Textbook on Theoretical Mechanics. 9786010428959

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AL-FARABI KAZAKH NATIONAL UNIVERSITY

Z. B. Rakisheva A. S. Sukhenko

TEXTBOOK ON THEORETICAL MECHANICS Second edition

Almaty «Qazaq university» 2017

UDC 531.011 (076) LBC 22.21 R 17 Recommended for publication by the decision of the Academic Council of the Faculty of Mathematics and Mechanics, Editorial and Publishing Council of Al-Farabi Kazakh National University (Protocol №1 dated 29.09.2017); Educational and methodical association on groups of specialties «Natural sciences», «Engineering and technology» of Republican educational-methodical council on basis Al-Farabi Kazakh National University (Protocol №2 dated 29.06.2017) Reviewers: Doctor of Technical Sciences, Professor E.S. Temirbekov Doctor of Technical Sciences, Professor Z.G. Ualiev Doctor of Physical and Mathematical Sciences, Associated Professor K.S. Zhilisbayeva Authors: Z.B. Rakisheva, candidate of physical and mathematical sciences, associated professor – lecture course A.S. Sukhenko, PhD, senior teacher – tasks and exercises

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Rakisheva Z.B. Textbook on Theoretical Mechanics / Z.B. Rakisheva, A.S. Sukhenko. – 2nd ed. – Almaty: Qazaq university, 2017. – 354 p. ISBN 978-601-04-2895-9 Textbook was prepared on the base of a compulsorysubjectof theoretical mechanics, read by the authors for the students of specialty «Mechanics». It contains a lecture course, covering all topics of the Typical Curriculum of 2016, on the discipline of «Theoretical Mechanics», and includes also thesections of rigid body dynamics and analytical mechanics.After each studiedsection, the practical tasks, tasks for independent work, control questions and assignments are given. In conclusion, tests on the studied material with the keys of the correct answers are given. The textbook is recommended for teaching students on the specialty «5В060300-Mechanics», as well as for the students of all related specialties of science and technology, where the mechanics is studied. Publishing in authorial release.

UDC 531.011 (076) LBC 22.21 ISBN 978-601-04-2895-9

© Rakisheva Z.B., Sukhenko A.S., 2017 © Al-Farabi KazNU, 2017

Content

CONTENT

INTRODUCTION ..................................................................................... 7 1. KINEMATICS ....................................................................................... 9 1.1. The subject of mechanics. Kinematics of a point. Problems of kinematic. Methods of the point’s motion setting.................................. 9 Questions ................................................................................................... 13 1.2. The velocity and acceleration. The decomposition of the velocity and acceleration. ................................................................ 13 Questions ................................................................................................... 18 Practice ...................................................................................................... 18 Self study of the student............................................................................. 22 1.3. Mechanical system. Basic movements of a rigid body. Translational motion of a rigid body.......................................................... 23 Questions ................................................................................................... 25 1.4. Rotational motion of a rigid body around a fixed axis. ....................... 25 Questions ................................................................................................... 30 Practice ...................................................................................................... 30 Self study of the student............................................................................. 33 1.5. Plane-parallel motion of a rigid body. Velocities of the points of a plane figure. Instantaneous center of velocity. .................................... 35 Questions ................................................................................................... 40 1.6. Acceleration of the points of a plane figure. Instantaneous center of acceleration. ................................................................................ 40 Questions ................................................................................................... 43 Practice ...................................................................................................... 43 Self study of the student............................................................................. 49 1.7. Compound motion of a point. Full and relative derivatives of the vector. Addition of velocities........................................................... 52 Questions ................................................................................................... 55 1.8. Theorem on the addition of accelerations (Coriolis theorem) ............. 56 Questions ................................................................................................... 58 Practice ...................................................................................................... 58 Self study of the student............................................................................. 65 1.9. Complex motion of a solid body. Addition of the translational velocities. Addition of the instant angular velocities. Addition of the instant angular and translational velocities. ..................................... 68 Questions ................................................................................................... 71

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Textbook on Theoretical Mechanics 1.10. Reduction of the system of sliding vectors. Principle vector and principle moment. Invariants of reduction .......................................... 71 Questions ................................................................................................... 73 2. STATICS ............................................................................................... 74 2.1. Statics. Basic definitions and axioms of statics. Constraints. Constraint reactions. Axiom of constraints. ............................................... 74 Questions ................................................................................................... 78 2.2. System of forces. Convergent system of forces. Parallel forces. Center of gravity. ....................................................................................... 79 Questions ................................................................................................... 87 Practice ...................................................................................................... 87 Self study of the student............................................................................. 93 2.3. Moment of force relative to center and axis. Theory of couples. ........ 94 Questions ................................................................................................... 98 2.4. Arbitrary system of forces. Reduction of spatial system of forces. Principle vector and principle moment. ..................................................... 98 Questions ................................................................................................... 101 2.5. Equilibrium conditions of arbitrary spatial system of forces. Special cases of the equilibrium conditions ............................................... 102 Questions ................................................................................................... 104 Practice ...................................................................................................... 104 Self study of the student............................................................................. 120 2.6. Conditions of equlibrium of a constrained solid body. Friction and constraints with friction ......................................................... 122 Questions ................................................................................................... 127 3. DYNAMICS OF THE MASS POINT AND THE SYSTEM ................ 128 3.1. The laws of Newton. Direct and inverse problems of dynamics. Motion equations ....................................................................................... 128 Questions ................................................................................................... 133 Practice ...................................................................................................... 133 3.2. Basic dynamic variables. Properties of internal forces of the system. ............................................................................................. 137 Questions ................................................................................................... 146 3.3. Theorem of change of linear momentum of a mass point and mechanical system. Theorem on the motion of the center of mass ..... 147 Questions ................................................................................................... 151 Self study of the student............................................................................. 151 3.4. Angular momentum theorem of a mass point and mechanical system ........................................................................................................ 154 Questions ................................................................................................... 160 Practice ...................................................................................................... 160 Self study of the student............................................................................. 165 3.5. Work of force. Work of potential force............................................... 169

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Content Questions ................................................................................................... 175 3.6. Work-Energy theorem for the mass point and mechanical system. Energy integral........................................................................................... 175 Questions ................................................................................................... 179 Practice ...................................................................................................... 179 Self study of the student............................................................................. 185 3.7. Rectilinear motion of a mass point. Harmonic oscillations of the mass point. Parameters of oscillations. Oscillations in a resistant medium. ..................................................................................................... 190 Questions ................................................................................................... 200 3.8. Forced oscillations in a medium without resistance and in a resistant medium. Resonance. ................................................................. 200 Questions ................................................................................................... 205 Practice ...................................................................................................... 206 3.9. Motion of mass point under the action of central forces. Law of areas. Binet formulas. ............................................................................ 210 Questions ................................................................................................... 215 3.10. Planetary motion problem. Kepler's laws. Derivation of Newton’s law of gravitaion from Kepler's laws..................................... 215 Questions ................................................................................................... 217 3.11. Motion of a non-free mass point. Concept of a constraint. Motion of a mass point over given curve. Motion of a mass point over given surface. Geodetic line. .............................................................. 218 Questions ................................................................................................... 225 3.12. Relative motion and equlibrium of a mass point. Equations of relative motion. Inertial forces of transportation motion, Coriolis inertial force ................................................................................. 225 Questions ................................................................................................... 229 3.13. Apparent weight of a body. Deflection of bodies falling on the Earth from the vertical. Work-energy theorem for relative motion. Relative motion and equilibrium .................................................. 230 Questions ................................................................................................... 233 4. DYNAMICS OF THE SOLID BODY .................................................. 234 4.1. Mass geometry. Inertia moment.......................................................... 234 Questions ................................................................................................... 238 4.2. Theorem of Guigens-Shteiner. Inertia moments relative to the axes of the beam, coming from this point. ....................................... 239 Questions ................................................................................................... 241 Practice ...................................................................................................... 241 Self study of the student............................................................................. 245 4.3. Rotation of solid body around fixed axis. Differential equations of motion. Axle pressure. Motion of absolutely rigid body with one fixed point. ................................................................................................. 248 Questions ................................................................................................... 253

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Textbook on Theoretical Mechanics Practice ...................................................................................................... 254 Self study of the student............................................................................. 262 4.4. Motion of absolutely rigid body with one fixed point. Euler kinematic and dynamic equations. ................................................... 264 Questions ................................................................................................... 267 Practice ...................................................................................................... 267 4.5. General formulation of the problem of heavy solid body with one fixed point. .................................................................................. 269 Questions ................................................................................................... 275 4.6. Special cases of integration and its geometrical interpretation: the case of Euler-Puanso, the case of Lagrange-Puasson, the case of Kovalevskaya............................................................................................. 276 Questions ................................................................................................... 277 5. ANALYTICAL MECHANICS ............................................................. 278 5.1. The notion of holonomic and nonholonomic systems. Actual and virtual displacement of the point. ............................................ 278 Questions ................................................................................................... 281 Practice ...................................................................................................... 281 Self study of the student............................................................................. 289 5.2. Virtual work of forces. Ideal constraints. Principle of virtual work. d'Alambert's principle for the point and system. General dynamic equation. .................................................................................................... 293 Questions ................................................................................................... 296 Practice ...................................................................................................... 296 Self study of the student............................................................................. 300 5.3. Generalized coordinates. Generalized forces. Equations of motion in generalized coordinates (Lagrange equations of 2-nd kind). . 304 Questions ................................................................................................... 307 5.4. Expression of kinetic energy in generalized coordinates. Lagrange equations of 2-nd type for the system under action of potential force. ...... 307 Questions ................................................................................................... 309 Practice ...................................................................................................... 309 Self study of the student............................................................................. 313 TESTS ....................................................................................................... 318 KEYS TO THE TEST ............................................................................... 351 CONCLUSION ......................................................................................... 352 REFERENCES .......................................................................................... 353

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Content

INTRODUCTION

Theoretical mechanics is the science about general laws of mechanical interactions between material bodies, as well as general laws of motion of bodies relative to each other. Mechanical interaction between material bodies is the simplest and at the same time common type of interaction between physical objects. Mechanical motion being the simplest type of motion is a fundamental property of matter. Theoretical mechanics taught in the university contains four sections: kinematics, statics, dynamics and analytical mechanics. Kinematics is a part of mechanics in which the dependences between quantities characterizing the state of systems motion are studied, however the causes changing the state of the motion are not considered. Statics is a study of equilibrium of bodies set of some reference system. Dynamics is the part of mechanics that considers an influence of forces to the state of motion of material objects system. Analytical mechanics is the part of theoretical mechanics and theoretical physics in which general principles of mechanics (differential and integral) are formulated and used, on this basis main differential equations of motion are derived and equations and methods of their integration are investigated. Purpose of studying a discipline «Theoretical mechanics» is a formation of a necessary base of knowledge to study other technical disciplines on a profile of future professional activity as a strength of materials and the theory of mechanisms and machines. Tasks of the discipline «Theoretical mechanics» are: – development of practical skills in solving problems of mechanics by studying methods and algorithms of constructing mathematical models of motion or state of the considering mechanical systems, also research methods of these mathematical models; 7

Textbook on Theoretical Mechanics

– education of the natural-science worldview on the basis of studying the basic laws of nature and mechanics. Textbook gives the basic materials on each of main sections of theoretical mechanics including the summary of lectures, tasks and examples of solution, tasks for independent work, and literature on theoretical mechanics. Work with this textbook allows students studying the main concepts, laws and theorems of theoretical mechanics, obtaining the skills of mathematical modeling of physical processes, using the obtained knowledge for the solution of practical tasks, analyzing the solutions of the tasks and making conclusions. This textbook was written on the base of the course of lectures, which are read by prof. Zaure Rakisheva during last 20 years for the students of the mechanical and mathematical faculty of al-Farabi Kazakh National University.

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1. Kinematics

1 KINEMATICS

1.1. The subject of mechanics. Kinematics of a point. Problems of kinematic. Methods of the point’s motion setting Theoretical mechanics is the science of the general laws of mechanical motion and interaction of material bodies. Motion is the mode of existence of matter. The simplest form of matter in motion is a mechanical movement. The mechanical motion is that the body changes its position in space in relation to other objects in the course of time. The notion of force is used in classical mechanics to account the mechanical interaction that occurs between the bodies. However, the nature of movement of the body depends on the forces as well as on inertia of the body. The measure of inertia of a body is its mass, which depends on the amount of the body. Thus, the basic concepts of classical mechanics are: moving matter (physical bodies), the space and time as forms of existence of matter in motion, mass, as a measure of the inertia of material bodies, the force as a measure of the mechanical interaction between the bodies. The relationships between the basic concepts of mechanics is defined by axioms or basic laws of motion, which were given by Newton. First law (the law of inertia). Every body stays in its state of rest or uniform rectilinear motion as long as it isn’t forced by the applied forces to change this state. Second law (the basic law of dynamics). Change of linear momentum is proportional to the applied driving force and occurs in the direction of the straight line which the force acts along: 9

Textbook on Theoretical Mechanics

 d (mv )  F dt

Third law (the law of action and reaction). Every action has an equal and opposite reaction. In other words the interaction of two bodies on each other are equal and opposite in direction:

  FA   FB Theoretical mechanics, as any science that uses mathematical methods, does not deal with the real material objects, but with their models. Such models are mass points, the system of mass points, perfectly rigid bodies, deformable continuum. Kinematics of the point Basic concepts. Kinematics studies the motion of bodies from a geometric point of view, excluding the causes of variation of this motion (forces). Mass point is the body, the size of which can be ignored when studying its motion. The mass of mass point is not taken into account in kinematics. Position of the body in space can be defined only in relation to an arbitrarily chosen another body, called the body or frame of reference. If the position of the body relative to the chosen reference frame doesn’t change with time, the body stays in rest relative to this reference frame. But if the position of the body is changed, this body moves relative to a given reference frame. Thus, the motion and the rest are relative concepts, and make sense only in relation to a particular reference frame. Form of trajectory depends on the reference frame. The motion of the body relative to the chosen reference frame will be known, if at any arbitrary moment of time its position can be determined (relative to this reference frame). The position of the point is determined by the appropriate parameters (coordinates), and 10

1. Kinematics

its motion (law of motion) determined by the equations that express these parameters as a function of time. Determining the methods setting the motion of the point is one of the problems of kinematics. The main problem is to get all kinematic characteristics of a given point (trajectory, velocity, acceleration) from the equations determining the law of motion of this point. Methods of the point’s motion setting. Setting the motion of the point is to determine its position relative to the chosen reference frame at any given time. There are three ways to set the motion of the point: natural, coordinate, vector. 1. The natural method requires the specification of the trajectory relative to the selected reference frame xyz . The origin and the positive direction of measuring distances S  OM should be set. The distance S (from O to M), measured along the arc of trajectory and taken with the appropriate sign, will uniquely identify the position of the point M on the trajectory, and therefore in the reference frame. Then, it is necessary to specify the time reference point. If the value S is known for each time t the motion will be determined, in other words it will be obtained the relation: S=f(t)

(1)

This relation is named as the law of motion of the point. Thus, when the natural method of point’s motion is considered it must be defined the following parameters: a) the trajectory of the point; b) the origin and the positive direction of measuring distances, initial moment of time; c) the law of motion along the trajectory S=f(t). Function f(t) must be unique, continuous and differential. 11

Textbook on Theoretical Mechanics

2. Coordinate method. When the coordinate method of point’s motion is considered it must be defined the following parameters: 1) any coordinate system associated with the reference body; 2) the coordinates of the moving point as a function of time. It is required three numbers q1 , q 2 , q3 that are named the coordinates of point because the space is three-dimensional. In general case the law of motion is given by the expression:

q1  q1 (t ) , q 2  q 2 (t ), q3  q1 (t )

(2)

The motion of the point in Cartesian coordinate system is set with the expression: (3) x  x (t ) , y  y (t ), z  z (t ) Each of the three equations (3) determines the motion of the projections of the point on the corresponding axis. Equations (3) define the law of motion of a point, as it allows to determine the coordinates at any time. On the other hand, these equations set the trajectory in a parametric form (parameter t). Except the Cartesian coordinate system other coordinate systems are used (spherical or cylindrical). 3. Vector method. The position of the point is set by the position  vector r drawn from the origin O of the chosen reference frame. As the position vector of the point is set by the expression:

    r  xi  yj  zk The law of motion of the point in vector form will have the form:      r  r (t )  x (t )i  y (t ) j  z (t )k .  And hodograph curve of r will be the trajectory of the point. In case of plane motion (when the trajectory is the plane curve) the law of motion is expressed with two equations. 12

1. Kinematics

According to the character of the trajectory the motion of the point can be linear and curvilinear in the dependence on the reference frame. Questions 1. What is the subject of mechanics? 2. What is the mass point? 3. How many methods of mass point's motion setting do you know? 4. What are the main methods of mass point's motion setting? 5. What are the basic laws of Newton?

1.2. The velocity and acceleration. The decomposition of the velocity and acceleration The vector of velocity of the point The velocity of the point characterizes the speed and direction of motion of the point and is equal to the time derivative of position vector of the point: dr  v r dt The vector of velocity is directed along the tangent to the trajectory. The vector of acceleration of the point The acceleration of the point is a vector quantity that characterizes the change of the magnitude and direction of the velocity with time. The acceleration of a point equals to the first time derivative of the vector of velocity or the second time derivative of the position vector of the point:    dv d 2 r . w  2 dt dt The vector of acceleration is always directed towards the concavity of trajectory. 13

Textbook on Theoretical Mechanics

Decomposition of velocity on the radial and transversal components Let’s consider the position vector in the form:

  r  rr0, 

where r 0 is the unit vector of direc tion r .  Position vector r changes by its  length and direction, therefore, r  and r 0 are the functions of time t. Let’s differentiate the position vector with respect to time:

   dr dr  0 dr 0 r  r  v dt dt dt  The first term characterizes the change of magnitude of r .  Direction of the second term is perpendicular to r 0 because the differential of unit vector is perpendicular to the direction of the 0 vector itself, dr  d . Thus, the second term characterizes the

dt

dt

change of direction of

 r:

    dr  0 d  0 , v  vr  v p  r r p dt dt where the first term is the radial component of the velocity an the second is the transversal component.

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1. Kinematics

Decomposition of acceleration on the radial and transversal components As the velocity is:

  dr  0 d  0 , v r r p dt dt than acceleration will be:      dv d 2 r  0 dr dr 0 dr d  0 d d p 0 . d 2  w p  r 2 p0  r  2 r   dt dt dt dt dt dt dt dt dt

Let’s find the magnitude and direction

of

the

vector

 dp 0 . dt

  d p 0  p 0 is the differential of the unite vector that is perpendicular to the direction of the vector itself and  directs opposite to r 0 (direction of 0 d p is obtained by the rotation of  p 0 at an angle 90° in the direction



0 of positive count of a angle). Besides, dp  d , therefore:

dt dt 0 Let’s substitute dp   d r 0 in the expression of dt dt

 w:

2   d 2 r  d    0  d 2 dr d   0 . w   2  r   r  r 2  2 p dt dt   dt    dt  dt 2   d 2 r  d    0 Radial component wr   2  r    r directs along the  dt    dt position vector.

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Teextbook on Theoreticaal Mechanics

 d 2  dr d   0 Transversal comp ponens w p  r  2  p is perpen2 dt dt   dt dicullar to the position n vector. Natural triheedral 1) Let’s draw a tangent t in the pooint M and definee the positive direcction with the unit vector  0 (in tthe direction of in ncreasing the curvilinear co oordinates S). Osculating plane p is the end position of the plane passing throu ugh any three points of thee curve when these points tend to M. IOW: Osculaating plane is the end possition of the plane passing g through the tangent M  and the point w M ' when M ' M . p through the point M and its plane will be 2) Let’s draw a plane perpendicular to the taangent  0 . This pplane is called the normal plane he trajectory at thee point M. All lines lying in this pllane is named of th norm mal, and the line of o intersection of nnormal plane to the osculating  plane is called the prin ncipal normal n 0 . 3) Straight line perpendicular p to the tangent and the principal  mal is called binorrmal. The unit vecctor b 0 . The positive direction norm  is selected so that  0 , b 0 , n 0 form a rigght-handed system m. This system is called the natural axxes. The plane pas ssing through 0 0 0 0 n plane, ( , n ) is called osculating plane, throough ( n , b ) – normal   ugh ( b 0 ,  0 ) – rectifying r plane. Trihedral with veertices at the throu poin nt M is a natural trihedral, it movees along a trajecttory with the poin nt M.

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1. Kinematiccs

Curvature of a curve Let’s identify wiith  the anglee  of co ontingence betweeen the tangent  0   and  0 ' , and arc leength ММ '  S , a curvaturee than   k  is the average ср S he segment ММ ' . lim   d  k is the curvature of the curve. on th S 0 S

dS

The value   1  dS is the radiuus of curvature. k d Decomposition of the acceleeration on the ax xes of natural trih hedral

   v  v  0  v 0    dv dv  0 d 0    v w dt dt dt The firs term direects along the tanggent. Let’s considder the second 0  term m d v . The vecctor d 0 directs along the normaal to n 0 . Its dt magn nitude equals:  d 0 d  dt dt

 sin   2  lim lim  lim ( liim m ).   t 0 t t 0 t t 0 t 0 t 2   0

0 2 But d  d dSS  1 v , then d v  v n 0 v  v n 0 . finally, d dt   dt dS dtt 

 dv  v2  w   0  n0 . dt  17

Textbook on Theoretical Mechanics

w 

v 2 – is the normal dv is the tangential component, wn  dt 

component.  Since wb  0 , than w  rectifying plane. Absolute magnitude of acceleration is: 2

2

w  w 

wn2

2 2  dv   v  .        dt    

 The angle  is the angle between w and the principle normal:   w tg    . If the angle between w and v is sharp then the motion is wn accelerated. If the angle is blunt then the motion is retarded. Questions 1. What is velocity? 2. What is acceleration? 3. How can be decomposed the vector of velocity and acceleration? 4. Give the notion of curvature. 5. How can be defined tangential and normal acceleration?

PRACTICE 1.1. Find the path equation of material point in coordinate form according the following equations and show it graphically. 1) x  3t  5, y  4  2t

2) x  2t , y  8t 2 3) x  5 sin 10t , y  3 cos10t 4) x  2  3 cos 5t , y  4 sin 5t  1 Answer: 1) half-line 2 x  3 y  2  0 with origin in x  5, y  4 ;

2) right branch of parabola y  2x 2 with origin in x  0, y  0 ; 2 2 2 2 3) ellipse x  y 1 with origin in x  0 y  3; 4) ellipse (x  2)  (y 1) 1

9

25 9

with origin in x  1, y   1 . 18

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1. Kinematics

1.2. The train is moving with the velocity of 72 km per hour and in case of braking it becomes slowdown of 0.4 m / sec 2 . Find the time and the distance when the train must begin its braking before it comes to the station. Answer: 50 sec, 500 m. 1.3. The train passed 600 meters in first 30 seconds with initial velocity of 54 km per hour. Define velocity and acceleration of the train in the end of 30-th second assuming its motion as uniformly variable motion and assuming that it takes place in condition of radius of curvature of trajectory of 1 km. Answer: v  25 m / sec, w  0,7 m / sec2 . 1.4. The projectile is moving in the vertical plane according the equations x  300t , y  400t  5t 2 (t – in seconds, x,y – in meters). Find velocity and acceleration of projectile in initial moment, height and distance of projectile, radius of curvature of projectile trajectory in initial and the highest points of its trajectory. Answer: v0  500m / sec, w0  10 m / sec2 , h  8 km, s  24 km, 0  41,67 km,   9 km . 1.5. Motion of material point is given with the equations: x  v0t cos 0 , y  v0t sin  0 

1 2 gt . Аxis Ox is horizontal, axis Oy 2

directs upwards along the vertical, v0 , g and  0 



2

are constants.

