo transformations: If nature does not limit the speeds, there are no problems with synchronizing. For the velocity transformation, we obtain {r′=r − V t t′=t(1.40) These transformations define the Galilean inertial frames we will use to study Newtonian mechanics. Lorentz-Poincaré transformations: If nature places a limit on speeds, we will have a maximum speed cthat we use for synchronization. We therefore obtain Fig. 1.21: Hendrik Antoon Lorentz (1853-1928) was a Dutch mathematician {r′=r −γ V t +(γ−1) V2( V·r) V t′=γt −1 c2γ V·r (1.41) where γ=1 √1−V2 c2 . Since cis the maximum speed there is no contradiction, so V < c and the squareroot are always real. These transformations define the Poincaré inertial frames upon which Albert Einstein’s special theory of relativity is based. Fig. 1.22: Henri Poincaré (1854-1912) was a French mathematician It is an experimental fact that there exists a maximum speed that coincides with that of light in a vacuum: c= 299792458 m/s. The consequences of this fact are known as the special theory of relativity, which has been experimentally proven. However, whenever we deal with systems that have velocities v≪cwe can use the c→ ∞ approximation and Newtonian mechanics for our calculations. In Newtonian mechanics, time is absolute: For every point A,tA=tO; that is, all 34 1.6 Newton's first law. Inertial reference frames points in the space of an observer can be synchronized with the same time t. Unless explicitly stated otherwise, we will study Newtonian mechanics. Regardless of whether or not we apply relativity, choosing an inertial reference frame for defining another one depends on the degree of approximation we are working with. In general, it is enough to consider the Earth as an inertial frame, although we know that it moves around the Sun and therefore is not strictly inertial. If more precision is needed, the inertial frame attached to the Sun can be used. For example, GPS (Global Positioning System) technology currently links its reference frames to quasars (quasi-stellar radio sources), which are extremely distant celestial objects. Fig. 1.23: Albert Einstein (1879-1955) was a physicist of German descent, later nationalized in Switzerland and the United States To end this section, we will solve a single relativistic problem that draws on a specific and real case to illustrate how relativistic mechanics explained a fact that caused surprise at the time, because it could not be explained by Newtonian mechanics. Problem 1.6.1. In the laboratory, low speed µ−muons have a half life of τ= 2.197 ×10−6s. Cosmic rays reach the atmosphere at a height of h= 2.5×106m and produce µ−, which are detected on the Earth’s surface. Calculate in the following two ways the (constant) speed at which muons reach Earth. a) without relativistic mechanics. b) using relativistic mechanics. Solution Solution to Problem 1.6.1 a) Time is the same for all observers. The µ−can “live”, at most, τ= 2.197×10−6s. According to the Sreference frame, at rest on the ground, µ−should travel at a minimum speed v=h τ=2.5×106 2.197×10−6= 1.138 ×1012 m/s, which far exceeds the maximum possible speed! b) Here, we give the relativistic explanation. Time depends on the observer. The time to consider for the muon µ−is its proper time, which in this case is the time of the reference frame S′that is traveling with the muon. With this interpretation, the muon can live at most ∆t′=τ= 2.197 ×10−6s , which is the time measured by a clock at rest relative to S′. In contrast, to measure the speed of the muon, v, the observer Suses his or her time ∆t,v=h ∆t. Taking increments to the relativistic Lorentz-Poincaré transformation of time (1.41), with  V=v,v ·r =vz,∆z=hand h ∆t=v, we have ∆t′=1 √1−v2 c2 ∆t−1 c2 1 √1−v2 c2 vh Dividing everything by hand simplifying, we have ∆t′ h=1 v√1−v2 c2 35 1 Physico-mathematical foundations of mechanics Therefore, the minimum speed vat which the muon must travel in order to reach the ground alive must satisfy (v c)2=1 1 + (c∆t′ h)2= 0.9999 that is, v < c, which is consistent with the fact that cis the maximum possible speed and, therefore, is consistent with relativity.  1.7 Point kinematics: position, trajectory, velocity and acceleration Position vector: The position vector of point P= (xP, yP, zP)relative to point O= (xO, yO, zO)is rP(O)= (xP−xO, yP−yO, zP−zO). If O= (0,0,0), it is written as rP= (xP, yP, zP)or, if there is no doubt, r = (x, y, z)(see Figure 1.24). Generally, we can write rP(Q)=rP−rQ. Fig. 1.24: Position vector with respect to the origin In general, two observers will assign different position vectors for a given point P: rP(O)=rP(O′) Position vector module: We denote this by |r|=r. Trajectory: The trajectory of a particle is the curve r(λ)that it follows. Temporal trajectory: The temporal trajectory of a particle is the time parametrized curve r(t)that it follows (see Figure 1.25). Fig. 1.25: Temporal trajectory of a particle Displacement vector: The displacement vector between two points is ∆r = r2−r1= (x2−x1, y2−y1, z2−z1). If the two points are infinitely close, then dr = (dx, dy, dz). Module of dr: Be careful to note that dr is not the module of dr, which we denote as dℓ, specifically |dr|=dℓ, where ℓis the length of the curve relative to some reference point on it (see Figure 1.26). It is only when the displacement dr is aligned with r that dℓ =dr is fulfilled. Fig. 1.26: Length of the curve ℓ, module of dr and dr Differential of U:Given a scalar U(r)the differential of U, is dU = ∇U·dr and it represents the infinitesimal variation of the Ufunction when the position varies infinitesimally in a specified direction of the space dr. If Uexplicitly depends on the time, U(r, t), and the position varies infinitesimally in an unspecified space direction dr spending an unspecified time dt the Uvariation is dU = ∇U·dr +∂U ∂t dt (1.42) Velocity vector: The particle velocity vector is the tangent vector to the trajectory, and its module indicates the rate of change of the particle position at every 36 1.7 Point kinematics: Position, trajectory, velocity and acceleration instant: v =dr dt =˙ r = (dx dt ,dy dt ,dz dt )(1.43) Relative velocity: The velocity of one Pparticle relative to another Qis: vP(Q)=drP(Q) dt =vP−vQ(1.44) Acceleration vector: The acceleration vector of a particle is the vector a =˙ v =dv dt =d2r dt2=¨ r =(d2x dt2,d2y dt2,d2z dt2)(1.45) Frenet trihedron The rate of change in velocity v of a particle is the acceleration a =˙ v. The Cartesian components of these vectors do not tell us much about what they represent. Returning to the definition of a vector as magnitude with modulus and direction, we can thus express the velocity vector with the form v =vˆv. In this way, each factor has a very clear physical meaning, with vbeing the modulus and ˆvbeing the direction. Now, we can derive to find the acceleration using a =dv dt =d(vˆv) dt =dv dt ˆv+vdˆv dt (1.46) with the first term corresponding to acceleration in the velocity’s direction. This is the tangent acceleration aT=dv dt ˆv. The tangential component of the acceleration is aT,aT=dv dt . To study what the second term vdˆv dt represents, note that ˆvis a unit vector, ˆv2= 1. Deriving this with respect to time, we have 2ˆv·dˆv dt = 0, meaning that dˆv dt is a vector normal to ˆv, that is, normal to the trajectory. The ˆn=dˆv dt dˆv dt −1vector is unitary and normal to the trajectory. We can still define a third vector, the binormal vector ˆ b, which is normal to both ˆvand ˆnas well as unitary: ˆ b= ˆv׈n. We will not use this third vector to describe the acceleration but instead define it only to complete the basis. What we are interested in analysing is the part of the acceleration aligned with the normal direction ˆn. This is the normal acceleration aN=vdˆv dt ˆn. To adequately describe the normal component of acceleration, it is necessary to know more about the trajectory. Figure 1.27 compares ˆvof two points that are very close in the trajectory. These are ˆv(t)and ˆv(t+dt), which form a triangle with two sides of length 1and a third of length |dˆv|. Because this is an isosceles triangle with an infinitesimal angle, it is comparable to an arc of radius 1. At the same time, the small section of the 37 1 Physico-mathematical foundations of mechanics trajectory is a curve shaped as a small arc with angle dϕ,radius of curvature R and an arc length of dℓ =vdt. We thus have the relationships Fig. 1.27: Radius of curvature R 1dϕ =|dˆv|;R dϕ =dℓ (1.47) which, by eliminating dϕ, allow us to obtain an expression of the radius of curvature Rin terms of the unit vector tangent to the trajectory ˆv. For convenience, the curvature is defined as ρ= 1/R. All in all, we have the following. Radius of curvature: At each point of a given trajectory with tangent unit vector ˆv, a radius of curvature Rand a curvature ρcan be associated according to ρ=1 R= dˆv dℓ (1.48) Fig. 1.28: Frenet trihedron and radius of curvature Frenet trihedron: At each point of a given trajectory with tangent unit vector ˆv, we can define a basis of vectors formed by the tangent, normal and binormal vectors to the trajectory called the Frenet trihedron; (see Figure 1.28) specifically according to ˆv; ˆn=Rdˆv dℓ ;ˆ b= ˆv׈n(1.49) Tangent and normal acceleration: The acceleration of a particle moving through a trajectory that is defined by the ˆvunit vector can always be written as a =aTˆv+aNˆn. The tangent aTand normal aN, which are acceleration components, can be expressed in terms of vand Ras Fig. 1.29: Jean Frédéric Frenet (1816-1900) was a French mathematician, astronomer and meteorologist aT=dv dt ;aN=v2 R(1.50) Problem 1.7.1. An inertial observer Omeasures the position of a particle (SI units) rP(O)= (6t2−4t)ˆı−3t3ˆȷ+ 2ˆ k. Another observer O′, with the same orientation, measures the position of the same particle rP(O′)= (6t2+ 3t)ˆı−3t3ˆȷ−3ˆ k. a) Determine the relative velocity of the reference frame O′with respect to O. b) Calculate the particle acceleration with respect to Oand O′. c) Is O′an inertial observer? 38 1.7 Point kinematics: Position, trajectory, velocity and acceleration Solution a) The velocity of the particle with respect to Ois: vP(O)=drP(O) dt = (12t−4)ˆı−9t2ˆȷ The velocity of the particle with respect to O′is: vP(O′)=drP(O′) dt = (12t+ 3)ˆı−9t2ˆȷ The relative velocity of O′with respect to Owill be  V=vO′(O)=vO′−vO=−vP+vO′+vP−vO=vP(O)−vP(O′)=−7ˆı Solution to Problem 1.7.1 b) aP(O)=dvP(O) dt = 12ˆı−18 tˆȷ From the previous section, we have vP(O′)=vP(O)− V, thus aP(O′)=dvP(O′) dt =dvP(O) dt −d V dt =aP(O)= 12ˆı−18 tˆȷ c) Yes, because O′is traveling at constant velocity with respect to an inertial observer O:vO′(O)=ct  Problem 1.7.2. A particle describes a rectilinear motion, travelling in space s= 4t3−3t2−6, with sin meters and tin seconds. If the particle starts from t= 0, calculate a) the time it will take to reach a speed of 6m/s. b) the value of its acceleration at the same instant. Solution v(t) = ds dt = 12t2−6t a) If the particle starts from t= 0, v(t) = 6 = 12t2−6t⇒2t2−t−1=0⇒t= 1s b) At this moment, the acceleration will be a(t= 1) = dv dt t=1 = (24t−6)|t=1 = 18 m/s2  Problem 1.7.3. A body describes a rectilinear motion with acceleration a= 4−t2 (ain m/s2and tin s). Calculate the velocity and the displacement as a function of time, if at t= 3 s, v= 2 m/s and x= 9 m. 39 1 Physico-mathematical foundations of mechanics Solution By integrating the expression of the acceleration we obtain the speed: v=∫adt =∫(4−t2)dt = 4t−t3 3+K1 Integrating this again we get the displacement x=∫vdt =∫(4t−t3 3+K1)dt = 2t2−t4 12 +K1t+K2 K1and K2are integration constants that depend on the initial conditions. With t= 3s, v= 2 m/s and x= 9 m, we have v(3) = 2 = (4·3−33 3+K1)⇒K1=−1 x(3) = 9 = (2·32−34 12 + (−1) 3 + K2)m⇒K2= 0.75 Finally, we obtain x=−t+ 2t2−t4 12 + 0.75 ; v= 4t−t3 3−1 Problem 1.7.4. The three-dimensional motion of a particle is defined by the vector position r =Rsin (ω t)ˆı+c t ˆȷ+Rcos (ω t)ˆ kwith R,ωand cconstants. a) Determine the particle’s magnitudes of velocity and acceleration. b) Calculate the radius of curvature, the tangent, the normal components of the acceleration and the unit normal vector. Solution a) v =dr dt =ωR cos ωt ˆı+cˆȷ−ωR sin ωt ˆ k a =dv dt =−ω2Rsin ωt ˆı−ω2Rcos ωt ˆ k=−ω2R(sin ωt ˆı+cos ωt ˆ k) Solution to Problem 1.7.4 b) v=√v ·v =√ω2R2(cos2ωt +sin2ωt) + c2=√ω2R2+c2 a=√a ·a =√ω4R2(sin2ωt +cos2ωt) = ω2R aT=dv dt = 0 With aT= 0, we can conclude that the entire acceleration is normal, that is, aN= ω2R. If aT= 0, we can solve for aNusing a2=a2 T+a2 N. However, we will not do it in this way but will instead first solve it by finding the curvature: ˆv=v v=ωR cos ωt ˆı+cˆȷ−ωR sin ωt ˆ k √ω2R2+c2 40 1.7 Point kinematics: Position, trajectory, velocity and acceleration dˆv dt =−ω2R(sin ωt ˆı+cos ωt ˆ k) √ω2R2+c2  dˆv dt =ω2R √ω2R2+c2 We will denote the radius of curvature by Rcurv so as not to confuse it with R, which is a constant of the problem statement. Rcurv =v dˆv dt  −1 =√ω2R2+c2 ω2R √ω2R2+c2 =R+c2 ω2R If c= 0 the curve does not advance in the ˆȷdirection, it is a circumference of radius Rin the x−zplane. In this case, Rcurv =R. We will calculate the normal acceleration aNusing the radius of curvature Rcurv aN=v2 Rcurv =ω2R2+c2 (ω2R2+c2 ω2R)=ω2R ˆn=Rcurv v dˆv dt =−sin ωt ˆı−cos ωt ˆ k Problem 1.7.5. A particle describes a circular (non-uniform) motion r(t) = R0(cos φ, sin φ, 0) where φ=φ(t)is an increasing function of time. Calculate: a) the velocity and the acceleration. b) the trajectory’s radius of curvature. c) the tangent ˆvand normal ˆnbasis vectors and the acceleration tangent aTand normal aNcomponents. Solution Solution to Problem 1.7.5 a) The velocity and acceleration. When deriving with respect to t, we will take into account that we are doing a total derivative and φis a function of t: v =R0˙φ(−sin φ, cos φ, 0) a =R0¨φ(−sin φ, cos φ, 0) + R0˙φ2(−cos φ, −sin φ, 0) The velocity module is v=R0˙φ. b) The ˆvvector of the Frenet basis and the curvature radius R ˆv=v v= (−sin φ, cos φ, 0) ; dˆv dt = ˙φ(−cos φ, −sin φ, 0) R=vdˆv dt −1=R0 c) The ˆnvector of the Frenet basis and the normal and tangent components of the acceleration ˆn=R v dˆv dt = (−cos φ, −sin φ, 0) aT=dv dt =R0¨φ;aN=v2 R=R0˙φ2 41 2 Dynamics of a particle e) a=−g⇒N= 0  Problem 2.1.2. A small object of m= 4 kg, is subject to the action of two forces,  F1= ˆı−2ˆȷand  F2= ˆı+ ˆȷ(N units). Calculate the acceleration, velocity and position vectors of the object at time t= 3 s if at t= 0 it is at rest at the origin of coordinates. Solution m= 4 kg. The resulting force is  F= F1+ F2= (2,−1). From the equation of motion, we obtain the acceleration:  F=ma ⇒a = F m=(1 2,−1 4) Integrating the acceleration, v =∫a dt =(1 2,−1 4)t+ C1 and taking into account the initial conditions at t= 0,v = 0 implies  C1= 0 and, as a result, v =(1 2,−1 4)t Integrating the velocity, r =∫v dt =(1 4,−1 8)t2+ C2 and taken into account the initial conditions at t= 0,r = 0 implies  C2= 0 and, as a result r =(1 4,−1 8)t2 Problem 2.1.3. A particle tied to a rope of negligible mass and length ℓdescribes a uniform circular motion. Find the minimum angular speed ωthat makes it possible. Figure for Problem 2.1.3 Solution Newton’s motion equations in vertical and normal path directions are: Tcos φ−mg = 0 Tsin φ=maN Solution to Problem 2.1.3 The normal acceleration is aN=v2 R=Rω2=ℓsin φ ω2. Resolving Tin the first and replacing it in the second, we have mg cos φsin φ=mℓ sin φ ω2 from which we get cos φ=g ℓω2. The cosine is positive and cos φ≤1must be fulfilled, thus ω≥√g ℓ 48 2.2 Force and momentum Problem 2.1.4. A particle tied to a rope of negligible mass and length ℓoscillates in a vertical plane. a) Find the equation of motion using Newton’s second law. b) Specify the result for small oscillations. Figure for Problem 2.1.4 Solution Newton’s motion equations in normal and tangent path directions are: T−mg cos φ=maN −mg sin φ=maT Solution to Problem 2.1.4 With the speed being v=ℓ˙φ, the tangent acceleration is aT=dv dt =ℓ¨φ. Substituting the tangent component of the equation of motion −mg sin φ=mℓ ¨φ, we can thus simplify it to: ¨φ+g ℓsin φ= 0 If φ≪1, expressed in radians, then sin φ≈φand we obtain ¨φ+g ℓφ= 0  2.2 Force and momentum Momentum. The linear momentum or quantity of motion (or simply momentum, if there is no confusion), p is defined as the vector p =mv (2.6) The SI unit of momentum is kg m/s. Newton’s law of motion can be written using the momentum as dp dt = F(2.7) Equation (2.7) allows interpreting the force as the cause of the momentum variation and thus states the following. Conservation of momentum theorem. If no forces act on a particle or their net force is zero, the momentum is constant. For a particle, this statement is immediate, because, in this case, the momentum is directly proportional to the velocity of the particle. We will see later in Section 3.3 where this concept is applied to Nparticles, that these expressions are also valid and that they will allow us to solve more complex situations. 49 2 Dynamics of a particle Impulse applied by a force. The impulse  Iapplied by a force  Fin the interval t1→t2is defined as the vector  I= t2 ∫ t1  Fdt (2.8) The impulse unit in SI is kg m/s. Note that in order to find  I, it may be necessary to know the trajectory r(t). The momentum theorem. The change in momentum is equal to the applied impulse  I  I= ∆p (2.9) Proof. One need only consider Newton’s law (2.7) in the form  F dt =dp and take the integral in the interval t1→t2 Problem 2.2.1. A1kg particle, for t≤0, moves rectilinearly at a constant speed of 100 m/s. At instant t= 0 and for a duration of 1s a force F= 1000 e−t(Fin N and tin s) acts in the same direction but opposite sign to the motion. Calculate the impulse applied by the force and the final momentum of the particle. Solution Since force and velocity have the same direction, the particle direction will not change. As we are working with a one-dimensional problem, the vector notation will be omitted. We have only two directions (let us say ±ˆı). We take the positive sign as that of the initial velocity. The impulse that the force supplies to the particle is I=∫t2 t1 F dt =∫1 0−1000 e−tdt = 1000 e−t 1 0 =−632kg m s The final momentum p2is ∆p=p2−p1⇒p2=p1+ ∆p=mv1+ ∆p= 100 −632 ⇒p2=−532 kg m/s This result means that the particle ends up moving at a constant speed of 532 m/s in a direction opposite to the initial velocity.  Problem 2.2.2. A particle of 2kg of mass moves at a certain instant with a velocity expressed by v = 5 ˆı+ 2 ˆȷ. Then, a force  F= 4 ˆȷis applied to it. Knowing that v and  Fare expressed in SI units, determine the momentum of the particle after applying the force for 3s. 50 2.3 Torque and angular momentum of a particle Solution m= 2kg. The initial momentum is p1=mv = 10 ˆı+4 ˆȷ. By using the momentum theorem and being p2the final momentum ∆p =p2−p1= t+3 ∫t  F dt = 3  F= 12 ˆȷ We note that the limits of integration, in principle, depend on time. The force is constant over time. Insulating p2, we obtain p2=p1+ 12 ˆȷ= 10 ˆı+ 16 ˆȷ in SI units.  2.3 Torque and angular momentum of a particle Fig. 2.5: Torque of force  Fwith respect to point A. Corkscrew rule Torque or moment of a force. The torque  M(A)of a force  Fapplied to a Q point with respecte point Ais defined as  M(A)=r(A)× F(2.10) where r(A)is the position vector of point Q, which is where the force is applied, relative to point A. Note that a consequence of the vector product is that the direction of the torque of the force is perpendicular to the plane formed by the vectors r(A)and  F. What is more, because the first vector of the (2.10) vector product is a position, the modulus can be calculated by means of the simple expression (see Figure 2.5): M(A)=F r(A)sin φ=F A  F(2.11) where A F=r(A)sin φis also the distance between point Aand the line of action of force a  F. The direction is given by the corkscrew rule. Problem 2.3.1. Calculate the torque of the  Fforce applied to the sphere at the point indicated in the figure, with respect to the ground contact point C. Figure for Problem 2.3.1 Solution The torque can be obtained directly by the vector product. Looking at the figure, r(C)= 2R(0,1,0) and  F=F(sin φ, −cos φ, 0), and thus  M(C)=r(C)× F= 2RF  ˆıˆȷˆ k 0 1 0 sin φ−cos φ0  =−2RF sin φˆ k Solution to Problem 2.3.1 We can also proceed by applying the corkscrew rule, which allows us to state  M(C)= −M(C)ˆ kand then calculate the modulus according to (2.11) M(C)=F C  F=F2Rsin φ 51 2 Dynamics of a particle Angular momentum. The angular momentum of a particle with respect to point A, L(A), is defined as the moment of momentum p applied to point Q,which is where the particle is (see Figure 2.6): Fig. 2.6: Angular momentum  L(A). Corkscrew rule  L(A) = r(A)×p (2.12) where r(A)is the position vector of point Q(where the momentum is applied) with respect to point A. It is important that Ais a fixed point in the reference frame. The unit of angular momentum in the SI is kg m2/s. Note that as a consequence of the vector product, the direction of angular momentum is perpendicular to the plane formed by the r(A)and p vectors. What is more, because the first vector of the (2.12) vector product is a position, the modulus can be calculated by the simple expression: L(A)=p Ap (2.13) where Ap is the distance between point Aand the line of action of the p momentum. The direction is given by the corkscrew rule. Newton’s law of motion implies that: d L(A) dt = M(A)(2.14) Proof. Multiplying both members of Newton’s law of motion md v dt = Fby r(A)× and taking into account thatr(A)×dp dt =d dt (r(A)×p)(since ˙ r(A)=v andv ×p = 0), we get the result.  We have shown Newton’s law of motion implies (2.14), but not that they are equivalent: (2.14)does not imply Newton’s law of motion. Equation (2.14) allows us to interpret that the torque is the cause of the variation in the angular momentum, and we can thus state the following. Conservation of angular momentum theorem. If the torque of a force acting on a particle with respect to one point Ais zero, the angular momentun relative to this point remains constant. We observe that both the angular momentum and the torque of a force depend on point Awith respect where it is calculated. It is possible that relative to Awe instead have  M(A)= 0, and relative to another point Bwe have  M(B)= 0. The null torque of a force and, therefore, the conservation of the angular momentum occurs if the net force applied to the particle is zero. However, this also occurs in 52 2.3 Torque and angular momentum of a particle other more interesting cases, such as when the r(A)vector and  Fforce are parallel. That is why, even in the case of a single particle, we will use Equation (2.14) in order to solve some interesting problems. As we will see later, the conservation of angular momentum theorem is especially important in the case of a particle system, particularly in rigid bodies. Angular impulse. The angular impulse  Yapplied by the torque  M(A)in the interval t1→t2is defined as the vector  Y(A)= t2 ∫ t1  M(A)dt (2.15) The unit of angular impulse in SI is kg m2/s. Angular momentum theorem. The increase in angular momentum is equal to the applied angular impulse  Y  Y(A)= ∆ L(A)(2.16) Proof. One need only consider (2.14) in the form of d L(A)= M(A)dt and integrate it into the interval t1→t2. Problem 2.3.2. Planets are known to have elliptical trajectories relative to the Sun (which we consider to be fixed) at one foci. Prove Kepler’s second law. Solution Solution to Problem 2.3.2 We take the torque with respect to the Sun (point A). The torque of the force is zero and, therefore, the angular momentum in conserved:  L=r ×p =ct Solution to Problem 2.3.2 As the direction of angular momentum is fixed, the relevant information is contained only in the modulus m r v sin θ=ct Taking in to account that the velocity modulus can be written as v=dℓ dt , it becomes rdℓ dt sin θ=ct Solution to Problem 2.3.2 Looking at the figure, the areas swept by the position vector for a time are triangles that are, usually, scalene, such as Atriangle in the figure. The length of the trajectory between the two vertices, ∆ℓ, does not match the corresponding side of the triangle. If we consider a time ∆t→dt, the planet will have travelled ∆ℓ→dℓ and we will have the Btriangle in the figure. Now, the length of the trajectory between the two 53 2 Dynamics of a particle vertices, dℓ, coincides with the corresponding side of the triangle. We can expand it in order to see it better, such as with the Ctriangle in the figure. The area dA that is swept by the position vector in dt is thus equal to the area of the Ctriangle in the figure. Because we know that the modulus of the cross product of two vector is the area of the parallelogram they form, the area of our triangle will therefore be dA =1 2r dℓ sin φ=1 2r dℓ sin θ Solution to Problem 2.3.2 Thus, comparing this result with that obtained from the conservation of angular momentum, we can conclude dA dt =ct which is Kepler’s second law: The areas swept by the position vector of the planet at equal times are equal.  Fig. 2.7: Johannes Kepler (1571-1630) was a German astronomer and mathematician 2.4 Work, kinetic energy and potential energy. Power Work and energy are closely related concepts, that go beyond mechanics and play a key role in the world of physics. Work done by a force Fig. 2.8: One particle moves rectilinearly under the action of a force in the same direction as the displacement vector Let us look at the simple case of one particle moving rectilinearly over a distance ∆xunder the effects of a constant force F, parallel to and with the same direction as the displacement and perhaps in the presence of a frictional force (see Figure 2.8). The work done by the force Fis defined by the product of the force Fand the displacement ∆x: W=F∆x(2.17) Fig. 2.9: One particle moves rectilinearly ∆r under the action of a constant force  F. Only the component of the force in the direction of the displacement causes the particle to advance along the trajectory If the motion is rectilinear and the applied force is constant but forms an angle θ with respect to the displacement (see Figure 2.9), the work is: W=F∆xcos θ= F·∆r (2.18) Observe that in this case we consider only the component of the force in the direction of the displacement, while the perpendicular component performs no work. In the general case where the force is not constant and the path Cfollowed by the particle is not rectilinear (see Figure 2.10), calculating the work between two P1and P2points requires considering the infinitesimal displacements dr along the path, and all infinitessimal work  F·dr must be added toghether. Work done by a force. We define the work Wdone by a force  Falong a path 54 2.4 Work, kinetic energy and potential energy. Power Cfrom P1to P2by means of the integral W= P2 ∫ C:P1  F·dr (2.19) Fig. 2.10: Work done by a force  F from P1to P2along the Cpath If we know the parametric expressions of the pathr(λ) = (x(λ), y(λ), z(λ)),P1= r(λ1)and P2=r(λ2), we can make the integral work explicit: W= λ2 ∫ λ1  F·dr dλdλ = λ2 ∫ λ1(Fx dx dλ +Fy dy dλ +Fz dz dλ)dλ (2.20) The concept of work is highly important from both theoretical and practical point of view. The ingredients for calculating work are a force  Fand a path Cwith two points on it. This force and path need not be related. The following two interpretations are highlighted below. 1) If force  Fcontributes to moving the particle along path C, then the path can be parameterized with time t.r(t)is the time path and, in the absence of other forces it, fulfils  F(r, ˙ r, t) = m¨ r. This interpretation has eminently practical utility, as in this case work becomes a measure of the effectiveness of the force that displaces the particle. There may be a lot of force and little work (little displacement) or little force and a lot of work (a lot of displacement). The work thus approximates our everyday concept of work, not so much as a measure of effort but as a measure of efficiency in transforming something in the environment (in our case, the displacement of the particle). Later we will see the conservative forces that are especially effective in accordance with this interpretation. 2) Path Cdoes not necessarily have to be a solution to the equation of motion with the force  F. That is, if r(t)is a time trajectory passing through Cand mis the mass of the particle to which force  Fis applied, it does not necessarily follow that  F(r, ˙ r, t) = m¨ r(t). If the force depends only on the position,  F(r), we can consider different paths to calculate the work. We do not even need to account for the presence of the particle. This interpretation, as we will see, is of great theoretical interest. The concept of kinetic energy is related to work in terms of interpretation 1, above, as we will see in the following. Instead, the concept of potential energy will require the most abstract interpretation 2. The concept of power can be defined as long as we know the temporal trajectory of the particle, as the interpretation 1. Power of a force. The power exerted by a force  Fat each instant is defined as 55 2 Dynamics of a particle the rate of work done by the force on the particle: P=dW dt = F·v (2.21) In general, the particle will be subjected to different forces. If we know the temporal trajectory caused by these forces, we can calculate the power of each one separately. Kinetic energy The kinetic energy of a particle of mass mmoving at a velocity v is defined as the capacity to do work associated with the fact that the particle is in motion. Thus, if we combine the definition of work, W=∫ F·dr, with Newton’s second law,  F=ma, and the definition of velocity, v =dr dt , from which we obtain dr =vdt, we can write W=∫ F·dr =∫ F·v dt =∫ma ·v dt (2.22) Now, taking into account that a ·v dt =dv dt ·vdt =d dt (1 2v2)dt =d(1 2v2), we obtain W=∫d dt (1 2m v2)dt =∫d(1 2m v2)(2.23) Kinetic energy. The kinetic energy, Ec, of a particle of mass m, moving with a velocity v is defined as Ec=1 2m v2(2.24) Work-energy theorem. The work done by a force on a particle is equal to the change in its kinetic energy. W= ∆Ec(2.25) Proof. It is enough to finish the final integral in (2.23) with the integration limits fi ∫ ini and take into account the definition of kinetic energy (2.24): fi ∫ ini d(1 2m v2)=Ec:fi − Ec:ini  Potential energy The potential energy associated with a particle subjected to a conservative force is the capacity of doing work and it is related with occupying a certain position in space. Consider the simple case of a particle moving in one dimension (we will use the xcoordinate) under the effects of a force that depends only on the position F(x). 56 2.4 Work, kinetic energy and potential energy. Power For this type of force, it is always possible to define a differentiable function U(x) in such a way that F=−dU dx (2.26) This is one example of conservative force. The potential energy is defined as Fig. 2.11: One dimensional case. Work done by a force  Ffrom xA to xBby along the only possible path U(x) = U(0) − x ∫0 Fdx (2.27) The work done by the conservative force F(x)when the particle moves between the initial xAand final xBis W= xB ∫ xA F(x)dx =− xB ∫ xA dU(x) dx dx =− xB ∫ xA dU(x) = U(xA)−U(xB)(2.28) The potential energy and conservative concepts can be generalized if we apply the appropriate definitions. Fig. 2.12: Possible closed paths Conservative force. One force  F(r)is conservative if, for all closed paths, the work done is null (see Figure 2.12): I F·dr = 0 (2.29) If  F(r)is conservative, the work done for going from one point to another does not depend on the specific path we use, as we can see by observing Figure 2.13: Fig. 2.13: We can go and come back to P1by a closed path that passes through P2, which we can interpret as formed by two different paths, C1and C2, going from P1to P2 I C1∪C2  F·dr = P2 ∫ P1:C1  F·dr + P1 ∫ P2:C2  F·dr = P2 ∫ P1:C1  F·dr − P2 ∫ P1:C2  F·dr (2.30) and if the force is conservative, we have P2 ∫ P1:C1  F·dr = P2 ∫ P1:C2  F·dr (2.31) As it does not depend on the path, the work integral can be expressed as a function that is dependent only on the departure and arrival points P1and P2: P2 ∫ P1  F·dr =f(P1, P2)(2.32) 57 2 Dynamics of a particle By integrating it twice we get r(t) =  C1+ C2t+1 2  F mt2and considering the initial conditions r(t0) = r0and ˙ r(t0) = v0 r(t0) = r0= C1+ C2t0+1 2  F mt2 0 ˙ r(t0) = v0= C2+ F mt0}⇒{ C2=v0− F mt0  C1=r0−v0t0+1 2  F mt2 0 we get the trajectory expression r(t) = r0+v0(t−t0) + 1 2  F m(t−t0)2(2.50) If we choose the reference frame so that the plane of motion is (x, y)and the y-axis in the direction of the force  F= (0, F )(see Figure 2.15) Fig. 2.15: Constant force in the direction of increasing y {x(t) = x0+vx0(t−t0) y(t) = y0+vy0(t−t0) + 1 2 F m(t−t0)2  F=ct is a conservative force, as we can find the associated potential energy: U=−∫ F·dr =− F·r +ct (2.51) Fig. 2.16: Constant gravitational field gin the direction of decreasing y An important case (but not the only one!) is that of a particle of mass m, subjected to the gravitational field g =constant, close to the Earth or some other planet (see Figure 2.16): ma = F=mg ⇒a = F m=g =ct (2.52) With the above-mentioned choice of the axis and choosing the positive direction of the yaxis in the opposite direction of the force g = (0,−g), it results in {x(t) = x0+vx0(t−t0) y(t) = y0+vy0(t−t0)−1 2g(t−t0)2 and the potential energy, by using (2.51), is U=m g y +ct. Another case of interest is that of a particle of mass mand electric charge qunder the action of a constant electric field  E=ct (see Figure 2.17): ma = F=q E⇒a =q E m=ct Fig. 2.17: Constant electric field Ein the direction of increasing y With the above-mentioned choice of the axis and by choosing the positive direction of the yaxis in the direction of the field  E, we have {x(t) = x0+vx0(t−t0) y(t) = y0+vy0(t−t0) + 1 2 q E m(t−t0)2 And by using (2.51), the resulting potential energy is U=−q E y +ct. 64 2.5 Some forces  F(r)unidimensional motion If the force has a fixed direction, with a suitable choice of axes we can express it as  F=F(r) ˆı. If the initial velocity has the same direction as the force, the force will not change the direction of velocity and the motion can be described with a single xcoordinate (see Figure 2.18): Fig. 2.18: One-dimensional motion m¨x=F(x) The force will always be conservative, since U(x) = −∫ F(r)dr =−∫F(x)dx and the integral can always be calculated. Because mechanical energy is conserved, we can take advantage of this fact to find the trajectory. E=constant and also E=1 2m(dx dt )2 +U(x) = ct which is a differential equation for finding the trajectory x(t). We can separate the variables xand tand integrate tbetween the limits t0to t, and xbetween the limits x(t0)to x(t). The integral in tis immediate: x(t) ∫ x(t0) dx √E−U(x)=√2 m(t−t0)(2.53) If we know explicitly the function U(x), we can always perform the integral in (2.53) and then isolate x(t). The two constants that appear in the trajectory are E and x0=x(t0). Although the problem is formally solved, it must be said that the remaining integral can be very complicated. One-dimensional harmonic motion This is a special case from the previous subsection F=−k x +F0 It is usually written as F=−k(x−x0), that is to say, x=x0is the position for which the force is null: x0=F0 k(see Figure 2.19). Fig. 2.19: One-dimensional harmonic motion Having this provides a new reference frame, x′, such that x′=x−x0. The force will be expressed as F=−k x′. For convenience, we rename the x′coordinate as x. In fact, the force we want to deal with is F=−k x 65 2 Dynamics of a particle The equation of motion is m¨x=−k x The potential energy is U(x) = −∫(−kx)dx =1 2k x2+ct And we can find the trajectory through the integral (2.53), which now is x(t) ∫ x(t0) dx √E−1 2k x2 =√2 m(t−t0) By taking the common factor k 2in the root of the denominator (for this we multiply and divide Eby k 2), the first member can be written as √2 k x(t) ∫ x(t0) dx √2E k−x2 Taking into into account that ∫dx √a2−x2=arcsin (x a) and once the integration limits have been replaced and matching the second member, we obtain √2 k{arcsin (√k 2Ex)−arcsin (√k 2Ex0)}=√2 m(t−t0) Now, isolating x, we find the trajectory x(t) = Asin (ω(t−t0) + φ0)(2.54) with A=√2E k, sinφ0=x0 A,ω=√k m. It mus be remembered that we have the relationship E=1 2m v2 0(t0)2+1 2k x2 0. Note that the relationship between Aand Ecan also be written as E=1 2kA2. Harmonic motions will be studied in depth in Chapter 6. Central forces Fig. 2.20: Central force Fig. 2.21: Central force caused by the tension in a rope passing through a small pulley Central forces have a line of action that always passes through the same point (see Figure 2.20). Some examples are the gravitational force that the Sun (or any other mass) exerts on a particle; electrostatic force of a fixed charge on a charged particle; and the tension of a rope passing through a small pulley with one end tied to a body (see Figure 2.21). In Figure 2.21, we see that tension TBcan be a constant force but tension Tis a central force! 66 2.5 Some forces All the central forces fulfil Kepler’s law. Proof. The torque  M(C)of the force with respect to point Cis always null. Taking into account that d L(C) dt = M(C)and if no other force acts, we deduce d L(C) dt = 0. The angular momentum with respect to Cis conserved throughout the motion and, as we have seen in Section 2.3, this is equivalent to Kepler’s second law.  Potential energy of a central force of the type  F=−F(r) ˆr(see Figure 2.22) Fig. 2.22: Vector expression of a central force Let us try to calculate the work integral regardless of the path: ∫ F·dr =−∫Fˆr·dr Now d(r ·r) = 2r ·dr dr2= 2rdr }⇒r ·dr =rdr ⇒ˆr·dr =dr Therefore, the integral involves a single variable rand we have not chosen any particular path. Central forces of type  F=−F(r) ˆrare conservative with an associated potential energy: U=∫F(r)dr +ct (2.55) For example, the force exerted by the Sun (fixed, of mass M) on a planet (particle of mass m) is  F=−GmM r2ˆr Substituting this force in (2.55) gives the potential energy: U=−GmM r+ct Non-conservative forces A force is called gyroscopic when the potential energy function cannot be defined, but dE dt = 0. From an energy point of view, they can be considered conservative, since they do not dissipate energy. A force is called dissipative when the potential energy function cannot be defined and dE dt <0. Gyroscopic forces The best known example of a gyroscopic force is the Lorentz force  FBdue to a magnetic field  Bacting on a charged particle qmoving at a velocity v:  FB=q v × B(2.56) 67 2 Dynamics of a particle Let us suppose that a particle is subjected to a conservative force  F, with potential energy U(r), and to a Lorentz force (2.56). According to what we have seen, in regard to energy and non-conservative forces, (2.46), we have dE dt =v ·(q v × B)= 0 Energy Eis conserved! Dry friction forces When a rigid object slides in contact with a non-smooth surface that we assume to be at rest, whether through gravity or due to any other force acting on it, the points of contact contain distributed forces that, if the object is small, can be decomposed into two: one perpendicular to the contact surface, which we call normal,N; and another one opposite to the motion, which we call dry friction,Ff. For surfaces that are sufficiently smooth and hard while the object is moving, the modulus of the dry friction force is Ff=µkN, where µkis the kinematic or dynamic friction coefficient, which depends on the nature of the surfaces in contact. The force Ff can be expressed in the following vector form (see Figure 2.23):  Ff=−Ff v v, Ff=µkN(2.57) Fig. 2.23: The dynamic friction force has the opposite direction of velocity Thus, it is clear that it is a force that depends on the velocity (not on the modulus, but on the direction!). It is only in the case that the surface causing the friction force is at rest that we can say it is always opposed to the motion. It is the dependence on the velocity that prevents us from finding the potential energy. When we want to calculate the work integral, we need to know the path followed by the object. Let us suppose that the particle is subjected to a conservative force,  F, with potential energy U(r)and to a dry friction force. According to what we have seen concerning the energy and non-conservative forces (2.46), we will have dE dt =v ·(−Ff v v)=−Ffv A dry friction force is dissipative. From a mechanical point of view, this energy disappears. A physical interpretation beyond mechanics can be found by applying conservation of energy or the first principle of thermodynamics: some of this energy increases the internal energy of the system and the rest dissipates in the form of heat/radiation. However, understanding this requires going a bit into thermodynamics which is beyond the scope of this course. We have seen the effect of dry friction when an object moves, generally due to the fact of being subjected to another force that exceeds the friction. But what happens if the object does not move? 68 2.5 Some forces If a force Fis applied to the block in Figure 2.24 and friction exists between the block and the surface supporting the block, the friction force evolves as follows. Fig. 2.24: Evolution of the friction force If the force is not great enough, the block does not move and Ff=F. If we increase F, a boundary value is reached, F=µsN. Even though the block does not move (again, Ff=F), any increase in F, no matter how small, will make it move. We say that it is in imminent motion. Coefficient µsis called the static friction coefficient. If we continue to increase F, then F > µsN, whereby the block moves and the friction force decreases to the value Ff=µkN. Coefficient µkis, already discussed above, the kinetic or dynamic coefficient of friction, which always µs> µk. The graph in Figure 2.25 illustrates the evolution of the friction force Ffas a function of the applied force F. Fig. 2.25: Evolution of the friction force Ffas a function of the applied force F Note: Although real problems must take into account this behaviour and, therefore, the existence of both coefficients, this course will use the friction coefficient µ without any specification (unless otherwise stated) and we will understand that µ=µs=µk. Problem 2.5.1. A particle of 2kg slides on an inclined plane 2m high. Starting from the top with null speed, it reaches the end at a speed of 5m/s. Calculate the energy loss due to the friction force and the value of this force considered to be constant, if the angle between the plane and the horizontal is 45◦. What is the value of the dynamic coefficient of friction? Solution The loss of mechanical energy is the increase in mechanical energy with changed sign: E=1 2mv2+mgz −∆E=−(Efi −Eini) = Eini −Efi = (1 2m v2 ini +mghini)−(1 2m v2 fi +mghfi)= 14.2J Solution to Problem 2.5.1 Solution to Problem 2.5.1 The energy variation is due to the work of the friction force: ∆E=Wf=∫fi ini  Ff·dr =−∫sfi sini Ffds =−Ff(sfi −sini) = −Ff hini sin 45◦ where ds =|dr|is the length travelled by the particle along the inclined plane and the friction force  Ffalways has an opposite direction to velocity. Finally, the friction force is Ff=−∆E (hini sin 45◦)= 5.02 N The dynamic coefficient of friction is Ff=µN ⇒µ=Ff N=Ff mg cos 450= 0.36  69 2 Dynamics of a particle Viscous friction forces We know through experimentation that the friction of a body in a fluid (aerodynamic or viscous friction) depends on the shape of the body, the material characteristics of the surface, the type of fluid and the velocity of the body (with respect to the fluid, which we will consider here to be at rest). A good approach to this type of friction is, in the laminar flow regime, (see upper part of Figure 2.26): Fig. 2.26: Upper: the laminar regime occurs for high viscosities and low relative velocities. Down: the turbulent regime is due to low fluid viscosities and high relative velocities  Fb=−bv b > 0(2.58) It is not conservative. Let us suppose that the particle is being submitted to a conservative force  F, with potential energy U(r), and to a viscous friction force. According to what we have seen concerning energy and non-conservative forces (2.46), we have dE dt =v ·(−bv) = −b v2 If the regime is turbulent (see lower part of Figure 2.26), the friction force can be approximated as  Fκ=−κv2ˆv κ > 0(2.59) Problem 2.5.2. A particle in a viscous medium is subjected to a constant force in addition to the frictional force due to viscosity. a) Prove that the particle ends up having a constant velocity and determine this velocity. b) Apply this result to the case of it falling in a viscous medium. Solution a) The force acting on the particle is  FT= F+ Fb=ma, with  Fbeing constant and  Fb=−bv , with b > 0being constant. Looking at the figure, we can see that  Fbtends to decrease the velocity component normal to  F, so that v and also  Fbitself will align with  F. This process will end when  FT= 0, that is, when the velocity reaches a value vLsuch that  F−bvL= 0. We thus obtain Solution to Problem 2.5.2 vL= F b Note that this is a limit value. v = F bis one possible solution to the equation of motion that needs an initial given velocity v0= F b.The general solution for the evolution of the velocity from an initial velocity v0must be obtained by integrating of ma =  F−bv, which we can write as ˙ v +b mv =b mvL. The solution to this equation is v(t) = (v0−vL)e−b mt+vL 70 2.5 Some forces We can see that for t→ ∞, we have v →vL. b)  F=mg ⇒vL=mg b Forces that only depend on time These are forces in the form of  Ft= Ft(t). They are not conservative because they explicitly depend on time. Let us suppose that the particle is subjected to a conservative force,  F, with potential energy U(r), and to a force  Ft= Ft(t). According to what we have seen concerning energy and non-conservative forces (2.46),we have dE dt =v · Ft If the particle is subjected only to  Ft, the formal integration of the equation of motion is simple: md2r dt2= Ft(t)⇒r(t) = 1 m∫ ∫ Ft(t)dt dt (2.60) The initial conditions r(t0) = r0and ˙ r(t0) = v0determine the integration constants. Problem 2.5.3. A particle of mass mat rest at r = (0,0,0) is subjected to the force depending on time  F= F0sin Ωt. a) Find the time trajectory. b) What is the initial velocity that makes the motion the same as that of a spring? Solution a) We find the velocity: v =1 m∫ F dt =1 m∫ F0sin Ωt dt =− F0 mΩcos Ωt+C1 We find the position: r =∫vdt =− F0 mΩ2sin Ωt+ C1t+ C2 We find the integrations constants: r(0) = (0,0,0) ⇒ C2= (0,0,0) v(0) = (0,0,0) ⇒ C1= F0 mΩ We obtain: r(t) =  F0 mΩ(t−1 Ωsin Ωt) 71 2 Dynamics of a particle b) For a spring r = Asin(ωt +φ0)⇒¨ r =− Aω2sin(ωt +φ0)⇒ F=−m Aω2sin(ωt +φ0) If we compare this force with the one given in the statement, we need that φ0=π, ω= Ω and  A= F0 mΩ2. Thus r = F0 mΩ2sin(Ωt+π)⇒˙ r = F0 mΩcos(Ωt+π)⇒v(0) = − F0 mΩ 72 3 Dynamics of Nparticles If we interpret the masses mias mi>0⇒to add, and mi<0⇒to subtract, we can treat a body composed of bodies and of simple holes (see Figure 3.11). Fig. 3.11: We can interpret the sum of masses in such a way that the positive sign means to ''add mass'' and the negative sign means to ''remove mass'' Problem 3.2.1. The following table shows the positions and velocities, at a given instant, of a three-particle system. i1 2 3 mi(kg) 2 3 5 ri(m) (−10,−10) (30,10) (10,20) vi(m/s) (10,30) (−20,−10) (10,−10) In this instant, determine: Table for Problem 3.2.1: masses, positions and velocities a) The position of the system’s centre of mass. b) The velocity of the system’s centre of mass. Solution a) rCM = 3 ∑ i=1 miri m=2(−10,−10) + 3(30,10) + 5(10,20) 10 = (12,11) m b) Solution to Problem 3.2.1 vCM =˙ rCM = 3 ∑ i=1 mi˙ ri m=2(10,30) + 3(−20,−10) + 5(10,−10) 10 = (1,−2) m/s  Problem 3.2.2. Given the flat homogeneous sheet shown in the figure, obtain the centre of mass. Figure for Problem 3.2.2 Solution to Problem 3.2.2 Solution It is a homogeneous plane surface. Because it is symmetrical, we have yCM = 0 . Also, since σ=ct, instead of masses miwe can use surfaces Si. Thus, we decompose the sheet into three pieces: piece 1 is a square sheet; piece 2 is a triangle sheet; and piece 3 is a hole sheet. See the figure. We have xCM =1 m N ∑ i=1 mix(i) CM =1 σS N ∑ i=1 σSix(i) CM =1 S N ∑ i=1 Six(i) CM Remember that the negative sign means it is a hole! xCM =1 S1+S2−S3{S1x(1) CM +S2x(2) CM −S3x(3) CM } S1= 4a2;S2=1 22a2a;S3=1 2πa2 x(1) CM =a;x(2) CM = 2a+1 32a;x(3) CM =4a 3π}⇒xCM = 1.96a 80 3.3 Momentum 3.3 Momentum The momentum  Pof a particle system is defined as the sum of all particle momentums.  