Find 1) the trajectory of material point 2) coordinates of its highest position. Answer: 1) Parabola y  xtg 0  2) x 

g 2v02

cos 2  0

v02 v2 sin 2 0 , y  0 sin 2  0 . 2g 2g

19

x2 ;

Textbook on Theoretical Mechanics

1.6. When moving from the station velocity of the train is increasing uniformly and its magnitude reaches the value of 72 km per hour in 3 minutes. Track of the train has curvature with the radius of 800 meters. Define tangential acceleration, normal acceleration and magnitude of acceleration in 2 minutes after moving from the station.

Answer: w  0,16

m m m . , wn  0,2 , 0,25 2 2 sec sec sec 2

1.7. Find the trajectory of the point M of coupler link of the slider-crank mechanism if 1 r  l  60 cm, MB  l ,   4t (t in 3

seconds) and define velocity, acceleration and radius of curvature of trajectory of the point when  0. Answer: Ellipse x2 y2   1, v  80 cm / sec, w  1600 2 cm / sec 2 ,   4 cm . 100 20

1.8. End A of the rod AB is moving along the rectilinear guide CD with the constant velocity v A . All the time the rod AB is moving through the swinging coupling O, which is situated at the distance a from the guide CD. Find velocity and acceleration of the point M situated at the distance b from the slider A assuming the point O as a pole.

Answer: v 

vA a

a 2 sin 2   r 2 cos 4  , w 

v t where r  a 2  v A2 t 2  b,   arctg  A  .  a 

20

v A2 b a

2

cos 3  1  3 sin 2  ,

1. Kinematics

1.9. The crank OA rotates with the constant angular velocity  . Find the velocity of the slider B and middle point of coupler link in dependence on time if OA=AB=a. а 2

Answer: VB  2a sin t , VМ   8sin 2t  1 . 1.10. Find the path equation of material point in coordinate form according the following equations and find the law of motion of the material point. 1) x  3t 5, 2 y  4t 2 2) x  3 sin t , y  3 cos t

Answer: 1) half-line 4 x  3 y  0; s  5t 2 2) circle x 2  y 2  9; s  3t . 1.11. Assuming the landing speed of the aircraft equal 400 km/h define its slowdown during the landing on the distance 1200 m. Answer: w=5,15 m/sec2 . 1.12. The train moves uniformly retarded along the arc of a circle of radius R=800m and pass the distance s=800 m with the initial velocity 54 km/h and final velocity 18 km/h. Determine the total acceleration of the train in the beginning and in the end of the arc and the time of motion along this arc. Answer: wb  0,307m / sec 2 , we  0,128m / sec 2 , T  80c. 1.13. The motion of the projectile is set by means of the equations: x  v0t cos  0 , y  v0t sin  0  0,5 gt 2 ( v0 , 0  const ). Find the radius of

curvature of the trajectory when t=0 in the moment of falling down. Answer:   v02 / g cos  0 . 1.14. The shot was fired from the weapon of coast artillery at angle 45 degrees to horizon with the initial velocity of the projectile v0  1000m / sec . Determine the distance from weapon at which the projectile will hit the mark situated at the sea level. 21

Textbook on Theoretical Mechanics

Answer: 102 km. 1.15. The motion of the point is set with the motion x  2t , y  t 2 (t in seconds, x,y – in cm). Determine the velocities and accelerations of the point in the moment of time t=1sec. Answer: v  2 2cm / sec, w  2cm / sec2 , (v,  x)  45 , (w,  x)  90 . 1.16. Find the magnitude and direction of acceleration and radius of curvature of trajectory of the point of the wheel rolling without sliding along the horizontal axis Ox if the point describes a cycloid in accordance with the equations: x  20t  sin 20t , y  1  cos 20t .

Determine the magnitude of radius of curvature when t=0. Answer: acceleration w=400m/sec2,   4sin10t,  0  0 . SELF STUDY OF THE STUDENT

Determine the path equation and its form in accordance with the given equations of motion of the mass point M for the moment of time t  t1 (sec) . Find the position of the point on the trajectory, its velocity, total, tangential and normal acceleration and radius of curvature of the trajectory. 22

1. Kinematics Number of variant 1 2 3 4 5 6 7 8 9 10 11 12

Equations of motion y  y (t ) , cm x  x(t ) , cm 2 5t  2t  3 4 /(t  1) 4t  4 3 cos(t / 3)  4 2 sin(t / 3) 3t 2  t  1 3 /(t  2)

5t 2  5t / 3  2 3t  6

4 cos(t / 3)

3 sin(t / 3)

3t

4t 2  1

t1 (sec)

1/2 2 1 1

7t  3

 2t 2  4 5t

7 sin(t 2 / 6)  3

2  7 cos(t 2 / 6)

2 1 1/2 0 1 1/4 1

3t 2  t  3

1

2  3t  6t

3  3t / 2  3t 2

2

6t 2

2

5t  5t / 3  3

1.3. Mechanical system. Basic movements of a rigid body. Translational motion of a rigid body Mechanical system is a set of material points, where the motion of each point depends on the position and movement of the other points of the system. Position of the system of n points is assumed to be known if it is known all the coordinates of all points of the system, ie: x1 , y1 , z1 , x2 , y2 , z 2 , ..., xn , yn , z n . The relationship between the movements of the points occurs as a result of the forces of interaction between them and due to the presence of constraints. Conditions that impose the restrictions on the movement of the points of system are called constraints. Constraints are expressed by equations, that coordinates or velocities of the points of system must satisfy: – f x1, y1, z1, x2 , y2 , z2 ,..., xn , yn , zn ; t   0 – geometric constraint; – x1, y1, z1, x2 , y2 , z2 ,..., xn , yn , zn ; x1, y1, z1, x2 , y2 , z2 ,..., xn , yn , zn ; t   0 – kinematic constraint. 23

Teextbook on Theoreticaal Mechanics

Let the system bee impose by the k – geometric consttraints: f   x1 , y1 , ...., z n ; t   0, (   1, k )

Then from 3n – coordinates will bbe (3n-k) – indeppendent coordinattes, that is, if it iss set any (3n-k) – coordinates, thee rest of them will be determined froom the constraint equations. Thesee independent coorrdinates are calledd the coordinates of the system. Onnly in case of geom metric constraintss the number of coordinates of thhe system is calleed the number of degrees of freedoom of the system. Mechanical systeem where the distaance between any two points is consstant is called absoolutely rigid (solidd) body. The basic motion ns of a rigid body.. There are five typpes of motion of a rigid body: 1) translational motion; m 2) rotational motiion; 3) plane-parallel or o plane motion; 4) spherical motioon; 5) general case off rigid body motioon. Translational andd rotational motion are simple, basiic motions of a rig gid body. Transslational motion oof a rigid body

Iff every line connnecting two pointss of the body movves parallel to itself this motion is called c translational motion. T Theorem. All po oints of the solid bbody executing a translational motionn describe identiccal and parallel traajectories and haave geometrically equal velocities and a accelerations aat each moment of time. Thus, the study of o translational m motion of a rigid body can be redu uced to the study of o the motion of a single point of thhis body, that is, to o the problem of kinematics k of the ppoint. 24

1. Kinematics

  Vector v and w are free, that is, they can be applied to any point of the body. They are called the velocity and acceleration of translational motion of a solid body. The points of a rigid body are moving with different velocities and accelerations for any other type of motion. The equations of translational motion of a solid body are the equations of motion of any point of body (usually the center of mass). xc  f1 (t )   yc  f 2 (t ) zc  f 3 (t ) 

The points of a solid body executing the translational motion can describe any trajectory including a straight line. If the velocity is constant, then all points of the system are moving rectilinearly and uniformly. That is, the system performs inertial motion. If the velocities of all points of the system are unchanged from each other only for some moment of time t, then we say that the invariable system has an instantaneous translational velocity in a given moment of time. Questions 1. Give the notion of mechanical system. 2. Give the notion of constraints. 3. Give the notion of absolutely rigid body. 4. How can be defined the translational motion of rigid body? 5. How many types of rigid body motion do you know?

1.4. Rotational motion of a rigid body around a fixed axis

If the body moves so that its two points A and B stay fixed, the motion of this body is called the rotational motion, and the line AB – the axis of rotation. Trajectories of all points of the body are circles with their plane perpendicular to the axis of rotation, and with the centers lying on this axis in this case. 25

Teextbook on Theoreticaal Mechanics

Let's suppose thaat the rotational axxis AB coincides with the axis z. To o determine the position of a rotatiing body let's draw w two planes throu ugh z-axis: Q – fixed plane annd P – fixed annd invariably assocciated plane withh the body. The ddihedral angle  between the fixed d plane Q and unssteady plane P is called the angle of o rotation of the body b m the end of  . Angle  assumed to bee positive if from poositive z axis it is seen directed oppposite clockwise from the fixxed plane. Position of thee plane P is deefined with the magnitude m and thee sign of angle  . Therefore thee position of the body is defineed with this angle. That is  caan be considered as a the angular cooordinate of the boody. Angle of rottation changes with time, thereffore it is the functtion of time:

   (t ) – the law of rotational motion. The body have one degree of ffreedom in case of rotational motiion. If the angle  beecomes incrementt  for the timee interval t  is called mean angular velocity off the body for the value v   t the given g time intervaal. Proceeding to thhe limit we will become: b

 d   – anguular velocity of rootation  t 0 t dt

  lim

of the solid bbody.

Angular velocityy changes with time in generaal case, i.e.    (t ) . If the incrrement of  forr t equals to  , than the 26

1. Kinematics

value    is the mean angular acceleration of the body for the t given time interval. Proceeding to the limit we will become:  d      – angular acceleration of the body. t 0 t dt

  lim

The angular velocity at the moment is characterized with the vector directed along the axis of rotation. Length of the vector  corresponds to the module  , and the direction will be such that the rotation of the end of the vector was opposite clockwise.  Angular acceleration is the vector  directed along the axis of   rotation. If the directions of  and  are the same the motion is accelerated. If the directions are opposite, the motion is retarded. If   const at all the time of motion, the rotation is called uniform. The law of such motion is:      t . 0

If   const at all the time of motion, this motion is called a uniformly accelerated motion. The law of this rotation is:  t 2 . And angular velocity varies according to the   0 0 t  2 law:      t . 0

The angle of rotation is often expressed with number of revolutions N. Then the angle  in radians, corresponding to N revolutions is defined as:   2 N . The moment of vector. The notion of moment relative to the center and moment relative to the axis can be introduced for stationary or sliding vector.  Let the vector a is attached at the point M. The position of point  M relative to Oxyz is defined with the position vector r .

    r  xi  yj  zk . 27

Teextbook on Theoreticaal Mechanics

 

Cross product r  a is called the m moment of vector the center: c

 i    momO a  r  a  x ax

 j y ay

 a

relative to

 k    z , momO a  r  a  sin( s r , a) az

The direction of the cross productt is defined by thee «right-hand screw w rule».  t the area of As any cross prodduct the moment momO a equals to   paralllelogram coverring the vectoors r and a , that is   a h  r a sin(r , a ) . Moments relativee to the axes: y  M x  mom x a  ay

z

 x , M z  mom mz a  ax ax

 z , M y  mom y a  az az

y

x

ay

Velocities Ve and acceelerations of the ppoints of rotating solid body

Any point M of rotating solid body makes circular m motion, that is: v  O M   . M



0

Then v  mo om M  , as O M is the arm 0



of vector  relaative to the point M. In accordannce with the defiinition of the   moment: v coiincides with mom m  by the





direction, that iss v  mom  . M M

M

 If О is the arbitraary point of the axis, to which a slidding vector   is ap pplied and r  OM then:

28

1. Kinematics

        v M  mom M   MO    OM    r      r . Thus, the velocity of any point of rotating solid body is determined by Euler's formula:

   v r Let's consider an arbitrary point О on the axis of rotation as the origin of the coordinate system Oxyz , Oz directs along the axis of rotation, then:    v x   y i j k     v    r  0 0    vy   x

x

y

z

 v 0  z

Acceleration of any point M of rotating body can be found in accordance with the formulas of circular motion:

w  O 0 M   , wn  O 0 M   2 , w   2   4 .

 wn always directs along the radius of circle and it is called  centripetal acceleration, w is called tangential acceleration.

If velocities of the points lying on the axis AB equal to zero all the time of motion this axis is called constant axis of rotation. If velocities of the points lying on any axis equal to zero at the given moment of time only, this axis is called the instantaneous axis of rotation. Velocities of all points in this case are also determined  by the Euler formula where  directed along the instantaneous axis of rotation is called instantaneous angular velocity of the body. Unlike the permanent axis, instantaneous axis changes its direction continuously as in the body as relative to the coordinate system.  Then   d will not coincide with the direction of angular velodt  city  . 29

Textbook on Theoretical Mechanics Questions 1. Give the definition of rotational motion of rigid body. 2. How many degrees of freedom has rotating rigid body? 3. Give the notion of moment of vector. 4. How can be defined the velocity of any point of rotating solid body? 5. How can be defined the acceleration of any point of rotating solid body?

PRACTICE 1.17. Axle begins to rotate uniformly accelerated from the state of rest. In the first 5 seconds it makes 12.5 turns. What is its angular velocity after these 5 seconds? Answer:   10 rad / sec . 1.18. The wheel having its fixed axis became initial angular velocity of 2 рад / сек . After 10 turns it stopped because of friction in bearings. Define angular acceleration  of the wheel assuming it constant. Answer:   0,1 rad / sec2 . 1.19. Point A of the pulley situated on its band is moving with the velocity of 50 centimeter per second. And point B taken on the same radius as point A is moving with the velocity 10 centimeter per second. Distance AB is 20 centimeter. Define angular velocity  and diameter of the pulley.

Answer:   2 rad / sec, d  50 cm . 1.20. Fly wheel with the radius of 2 meters rotates uniformly accelerated from the state of rest. Points situated on the band become linear velocity v  100m / sec in 10 seconds. Find the velocity, tangential and normal acceleration of the points of the band for the moment t  15sec . 30

1. Kinematics

Answer: v  150 m / sec, wn  11250 m / sec 2 , w  10 m / sec 2 . 1.21. Axel with the radius of R  10 centimeters is implemented in rotational motion with the load P hung on it with a band. Motion of the load is expressed with the equation x  100t 2 , where x is the distance between the load and the place where the band hangs from the axel, t is time in seconds. Define angular velocity  and angular acceleration  of the axel and magnitude of acceleration w on the surface of the axel in moment t. Answer:   20t rad/sec,   20 rad / sec2 , w  200 1  400t 4 cm / sec2 . 1.22. The body makes 3600 turns in first 2 minutes when it begins to rotate uniformly accelerated from the state of rest. Define its angular acceleration. Answer:    rad / sec 2 . 1.23. Airscrew rotating with angular velocity 40 rad / sec made 80 turns before its stop after turning off the engine. How much time it was passed from the moment of turning off the engine before the stop? Assume rotation of the airscrew uniformly retarded. Answer: 8 sec. 1.24. Define velocity and acceleration of the point situated on the surface of the Earth in St. Petersburg taking into account only rotation of the Earth around its axis. Latitude of St. Petersburg is 60 degrees, radius of the Erath is 6370 km. Answer: v  232 m / sec, w  0,0169 m / sec 2 . 1.25. Pendulum is oscillating in vertical plane around fixed horizontal axis O. It becomes its maximum deviation    / 16 rad in 2 / 3 seconds. 1) Find the equations of oscillation of pendulum assuming that it makes harmonic oscillations 2) What is the maximum angular velocity of pendulum? In what position the pendulum has its maximum angular velocity? 31

Textbook on Theoretical Mechanics

Answer: 1)    / 16 sin 3 t rad 2) in plumbness  max

4 3   2 rad / sec 2 . 64

1.26. Find horizontal velocity v which is required to give to the body situated on the equator of the Earth so that it has free fall acceleration when it is moving uniformly around the Earth along the equator. Define the time T after what the body will turn back to its initial position. Radius of the Earth is 6370 km, free fall acceleration is 9,87 cm / sec2 . Answer: v  7,9 km / sec, T  1,4 h. 1.27. Find the equation of rotation of the disc of steam turbine at the beginning of motion if it is known that the rotation angle is proportional to the cube of time and angular velocity of disc equals   27 rad / sec when t=3 sec.

Answer:   t 3 rad . 1.28. Inclination angle of total acceleration of the point of rim of wheel to the radius equals 60 degrees. Tangential and normal acceleration at the present moment equals w  10 3 cm / sec 2 . Find the normal acceleration of the point situated at the distance r=0,5 m from the rotation axis. Radius of the wheel is R=1m. Answer: wn  5 cm / sec 2 . 1.29. Fly wheel of radius 0,5 m rotates uniformly around its axis, velocity of the points lying on its rim equals 2m/sec. How many revolutions per minute make the wheel? Answer: n=38,2 rev/min. 1.30. Clock balance wheel performs torsional oscillations with the period T=1/2sec. The maximum deviation angle of the point of the rim of wheel from the position of equilibrium is    / 2 rad . 32

1. Kinematics

Find the angular velocity and angular acceleration of balance wheel in 2 seconds after the moment when balance wheel pass the position of equilibrium. Answer:   2 2 rad / sec,   0 .

SELF STUDY OF THE STUDENT

The motion of the load 1 must be described by the equation: x  c 2 t 2  c1t  c0 where t is the time in seconds, c02 – some constants. Coordinate of the load must be x0 and velocity v0 at the initial moment (t=0). Besides, it is necessary that coordinate of the load were equal x 2 at the time t  t 2 . Determine the coefficients c0 , c1 , c2 at which the required motion of the load 1 takes place. Determine the velocity and acceleration of the load and the point M of one of wheels of mechanism at the moment of time t  t1 . Number of variant 1 2 3 4 5 6 7 8 9 10 11 12

Radii, cm

R2 60 58 80 100 45 40 20 15 15 20 15 40

r2 45 45 60 35 25 15 10 10 15 10 30

R3

r3

Coordinates and velocities Moments of the load 1 of time, sec x0 , cm v0 , cm / sec x2 , cm t2 t1

36 60 45 30 105 20 10 15 20 10 20 30

30 15

2 4 3 7 8 9 5 5 8 3 5 5

33

12 4 15 16 5 8 10 2 4 12 10 9

173 172 102 215 124 65 179 189 44 211 505 194

3 4 3 4 4 2 3 4 2 4 5 3

2 3 2 2 3 1 2 2 1 1 3 2

Textbook on Theoretical Mechanics

34

1. Kinematics

1.5. Plane-parallel motion of a rigid body. Velocities of the points of a plane figure. Instantaneous center of velocity Plane-parallel motion of a absolutely rigid body.

When each point of a body moves in the plain that is parallel to some fixed plane this motion is called plane or plane-parallel motion of a rigid body. A plane figure formed by the section of the body by this stationary plane Q stays in this plane all the time of motion. Let’s consider two points of the body located on one perpendicular to the fixed plane Q. Point M1 moves in the plane Q1||Q, the point M2 moves in the plane Q2||Q. The segment M1M2 stays perpendicular to Q when moving, therefore, it stays parallel to itself. 35

Teextbook on Theoreticaal Mechanics

Thus alll points of the perpendicular p M1M2 ddescribe identicall and parallel to eachh other trajectoriies and have geometrrically equal veelocities and acceleraations. That is, trajectories A1B1, A2B2, AB are identical i and parallel and their velocitties and acce   lerationns equal to v  v  v and 1 2    w1  w 2  w respectiveely at the mo-

ment. Thuus, the motion of each point of the plaane figure in a fixed plane determiines the motionn of all the poin nts of a rigid body y, located on the pperpendicular to thhe plane Q. It allow ws to reduce the study of plane m motion of a rigidd body to the study y of the motion off a plane figure inn its plane. As the position of o a plane figuree on the plane deefined by the posittion of two pointss or a segment conntaining two pointts, the motion of a plane figure can be b studied as a m motion of rectilineaar segment in p this plane. d rotational motionn are the main typpes of motion Translational and of fig gure in its plane. f moves When every linee in the plane oof the moving figure paralllel to itself this motion m is called trranslational motioon of a plane figurre. t center of When one pointt of the moving figure, called the rotattion, is fixed this motion is called rrotational motionn of the figure in itss plane. Here all the t points of the figure moves alonng the circles with h centers at the cen nter of rotation. And v М  OM   , w  OM   2   4 , tg  w   M wn  2 (Accceleration deviatees from the radiuus of rotation at an angle  whicch is the same for all points). 36

1. Kinematiccs

Theorem 1. An ny displacementt of a plane figuree in its plane can be b composed of translational t motiion and rotation n about an arbittrary center (pole) (see the figurre). Theorem 2. Everry displacementt of a plane figuree in its plane that is not translatio onal can be impllemented by one turn around the appointed a center, called the centerr or pole of finite rotation (see the figure). f These theorems inform about thhe displacement of the plane figurre from one fixeed position to annother. But one can c have the geom metrical descriptiion of the plane figure motion. Any motion, inccluding the motiion of plane fig gure in its plane can be treated as a continuou us sequence of elementary displlacements, which can be represen nted in two ways. 1) According to Theorem 1 elem mentary displacement can be obtaiined by an infinitesimal translationnal displacement with w arbitrary seleccted pole and turn n at an infinitely small angle arou und this pole. This implies, that any motion of a planee figure can be co onsidered as a comb bination of translational motion ddetermining by th he motion of arbittrary selected polee and rotational m motion around this pole. 2) According to Theorem 2, anyy elementary disp placement of figurre can be achieveed by turning onn only one infinittesimal angle arou und a specified cen nter, called the innstantaneous centeer of rotation. Therrefore, any motion n of a plane figurre that is not tran nslational can be considered c as a continuous c sequeence of infinitesim mal rotations abou ut the instantaneeous center of rrotation (the possition of the instaantaneous center of o rotation is channging continuously y). Rotation around any pole or aroound the instantaaneous center occu urs with the saame angular veelocity for this moment t: 37

Teextbook on Theoreticaal Mechanics

 (indepeendent of the t chooice of pole).  is i the angular velocity of the fiigure at the i mooment or the instantaneous anggular velocity. Instantaneous ceenter of rotationn is the end position of the cennter of end rotattion О when possition П2 tends too position П1. Thatt is, the figure mo oves from the givven position to innfinitely close adjaccent position by elementary e rotatioon around the poiint that is the instaantaneous center of o rotation. Velocities of all the points А1, А2,,…, Аn of plane figure f will be perpendicular to the radius of rotatiion А1P, А2P,…,, АnP at the ment. Velocity of a point of plane fi figure which coinccides with the mom poin nt P, will be equall to zero at the m moment. This poinnt of figure is calleed the instantaneous center of veloccities. Knowing thhe position P, the direction of vellocity of any pooint of plane figgure can be deterrmined at the giveen moment t. To determine the position of instanntaneous center of o velocity we mustt know the directiion of velocity off any two points of the figure. Resttoring the perpen ndiculars from thhese points we will get the instaantaneous center of o rotation P at thee point of their inttersection. Centroids. Durin ng the motion tthe position of instantaneous i centeer of rotation of plane figure channges continuouslyy both on the fixed d plane and on a plane associated with a movingg figure. The locus of the instantan neous centers of rotation on a fixeed plane is a hich is called a fixxed centroid. Thee locus of the contiinuous curve, wh instaantaneous centers of rotation on a moving plane, related r to the mov ving figure is also a continuous curvve, which is called the moving centrroid. t when the fiigure is moving the moving This implyies, that centrroid is rolling without w sliding onn a fixed plane (the ( point of contact, which is the instantaneous cennter of velocity for fo the figure, has a velocity equal to o zero).

  lim

t 0

38

1. Kinematiccs

Velocitiies of the points oof a plane figure Let the plane fig gure moves relatiive to the origin of reference fram me  . Point А is the po ole and  is its A

posittion vector, po oint М is the  arbittrary point of the figure and  M

its position vector. Then at any    mom ment of time t:     r . M A Differentiating  time t d M

with respect  d A dr to   . As dt  dt dt AM=const, then r changes only by the directioon, therefore

 dr   . r dt  Hence, v

M

       v A    r or v M  v A  v MA

Velocity of any point M is compposed of velocity y of arbitrary seleccted pole A which h is common to alll points of the fig gure (velocity    of trranslational motiion of the body)) and velocity vMA    r ,



whicch occurs due to the rotation of figure around the pole p A. v MA direccts perpendicularly to МА in the dirrection of rotation n of figure.

v MA    AM . Thus, knowing the velocity of any point A of pllane figure and angu ular velocity of figure  , we can find the velocity of any a point of the figurre. Theorem. If it is known the velocity of any point of plane figure aand direction of velocity v of its otherr point, it is possiible to determine the velocity of any a point of a figurre using the instan ntaneous center off rotation. 39

Textbook on Theoretical Mechanics

Theorem. The projections of the velocities of the ends of invariable segment on its direction are equal. Questions 1. Give the definition of plane parallel motion. 2. Describe the two main approaches of setting the plane parallel motion. 3. What is the instantaneous center of rotation? 4. How can be defined the instantaneous center of velocity? 5. How can be defined the velocities of the points of a plane figure.