P= N ∑ i=1 pi= N ∑ i=1 mivi=mvCM (3.9) The CM equation can be written as  F=d P dt (3.10) Conservation of momentum theorem. If the net force of a system of particles is zero, the momentum  Premains constant over time. d P dt = 0 (3.11) This can also be written using the integrated expression between the two instants tini and tfi  Pini = Pfi (3.12) In usual parlance, it is said that the momentum is conserved. Note that as long as no external forces act on the system, the momentum is conserved. Problem 3.3.1. A cobra of length Land uniformly distributed mass mlies stretched out on the ground. The cobra decides to rise vertically with a uniform velocity v (the tip of the tail does not move at any time). What is the reaction force of the ground on the part of the body in contact with it? Give the result as a function of L,m,vand the gravitational field g. Figure for Problem 3.3.1 Solution Solution to Problem 3.3.1 Let xbe the portion of the snake in contact with the ground and ythe vertical part. Note that both xand ywill vary as the snake rises. These are functions of time t(see the figure). At any instant t, the total length of the snake is L; thus, x+y=Land, since this expression is valid for all t,˙x+ ˙y= 0. If, in addition, we take into account that vis the constant velocity in the direction of y, we have the relations ˙y=v; ˙x=−v; ¨x= ¨y= 0 (3.13) We can calculate the position of the centre of mass, for all t, as a function of xand y. At any instant, the snake is a linear body formed by two straight homogeneous segments of length xand y, therefore, rCM = (xCM , yCM ) = 1 L(x 2x+xy, y 2y)(3.14) 81 3 Dynamics of Nparticles If we derive (3.14) with respect to time and take into account (3.13), ˙ rCM =1 Ly v (−1,1) (3.15) The momentum of the system can be written as  P=m˙ rCM =m Ly v (−1,1) (3.16) The net forces acting on the snake will be due to the weight and the reaction force of the ground,  Fi.e., m(0,−g) +  F(see Figure). Thus, we have Solution to Problem 3.3.1 d P dt =m Lv2(−1,1) = m(0,−g) +  F⇒ F=(−m Lv2, mg +m Lv2) 3.4 Angular momentum The angular momentum  L(A)of a system of particles with respect to a fixed point A (see Figure 3.12) is defined as the sum of the angular momentum of each particle with respect to this point  L(A)= N ∑ i=1  L(A)i= N ∑ i=1 ri(A)×mivi(3.17) Fig. 3.12: Angular momentum of a system of particles with respect to a fixed Apoint Then we take the time derivative: d L(A) dt = N ∑ i=1 (˙ ri(A)×mivi+ri(A)×miai)(3.18) Now, taking into account that point Ais fixed, ˙ r(A)i=vi, the first vector product of (3.18) becomes zero. For the second vector product, we will take into account the equations of motion for the Nparticles, miai= Fi+ N ∑ j=1  Fji. Thus, we have d L(A) dt = N ∑ i=1 (ri(A)×[ Fi+ N ∑ j=1  Fji]) (3.19) Assuming that the action–reaction forces have the same line of action, the net torque of the internal forces cancels out (see Figure 3.13): Fig. 3.13: Net torque of the internal forces ri(A)× Fji +rj(A)× Fij = (rj(A)−ri(A))× Fij = 0 Therefore, we obtain d L(A) dt = M(A)(3.20) where  M(A)is the net torque of the forces. We observe that only external forces are involved in  M(A)  M(A)= N ∑ i=1 ri(A)× Fi(3.21) 82 3.4 Angular momentum Conservation of angular momentum theorem. If the net torque of the external forces with respect to point Ais zero,  M(A)= 0, the angular momentum  L(A) remains constant over time: d L(A) dt = 0 (3.22) Using the integrated expression between two instants tini and tfi it can be written as  L(A)ini = L(A)fi (3.23) In usual parlance, it is said that the angular momentum is conserved. If the particle system moves only by translation,vi=v, the angular momentum of the system,  L(A), can be expressed as  L(A)= N ∑ i=1 ri(A)×mivi=(N ∑ i=1 ri(A)mi)×v =rCM(A)× P(3.24) Generally, if we use ri(A)=rCM(A)+ri(CM), we can write the angular momentum with respect to any fixed point Aas  L(A)= L(CM)+rCM(A)× P(3.25) The expressions (3.20) and (3.21) are also valid if A=CM, although CM is in motion. Care must be taken to not take moving points A=CM. In this section it is better to consider Afixed. Later, when we study the rigid body, we will prove and use A=CM. Problem 3.4.1. In the following table, we have the positions and the velocities of a three-particle system at a given instant. Table for Problem 3.4.1: masses, positions and velocities i1 2 3 mi(kg) 2 3 5 ri(m) (−10,−10) (30,10) (10,20) vi(m/s) (10,30) (−20,−10) (10,−10) At this instant, determine: a) The total momentum. b) The total angular momentum with respect to the origin. Solution Solution to Problem 3.4.1 a) From Problem 3.2.1, we have vCM = (1,−2) m/s and, thus,  P=mvCM = 10(1,−2) = (10,−20) kg m/s 83 3 Dynamics of Nparticles b) Be careful! We cannot use  L=rCM × P, because the velocities of the particles are different. We have to use  L= 3 ∑ i=1  Li:  L1=r1×p1= 2  ˆıˆȷˆ k −10 −10 0 10 30 0  =−400ˆ k  L2=r2×p2= 3  ˆıˆȷˆ k 30 10 0 −20 −10 0  =−300ˆ k  L3=r3×p3= 5  ˆıˆȷˆ k 10 20 0 10 −10 0  =−1500ˆ k We obtain  L= 3 ∑ i=1  Li=−2200 ˆ kkg m2 s Problem 3.4.2. A truck transports a homogeneous rectangular box that is 2m high and 1m long. The friction coefficient of the box and truck is 0.4. Calculate: Figure for Problem 3.4.2 a) The maximum acceleration that the truck can have without the box slipping (if we know that before the box does not overturn). b) The maximum acceleration that the truck can have without the box overturning if we attach the box to the truck by a frictionless axle A. Solution a)  F=ma ⇒(µN, N −mg) = m(a, 0) ⇒a=µg = 0.4×9.8 = 3.92 m/s2 Solution to Problem 3.4.2 b) We use d L(O) dt = M(O)with the angular momentum and torque calculated with respect to the origin (see the figure). We can use the corkscrew rule. For the angular momentum, note that the motion is translational. It is enough to use the velocity vector at the CM (see the figure). We obtain  L(O)=−1mv ˆ k. For the moment of force, we analyse the instant in which the box is about to overturn. At this instant, the normal passes through point A, since the only contact between the box and the truck is at point A. We obtain  M(O)=−{(OA + 0.5)mg −OA N}ˆ k=−0.5mg ˆ k Thus, deriving with respect to time  L(O)and making it equal to the torque −1maˆ k=−0.5mg ˆ k Therefore, a= 0.5g= 4.9m/s2Solution to Problem 3.4.2 84 3.5 Work, kinetic energy and potential energy 3.5 Work, kinetic energy and potential energy Work The concept of work, according to Interpretation 1of Section 2.4, can be applied here for each force separately. The same must be done with the concept of conservative force and potential energy associated with a force. In order to define the concept of energy for a system, we extend Interpretation 2of Section 2.4 to all particles and forces in the system. We call the configuration of the system at a given instant the set of positions for all particles. The work Wperformed by all the forces on each particle,  Fi+ N ∑ j=1  Fji, when particles go from one configuration P1={Pi1}to another P2={Pi2}through a path C={Ci}, is (see Figure 3.14) W= N ∑ i=1 P2 ∫ C:P1( Fi+ N ∑ j=1  Fji)dri(3.26) Fig. 3.14: The system goes from configuration P1to configuration P2. There are many paths C={Ci}that go from one configuration to the other Kinetic energy or ability to work due to speed Using a similar calculation to the one in Section 2.4 leads us to the kinetic energy concept W= N ∑ i=1 P2 ∫ C:P1( Fi+ N ∑ j=1  Fji)·dri= N ∑ i=1 P2 ∫ C:P1 miai·dri =... = P2 ∫ P1 d(N ∑ i=1 1 2miv2 i) Consequently, we define the kinetic energy Ecof a system of Nparticles as Ec= N ∑ i=1 1 2miv2 i(3.27) and we can state the following theorem: Work-energy theorem: W= ∆Ec(3.28) This is also known as the theorem live forces and as the kinetic energy theorem. 85 3 Dynamics of Nparticles Potential energy or ability to work due to position The forces on each particle are the external forces and the interaction forces between particles:  Fi+∑ j=1  Fji. If all the forces are conservative, they will have potential energies Ui, Uij that fulfil the relationships  Fi=−∂Ui ∂ri ; Fji =−∂Uji ∂ri , Uii = 0 (3.29) In addition to the forces being conservative, and as a consequence of the actionreaction principle, the forces depend only on the position of the particles in accordance with  Fji =f(ji)(ri−rj), where f(ji)=f(ij)depends only on the distance between particles iand j1. This allows choosing Uij which satisfies (3.29), de1Looking at the gravitational and electrostatic cases we see that they meet these conditions pends only on the distance between particles and is symmetric Uji =Uij . We denote this as (potential) energy of interaction. If the system goes from one position configuration P1to another one, P2, through a path C(see Figure 3.14), the work it does is W= N ∑ i=1 P2 ∫ C:P1( Fi+∑ j=1  Fji)·dri= N ∑ i=1 P2 ∫ C:P1  Fi·dri+ N ∑ i=1 N ∑ j=1 P2 ∫ C:P1  Fji ·dri (3.30) The work of the conservative forces is equal to the decrease in potential energy: W=−∆U(3.31) with the potential energy Ufor the system of Nparticles being defined as U= N ∑ i=1 Ui+ N ∑ i=1 N ∑ i<j Uij (3.32) Proof. The first term of the right side of (3.30) can be written as N ∑ i=1 P2 ∫ C:P1  Fi·dri=− N ∑ i=1 P2 ∫ :P1 dUi(3.33) Concerning the second term of the right side, we have to be much more careful. First, we exchange ifor j,N ∑ i=1 N ∑ j=1 P2 ∫ C:P1  Fij ·drj, and add the result to the original term. Since the added term is the same as the old one, we divide the sum by 2 N ∑ i=1 N ∑ j=1 P2 ∫ C:P1  Fji ·dri=1 2 N ∑ i=1 N ∑ j=1 P2 ∫ C:P1 ( Fij ·drj+ Fji ·dri) = ... (3.34) 86 3.5 Work, kinetic energy and potential energy Now we substitute the forces as a function of their potentials according to (3.29) ... =−1 2 N ∑ i=1 N ∑ j=1 P2 ∫ C:P1 (∂Uij ∂rj·drj+∂Uji ∂ri·dri) = ... (3.35) Remember that Uji =Uij . In addition, we will take into account that Uij depends only on riand rjand, therefore, the differential is dUij =∂Uij ∂rj·drj+∂Uij ∂ri·dri. We can write (3.35) as ... =−1 2 N ∑ i=1 N ∑ j=1 P2 ∫ C:P1 dUij =... (3.36) Finally, in order to not add any more terms than necessary, we restrict the sum of jto only terms i<j, and thus the factor 1 2disappears. ... =− P2 ∫ C:P1 d(N ∑ i=1 N ∑ i<j Uij )(3.37) Considering the above expressions (3.30,3.33-3.37), the potential energy Uof the system of Nparticles can be defined as (3.32), and we obtain (3.31).  Mechanical energy The mechanical energy Eof a conservative system of Nparticles is defined as E=Ec+U(3.38) Deriving with respect to time and following steps that are similar to those seen in Section 2.4, we obtain the following, as expected. Conservation of mechanical energy theorem. The numerical value of mechanical energy in a conservative system of Nparticles remains constant over time. dE dt = 0 (3.39) It can also be written using the integrated expression between two instants tini and tfi Eini =Efi (3.40) In usual parlance, it is said that the mechanical energy is conserved. If there are non-conservative forces (NC), these are not included in the energy (they have no associated potential energy). However, the following can be applied. Mechanical energy theorem. The work of the non-conservative forces is equal to the increase in the mechanical energy of the system: WNC = ∆E(3.41) 87 3 Dynamics of Nparticles We observe that energy can still be conserved if we restrict ourselves to motions for which the NC forces “do not work”, that is, WNC = 0. Later, when we deal with constrained systems, we will use this condition. As we will see later in Section 3.11 for the case of a rigid body, the work of the internal forces is zero; so, W= ∆Ec, where Wis the work of the external forces. If, in addition, the motion is only translation,vi=vCM and we have Ec= N ∑ i=1 1 2miv2 i=1 2m v2 CM . Thus, the mechanical energy of a rigid body restricted to translational motions is E=1 2m v2 CM +U(3.42) where, in this case, Uis the sum of potential energies of the body’s conservative external forces. Problem 3.5.1. Two particles of masses 2kg and 3kg are bound together with a rope running through a spring, as seen in the figure. Both the spring and the rope have negligible mass. The spring has a constant of 12000 N/m and is compressed to a length of 10 cm. We cut the rope. Find the velocity of each particle. Figure for Problem 3.5.1 Solution For convenience, we define m1= 2 kg, m2= 3 kg, k= 12000 N/m and L= 0.1m. The causes of the motion are aligned along the x-axis; so, we can apply the symmetry principle (see the Section 1.4) to deduce that the velocities of both particles will be aligned with the x-axis: v1= (v1,0,0) and v2= (v2,0,0). Applying the momentum conservation between the instants just before and just after cutting the rope, we have: Solution to Problem 3.5.1 0 = m1v1+m2v2⇒2v1+ 3v2= 0 We can do the same with energy: 1 2kL2=1 2m1v2 1+1 2m2v2 2⇒60 = v2 1+3 2v2 2 from which we get v1=±6and v2=∓4. Using the fact that the spring decompresses (this fact has not been used yet!), the signs are determined and we obtain Solution to Problem 3.5.1 v1= (−6,0,0) m/s;v2= (4,0,0) m/s Problem 3.5.2. The following table shows the positions and velocities of a threeparticles system at a given instant. At this instant, determine: 88 3.6 Collisions i1 2 3 mi(kg) 2 3 5 ri(m) (−10,−10) (30,10) (10,20) vi(m/s) (10,30) (−20,−10) (10,−10) a) The kinetic energy of the system. Table for Problem 3.5.2: masses, positions and velocities b) The kinetic energy associated with the motion of the centre of mass (all the mass concentrated in the CM together with its velocity) Solution a) Solution to Problem 3.5.2 Ec=1 2 3 ∑ i=1 miv2 i=1 2{2(102+ 302) + 3(202+ 102) + 5(102+ 102)}= 2250 J b) From Problem 3.2.1, we havevCM = (1,−2) m/s. The kinetic energy of a particle of mass m= 10 kg moving with the velocity of the CM is 1 2m v2 CM =1 210 ·(12+ 22) = 25 J which generally does not match the kinetic energy of the system.  3.6 Collisions Acollision is a process in which two or more particles with very short range interaction, meet at a point and at an instant and, consequently, modify their velocities (see Figure 3.15). Fig. 3.15: Forces exist only at the instant of collision We will restrict ourselves to two particles while considering two instants in which the particles are far enough apart and do not interact: the initial instant (ini), just before the collision; and the final instant (fi), just after the collision. We will assume that the external forces are much weaker than those caused by the collision, such that the external impulse between (ini) and (fi) is negligible. Fig. 3.16: Collision plane We will also restrict ourselves to frictionless collisions. This means that when contact occurs between the two particles, which we can visualize as spherical bodies (see Figures 3.16 or 3.17), there are no forces in the tangent to the contact plane or collision plane and, therefore, there is no exchange of angular momentum. If the particles or bodies were to had an initial rotation, they would maintain this rotation after the collision. We can apply the theorem of conservation of total momentum, giving us:  P(ini)= P(fi)(3.43) 89 3 Dynamics of Nparticles where we observe that we have also taken into account the dt variation, since the possible displacement takes some time to be carried out. We can use the parametric form of the constraints to find the possible dridisplacements in terms of the L qaparameters: dri= L ∑ a=1 ∂ri ∂qa dqa+∂ri ∂t dt (3.59) We define the δrivirtual displacements of the system as displacements that are carried out without variation in time, i.e. by freezing time. If we imagine that we are filming a movie scene in which the pivot of an oscillating pendulum is moved by a hand (see Figure 3.27), then the possible displacement are what we are recording and can see later in the finished movie. Virtual displacement would correspond to stopping the recording, moving the particle without moving the hand and, at the end of this displacement, continue recording. Fig. 3.27: The subtle difference between possible and virtual displacement From the mathematical point of view, virtual displacements fulfil N ∑ i=1 ∂fa ∂ri·δri= 0 (3.60) or, in parametric form, δri= L ∑ a=1 ∂ri ∂qa dqa(3.61) Note that if the constraints do not depend on time, then δri=dri. This course does not deal with time-dependent constraints, nor will we make a special distinction between possible and virtual displacements. The one exception will be in Chapter 9, where we will consider time-dependent constraints and then δri=dri. Reactions The particles move according to the constraints, which exert forces on the particles. These are the so-called reaction forces  Ri. If, in addition to  Ri, the force  Fi(generally known) acts on each iparticle in the system, we can write the equation of motion (Newton’s second law)  Fi+ Ri−miai= 0 i= 1...N (3.62) These equations have to be completed with the constraints (3.56) or (3.57). Fig. 3.28: In the case of the pendulum, the  Rireaction force is the one exerted by the rod The problem we find here is that the constraints are generally known, but their reactions  Riare not. As we have mentioned, the reaction forces  Riare the forces exerted by the constraints, such that the particles move according to their orders 96 3.8 Constraints and reactions. Possible displacements and virtual displacements... and, of course, also according to Newton’s laws (3.62). In equation (3.62),  Riare unknown. Only one part of ri(t)is unknown since, as the constraints are known, we will know part of the trajectory of the particles. For example, in the case of the pendulum (see Figure 3.28) restricted to the vertical plane, we already know where the particle passes. We just need to know its speed or, more specifically, we only need to know θ(t). In the case of a particle passing through the wire (see Figure 3.29), we also know the trajectory. If the wire is frictionless, we can say that the reaction  Ris normal to it. Fig. 3.29: In the absence of friction, the reaction of the wire is normal to it Ideal reactions A very interesting property that we observe in the cases represented in Figures 3.28 and 3.29 is that the reaction is normal to the displacement. In the case represented in Figure 3.30, the reaction is normal to the virtual displacement,  R·δr = 0, but not to the possible displacement,  R·dr = 0. Fig. 3.30: The Rtension is normal to the δr virtual displacement but not to the possible displacement dr The fact that the reaction is normal to the displacement means that the work of the force is zero when the system performs the displacement. In fact, what interests us is not that the reactions do not work individually. Although they could work separately, it is enough for us that they do not work together. Driven by this interest, we define the ideal reaction sets below. Ideal reaction set. A set of  Rii= 1 ... S reactions is called a set of ideal reactions if they fulfil S ∑ i=1  Ri·δri= 0 (3.63) This important property is satisfied in many situations. In general terms, reactions are ideal in the absence of dissipative friction. Let us see a couple of cases. Problem 3.8.1. Prove that the reaction forces in a frictionless pivot (also called pivot point) are ideal. Figure for Problem 3.8.1 Solution The point where the pivot is located is occupied by one particle from body 1at position r1and another particle from the body 2at position r2. As a result of the constraint, they move the same, r1=r2. If we differentiate we get dr1=dr2, which can be written as dr1−dr2= 0. Solution to Problem 3.8.1 The reaction forces  R1and  R2fulfil the law of action-reaction  R2=− R1. If now we calculate the work  R1·dr1+ R2·dr2= R1·(dr1−dr2) = 0 97 3 Dynamics of Nparticles then the set of forces  R1and  R2becomes a set of ideal reactions.  Problem 3.8.2. Prove that the reaction forces with frictionless contact between two moving solids are ideal. Figure for Problem 3.8.2 Solution The bodies have velocitiesv1andv2with respect to an observer at rest. At any instant, body 1moves at a velocity v1(2) relative to body 2and in the direction of the contact plane. The reaction forces  R2=− R1are normal to this plane and, therefore, they are normal to the relative velocity, that is to say,  R1·v1(2) = 0. We can write Solution to Problem 3.8.2 0 =  R1·(v1−v2) = 1 dt  R1·(dr1−dr2) = 1 dt ( R1·dr1− R1·dr2)=1 dt ( R1·dr1+ R2·dr2) where dr1and dr2are the possible displacements. Therefore, the contact reactions are a set of ideal reactions.  3.9 The general equation of dynamics also known as d'Alembert's principle We want to write some equations of motion in which the constraints appear with out the reaction forces. That is, we want to rewrite Newton’s second law while taking into account that the constraints are part of the data while their reactions are unknown. The set of forces acting on the system is classified into two subsets: the reactions  Riand the other forces,  Fi, which we will call directly applied forces. General equation of dynamics. If all the reactions of the system are ideal, Newton’s equations of motion are equivalent to N ∑ i=1 ( Fi−miai)·δri= 0 (3.64) We note that in the equations extracted from (3.64): 1) Reaction forces do not appear. 2) We obtain as many independent equations as we do degrees of freedom. The first version of the general equation of dynamics came from Jean le Rond d’Alembert and was given the name of d’Alembert’s principle, stated in the following way: Any position of a moving system is in equilibrium if inertial forces are added to external forces. Fig. 3.31: Jean le Rond D'Alembert (1717-1783) was a French mathematician and philosopher 98 3.10 Conservative system with constraints. Energy conservation This statement will be later understood in the context of statics and in relation to the principle of virtual works. However, we will not use this interpretation. The proof of the general equation of dynamics (3.64) is not difficult from a mathematical point of view, although it may be conceptually surprising. Proof. We want to prove that N ∑ i=1  Ri·δri= 0 (3.65) and  Fi+ Ri=miai(3.66) are equivalent to N ∑ i=1 ( Fi−miai)·δri= 0 (3.67) Proof (3.65,3.66)⇒(3.67): If we multiply (3.66) by δriand sum over index i, we obtain 0 = N ∑ i=1 ( Fi+ Ri−mia)·δri Thus, taking into account (3.66), we obtain (3.67). Proof (3.67)⇒(3.65,3.66): We now observe that the reactions are not determined. They have to enforce the system’s motion through the constraints. Thus, they are determined as  Ri=miai− Fiand, obviously, (3.66) is fulfilled. If we multiply (3.66) by δriand sum over index i, we obtain N ∑ i=1 ( Fi+ Ri−miai)·δri= 0. Now, taking into account (3.67), we obtain (3.65).  3.10 Conservative system with constraints. Energy conservation Here we consider a conservative system with constraints: a particle system with time-independent constraints that all have ideal reactions  Riand, furthemore, the other forces  Fi(internal or external) are conservative. This means that the system will have the potential energy function U: U=−∫N ∑ i=1  Fi·dri(3.68) Conservation of mechanical energy theorem. If we define the mechanical energy in a constrained conservative system of Nparticles as E=1 2 N ∑ i=1 miv2 i+U(3.69) 99 3 Dynamics of Nparticles where Uis defined according to (3.68), then the numerical value is maintained constant over time. dE dt = 0 (3.70) Proof. We apply the general equation of dynamics by taking as possible displacements the actual displacements of the system (the displacements that fulfil the general equation of dynamics) and divide by elapsed time, dt: N ∑ i=1 ( Fi−miai)·dri dt = N ∑ i=1  Fi·dri dt − N ∑ i=1 miai·vi(3.71) =−(dU dt +d dt 1 2 N ∑ i=1 miv2 i)= 0 (3.72)  It is usually said that mechanical energy is conserved. Energy conservation can also be written using the integrated expression between two instants tini and tfi Eini =Efi (3.73) We must remember that the constraints must be time-independent in order for the energy to be conserved. It is not enough that the directly applied forces are conservative and the reactions are ideal. The virtual displacements and possible displacements must also coincide: δri=dri. Throughout the book, we will consider primarily time-independent constraints, and only Chapter 9will deal with time-dependent constraints. In the case of conservative systems with one degree of freedom, this result is sufficient for finding the temporal trajectory of the system. It should be noted that drifrom (3.68) are the possible particle displacements to which the forces are applied, but they do not need to coincide with the point displacements in the space where the forces are applied. We can see this in the following example. Figure for Problem 3.10.1 Problem 3.10.1. In the figure, spring is relaxed when the lower face Aof a prism with section Sand mass m, which can slide along a fixed track, is at the level of a liquid with density ρ. If we let it go from the described position, what will be the depth ywhat Awill reach? Note: The container is large enough so that the change in the liquid level is negligible. 100 3.10 Conservative system with constraints. Energy conservation Solution It is rigid body that can be moved only by translation. We take the y-axis in the direction of g with the origin at the fixed level of the liquid. Under these conditions, the point Acoordinate is y. When y= 0, the spring exerts no force and Archimedes’ buoyant force is null. The possible infinitesimal displacement of any particle of the body is dr =dyˆȷ. The system is conservative and, as a result, we can apply energy conservation between the initial and final situations. To be able to write the different potential energy terms, we will consider point Ato be sunk to y: Ug: the force is  P=mgˆȷand the associated potential energy Ug=−∫(mgˆȷ)·(dy ˆȷ) = −mgy. Uk: the force exerted by the spring is  Fk=−kyˆȷand the associated potential energy, Uk=−∫(−kyˆȷ)·(dy ˆȷ) = 1 2ky2. UE: Arhimedes’ buoyancy (see Section 4.3) is  E=−ρgSy ˆȷand the associated potential energy, UE=−∫(−ρgSyˆȷ)·(dy ˆȷ) = 1 2ρgSy2. Let us note that the displacement is always that of the particle in the body where the corresponding force is applied. In particular, we do not need to know, at each instant, which is the point where Archimedes’ buoyant force Eis applied, which we will study in the next chapter. We will see that when Ais at y, the buoyancy Eis applied to y/2 and, therefore the displacement of the Eapplication point is dy/2. However, we now need the possible displacement dy of the particle of the body where Eis applied, although this particle may be different during the process. The mechanical energy of the system is E(y, ˙ y) = 1 2m˙y2−mgy +1 2ky2+1 2ρgSy2=1 2m˙y2−mgy +1 2(k+ρgS)y2 applying its conservation, we have E(0,0) = E(y, 0) ⇒y=2mg k+ρgS  Problem 3.10.2. A particle tied to a rope of negligible mass and that at all times remains tense, oscillates in a vertical plane. Find the motion equation using energy conservation. Figure for Problem 3.10.2 Solution We observe that the tension Tof the rope is always normal to the possible particle displacement. This is an ideal reaction. Aside from the reactions, we have the gravity force, which is conservative. We can express the constraint in the parametric form as x=ℓsin θ y=ℓcos θ The energy system is E=1 2mv2−mgℓ cos θ, where v=√˙x2+ ˙y2which, in terms of θ, is v=ℓ˙ θ. Thus, we obtain Solution to Problem 3.10.2 E=1 2mℓ2˙ θ2−mgℓ cos θ 101 3 Dynamics of Nparticles and applying energy conservation, we have 0 = dE dt =mℓ2˙ θ¨ θ+mgℓ sin θ˙ θ Taking into account that we look for solutions allowed by the constraints ˙ θ= 0, we obtain the motion equation ¨ θ=−g ℓsin θ Problem 3.10.3. In the figure, nail Cis at a vertical distance dfrom where a rope attached to a ball of mass mis fixed. By letting the ball go so that it can make a complete revolution in a circle centred on the nail, prove that it is necessary that d≥0.6L. Figure for Problem 3.10.3 Solution It is a constrained conservative system with ideal constraints. We will use energy conservation Eini =Efi between the ini position, with v= 0 and y= 0 ⇒Eini = 0, and the fi position, with yand v⇒Efi =1 2mv2−mgy. At the red point in the figure, we have y=d−(L−d). Substituting yin Eini =Efi and isolating the speed, we have v=√2g(2d−L)(1) At the same time, this speed must be sufficient so that the rope does not slacken. Thus, at the point where it has the least speed, the tension Tof the rope must keep it stretched, that is, T≥0. By writing the motion equation in the tension direction and taking into account that the normal acceleration is an=v2 L−d, Solution to Problem 3.10.3 mg +T=mv2 L−d Isolating vand imposing that T≥0, we have v≥√g(L−d)(2) Combining (1) and (2): √2g(2d−L)≥√g(L−d)⇒4d−2L≥L−d⇒d≥3 5L Problem 3.10.4. The blocks in the figure are released from rest. There is neither friction on the horizontal ground nor in the pulleys nor in the rope, which are of negligible mass. Determine the acceleration of the blocks and the tension in the rope. Figure for Problem 3.10.4 102 3.10 Conservative system with constraints. Energy conservation Solution It is a constrained conservative system. Given that when we make a possible displacement dx, once the section of rope that advances dx past the pulley must be divided between two rope sections, thus giving us dy =1 2dx and, dividing by dt,˙y=1 2˙x. We can write the energy of the system, directly using the numerical values of the masses, as E=1 2700 ˙x2+1 2300 ˙y2−300 ×9.81 y= 387.5 ˙x2−2943y Deriving with respect to time, we have Solution to Problem 3.10.4 ˙ E= 2 ×387.5 ˙x¨x−2943 ˙y Substituting the constraint ˙y=1 2˙ xand imposing energy conservation give us (775 ¨x−1471.5) ˙x= 0 Taking into account that we are looking for solutions compatible with the constraints, ˙x= 0, we obtain ¨x= 1.8987 m/s2¨y= 0.94935 m/s2 Now applying Newton’s second law to each block: T1= 700 ¨x⇒T1= 1329.1N 300 ×9.81 −T2= 300 ¨y⇒T2= 2658.2N  Problem 3.10.5. The 6kg block Bis left to slide down on the 15 kg wedge A resting on the frictionless horizontal ground. Calculate the accelerations of Aand B. Figure for Problem 3.10.5 Solution There is no friction and there are no external forces on the horizontal ground direction. The conservation of momentum imposes mA˙xA+mB˙xB= 0 ⇒ ˙ xA=−mB mA ˙xB=−2 5˙ xB(1) In the figure, we see that the velocity of Brelative to Ahas an inclination of 30◦(We have to take into account that the inclined plane moves!). vB(A)= ( ˙xB−˙xA,˙yB−˙yA)⇒˙yB ( ˙xB−˙xA)=tan 30◦⇒ ˙yB=7 5√3˙xB(21) Now we use energy conservation. To do this, we write the energy as a function of ˙xB and yB, derive it and equal to zero Solution to Problem 3.10.5 103 3 Dynamics of Nparticles E=1 2mA˙x2 A+1 2mB( ˙x2 B+ ˙y2 B) + mBgyB= 6.16 ˙x2 B+ 58.86yB ˙ E= 0 = 12.32 ˙xB¨xB+ 47.576 ˙xB⇒¨xB=−47.576 12.32 =−3.86 m/s2 Substituting in (1) and (2) derived with respect to time, it follows that ¨yB=−3.12 m/s2¨xA= 1.545 m/s2  Problem 3.10.6. A mass m= 0.5kg slides without friction in a vertical plane along a wire (d= 0.8m). The spring has a natural length of ℓ= 25 cm and k= 600 N/m. If the mass is released with no initial velocity when b= 30 cm, determine: a) The velocity when it reaches C. b) The velocity when it reaches B. Figure for Problem 3.10.6 Solution It is a conservative system with an ideal constraint. The conservative forces are the weight and the force exerted by the spring. The energy as a function of x,yand v can be written as E(x, y, v) = 1 2mv2+mgy +1 2k(ℓ−√x2+y2)2 where xand yhave to be on the wire which is the constraint. Let us impose energy conservation between the initial point and the points indicated in each part. The energy at the initial point is Solution to Problem 3.10.6 E=1 2m02+mg(−0.3) + 1 2k(ℓ−√0.42+ (−0.3)2)2= 17.2785 J a) 1 2m v2 C+mg 0 + 1 2k(ℓ−0.4)2= 17.2785 ⇒vC= 6.48953 m/s b) 1 2m v2 B+mg 0.4 + 1 2k(ℓ−0.4)2= 17.2785 ⇒vB= 5.85372 m/s 3.11 Rigid body A rigid body is a system of iparticles of ripositions with geometrical constraints (ri−rj)2=ct (3.74) 104 3.11 Rigid body Ideal reactions The cohesive forces of a rigid body form a set of ideal reactions. Fig. 3.32: In a rigid body, the internal forces are the reaction forces of the constraints that keep the distance between particles constant Proof. We will analyse any two of the many particles in the body, 1and 2, with position vectors r1and r2, respectively (see Figure 3.32). The cohesive forces of the body (internal forces) are the constraint reactions. The forces  F12 and  F21 will satisfy Newton’s third law:  F21 =− F12 ; F12 ∝(r2−r1)(3.75) The possible (= virtual) dr1and dr2displacements will fulfil 0 = d(r2−r1)2= 2(r2−r1)·(dr2−dr1) = 0 (3.76) The condition for ideal constraints for these two particles is  F21 ·dr1+ F12 ·dr2= 0 and that is what we must prove. Taking into account (3.75) and (3.76), we have  F21 ·dr1+ F12 ·dr2=− F12 ·dr1+ F12 ·dr2=  F12 ·(dr2−dr1)∝(r2−r1)·(dr2−dr1) = 0 We can extend this result to all the particle pairs in the body and, thus, the set of cohesive forces in a rigid body is ideal.  Possible displacements From a body, we take a reference point C, of position rC(see Figure 3.33). The position of any other point in the body can be written as ri=rC+ri(C). We can thus decompose the possible displacements drias Fig. 3.33: Translation without rotation dri=drC+dri(C)(3.77) If we substitute ri=rC+ri(C)into condition (3.74), the result does not depend on rC. Thus, any drCis possible. When the possible displacement has the form dri=drC(3.78) we say that it is a translation. It remains to be seen what restrictions are imposed by (3.74) on dri(C). The condition is d(ri−rj)2= 2(ri(C)−rj(C))·(dri(C)−drj(C)) = 0 (3.79) Fig. 3.34: Rotation without translation The solution to these equations gives us the expression of the possible displacements with respect to point C,drj(C), which we denote as rotations with respect to C(see Figure 3.34). 105 3 Dynamics of Nparticles Problem 3.13.1. A car of mass mwith wheel width 2ais travelling at a constant velocity modulus von a horizontal track, such that its CM at a height habove the track, describes a circle of radius R≫a. Find the maximum speed at which it can travel without overturning. Solution Solution to Problem 3.13.1 The frictional force represented in the figure is the resultant of the frictional forces where each wheel has contact with the road, which points radially inward due to the fact that the speed is constant. The translational motion of CM is determined by:  F=ma ⇒N=mg ;Ff=maN=mv2 R The movement is at a uniform speed. As long as the car does not overturn, the angular momentum will be constant. In other words, the car does rotate, but uniformly. Thus, if  L(CM )is constant, ⇒d L(CM ) dt = 0 It is sufficient to calculate the momentum of the forces with respect to the CM and make it equal to zero  M(CM )= 0. Looking at the figure, we can formulate  M(CM)= 0 = (Na −Ffh, 0,0) ⇒Na −Ffh= 0 and, combining this with the translation equations, we get v=√gRa/h Problem 3.13.2. What are the conditions that a frictionless pulley must meet in order for the tension of a rope, with negligible mass, to be equal on both sides? Figure for Problem 3.13.2 Solution The moments are calculated with respect to the CM, located at the centre of the pulley:  L(CM)=∑ i ri(CM)×mivi We decompose the velocity into rotational and translational velocities: vi=vCM + vi(CM) The translation term is null: (∑ i miri(CM)) | {z } =0 ×vCM = 0 Thus, we have  L(CM )=∑ i ri(CM)×mivi(CM)⇒d L(CM) dt =∑ i ri(CM)×miai(CM) With regard to the moments of the forces with respect to the CM:  M(CM)= (T′R−T R)ˆ k 112 3.14 Torque Equal tensions ⇔ M(CM )= 0 ⇔∑ i ri(CM)×miai(CM)= 0 This cancellation can be due to: 1) A pulley of negligible mass (mi→0). It can rotate and move arbitrarily. 2) A pulley in uniform rotation (null tangent accelerations): ri(CM)×ai(CM)= ri(CM)×aN i(CM)= 0 since ri(CM)||aN i(CM). It can have arbitrary dimensions and mass, and it can move arbitrarily. If the pulley has a negligible radius, care must be taken: the rotation can become very great, so much that the product does not needs to be zero!  3.14 Torque Torque is a system of forces applied to a rigid body and equivalent to two equal forces, in opposite directions and with parallel lines of action; hence, of zero net force and non-zero net moment (see Figure 3.40). We will call torque the resultant moment of the forces. The torque is independent of the point in space used to calculate it. Proof. We can see this by calculating the net moment: considering Figure 3.40,  M=r+× F−r−× F= (r+−r−)× F=r∓× F Fig. 3.40: Torque The modulus of the torque can be expressed as M=Fd (3.97) where dis the distance between the lines of action of the torque forces. This distance is called the torque arm. Therefore, the torque depends on the distance between the forces (arm) and the value of the forces. Fig. 3.41: A force  Fapplied to the CM is the cause of a pure translation: If the body is initially at rest, the effect of the force  Fis to move the CM without rotating the body It is very common to know the torques,  Mi, applied to a body and not the forces that constitute them. This does not prevent us from determining the motion of the body using equations (3.91) and (3.92). The torques  Miare part of the additive terms in the determination of the total moment  M(C). A torque normal to the plane of the paper can be represented by the symbols xor y, which determine the sign of the torque. Pure translation. In Figure 3.41 only one force is applied to the CM of a body. According to the equations of motion (3.91) and (3.92) with C=CM, the effect on the body, if it is initially at rest, is a translation of the CM without any rotation, which we call pure translation.Fig. 3.42: A torque  Mis the cause of a pure rotation: If the body is initially at rest, the effect of the torque  Mis to make it rotate around CM. The CM remains at rest Pure rotation. In Figure 3.42 only one torque  Mis applied to the same body. Taking into account the equations of motion, the effect on the body, if it is initially at rest, is a rotation around the CM without it moving, a motion we call pure rotation. 113 3 Dynamics of Nparticles Work, power and energy of a torque The work done by a torque  Mapplied to a rigid body is the work done by the constituent forces of the torque. If we express the torque with two forces  F+= F and  F−=− Fapplied to the points in the body r+and r−, we have  M= (r+− r−)× F. The work will be W=∫( F+·dr++ F−·dr−)=∫ F·(dr+−dr−) where we have omitted the specification of the limits of integration. The torque forces are applied to points on the body and these points can change. Now, as we have already seen in Problem 3.10.1, the displacements dr±are displacements of the particles in the body, in this case in a rigid body. Considering (3.83), we can write dr+−dr−=dφ ×(r+−r−)and  F·(dr+−dr−) =  F·(dφ ×(r+−r−))=((r+−r−)× F)·dφ. We obtain the expression for work done by and the power of a torque when applied to a rigid body: W=∫ M·dφ ;P= M·ω (3.98) where ω =dφ dt is the angular velocity of the body. If the torque has the constant direction ˆu, M=Mˆu, and the body moves along the same axis, dφ =dφ ˆu, the expression for the work Wand the power Pare W= φfi ∫ φini M dφ ;P=Mω (3.99) If the torque Mdepends only on φ, we can write the potential energy associated with the torque as follows U=−∫M(φ)dφ (3.100) If the torque is constant, we have U=−Mφ (3.101) Fig. 3.43: The spring of recovery constant κgives the body a torque of M=−κφ when it is deformed by an angle φ If the torque has the form M=−κφ, as in the case of the spring shown in Figure 3.43, where κis the spring’s recovery constant (units N m rad−1), the resulting corresponding potential energy is U=1 2κφ2(3.102) 114 4 The statics of rigid bodies Introduction In this chapter, we will deal with the statics of rigid body and see what conditions must be met for a rigid body to be at rest. Using equivalence of force systems, we will learn how to represent the most common force systems acting on rigid bodies (see Figure 4.1) and by nulling the net moment, we will find the point at which the net force must be applied. We will apply the motion equations by inverting what we usually take as data and unknowns. The data will be, apart from some forces, the motion of the body that we want to be at rest. We will also see how the general equation of dynamics becomes the general equation of statics, also known as the principle of virtual works. If the constraints are ideal and the forces are conservative, we will see that the principle of virtual work can reduce the problem of equilibrium and its stability to simply studying the geometry of the potential function. Fig. 4.1: Rigid body subjected to forces applied at different points 4.1 The statics of a body: Equilibrium conditions A body is said to be in equilibrium when the accelerations of translation and rotation are zero. Thus, a body at rest is in equilibrium, but so is a body that moves with uniform velocity and/or has a uniform rotation. What is important is that if the initial conditions of the body are at rest and the body is at equilibrium, it will be at rest. In the previous chapters we have posed the dynamics problem by trying to find out what effects or motions caused the known forces and, therefore, the motion was unknown to us. What we want to know now is what forces cause a known effect: zero acceleration. The forces are now unknown to us, although not all of them, of course. In general, we will try to find the unknown forces that, together with the known forces and the geometric conditions (or constraints), cause the body to be in equilibrium. We will begin with the motion equations of a rigid body (3.91,3.92), which we 4 The statics of rigid bodies have already seen in Chapter 3. If the body is in equilibrium, we have  F= N ∑ i=1  Fi= 0 ;  M(O)= N ∑ i=1 ri(O)× Fi= 0 (4.1) Thus, we can state: Equilibrium condition of a rigid body. The necessary and sufficient condition for a rigid body to be in equilibrium is that the net force and torque acting on the body are zero. Properties and relationships In the following we will see a series of properties and relations that will be useful to us. a) The point Owhere we calculate the moments in equilibrium equation (4.1) can be any fixed point in space. This is a consequence of the motion equations of a body, derived in Section 3.11. There, we commented that point C, if it existed, could be a fixed point point attached to the body or, in general, the centre of mass. In the case of statics, the body is at rest and, therefore, Ccan be any point in a fixed reference frame. b) Forces on a body are sliding vectors. This property is fulfilled whether the body is in equilibrium or not, although in the latter case it must be specified that it is fulfilled in each instant. We can make the force slide along its line of action by moving the point of application to any other point on this line, since the moment with respect to a fixed point Owill be the same (see Figure 4.2). Proof 1 The proof can be made by beginning with the definition of the moment of a force r1(O)× F= (r1(O)+r2(O)−r2(O))× F=r2(O)× F+(r1(O)−r2(O))× F=r2(O)× F  Fig. 4.2: The forces on a rigid body are sliding vectors Proof 2 As we have seen in Chapter 2, we can calculate the moment using the corkscrew rule and (2.11). In our case, it would be M(O)=F O  F Since the force  F, the points of application between its line of action and point Oare in the same plane, the direction of the momentum will be the same and the modulus 118 4.1 The statics of a body: Equilibrium conditions will depend only on O F, that is, on the distance between point Oand the line of action of  F, which is independent of the point of application of  Fon this line (see Figure 4.2).  Fig. 4.3:  F1and  F2forces not known c) If only two forces act on a body in equilibrium, they must necessarily be of equal modulus, in opposite direction and have the same line of action (see Figures 4.3 and 4.4). Fig. 4.4: The equilibrium conditions impose this configuration for forces  F1and  F2 Proof. We want the body in Figure 4.3 to be in equilibrium. The net force must be zero,  F2=− F1. The system in Figure 4.3 is therefore a torque. The net moment must also be zero. According to the expression for the torque (3.97), the distance between the lines of action of the forces, d, must be zero. That is, the two forces have the same coincident line of action: The force system in Figure 4.3 must be the same as the one in Figure 4.4. Figure for Problem 4.1.1 Problem 4.1.1. Calculate the reaction at point C(force and torque) required to keep the sphere, of negligible weight, in equilibrium. Solution Let  Cbe the reaction force at point Con the sphere and MCthe reaction torque. With the axes in the figure the equilibrium equations for the forces are Solution to Problem 4.1.1 Fsin φ+Cx= 0 −Fcos φ+Cy= 0 and the equilibrium equation for the moments with respect to point C(remembering that we can do this for any point) MC−Fsin φ2R= 0 We find  C=−Fsin φˆı+Fcos φˆȷ MC= 2RF sin φˆ k Problem 4.1.2. The figure shows the forces applied to a rigid body, of negligible weight, which is found to be in equilibrium. What is the value of d? Figure for Problem 4.1.2 Solution We do not have to calculate the force F. Let us express the equilibrium condition of the moments by taking moments with respect to the point of application of the force F 90 (d−6) −50 d= 0 and isolate d d= 13.5m 119 4 The statics of rigid bodies 4.2 Weigth and centre of gravity The weight of a body is the force exerted by the Earth’s gravitational field on each of its particles (see Figure 4.5). The Earth’s gravity g is considered constant due to the relatively small dimensions of the body to be treated with respect to the Earth. Unless stated otherwise, we will use the standard value g= 9.81 m/s2 Fig. 4.5: The gravitational field acts on every particle in a body The moment of the weight forces on a body with respect to the CM is zero. Proof. The position of CM relative to itself is obviously zero, rCM(CM)= 0. Therefore, 1 m∫r(CM)dm = 0. Thus, we have  M(CM)=∫r(CM )×d P=∫r(CM)×g d m =(∫r(CM)d m)×g = 0 where we have used g as a constant by taking it out of the integral.  This important property of the CM means that in this context it is referred to as the centre of gravity,CG, of the body. For mechanical purposes, and in accordance with what we know about equivalent force systems explained at the end of Section 3.11, we can substitute the forces of each of the body parts by the net weight force applied to the CM or CG (see Figure 4.6). This weight force will have the same net force and the same (zero) net torque with respect to the CM as the weight force system d P=gdm, of each differential mass, dm, of the body. We must remember that this substitution can be made because the body is a rigid body. Fig. 4.6: A very simple system of equivalent forces is that of the weight applied to the CG =CM 4.3 Forces on bodies due to gravitating fluids: Archimedes principle In this section, we will study the forces that fluids exert on bodies and their application points. Fig. 4.7: Body immersed in a fluid Consider a body immersed in a fluid (see Figure 4.7). As with any contact, there will be normal forces and tangential forces. In a general situation, static or not, if the fluid is non-viscous, the different fluid layers slide; there is no friction either between the fluid layers or between the fluid and the body. Looking at Figure 4.8, although the surface of the body is not smooth, the non-viscosity of the fluid means that there is no frictional force between the body and the fluid. Therefore, tangential forces at the contact surface are zero and we take into account only the normal forces that act in a distributed manner on the contact surface (see Figures 4.8 and 4.9). If the situation we are dealing with is static and even though the fluid is viscous, the viscous frictional forces (proportional to the velocity) will be zero and the force will be normal on the surface. Fig. 4.8: Even if the surface of the body is rough, if the fluid is non-viscous it will not cause tangential forces 120 4.3 Forces on bodies due to gravitating fluids: Archimedes Principle Pressure concept We define a fluid’s pressure pas the normal force per unit area acting on the surface of a body immersed in the fluid (see Figure 4.9) p=dF dS (4.2) Fig. 4.9: Distribution of normal forces on a body caused by a gravitating fluid If we consider smaller and smaller cubes, the concept of pressure is independent of the existence of a material surface. It will depend on the characteristics of the fluid and the point where we want to calculate it (see Figure 4.10). In Figure 4.10, the pressure at the cube’s location is p, even if the cube is not there. Fig. 4.10: Pressure exists even if there is no submerged body Units of pressure In the International System of Units, pressure is measured in pascals (Pa) 1Pa = 1 N/m2 Other widely used units are: milibar (mbar): 1mbar = 1 hPa atmosphere (atm): 1atm = 1.013 ×105Pa millimetres of mercury (mmHg): 750 mmHg = 1 atm Fluid pressure in a uniform gravitational field Consider a differential volume of fluid in the form of a vertical cylinder of base S and height dz (see Figure 4.11). zis the depth, a coordinate that has the direction of g with origin at the fluid level. The force that this portion of fluid exerts on the base of the cylinder is equal to the weight of the fluid inside the cylinder. Fig. 4.11: Differential fluid volume in the form of a vertical cylinder dF =dm g =ρSdz g Therefore, dp =ρg dz (4.3) If the fluid is liquid, we can consider it as incompressible, i.e., its density ρwill be constant at all points within the liquid. Integrating the above equation, we obtain the difference of pressures between two points at different heights or depths: ∆p=ρg ∆z(4.4) 121 4 The statics of rigid bodies value of the modulus (as well as the direction). In the case of  F3, we know it as a function of the point of application r3. What is unknown in