1.6. Acceleration of the points of a plane figure. Instantaneous center of acceleration

The motion of plane figure in its plane can be considered as a combination of two movements: translational motion of figure together with its pole and its rotation around the pole. Theorem. Acceleration of any point of a plane figure is the geometric sum of the acceleration of a pole and the acceleration of this point in the rotational motion around the pole.

   wM  w A  w

MA

  rot  cen wMA  wMA  wMA – total acceleration of the point M in the rotation around the pole A. by magnitude: rot cen    r    AM ; wMA   2  r   2  AM , wMA rot 2 cen 2 wMA  wMA  wMA  MА  2   4 ; tg 

rot wMA cen wMA



 . 2

 rot The direction of wMA always corresponds to the direction of  . You should not mix the notions of normal acceleration of the point with the notion of centripetal acceleration around the pole, and the notion of tangential acceleration with the notion of rotational acceleration around the pole. 40

1. Kinematiccs

 pend on the polee and is perpend dicular to the wn does not dep

velocity, that is, it is directed along thhe instantaneous radius to the  instaantaneous center of velocity. wcenn depends on thee pole and is  alwaays directed radiaally towards the ppole. w is directted along the  velocity of the point or o opposite to v , that is, it does not n depend on  p ds on the choice of the pole and is perpendicular p the pole, wrot depend to th he line AM. Instan ntaneous center oof acceleration Theorem. At each h moment of mottion of plane figurre in its plane (if   0,   0 ) theree is a point of a pllane figure whosee acceleration equaals to zero at this time. t Point Q, which haave the acceleratioon equal to zero at a this time, is calleed the instantaneouus center of accellerations. If we assume thhe instantaneous centeer of accelerationns as a pole then the acceleration a of anny point of plane figurre can be determined d as:

   wM  wQ  wMQ . Similarly for any 







poin nt B, K, etc. wB  wQB, wK  wQK . By magnitude: w A  QA  2   4   w QB wK QK K wK  QK K  2   4  B  ;  w A QA w A QA A wB  QB  2   4  

That is, acceleraations of the poiints of plane fig gure are proportiional to the distannces of these poinnts to the instantaaneous center of accelerations at each moment of time. The angle between the p and the seggment connecting g it with the acceleration of the point o acceleration is the same for all points of the instaantaneous center of figurre and equal to  . 41

Textbook on Theoretical Mechanics

In general case, the instantaneous center of velocity and instantaneous center of acceleration are different points. If the figure performs only rotational motion, the centers P and Q coincide with the fixed center of rotation of a plane figure. Particular cases of finding the instantaneous center of acceleration

1. Let

  0,   0 (  const , or   min ) , then max

tg 

  0    0. 2

Consequently, the instantaneous center of acceleration Q is located at the point of intersection of straight lines, along which the accelerations of points of plain figure are directed. All accelerations  are directed to Q, as wQrotA, B ,C  0 , that is:

   rot  cen   cen w А  wQ  wQA  wQA  w А  wQA 2.   0,   0 . This is possible when it is the instant translational motion. Then tg         . This angle is measured 2  2  from w in the direction of  . Instantaneous center of acceleration lies at the point of intersection of perpendiculars to accelerations.

   rot  cen  rot w A  wQ  wQA  wQA  wQA . Methods of calculating the angular acceleration

1. Rotation angle  or angular velocity  in dependence of time t is known, then:



d d 2 .  2 dt dt 42

1. Kinematics

2. Angular velocity is found with the help of instantaneous center of velocity:   v A . Differentiating in terms of time:

AP



d d  1 . 1 dv A   vA   dt AP dt dt  AP 

а) AP  const all the time of motion, therefore, 1 dv A 1 .   wA AP dt AP б) AP  AP(t ) is a variable value all the time of motion. Than  can be found by forming the equation:

   rot  cen wM  w A  wMA  wMA . 

rot Here wMA    MA . Hence, we find

.

Questions 1. How can be defined the acceleration of the points of plane figure? 2. How can be defined the rotational and centripetal acceleration? 3. What is the difference between normal and centripetal acceleration? 4. What is the difference between rotational and tangentional acceleration? 5. What is angular acceleration and how it can be defined?

PRACTICE 1.31. In crank mechanism the length of the crank OA=40 cm, length of link AB=2 m; crank rotates uniformly with angular velocity 6π rad/sec. Find the angular velocity ω of the link and the velocity of its middle point M at four positions of crank for which the angle AOB respectively equals to 0, π/2, π, 3π/2. 43

Textbook on Theoretical Mechanics

6 5 II.   0 rad / sec, v M  754m / sec 6 III.   rad / sec, v M  377m / sec 5 IV.   0 rad / sec, v M  754m / sec . The minus sign in the expres-

Answer: I.    rad / sec, v M  377m / sec

sion shows that the link rotates in direction opposite to cranks'. 1.32. Define the velocity of the piston E of pump drive mechanism in the position showed in a picture if OA=20 cm, O1 B=O 1 D . Crank OA rotates uniformly with angular velocity of 2 rad/sec.

Answer: 46,2 cm/sec. 1.33. Sliders B and E of the twinned crank-slider mechanism joined with bar BE. The leading crank OA and the driven crank OD swing about common fixed axis O that is perpendicular to the picture plane. Define the instantaneous angular velocities of the driven crank OD and of the link DE at the time when leading crank OA that has an instantaneous angular velocity ω 0=12 radpsec. is perpendicular to the slider guide. Sizes given: OA = 10 cm, OD = 12 cm, AB = 26 cm, EB = 12, DE= 12 3 сm .

Answer: OD  10 3 rad / sec,  DE  44

10 3 rad / sec . 3

1. Kinematics

1.34. To the middle D of a bar AB of a parallel link mechanism OABO1 connected (with the help of joint D) bar DE that sets the slider K in reciprocating motion. Define the velocity of slider K and the angular velocity of the bar DE in position showed in a picture if OA=O1 B=2DE=20 cm and the angular velocity of the link OA equals at the moment to 1 rad/sec. Answer: υ K =40 cmpsec , ω DE =3,46 radps 1.35. The wheel rolls without sliding in a vertical plane against the inclined straight way. Find the acceleration of the ends of two orthogonally related diameters of the wheel (one of diameters is parallel to the rail) if at concerned period of time velocity of center of the wheel is vo  1m / sec , acceleration of the center of the wheel wo  3m / sec 2 , radius of the wheel is R = 0,5 m. Answer: w1 = 2 m/sec2, w2 = 3,16 m/sec2, w3 = 6,32 m/sec2, w4 = 5,83 m/sec2 1.36. Square ABCD with side а makes plane motion in a plane of picture. Find the position of instantaneous center of acceleration and the accelerations of its vertices C and D if known that at the moment the accelerations of two tops A and B are equal to 10cm / sec 2 . Direction of accelerations of points A and B coincides with the sides of the square as shown in the picture. 45

Textbook on Theoretical Mechanics

Answer: wC  wD  10cm / sec 2 and directed against the sides of the square. Instantaneous center of accelerations is in the point of intersection of diagonals of the square. 1.37. The bar OB rotates about the axis O with constant angular velocity − 1 and sets in motion the bar ω=2 c AD which points A and C move along the axes: A moves along the horizontal axis Ox, C moves along the vertical axis Oy. Define the velocity of point D of a bar at   45 and find the equation of the path of this point if AB=OB=BC=CD=12cm. 2 2 x y Answer: υ D=53,66 cmpsec , ( ) + ( ) =1 12 36 1.38. Onto the axis O there are pinned a tooth-wheel K of diameter 20cm and the crank OA of length 20 cm that are not connected with each other. The tooth-wheel L of diameter 20cm is tightly connected with the link AB of length 1m. The wheel K rotates uniformly with angular velocity that is equal to 2 π radpsec . Taking teeth of the wheel L wheel K sets in motion link AB and crank OA. Define the angular velocity ω1 of the crank OA in 4 positions: two horizontal and two vertical.

Answer: I . ω1=

10 10 π radpsec , III . ω 1= π radpsec , II . ω 1=π radpsec , IV . ω 1=π radpsec 11 9

46

1. Kinematics

1.39. The load K that is connected via non-stretching thread with the coil L moves down vertically 2 under the law x  t m . The coil L rolls without sliding against the fixed horizontal rail. Define the accelerations of points A, B and D lying on the rim of the coil, its angular velocity and angular acceleration at the period of time t=0,5sec in the position shown in the picture; OD=2 AD  OB OC=0,2 m. Answer:

wA=20,9 cm/sec2, wB=22,4 cm/sec2, wD=20,1 cm/sec2, ω=10 rad/sec, ε=20 rad/sec

1.40. The bar AB of length 0,2 m makes plane-parallel motion. Accelerations of is ends A and B are perpendicular to AB and directed at opposing sides and its magnitude equal to 2m / sec 2 . Find the angular velocity, angular acceleration of the bar and the acceleration of its middle C. Answer:   0,   20rad / sec, wC  0 1.41. The crank OA rotating with angular velocity 0  2,5rad/ sec around the axis O of fixed wheel of radius r2  15cm sets in motion the gear of radius r1  5cm stuck on the end A of the crank. Determine the magnitude and direction of the velocities of the points A,B,C,D,E of the mobile gear if СE  BD . Answer: v A  50cm / sec, vB  0, vD  100 cm / sec, vC  vE  70,7cm / sec .

47

Textbook on Theoretical Mechanics

1.42. The wheel rolls along the plane making an angle 30 degrees with the horizon. Center O of the wheel moves in accordance with the law x0  10t 2 cm, where x – axis directed parallel to the inclined plane. The rod OA=36cm is suspended to the center of the wheel O and oscillates around the horizontal axis O, perpendicular to   the plane of the picture in accordance with the law   sin t rad . 3

6

Find the velocity of the end A of the rod AO in the moment of time t=1 sec. Answer: velocity equals 2,8m/sec and directed down parallel to the inclined plane. 1.43. Straight line AB moves in the plane of the picture so that its end A situated on the half-circle CAD all the time and the line itself pass through the fixed point C of diameter CD. Determine the velocity vC of the point of line coinciding with the point C in that moment when radius OA is perpendicular CD and it is known that velocity of the point A in this moment equals 4m/sec.

Answer: vC  2,38m / sec . 1.44. The load K connected by means of the unstretched thread with the coil L descends vertically down under the law x  t 2 m . The coil L rolls without sliding along the fixed horizontal rail. Determine the velocities of the points C,A,B,O,E of the coil in the moment t=1 sec in the position showed in the picture. Determine angular velocity of the coil if AD  OE , OD=2OC=0,2m. Answer: vC  0,vA  6m/ sec,vB  4m/ sec,vO  2m/ sec,vE  4,46msec,  20rad/ sec. 48

1. Kinematics

1.45. The bar OA of the fourbar linkage OABO1 rotates at the constant angular velocity ω 0 . Define the angular velocity, angular acceleration of the bar AB and also the acceleration of the joint B in position shown in the picture if AB=2 OA=2 a .

Answer:   0,  

3 2 3 0 ,  B  a 02 . 6 6

1.46. The crank OA=20cm rotates around the fixed axis O perpendicular to the plane of the picture with angular velocity 2 rad/sec. A gear 2 of radius 10cm and fastened internally with the fixed wheel 1 is stuck to the end A of the crank. Determine the velocities of the points B,C,D,E lying on the rim of the gear 2 if BD  OC .

Answer: vC  0, vB  vD  40 2cm / sec, vE  80cm / sec . SELF STUDY OF THE STUDENT

Find the velocities and accelerations of points B and C for a given position of mechanism and also an angular velocity and angular acceleration of the link to which these points belong. Note. ωOA And ε OA are angular velocity and angular acceleration of the crank OA at specified position of the mechanism; ω1 is an angular velocity of the wheel I (constant); υ A and a A are velocity and acceleration of the point A. Rolling of wheels is without sliding. 49

Textbook on Theoretical Mechanics Number of variant 1 2 3 4 5 6 7 8 9 10 11 12

Dimensions, cm OA 30 35 25 35 25 40 12 10 20 40 -

r 15 15 15 15 50

AB 75 80 30 35 10 -

AC 8 45 20 60 20 15 8 15 5 10 8 -

I , rad / sec rad / sec

 OA ,

3 4 1 5 3 4 2 1 2 -

2 8 1 10 0 6 6 0 2 -

ОА ,

12 2,5 -

50

rad / sec

2

vA ,

wA ,

cm / sec

cm / sec 2

10 10 50

0 0 100

1. Kinematics

51

Teextbook on Theoreticaal Mechanics

1.7 7. Compound mo otion of a point. F Full and relative derivatives n of velocities of thee vector. Addition

Motion of a poinnt relative to two or several referennce frames is calleed compound mootion. For examplle, a boat crossingg the river, a passeenger moving in the carriage of tthe moving train,, a passenger mov ving along the deckk of the ship, etc. Suppose that the point is moving rrelative to a movinng coordinate system Oxyz, which is moving relative to the main (fixed) coordinate system  . The mottion, velocity   and accceleration of the point p ( vr , wr ) consideered in relation too the system Oxyz iss called relative. The motion, velocityy and accelerationn of the point relative to the system  is called    absolutee ( va , wa ). The motion m of the mov ving coordinate syystem Oxyz relativve to a fixed coorddinate system   is called transpportation. In transsportation motionn velocity and acceleration is consid dered relative to the point wheree the moving poin nt is located at the moment. In otherr words transportaation velocity and acceleration is th he velocity and accceleration of thee point which doessn't execute the rellative motion. For example, if a person walks aalong a radius off the rotating platfform, the platform m can be associaated with a moviing reference fram me, and the surfacce of the earth w with the fixed refeerence frame. Then n motion of the platform p is transpportation, motionn of a person relattive to the platforrm is relative, mootion of a person in relation to the Earth E is the absoluute motion. The main problem m of studying thee compound motion is to establish h the relationship p between relativee, transportation and absolute velocities and acceleraations of the pointt. Suppose that thee point is movingg relative to a moving m frame r to the baasic (fixed) coorddinate system Oxyzz that moves in relation   . 52

1. Kinematiccs

Then the motion, velocity, and accceleration of a poin nt considered in reelation to the systtem Oxyz are callled relative, in reelation to the systeem  – absoolute (the motion is also called com mpound). The motiion of the movingg system Oxyz reelative to the fixeed coordinate systeem  is callled transportationn motion. Velociity and accelerattion of a point whhere the moving ppoint is located att the moment are called c transporatioon. Addition of vellocities

t compound Let's consider the motiion of the point М. This point execcutes relative displacement (vecttor MM  ) along the trajectory АВ for f a period of tim me t  t1  t . Curv ve АВ transferress to the curve o time t . At A1B1 for the period of the same time the point p M of the curv ve АВ where the point p M located at th he moment t execuuting the transportaation displacemennt mm1  M M 1 . As result the p point M will comee to the positionn M 1 and execuute the absolute displacement MM M1 for the time t :

MM 1  Mm1  m1M 1 a proceeding too the limit: Dividing by t and MM 1 Mm1 mM  lim  lim 1 1  t  t t 0 t 0 t 0 t lim

By definition, ass when t  0 tthe curve A1 B1  AB , then we have: h 53

Textbook on Theoretical Mechanics



MM  = v m1M 1 = rel lim lim  t t 0 t 0 t



Mm1

  vr



= vtr  vt lim t 0 t MM 1  vabs  va lim t 0 t As result: v  v  v ( vr , v , v directed along the tangent to the t a a t r appropriate trajectories). Theorem on the addition of velocities: Absolute velocity equal to geometric sum of relative and transportation velocity in compound motion. By magnitude va  ve2  vr2  2vr2ve2 cos  , where α is an angle





between the directions of v and v . t r Absolute and relative derivatives of the vector

Suppose that moving and fixed coordinate system have a common origin O.  is the instantaneous angular velocity of the moving coordinate system relative to the fixed coordinate system. Consider a point M,  executing a motion, regardless of Oxyz. Then vector r  OM chan ges differently in the both systems. During the time t vector r will have different increment in different systems:

 r

( O )

~ and  r

(Oxyz )

dividing by t and, proceeding to the limit, we will get: ~ ~   dr r dr r  lim  lim – absolute or total derivative and dt t  0 t dt t  0 t

– relative or local derivative. 54

1. Kinematics

Let's find the relationship between these derivatives. From the definition of v and v it follows that, r a

~  dr vr  , dt

     dr va  , as va = vr + v t dt



where v is velocity of the point steadily associated with Oxyz, t   where the point M is located at the moment. Then v e    r (Euler formula for any point of rotating solid body).  ~  da d~a  dr d r       r or, for arbitrary vector a :  a . dt dt dt dt     Let's deduce this formula in another way: a  axi  a y j  az k . Differentiating with respect to t :

    da x  da y  da z  da di dj dk (  a y  az ) i j k )  (a x dt dt dt dt dt dt dt

   di , dj , dk – velocities of the ends of unit vectors, that is velodt dt dt cities of coordinate trihedral Oxyz, therefore, in accordance with     di Euler formula: =   i , d j =   j , dk =   k . dt dt dt Then:  ~  d~a  d~a     da da    ax   i  ay   j  az   k     (axi  ay j  az k )     a dt dt dt dt 

~

dt

dt

Finally we have: da  d a    a . Questions 1. What is compound motion of a mass point? Give the examples.

55

Teextbook on Theoreticaal Mechanics 2. What is relative mootion of mass point? G Give the examples. 3. What is transportaiional motion? Give thhe examples. 4. How can be defineed the absolute velocitty in the compound motion? m 5. How can be definned the magnitude off absolute velocity inn the compound motion?

1.8. Theorem on the addition of acceleratio ons (Coriolis theoorem) Addition of accellerations

Abbsolute accelerattion of the point М equals to the derivative of absolutte velocity of the point p M:

va 

d d 0 dr ,   dt dt dt

r  xi  y j  z k ,  dva  wa  dt

Differentiating with w respect to t, w we will get:            dj d 2i d 2k dk (1) di d2 j  wa  0  x 2  y 2  z 2  xi  yj  zk  2( x  y  z ) dt dt dt dt dt d dt

Let’s divide the terms of the right side of this equation into three grou ups. The first group comprises thee terms containin ng derivatives only y on the relative cooordinates x, y, z, but do not contaiin derivatives of th he vectors, that is:     wr  xi  yj  zk The second groupp comprises the terms which only contain deri    vativ ves of vectors  0 , i , j , k but do not contain the deriv vatives of the relattive coordinates x,, y, z, that is: 56

1. Kinematics

    d 2i d2 j d 2k .   wt   0  x 2  y 2  z 2 dt dt dt

There is one group of terms, let's denote it by:

    di dj dk wc  2( x  y  z ) dt dt dt Each group represents a certain acceleration. Let us find out their physical meaning.  The acceleration wr is calculated as if the moving system Oxyz  rested, and the point M moved. Therefore wr is relative acceleration  of the point M. The acceleration we is calculated on the assumption that the point M rests relative to the moving coordinate system Oxyz and moves together with this system with respect to the fixed  coordinate system  . Therefore we is transportation acceleration  of the point M. The third group of terms does not refer to w r or to wt .     Accounting that di =   i , dj =   j , dk =   k the third accedt dt dt leration can be transformed to:

 wc  2  v r

(2)

This acceleration is called Coriolis acceleration, as it appears in case of rotation of a moving coordinate system. From a physical point of view, the appearance of the Coriolis acceleration of the point takes place due to the mutual influence of transportation and relative motions. Equation (1) now takes the form:

    wa  we  wt  wc Equation (3) represents the acceleration addition theorem: 57

(3)

Textbook on Theoretical Mechanics

Coriolis theorem: Absolute acceleration of the point equals to the vector sum of transportation, relative and Coriolis acceleration. From (2) it follows, that w  2 v sin( , v ) . c e r e r The direction of the Coriolis acceleration w is determined by c the corkscrew rule. Let's consider the cases when Coriolis acceleration is equal to zero:  =0; 1. If transportation motion is translational then  e =0  w c  2. If v e , i.e. relative motion of the point has a direction

r

parallel to the axis of transportation rotation then w =0; c

3. Point don't move relative to the moving coordinate system   then v =0  w =0. r c Questions 1. How can be defined the absolute acceleration of a mass point in compound motion? 2. What is Coriolis acceleration? 3. Give the examples of compound motion where Coriolis acceleration takes place. 4. When Coriolis acceleration equals to zero? 5. What is the formulation of Corilolis theorem?

PRACTICE 1.47. The passenger of a car that moves with the velocity of 72 km per hour along the horizontal highway sees through the window paths of raindrops that are inclined against the vertical under the angle of 40 O . Define the absolute velocity of falling of raindrops neglecting the friction of drops against the glass.

Answer: v a 

ve tg 40 

 23.8 m / sec .

58

1. Kinematics

1.48. The ship goes to the south with the velocity 36 2 km/ hour . The second ship goes to the south-east with the velocity 36 km/hour. Find the magnitude and the direction of velocity of second ship that is defined by observer from the first ship. Answer: v r  36 km / hour directed at north-east. 1.49. The balls of Watt centrifugal governor that rotates around the vertical axis with angular velocity ω=10 radpsec move away from this axis due to the change of the load. The balls get in that position at angular velocity of ω1 =1,2 radpsec . Find an absolute velocity of the governor balls at the concerned time if the length of bars is l=0,5 m , distance between axes of its suspension is 2 e=0,1 m , angles made with an axis by governor bars O α 1=α 2=α=30 . Answer: v  3.06 m / sec . 1.50. In the link gear the crank OC swings about the axis O that is perpendicular to the picture plane, the slider A moving along the crank OC sets in motion the bar AB that moves in vertical ways K. The distance OK =l . Define the velocity of slider A relative the crank OC as the function of angular velocity ω and rotation angle φ of the crank.