this approach is the system configuration. In the case of the figure (this is a system with two degrees of freedom), the unknowns are the angles q1and q2formed by the two bars with the vertical. The solution can be found by applying the static equations to the two bodies which would reveal all the reaction forces as new unwanted unknowns. Remembering that ideal reaction forces do not appear in the general equation of dynamics, we can use this in our approach to the case of statics. General equation of statics. The necessary and sufficient condition for a system of bodies with ideal constraints to be in equilibrium in a given position is that, for any virtual displacement from this position, the sum of the virtual work of the directly applied forces (those which are not reactions to the constraints) is zero: ∑ a=1  Fa·δra= 0 (4.17) Proof. Starting from the general equation of dynamics (3.64) and taking into account that at equilibrium ai= 0, (4.17) is verified. Note that instead of using the index i= 1...N, we use the index a. This is because suppressing the accelerations eliminates the need to account for what happens in all the mi. Since the forces will generally be applied to only a few particles, we can use a, which labels only these forces and the particles on which they act, ra. Expression (4.17) is commonly referred to as the virtual work principle (VWP). In this equation, external forces are included, e.g., the weights applied to the CM of each body, but not the ideal reaction forces, as discussed in Section 3.9. If we know the constraints, we can know the possible motions and therefore the degrees of freedom, L, of the system. The number of degrees of freedom will be equal to the number of parameters or independent variables. Thus, looking at Problem 4.6.2 on page 129 as an example, because it is a set of two rods joined at a fixed end, the system has two degrees of freedom and we can therefore give its position with the values anglesαand β, which are the parameters or variables that must be taken into account in the equation. If the system has Lindependent parameters qi, with i= 1...L, and we know the constraints in the form of ra=ra(q1, q2, ...), the expression for the virtual displacements will be δra=dra= L ∑ i=1 ∂ra ∂qi dqi Applying this to the VWP equation (4.17), we obtain a number of Lequilibrium 128 4.6 Virtual work principle equations that is equal to the number of unknown qi. ∑ a=1  Fa·∂ra ∂qi = 0 (4.18) The solutions qi=qeq i are the equilibrium positions of the system. Problem 4.6.1. If the system in the figure is in equilibrium, determine the relationship between the forces Fand Pas a function of the angle θof equilibrium. Figure for Problem 4.6.1 Solution The system is conservative because the forces  Pand  Fare constant and because the constraints have ideal reactions. Thus, the VWP is expressed as  P·drP+ F·drF= 0 (1) where drPand drFare the virtual displacements of the points of application of the forces  P= (0,−P)and  F= (−F, 0). The system has a single degree of freedom expressed by the angle parameter θ. Looking at the figure and taking into account the chosen axes, we see that rP= (Lcos θ, L sin θ)and rF= (2Lcos θ, 0). By differentiating, we get the possible displacements Solution to Problem 4.6.1 drP= (−Lsin θ, L cos θ)dθ drF= (−2Lsin θ, 0) dθ Substituting the forces and displacements in (1), we have (−P L cos θ+F2Lsin θ)dθ = 0and as dθ is any possible displacement, we are given −P L cos θ+F2Lsin θ= 0 from which we have tan θ=P 2F Problem 4.6.2. The two homogeneous bars have different lengths and masses. The joints Aand Band the pulley Dare frictionless. CD ≫ℓ1+ℓ2and therefore the rope CD is always horizontal. Find the angles αand βin the equilibrium configuration of the system using the virtual work principle. Figure for Problem 4.6.2 Solution Taking into account the body hanging from the rope, the tension of the rope is T= m3g. We will use the (x, y)reference frame such that the origin is at point A, the x-axis is horizontal and extends to the right, and the y-axis is vertical and extends downwards. The VWP expression will be  F1·dr1+ F2·dr2+ T3·dr3= 0 which, making explicit  F1= (0, m1g), F2= (0, m2g)and  T= (m3g, 0), results in m1dy1+m2dy2+m3dx3= 0 (1) 129 4 The statics of rigid bodies where we have already eliminated gfrom the expression. The displacements dr1,dr2 and dr3are not independent, but they depend on the two angles αand β: y1=ℓ1 2cos α⇒dy1=−ℓ1 2sin α dα y2=ℓ1cos α+ℓ2 2cos β⇒dy2=−ℓ1sin α dα −ℓ2 2sin β dβ x3=ℓ1sin α+ℓ2sin β⇒dx3=ℓ1cos α dα +ℓ2cos β dβ Substituting in (1), we obtain −m1 ℓ1 2sin α dα +m2(−ℓ1sin α dα −ℓ2 2sin β dβ) +m3(ℓ1cos α dα +ℓ2cos β dβ)=0 (2) If we now take into account that dα and dβ are independent possible displacements, we can consider (2) with dα = 0 and dβ = 0: −m1 1 2sin α−m2sin α+m3cos α= 0 (3) and also dα = 0 and dβ = 0 −m2 1 2sin β+m3cos β= 0 (4) From (3) and (4), we obtain tan α=m3 m1 2+m2 and tan β=2m3 m2 , a result already obtained in Problem 4.5.2 using the moment equilibrium equations.  4.7 Equilibrium and stability in conservative systems Equilibrium Let us now consider that we have a conservative system, that is, with the same conditions as in the previous section and, in addition, all the forces applied directly  Faare conservative, that is, each force has an associated potential energy and, therefore, ∑ a=1  Fa·dra=−dU where Uis the potential energy of the system. The VWP can now be written as dU = 0 (4.19) Making this explicit, the equilibrium equations can be written as a geometric condition in the potential function: The equilibrium position qeq ={qeq1, qeq2...qeqL}of a conservative system corresponds to the point at which the potential energy is extreme: ∂U ∂qiq=qeq = 0 i= 1...L (4.20) 130 4.7 Equilibrium and stability in conservative systems Stability If a system is in the equilibrium position qeq, with Ec= 0, its energy is E=Uext, where ext stands for extreme. If we provide kinetic energy Ecini, which is always positive, the system will move with mechanical energy E=Uext +Ecini =constant. In Figures 4.23 and 4.24, we can see the graphs of the potential energy (blue) and mechanical energy (pink) around an extreme for any system. It could be a roller coaster. In this case, the shape of the guide coincides with the shape of the potential function. Fig. 4.23: The carriage is located at the bottom of the roller coaster. If we push it a little, i.e., if we give it a little Ecini, the mechanical energy will be E=Umin +Ecini and it will be constant. The carriage will move around the minimum between the two positions where E=U, because, beyond that, it would have to be Ec<0to fulfil E= constant, which cannot be the case Depending on the type of equilibrium qeq position, we can say: qeq is a stable equilibrium position when U(qeq)is a minimum. In this case we have that U > Umin and since E=Umin +Ecini =constant, the system cannot move very far away from position qeq, it can go until Ec= 0. qeq is an unstable equilibrium position when U(qeq)is either a maximum or an inflection point. If it is a maximum, we have U < Umax and since E=Umax + Ecini =constant, the system moves far away from position qeq,Ec= 0 and increase. If it is an inflection point and goes towards zone U > Uinf, it behaves as in the case of the minimum, reaching point Ec= 0 and returning to zone U < Uinf and from here the same thing happens as in the case of the maximum: Ec= 0 and it increase. qeq is an indifferenct equilibrium position when in a finite environment of qeq, Uis constant. For a system with a single degree of freedom q,q=qeq, the equilibrium condition (4.20) reduces to dU dq q=qeq = 0 (4.21) Fig. 4.24: The carriage is located at either a maximum or an inflection point on the roller coaster. If we push it a little, i.e., if we give it a little Ecini, the mechanical energy will be E=Umax/inf +Ecini. If it is a maximum, Ecwill increase as it will be in a zone where U < Umax and E=constant must be fulfilled. If it is an inflection point and it goes towards a zone where U > Uinf, it will not be able to go beyond E=U, that is, Ec= 0 and it will return downwards, now increasing its Ec To study the stability of the position qeq, we analyse the second derivative at point q=qeq. If it happens that d2U dq2q=qeq >0(4.22) then q=qeq is a stable equilibrium point. If it happens that d2U dq2q=qeq <0(4.23) then q=qeq is an unstable equilibrium point. 131 4 The statics of rigid bodies If it happens that d2U dq2q=qeq = 0 (4.24) then the sign of the higher order derivatives must be examined: The equilibrium is stable if the order of the first non-zero derivative is even and if its sign is positive. The equilibrium is indifferent if all successive derivatives are zero. The equilibrium is unstable in all other cases. Problem 4.7.1. The two homogeneous bars have different lengths and masses. The joints Aand Band the pulley Dare frictionless. CD ≫ℓ1+ℓ2and therefore the rope CD is always horizontal. Find the angles αand βin the equilibrium system configuration, bearing in mind that the system is conservative. Figure for Problem 4.7.1 Solution The potential energy of the system as a function of the two degrees of freedom, angles αand β, is U=−m1gℓ1 2cos α−m2g(ℓ1cos α+ℓ2 2cos β)−m3g(ℓ1sin α+ℓ2sin β)(1) The two equilibrium equations are ∂U ∂α = 0 ⇒1 2m1sin α+m2sin α−m3cos α= 0 ∂U ∂β = 0 ⇒1 2m2sin β−m3cos β= 0 from which we obtain tan α=m3 m1 2+m2 and tan β=2m3 m2 , a result already obtained in Problems 4.5.2 and 4.6.2 by other methods.  Figure for Problem 4.7.2 Problem 4.7.2. In the system shown in the figure, the bars have negligible mass. Neither the joints Aand Bnor the roller C, of negligible mass, have friction. The spring, of negligible mass, has a recovery constant kand a natural length Lk= 0.6m. From joint Ahangs a block of mass m= 75 kg. Find the value of kso that the equilibrium angle θis θeq = 35◦.Other data: L= 1.2m, L0= 0.85 m Solution The potential energy of the system is U(θ) = mgL cos θ+1 2k(2L0sin θ−Lk)2 and the equilibrium condition is dU(θ) dθ =−mgL sin θ+k(2L0sin θ−Lk)2L0cos θ= 0 132 4.7 Equilibrium and stability in conservative systems The spring recovery constant kmust satisfy this condition for the angle θeq = 35◦. We obtain: k=mgL tan θeq 2L0(2L0sin θeq −Lk)= 969.54 N/m Problem 4.7.3. The homogeneous bar of length Land mass mcan slide without friction. The spring is relaxed when θ= 90◦. What value must the spring recovery constant khave so that θ= 45◦is an equilibrium position? Figure for Problem 4.7.3 Solution The potential energy of the system is U(θ) = mg L 2sin θ+1 2k(L−Lsin θ)2 and the equilibrium condition is dU(θ) dθ =mg L 2cos θ−kL2(1 −sin θ)cos θ= 0 The spring recovery constant kmust satisfy this condition for the angle θeq = 45◦ dU dθ (θeq) = 0 We obtain: k=mg 2L(1−1 √2) 133 5 Dynamics of a rigid body in a plane Introduction Fig. 5.1: Motion of a rigid body in a plane A body is in plane motion if the direction of the axis of rotation remains constant. Every particle of the body moves in a plane normal to the axis of rotation. Of all these planes, the one containing the body’s centre of mass is called the plane of motion (see Figure 5.1). According to this definition, the angular velocity ω and angular acceleration α in a body’s plane motion will at all times be parallel to each other and perpendicular to the plane of motion. The motion of a rigid body is in a plane because: Fig. 5.2: Rigid body rotating about an axis a) There are external constraints that force it to move in a plane. Example 1: A pulley rotating about a fixed axis (see Figure 5.2). Example 2: A flat sheet moving in a plane (see Figure 5.3). Fig. 5.3: Flat sheet moving in a plane b) The body has plane symmetry and the external forces are on its plane of symmetry as well as the initial velocities. Example 3: A flat sheet thrown into a vertical plane coincident with its own plane (see Figure 5.4). Example 4: A homogeneous sphere moving down an inclined plane starting from rest (see Figure 5.5). c) The forces external to the body are equivalent to a net force passing either through the CM and a zero net moment or in the direction of the axis of the body. The initial rotation of the body has the direction of this axis. Fig. 5.4: Flat sheet moving in a vertical plane Example 5: A disc thrown with an initial rotation about the axis of the disc in the presence of gravity. Its CM performs a parabola and the direction of rotation is constant (see Figure 5.6). Example 6: in the presence of gravity, a spinning top rotates along a vertical axis 5 Dynamics of a rigid body in a plane passing through the CM and through the vertex of contact with the ground. The net force and the net moment are both zero. Its CM will move with uniform motion and the direction of rotation will be constant (see Figure 5.1). Fig. 5.5: Sphere moving down an inclined plane In all these cases, and in accordance with what has been seen in Section 3.11, the possible displacements of a rigid rigid body’s particles will be a subset of dri=drC+dφ ×ri(C)(5.1) The constraint we have to impose is keeping the axis of rotation constant. If ˆuis the unit and constant vector in the direction of the axis of rotation, the possible rotational displacements will go from three degrees of freedom, represented by the vector character of dφ, to one degree of freedom, represented by dφ according to: Fig. 5.6: A disc maintains the direction of rotation while the CM traces a parabola dφ = ˆu dφ (5.2) In this course, we will deal exclusively with the motions of the body in a plane. Valid relations in general will be indicated as 3D and those that are valid only for plane motions will be indicated as 2D. It should be noted that this chapter will use the nomenclature of the particle system seen in Chapter 3although, for continuous bodies, the expressions in the form of summations are converted into integrals over the whole body (expression (3.1) in the Chapter 3introduction). 5.1 Translation equation in 3D According to equation (3.91) in Section 3.13, the equation of motion for the translation of a rigid body is  F=d P dt =maCM (5.3) where  Fis the resultant of the forces external to the body. The equation of the translational motion of the body is therefore that of the motion of its CM (Figure 5.7). Fig. 5.7: Translational dynamics of a body 5.2 Rotation equation for (2D) plane motion. Moment of inertia Since the motion is in a plane, the rotation of a rigid body is reduced to a single component in the direction of the rotational angular velocity vector ω, which is a fixed direction of space ˆuand is perpendicular to the plane of motion. In Section 3.13, the general equation of the rotational dynamics of a rigid body was derived: d L(C) dt = M(C)(5.4) 136 5.2 Rotation equation for (2D) plane motion with  L(C)= N ∑ i=1 ri(C)×mivi(5.5)  M(C)=∑ a ra(C)× Fa(5.6) where Ccan be either the centre of mass, CM, or a fixed point of the body, if it exists, and ra(C)is the position vector of the point of application of the external force to the body  Fawith respect to C. The rotation equation of the body in its plane of motion can be obtained by projecting the vector equation (5.4) onto the direction of its axis of rotation defined by its vector ˆu. This is so because the angular velocity can be written as ω =ωˆu, where ωis the component of the vector ω in the direction of ˆu.ωis the only degree of freedom of rotation that the body has. In terms of the possible displacements, we have dφ =dφˆu. Now, taking into account the general equation of dynamics for rotations, equation (3.94), we obtain, for rotations with axis dφ =dφˆu: d L(C) dt ·ˆu= M(C)·ˆu(5.7) Since ˆuis a constant vector, (5.7) can be expressed as follows: dL(C) dt =M(C)(5.8) where L(C)= L(C)·ˆu=(N ∑ i=1 ri(C)×mivi)·ˆu(5.9) M(C)= M(C)·ˆu=(∑ a ra(C)× Fa)·ˆu(5.10) are, respectively, the angular momentum and the momentum of the forces with respect to the fixed axis of rotation ˆupassing through C. According to (3.84), with ω =ωˆu, we have vi=vC+ωˆu×ri(C) and substituting in L(C)of (5.9) we have L(C)=(∑ i ri(C)×mi[vC+ωˆu×ri(C)])·ˆu and developing L(C)=(∑ i ri(C)×mivC)·ˆu+(∑ i miri(C)×[ ˆu×ri(C)])·ωˆu 137 5 Dynamics of a rigid body in a plane (3) Rotation of the wheels: M−FfR=Iα (4) Horizontal translation of the body: F= (M−m)a (5) Vertical translation of the body: N′−(M−m)g= 0 (6) Rotation of the body: M+F d cos φ−N′dsin φ= 0 Solution to Problem 5.2.3 In order to solve our problem, we have enough with (1), (3), (4) and (6) and the condition of rolling without sliding. If we solve for Fand Fffrom (1) and (4), we find Ff=Ma. We can also obtain this equation by considering the entire robot (even though it is not a rigid body). Substituting in (3) with α=a/Rand solving for a, we obtain: a=MR I+M R2= 1.25 m/s2 Substituting Fwith the previous result in (6) and solving for φ, we obtain: sin φ= 0.165048 ⇒φ= 0.165807 rad = 9.5◦ 5.3 Kinetic energy of rotation and translation. Energy conservation In this section, the aim is to derive a very simple expression for the kinetic energy of a rigid body in plane motion. We start from the expression for the kinetic energy of a system of Nparticles (3.27): Ec= N ∑ i=1 1 2miv2 i= N ∑ i=1 1 2mivi·vi(5.19) According to (3.84), vi=vC+ω ×ri(C), then (5.19) can be written as Ec=1 2mv2 C+vC·(ω × N ∑ i=1 miri(C))+ N ∑ i=1 1 2mi(ω ×ri(C))2(5.20) The first term is zero if Cis a fixed point, since vC= 0; and it is the translational kinetic energy if C=CM. The second term is null in both of the following situations: if Cis a fixed point, since vC= 0, and if C=CM, since N ∑ i=1 miri(CM)= 0. To deal with the last term, note that ω ×ri(C)=ω ×ri(|C), since the component of ri(C)parallel to ω does not contribute to the vector product with ω. Considering that ω andri(|C)are perpendicular and, therefore, |ω ×ri(|C)|=ωri(|C), we have (ω ×ri(|C))2= (ω ×ri(|C))·(ω ×ri(|C)) = ω2r2 i(|C)(5.21) Substituting (5.21) into (5.20) after cancelling the second term, we have Ec=1 2mv2 C+1 2 N ∑ i=1 mir2 i(|C)ω2=1 2mv2 C+1 2I(C)ω2(5.22) 144 5.3 Kinetic energy of rotation and translation. Energy conservation If C=CM, then (5.22) is Ec=1 2mv2 CM +1 2I(CM )ω2(5.23) and the kinetic energy of the body has two terms: the first one associated with the CM translation and the second one with the rotation around the CM. If Cis a fixed point, then (5.22) is Ec=1 2I(C)ω2(5.24) and the kinetic energy of the body has only one term associated with the rotation around C. The expressions (5.23,5.24) are also valid for 3D situations, although in this case I(CM)(or I(C), if Cis fixed) will not be constant, since the direction of the axis of rotation will not be fixed. For 2D situations, the direction is fixed and I(CM )(or I(C), if Cis fixed) is constant. Finally, it is not difficult to prove the corresponding conservation of mechanical energy theorem: Energy conservation. For a rigid body subjected to external forces that are either conservative with joint potential energy Uor they are constraints with ideal reactions, its mechanical energy E=Ec+Uis conserved throughout the motion. In the case of rigid body systems whose forces have the same characteristics, as in the previous case, the energy will be the sum of the energies of each body and it will also be conserved throughout the motion. Figure for Problem 5.3.1 Problem 5.3.1. A mass of m1= 1 kg hangs from the end of a rope of negligible weight passing through a frictionless pulley (see Figure). The rope is wrapped around a homogeneous cylinder of mass m2= 8 kg and radius R= 10 cm, which rotates without sliding in a horizontal plane. Find: a) The acceleration of the mass m1. b) The rope tension. c) The angular acceleration of the cylinder. Solution a) Bearing in mind that all the forces acting on the system are either conservative or they do not work in their displacement, the principle of conservation of mechanical energy Ecan be applied to it. This can be written as E=1 2m1v2 1+1 2m2v2 2+1 2Iω2−m1gy1 145 5 Dynamics of a rigid body in a plane where, being a solid cylinder, I=1 2m2R2. Solution to Problem 5.3.1 Since the cylinder rolls without sliding on the horizontal surface and the rope does not slide, v2=ωR ;v1= 2ωR ⇒v2=v1 2 and the time derivatives are α=a1 2R;a2=a1 2 Let us apply conservation of energy by imposing that its time derivative is zero: ˙ E= 0 = (m1+m2 4+I 4R2)v1a1−m1gv1 from which it follows a1=m1g m1+3m2 8 = 2.45 m/s2 b) To determine the tension in the rope, the equation of translation can be applied to m1 Solution to Problem 5.3.1 m1g−T=m1a1⇒T=m1g−m1a1= 7.35 N c) The angular acceleration of the cylinder is α=a1 2R= 12.25 rad/s2 Problem 5.3.2. At a given instant, a homogeneous cylinder of mass Mand radius Ris left to roll from rest on the top of an inclined plane at an angle θwith the horizontal. Knowing that it rolls without sliding, determine: Figure for Problem 5.3.2 a) The downward acceleration of the cylinder. b) The velocity at the end of the plane after rolling over it for a distance L. c) The minimum coefficient of friction µbetween the cylinder and the plane that is compatible with rolling without slipping. Solution to Problem 5.3.2 Solution a) Since the forces acting on the cylinder are either conservative (its weight) or do not work during the motion (the normal of the plane and the friction force in the rolling displacement without sliding), the principle of conservation of mechanical energy E can be applied. Taking the base of the inclined plane as the origin of the gravitational potential energy, mechanical energy can be expressed as follows E=1 2Mv2+1 2Iω2+Mgh From the figure, it can be seen that the height hof the centre of mass of the cylinder at any instant of its motion can be expressed as h= (L−x)sin θ+Rcos θ 146 5.3 Kinetic energy of rotation and translation. Energy conservation Therefore, the mechanical energy of the cylinder can be written as E=1 2Mv2+1 2Iω2−Mgx sin θ+constant and the nullity of its time derivative, using ˙x=v=ωR,¨x=a=αR and I= 1 2MR2, where Iis the moment of inertia of the cylinder with respect to the axis passing through its CM and perpendicular to the plane of the figure, gives us ˙ E=0=3M 2va −Mgv sin θ from which it follows a=2 3gsin θ b) The centre of mass of the cylinder describes a uniformly accelerated motion as it descends the inclined plane, with acceleration aand zero initial velocity. Therefore, the velocity it will reach at the end of the inclined plane, after having travelled a distance L, will be v=√2aL =√22 3gL sin θ=√4 3gL sin θ c) This requires finding the frictional force acting between the cylinder and the inclined plane. If we take into account Newton’s law for the translation in the normal direction and the rotation, we can write Solution to Problem 5.3.2 N−Mg cos θ= 0 FfR=Iα from which we obtain N=Mg cos θ Ff=Iα R Now, the coefficient of friction µis µ=Ff N if we substitute Ff,N,I=1 2MR2and α=a R, with the acceleration found in a), we obtain µ=Ia R2Mg cos θ=1 3tan θ Problem 5.3.3. A circular disk at rest with a radius of 0.5m and a moment of inertia of 4kg m2can rotate about a fixed axis that passes through its centre and has a rope wound around its periphery. The rope is stretched with a constant force of 2N for 10 s. Assuming no friction, calculate the length of rope that becomes unwound within this time. Figure for Problem 5.3.3 147 5 Dynamics of a rigid body in a plane Solution In addition to the force F, the disc is acted on by the weight and the reaction of the axis. Both have their point of application at the centre of the disc, which is a fixed point. Therefore, they do not work. The vertical force Fapplied at the periphery of the disc is constant; therefore, it is conservative. The mechanical energy of the disc, which consists of the kinetic and potential energies associated with F, will remain constant during its rotation. Taking into account that, at any given instant of the movement, the position vector of the point of application of the force Fis r = (R, y) and, in addition, that  F= (0, F ), the potential energy associated with Fis Solution to Problem 5.3.3 U=− F·r =−F y The mechanical energy Eof the disc will be E=1 2Iω2−F y and the time derivative, taking into account that ˙y=vand v=ωR, will be ˙ E=Iωα −F ωR from which, with ˙ E= 0, the angular acceleration of the disk is obtained α=F R I Finally, the acceleration awhich lowers the point of application of the force Fis also obtained: a=αR =F R2 I= 0.125 m/s2 Assuming that this point starts from rest and describes a uniformly accelerated rectilinear motion with acceleration a, the length Lof the unwound rope for 10 s is L=1 2at2= 6.25 m Problem 5.3.4. Two masses of 1and 2kg are connected by an inextensible rope with no mass, which passes, without slipping, through a 1.35 kg cylindrical pulley with a fixed axis. Calculate the tensions of the rope. Figure for Problem 5.3.4 Solution We take into account the following. First, the reaction forces of the pulley axis and its weight are forces whose application points are not displaced in the rotational motion. Second, m1gand m2gare conservative forces. Therefore, we can apply the principle of conservation of mechanical energy Eof the system. This energy can be written in the following way (see the considered reference frame in the figure): E=1 2m1v2−m1gy1+1 2m2v2−m2gy2+1 2Iω2 Under the problem conditions, ˙y1=−v,˙y2=v,v=ωR and a=αR, where ωand αare, respectively, the angular velocity and angular acceleration of the pulley, and 148 5.3 Kinetic energy of rotation and translation. Energy conservation I=1 2MR2is the moment of inertia of the pulley with respect to the axis of rotation, which passes through its centre and is perpendicular to the plane of the figure (Mis the mass of the pulley and Rits radius). Solution to Problem 5.3.4 m1= 1kg,m2= 2 kg E=1 2(m1+m2+I R2)v2−m1gy1−m2gy2 Taking into account that Eis conserved, its derivative will be zero: ˙ E= 0 = (m1+m2+I R2)va −(m2−m1)gv From this last expression, substituting also I, the acceleration acan be found: a=(m2−m1)g m1+m2+M 2 = 2.67 m/s2 Solution to Problem 5.3.4. Free body diagram of m1 Although equal in modulus, the accelerations of both bodies have opposite directions as vectors. In order to calculate the stresses in the rope, the dynamics of m1and m2 must be analysed separately. With respect to m1, taking into account that its motion is upward, its dynamics is reflected in the following equation: T1−m1g=m1a from which we obtain T1=m1(g+a) = 12.47 N As far as m2is concerned, taking into account that its motion is downward, its dynamics is reflected in the following equation: Solution to Problem 5.3.4. Free body diagram of m2 m2g−T2=m2a from which we obtain T2=m2(g−a) = 14.47 N  Problem 5.3.5. A wheel of radius 6cm has an axis of radius 2cm. The assembly has a moment of inertia of 0.004 kg m2and a mass of 3kg. The wheel rests on the ground and does not slip. Determine the direction of rotation of the wheel and its acceleration if, starting from rest, we pull horizontally on a rope wound around the axis with a force of 5N, as shown in the figure. Figure for Problem 5.3.5 Solution All the forces that act on the wheel are either conservative, or do not do any work in its rolling displacement without slipping. It is therefore a conservative system. Its mechanical energy Ecan be expressed in the form E=1 2Mv2+MgR +1 2Iω2− F·rA+C 149 5 Dynamics of a rigid body in a plane where rA= (xA, R)and, therefore, UF=− F·rA=−F xAis the potential energy associated with the horizontally applied force F= 2 N, with Cbeing a constant. Solution to Problem 5.3.5 Since Eis conserved, its time derivative will be zero: ˙ E= 0 = Mva +Iωα −F˙xA where ˙xAis the velocity of the points on the rope, as A,vA= ( ˙xA,0,0). If the rope does not slide on the axis, this is also the velocity at the point of contact between the wheel and the rope (see the figure for the solution). If we use vA=v +ω ×(rA−r), where, observing the figure, r = (x, R, 0) and v = (v, 0,0) are, respectively, the position vector and the velocity of the cylinder’s centre, and ω = (0,0,−ω), we find ˙xA=v−ωr If this expression is substituted into the one for mechanical energy conservation, together with v=ωR and a=αR, we find ˙ E= 0 = (M+I R2)va −F(1 −r R)v and finally we get a=F(1 −r R) M+I R2 = 0.81 m/s a > 0implies that α=a R>0. Taking into account that we have taken the positive direction to the right and zero initial v, then, according to the figure, the wheel moves clockwise to the right.  150 6 Small oscillations Introduction Oscillatory phenomena are very important. Everything that surrounds us tends to be in a position of stable equilibrium. The sea is in stable equilibrium and its waves are small oscillations around that equilibrium. A structure, such as a building, is in stable equilibrium and any perturbation that does not break it will also cause small oscillations around that equilibrium. The basic approximation of solid matter’s internal behaviour is rigidity, which is a configuration of stable equilibrium. The first approximation to the internal motion of solid matter is the small oscillations around its rigidity configuration. The study of small oscillations is therefore a first approach to studying of the dynamic behaviour of many systems that at first sight appear to be immovable but, for whatever reason, then begin to wobble. It is quite remarkable that we can carry out this study without going into detail about the causes that alter the state of equilibrium. 6.1 Small oscillations around a stable equilibrium position Let us consider a conservative system with one degree of freedom xof potential energy U(x), which has a stable equilibrium position x0. This means that U(x) fulfils dU dx (x0) = 0 ,d2U dx2(x0) = k > 0(6.1) Fig. 6.1: A small ball strung on a wire performs harmonic oscillations around the stable equilibrium position We redefine the reference frame so that the equilibrium position is x0= 0 and consider small deviations xaround the equilibrium position x0= 0. We perform a Taylor serie of U(x)and remain at the first significant order, i.e., at the first order where we get a non-zero result: U(x) = U(0) + dU dx (0) x+1 2 d2U dx2(0) x2+O[x3]≈1 2k x2+constant (6.2) P 4 Problems and questions Problem 4.2.1. The piece in the figure is made of two different wires. The horizontal one has a linear density which is double that of the semicircular part. What is the ratio between aand rif, when hanging the piece from point P, it remains in equilibrium, as shown in the figure? Figure for Problem 4.2.1 Solution: a=√2r Question 4.3.1. The piston in the figure with surface area Sis frictionless and in equilibrium under the action of the force exerted on it by the water. The gas inside the container has a pressure P. Knowing that the density of the water is ρ, and if Pais the atmospheric pressure, the weight of the piston is: Figure for Question 4.3.1 a) (H−h)ρgS b) (P−Pa)S+hρgS c) (P−Pa)S+ (H−h)ρgS d) HρgS e) hρgS Question 4.3.2. The rigid body in the figure is made up of two homogeneous cubes, Aand B, of density 900 kg/m3and welded together so that their centres of mass are on the same vertical. The volumes of Aand Bare 1m3and 8m3, respectively. The body is in equilibrium floating in water with cube Btotally submerged, as can be seen in the figure. The centre of buoyancy in this equilibrium situation is at a distance from the free surface of the water, which is: Figure for Question 4.3.2 a) 1.115 m Problems and questions b) 1.087 m c) 1.057 m d) 1.500 m e) 1.322 m Question 4.3.3. A cube of edge aand density 600 kg/m3floats in a liquid with one third of the edge submerged. If we add a cylinder of the same density but with a volume of 2m3to the cube, we know that the cube will be completely submerged with its top base flush with the surface of the liquid. What is the length of the edge a? Figure for Question 4.3.3 a) 3m b) 1.5m c) 1m d) 0.5m e) 2.5m Question 4.3.4. In the vessel shown in the figure, the gauge pressure of gas 2is P2= 51300 Pa. Knowing that the densities of the liquids are ρ1= 3 g/cm3and ρ2= 1 g/cm3, the pressure of gas 1will be (g= 9.81 m/s2): Figure for Question 4.3.4 a) 60129 Pa b) 72240 Pa c) −20340 Pa d) 54328 Pa e) 129354 Pa Question 4.3.5. The 1.5m wide body in the figure consists of two bodies of different densities, as shown in the figure. The centre of buoyancy is at a depth of: Figure for Question 4.3.5 a) 4.96 m b) 6.05 m 258 Problems and questions c) 4.74 m d) 5.22 m e) 5.47 m Question 4.3.6. The spherical body in the figure is made up of two solid hemispheres, of which one density is twice than of the other. We tie it to a rope and immerse it halfway in a liquid, as shown in the figure. In this situation, the distance between the body’s centre of buoyancy and centre of mass is: Figure for Question 4.3.6 a) R/7 b) R/2 c) R/6 d) R/4 e) R/8 Problem 4.3.4. We pour a 13 g drop of mercury into a beaker full of water 20 cm deep. Calculate how long it will take for the drop to reach the bottom of the beaker under the following conditions: a) If we do not take into account the Archimedean buoyancy. b) Taking into account the Archimedean buoyancy. Data: ρHg = 13 ×103kg m−3;ρwater = 103kg m−3 Solution: a) a=g,t= 0.202 s; b) a=g(1 −ρwater/ρHg);t= 0.210 s Problem 4.3.5. The sewer network of two towns is linked by a series of underground conduits. One day the atmospheric pressure in town Ais PA= 1.035 × 105Pa and that in town Bis PB= 1.03×105Pa. Recreate the diagram in the figure indicating in which well the height of the water will be higher and why. Calculate the difference in water height between the wells of town Aand town B.Figure for Problem 4.3.5 Solution: 0.051 m Problem 4.3.6. When a boat like the one in the figure is on the water, part of it is submerged. If we call the height of the water relative to the bottom of the boat h, will height hbe greater if the boat is in fresh water or in salt water? Figure for Problem 4.3.6 259 Problems and questions The boat in the figure has a total volume of 100 m3and a mass of 2000 kg. Calculate what percentage of the volume of the boat is in the sea. If the same boat is in river water, calculate what percentage of the volume will be under water. Data: ρmar = 1.025 ×103kg m−3,ρriu = 103kg m−3 Solution. The percentage of boat submerged in sea water is 1.95% and in river water it is 2.00% Problem 4.3.7. A container holds water and air, as shown in the figure. What is the gauge pressure, Pm, at points A,B,Cand D? Note: Pm=P−Pat. Figure for Problem 4.3.7 Solution.PmA = 11760 Pa, PmB =PmC =−2940 Pa, PmD =−17600 Pa Question 4.5.1. The cylinder in the figure (of mass M) is kept in equilibrium by the force Fof the rope (attached to the cylinder) and the friction between the cylinder and the wall (of coefficient µ). Indicate which of the following statements is correct. Figure for Question 4.5.1 a) In equilibrium it is verified that F < Mg. b) In equilibrium it is verified that F > Mg. c) There will only be equilibrium if µ > 1. d) If Fis much greater than Mg, the frictional force and the weight of the cylinder have the same direction. e) Equilibrium is impossible because there are three non-concurrent coplanar forces acting on the cylinder. Question 4.5.2. The homogeneous triangular block in the figure of weight Phas height hand base b. The coefficient of friction between the ground and the block is µ. Under conditions of sliding and imminent overturning, which of the answers is true? Figure for Question 4.5.2 a) µ=b 3h b) µ=2b 3h c) If Q= 0, the block can never be in equilibrium. d) If h > b, the block will always rotate before sliding. 260 Problems and questions e) The above statements are wrong. Question 4.5.3. We want to overturne a homogeneous block of weight W, without it sliding, by applying force F, as shown in the figure. The friction coefficient µ between the block and the horizontal floor must be: Figure for Question 4.5.3 a) µ < a 2b b) µ=a 2b c) µ > a 2b d) µ=F W e) None of the above. Question 4.5.4. Based on the figure, indicate which of these answers is correct: Figure for Question 4.5.4 a) The force of the water on the wall AB of the tank in the figure is horizontal and is applied at a height of h/3. b) The buoyant force of the water on a body floating in equilibrium can have an arbitrary direction, depending on the shape of the body. c) None of the other answers are correct. d) If the barrier CD in the figure rests without friction on the bottom at C, it can be kept in equilibrium by applying an appropriate force Fat a height of h/2. e) The force of the water on the bottom BC of the tank has a horizontal component to the left. Question 4.5.5. The wall of a reservoir has the trapezoidal cross-section shown in the figure. The height is hand the length in the transverse direction to the figure is ℓ. If gis the acceleration of gravity and ρis the density of water, the modulus of the force exerted by the water on the wall is Figure for Question 4.5.5 a) F=ρgh2ℓ 2sin θ b) F=ρgh2ℓ 3 c) F=ρgh2ℓ 2 d) F=ρgh2ℓ 2cos θ 261 Problems and questions e) It cannot be calculated from the data provided. Question 4.5.6. A ball of mass 0.3kg is attached to point Aon a wire with the shape of a semicircle of radius 50 cm and centre C. The wire is homogeneous and has a mass of 0.2kg. When the assembly is hung in the manner shown in the figure and once the equilibrium position is reached, the angle θwill be: Figure for Question 4.5.6 a) 14.4◦ b) 17.6◦ c) 40.5◦ d) 32.5◦ e) 30.3◦ Question 4.5.7. A homogeneous rigid sphere of weight Wand radius Ris suspended from a wall by a wire of negligible mass and length 5/3R, as shown in the figure. The junction Aof the sphere with the wire is on the vertical passing through the centre of the sphere. There is no friction where the wall and the sphere are in contact at point B. If Tis the tension of the wire and Bis the reaction in the wall, we can affirm that: Figure for Question 4.5.7 a) T=5 3W b) B=4 3W c) B=3 4W d) The wall reaction Bis zero, taking into account that there is no friction. e) The sphere cannot be in equilibrium in the position indicated in the statement. Question 4.5.8. A thin homogeneous ring of radius Rand weight Pis placed on the inclined plane in the figure, supported by a cable CB running parallel to the inclined plane, with Tbeing its tension. If µis the coefficient of friction between the plane and the ring, and the ring is in a condition of imminent motion, indicate which of the following answers is true: Figure for Question 4.5.8 a) µ= 0.29 b) µ= 0.45 262 Problems and questions c) µ= 0.84 d) µ=tan 60◦ e) µ=tan 30◦ Question 4.5.9. The maximum tension supported by the cable in the figure is 600 N. The bar, joined at point Ato the wall, has a weight of 800 N and a length of 8m. To keep it in equilibrium in a horizontal position, the maximum distance from its CM to the end Amust be: Figure for Question 4.5.9 a) 4m b) 3.51 m c) 4.63 m d) 3.25 m e) 3.86 m Question 4.5.10. Two horizontal forces of modulus F= 160 N are applied to a homogeneous cubic block of 3m length and weight P. Their lines of action are parallel and displaced 0.3m from the position of the centre of mass of the block, as shown in the figure. The coefficient of friction with the ground is 0.2. Indicate which of these statements is true: Figure for Question 4.5.10 a) In order for the block to be in equilibrium, the minimum value of Pmust be 192 N. b) In order for the block to be in equilibrium, the minimum value of Pmust be 64 N. c) In equilibrium, the normal component of the ground contact force passes through the centre of mass of the block. d) If P= 80 N, the block is in equilibrium and the normal component of the ground contact force passes through point A. e) If P= 80 N, the block is in equilibrium and the frictional force is 16 N. Question 4.5.11. The system in the figure is in equilibrium. If the bar AC has negligible mass relative to 100 kg, the tension of the horizontal wire BC (in N) is: Figure for Question 4.5.11 263 Problems and questions a) 735.75 b) 981.00 c) 490.50 d) 1226.25 e) None of the above. Question 4.5.12. If a beam is simply supported at ends Aand B, supporting two vertical loads of 1000 N, one applied at a midpoint and the other at support point A, it is true that: a) The beam cannot be in equilibrium because the net moment with respect to B is not zero. b) The beam is not in equilibrium because the net moment with respect to Ais not zero. c) The reaction at Ais three times that at B. d) The reaction at Ais twice that at B. e) The reaction at Bis 1000 N. Question 4.5.13. A frictionless hinged gate at point A, width a, separates two liquids of densities ρand ρ/2. What must the ratio be between heights H1and H2 of the liquids so that the gate does not move? Figure for Question 4.5.13 a) H2 H1= 2 b) H2 H1=√2 c) H2 H1=3 √2 d) H2 H1=1 2 e) H2 H1=1 3 √2 Question 4.5.14. A square hinged frictionless gate of mass mis held in equilibrium at an inclined angle θby a fluid of density ρ. In this situation, tan θis: Figure for Question 4.5.14 a) 3m 2ρL3 264 Problems and questions b) 3m ρL3 c) m ρL3 d) 2m 3ρL3 e) m 3ρL3 Question 4.5.15. The rope in the figure holds a gate of negligible weight, width L, and hinged without friction at the lower edge. What is the minimum tension of rope required to ensure that the gate can withstand the liquid of density ρin the tank that is open at the top, as shown in the figure? Figure for Question 4.5.15 a) T=3 4ρgL3 b) T=ρgL3 c) T=1 2ρgL3 d) T=√3 2ρgL3 e) T=1 3ρgL3 Question 4.5.16. The homogeneous ladder in the figure is supported by point A on the smooth wall and by point Bon the rough floor. It is known to slide when θ≤30◦. The coefficient of friction with the ground is: Figure for Question 4.5.16 a) µ= 0.766 b) µ= 0.866 c) µ= 0.666 d) µ= 0.566 e) µ= 0.966 Question 4.5.17. Given the system in the figure, determine the coefficient of friction µthat causes body 2to be in a condition of imminent downward motion if m1=m,m2= 2mand θ=π 3. The rope has negligible mass and the pulley shaft is frictionless. Figure for Question 4.5.17 a) 1.23 265 Problems and questions Solution: P F= 2 tan θ Problem 4.7.16. The roly-poly toy in the figure has a symmetry of revolution and a hemispherical base, such it always returns to its initial vertical position after striking it. What condition must the assembly’s centre of mass fulfil? Figure for Problem 4.7.16 Solution: The CM must be below the base of the hemisphere. Problem 4.7.17. The system shown in the figure consists of two springs of recovery constants k= 100 N/m and a bar of mass 30 kg and length 2m. The end Aof the bar can move without friction in the horizontal direction xand the other end Bcan move on the vertical wall, which is also frictionless. Both springs have the same natural length corresponding to the vertical position of the bar. Calculate: Figure for Problem 4.7.17 a) The values of xcorresponding to the two equilibrium positions presented by the system. b) The work of the spring on the right (WSR) when the bar moves, in the negative direction of the x-axis, between these two positions. c) The work of the gravitational force (WG) in this displacement. Solution: a) 0and 1.8598 m; b) 172.92 J; c) −186.03 J Problem 4.7.18. A small trolley of mass M(wheels of negligible mass) with an overload M1and M2is placed on an inclined plane and attached at one end to a spring (recovery constant kand natural length ℓn), while the other end is tied to a rope (always taut) passing through a pulley (of negligible mass and no friction), from which hangs a mass m. Figure for Problem 4.7.18 a) Write the expression for the mechanical energy of the system as a function of the y-coordinate (the energy may have an additive constant, independent of y). b) Find the equilibrium position y0. c) Imposing conservation of energy, find the equation of motion. 272 Problems and questions d) Write the equation of motion as a function of the coordinate x, for which the equilibrium position is x= 0. e) How does the trolley move if we release it from 5cm downwards from the equilibrium position? What if we release it from 5cm upwards? f) What is the maximum distance (with respect to the equilibrium position) from which we can release the trolley without the rope becoming slack? Data: M= 0.487 kg;,M1= 0.494 kg, M2= 0.289 kg;,L= 60 cm, H1= 8.13 cm, H2= 1.22 cm, m= 105 g, k= 3.63 N/m and ℓn= 9.5cm Solution: a) E=1 2(M+M1+M2+m) ˙y2−(M+M1+M2)gy sin α−mgy +1 2k(y−ℓn)2 sin α= 0.115 b) y0=((M+M1+M2)sin α+m)g+kℓn k. c) (M+M1+M2+m) ¨y= ((M+M1+M2)sin α+m)g−k(y−ℓn) d) (M+M1+M2+m) ¨x=−k x e) x(t) = 0.05 cos ωt;x(t) = −0.05 cos ωt;ω=√k M+M1+M2+m f) distance d=A < g ω2 Problem 4.7.19. U(x)is the potential energy function of a force as a function of the Cartesian x-coordinate. Figure for Problem 4.7.19 a) Draw the attached graph by schematically indicating the modulus, direction of the forces at different points on the curve. Using the graph, indicate: b) The point where the force will be maximum, zero, attractive and repulsive. c) The points of equilibrium and what these are points like. If we assume that this curve represents the interaction potential between two atoms: d) What energy will hold both atoms together? Problem 4.7.20. A homogeneous rod of mass mand length R√3rests without friction inside a spherical cavity of radius R. A mass m/2 is fixed at one end of the rod. What is the angle αof equilibrium? What kind of equilibrium is it? Figure for Problem 4.7.20 Solution: α= 30◦, stable 273 Problems and questions Problem 4.7.21. Determine the equilibrium positions of a homogeneous bar of length L= 1 m and weight 10 N, on which a spring of recovery constant k= 10 N/m is acting. ℓNis the natural length of the spring, and the contacts with the wall and floor are smooth. Figure for Problem 4.7.21 Solution: x= 0 and x= 0.866 m 274 P 5 Problems and questions Question 5.2.1. Which of the following statements is generally true about rigid body kinematics? a) If a rigid body is in pure translational motion, no point on the body can describe a curvilinear motion. b) Two points on the same body can have different angular velocities. c) The kinematic condition of rigidity applies only to plane motion. d) The relative velocity of two points on a moving body can never be zero. e) All of the above statements are false. Question 5.2.2. The figure shows a disc of radius Rrolling on a flat surface without sliding. The velocity of the centre Oof the disc is constant and is v0. Which of the following statements is false? Figure for Question 5.2.2 a) The angular velocity of the disc ω=v0 R. b) The velocity of point Aon the body is zero. c) The velocity of point Bis vB=√2v0. d) The acceleration of point Aon the body is zero. e) Although there may be friction at contact point A, it does no work. Question 5.2.3. We know that points Aand Bon a rigid body (see Figure) moving in the x, y plane have, at a given instant, the following positions and velocities (in m and m/s): Figure for Question 5.2.3 Problems and questions rA= (0,0) ; vA= (0,3) ; rB= (3,2) ; vB= (vB,0) With respect to the angular velocity ωof the body (in rad/s) and vB(in m/s), which of the following answers is true? a) vB= 3 and ω= 2 b) vB= 2 and ω= 2 c) vB= 2 and ω= 1 d) vB=√13 and ω= 1 e) Without prior knowledge of vB, nothing can be deduced. Question 5.2.4. A disc of mass mrolls down an inclined plane for a distance d without sliding. If the angle of the plane to the horizontal is αand the disc-plane friction coefficient is µ, the energy dissipated by the frictional force is given by: a) 0 b) dµmg cos α c) dµmg sin α d) dµmg tan α e) dµmg Question 5.2.5. An equilateral triangle of total mass mis made up of three rods of length band is arranged as shown in the figure. Its moment of inertia about the z-axis is Figure for Question 5.2.5 a) 3mb2 7 b) mb2 3 c) mb2 2 d) 3mb2 4 e) mb2 4 278 Problems and questions Question 5.2.6. A ring of radius rand mass mrolls down an inclined plane at an angle αwithout sliding. At a given instant, its centre of mass has a velocity v. At this instant, the modulus of the angular momentum of the ring with respect to point O(see Figure) is: Figure for Question 5.2.6 a) mvr b) 3mvr 2 c) 2mvr d) mvr 2 e) mvr sin α Question 5.2.7. A dancer who wants to increase her rotation speed has to bring her arms closer to her body because this: a) increases the angular momentum. b) reduces the effort. c) increases the momentum. d) reduces the moment of inertia. e) increases the physical resistance. Question 5.2.8. Four particles, m1=m3= 3 kg and m2=m4= 4 kg are at the vertices of a square, joined by rods of negligible mass. The length of the side of the square is L= 2 m. The moment of inertia with respect to an axis perpendicular to the plane of the particles and passing through m4is: Figure for Question 5.2.8 a) 88 kg m2 b) 40 kg m2 c) 56 kg m2 d) 20.5kg m2 e) 19.8kg m2 279 Problems and questions Question 5.2.9. A bar of length Lis held horizontally at a height Habove a table (see Figure). It is released and falls while maintaining its orientation. When it has descended a distance H, one of its ends touches one end of the table and the bar begins to rotate without friction as the end remains fixed to the table. The angular velocity at which the bar starts to rotate is: Figure for Question 5.2.9 a) √2gH 3L b) √gH L c) 3√2gH 2L d) √2gH L e) 3√gH 2L Question 5.2.10. A weight lifter lifts weight m2in order to change