Answer: v r  l tg . cos 

1.51. A truck on which an engine is placed moves horizontally to the right with constant acceleration w  0.4m / sec 2 . The engine 59

Textbook on Theoretical Mechanics

rotates under the law φ=1 /2 t 2 . Define the absolute acceleration at time t  1 sec for four points M 1 , M 2 , M 3 , M 4 of the rotor being at the distance l  0,2 2 m from the rotor axis and taking the position shown in a picture. Answer: w1  0,4 2 m / sec2 , w2  0 m / sec2 , w3  0,4 2 m / sec2 , w4  0,8m / sec2 .

1.52. A disc rotates about the axis that is perpendicular to the plane of disc clockwise uniformly accelerated at angular velocity of 2 1 radpsec ; at time t =0 its angular velocity equals to zero. A point M oscillates along one of diameters of disc so that coordinate of the point is ξ =sin π t m , where m – meters, t is in seconds. Define the projection of absolute acceleration of the point M at the axis ξ , 2 η connected with a disc at time t  1 sec . 3 Answer: ω ξ =10,95 mpsec 2 ,ω η=− 4,37 mpsec3

1.53. The point moves uniformly with the relative velocity υ r along the disc chord. The disc rotates about its axis O that is perpendicular to the disc plane with constant angular velocity ω . Define the absolute velocity and acceleration of the point at time when it is in a shortest distance h from the axis under the assumption that relative motion of the point is sideways the disc rotation. Answer: υ=υr + h ω ,ω=ω 2 h+ 2 ω υ r 1.54. The bicyclist moves along the horizontal platform that rotates about the vertical axis with constant angular velocity   1 / 2rad/sec ; the distance of the bicyclist to the platform rotation axis is constant and equals to r =4 m . The relative velocity of the 60

1. Kinematics

bicyclist is υ r =4 mpsec and it is directed at the side opposite to the transportation velocity of the appropriate point on the platform. Define the absolute acceleration of the bicyclist. Find at what relative velocity he should move in order that his absolute acceleration equals to zero. Answer: 1) w  1m / sec 2 directed along the radius to the disc center; 2) υ r =2 mpsec . 1.55. A compressor with straight channels rotates uniformly with the angular velocity of ω about the axis O that is perpendicular to the picture plane. The air flows along the channels with constant relative velocity υ r . Find the projections of absolute velocity and acceleration on the coordinate axis for the air particle that is in the point C of channel AB according to next data: the channel AB is inclined to the radius OC at an angle 45 o , OC = 0,5 m, ω=4 π radpsec , υ r=2 mpsec . Answer: v  7,7 m / sec,v  1,414m / sec, w  35,54m / sec2 , w  114,5m / sec2 . 1.56. A square ABCD with side 2a m rotates about the side AB with constant angular velocity    2 rad / sec . A point M executes harmonic oscillations along the π diagonal AC under the law ξ =a cos t m , 2 where m – meters. Define the magnitude of absolute acceleration of the point at t=1sec and t=2 sec . Answer: wa  a 2 5 m / sec2 , wa  0,44a 2 m / sec2 . 61

Textbook on Theoretical Mechanics

1.57. A point M moves along the radius of the disc in the direction from the disc center to its rim under the law OM =4 t 2 cm . The disc rotates about the axis O1 O2 at angular velocity ω=2 t radpsec . Radius OM makes with the axis O1 O2 an angle 60o . Define the magnitude of absolute acceleration of the point M at time t =1 sec .

Answer: w м  35,55 сm / sec 2 . 1.58. The ball P moves along the disc chord AB with the velocity 1,2 m/sec from A to B. The disc rotates around an axis that is perpendicular to the plane of the disc and situated in its center. Find the absolute acceleration of the ball when it is at its shortest distance from the center of the disc equals to 30 cm. At this moment angular velocity of the disc is 3 rad/sec, angular acceleration is 8 rad/sec2. Answer: wа  10,18 m / sec2 . 1.59. In the hydraulic turbine water from the wicket gate gets into rotating working wheel the blades of which are set to avoid the hit of entered water so that relative velocity υ r touched the blade.

Find the relative velocity of water particle on an outward rim of the wheel (at time of entering) if its absolute velocity at entrance is 62

1. Kinematics

υ=15 mpsec , the angle between absolute velocity and radius is o α=60 , entrance radius R=2m, angular velocity of the wheel is

π radpsec 

Answer: vr  10,106 m / sec, (vr , R)  41 50 ' . 1.60. A stone А of the rocking link of the planing machine is set in motion by tooth gear that consists of rackwheel D and rackwheel E and carrying on itself the stone axis A as a finger. Rackwheels radii are R=0,1 m , R1=0,35 m ,O 1 A=0,3 m ,

distance between the axis O1 of the rackwheel E and center B of the link rolling O1 B=0,7 m . Define the angular velocity of the link at moments when segment O1 A either vertical (upper and lower positions) or perpendicular to the link AB (left and right positions) if the rackwheel has angular velocity ω=7 radpsec . Points O1 and B are situated in one vertical. Answer: ω1 =0,6 radpsec , ω II =ω IV =0 , ω III =1,5 radpsec . 1.61. Hollow ring of radius r hard connected with the axle AB so that the axis of the axle is in the plane of the ring axis. The ring is filled with liquid that moves in it in the direction of arrow at constant relative velocity u. The axle AB rotates in the direction of motion of clock hand when looking along the rotating axis from A to B. Angular velocity of the axle ω is constant. Define magnitudes of absolute accelerations of liquid particles located in points 1, 2, 3, 4. 63

Textbook on Theoretical Mechanics

2 Answer: w 1=r ω −

u2 u2 u2 2 2 , w3 =3 r ω + , w 2=w4 =2 r ω + r r r

1.62. In accordance with the conditions of previous problem. But now the axis of ring is perpendicular to the axis AB, define the same magnitudes in two cases: 1) transportation and relative motion of one direction; 2) transportation and relative motion are opposite in direction.

Answer: w1  r 2  u 2 / r  2u, w2  w4  (u 2 / r  2u   2 r ) 2  4 2 r 2 ,

1) w3  3r 2  u 2 / r  2u 2) w1  r2  u2 / r  2u, w2  w4  (u2 / r  2u  2r)2  44r 2 , w3  3r2  u2 / r  2u

64

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1. Kinematics

SELF STUDY OF THE STUDENT

Point M moves relative to the body D. According to specified equations of relative motion of the point M and motion of body D define the absolute velocity and absolute acceleration of point M for the time t=t1. Number of variant

Equation of relative motion of the point

Equation of motion of the body

M OM  s r   s r (t ), cm

 e   e (t ),

xe  xe (t ),

rad

cm

1

20 sin(t )

0,4t 2  t

2

6t 3

3

6(t  0,5t 2 )

4

5/3

20

-

-

0,5t 2  2t -

2

-

30

-

t 3  5t

-

2

-

-

30

10(1  sin 2t )

4t  1,6t 2

-

1/8

-

-

-

5

15t 3 / 8

5t  4t 2

-

2

30

30

-

6

5 2 (t 2  t )

0,2t 3  t

-

2

-

60

45

7

20 sin(t )

t  0,5t 2

-

1/3

-

20

-

8

75 (0,1t  0,3t 3 )

2t  0,3t 2

-

1

30

-

-

9

15 sin(t / 3)

10t  0,1t 2

-

5

-

-

-

t  4t

2

48

-

-

-

1/2

25

-

-

-

2/3

30

-

-

10

4t

-

2

3

11

25 (t  t )

2t  4t

12

10 sin(t / 4)

4t  0 , 2 t 2

2

-

R, a,  , sec cm cm deg t1 ,

2

Notes. For each variant the position of point M on the scheme corresponds to the positive value s r ; in variants 5, 8, 10 OM =s r is a an arc of a circle; on schemes 5, 8 OM is an arc that corresponds the smallest central angle.

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1.9. Complex motion of a solid body. Addition of the translational velocities. Addition of the instant angular velocities. Addition of the instant angular and translational velocities

The solid body moves relative to the coordinate system Oxyz that moves relative to the fixed coordinate system Oa . Suppose





that v1M is the relative velocity of any point M of the body, v2M is the transportation velocity (velocity of the point of space associated with the system Oxyz where the point M is located at the moment). The absolute velocity of the point M in the compound motion can be represented by the formula:    v M  v1M  v2M

The terms of motion are commutative (the velocity distribution does not change if we transpose relative and transportation velocity). We will determine the type of resultant compound motion at each moment t if we have an assumptions about the nature of relative and transportation motion. All the above is distributed to the case when the motion of a body is considered in relation to n – coordinate systems. Addition of translational velocities



Suppose that v is the velocity of instant translational motion of 1 a rigid body relative to the coordinate system Oxyz and v is the 2 velocity of instant translational motion of the coordinate system Оxyz relative to the system O  . For an arbitrary point M of the body a we have v = v + v in accordance with the theorem of addition of a r t velocities, or in our case we have v  v  v . 1 2 68

1. Kinematics

М is an arbitrary point  all point of the body have equal  velocity v at the moment and  compound motion of the body is instantly translational. n For the case of n – terms: v   v . i i 1 Addition of instant angular velocities

Let the body executes the instant rotation relative to the coordinate system Оxyz, and the coordinate system Оxyz execute instant rotation relative to the coordinate system O  . a

The possible cases: 1. instantaneous angular velocities intersect at one point, then the instantaneous angular velocity of the resulting motion will be: n    i i 1

2. instantaneous angular velocities are parallel and have the same direction:

   1   2 3. instantaneous angular velocities are not parallel:

   1   2

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Teextbook on Theoreticaal Mechanics

Addition of instantaneouss angular velocityy and translationall velocity

1. T The translationall velocity is perpendicular to the instan ntaneous axis  of rotatioon. Instead of v we w take a pair (  ' , ' ). Then  annd  ' are destroyedd. It remains onlly a momentary rotaation about anoth her axis Bb with anguular velocity  ' =  . Thuss, when adding the instantaneous rottational motion an nd translational motioon resulting motio on represents the instaantaneous rotation n with same angular velocity, but aro ound another axis displaced on a distancce d  v .



2. Trranslational veloccity is parallel to the axxis of rotation (the ( resulting motion iss a permanent screw motion or instantanneous screw motio on). 2.1. Permanent screw w motion.   v : Both motioons are uniform. Thhe resulting motio on is called a permanennt screw motion,, the axis of rotaөtion is called the axis of the screw. 2.2. IIf instant rotation n with angular velocity  is added to translational motion w with the velocity v  then instantaneeous screw motiion will take place. Thhe resulting motiion is instantaneous sscrew motion, thee axis of this screw w is the instant sscrew axis. Positioon of the instantaaneous screw axis changes with tim me. 70

1. Kinematics

3. Translational velocity forms an arbitrary angle with the instantaneous axis of rotation. Resulting motion will be the instantscrew motion. Questions 1. How can be defined the complex motion of a solid body? 2. How can be defined the velocity of the point of solid body in complex motion? 3. What equals the instantaneous angular velocity of a rigid body in complex motion when all instantaneous angular velocities intersect at one point? 4. What equals the instantaneous angular velocity of a rigid body in complex motion when instantaneous angular velocities are parallel and have the same direction? 5. What equals the instantaneous angular velocity of a rigid body in complex motion when instantaneous angular velocities are not parallel?

1.10. Reduction of the system of sliding vectors. Principle vector and principle moment. Invariants of reduction Reduction of the system of sliding vectors. Principle vector and principle moment

Consider the most general case when the body is involved in the k rotational movement with instantaneous angular velocities  1 ,…,

 k and m translational movement with velocities v1 ,…, v m . But as it is known any vector v is reduced to the pair, consequently, the instant angular velocities  1 ,…,  n addition represents the general case. Lemma. Every sliding vector  attached at point A is possible to move to any point B without changing its action by adding a couple with the moment equal to the moment of the vector  with respect to point B. If we now consider a system of sliding vectors  1 ,…,  n randomly located in the space they can removed to the point O (center of reduction) by adding the corresponding pairs. 71

Textbook on Theoretical Mechanics

For an arbitrary point Ai where the vector  i is attached by its '

transfer to the point O we obtain the vector  i   i and couple with the torque momO  i  OAi   i  r i   i . As a result we have a system of vectors  i and system of vectors r i   i . Summing up we get: n

    i – principle vector of this system of sliding vectors, i 1 n

v   (ri   i ) – principle moment of this system of sliding i 1

vectors. Change of center of reduction. Invariants of reduction

Let's reduce the same system of sliding vectors  1 ,…,  n to another center O . Then we get:

 n  n     i and v   (ri   i ) , i 1

i 1

where r i   OAi .

 Obviously, that    (i.e. the principle vector is not changed when you change the center of reduction – this is the first invariant of the system). Let us find the change of the principle moment. As ri   ri  OO  ri  OO

,

then v   (r i  OO)   i   (r i   i )  OO    i  v  v  OO   . i

72

1. Kinematics

I.e. the principle moment is changed by an amount equal to the moment of vector  relative to the new center of reduction. The second invariant of the system of sliding vectors is:

v    v cos(v, ) : Proof:

v    (v  O O  )    v    (O O  )    v   . As  is the first invariant then the second invariant is the value: v cos(v, ) . Thus, the projection of v onto the direction  is a constant value for a given system of sliding vectors and it is independent on the center of reduction. Questions 1. What can be the sliding vectors reduced to? 2. What is the principle vector? 3. What is the principle moment? 4. How will change the principle vector and principle moment if the center of reduction is changed? 5. What are the main invariants of reduction of vectors?

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2 STATICS

2.1. Statics. Basic definitions and axioms of statics. Constraints. Constraint reactions. Axiom of constraints Statics is the branch of mechanics that is concerned with the analysis of loads (force and torque, or «moment») on physical systems in static equilibrium, that is, in a state where the relative positions of subsystems do not vary over time, or where components and structures are at a constant velocity. Equilibrium means that the object is said to be in a state of equilibrium if all the forces that act upon an object are balanced. Force is the action of one body on another. A force is either a push or a pull. A force tends to move a body in the direction of its action. The action of a force is characterized by its magnitude, by the direction of its action, and by its point of application. Thus force is a vector quantity, because its effect depends on the direction as well as on the magnitude of the action. Forces are classified as either contact or body forces. A contact force is produced by direct physical contact; an example is the force exerted on a body by a supporting surface. A body force is generated by virtue of the position of a body within a force field such as a gravitational, electric, or magnetic field. An example of a body force is the weight of a body in the Earth's gravitation field. In addition to the tendency to move a body in the direction of its application, a force can also tend to rotate a body about an axis. This rotational tendency is known as the moment (M) of the force. Moment is also referred to as torque. 74

2. Statics

Moment about a point The magnitude off the moment of a force at a point O, O is equal to the perpendicular diistance from O to the line of action of F, multtiplied by the mag gnitude of the forcce: M = Fd, wheree F – the force applied, d – the peerpendicular distance from the axis to the line of acttion of the force. This perpendicular distance is calleed the moment arm m.

The moment cou unts as positive iif force tends to turn the arm coun nterclockwise. In vector formaat, the moment can be defined as the cross prod duct between the radius r vector, r (thhe vector from point O to the line of action), and thee force vector, F:

Mo  r  F Force couple is a pair of forcess, equal in magn nitude, oppositely y directed, and diisplaced by perpendicular distancee or moment. The simplest kin nd of couple conssists of two equaal and opposite forrces whose lines oof action do not coincide. c The forrces have a turninng effect or mom ment called a torq que about an axiis which is norm mal to the plane off the forces. If thee two forces 75

Textbook on Theoretical Mechanics

are F and −F, then the magnitude of the torque is given by the following formula:

  Fd where τ is the torque F is the magnitude of one of the forces d is the perpendicular distance between the forces, sometimes called the arm of the couple/ Equilibrium equations

The static equilibrium of a particle is an important concept in statics. A particle is in equilibrium only if the resultant of all forces acting on the particle is equal to zero. In a rectangular coordinate system the equilibrium equations can be represented by three scalar equations, where the sums of forces in all three directions are equal to zero. Statics for solids. Statics is used in the analysis of structures, for instance in architectural and structural engineering. Strength of materials is a related field of mechanics that relies heavily on the application of static equilibrium. Statics for fluids. Hydrostatics, also known as fluid statics, is the study of fluids at rest (i.e. in static equilibrium). The characteristic of any fluid at rest is that the force exerted on any particle of the fluid is the same at all points at the same depth (or altitude) within the fluid. If the moment of force is greater than zero the fluid will move in the direction of the resulting force. Axioms of statics

Statics theory is based on four axioms: 1. A system of two mutually opposing forces that are equal in magnitude and attached at one point, is in equilibrium. 2. The system of two equal by magnitude mutually opposing forces applied to any two points of a rigid body and directed by a 76

2. Statics

straight line connecting the point of their application, are in equilibrium. (This axiom is valid only for a rigid body, where the distance between two points is not changed.) 3. Any system of forces (s 1 ) can be replaced with another system (s 2 ) equivalent to it without changing its action. The result: Every force applied at any point of a rigid body can be transferred without changing its action to any other point lying on the line of action of this force. 4. Two systems of forces that differ by a system of forces , equivalent to zero, are equivalent. 5. The equilibrium of a mechanical system in rest is not disturbed by imposing the new constraints. 6. System of two forces applied at one point, is equivalent to a single force applied to the point M and equal to the geometric sum of these forces.

a)

b)

Constraints. Constraint force (reaction)

If each mass point of c) d) e) material system can take up an arbitrary position in space and have arbitrary velocities the material system can be named as free, in opposite case it could be named as f) g) h) unfree. Conditions that impose restrictions on a motion of the material system have the name of constraints. 77

Textbook on Theoretical Mechanics

There are geometrical constraints, kinematic re-lations, stationary, nonsta-tionary. Сonstraints are usually represented as various bodies that hinder the movements of the points of material system. Action of constraints can be replaced by the appropriate forces that has the name of constraint forces (reactions). Constraint forces can not impart the motion to the body. Constraint force is the force applied in the point where the constraint contacts with the body. And direction of the constraint force coincides with the direction along which the constraint hinder the movements of the body. Examples of constraints. a) bearing on a smooth surface – reaction is directed along the normal to it, that is perpendicularly to the tangent (normal reaction); b) one of the connecting surfaces is a point (angle), reaction is directed along the normal to another surface; c) thread – reaction directed along the thread; d) fixed joint – replaced with two orthogonally related components when solving the problem; e) moving joint (joint on rollers) – reaction directed perpendicularly supporting plane; f) spherical joint (in space) –replaced with three orthogonally related components when solving the problems. g) weightless bar – reaction directed along the bar; h) dumb end restraint (cantilever bar) – appears an arbitrarily directed reaction (decompositions in two components) and reactive moment. Axiom of constraints: Any unfree body can be released from constraints by replacing the constraints with its reactions and consider this body as free body under action of active and constraint forces. Questions 1. What is the main problem of statics? 2. When we can say that the object is in equilibrium? 3. What type of constraints do you know? 4. How you understand the notion of free and unfree body? 5. What is the result of acting of force couple?

78

2. Statics

2.2. System of o forces. Converggent system of fo orces. Para allel forces. Centter of gravity System m of forces applieed in one point

d in one point If the body is acteed by two forces P and Q applied А th hen resultant of th hese forces R equuals to a vector su um of P and Q:

R = P +Q Resultant modulee is: = P 2 + Q 2 + 2 PQcos | R |= c P,Q  . before cosine is a «plus» sign thou ugh according to cosine theorem: R 2  P 2  Q 2  2 PQ P cos  ,





but as a   180 0    cos   cos 180 00     cos  .

 

Direction R is deetermined throughh angles   P, R and   Q,R . According to sinu us theorem: P R , Q = = sin R,Q sin P,R sin

 

wherre



 







sin = sin 180 0  γ = sinn γ = sin Q , P .

If we have a system of forces F1,..., Fn , applied in a point О then apply ying sequentially y an axiom of paarallelogram of fo orces we will get: 79

Textbook on Theoretical Mechanics n

R  F1  ...  Fn   Fi i 1

Thus, the system of forces applied to one point is equivalent to one force that has a resultant equal to a vector sum of all forces of a system and applied to the same point.

R can be obtained either graphically (according to vector polygon method) or analytically: n n n   , , R F R R R Riz       ix y iy z  x  i 1 i 1 i 1     2 2 2  R  Rx  R y  Rz    cos R, x  Rx ; cos R, y  R y ; cos R , z  Rz  z  R R R 

 

 





Resultant of a convergent system of forces

System of acting on absolutely solid body forces the lines of action of which intersect in one point is called a convergent system of forces. Obviously all the forces along the line of their action can be transferred and applied in one point О, consequently, the convergent system of forces is brought to the resultant equal to the sum of all these forces and passing through the point О. Then for the equilibrium of that system is necessary and sufficient that the resultant n

R  0 , that is: R   Fi  0 – equilibrium condition in vector form. i 1

In analytical form: n

n

n

i 1

i 1

i 1

Rx   Fix  0; R y   Fiy  0; Rz   Fiz  0. As when R  0 the polygon of forces will lock itself we get an equilibrium condition in geometric form: for the equilibrium of a 80

2. Statics

conv vergent system off forces it is neccessary and sufficcient that the vecto or polygon built from f forces of a syystem was locked d. Three forces theeorem. If a planee system of threee nonparallel forcees is in equilibriium then the linnes of action of these forces interrsect in one point. n equilibrium Proof. Given: F1, F2 , F3 – plane syystem of forces in in po oints A1, A2 , A3 . Let us assume that the lines of acttion a i a point О. Lett us F1 and F2 intersect in transsfer them in point p О and ffind quently we have an R12  F1  F2 . Conseq equiv valent system F1 , F2 , F3  R12 , F3  . But the t system of two o forces is in equilibrium m only in case wheen these forces dirrected along a one straight line, consequenntly, the line of action F3 should s coincide w with the line of acttion R12 , that is sh hould pass throug gh the point О. Thus, for equilib brium of the systeem of three forcees lying in one pplane, it is necesssary (but not suffiicient) that the lin nes of action of tthese forces interssected in one poin nt.





Parallel forrces

Let us consider a system of two pparallel forces directed to one side and acting on an absolutely solid body. In pooint А th he force P is app plied, in point В tthe forcee Q is applied. t points А andd В Let us connect the and apply a in point А the t force S , and tthe forcee S ' in point В.

S =- S ' ; | S |=| S ' |. Consequently,  S , S '  0 .   81

Textbook on Theoretical Mechanics

Now

let

us

combine

forces:

P, Q   P, Q, S ,S   P, Q   R , R . 1

P + S = R1 ;

' Q + S = R2 .

2

Lines of action of forces R1 and R2 intersect in point О. Let us transfer these forces in point О. Let us decompose them in a point О: R1 = P + S ; R2 = Q + S ' . We will get the system of forces  P ,Q , S,S '  applied in one point О.

   S , S '  0 , consequently, these forces can be rejected. There are two    

forces left P , Q , directed in one side and acting along one straight line that is parallel to lines of action P and Q . Their resultant R = P + Q is directed parallel to given forces. Let us find how the line of action R is situated related to points А and В. From triangles similarity we have: P S Q S  ,  OC AC OC CB .

Let us divide first equality by second one: P CB P Q .    Q AC CB AC

Thus point С AB divides it internally into parts inversely proportional to forces. From last proportion let us make: PQ P Q   AC  CB CB AC

or R P Q .   AB CB AC 82

2. Statics

From here it is easy to determine AC , CB. System of two parallel forces directed to one side has a resultant with the magnitude equal to sum of magnitudes of given forces, parallel to them and directed to the same side. Line of action of the resultant R passes through the point which divides the segment AB into parts inversely proportional to given forces internally. Let us consider the inverse problem about decomposition of R into two parallel forces. For distinctness of the problem we need to set the module and the line of action of one force, or the line of action of both forces. The module and the line of action of one force are given. We need to decompose the resultant in such a way that one force was applied in point А and its module was equal to P . We find the module of second force Q = R  P and the position of point В from equality: P  Q  CB  P AC . CB AC Q Resultant R and positions of points A, B,C (that is distances AC,BC ) are given. From the equality CB    P  R AB  P Q R     CB AC AB Q  R AC   AB 

The distance from point C to the line of action is called arm of force relative to the point C. Product of force and arm is a module of value called moment of force relative to given point. System of two anti-parallel forces (parallel forces directed in opposite sides)

Given: force P is applied in point А, force Q is applied in point B . Q . P  Q ; P  Q . 83

Teextbook on Theoreticaal Mechanics

Let us decomposee the bigger forcee P into two paraallel forces R and Q1 , so that | Q1 |=| Q | and Q1 w was applied in po oint В. Then according to the known formulas we module of R and d its point of find the m applicationn С:

P  R  Q1  R  P  Q R P Q .   C AB CB AC

Force R will be the resultant oof the system off anti-parallel



 

 



forcees P and Q . Indeeed, P, Q  R, Q1 , Q  P, Q  R . System of two an nti-parallel forcess has a resultant th hat equals by magn nitude to the diffference of modulles of these forcees, parallel to them m and directed to o the side of biggger force. Line of action R passees through the point that lies on ann extension  AB and a divides it exterrnally into parts in nversely parallel tto forces. Inverse problem – decompositionn into two anti-paarallel forces beco omes definite in th he same cases as for parallel forcees. Should be know wn: module and position p of the linne of action of onee of forces or the line of action of bo oth forces. System m of multiple para allel forces

If theere is a system of multiple parallel fo forces then it red duces in the same wayy to one resultan nt. The point through w which passes the resultant of the system m of parallel forcees directed to one side iis called the center of parallel forces. 84

2. Statics

Center of gravity

The force with which the body is pulled to the ground is called the force of gravity. The numerical value of this force equal to the weight of the body. Let's consider a small body. Let's break it down into a set of elementary particles. Each particle is acted by the force of gravity applied to a point coinciding with the particle . All the forces of gravity will be parallel. The resultant of forces is equal to the weight of the body, and its line of action will pass through the point that coincides with the center of parallel forces of gravity of the particles. The point that is the center of parallel forces of gravity of the particles of the body is called the center of gravity of this body. Usually denoted by the letter C. Assume that V is the volume of the body, V is the volume of any particle of the body, P is the weight of this article. The value   lim P  dP is called a weight per unit of V  0 V dV volume of the body at this point, and the value    represents the

g

density of the body (weight of unit of volume) at this point. If the body is homogeneous, and  and g are constants weight of any particle

i

with the volume Vi will be:

Pi   i Vi  gi i Vi . If all forces of gravity of the particles are parallel, then its resultant is equal to the sum of the weights of all the particles, i.e. weight of the body. The radius vector of the point of application of this force is determined by the radius vector of the center of parallel forces:

rc 

 Pi ri   i Vi ri .    i Vi  Pi 85

Textbook on Theoretical Mechanics

As

 i  i gi , then

rc 

  i g i Vi ri .   i g i Vi

If the body is sufficiently small, it is possible to reduce gi , as acceleration of force of gravity for all the points of the body is the same. Then we will get: rc 

 i Vi ri ,  i Vi

wherе  i Vi is the mass of the body. i

This formula determines the position vector of the center of mass of the body. Center of gravity depends on the mass distribution in the volume occupied by the body. The concept of the center of mass – is more general than the concept of center of gravity. Furthermore, this concept is not connected with the fact that the body is located in the field gravity forces. For a body located in a uniform field of gravity the center of gravity coincides with the center of mass. The center of gravity is a geometric point, it may not coincide with any point of the body (for example – the center of gravity of a donut). If γ and  are continuous functions of coordinates, than the amounts represent the volume integral. If the body is homogeneous, than  can be reduced:

rdV – position vector of the center of gravity of the body. rc   V The value in numerator  rdV is called the static moment relative to the point O. 86

2. Statics

The projections of the static moment on the axis of a Cartesian coordinate system: xc  

xdV , V

yc  

ydV , V

zc  

zdV V

When looking for the center of mass of plane figures, then, instead of V it is taken S. Then the weight of the particle is equal to  S , where   is the weight per unit of area.

rc 

  S r   S

rdS . If the body is homogeneous than rc   S

When looking for the center of mass of a material line, instead of V it is taken l . Then the weight of the unit of length is equal to   . We break the length down to the line elements of length l . Then   l r . rc    l If the line is homogeneous than rc   rdl . l Questions 1. What is convergent system of forces? 2. What is the center of gravity? 3. How can be defined the center of gravity?

PRACTICE 2.1. Determine the position of center of gravity of homogeneous disc with the round hole if the radius of disc is r1 and radius of the 87

Textbook on Theoretical Mechanics

hole is r2 . Center of this hole is situated at the distance

r1 from the 2

center of the disc.

Answer: xc  

r1r22 2(r12  r22 )

2.2. Determine the coordinates of the center of gravity of the figure shown in the picture.

Answer: xc  1,6а

88

2. Statics

2.3. Find the distance of the center of gravity of the section ABCD from its side AC if the height BD=h, width AC=a, thickness of the cap is d and thickness of the wall is b.

Answer: yc 

ad 2  bh 2  bd 2 2(ad  bh  bd )

2.4. Determine the position of the center of gravity C of the area, restricted by the half-circle AOB of radius R and two straight lines AD and DB of equal length. OD = 3R.

Answer: OC 

3  16 R  0,19 R 3  12

2.5. Determine the coordinates of the center of gravity of the system of loads situated in the vertices of rectangular parallelepiped with the links AB=20cm, AC=10cm, AD=5cm. Weights of the loads in the vertices A,B,C,D,E,F,G,H are equal to 1N, 2N, 3N, 4N, 5N, 3N, 4N, 3N accordingly. 89

Textbook on Theoretical Mechanics

Answer: xc  3,2cm, yc  9,6cm, zc  6cm 2.6. Determine the coordinates of the center of gravity of the contour of rectangular parallelepiped with the links considered as the homogeneous bar with the length: OA =0.8 cm, OB = 0.4 cm, OC = 0.6 cm. The weights of the bars are equal to POA = 250N, POB = POC = PCD = 75N, PCG = 200N, PAF = 125N, PAG = PGE= 50N, PBD = PBF = PDE = PEF = 25N.

Answer: xc  0,263cm, yc  0,4cm, zc  0,105cm 2.7. Let’s consider the homogeneous tetrahedron ABCDEF which is truncated parallel to its base. Area of ABC is equal to a, Area of DEF is equal to b, distance between them is h. Find the distance z of center of gravity of this truncated tetrahedron from the base ABC.

90

2. Statics

Answer: z 

3(a( a  b )  3b) 4(a  ab  b)

2.8. Find the center of gravity of the contour shown in the picture.

Answer: xc  9cm 2.9. Determine the position of the center of gravity of the ring shown in the picture.

Answer: xc  yc  1,38cm 91

Textbook on Theoretical Mechanics

2.10. Determine the coordinates of the center of gravity of the body considered as the chair and consisted from the rods with equal length and weight. Length of the rod is 44cm.

Answer: xc  22cm, yc  16cm, zc  0cm 2.11. A thin homogeneous paper is curved in the form of two triangles and square as shown in the picture: isosceles triangle lies in the plane xy, rectangular triangle OAB lies in the plane yz, square OBKE lies in horizontal plane. Determine the coordinates of the center of gravity of the curved paper.

Answer: xc  3,33cm, yc  0,444cm, zc  3,55cm

92

2. Statics

SELF STUDY OF THE STUDENT

Find the coordinates of center of gravity of the figures shown in the pictures. All dimensions are shown in the pictures in centimeters. 1

2

3

4

5

6

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8

9

10

2.3. Moment of force relative to center and axis. Theory of couples Moment of force relative to center

  Let F – is a force applied in point А (end of vector F situated in point В), point O – is some center. Then the moment of force relative to center O will be vector applied to the center O , directed perpendicularly to the plane of triangle OAB to the side from which the turn made by the force is seen counterclockwise and numerically equal to double square of that triangle:    mom0 F  r  F



 Module of that vector equals to product of F on distance h from center O to the line of action of force ( h is called arm of the force relative to center O ), that is 94

2. Statics

  r  F  Fh If center O bellongs to the linee of actio on

of

force,

then

 mom0 F = 0

(as h =0). If through the ceenter O we draw w some system of o rectangular axes Oxyz , then wee can express thee moment of forcce relative to centeer as: i    mom0 F  r  F  x Fx

j

k

y Fy

z Fz

   Let F1 ,......, Fn is some system of fforces. Then vecto or M 0 equals to su um of moments off all these forces rrelative to center О, О that is n   M 0   mom m0 Fi   (ri  Fi ) i 1

calleed principle mom ment of system oof these forcees relative to cen nter О. If all forrces are appliied in one point, then t n n  M 0   ( r  Fi )  r   Fi , i 1

i 1

it fo ollows that the moment m of sum off forces appliied to one point relative to somee center equaals to sum of mo oments of these vectors relattive to the saame center (Vaarignon theorrem). 95

Teextbook on Theoreticaal Mechanics

Mo oment of force rellative to axis

Moment of forcee relative to axis is a projection of o moment of forcees taken relative to o any point on axis onto that axis, that t is momx F  (r  F ) x

Let us proof thaat point O on axis x relative to which is



conccerned a projection of moment of fforce F can takee any position on ax xis. Can be writteen as: momx F  (r  F ) x  (r  F )  x 0  ( x 0  r )  F

 F stays constannt, (x 0  r ) is alsoo a constant valuee equal to 2S of a triangle with base x 0

and altitude rsin (x 0 , r ) = d

and,

conssequently, the vaalue (r  F )x doees not depend on n position of poin nt O .



We can give ano other definition. M Moment of force F relative to



any axis x is a moment m of projecction of force F on a plane ntersection of perpendicular to axis x taken relative to point O1 of in axis x and that plane. At the same tiime the moment of projection F A relative to point O1 is conssidered as scalar value. If the turn made by force F A is seen from m the positive en nd of axis х clockkwise then the moment is takenn with «minuss» sign, if counnterclockwise then n with «plus» sign:

mom x F  momO1 FA   FA h1 (scalar valuees). 96

2. Statics

Moment of force F relative to x axis equals to zero, if the line of



action of F intersects the axis x (in that case arm h = 0 ) or it is parallel to it (in that case projection F A onto a plane equals to zero). Projections of moment of force relative to Cartesian coordinate system may be written as: mom x F  (r  F ) x  yFz  zFy

mom y F  (r  F ) y  zFx  xFz momz F  (r  F ) z  xFy  yFx Theory of couples

Moment of couple. A couple of forces is a system of two equal by magnitude and opposite by direction forces acting on a solid body. ' (F , F ' ) is a couple if F = F , F  F ' .

Distance d between lines of action of forces of a couple is called an arm of a couple. 1) A couple of forces does not have a resultant. A couple of forces acting on a body tends to impose to it some rotation. The turning effect is characterized by the moment of a pair (moment equals to F  d ) and direction of rotation. Moment of couple (of free rotation) is a vector perpendicular to the plane of action of a couple directed according to right-hand screw rule and numerically equal to product of one of forces of a couple on an arm. Equivalence of couples

Theorem 1. Action of a couple on an absolutely solid body does not change if the couple is transferred to another position in a plane of its action. 97

Teextbook on Theoreticaal Mechanics

Theorem 2. Actio on of a couple on an absolutely sollid body does not change c if the plane of its action is trransferred paralleel to itself. Theorem 3. Actio on of a couple on an absolutely sollid body does not change c if the forcces and arm of a ccouple will modiffy keeping in consstant their productt, that is moment oof a couple. Addition of coouples

Theorem. System m of couples actinng on an absolutelly solid body is eq quivalent to one couple the momeent of which equ uals to vector sum of moments of th hat couples. Questions 1. How can be defineed the moment of forcce relative to the centeer? 2. How can be defineed the moment of forcce relative to the axis?? 3. How can be determ mined the moment of couple? 4. What is the couplee equivalent to?

2.4. 2 Arbitrary sysstem of forces. Reeduction of spatiial system of forces. Prrinciple vector an nd principle mom ment Lemm ma (main). Any force applied to the absolutely solid bod dy at a given point A, iis equivalent to th he same force applied too another point B, and the pair of forcess with the momeent, which is equal to tthe moment of forrce applied at point A w with respect to poiint B. Proof. Assume th hat the force FA is apllied at the point p A. Let's apply y two mutually op pposing forces FB   FB at the po oint B. These forcees are equal by maagnitude to the foorce FA and paralllel to it. Then the forces f FB and FA give a torque ccouple equal to thhe moment of

this force FA with resspect to point B, ii.e. BA  F A . Wh hich was to be prov ved. 98

2. Statics

Reduction of spatial system of forces. It's given a system of arbitrary forces: F1 ,....., Fn . Let' s choose an arbitrary point O and transport all the forces to this point. In accordance with the fundamental lemma, each force will give an additional force couple with the moment equal to the moment M i of the transferred M i relative to the selected center O. Adding all the forces, we get a resultant force which is called the principle vector: n

 Fi  R .

i 1

Adding the moments of couples we get the principle moment: n

n

i 1

i 1

 (ri  Fi )   M i  M 0 .

As the forces are located randomly in space, the principle moment M 0 is directed with respect to R at an arbitrary angle. Thus, any spatial system of forces, reduced to a certain center O is replaced by one principle vector R applied to the point O, and the resulting pair with the principle moment M 0 . Change of center of reduction. Suppose that an arbitrary spatial force system is reduced to the center O, then we have the principal vector and the principal moment – R , M 0 . Let's take any other

center O  and reduce all the forces of the system to this center. It is obvious that R    Fi  R (i.e., the resultant force does not change with the change of center of reduction. This is the first invariant of the system). Let's find the change of M 0 . Let's denote the principle moment relative to the point O with M and relative to the point O  with M  . then:

M    (ri Fi ). 99

Teextbook on Theoreticaal Mechanics

The position-vecctor of the point of application o of one of the forcees of the system, drawn from the new center is: ri  ri  OO  . Subsstituting this expreession, we obtain::

M    (ri  OO )  Fi    ri  Fi   OO   Fi   M   M  OO    Fi  M  OO   R  M  O O  R . That is, the principle momen nt is changed by an amount eequal to the mom ment of vector R relative to thhe new center of reeduction. The secondd invariant is the quantity R M  R M cos(R R, M ) or the projeection of the vector M on thee direction R . Proof: Forr the center O, we have: R   Fi , M   (ri  Fi ) . For the new center O  : R   R   Fi ,

M   M  OO  R ,  R  M   R(M  OO R)  RM  R  (OO R) . But R  (O O  R ) iss equal to zero, aas it have to equaal multipliers

R  (O O  R )  О О ( R  R )  ( R  R )  0, as. sin O o  0

 R   M   R  M  const is the secoond invariant. Thus, the projecttion of M on thhe direction R is the constant value for a given systeem of forces and does not depend on o the choice he center of reducttion. of th Possiblee location of the vvectors M and R

1) R  M  0 а) R  0 ( R liees in the plane oof the pair ( F , F  ), which is replaaced by M ). The system of forces is reduced to onee force, i.e. to the resultant r force:

R   Fi . 100

2. Statics 

It is possible because of R  M  0  p  R  M  0  M  M  0. R R2 b) R  0 , but M 0 M  M  cons, as M   M  OO  R  0 . That is the principle moment doesn't depend on the choice of center of reduction. The system of forces is reduced to the pair of forces with the moment:

M   (ri  Fi )   mom0 Fi , where О is an arbitrary center. c) R  0 , M  0 – the system of forces is in equilibrium. 2) If R  M  0 , then any system of forces is reduced to the wrench (screw) with the parameter:

p

R M R2

or to another two forces in different planes. Wrench (dynamic screw) is formed with the principle vector R directed along the central axis, and a pair with the minimal moment M  collinear to R . Two forces are formed, if instead of the principle moment we give the pair of forces M  ( F , F ) when one of them pass through the point O. Adding the principle vector with this force, we get a force R  F  Q passing through the point O, and finally we have two forces Q and F  that lie in different planes. Questions 1. What is the force applied to the absolutely solid body is equivalent to? 2. What is the principle vector and principle moment? 3. What are the main invariants of reduction of system of forces? 4. When the system of forces can be reduced to the wrench? 5. When we can say that the system of forces in equilibrium?

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2.5. Equilibrium conditions of arbitrary spatial system of forces. Special cases of the equilibrium conditions Conditions of equilibrium of systems of forces

The system is in equilibrium if R  0 , M  0 . Common case Arbitrary spatial system of forces. The principle vector and principle moment give three projections, therefore, we have six equilibrium conditions:

Rx  0,  momx Fi  0,

R y  0,

Rz  0

 mom y Fi  0,

 momz Fi  0

1. Arbitrary plane system of forces Three equilibrium conditions. There are three set of equivalent equilibrium conditions. А) The principle vector R gives two projections in a plane, principle moment perpendicular to this plane gives one equation:

Rx  0,

R y  0 ,  momz Fi  0 .

Б) momA Fi  0,  momB Fi  0,  ( Fi )l  0i , where the straight line l must not be perpendicular to the line AB. В)  mom A Fi  0,

 momB Fi  0 ,

 momC Fi  0 . Points A,B,C must not lie on one straight line. 102

2. Statics

2. Plane system of parallel forces. R gives one projection. Then there is one equation. M is perpendicular to this plane, this implies another equation. As a result, we obtain two equilibrium conditions. There are equivalent equilibrium conditions. А)  Yi  0  Mi  0 Б) momA Fi  0,  momB Fi  0, where the points A and B don't belong to the straight line which is parallel to the forces. 3. The spatial system of parallel forces. Three equilibrium conditions. R gives one projection. Then there is one equation. The principle moment M lie in the plane that is perpendicular to R , therefore provides two projections:

 Rz  0  mom x Fi  0  mom y Fi  0 4. Convergent system of forces R gives two projections in space. M  0 , as r  R  0 , r  0 . Therefore, we have three equilibrium conditions .  X i  0,  Yi  0,  Z i  0 If the system of forces is located in the plane than R gives two projections. Therefore there is two equilibrium conditions:

 X i  0,  Yi  0 .

103

Textbook on Theoretical Mechanics Questions 1. How many equilibrium conditions is there for arbitrary spatial system of forces? 2. What types of equilibrium conditions can be constructed for arbitrary plane system of forces? 3. What are the equilibrium conditions for plane system of parallel forces? 4. What are the equilibrium conditions for spatial system of parallel forces? 5. What are the equilibrium conditions for convirgent system of parallel forces?

PRACTICE 2.12. The street lamp is suspended in the point B to the middle of the cable ABC fixed with its ends to the hooks A and C that are on one horizontal. Define the strains T 1 and T 2 in the parts AB and BC of cable if the weight of the lamp equals to 150 N, the length of the whole cable ABC equals to 20 m and the deviation of the point of the suspension from the horizontal BD equals to 0,1 m. Neglect the weight of cable.

Answer: T1 = T2 = 7,5 kN. 2.13. A homogeneous ball O of weight 60 N On lies on two orthogonally related smooth inclined planes AB and BC. Define the pressure of the ball on each plane knowing that the plane BC makes with the horizon the angle 60°.

Answer: ND=52N, NE =30N. 104

2. Statics

2.14. A homogeneous rod AB of weight 160 N, length 1,2 m is suspended in point C on two cables AC and CB of equal length of 1 m. Define the tension of the cables. Answer: The tension of each cable equals to 100 N. 2.15. Homogeneous rod AB with the length of 1m and weight of 20N is hung horizontally on two parallel ropes AC and BD. The load P=120N is hung to the rod in the point E at the distance AE=1/4m. Determine the tension of the ropes TC and TD . Answer: TC  100 N , TD  40 N . 2.16. Horizontal beam supporting the balcony is exposed to the action of uniformly distributed load of intensity q=2kN/m. The load of the column P=2kN is transmitted to the beam. Distance between the axis of a column and the wall is l=1,5 m. Determine the reactions of end restraint.

Answer: R = 5 kN, M = 5.25 kN m 2.17. Force with the moment M=6kNm acts on the horizontal cantilever. Vertical load P=2kN acted at the point C. The length of the beam AB=3,5m, BC=0,5m. Determine the reactions of the supports.

Answer: R A  2kN , RB  4kN . 105

Textbook on Theoretical Mechanics

2.18. The bar of length l is under action of distributed force shown at the picture. Intensity of the load is equal to q N/m at the ends of the bar A and B, and in the middle of the bar it is equal to 2q N/m. Find the reactions of the supports D and B neglecting the weight of the bar.

Answer: RD  2ql N , RB  ql N . 2.19. Determine the intensity qmax of distributed load which causes the moment in end restraint А equals to 270 N m. The dimensions are: АВ =1m, AС = 4m.

Answer: 60N 2.20. Determine the reaction of the support D in kN if the moment of force couple М = 13 kN•m, intensity of distributed load is qmax = 8 кN/m and dimensions are: АВ = ВС = 3 m.

Answer: 10N 2.21. Define the reactions of supports A and B of the beam

under action of one concentrated force P1 of 4 kN and couple of forces P2 of 6kN. 106

2. Statics

Answer: X A  2kN , Y A  4,32kN , YB  7,78kN . 2.22. Define the reactions of supports A and B of the beam under action of two concentrated forces P1 and P2 of 6 kN and 8 kN accordingly and distributed load q=3kN/m.

Answer: X A  2,6kN , Y A  4,2kN , X B  15,6kN 2.23. Determine the reactions of supports A,B,C and the joint D of the composite beam showed in the picture.

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Textbook on Theoretical Mechanics

Answer: X A  2.8kN,YA  4.4kN,YB  22.2kN,YC  5kN, X D  0,YD  5kN 2.24. Two loads C=2kN and D=1kN are put on the horizontal beam lying on two supports so that reaction of the support A is two times as much as the reaction of the support B if the weight of the beam is neglected. Distance CD between the loads equal to 1 m. What is the distance x of the load C from the support A?

Answer: x=1m. 2.25. The beam AB of length 10 m and weight 2kN is lying on two supports C and D. Distance between the support C and the end A is 2 m, distance between the support D and the end B is 3m. End of the beam A is pulled up vertically by means of the rope thrown over the block. The load Q with the weight of 3kN is hung to this rope. The load P with the weight 8kN is hung to the beam at the distance of 3 m from the end A. Determine the reactions of the supports neglecting the friction on the block.

Answer: RC  3kN , RD  4kN . 2.26. Two rods AB and OC which weight of unit of length is equal to 2p are connected orthogonally in the point C. The rod OC can rotate around the horizontal axis O. AC=CB=a, OC=b. Two 108

2. Statics

loads with the weights P1 and P2 are suspended to the rod. P2  P1 . Determine the inclination angle  of the rod AB to the horizon in the position of equilibrium.