it, keeping m1touching the ground. When the barbell forms an angle of 30◦, it falls from his hands. What is the angular acceleration αof the assembly at this instant? Figure for Question 5.2.10. Data: The weights can be considered as point-like, the mass of the bar is negligible and L= 2 m. a) 2.5rad/s2 b) 9.8rad/s2 c) 4.2rad/s2 d) 19.2rad/s2 e) 4.9rad/s2 Question 5.2.11. Which of the following statements is true? a) If two bodies have the same dimensions and the same mass, their moment of inertia about the same axis is equal. b) Steiner’s theorem shows that the moment of inertia about an axis through the CM is less than it is about any other axis parallel to it. c) Like mass, the moment of inertia is a characteristic quantity of the body. d) The moment of inertia of a body about an axis depends on the angular velocity of the body. 280 Problems and questions e) Two bodies of different masses always have different moments of inertia about the same axis. Question 5.2.12. The homogeneous cubic block in the figure slides without friction at a velocity valong the horizontal floor until it encounters a chock and becomes hooked to it, although this does not prevent it from rotating freely. The angular velocity of the block just after colliding with the chock is: Figure for Question 5.2.12 a) v 2b b) 5v 7b c) 9v 7b d) v b e) 3v 4b Question 5.2.13. A disc of radius R= 25 cm rotates about an axis of symmetry, fixed and frictionless, with an angular velocity of ω0= 37 rad/s. The moment of inertia about the axis is I= 0.5kg m2. To stop it, we apply a force of F= 2 N to a brake pad of friction coefficient µ= 1.5. The time it takes for the disc to stop rotating is: Figure for Question 5.2.13 a) 34.7s b) 54.7s c) 14.7s d) 44.7s e) 24.7s Figure for Question 5.2.14 Question 5.2.14. A homogeneous bar of length h= 0.8m and mass m= 0.40 kg can rotate without friction about a fixed point A. A point object of mass mb= 0.05 kg and velocity vb= 10 m/s impacts it horizontally and becomes embedded in the upper end of the bar. The modulus of the velocity of the lower end of the bar just after impact will be: a) 1.73 m/s b) 3.73 m/s 281 Problems and questions e) With the initial data at t= 0,y1(0) = y10,y2(0) = y20 and ˙y1(0) = 0,˙y2(0) = 0, and taking into account that the accelerations are constant, we have y1=y10 +1 2¨y1t2 y2=y20 +1 2¨y2t2 from which we obtain y2−y1=1 2(¨y2−¨y1)t2 Substituting the accelerations and t= 0.8s: |y2−y1|= 1.24383 m Problem 5.3.8. Calculate the final CM velocity of a homogeneous sphere that is allowed to roll without sliding down an inclined plane to a drop h. Calculate it in the following ways. a) Applying the equations of motion of the rigid body, b) Applying the conservation of energy. Solution: vCM =√10 7gh Problem 5.3.9. A homogeneous disc of radius 20 cm and mass 5kg can rotate without friction in a vertical plane about its fixed axis. A rope of negligible mass is wound around it and a mass of 2kg hangs from it and is dropped. The rope does not slide. Calculate the angular acceleration of the disc and the acceleration at which the 2kg mass falls. Solution: α= 21.8rad/s2;a= 4.36 m/s2 Figure for Problem 5.3.10 Problem 5.3.10. With an initial angular velocity ω1, we drop (rolling without sliding) the homogeneous reel in the figure, of mass mand moment of inertia I(CM)=1 3mR2, down the guide (1) from a height y= 3R. At all times it rolls without sliding and there is no dissipative friction. What velocity of CM,v2, and 288 Problems and questions angular velocity ω2, will the reel have after coming into contact with the horizontal plane (2)? Note: Express the results as a function of ω1, R and gravity g. Solution: v2=√3gR +7 16 R2ω2 1;ω2=√3g R+7 16 ω2 1 Problem 5.3.11. A small ball of mass m= 100 g bounces elastically and horizontally against the lower end, B, of a bar of mass M= 6 kg and length L= 50 cm, which is hinged without friction at the upper end A. The bar is initially at rest. If the velocity of the ball just before the collision is 30 m/s and it leaves in a horizontal direction, determine: Figure for Problem 5.3.11 a) The modulus of the velocity of the ball and the angular velocity of the bar just after bouncing. b) The maximum angle the bar reaches, with respect to the vertical (the initial angle is zero). c) The modulus of the centre of mass velocity of the bar as it passes again through the initial position. Solution: a) 27.143 m/s, 5.714 rad/s; b) 63.56◦; c) 1.43 m/s Problem 5.3.12. A Maxwell wheel is a device as shown in the figure. The rope has negligible mass; the axis has a radius r; the moment of inertia of the wheel and axle is I; and its mass is m. If starting from rest it descends a height h, how much are the final speeds of rotation and translation? Figure for Problem 5.3.12 Solution: v=√2mgh m+I/r2and ω=√2mgh mr2+I Problem 5.3.13. Find the equation of motion of the homogeneous bar of mass m and length Lwhen it oscillates about an axis passing at a distance dfrom its centre of mass. Figure for Problem 5.3.13 Solution: ¨ θ+gd 1 12 L2+d2sin θ= 0 289 P 6 Problems and questions Question 6.2.1. A particle of mass mis dropped from a height honto the pan of a scale, where it remains attached (see Figure). The pan has mass mand the balance is equivalent to a spring of recovery constant k. The period of the subsequent oscillations will be Figure for Question 6.2.1 a) T= 2π√m k b) T=π√8m k c) T= 2π√mh kg d) T= 2π√h g+2m k e) T= 2π√m k+h g Question 6.2.2. The three systems in the figure consist of springs of negligible mass with the same recovery constant and attached to the same body of mass m. There is no friction in any contact. In all cases, the body of mass mcan perform a simple harmonic motion. If we designate T1,T2and T3as the periods of oscillation in each case, it will be fulfilled that: Figure for Question 6.2.2 a) T1> T2=T3 b) T1=T2< T3 c) T1=T2=T3 d) T1< T2=T3 e) It is necessary to know the value of the recovery constant to order the periods. Problems and questions Question 6.2.3. A mass hanging from a spring oscillates as shown in the graph, which represents the elongation as a function of time. At time t1, the mass has: Figure for Question 6.2.3 a) positive velocity ˙xand positive acceleration ¨x. b) positive velocity ˙xand negative acceleration ¨x. c) negative velocity ˙xand positive acceleration ¨x. d) negative velocity ˙xand negative acceleration ¨x. e) positive velocity ˙xand zero acceleration ¨x. Question 6.2.4. The mass M= 10 kg in the figure oscillates on a frictionless horizontal platform attached to two springs of recovery constants k1= 100 N/m and k2= 50 N/m. Which of the following statements is true? Figure for Question 6.2.4 a) The period of oscillation is 2.81 s. b) The period of oscillation is 3.44 s. c) The period of oscillation is 1.86 s. d) If we tilt clockwise the platform by rotating it 30◦with respect to the horizontal, the period of oscillation will be the same as when it is horizontal. e) None of the other four answers is correct. Question 6.2.5. When a simple pendulum oscillates: a) The longer the length, the longer the period. b) The shorter the length, the longer the period. c) The longer the length, the shorter the period. d) The period does not depend on the mass, unless the mass is small; thus, it increases with the mass. e) All of the above answers are false Problem 6.2.1. A homogeneous sphere of radius rand mass mrolls without sliding in a vertical plane on the inner surface of a fixed semi-cylinder of radius R > r. Find the mechanical energy and the period of the small oscillations as a function of R,rand g.Figure for Problem 6.2.1 292 Problems and questions Solution This is a conservative system with one degree of freedom: E=1 2mv2+1 2Iω2−mg (R−r)cos θ with I=2 5mr2and v=ωr = (R−r)˙ θ. We obtain E(θ, ˙ θ) = 7 10m(R−r)2˙ θ2−mg (R−r)cos θ The equation of motion can be extracted from ˙ E= 0 = 7 5m(R−r)2˙ θ¨ θ+mg (R−r)sin θ˙ θ which, for small oscillations, we can write as ¨ θ+5g 7 (R−r)θ= 0 Comparing this with the canonical expression of the SHM, ω0=√5g 7(R−r) it corresponds to a period: T0= 2π√7 (R−r) 5g Problem 6.2.2. A body has a simple harmonic motion with an amplitude of 5.2cm. When the elongation is 3.4cm, the velocity is 49.8cm/s. Do we have enough information to find the phase of the motion? How about for calculating the period? Find as much as you can from the information we have. Solution: T= 0.50 s; θ= 0.7121 rad (x=Asin θ) Problem 6.2.3. A2kg body is at rest on a smooth horizontal plane and is subjected to two horizontal springs of recovery constants k1= 100 N/m and k2= 200 N/m. The length of each of the two non-deformed springs is 40 cm. The free ends of the springs are stretched and attached to two fixed walls 120 cm apart. Determine the equilibrium position of the body. What is the frequency of oscillation about the equilibrium position? Solution: 66.7cm; f= 1.95 Hz 293 Problems and questions Problem 6.2.4. A particle moving in simple harmonic motion has a velocity of 16 cm/s and 12 cm/s when it passes, respectively, 3cm and 4cm from the centre of vibration. Calculate its amplitude and period. Solution: A= 5 cm; T= 1.57 s Problem 6.2.5. A point mass performs a simple harmonic motion. When the elongation is +10 cm moving towards the equilibrium point, its kinetic energy is 10−5J and its potential energy is also 10−5J. If the mass is 2g, find the amplitude, period and phase of the motion. Solution: A= 14.1cm; T= 6.28 s; θ= 3π/4 rad (x=Asin θ) Problem 6.2.6. A cylindrical buoy of height 4m, radius 2m and mass 40000 kg floats vertically in water. If it makes small vertical oscillations, find its period of oscillation. Solution: T= 3.58 s Problem 6.2.7. Two identical cylinders of mass Mand cross-section Sare arranged as shown in the figure and are partially immersed in water. Neglecting the mass of the pulley, the frictions and the inertia of the water, determine the period of oscillation of the system of weights when they are slightly separated from their equilibrium position. Figure for Problem 6.2.7 Solution: T= 2π√M Sgρ Problem 6.2.8. In a U-shaped tube of constant cross-section, with both branches in a vertical position, a liquid occupying length Lof the tube is introduced. Initially, it becomes unbalanced and consequently starts to oscillate around its equilibrium position. Assuming that the liquid is incompressible and that there is no friction, show that the liquid will oscillate with harmonic motion and determine the corresponding period. Solution: T= 2π√L 2g Problem 6.2.9. A100 g body hangs from a long spring. If we stretch it by lowering it 10 cm below its equilibrium position and then release it, it vibrates with a period of 2s. a) With what speed does it pass through its equilibrium position? b) What is its acceleration when it is 5cm above this position? 294 Problems and questions c) In the upward motion, how long does it take to move from a point 5cm below its equilibrium position to a point 5cm above it? d) How much will the spring shorten when the body is unhooked? Solution: a) direction ↑and modulus v= 31.4cm/s; b) direction ↓and modulus a= 49.3cm/s2; c) t= 0.33 s; d) ∆ℓ= 99.3cm Problem 6.2.10. A12 kg body hangs from a spring. If we stretch it until its length increases by 10 cm and release it, an oscillatory motion of period 1.45 s starts. Answer the following. a) When oscillating and moving downwards, how long does it take the body to move from a point 3cm above its equilibrium position to a point 6cm below its equilibrium position? b) What is the velocity of the body passing through this latter position? c) Once at rest, how much will the spring shorten if we unhook the 12 kg body? Solution: a) t= 0.22 s; b) v=−0.347 m/s; c) ∆ℓ= 0.52 m Problem 6.2.11. Two springs of recovery constants k1= 1200 N/m and k2= 600 N/m are joined in series. The free end of k1hangs from a point and a body of mass m= 10 kg hangs from the free end of k2. Find: a) The period of the free oscillations that the body can make. b) Calculate this while also assuming that the springs are connected in parallel. Solution: a) T= 0.99 s; b) T= 0.47 s Problem 6.2.12. An inextensible wire of negligible mass passes through the throat of a pulley whose mass is concentrated at its periphery. A mass Mhangs from one end and the other is attached to a vertical spring fixed to the ground (see Figure). If the mass of the pulley is m= 800 g, the mass of the hanging body is M= 200 g and the spring has a negligible mass and a recovery constant k= 16 N/m, calculate the period of the small oscillations in the system. Figure for Problem 6.2.12 Solution: 1.57 s Problem 6.2.13. By hanging a mass Mfrom a spring (which we assume to be massless and initially non-deformed), it elongates by 2.5m. In this situation, we push it upward at a velocity of v= 2 m/s. Find the trajectory of the mass M. Solution: y= 1.01 sin(1.98t)(SI) 295 Problems and questions Problem 6.2.14. A1kg body attached to the end of a spring starts its motion when at position x= 1 m with an initial velocity of v= 2 m/s. If the period (T) of the movement is πs, calculate the maximum elongation of the spring (A) and find its trajectory. Calculate the maximum velocity and acceleration and the positions at which they occur. Solution: (SI) A=√2;x=√2sin(2t+π/4);vmax =±2√2 (x= 0);amax = ±4√2;(x=±√2) Problem 6.2.15. A simple model currently used to describe the proteins present in our body consists simply of balls representing the amino acids (of approximately 10−24 kg), held together by springs of recovery constant k= 5 ×10−20 N/m. Calculate the vibration frequency of the amino acids if we assume that this model is valid. Figure for Problem 6.2.15 Solution: f= 71 Hz Problem 6.2.16. We build a pendulum from two identical uniform rods aand b, each of length Land mass m, joined at right angles in the form of a T, with the centre of rod ajoined to the end of b. We hang the pendulum from the free end of rod b, swinging it in a vertical plane. a) Calculate the moment of inertia with respect to the axis of rotation. b) Find the expressions for the kinetic and potential energy as a function of the angle of the axis of the pendulum with respect to the vertical. c) Derive the equation of motion. d) Find the period for the small oscillations. Solution: a) I=17 12 mL2; b) Ec=17 24 mL2˙ θ2,U=3mgL 2(1 −cos θ); c) ¨ θ+ 18 17 g Lsin θ= 0; d) T= 2π√17L 18g Problem 6.2.17. The motion of a simple harmonic oscillator is described by the equation x(t) = 4 sin (0.2t+ 0.3), with xin m and tin s. a) Calculate the amplitude, period, frequency and initial phase of the motion. b) Determine the velocity and acceleration as a function of time, as well as the initial conditions. c) What is the phase difference between the elongation and the velocity? Between the elongation and the acceleration? d) Calculate the position, velocity and acceleration at t= 5 s. 296 Problems and questions Solution: a) A= 4 m; φ0= 0.3rad; f0= 0.032 s−1; b) v(t) = 0.8cos (0.2t+ 0.3); a(t) = −0.16 sin (0.2t+ 0.3) (SI units); x(0) = 1.18 m; v(0) = 0.76 m / s; c) π/ 2;π; d) x(5) = 3.85 m ;v(5) = 0.21 m / s;a(5) = −0.154 m / s2 Problem 6.2.18. A1kg particle performs a simple harmonic motion with an amplitude of 0.5m. At instant t= 0, it passes through the equilibrium position with a velocity of ˙x(0) = +2 m/s. a) Calculate the frequency and period. b) Determine the elongation and velocity as a function of time. c) Calculate the force and the kinetic and potential energies when the particle is at 0.2m from its equilibrium position. Solution: a )f0= 0.637 Hz; T0= 1.57 s; b) x= 0.5sin (4t);v= 2 cos (4t); c) F= 3.2N; Ec= 1.68 J; U= 0.32 J Problem 6.2.19. In the mechanism shown in the figure, the spring recovery constant is k= 100 N/m; the mass of the homogeneous cylindrical pulley is M= 4kg; and the radius R= 30 cm. The mass of the block is m= 1 kg. The rope does not slide at any time and there is no friction on the axis. Find the equation of motion and the period. Figure for Problem 6.2.19 Solution: ¨x+ 33.33 x= 0 (SI units); 1.088 s Problem 6.2.20. The pulley, the springs (of recovery constants 3kand k) and the inextensible rope in the figure all have a negligible mass. The bar, of mass m, moves slightly and vertically from the horizontal equilibrium position. What is the period of the oscillations? Figure for Problem 6.2.20 Solution: 2π√m 3k Question 6.3.1. The vibration amplitude of a damped oscillator decreases from 75 mm to 70 mm in one cycle. The oscillating mass is 1.2kg and the time it takes to go from the centre to the end of the oscillation is 0.5s. The damping constant of the viscous friction force is: a) 82.8×10−3N s/m b) 165.6×10−3N s/m c) 34.5×10−3N s/m 297 Problems and questions b) If Ω<1rad/s, the system does not oscillate. c) The mechanical impedance of the system is 2N s/m when we have amplitude resonance. d) The elongation is out of phase by π 3rad with respect to the force applied by the external agent when we have amplitude resonance. e) The elongation is in phase with the force applied by the external agent when we have amplitude resonance. Problem 6.4.1. A1kg mass is attached to a structure by an elastic spring and subjected to a force F=F0sin(Ωt), with F0= 2.5N and Ωvariable. Using the observed relationship indicated in the table between Ωand the amplitude AP, estimate the spring recovery constant kand the damping constant b. Table for Problem 6.4.1 Ω(s−1)14 20 26 32 36 40 Ap(cm) 0.31 0.42 0.78 1.10 0.85 0.41 Solution: k= 1089 N m−1;b= 6.9N s m−1 Problem 6.4.2. A mass of 3g is subjected to a restoring force of 1N/m and a damping force of 0.1N s/m. If a force F= 0.1cos(10πt)(SI units) is applied, calculate the amplitude and the phase difference between the force and the velocity. Also calculate the mechanical impedance and find the velocity resonance frequency. Solution: AP= 2.70 cm; θ= 32.0◦;Z= 0.12 N s m−1;fRV = 2.91 Hz Problem 6.4.3. A10 g particle is subjected to the action of a restoring force of 0.05 N/m and a damping force of 0.03 Ns/m. If a periodic force of 50 rad/s of pulsation and 0.001 N of amplitude acts on this particle, find, in steady state: a) The mechanical impedance. b) The maximum velocity. c) The velocity resonance frequency. d) The velocity amplitude in this case. Solution: a) Z= 0.5N s m−1; b) vmax = 0.20 cm s−1; c) fRV = 0.36 Hz; d) vmax = 3.33 cm s−1 304 Problems and questions Problem 6.4.4. A body of mass 5kg is hanging from the end of a spring. It is separated from its equilibrium position and is observed to perform a vertical SHM, which takes 0.4s to go from one end of the oscillation to the other. The mechanical energy of the particle is 100 J. a) Calculate the time taken by the body to go in a downward motion from the position 0.5m above the centre of oscillation to 0.2m below the centre. If we oscillate the above system in a viscous medium, the frequency becomes 90% of what it was in the SHM. b) Calculate the amplitude reduction factor over a 0.5s interval. We apply to the system a harmonic force of the same frequency as that of the damped oscillation. It is observed that, in steady state, the system reaches a maximum velocity of 1.5m s−1. c) What is the amplitude of the applied force? Solution: a) t= 0.117 s; b) A A0= 0.181; c) F0= 52.8N Problem 6.4.5. A mass of 3kg undergoes a simple harmonic motion in the direction of the x-axis with amplitude 10 cm and a period of 3s. For t= 2.5s, the mass passes through the equilibrium position, x= 0, with positive velocity. a) Determine the elongation and velocity at t= 0. At an instant when the mass passes through the equilibrium position while moving in the positive direction of the x-axis, a damping device is set in motion which provides a viscous friction force proportional to the velocity and with a coefficient of 10 N s/m. b) Taking as a new time origin (t= 0) the instant when the damper is started, write the expression of the trajectory and determine all the parameters involved. c) Calculate the kinetic energy of the system at the end of the first oscillation cycle. Finally, a periodic force F= 5 cos(Ωt)(SI units) is applied to the mass, also in the direction of the x-axis. d) What would the amplitude of the forced oscillations be if the system were in velocity resonance? Solution: a) x0= 0.087 m, v0= 0.105 m/s b) x= 0.165 e−1.67tcos (1.27t+ 3π/2); c) Ec= 4.4×10−9J; d) A= 0.239 m 305 Problems and questions Problem 6.4.6. A block of mass m= 3 kg on a frictionless horizontal plane is attached to two springs of recovery constants k1= 7 N/m and k2, and to a damper of constant b= 10 N s/m. A harmonic force F= 4 sin(2t), in SI units, is applied to the block. Figure for Problem 6.4.6 a) Determine the value of k2so that in the oscillatory motion of the block the phase difference between the harmonic force and the elongation is π 2rad. We eliminate the spring of recovery constant k2. b) Write the equation of the elongation as a function of time and determine all the involved parameters. c) What is the maximum speed reached by the block in its oscillation? When the block is at the right end of the oscillation we eliminate the harmonic force. d) What kind of damped motion will the block make? e) Determine the equation that gives the position of the block as a function of time and calculate all the involved parameters. Solution: a) k2= 5 N/m; b) x= 0.194 sin (2t−1.82); c) vmax = 0.388 m s−1; d) the system is overdamped; e) x= 0.340 e−t−0.146 e−2.33t Problem 6.4.7. AU-shaped tube with a cross-section of 1.8cm in diameter contains 120 g of ethanol (ρ= 787.4kg/m3). In one of the branches, a small displacement of the liquid is provoked. a) If there is no friction between the liquid and the walls of the tube, find the natural pulsation of its oscillations. When these oscillations are observed, it is found that their amplitude decreases by 3.5% in each period. b) What is the period of the damped oscillations? c) With what frequency should the liquid in the tube be blown so that the oscillating movement has the maximum amplitude? Solution: a) 5.72 rad s−1; b) 1.1s; c) 0.91 Hz Problem 6.4.8. A mass m1has been added to a disc Dof mass mD, which is attached to a spring (of recovery constant kand equivalent mass mu) and, through a rope and a pulley P2, to a second mass m2. The disk causes an aerodynamic friction of coefficient b. The system can be forced to oscillate by an engine that 306 Problems and questions moves the spring harmonically with a pulsation Ω. The pulleys P1and P2have negligible mass. Figure for Problem 6.4.8 With the engine stopped, we move the mass m2vertically and release it: a) Write the equation of motion using the y-coordinate. We start the engine so that ym=ymo +Rsin(Ωt+θ0). b) Find the new equation of motion. c) Write the equation of motion in canonical form (call xthe new coordinate) and identify all the parameters involved. d) Plot the amplitude of the forced stationary oscillations as a function of engine pulsation. Data: m=m1+mD+m2+mu= 0.80 kg; k= 9.7N/m; R= 4 cm; b= 1.5N s/m Solution (in SI units): a) m¨y+b˙y+k(y−ym−ℓeq) = 0, with yeq −ym−ℓeq = 0 b) m¨y+b˙y+k(y−ymo −ℓeq)−kR sin(Ωt+θ0) = 0 c) ¨x+ 2 (b 2m) | {z } γ ˙x+(k m) |{z} ω2 0 x=(kR m) | {z } B sin(Ωt+θ0) d) Ap(Ω) = 0.48 √3,52Ω2+(Ω2−12.13)2 Problem 6.4.9. A block of 2kg mass moves in a horizontal plane in the direction of the x-axis, under the action of a restoring force of recovery constant 18 N/m and in the presence of a viscous friction force of coefficient 16 N s/m. By means of an engine, a harmonic force F=F0sin Ωtis applied to the block, also in the xdirection. Under stationary conditions, the expression for the trajectory of the block is x= 0.01 sin(Ωt−π/2) (xin m and tin s). a) Calculate the pulsation Ω, the mechanical impedance and the amplitude of the force. b) Express the velocity of the block as a function of time and calculate the phase difference between the velocity and the force. Now, the engine is switched off: c) What kind of motion does the block make? Give reasons and numerical 307 Problems and questions justification for the answer. Solution: a) 3rad/s; 16 kg/s; 0.48 N; b) 0.03 sin 3t;0 Problem 6.4.10. Consider the system in the figure. When in equilibrium, ℓ=ℓeq. The pulley rotates about a fixed axis passing through Owith moment of inertia I. It is affected by viscous friction with a moment about the shaft Mβ(O)=−β˙ ϕ, where ˙ ϕis the angular velocity and βis a constant. Figure for Problem 6.4.10 We assume that, at all times, the rope remains taut and does not slip. With the engine stopped, L=ct, we move the mass mvertically and release it: a) Write the equation of motion using the y-coordinate. We start the engine in such a way that Lis no longer constant and can be expressed as L=COeng +rsin(Ωt+θ0)due