P2  P1 a b P  P   2 1  p ( 4a  b)

Answer: tg   

2.27. The link BC is parallel to the fixed link AD in the four-bar linkage mechanism. AB=h and AB  AD. Horizontal force P is applied in the middle of AB. What horizontal force Q should be applied to the link CD in the point E so that mechanism was in equilibrium? CE=CD/4. Find the reaction in the joint D. Weight of the links can be neglected.

Answer: Q 

2 1 P, RD  P and directs along AD to the right. 3 6

2.28. Horizontal beam of a crane of length l is fastened with a joint at one of the ends and it’s another end B suspended to the wall by means of a rod BC which makes an angle α with the horizon. Load P can move along the beam. Position of the load is defined with the variable distance x to the joint A. Define the tension T of the rod 109

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BC in dependence on the position of the load. Weight of the beam can be neglected.

Answer: Т  Px l sin  2.29. Homogeneous sphere of the weight Q and radius a and the mass of the weight P are suspended with the strings in the point O as shown at the picture. Distance OM=b. Define the angle  between the straight OM and vertical when the system is in equilibrium.

Answer: sin  

a P b PQ 110

2. Statics

2.30. When the bridge was constructed its part of bridge truss ABC was lifted up by means of a rope as shown at the picture. Weight of this part of the truss is 42kN, center of gravity is in the point D. Appropriate distances are AD=4m, DB=2m, BF=1m. Find the tension of the ropes if the straight AC is horizontal.

Answer: Т A  18kN , TB  17,57kN , TC  12,43kN . 2.31. The crane for lifting the loads consists of the beam AB, its lower end is connected with the wall by means of the joint A, its upper end is supported by the horizontal rope BC. Define the tension T of the rope BC and pressure on the support A, if it is known that the weight of the load P=2kN, weight of the beam AB is 1kN which is applied in the middle of the beam and angle   45 . Answer: Т  2,5kN , X A  2,5kN , Y A  3kN . 2.32. Define the reactions of the bearings A,B,C and reaction of the joint D of the composite beam under action of distributed force q with the magnitude 1,75 kN/m, concentrated force P1 of 6 kN and concentrated force P2 with the magnitude 5 kN shown at the picture.

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Answer: X A  3kN , Y A  13,8kN , YB  6,6kN , YC  10kN , X D  0, YD  5kN .

2.33. The street lamp of weight 300 N is suspended to the vertical pole with the horizontal crossbeam AC=1,2 m and brace BC=1,5 m. Find the forces S1 and S2 in rods AC and BC assuming the bracing in points A, B and C as a joint. Answer: S1=400N, S2=-500N. 2.34. The rods AC and BC are connected with each other and with vertical wall by means of the joints. Vertical force P=100n acts on the joint C. Angles made by the rods with the wall are equal accordingly to   30  ,   60  . Determine the reactions of this rods on the joint C. Answer: 866N, 500N. 2.35. The rope CAEBD is thrown over two blocks A and B situated on one horizontal line AB=l. Loads with the weight p are hung to the ends C and D of the rope. Load E of weight P is hung to the point E. Determine the distance x between the point E and the line AB in the position of equilibrium. Weight of the rope can be neglected.

Answer: x 

Pl 2 4 p2  p2

.

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2.36. Area of the plunger of internal-combustion engine is 0,02m 2 . Length of the crank BC=6 cm. Length of the connecting rod BC=6 cm. Pressure of the gas over the plunger is P=1000kPa. Pressure of the gas under the plunger is P=200kPa. Find the force T acting on the crank BC from the side of the connecting rod AB caused by the pressure difference of gas. ABˆ C  90 . Friction between the plunger and cylinder can be neglected. Answer: T=16kN. 2.37. Homogeneous sphere O is hung with the rope AC to the smooth vertical wall AB. The rope make an angle  with the wall. Weight of the sphere is P. Determine the tension of the rope T and pressure of the sphere to the wall Q.

Answer: T 

P , Q  Ptg . cos 

2.38. The rods are connected between each other, with the ceiling and walls with the help of joints. The load Q=1000N is suspended to the joint B. Define the forces in the rods if     45 . Answer: S1=S2=707N 2.39. The rods are connected between each other, with the ceiling and walls with the help of joints. The load Q=1000N is suspended to the joint B. Define the forces in the rods if   30 ,   60 . Answer: S1=577 Н, S2=-1154N 113

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2.40. The mechanism of antiparallelogram ABCD consists of absolutely rigid rods AB, BC and CD connected between each other in points B and C with the help of joints and fixed with the help of joints A and D to the link AD. AD=BC, AB=CD. Horizontal force Fc=10kN is applied to the joint C. Define the magnitude of force applied to the joint B and directed vertically down if the mechanism is in equilibrium and in position shown in the picture so that angle ADC = 90°, angle BAD = 30°. Neglect the weight of rods. Answer: FB  4 3FC  23,1kN 3 2.41. Homogeneous ball of weight P=20kN leans in the point A on a smooth inclined plane that makes an angle 60° with the horizon and in the point B on the hill that is in one horizontal with the point A. Define the reactions of supports of the inclined plane and the hill.

Answer: RА = RВ = P= 20 kN. 2.42. The cable is thrown over the nail driven into in the wall. One end of the cable is fixed to the floor at an angle 30° to the horizon. The load with the weight is P=100kN is suspended to the other end of the cable. Define the magnitude of the reaction of the wall where the nail is driven. The cable is situated in the vertical plane. Answer: R  1723kN 2.43. Homogeneous rod AB is fixed to the vertical wall with the joint A and is held at the angle 60° to the vertical by the cable BC 114

2. Statics

making the angle 30° with it. Define the magnitude and direction of reaction R of the joint if it is known that the weight of rod equals to 20 N.

Answer: R = 10N, angle (R,AC) = 60° 2.44. Beam AB is held in horizontal position by the rod CD; mounts in A,C and D are joints. Vertical force F=5kN acts on the end of the beam. Define the reactions of mounts A and D. Neglect the weight.

Answer: RA = 7,9 kN, RD = 10,6 kN 2.45. For three-hinged arch determine the reactions of the supports A and B appearing under the action of the force P. Weight of the arch can be neglected. 115

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Answer: R A  RB  P

2 . 2

2.46. The bar AB is fastened at the support A by means of a joint. It is put on the rollers at the end B. Force P=2kN acts in the middle of the bar at an angle of 45 degrees. Determine the reactions of supports for the cases a) and b). Answer: a) R A  1,58kN , RB  0,71 kN b) R A  2,24kN , RB  1 kN . 2.47. Define the reactions of cantilever under action of couple of forces P with the moment 4kNm and distributed force q=1,5 kN/m as shown at the picture.

Answer: X  9kN , Y  0kN , M  40kNm . 116

2. Statics cm

cm

cm

2.48. Load with the weight P=25 N is suspended to the end of horizontal bar AB. Weight of the bar of Q=10 N is applied at the point E. The bar is fastened to the wall with the joint A and supported by the rod CD, to which it is also fastened with the joint. Weight of the rod CD is neglected. Dimensions are shown at the picture. Define the reactions of the joints A and C. Answer: X A  30 N , Y A  17 N , RC  60 N . 2.49. Two homogeneous rods AB and BC of equal cross-section are connected by its ends making an angle 60°. BC=2AB. End A is suspended to the thread AD. Determine the inclination angle α of the rod BC to the horizon when the system is in equilibrium. Neglect the lateral dimension of the rods.

Answer: tg 

1 3 ,  195 3

2.50. Homogeneous horizontal bar of length 4m and weight 5kN is built in the wall of thickness 0.5 m so that it bears on in two points A and B. Determine the reactions in this points if the load P of weight 40 kN is suspended to the free end of the bar. Answer: R A  340kN , RB  295kN 2.51. A homogeneous ball of weight 20N is held on the smooth inclined plane with the cable that is connected to the spring-balance fixed over the plane; registration of 117

Textbook on Theoretical Mechanics

the spring-balance is 10N. The angle of slope of the plane to the horizon equals to 30°. Define the angle α that is made by the cable direction with the vertical and the pressure Q of the ball on the plane. Neglect the weight of the spring-balance. Answer: α = 60°, Q = 17,3 N. 2.52. The boiler with the weight P=40kN that is uniformly distributed along the length and radius R=1 m lies on the hills of stonework. The distance between walls of the stonework l =1,6 m. Neglecting the friction find the pressure of the boiler on the stonework in points A and B. Answer: NA = NB = 33,3 kN. 2.53. Horizontal beam AB of the weight 100N can rotate around the fixed axis of the joint A. The end B is pulled up by means of the rope thrown over the block. The load of weight P=150N is hung to this rope. The load Q of weight 500N is applied at the point which is situated at the distance 20 cm from the end B. What is the length x of the rod AB if it is in equilibrium?

Answer: 25 cm. 2.54. The force couple (P,P) acts on the double cantilever. Distributed force with the intensity q acts on the left cantilever. Vertical load Q acts in the point D. Determine the reactions of the supports if P = 1 kN, Q = 2 kN, q = 2 kN/m, a = 0,8 m.

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2. Statics

Answer: R A  1,5kN , RB  2,1kN . 2.55. Determine the intensity of the load q which causes the end restraint moment А equals to 400 N•m. The dimensions are: АВ = 2m, ВС = 4m. Answer: 25N 2.56. The moment in end restraint A equals to 180 N m. Dimension of AC equals АС = 2 m, and intensity q equals to q = 30 N/m. Determine the length of BC. Answer: 2m 2.57. Define the reactions of cantilever under action of concentrated force P1 with the magnitude 2 kN and couple of forces P2 with the magnitude 3 kNm shown at the picture.

Answer: X=1 kN, Y = 1,73 kN, M = 0,47 kN m.

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2.58. Define the reactions of cantilever under action of distributed force q with the magnitude 1,5 kN/m, concentrated force P1 of 4 kN and couple of forces P2 with the magnitude 2 kNm shown at the picture.

Answer: X=2,8kN, Y=1,7kN, M=-5,35 kNm. SELF STUDY OF THE STUDENT

Find the reactions of support of composite structure consisting of three bodies connected in the point D. Scheme designs presented in Figures (dimensions in m), the load is shown in the table. The component parts are connected by a hinge at the figures 1 – 5, 8 – 10, and with help of the sleeve at the figures 6 – 7. Variant number 1 2 3 4 5 6 7 8 9 10

P1, kN 6 11 10 16 13 13 11 11 10 12

P2, kN 8 14 10 12 8 14 -

M1, kN m 25 34 30 25 26 38 34 30 34

120

M2, kN m 20 30 34 -

q, kN/m 0.8 1 0.9 1.5 1.4

2. Statics 1

2

3

4

5

6

7

8

9

10

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2.6. Conditions of equlibrium of a constrained solid body. Friction and constraints with friction Problem about equilibrium in the presence of friction

Up to here we were considering the equilibrium of ideal mechanical systems assuming that the surfaces of bodies in contact are absolutely smooth, there is no friction between them, and bodies themselves are absolutely solid. But this just approximately represents the facts. In real problems it is impossible to fully exclude the influence of friction force, otherwise we will get the results that slightly represent the facts. Friction forces essentially differ from all other forces. They appear in that cases when active forces are able to create a relative motion of bodies in contact. The contact of bodies never happens in one point as the bodies are deformed and no matter how small are the deformations the contact of bodies happens on some area the sizes of which can be usually neglected. Let us consider two solid contacting bodies А and В and let О be the point of contact. The instant motion of body B relative to body A always can be reduced to the instant-translational motion with relative velocity v0 of a point О of body B and to the instant-rotational motion of body B with relative angular velocity  the line of action of which passes through point О. Let (π) be the common tangent plane. Let us decompose vector  into two vectors, the first of which lies in plane (π), and the second is perpendicular to it:

  1  2 1 is called an angular velocity of rolling, 2 – an angular velocity of spinning. 122

2. Statics

An instant motion of body B relative to body А now can be prescribed as a set of three motions: sliding, rolling and spinning. The set of active forces acting on a body B in equilibrium can be reduced to the main vector F with line of action passing through the point О and to the main moment M . The action of that system of forces gets balance by the reaction forces from the side of point А which are reduced to the resultant force R and to a resultant couple with moment M 1 that meets the condition: F  R  0,

M  M1  0 .

Let us decompose the force R and the moment of a couple M 1 into components situated in the plane (π) and perpendicular to the plane (π):

R  R  Rn ,

M 1  M 1  M 1n .

Rn is directed along the normal to the plane (π).Let us call it a normal reaction. This force interferes the interpenetration of bodies. The component R lying in plane (π) we shall call the force of sliding friction. This force interferes the slippage of body B at body А. Moment M1n is perpendicular to (π) and interferes the spinning of a body. Let us call it as a couple of spinning friction. Component M1 lying in a plane (π) and interfering the rolling of a body let us call a couple of rolling friction. Note about rolling friction

The rolling friction appears by rolling of one body over the other. The occurrence of rolling friction can be roughly explained as by the contact of bodies the surfaces deform. In most cases the rolling friction is significantly less than the sliding friction and by solving problems it can be neglected. 123

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Note about spinning friction

Let us consider a heavy sphere lying on a horizontal plane and contacting it in point С in such a way that СО becomes a vertical radius of a sphere. Rotation about the vertical radius is called spinning. Reducing the system of active forces to the point С we will get the principle vector F and the principle moment M . Let for the simplicity vector M be parallel to ОС. Let us decompose the principle vector into vector components the first of which is perpendicular and the second is parallel to the plane (π): F  F1  F2 . Force F1 are balanced by normal reaction of a plane, force F2 – by force of sliding friction, and for the global equilibrium of the sphere the couple should be balanced. As it is know from the experiment if the moment of couple that tends to spin the sphere is small enough, then the sphere will not start to spin. To the action of active force in this case interferes some couple of forces reaction called the spinning friction. The limit moment of spinning friction can be expressed as a product of some coefficient k called as spinning friction coefficient and defined experimentally on a normal component of resultant active force, that is kF1 . The coefficient of spinning friction is usually a small value 5-10 times smaller than the coefficient of rolling friction. The conditions of equilibrium reduced to two inequalities: fF1  F2 , kF1  M . Problem about equilibrium of constrained solid body

Any constrained body can be considered as free rejecting the constraints and substituting their actions with reactions. Conditions of body equilibrium are the equations that bind active forces or parameters that define the position of the body and do not contain unknown constraint reactions. The number of independent displacements that can have a body is called a number of degrees of freedom. Free solid body has six 124

2. Statics

degrees of freedom. When the body is in equilibrium the forces should satisfy such kind of conditions that they could not impose to body the motions allowed by constraints that is why the number of equations equals to the number of degrees of freedom. Lever equilibrium. Lever is a solid body that can spin about the fixed axis under the action of forces located in plane that is perpendicular to that axis. P1 ,..., Pn  are active forces lying in plane xОy. Reaction of R also lies in that plane. As we have a plane system of forces, then we will have three conditions of equilibrium:

P  R  0, P  R  0,  mom P  0 . ix

x

iy

y

0

i

First two equalities give force R , so that we get Condition of lever equilibrium: sum of moments of all forces relative to the axis of rotation should be equal to zero. 2) Equilibrium of solid body that has one fixed point. There is a body with a spherical hinge in point О; P1,..., Pn  active forces. R is applied in point О and has an optional direction in space. Let us make up the conditions of equilibrium:

P

ix

 Rx  0 ,

P  R  0, P  R  0,  mom P  0 ,  mom P  0 , iy

y

iz

z

x

i

y

i

 mom P  0 . z

i

First three equalities serve for defining the reaction, so that. 125

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Condition of equilibrium of solid body having a fixed point: sums of moments of all forces relative to each of three orthogonally related axes passing through the fixed point should be equal to zero. 3) Equilibrium of the body having an axis of spinning-sliding. Let the body has the rotation axis along which it can slide. Let us choose coordinate axes. Reactions in points А and В have components YA, , Z A , YB , Z B . Let us make up conditions of equilibrium:

P

 0,

ix

P Y Y  0, P  Z  Z  0,  mom P  0 ,  mom P  mom Z  0 , iy

A

iz

A

x

y

i

B

B

i

y

 mom P  mom Y z

i

z

B

B

 0.

Only the first and the fourth equations do not contain unknown reactions of constraint. Consequently, for the body having the axis of spinning-sliding there are two conditions of equilibrium: sum of projections of all forces on a given axis and a sum of their moments relative to that axis equals to zero. 4) Equilibrium of the body having fixed axis of rotation. If one of hinges is made spherical, then the rotation axis becomes fixed, and here appears one more component of reaction directed along the z axis. Consequently, now we will have only one equality containing no unknown reactions:

 mom x Pi  0 . Then for the body having fixed rotation axis the condition of equilibrium is only one: sum of moments of all forces relative to that axis should be equal to zero.

126

2. Statics Questions 1. What is the rolling friction? 2. What is the spinning friction? 3. What are the equilibrium conditions for the lever? 4. What are the equilibrium conditions for the solid body with one fixed point? 5. What are the equilibrium conditions for the body having an axis of spinning-sliding? 6. What are the equilibrium conditions for the body having fixed axis of rotation?

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3 DINAMICS OF THE MASS POINT AND THE SYSTEM

3.1. The laws of Newton. Direct and inverse problems of dynamics. Motion equations A material point is called free if it сan move under the influence of applied forces in any direction in accordance with the basic laws of dynamics. If several forces F1 ,....., Fn are acting on the free material point of mass M simultaneously then the equation expressing the fundamental law of dynamics will take the form:

mw   Fi or mw  F

(1)

where F is the resultant of the forces applied to the point, and w is the acceleration directed along the line of action of the resultant force. dv d 2 r As w  , then equation (1) will take the form:  dt dt 2

mr  F

(2)

This equation is called the differential equation of free motion of a point in the vector form. Projecting both parts of the vector equation (2) on the axis of a coordinate system, one can obtain the differential equation of motion of free point in this coordinate system. Cartesian coordinate system or the natural trihedral axes are used most often. 128

3. Dinamics of the mass point and the system

If the rectangular Cartesian coordinate system Oxyz is in the state of rest, then, projecting equation (2) on its axis, we will get:  y  F y , mz  Fz mx  Fx , m

(3)

where x, y , z are the coordinates of a moving point, Fx , Fy , Fz are the projections of the acting force ( resultant force ) at the respective axis. Equations (3) are called differential equations of curvilinear motion of free mass point in the projections on the axis of a Cartesian coordinate system. Various forces can act on the mass point: constant forces (its magnitude and direction are constant, for example, the force of gravity close to the Earth) or variable forces (its magnitude and the direction changes during the motion). Variable forces can depend: A) only on time (for example, driving force of electric locomotive) B) only on the coordinates of the point (for example: the force of elasticity); B) only the velocity of the point (for example, the resistance force of the environment); In general, the force can depend on time t , on the positionvector r and velocity v of the point, i.e.  , equations (3) in the general form are:

mx  Fx (t; x, y, z; x, y , z) my  F y (t ; x, y , z ; x , y , z )

(4)

mz  Fz (t; x, y, z; x, y , z) Let’s find the differential equations of motion of a point in the projections on the axis of the natural trihedral, i.e. on the direction of tangent (  ), principal normal ( n ) and binormal ( b ) to the trajectory in the current position of the moving point. Projecting both sides of the vector equation (2) on this axes, we will get: 129

Textbook on Theoretical Mechanics

mw  F , mwn  Fn , mwb  Fb But from kinematics it is known, that:

v2 w  S, wn   , wb  0



Thus, finally we will find:

mS  F , m

v2



 Fn ,0  Fb

(5)

From the third equation follows, that the force as well as the acceleration lie in the osculating plane (  , n ). Equations (5) are called differential equations of curvilinear motion of free mass point in the projections on the axis of the natural trihedral. There are two main problems of dynamics of points: 1) Determine the action force knowing the mass of the point and its motion, i.e., coordinates of the point as a function of time; 2) Determine the law of motion of the point knowing its mass, acting forces, initial position and velocity; Solution of the first problem of dynamics. Knowing the law of motion, i.e. kinematic equations:

x  x(t ), y  y (t ), z  z (t ) ,

(6)

find the active force, i.e. Fx , F y , Fz . The problem can be easily solved with the help of equations (3) and reduced to the computation of second derivatives from given functions with respect to time. Example. Let the mass point moves under the law:

x  a sin kt , y  b cos rt , z  0 130

(a)

3. Dinamics of the mass point and the system

Trajectory of this point is an ellipse with semiaxes a and b .

Fx  mx   mak 2 sin kt , Fy  my  mak 2 cos kt , Fz  0 Or:

Fx   mk 2 x , Fy  mk 2 y или F   mk 2 r . This result gives the law of variation of force, under the influence of which the point can circumscribe any ellipse of the family (a). As we can see, such a motion is possible under action of a central force directed toward the center of an ellipse and changing proportionally to the distance of a point from the center. Solution of the second problems of dynamics. Knowing the acting force F , find the law of motion of a point, i.e. kinematic equations (6). Differential equations has the form (4):

mx  Fx (t; x, y, z; x, y , z) my  F y (t ; x, y , z; x , y , z )

mz  Fz (t; x, y, z; x, y , z) Finding the law of motion is reduced to the integration of the system (4), i.e. a system of three differential equations of second order. Integrating this system, we obtain x, y , z as a function of time and six arbitrary constants, i.e. we will find the general solution of (4) in the form:

 x  x(t , c1 ,......., c6 )   y  y (t , c1 ,........, c6 )  z  z (t , c ,.........., c ) 1 6 

131

(7)

Textbook on Theoretical Mechanics

Constants in the equations (7) show us that under the action of the force point can make not one specific motion, but a whole class of motions at different values c1 ,.....,c6 . Physically, this result is explained by the fact that the point which is influenced by a certain force will move in different ways, depending on the initial conditions, i.e. on the initial position and initial velocity of the point. For example, the motion of a free point under action of force of gravity can be straight or curved depending on the direction of its initial velocity. Initial conditions are usually represented by the equations: Initial position of the point x  x0 , y  y0 , z  z 0 . Initial velocity of the point x  x 0 , y  y 0 , z  z 0 . The constants of integration c1 ,.....,c6 can be determined with the help of these initial conditions. Taking the derivative of equation (7), we find the projections of the velocity:

 x  x (t , c1 ,......., c6 )   y  y (t , c1 ,........, c6 )  z  z (t , c ,.........., c ) 1 6 

(8)

Substituting the initial data in equation (7) and (8) we obtain the values t , x0 , y 0 , z 0 , x 0 , y 0 , z 0 on the left side, the value t 0 on the right side and unknown constants c1 ,.....,c6 . Solving this system of equations, we get the values of the constants corresponding to the given initial conditions, i.e.:

c k  f k (t , x0 , y 0 , z 0 , x 0 , y 0 , z 0 ), (k  1,6)

(9)

Replacing all ck in (7) with its values (9), we obtain a particular solution of the system of differential equations (4) satisfying the given initial conditions in the form: 132

3. Dinamics of the mass point and the system

x  x(t , x0 , y0 , z 0 , x 0 , y 0 , z 0 ) y  y(t , x0 , y 0 , z 0 , x 0 , y 0 , z 0 ) z  z (t , x0 , y0 , z 0 , x 0 , y 0 , z0 )

(10)

Equations (10) determine the motion of the point under the action of applied forces for a given initial conditions. Questions 1. How can be derived the differential equations of motion of a mass point form the fundamental law of dynamics? 2. What are the two main problems of dynamics? 3. What is the concept of first problem of dynamics? 4. What is the concept of second problem of dynamics? 5. Derive the differential equations of motion of a point in the projections on the axis of the natural trihedral.

PRACTICE 3.1. A stone with mass 0,3 kg tied to a thread with length 1m circumscribes a circle in a vertical plane. Define the lowest angular velocity ω of a stone at which the thread tears if the resistance to its break equals 9N. Answer: ω min =4,494 radps 3.2. Sporting airplane of mass 2000kg flies horizontally with acceleration 5 m/sec2 and now has velocity of 200 m/sec. Air resistance is proportional to square of velocity and when the velocity has the value of 1m/sec it equals to 0,5N. Assuming that resistance force is directed at the side opposite to velocity, define the thrust of a propeller if it makes an angle of 10° with direction of flight. Define the magnitude of lifting force now. Answer: Thrust equals to 30 463N, lifting force equals to 14 310N. 3.3. Define the motion of heavy ball along the imaginary channel passing through the Earth center if we assume that the 133

Textbook on Theoretical Mechanics

attractive force inside the globe is proportional to the distance of a moving point from the Earth center and directed to that center; the ball is dropped into the channel from the Earth surface without initial velocity. Indicate the velocity of the ball when it cross the Earth center and time of traveling to that center. Earth radius equals R = 6,37·106 m? g = 9,8 m/sec2. Answer: The distance of ball from Earth center changes under a law x  R cos

g t , v  7,9  103 m / sec, T  1266,4 sec  21,1 min . R

3.4. The airplane A flies at a height of 4000 m over the ground with a horizontal velocity of 140 m/sec. At what distance х measured along a horizontal straight line from a given point B should any load be dropped from an airplane without an initial relative velocity in order that the load fell on that point? Neglect the air resistance. Answer: х = 4000 m. 3.5. Stone falls into the mine without an initial velocity. The sound from the hit of the stone against the mine bottom is heard in 6.5 sec from the moment of its falling start. Velocity of sound equals to 330 m/sec. Find the mine depth. Answer: 175 m. 3.6. Heavy point ascends along the rough inclined plane that makes an angle a = 30° with the horizon. At the initial moment velocity of the point equaled to v0 = 15 mpsec. Coefficient of friction is f = 0.1. What way passes the point until it stops? During what time the point passes that way? Answer: s=19.57 m, t= 2.61 sec. 3.7. From a gun that is in the point O was fired a shot at angle a to the horizon at initial velocity v0. Simultaneously from the point A that is at the distance l horizontally from the point O was fired a shot vertically up. Define at what initial velocity v1 a second shell should be fired in order that it hit the first shell if velocity v0 and point A lie in one vertical plane. Neglect the air resistance. 134

3. Dinamics of the mass point and the system

Answer: v 1=v 0 sin a 3.8. River depth measurment is being taken with the help of the load that is dropped on a cable into water till the river bottom. By dropping the load at v0 the cable ruptured and the load reached bottom in T seconds after the moment of cable rupture. Define the way H passed by the load to the river bottom if the projection onto the x axis of the the water resistance force to the load motion equals to R x  kmx where m is load mass, x is the projection of its velocity on the axis x, k is constant coefficient. X axis is directed vertically down. Neglect the force of ejection of the load from the water. Answer: H 

g  kv0 g t (1  e  kt ) k k2

3.9. Load of weight P that was in rest on a smooth horizontal plane starts to move under the action of horizontal force which projection on a directed horizontally to the right x axis equals to F x = H sin kt where H and k are constant magnitudes. Find the law of load motion. 3.10. Define the slope angle of the gun tube to the horizon if the aim is detected at the distance of 32 km and the initial velocity of the projectile is v 0=600 mpsec. Neglect the air resistance. o o ' Answer: α 1=30 18 ' , α 2=59 42 3.11. A weight with mass 0.2 kg is hung to an end of a thread with length 1m. In the result of a push the weight got horizontal velocity of 5 m/sec. Find tension of the thread right after the push. Answer: 6,96N 3.12. A point with mass m starts moving from the state of rest from the position x0 = a straight under the attractive force that is proportional to the distance from the origin of coordinates: F x =− c 1 mx and the repulsive force that is perpendicular to the 135

Textbook on Theoretical Mechanics

cube of a distance: Q x =c 2 mx 2 . In what correlation of c 1 , c 2 , a the point will reach the origin of coordinates and stop? Answer: c 1=1/2 c 2 a 2 . 3.13. A plane A flies over the ground at a height h with a horizontal velocity v1. From a weapon B it was given a shot against the plane when it is on a same vertical with a weapon. Find: 1)What condition should an initial velocity v0 satisfy in order that weapon could hit the plane, and 2)At which angle α to a horizon should the shot be made. Neglect the air resistance. 2 2 Answer: 1. v 0≥ v 1+ 2gh ; 2. cosα=v 1 / v 0 . 3.14. Heavy body descends along the smooth plane inclined at an angle 30° to the horizon. Find for what time body travels 9.6 m if at the initial moment its velocity equaled to 2 m/sec. Answer: 1.61 sec 3.15. For what time and at what distance may be stopped by stopper a street car that goes along the horizontal way at 10 m/sec if running resistance that develops while stopping is 0.3 of car's weight. Answer: t = 3.4 sec, s = 17 m 3.16. Stone is thrown vertically up at v0. Define at what height H from the Earth surface the stone velocity reduces twice if the projection onto the x axis of a running resistance force r equals to: R x  kmx 2 where m is the stone mass, x is the projection onto x 2

axis of its velocity, k is the constant coefficient. X axis is directed vertically up. g  k 2 v02 Answer: H  1 2 ln 4 2 2 2k

4 g  k v0

3.17. The greatest horizontal range of a missile equals to L. Equations of motion of missile are: x  v0t cos , y  vo t sin   gt 2 / 2 .

136

3. Dinamics of the mass point and the system

Define its horizontal range l at throwing angle α = 30° and height h of trajectory in that case. Neglect the air resistance. L Answer: l  L 3 / 2 , h= 8 3.2. Basic dynamic variables. Properties of internal forces of the system The set of mass points or bodies where the position or motion of each mass point or body depends on the position and motion of other points or bodies is called mechanical system of mass points or bodies. Consider the mechanical system of n mass points or bodies. Internal forces are the forces with which the points or bodies of this mechanical system acting on each other (for example, the forces of mutual attraction of planets in the solar system). The forces with which the points or bodies that are not contained in mechanical system act on the points or bodies of mechanical system are called external forces (for example, the forces with which the stars and stellar cluster acting on the planets of solar system). Properties of internal forces of mechanical system 1. Geometric sum (principal vector) of all the internal forces of the mechanical system is equal to zero, i.e. n

R ( i )   Fk( i )  0 k 1

2. Geometric sum of moments (principal moment) of all internal forces of a mechanical system with respect to a fixed center O is equal to zero, i.e. n

M 0( i )   mom 0 Fk( i )  0 . k 1

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But it does not mean that the internal forces balance each other and do not affect on the motion of mechanical system, because these forces are applied to various points of the system and can cause the relative displacement of these points. The internal forces are balanced for an absolutely rigid body. Motion of mechanical system depend both on its mass or the mass distribution in the system and the acting forces. Mass of the mechanical system is equal to the arithmetic sum of the masses of all the points contained in this system: n

M   mk . k 1

The mass distribution is characterized by the position of the center of mass, or center of mass of the mechanical system. The center of mass is called a geometric point C, whose position relative to the selected reference system is defined by the position-vector: n

rc 

m r

k k

k 1

M

,

where mk (k  1, n) is the mass of a mass point, rk (k  1, n) is the position vector of these points. Position of center of mass does not depend on the forces acting on the system (this can be seen from the formula) and the coordinate system. Proof: Consider a second system of coordinates. The position vector of k -th point is rk :

rk  rk r 0 Let’s consider the position vector rc (the position vector of the center of mass in the new coordinate system): 138

3. Dinamics of the mass point and the system

rc 

1 M

n

 mk rk  k 1

1 M

n

m k 1

k

(rk  r0 ) 

m r  m r k k

M

k

M

0

 rc  r0

Thus, rc  rc  r0 and the ends of vectors coincide, i.e. they define the same point C – the center of mass. Consider the motion of n mass points in an inertial coordinate system. Assume that m is the mass of any point of the system, r its position vector. In the general case the point M  are influenced by the external and internal forces, which can be active and passive. Let’s denote: F e is the resultant of all external forces (active and passive ).

F i is the resultant of all internal forces. On the basis of axiom of coupling we can consider this point as free, so the equation of motion of this point (and all other points) will be: (11) m w  F e  F i (  1, n) where w is the acceleration of the point in inertial coordinate system. To study the motion we should integrate the system of equations (11) with the given initial conditions and find the dependence r on t . This is impossible in most cases, particularly if the number of equations is large. However, in the process of practical studies of motion it is often not necessary to study the system (11). It is enough to know the variation of some variables with time that are common to the whole mechanical system and are the functions of the coordinates and velocities of the points of system (and maybe the time). If such a function remains constant during the motion, it allows to simplify the problem and sometimes solve it to the end. The most common method of obtaining first integrals of equations (11) is based on a study of behavior of the main dynamic quantities of the system: linear momentum, angular momentum, 139

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kinetic energy. Changing these values over time is described with the help of the main theorems of dynamics, that are the direct consequence of equations (11). The main dynamical quantities. König's theorem Linear momentum is a vector quantity equal to the product of the mass point on its velocity.

q  mv . Let’s consider a mechanical system of n mass points in motion. Vector of linear momentum can be associated with each point of the system. This system of sliding vectors is reduced to the principal vector and principle moment. The principle vector of linear momentum of the system will be equal to the geometric sum of the linear momentum of all points in the system: n

Q   m v .  1

From the formula for determining the position vector of the n

center of mass of the system, it follows that Mrc   m r , as  1

n dr , we can write: Mv c   m v  Q ; v dt  1 Thus, Q  Mvc , Q  Mvc

That is, the linear momentum of the system is equal to the mass of the system, multiplied by the velocity of its center of mass. The vector Q according to the system C x y z  with the origin in the center of mass is equal to zero, because v c  0 relative to these axes. The principle moment of linear momentum (angular momentum) of the system relative to the center О : 140

3. Dinamics of the mass point and the system

G0   (r  m v ) 

i j k G0  x y z mx my mz

The angular momentum of the system relative to the axis is the projection of linear momentum of the system on this axis with respect to any selected center of the given axis. Let’s transform the expression for the angular momentum of the system: dr G0   ( rc  m v c )   ( r  m  ) . dt   Assume that C is the center of mass of the mechanical system. Assume that the point C is the origin of coordinate system Cx y z  moving steadily relative to the inertial coordinate system Oxyz . Let’s denote: M  is any point with the mass m ,

r is its position vector in the system Oxyz , r is its position vector in the system Cxy z  ,

rc is the position vector of the center of mass. The system is moving, it follows that all the position vectors r ,

rc and r are the functions of time t . r  rc  r is valid for any moment of time, therefore: 141

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dr drc dr .   dt dt dt Let’s substitute these relations in the expression for the angular momentum of the system:  dr    dr dr    G0   rc  r   m  c      rc  m vc     rc  m    dt dt dt         dr      r  m vc     r  m   dt    

m  

Note, that

 M ,  m r  Mrc  0 (as the position vector

of the point C in the axes Cx y z  is equal to zero). Then we have the following equations:

r  

c



r   

c

 m vc   rc  Mvc  m

dr  dr  d   rc   m   rc   m r  0 dt dt dt  

r  m v    m r  v    c



 r  m   

c

0

dr     r  m v  dt  

where v is the velocity of the point M  relative to the axis Cx y z  . Then the linear momentum will take the form:



G 0  rc  Mv c   r  m v 



rc  Mv c is the angular momentum of the center of mass on the assumption that mass of the entire system is concentrated in it.  r  m v  is the angular momentum about the center of mass. 

142

3. Dinamics of the mass point and the system

Hence:

G 0  rc  Mv c Gc or G0  rc  Mv c Gc , where Gc   r  m v  . 

Indeed as v  vc  v we have:

r  m v    r  m v    r  m v    r  m v  ,      c

i.е. Gc  Gc Thus it was proved the following theorem. Theorem. The angular momentum of the system relative to some fixed center is equal to the sum of the moment relative to this center and the angular momentum of the system relative to the center of mass in its motion relative to the moving coordinate system that executes the translational motion with the center of mass. G0  rc  Mvc  Gc

Angular momentum don't change when the center of reduction is changed. Let's consider two different '

centers A and B. Let r and r are the position vectors of the points A and B respectively. Then: n

n

n

 1

 1

 1

GB  r '  m v  (r  BA)  m v  r  m v  BA m v  GA  BA Q

Thus, GB  GA  BA  Q 143

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Angular momentum of the rotating body relative to the axis of rotation

Assume that absolutely solid body rotates around axis x with angular velocity  . GX is the sum of the angular momentum of all the points of the body relative to the axis x . As all velocities are perpendicular to Ox , then for any point M  angular momentum is equal to r  m v  h  m v  m v h . This implies, that Gx  then G x   m h 2    m h2 . 

Value

m h  

2

m v h , as v  h,  



 J x is called the moment of inertia about the

axis x . then Gx  J x . The value T is called the kinetic energy of the system:

T

1 n  m v2 2  1

Assume that C is the center of mass of the mechanical system. Consider the coordinate system Cx y z  executing translational motion relative to an inertial coordinate system Oxyz . then

v  vc  v Substituting this value in the expression for kinetic energy we will get: n 1 n 1 1 n T   m (vc  v ) 2   m vc2   m vc  v   m v 2 . 2  1 2 2  1  1 144

3. Dinamics of the mass point and the system

Let's transform the right side: 1 n 1 m v c2  Mv c2 ,  2  1 2 n n dr  d d m vc  v  vc   m v  vc  m   vc  m r  vc  Mrc  0 ,  dt dt dt  1   1

as rc  0 . Finally we will get:

T

1 1 Mvc2   m v 2 . 2 2 

Thus the theorem is proved. König's theorem. Kinetic energy of the system is equal to the sum of the kinetic energy of the center of mass (where the mass of the system is concentrated) and the kinetic energy of the system as it moves relative to the moving reference frame (moving translational with the center of mass). Kinetic energy of the rigid body

Absolutely rigid body is the body in which deformation is neglected. In other words, the distance between any two given points of a rigid body remains constant in time. Firstly we find the kinetic energy of a rigid body that executes the translational motion. Then for each point of the body v  vc , where vc is the velocity of the center of mass and T

1 n 1 1 m v2   m v c2  Mv c2 .  2  1 2 2 145

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Now we calculate the kinetic energy of a absolutely rigid body rotating about the axis Ox with angular velocity  . It is known, that the velocity of any point of the rotating body may be calculated with the help of formula:

v  h , where h is the distance of the point m from the rotation axis Ox . Value J x  1  m h2 is the inertia moment of a body about the 2

axis of rotation. This formula is also true in the case when axis Ох is the instantaneous axis of rotation. And J x will be the moment of inertia of the body relative to the instantaneous axis of rotation. In general case of motion kinetic energy of a rigid body is computed by Koenig theorem. In accordance with Schall theorem, motion of a rigid body with respect to the moving axes Cx y z  composed of instantaneous rotations about the axis passing through the point C. Then Koenig theorem gives:

T

1 1 Mv c2  J cx w 2 , 2 2

where J cx is the moment of inertia about the instantaneous axis of rotation passing through its center of mass, and  is the instantaneous angular velocity of the body. Questions 1. Give the definition of mechanical system. 2. What is the main property of internal forces? 3. What are the main dynamical quantities? 4. How can be defined the linear momentum of mechanical system? 5. How can be defined the angular momentum of mechanical system? 6. How can be defined the kinetic energy of mechanical system and rigid body?

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3. Dinamics of the mass point and the system

3.3. Theorem of change of linear momentum of a mass point and mechanical system. Theorem on the motion of the center of mass Theorem of change of linear momentum of a mass point

Let’s write the fundamental law of dynamics for the mass point:

mw  F ,

(1)

where F is the resultant of all forces, acting on the point. As w  v , m  const is the weight, equation (1) can be rewritten: d ( mv ) (2) F dt As it is known, a vector m v  q equal to the product of mass and velocity, and having the direction of velocity is called a linear momentum of mass point. Equation (2) expresses the theorem of change of linear momentum of a mass point in a differential form: the time derivative of the linear momentum of a mass point is equal to the force acting at this point. If we multiply (1) by dt , we will get:

d ( mw )  F dt

(3)

The value F dt is called an elementary impulse of force acting on the mass point for the elementary interval dt . Suppose that the mass point has the position M 0 and velocity

v0 at a time t0 and the position

M and velocity v at a time t .

Taking a definite integrals of (3) with the limits according to the motion of a point from M 0 to M , we will get: 147

Textbook on Theoretical Mechanics t

mv  mv0   F dt

(4)

t0

t

Integral  F dt is called a total impulse of acting force for the t0

finite time interval t  t 0 . Equation (4) is a theorem of change of linear momentum in integral form: change of linear momentum of a mass point for a finite period of time equal to the total impulse of the force acting on this mass point for the same period of time. If the force acting on the point is constant in magnitude and direction or depends only on time, the right-hand side of (4) can be easily integrated. If there are no forces acting on the point or they are equal to zero, then from (2) we will get: d ( mv )  0  mv  C  const dt

(5)

Or according to (3):

mv  mv0

(6)

Equation (5) or (6) is called the law of conservation of momentum of the mass point. From (5) it follows that the mass point executes the uniform and rectilinear motion (the law of inertia). Theorem of change of linear momentum of the mechanical system n

Vector Q   m k v k is called the linear momentum of mechak 1

nical system. Summing the equations of motion of mass points of the mechanical system:

 mw

k

  Fk( e )   Fk(i ) , 148

3. Dinamics of the mass point and the system

It is known that the sum of internal forces for mechanical system is equal to zero:

F

(i ) k

0

Let’s transform the left side:

 mw   m k

k

dVk d dQ ,  (  m k Vk )  dt dt dt

Finally we will have:

dQ   Fk(e ) dt

(7)

This equation expresses the theorem of change of linear momentum of mechanical system in the differential form: the time derivative of the vector of linear momentum of mechanical system is equal to the sum of all acting external forces (or principal vector of all the external forces of the system). This theorem can be represented in the integral form. Integrating both sides of (7) from t 0 to t , we will get: t

Q  Q0    Fk( e ) dt t0

Theorem in the integral form: change of linear momentum of mechanical system for a finite period of time is equal to the total impulse of the principal vector of all acting forces for the same period of time. In some cases, these theorems can give first integrals of motion. A). If there is no acting external forces or the resultant of external forces is equal to zero, then: 149

Textbook on Theoretical Mechanics

dQ  0  Q  const  Q0 dt Or, as Q  Mv c , then vc  const . In this case the center of mass of mechanical system is coasting and liner momentum Q is constant. Projecting the vector Q on the axis of coordinate, we obtain the first three integrals from the law of conservation of linear momentum: Q x  c1 , Q y  c 2 , Q z  c3 , or x c  c1 , y c  c2 , z c  c3 B). If the projection of principal vector of external forces on some fixed axis (for example on the axis Ox ) is equal to zero Fkx( e )  0 , then Qx  const  Qox or x c  const . Theorem on motion of the center of mass of mechanical system

If the number of points of mechanical system is large, it is difficult to count Q 

n

m v k 1

k

k

. Let’s find a different formula from

the definition of the radius vector of the center of mass: n

m r

k k

k 1

M

n

 rc   mk rk  Mrc k 1

Let’s differentiate the both sides with respect to time: n

 mk k 1

n drk dr  M c or MvC   mk v k , dt dt k 1

where vC is the velocity of center of mass of mechanical system. 150

3. Dinamics of the mass point and the system

As the left side is the linear momentum Q then:

Q  MvC

(8)

That is, the linear momentum of mechanical system is equal to the product of the mass of the system by the vector of velocity of its center of mass. We substitute (8) in the theorem of change of linear momentum of mechanical system:

dv C d  ( mvC )   Fk( e ) or M dt dt

n

 Fk(e )  MwC   Fk(e ) k 1

where wC is the acceleration of center of mass of the system. This equation represent the theorem on motion of center of mass of mechanical system: center of mass of mechanical system moves as a mass point with the mass of whole mechanical system which the principle vector of external forces is applied to. Thus, the internal forces do not affect the motion of center of mass of mechanical system. Problems of the dynamics of translational motion of solid body are solved with the help of this theorem. Questions 1. What is the linear momentum. 2. What is the concept of theorem of change of linear momentum of a mass point in a differential and integral form? 3. What is the impulse of force? 4. When the momentum theorem can give first integrals of motion? 5. Give the formulation of the theorem on motion of the center of mass of mechanical system.

SELF STUDY OF THE STUDENT

1. Truck of weight P=8500N together with useful load moves along the horizontal way and sustain the resistance the magnitude of which is 0.01 from all vertical loads. Worker pushes the truck with 151

Textbook on Theoretical Mechanics

the force F=170N. In what period the worker will impose velocity v =0,6 mps to the truck? 2. Define the time during which the body thrown angularly α0 to the horizon with initial velocity v0 reaches maximum height. Use the momentum theorem. Consider that body is affected by air resistance the magnitude of which expressed by formula R = kPv where k – constant coefficient, P – weight of body, v – its velocity. 3.