to the fact that COeng ≫r(Cis the rope-pulley contact point, which in this approximation is at rest). b) Find the new equation of motion. c) Write the equation of motion in canonical form (call xthe new coordinate) and identify all the parameters involved. d) Plot the amplitude of the forced stationary oscillations as a function of engine pulsation Ω. Data: R= 14 mm;I= 6.0×10−5kg m2;r= 20 mm m= 107 g;k= 3.50 N/m;β= 1.81 ×10−4N m s Solution (in SI units): a) (I R2+m)¨y+β R2˙y+k(y+ℓeq −L) = 0 b) (I R2+m)¨y+β R2˙y+k(y+ℓeq −COeng)−kr sin(Ωt+θ0) = 0 c) ¨ x+ 2 (β 2R2(I R2+m)) | {z } γ ˙ x+(k I R2+m) | {z } ω2 0 x=(kr I R2+m) | {z } B sin(Ωt+θ0); x=y+ℓeq −COeng d) Ap(Ω) = 0.169 √5,00Ω2+(Ω2−8.47)2 Problem 6.4.11. In a liquid of density ρ= 103kg/m3, a bottom-ballasted buoy of total mass m= 20 kg is held in equilibrium at y= 0, where yis a vertical coordinate. To avoid excessive oscillations, it is designed to have a damping of 308 Problems and questions −80 ˙y(SI units). The cylindrical part of the buoy, of radius R= 0.25 m, always touches the water level. Determine: Figure for Problem 6.4.11 a) The differential equation of motion for the y-coordinate. b) The period of the oscillations. c) The position as a function of time, y(t), if we hit it when it is in equilibrium so that the initial velocity is ˙y0=−10 m/s. Due to a smooth and persistent swell, it receives a vertical excitation force F= 100 sin(10t)(SI units). Determine: d) The position as a function of time, y(t), for the stationary motion. Solution: a) ¨y+4 ˙y+96.3y= 0; b) T= 0.654 s; c) y(t) = 1.041 e−2tsin(9.60 t+ π); d) y(t) = 0.1244 sin(10t−1.66) Problem 6.4.12. A2kg particle attached to a spring of recovery constant of 18 N/m and a damper with damping parameter γ= 3 s−1moves in a rectilinear motion in the x-direction, where x= 0 is its equilibrium point. a) Write the equation of motion. At the initial instant, the particle is at position x= 0.3m, with velocity ˙x= −0.2m/s. b) Write the expression of its trajectory We apply a harmonic force in the x-direction to the particle so that it oscillates with a constant amplitude of 0.1m and takes 0.25 s to go from one end of the oscillation to the other. c) What must the value be for the amplitude of this force? d) What is the maximum speed of the particle under these conditions? Solution: a) ¨x+ 6 ˙x+ 9x= 0; b) x(t) = (0.3 + 0.7t)e−3t; c) F0= 33.383 N; d) 1.257 m/s 309 P 7 Problems and questions Problem 7.2.3. The photograph of a wave pulse on a string at the instant t= 0 indicates that the shape of the pulse is (SI units) y(x, 0) = 18 ×10−3 8 + x2 If the elongation of the string at the point x= 5.20 m and at the instant t= 0.40 s is 2.25 mm, at what speed does this wave propagate? If the linear density of the string is 0.280 kg/m, to what tension is it subjected? Solution: 13.0m/s, 47.3N Problem 7.2.4. If A,B,C,k,ωand ϕare constants, which of the following functions represent waves? What is the propagation speed in the positive cases? a) y(x, t) = Acos2(kx −ωt +ϕ)b) y(x, t) = Acos kx cos ωt c) y(x, t) = A (B x+C t)2+1 d) y(x, t) = A (B x2−C t2)+1 Solution: The functions a,band care waves. The velocities are: for aand b, v=ω k; for c,v=C B. Problem 7.2.5. Answer the following related questions: a) Write a harmonic wave propagating towards decreasing x,8.0mm in amplitude, 230 Hz in frequency and 145 m/s in speed. b) What is the distance between two points which, at a given instant, are out of phase by π/3 rad? c) What is the phase difference of the elongation at the same point between two instants of time separated by 1.5×10−3s? Solution: a) y= 8.0×10−3sin(9.966x+ 1445t); b) 0.1051 m; c) 2.168 rad Problems and questions Problem 7.2.6. A harmonic wave passes through two points on a string, x1and x2, which are separated by 1.20 m and vibrate, respectively (in SI units), y1= 0.020 sin π(3t−1 2), y2= 0.020 sin π(3t−1) Calculate the velocity at which the wave propagates, the wavelength and the wave function. Solution: 7.20 m/s, 4.80 m; y(x, t) = 0.020 sin 3π(t−x/7.20 −1/6) Problem 7.2.7. Consider the wave y(x, t)=4cos [2π(t 6+x 240 )], where yand xare expressed in cm and tin s. Calculate: a) The phase difference, at a given instant, between two particles in the medium separated by 210 cm. b) The phase difference between two positions and the same instant knowing that the particle in the medium takes 1.0s to go from one to the other of these positions. c) If, at a given instant, a given particle has an elongation of 3.0cm, what will its elongation be 2.0s later? Solution: a) 7/4πrad; b) π/3 rad; c) −3.79 cm Problem 7.2.8. A transverse harmonic wave propagates along an indefinite string with a speed of 4.0m/s. At all times, the minimum distance between two points in phase is 20 cm. It is known that, at the origin x= 0 and at the initial instant t= 0, the elongation is maximum with a value of 20 cm. Find: a) The amplitude, wavelength and period. b) The elongation and velocity of a point x= 0.25 m, after t= 5/16 s. c) The minimum distance between two points with a phase difference of π/3 rad. d) The phase difference between two points separated by ∆x= 5 cm, Solution: a) 20 cm; 0.20 m, 0.05 s; b) 20 cm; 0m/s; c) 3.33 cm; d) π/2 rad Problem 7.2.9. A harmonic plane wave travels with a propagation speed of 32 m/s. The amplitude is 2.3cm and the frequency is 60 Hz. Assuming that at the origin x= 0 and at the initial instant t= 0 the elongation is maximum, what are the values of the elongation, velocity and acceleration at a point x= 15.3m, after t= 2.60 s have elapsed. 312 Problems and questions Solution: −0.88 cm; −8.01 m/s; 1251 m/s2 Problem 7.3.2. A long string, of linear density 0.10 kg/m and subjected to 25 N of tension, is made to vibrate at a frequency of 20 Hz and causes the propagation of a harmonic wave of amplitude 1cm. a) Calculate the speed at which the wave propagates and its wavelength. b) Write the wave function of this wave, knowing that at the initial instant t= 0 the elongation of the string is 0.50 cm at the origin point x= 0. c) At t= 4 s, what is the elongation, velocity and transverse acceleration of the point on the string at x= 90 cm? Solution: a) 15.81 m/s; 0.7905 m; b) y= 1×10−2sin(ωt−kx+π/6),125.66 rad/s; 7.948 rad/m; c) −0.3383 cm; 118.3cm/s; 53.42 m/s2 Problem 7.3.3. Using dimensional analysis, find the expressions giving the propagation velocities vof the following two types of waves: a) Waves on a very long string, of linear density µand subjected to a tension F. b) Waves on the surface of a lake or sea caused by the weight of the water when the vertical amplitude of the waves is much smaller than the depth hof the water. The quantities on which these surface waves may depend are the water density, ρ, the acceleration of gravity gand the depth, h. Solution: a) v=k√F µ; b) v=k√gh, where kis a dimensionless constant (a detailed physical study shows it to be 1). Problem 7.3.4. A long, heavy chain of length Land mass mis hung from the ceiling. One end is struck by a hand, causing a wave pulse that rises upwards, reaches the ceiling, is reflected and returns back to the original end. Calculate how long it will take for the pulse to go up and down. Solution: ∆t= 4√L g Problem 7.3.5. The speed of sound waves in air is given by (7.29): v=√γRT M where γ= 1.40,R= 8.314 J/molK and M, the molar mass of air, is 0.0290 kg/mol. Making the necessary approximations, find a simple expression for the speed of 313 Problems and questions the impedances for air aluminium and glycerine are, respectively, 418 rayl and 13.7×106rayl, 2.46 ×106rayl. Solution: a) 1.41 ×10−5%; b) 48.2% Problem 8.3.10. Consider two cables of cross-sections S1and S2and linear densities µ1and µ2, welded at a point and subjected to a tension F. The expressions for the transmitted and reflected amplitudes and powers of the transverse waves are governed by the same expressions as in the longitudinal case, with the corresponding impedance: Zi=ρivi=µi Si√F µi=√F√µi Si. Apply this knowledge to the following case: Two wires, one of copper and the other of steel, each with a radius of 1mm, are joined together to form a longer cable. The tension of the assembly is 50 N. A 10 Hz wave propagates from the copper to the steel with an amplitude of 2.0mm. a) Calculate the wavelength of the wave in each wire. b) Calculate the transmission and reflection coefficients. Data: Density of copper: ρCu = 8900 kg/m3; density of steel: ρAc = 7800 kg/m3 Solution: a) 4.23 m; 4.52 m; b) 1.033;0.033 Problem 8.4.3. A harmonic wave of 1cm amplitude is superimposed on another wave of 2cm amplitude, out of phase with respect to the first wave by −π/3 rad. What are the amplitude and phase difference of the resulting wave with respect to the first? Solution: 2.646 cm, −0.7137 rad Problem 8.4.4. At the points S1= (0,3) and S2= (4,0) (SI units), there are two coherent sources of 100 Hz spherical sound waves. The amplitudes at a distance of 1m from the sources are 1×10−3Pa and 3×10−3Pa, respectively. The sound propagates at 340 m/s. a) If the sources emit in phase, what is the amplitude of the sound pressure at the origin (0,0)? b) By how much would the second source have to advance relative to the first one in order for there to be constructive interference at the origin (0,0)? Solution: a) 0.190 ×10−3Pa; b) 1.85 rad 320 Problems and questions Problem 8.4.5. As shown in the figure, the sound of a 440 Hz tuning fork enters a tube at A, bifurcates into two waves that follow the two paths ABD and ACD, which then meet again at point D, where they interfere with each other. The length of the ABD path is 250 cm and, although the initial maximum length of the ACD path is also 250 cm, it slowly decreases to 75 cm while the tuning fork continues to sound. If the speed of sound is 340 m/s, through how many minima and maxima does the resulting sound intensity pass during the reduction in the length of ACD? Figure for Problem 8.4.5 Solution: Maxima, 2: ACD = 173 cm; 95 cm; minima, 2: ACD = 211 cm; 134 cm. Problem 8.4.6. Two loudspeakers aligned with a person coherently emit plane sound waves of the same frequency: 440 Hz. The speed of sound is 340 m/s. a) If they emit in phase, how far apart do the speakers have to be from each other for the person to hear nothing? b) If they are still emitting coherently but now the phase of the closest speaker is advanced π/3 with respect to the other one, by how much do the above distances change? Solution: a) 0.39 m, 1.16 m; 1.93 m...; b) All are reduced by 0.13 cm Problem 8.4.7. In a laboratory where the speed of sound is 340 m/s, two loudspeakers, Aand B, emit plane sound waves of 791 Hz while facing each other at a distance of 11 m, as shown in the figure. A sound level meter Cis located on the line between them, 5m from B. Speaker Aprovides an intensity of 0.75 W/m2 and speaker Bof 0.25 W/m2. a) If the two loudspeakers emit without coherence, what is the value of the intensity level β0recorded by sound level meter C? 321 Problems and questions b) If the loudspeakers emit coherently and in phase, what will the phase difference be between the sound from Aand Bwhen it reaches sound level meter C? Figure for Problem 8.4.7 c) What minimum distance δ1should we move the sound level meter Ctowards Bin order for the intensity to experience a maximum? And what δ2to experience a minimum? d) What level of intensity will the sound level meter measure at the maximum and at the minimum? Solution: a) 120 dB; b) 2.051 rad; c) δ1= 14.5cm; δ2= 3.7cm; d) 122.7dB; 111.3dB Problem 8.4.8. Two loudspeakers S1and S2are located on the y-axis at y=±d/2, as shown in the figure. Using an audio-frequency amplifier, they are made to emit sounds of frequency fin phase. Figure for Problem 8.4.8 An observer moves from y= 0 along a line parallel to the y-axis and situated at a great distance Dfrom this axis. If d≪Dand y≪D: a) Demonstrate that the observer will perceive the first maxima of sound intensity at the distances yconst =nD d v fwith n= 0,1,2, ... where vis the speed of sound. 322 Problems and questions b) With d= 2 m and D= 80 m, and assuming v= 340 m/s, what is the frequency for which the distance between two consecutive maxima of intensity is 3.0m? Solution: 4.53 kHz Problem 8.4.9. Two point sources A= (0,4,0) and B= (8,0,0) (SI units) and of, respectively, 1mW and 2mW power, emit spherical waves in phase in open air. If the speed of sound is 340 m/s, find out: a) The phase difference at which the two waves arrive at the origin if the frequency of both sources is 200 Hz. b) The frequencies between 200 Hz and 1500 Hz at which the two foci would have to emit simultaneously in order for there to be destructive interference at the origin. c) The intensities at which the two waves arrive—separately— at the source. d) How much the intensity of sound is if the foci emit at a frequency that cause destructive interference at the origin. Find the same when there is constructive interference. Solution: a) 14.784 rad; b) 212.5Hz; 255.0Hz ... 1.445 kHz; 1.488 kHz; c) 4.97 µW/m2;2.49 µW/m2; d) 0.427 µW/m2;14.50 µW/m2 Problem 8.5.4. The fundamental frequency of a given violin string of length Lis 196 Hz. If the violinist wants to obtain a fundamental frequency of 440 Hz, how long does the string have to be? Solution: 0.445L Problem 8.5.5. The first and last strings on a piano are tuned to 33 Hz and 4186 Hz, with lengths of, respectively, 198 cm and 5.1cm. If the two strings are under the same tension, how much is the quotient of the effective linear densities of the two strings? Solution: 10.68 Problem 8.5.6. The figure shows a rod Fmaking sinusoidal vibrations at a frequency of 100 Hz. This excites the horizontal string AB of length 120 cm, tensioned by mass M, which has a weight of 2.25 N, and remains practically immobile. A system of standing waves is thus obtained, with a node in the immediate 323 Problems and questions vicinity of end Aon the rod and another node at point B, which is in contact with the pulley. Between these two nodes are four antinodes. The amplitude of the vibrations of the antinodes is 10 mm. Determine: a) The wavelength of the vibrations and their propagation velocity. b) The maximum velocity of a point on the string corresponding to an antinode. Figure for Problem 8.5.6 c) The amplitude of the vibrations of the point on the string located 35 cm from end A. d) The weight of a new mass M′hanging from the string if we wanted to obtain three antinodes instead of four. Solution: a) 0.60 m; 60 m/s; b) 6.28 m/s; c) 5mm; d) 4N Problem 8.5.7. A standing wave on a 6.00 m long elastic string is described by the function (SI units): y(x, t) = 2.0×10−2sin (π 2x)cos(πt) a) Schematically represent the standing wave, indicating the positions of the nodes and of the bellies. In which harmonic does this wave vibrate? b) Calculate the propagation speed of the plane wave. c) Write the harmonic wave functions that generate this standing wave. d) At what instants will the string be completely straight? e) At which points on the string and for which instants will the transverse velocity be maximum? What value will this velocity have? f) When will the transverse velocity of the wave be zero? g) By what percentage should the tension be increased to form standing waves with one less belly? Solution: a) 3dharmonic; b) v= 2.00 m/s; c) y+(x, t) = 1.0×10−2sin(x/2−πt); y−(x, t) = 1.0×10−2sin(x/2 + πt); d) t=n+ 1/2 s; n= 0,1,2,3...; e) In 324 Problems and questions the antinodes, xv= 1,3,5m; for t=n+ 1/2 s ; v= 6.28 m/s; f) t=ns; g) 125% Problem 8.5.8. Two identical strings of mass 100 g and length 1m are fixed at the ends and subjected to tensions of, respectively, 200 N and 205 N. a) If the strings vibrate in the third harmonic, calculate the frequency of the beats resulting from the superposition of the sounds generated by each string. Then, only the 200 N tension string is made to vibrate at the fundamental harmonic. If the amplitude at a point 20 cm from one end is 1.0cm, determine: b) The maximum velocity of the transverse motion at this point on the string. c) The elongation at this point at the instant t= 0, knowing that at this instant the elongation at the centre of the string is maximum. Solution: a) 0.833 beats/s; b) ˙y= 1.41 m/s; c) y= 1.00 cm Problem 8.5.9. The end Aof a horizontal string is fixed to the wall and the other end Bpasses through a frictionless pulley and is attached to a body of mass M, which hangs from it. Figure for Problem 8.5.9 The frequency of the fundamental sound emitted by the string is 392 Hz. If the body is completely immersed in water (see Figure), the frequency drops to 343 Hz. Calculate the density of the body. Solution: 4270 g/cm3 Problem 8.5.10. A string of 28.28 cm length and 0.050 kg/m linear density is attached to a second string with a linear density that is half that of the first. 325 Problems and questions Figure for Problem 8.5.10 One end is fixed to a wall and the other end passes through a pulley and is attached to a weight of 100 N, which hangs from it. Between joint Sand the pulley P, the length of this second string is 100 cm. We want standing waves to form along both strings so that there is a node at junction S, as shown in the figure. What is the lowest frequency we can apply to the strings? In this case, how many antinodes will there be along the two strings? Solution: 158.1Hz; 7antinodes Problem 8.5.11. A string of 1.40 m length and 2.00 g mass is attached to another string of 1.00 m length and 4.98 g mass. The assembly is made to vibrate at 120 Hz. a) What is the maximum tension they must be subjected to so that standing waves form when the weld is at a node? How many antinodes will each string have in this case? b) What other lower tensions will also satisfy this condition? Solution: a) 17.9N; 3and 4antinodes; b) 4.48 N; 1.99 N... Figure for Problem 8.5.12 Problem 8.5.12. We want to measure the speed of sound in the air. To do this, we use a setup as shown in the figure, where we have a loudspeaker A, that emits harmonic sound that we can control by maintaining the frequency anywhere we want between 400 and 1200 Hz. The speaker faces the mouth of a 1m long tube T filled with water up to a height that we can modify by raising or lowering the tank D. The length of the air in the tube, y, is measured with a ruler parallel to the tube. 326 Problems and questions With the loudspeaker operating at 700 Hz, we lower the water level and observe that, when y1= 20.3cm, there is resonance. If we continue lowering the water, the next level at which there is resonance is y2= 44.6cm. a) How much is the speed of sound in the air inside the tube? b) With a frequency of 1000 Hz and with the water completely filling the tube, we lower the level until we reach the first resonance position and y1= 10.0cm. Find all the other positions at which the tube will resonate. c) If the tube is then left with a fixed column of air, y= 50 cm, at what frequencies will the tube resonate? Solution: a) v= 340.2m/s; b) y2= 27.0cm; y3= 44.0cm; y4= 61.0cm; y5= 78.0cm; y6= 95.0cm; c) 510 Hz, 851 Hz; 1191 Hz Problem 8.5.13. We have a cylindrical tube open at both ends that we excite with sound waves of varying frequency. Two consecutive resonance frequencies are observed: 360 Hz and 540 Hz. a) If we close one of the ends and excite it again, what are the first three frequencies at which it will now resonate? b) If the tube is 0.950 m long, what is the temperature of the air inside? Solution: a) 90 Hz; 270 Hz; 450 Hz; b) 18.2◦C Problem 8.6.2. On a day when the temperature is 35◦C, the driver of an express train, travelling at 110 km/h, sees a commuter train travelling on the same track further ahead. To determine the speed at which the commuter train is travelling, the driver sounds his 1000 Hz whistle and listens to the 1060 Hz frequency echo. a) At what speed does the sound propagate? b) Assuming there is no wind, at what speed does the commuter train travel? Solution: a) 352 m/s; b) 73.3km/h Problem 8.6.3. On a windless day when the speed of sound is 340 m/s, a siren emitting a sound of 1000 Hz moves away from an observer at rest and towards a cliff at a speed of 36.0km/h. What is the difference between the two sound frequencies —direct and reflected from the cliff— that will reach the observer? How fast should the siren be travelling so that 8.0Hz pulses can be heard? 327 Problems and questions Solution: 58.9Hz; 4.90 km/h Problem 8.6.4. An object motion detector could consist of a source of harmonic waves of frequency f0and a detector of the frequency of the beats fbobtained by superimposing the direct waves of the source with the waves reflected by the object. If the object does not move with respect to the source, fb= 0. If it approaches the source or moves away at a velocity vthat is much smaller than the velocity cof the waves in the medium, show that the beat frequency is fb= 2f0 v c. Problem 8.6.5. A supersonic aircraft travelling horizontally at Mach = 1.50 11Mach =Mis the ratio of the velocity of an object,v, to the velocity of the waves in the medium, c: M=v c passes through the vertical of an observer in open air on a day when the average speed of sound is 335 m/s. If this observer has to wait 3.20 s to hear the plane from the time he sees it pass overhead, at what height is the plane flying? Solution: 1438 m 328 Annexes Derivatives Function: F(x)Derivative: dF (x) dx a f(x) + g(x)adf(x) dx +dg(x) dx f(x)g(x)df(x) dx g(x) + f(x)dg(x) dx f(x) g(x)=f(x)g(x)−1df(x) dx g(x)−f(x)dg(x) dx g(x)2 f(x)nn f(x)n−1df(x) dx f(g(x)) df(g) dg (x)dg(x) dx f−1(x) f−1(x)is the inverse function of f; e.g. y= arcsin xis the inverse of x=sin y 1 df(y) dy y=f−1(x) exex axaxln a sin xcos x cos x−sin x tan x1 cos2x ln x1 x log(a)x1 xln a Integrals Function: F(x)Integral: ∫F(x)dx a f(x) + g(x)a∫f(x)dx +∫g(x)dx xnwith n=−11 n+1 xn+1 1 x=x−1ln x exex axax ln a sin x−cos x cos xsin x tan x−ln(cos x) ln x−x+xln x 1 (x−a)(x−b)1 a−bln x−a x−b 1 √a2−x2arcsin x a 1 √x2−a2ln(x+√x2−a2) 336 Annexes Centre of mass rCM of simple homogeneous bodies with respect to the given reference frames Arc of circumference L= 2Rα rCS =(Rsin α α,0) Circle sector S=R2α rCS =(2Rsin α 3α,0) Arbitrary triangle S=1 2b h The distance from a base i to the CS is 1 3of the corresponding height hi Semi-sphere (without the bottom cap) S= 2πR2 rCS =(0,0,R 2) Semi-sphere (solid) V=2πR3 3 rCS =(0,0,3R 8) Cone (solid) V=πR2h 3 rCS =(0,0,h 4) 337 Annexes Support reactions Cable or tensioned rope. The force generated by a cable is in the direction of the cable. The direction is always on the side of the cable or the force is cancelled out (cable not taut). Smooth regular contact. Reaction Nnormal to the tangent plane to the regular surface at the point of contact. Regular/singular smooth contact. Reaction Nnormal to the tangent plane to the regular surface at the point of contact. Rough contacts. A frictional force Ffmust be added to the normal reaction N. When the bodies do not move, |Ff| ≤ µ|N|, where µis the coefficient of friction. The maximum value of |Ff|is reached when movement is imminent. Extensive contacts.N-reaction normal to the contact surface. It does not have to pass through the centre of mass. Extensive contacts (imminent overturning).Nreaction normal to the support surface applied at the point where contact is concentrated. Rollers (frictionless). This is the same case as smooth contacts. Joint. R-reaction, generally unknown, which in 2D means two components. If there is friction (between the surfaces in contact in the area of the joint) or no joint but instead jamming, a torque of Mmoment must be added. Its maximum value will depend on the nature of the contact. Guides. Normal reaction to a guide. If the contact is rough, friction Ffmust be added and perhaps also a friction torque Mf. 338 Annexes Moments of inertia Iof simple homogeneous bodies with respect to the indicated axes. Cylinder thick: I=1 2m(R2 1+R2 2) disc: H= 0 solid: R1= 0 thin: R1=R2 Sphere solid: I=2 5mR2 empty: I=2 3mR2 Orthohedron orthohedron / rectangle: I=1 12m(a2+b2) bar: a= 0 ic= 0 339 Annexes Constants Name of the constant: Value: π−number 3.1415926536 e−number, e=lim n→∞ (1 + 1 n)n2.718281285 Coulomb constant, k=1 4πε08.987551788 ×109Nm2 C Elementary charge , e1.602177 ×10−19C Avogadro’s number, NA6.022137 ×1023 Boltzmann constant, k1.380658 ×1023 J K Ideal gas constant, R=NAk8.31451 J mol K Gravitational constant, G6.6726 ×10−11 N m2 kg2 Electron mass, me9,109390 ×10−31 kg Proton mass, mp1.672622 ×10−27 kg Neutron mass, mn1.674929 ×10−27 kg Speed of light, c2.99792458 ×108m s Gravity acceleration at the Earth’s surface, g Standard value: 9.81 m s2. In Barcelona: 9.804 m s2 Earth radius, RT6370 km Earth mass, MT5.98 ×1024 kg 340