A

force

which

projections

onto

coordinate

axes

X  6 cos 2t , Y  6 sin 2t , Z  6 sin 2t acts on a mass point with weight

sec m=2kg. Define velocity v 2 of the point at the time t 2    3,14 сек  if at the time t1  сек sec its velocity v1 equals in absolute value to 2

2m/sec and makes angles 30°, 60°, 90° with the coordinate axes x, y, z correspondingly. 4. Point with mass m starts to move from the state of rest rectilinearly under the force changing under the law F  F0 (1  kt ) . Find the velocity of the point when the force turns into zero and define the time different from initial time when velocity of the point turns into zero. 5. A vertical force F increasing from zero proportional to time acts on a body with the weight P=20N that is in rest in a horizontal plane. Coefficient of proportionality equals 2N/sec. In what time after the start of action of force F the body begins to move? Find the law of that motion. 6. The load descends along the rough inclined plane located angularly α to the horizon, f – coefficient of sliding friction of the load. Velocity of the load equals v at the initial time. In what period of time the velocity of load doubles? 152

3. Dinamics of the mass point and the system

7. Solid body with weight P begins to move from the state of rest along the rough horizontal plane under the action of force F that is proportional to time: F = at where a – constant. What velocity has the body in t – seconds after the beginning of motion if the coefficient of sliding friction on a horizontal plane equals f. 8. A boy with mass of 40 kg stands on the sledge runners the mass of which equals 20 kg. The boy every second makes a push with impulse 20 Ns. Find the velocity that gains the sledge in 15 sec if the coefficient of friction is f = 0,01. 9. The ball is moving along the tube from the position A. Find the velocity of the ball in the positions B and C. Neglect the friction on the curvilinear parts of the trajectory. Mass of the ball is 0.5 kg, initial velocity of the ball in the position A is 20m/sec, the time of moving of ball along the part AB is 2 sec, friction coefficient is 0.2. Besides R=2m, α = 30°, β = 45°. 10. Mechanical system consists of the load A with the mass m1 lying on the inclined plane of the wedge B of mass m2 that makes an angle α with the horizon. The system was in rest at the inital time. After that the load A began to slide along the inclined plane with the relative velocity u. Define the velocity of the wedge B. Neglect the friction force. 11. The ball is moving along the tube from the position A. Find the velocity of the ball in the positions B, C and D. Neglect the friction on the curvilinear parts of the trajectory. Mass of the ball is 0.4 kg, initial velocity of the ball in the position A is 153

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2 m/sec, the time of moving of ball along the part AB is 0.2 sec, friction coefficient is 0.4. Besides R=2m, α = 30°. 12. The ball is moving along the tube from the position A. Find the velocity of the ball in the positions B, C and D. Neglect the friction on the curvilinear parts of the trajectory. Mass of the ball is 0.2 kg, initial velocity of the ball in the position A is 5m/sec, the time of moving of ball along the part AB is 0.5 sec, friction coefficient is 0.1. Besides R=1 m, α = 45°. 3.4. Angular momentum theorem of a mass point and mechanical system Angular momentum theorem of a mass point

Let’s consider a mass point of mass m moving under action of the force F . Let’s write the fundamental law of dynamics for the point: dv m F dt Multiply on the left by the vector r : r  m dv  r  F , where dt r  F  mom0 F is the moment of force F relative to the fixed center O. Let’s transform the right side of the expression:

r m

dv d  ( r  mv ) , dt dt 154

(1)

3. Dinamics of the mass point and the system

As

d dr dv ,  mv  r  m ( r  mv )  dt dt dt dr  mv  v  mv  0 dt Then we will get:

d ( r  mv )  r  F dt

(2)

vector K O  mom O ( mv )  r  mv is called the angular momentum or the angular momentum of a mass point relative to the fixed center O. Magnitude of this vector is equal to: K O  r  mv  sin   mvd Equation (2) can be written: dK o  momo F dt

(3)

The theorem of change of angular momentum in the differential form: time derivative of the angular momentum of a point relative to some center is equal to the moment of acting force relative to the same center. If we project it onto the coordinate axes: dK z dK y dK x  momz F  momx F ,  mom y F , dt dt dt

That is, the time derivative of the angular momentum of a mass point relative to the axis is equal to the moment of force acting relative to the same axis. 155

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Let’s consider special cases. a) case of a central force. If the line of action of F passes through the fixed center O throughout the motion, the force F is called a central force, and the point O is the center of this force. In this case, mom O F  0  dK o  0  K o  const , dt

i.e. the angular momentum of a mass point relative to the center of force stays constant in magnitude and direction in the case of central force. This is the law of conservation of angular momentum of a point relative to this center. b) case when the moment relative to the axis is zero. Assume dKz that mom z F  0 , тогда  0  Kz  const. That is, if the moment dt of force acting on the mass point relative to some fixed axis is equal to zero all the time, the angular momentum of the mass point relative to this axis remains constant. This is the law of conservation of angular momentum of a mass point relative to this axis. Angular momentum theorem for mechanical system

Let’s consider the differential equations of motion of the system:

m

d 2 r  Fe  Fi (  1, n ) dt 2

and multiply both sides of the equation by the vector r , summing up, we have:

 d 2 r  r m     dt 2  

    r  F e   r  Fi    



156







(1)

3. Dinamics of the mass point and the system

As the internal forces are equal and have mutually opposite directions the sum of the moments of these forces about any center is equal to zero. This implyies:

r  F   0   i

Taking into account that d 2 r dr  d    r    , 2 dt  dt  dt dr dr dr d 2r d 2r d as  r         r  2  r  2 , dt  dt  dt dt dt dt r 

Rewrite (1) in the form:



dr  d  e  r  m     r  F  dt   dt  



(2)

Denote the angular momentum relative to the center O as dr   G O    r  m   , dt   

r  F   M   e

e 0

is the principle momentum of all external

forces acting on the mechanical system. Then we can write (2) in the form: e dG0  MO. dt

(2’)

Equation (2) or (2 ') expresses the angular momentum theorem of a mechanical system: time derivative of the angular momentum of the system in relation to a fixed center is equal to the sum of the moments of all external forces acting on the system relative to the same center. 157

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Obtain a theorem in integral form, if we integrate (2 ') from t1 up to t 2 : t2

G0  G02  G01   M 0e dt

(3)

t1

t2

where

M

e 0

dt is the impulse of moments of external forces for the

t1

period of time t 2  t1 . Thus, increment of angular momentum of the system relative to the center О for a finite period of time equal to the impulse of moments of external forces relative to this center during that time. In some cases, the theorem gives the first integrals. A). Assume that external forces do not act, then

r  F   0  ddtG   e

0

 0  G0  const .

A plane perpendicular to the direction G 0 that will have a constant direction in space is called invariable plane of the Laplace. Making a projections we will get the first three integrals:

G ox  c,1 G oy  c 2 , G oz  c 3 B). If the sum of all moments of external forces with respect to one of the axes (for example axis X) is equal to zero, we obtain a first integral: G x  const . We have considered the center O as fixed till now. Let's see what will change if we take a point O  for the center, which is moving relative to the main coordinate system: 158

3. Dinamics of the mass point and the system

GO   r  m v , 



r  r  OO 





GO   r  OO  m v   r  m v   OO   m v 





As Q   m v , 

 GO  GO  OO  Q , Denote GO  G ,

(4)

GO  G .

Differentiating (4) with respect to time and taking into account that d OO  v  is the velocity of the center O  relative to the main dt

coordinate system we will get: dG  dG dQ ,   v   Q  OO  dt dt dt

It is known that dQ   Fe . Then we can write: dt



OO  

Let express for the theorem:

dQ  OO    F e . dt 

dG by formula and substitute it in the expression dt





dG   r  Fe , then we will get: dt





dG   v   Q   r  OO   F e , dt 

159

Textbook on Theoretical Mechanics





Or taking into account that r  OO   r , finally we will have: dG  (5)  v   Q   r  F e , dt  This equation expresses the theorem of momentum of the system relative to the center to the main coordinate system with the velocity

change of angular O  , moving relative v.

Questions 1. What is the angular momentum. 2. What is the concept of angular momentum theorem for the mass point and the system? 3. When the angular momentum theorem can give first integrals of motion? 4. What are the special cases of angular momentum theorem? 5. How will change the formulation of angular momentum theorem if the point assumed as the center is moving relative to the main coordinate system?

PRACTICE 3.18. The train moves along the horizontal and rectilinear section of the railway. Resistance force equal to 0.1 of weight of the train appears during the braking. Velocity of the train equals to 20 m/sec at the moment of braking beginning. Find the time of braking and braking distance. Answer: 20.4 sec, 204 m 3.19. A heavy load moves down along the rough inclined plane that makes an angle α = 30° with the horizon. Define during what time T the body passes the way of length l = 39.2 m if the coefficient of friction is f = 0.2. Answer: T = 5 sec. 3.20. A bullet of mass 20g flies out of the rifle barrel with the velocity v = 650 mpsec running the bore of the barrel for the time l = 160

3. Dinamics of the mass point and the system

0.00095 sec. Define the mean magnitude of the gases pressure that throw the bullet if the area of bore section is S = 150 mm2 . Answer: mean pressure is P  9.12 10 4 N / mm 2 3.21. Find the impulse of a resultant of all forces that act on a shell during the time when the shell from the initial position O goes into the highest position M. Given that v0 = 500 m/sec, α0 = 60°, v1 = 200 m/sec, mass of shell is 100kg. Answer: Projections of resultant impulse: Sx = – 5000 N·sec, Sx = – 43300 N·sec. 3.22. The train with mass 4  105 kg goes to a rise i=tg α=0,006 (where α – rise angle) at 15m/sec. Coefficient of friction in motion equals 0,005. The velocity of the train decreases to 12,5m/sec in 50 sec after the entry to a rise. Find the thrust of a diesel locomotive. Answer: 23120N. 3.23. At the time when velocity of motor ship equals v 0 the engine switches off and ship moves sustaining water resistance the magnitude of which is proportional to velocity. The coefficient of proportionality equals μ: Mass of ship equals m. In what period of time the velocity of ship will decrease twice?

Answer:

m



ln 2 .

3.24. The boat is in the state of rest. Two people are sitting on the middle bench of the boat. One of them having a mass M1=50 kg moved to the right to the nose of the boat. In what direction and at what distance should move another man of mass M2=70 kg so that the boat stayed in rest? Length of the boat is 4m. Neglect the resistance of water. Answer: to the left at the distance 1.43 m. 3.25. An acrobat performing a somersault and pushing off from the ground imposes himself at the initial time an angular velocity 161

Textbook on Theoretical Mechanics

1  1 rev / sec around the horizontal axis passing through his center of gravity. The moment of inertia of an acrobat relative to the axis equals J1  1,5 kg m2 . In order to increase the angular velocity in flight an acrobat tucks his legs and arms to his body thereby decreasing the moment of inertia to the value of J 2  0,5 kgm2 . Define the acrobat's angular velocity of rotation  2 around the horizontal axis in flight. Resistance forces should be neglected. Answer: 2  3 rev / sec .

3.26. At start up of electric crab rotative moment mrot  at was attached to the axle A. Rotative moment is proportional to time t, a is constant. Load B of mass M1 rises by means of a cable twisted round the axle A of radius r and mass M2. Define the angular velocity of the axle assuming it as solid cylinder. The crab was in rest at the initial time.

Answer: ω=

(at − 2 M 1 gr )t r 2 (2 M 1+ M 2 )

.

3.27. Homogeneous round disc with radius r executes oscillations around fixed horizontal axis that is perpendicular to the plane of a disc and passes through the point O. The distance from the point

O to the gravity center C of the disc equals

r . Find the law of the 2

disc motion under small-amplitude oscillations and the period of that oscillations. At the initial time deviation angle of the disc  from equilibrium position equals to  0 and its initial angular velocity equals to zero. Note: when solving this problem it can be assumed that sin    due to the small-amplitude oscillations. Answer:    0 cos kt , T  2

3r 2g

162

3. Dinamics of the mass point and the system

3.28. Round horizontal platform rotates without friction around the vertical axis passing through its center of mass with the constant angular velocity 0 . Four people of the same mass stand at the platform. Two of them stand on the border of the platform, another stand at the distance equal the half of radius of the platform from the axis of rotation. How will change the angular velocity of the platform if the people standing at the border move along the circle sideways the rotation of platform with the velocity u and people standing at the distance equal the half of the radius of the platform move in the direction opposite to the relative velocity 2u? Consider the people as the point mass and the platform as the round homogeneous disc. Answer: the platform will rotate with the same angular velocity. 3.29. The point moves steadily round a circle at v  0,2m / sec making a complete revolution within T = 4 sec. Find impulse S of forces that act on a point during the time of one half-cycle if mass of a point m = 5 kg. Define the mean value of the force F. Answer: S  2 N  sec, F  1N . 3.30. Define the mass of the Sun using next data: Earth radius R  6,37  106 m , average density 5,5t / m3 , semi-major axis of earth

orbit a  1,49 1011m , the time of the Earth rotation about the Sun T =365, 25 days . Universal gravitation force between two masses that 2 equal to 1 kg in a distance of 1 m assume equal to gR N where m –

m

mass of the Earth. It follows from Kepler laws that the attractive 2 3 force of Earth by the Sun equals 4 π a m where r – is a distance T 2 r2 between Earth and Sun. Answer: M  1,966  1030 kg . 3.31. During the flying of the weapon its rotation about the symmetry axis slows down under the action of moment of force of air resistance that is equal to kω where ω is angular velocity of 163

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weapon rotation, k – constant coefficient of proportionality. Define the law of decrease of angular velocity if the initial angular velocity equals to ω 0 and the moment of inertia of the weapon relative the symmetry axis equals J. Answer:   0e

k  t J

.

3.32. The rotary crane tram moves with the constant velocity v relative to the arm. The engine rotating the crane generates rotational moment that equals to m0 in the acceleration period. Define the angular velocity  of the crane rotation depending on distance x from tram to rotation axis AB if mass of the tram with the load equals to M and J is a moment of inertia of the crane (without tram) relative to the rotation axis. Rotation starts when tram is in a distance of x0 from AB axis.

Answer:  

m0 x  x0 . J  Mx 2 v

3.33. Barrel A of mass M1 and radius r is set in rotation by means of the load C of mass M2 tied to the end of nonstretching rope. The rope is thrown over the block B and reeled on the barrel A. Resistance moment mc which is proportional to the angular velocity of the barrel is applied to the barrel A. Coefficient of proportionality equals α. Determine the angular velocity of the reel if the system was in rest at initial time. Neglect the mass of the rope and block B. Consider the barrel as the solid cylinder. 2 M gr Answer:   2 1  e  t where   2 . r ( M  1  2M 2 )





3.34. Define the angular acceleration of the car driving wheel of a mass M and radius r if the rotative moment mrot was attached to the wheel. The wheel moment of inertia relative to the axis passing 164

3. Dinamics of the mass point and the system

through the mass center C perpendicular to the plane of solid symmetry equals J c ; f k – coefficient of rolling friction, F fr is a force of friction. Find the magnitude of rotative moment at which the wheel rolls with constant angular velocity. Answer:  

mrot  Mgf k  F fr r Jc

, mrot  Mgf k  F fr r .

3.35. Define whit what angular velocity  falls the sawed off tree of mass M if its mass center C is located at a distance h from the plane and air resistance forces create the moment of inertia mc so that mcz   2 where α is constant. The moment of inertia of the tree relative to z axis coinciding with an axis around which the falling tree rotates, is equal to J.

Answer:  

2Mghl 2

J  4

2

(e



 J



2 ) J

SELF STUDY OF THE STUDENT

1. Solid body begins to rotate from the state of rest under the force couple with the moment dependent on an angular velocity  of a body rotation: mz  m0  a 2 , where m0 and a – constants. Here m0 is a rotational moment, and a – retarding torque. Define the angular velocity of rotation of a solid body, if its moment of inertia relative to the rotation axis z is equal to J z . Center of gravity of the body is situated on the rotation axis. 2. Solid cylinder with the radius r, placed onto cylindrical rollers A and B as shown in the picture is rotating with the angular velocity 0 around its horizontal axis O. At some moment rollers slow down. 165

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In what time after that cylinder stops if the friction ratio between the cylinder and rollers equals to f and the corner AOB 2 ? 3. Homogeneous round disk with weight P and radius r is suspended to the thin elastic wire in one of the points situated onto its rim. When the disc rotates to the angle  around the axis z passing along the wire and the vertical diameter of the disc, a couple of forces with elastic moment m z that is proportional to the angle of twist  arise in the wire. mz  c where c is a moment of couple that is needed for twisting the wire on one radian. Define the motion of the disc if at initial time to the disc being in rest was imposed the initial angular velocity 0 . Air resistance should be neglected. 4. Homogeneous round disc with weight P and radius r , fixed to the ceiling and floor with the help of two elastic wires that locate on one vertical, executes torsional oscillations. When disc rotates elastic moments that are proportional to the twist angle arise in wires; с1 is a coefficient of elasticity of the upper wire, с2 is the coefficient of elasticity of lower wire. A couple of forces with moment mz  m0 sin t ( m0 ,  are constants) was applied to the disc. Define the law of motion of the disc if at initial time it was in rest by the stress-free state of wires. The resistance force should be neglected. z axis directed along the wires. 5. Solid body starts rotating around the fixed axis z and from the state of rest under the couple of forces with moment depending on the rotation angle  of the rotating body m z  a  b 3 where a and b are constants. Define the angular velocity of the solid body depending on a rotation angle  if its moment of inertia relative to the rotation axis 166

3. Dinamics of the mass point and the system

z is equal to J z . Gravity center of the solid body is located upon a rotation axis. 6. The rotary crane tram moves with the constant velocity v relative to the arm. The engine that rotates the crane in the acceleration period generates rotational moment that equals to m0   where m0 and α are positive constants. Define the angular velocity  of the crane rotation depending on distance x from tram to rotation axis AB if mass of the tram with load equals to M and J is a moment of inertia of the crane (without tram) relative to the rotation axis. Rotation starts when tram is in a distance of x0 from AB axis. 7. Round horizontal platform rotates without friction around the vertical axis that passes through its mass center with constant angular velocity  0 . On the platform there are 4 people with the same weight: two of them stay on edges of the platform a two other in a distance from the rotation axis. The distance equals to the half of radius of the platform. How changes the angular velocity of the platform if people staying on the edge move round a circle sideways of the platform rotation with relative linear velocity u and people staying in a distance of half of circle radius move round a circle sideways to rotation of platform with relative linear velocity of 2u ? People are counted as point masses and platform as round homogeneous disk. Platform radius is R, its mass four times exceeds the mass of each person and evenly distributed throughout the plane. Also find out what relative linear velocity u should be equal to in order to make platform stop its rotation. 8. Two people of equal weight G1 = G2 are hanging on a rope thrown over a solid pulley with the weight G3 = G1/4. At some time one person began to climb up the rope with relative velocity u. Define the velocities of lifting of each persons. 9. The rope is thrown over the pulley. The load is fastened at one end of the rope. What will be the velocity of the load if a person will 167

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climb up at another end of the rope with the relative velocity v. Note that mass of a person is four times exceeds the mass of a pulley. 10. Horizontal tube CD can rotate around the vertical axis AB. The ball M is located on the distance MC=a from the axis inside the tube. At some moment of time the tube was imposed tie initial angular velocity  0 . Define the angular velocity of the tube at the moment when the ball will fly out the tube. Inertia moment of the tube relative to the axis of rotation is equal to J., L is the length of the tube, mass of the ball is m, neglect the friction. 11. The homogeneous rectangular plate ( a  0.6m ) of mass m1  50kg is rotating around the vertical axis z with the constant

velocity  0  3rad / sec . The mass point K of mass m2  12kg is located at the point O of the channel AB at the distance AO=0.2m from the point A. At some moment of time t=0 the pair of forces with the moment M z  27t 2 begins to act at the system. Action of forces will stop at the moment of time   2 sec . Define the angular velocity  of the plate at the moment t   . At the moment of time t1  0 the mass point K begins to move along the channel AB in accordance with the law OK  s  0.4t1 . Define the angular velocity T of the plate at the moment T  2 sec . Note that   60  .

12. The homogeneous round plate (R=1.6m) of mass m1  50kg is rotating around the vertical axis z with the constant velocity  0  2 rad / sec . The mass point K of mass m2  10kg is located at the point O of the channel AB at the distance AO=0.6m from the point A. At some moment of time t=0 the pair of forces with the moment M z  69t begins to act at the system. Action of forces will 168

3. Dinamics of the mass point and the system

stop at he moment of time   4 sec . Define the angular velocity  of the plate at the moment t   . At the moment of time t1  0 the mass point K begins to move along the channel AB in accordance with the law OK  s  0.6t1 . Define the angular velocity T of the plate at the moment T  2 sec . Note that   30  .

3.5. Work of force. Work of potential force

A force is said to do work when it acts on a body when there is a displacement of the point of application in the direction of the force. Projecting the fundamental law of dynamics on a tangent vector τ:

mw  F But w 

   dv dv dS dv      v , then the basic law of dynamics dt dS dt  dS v

will take the form: m

 dv   v  F => mvdv  F dS , dS

where dS is the increment of arc constant (the algebraic value).

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Textbook on Theoretical Mechanics

   2 mv dv  F dS or d  mv   d ( A) , i.e. the elementary work is  2   

equal to F dS .

 But F is the projection of F on the direction of τ, i.е. F  F  CosF ,  Then, the elementary work force d A on the dS displacement is equal to:  d A  F  dS  CosF ,  (only the tangent component F of force F

does work on the displacement dS , work of normal component Fn of force F is equal to zero). Elementary work can be expressed in the form of dot product:   d A  F  dr ,

  but F  Fx , Fy , Fz , dr  dx, dy, dz  d A  Fx dx  Fy dy  Fz dz

Work of force on a finite displacement M 0 M is calculated as:



  A   Fdr   F dS   Fx dx  Fy dy  Fz dz    M 0M

M 0M



M 0M

In general case the work of force depends only on the type of trajectory and law of motion. We will consider a special case – positional forces, which depend only on the coordinates:

  F  F  x, y , z  .

The region of space where on each mass point placed there is acted a certain force, which is a unique, limited-differentiable function of the coordinates of this point is called the force field. Potential force field

Consider such kind of force field, for which the elementary work is a total differential of a function of the coordinates U, i.e.: 170

3. Dinamics of the mass point and the system

  d A  F  dr  Fx dx  Fy dy  Fz dz  dU  x, y, z 

When the differential of function U  x, y, z  is equal to the elementary work it is called potential function or force function. As the total differential is: dU  Fx 

U U U dx  dy  dz , then x y z

U U U , Fy  , Fz  z x y

(1)

i.e. projections of forces are equal to the partial derivatives of force  function in potential force field, then the vector F is the gradient of scalar function U:    U  U  U  i j k  Fx i  Fy j  Fz k  U x y z       k j i z y x  F  grad (U ) grad (U ) 

Let’s find the conditions which must satisfy the forces of field, so it was a potential. For this purpose we take the partial derivatives of (1) with respect to the corresponding coordinates. Taking into account that:  2U  2U ….,  xy yx

We will get necessary and sufficient conditions for potentiality of force field:  Fy Fx 0   y  x  Fz Fy  0  z  y  Fx Fz 0   x  z

2 2 F F , as Fx   U   U  y  y  Fx  0 .

y

xy

yx

171

x

x

y

Textbook on Theoretical Mechanics

Example: Gravity force field

 P  Px , Py , Pz   0,0,mg

Let’s check up the conditions of potentiality:  Px Py     0  0  0  y x y x  Py Pz      0    mg   0  0  0    y z y z   Pz Px        mg    0  0  0  0  z x z x  

gravity force field is potential. Conditions of potentiality can be written in another form. It is known that:  i

 j

   rot ( F )  x y Fx Fy

 k

rotF  

x









   rotF , z  Fz rotF

y

z

 Fz Fy   0 y z   Fx Fz    0  z x  Fy Fx     0 x y  



conditions of potentiality can be written as: rotF  0

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3. Dinamics of the mass point and the system

  dU  x, y, z   F  dr  Fx dx  Fy dy  Fz dz ,

then potential function can be defined as

U  x, y, z    Fx dx  F y dy  Fz dz  сonst . Surface U  x, y, z   с is called level surface. The main property is: work of potential force on a finite   displacement AM M   F  dr   dU  U M  U M is equal 0 0

 M 0M

 M 0M

to the difference between the values of the force function in the initial and final points of the path. It depends only on the position of the start and end points, and does not depend on the type of trajectory along which moves the point. This is the basic property of a potential force field. If motion occurs in a closed circuit (or at the same level surface), work of potential force is equal to zero. Example: gravity force field. 1)

Concept of the potential energy V can be introduced for a potential force field. Potential energy V can be considered as a stock of work that can do the forces of a force field when the mass points are moving from the current position to any level surface that is taken for null level surface conventionally. Let’s choose such const so that U  x, y, z   0 on the null level surface: U Н  0 , then, by definition, the potential energy V  U Н  U  U  and V x, y, z   U x, y, z  . 173

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Example: gravity force field. As P  {0, 0,  mg} , then for the force of gravity we have dU   mgdz . Integrating we will get: U   mgz  const . Assume that U  0 when z  0 (i.e. C  0) , then U   mgz  V  mgz Work of various types of forces

А) work of gravity force on a finite displacement M 0 M

AM



0M

 U M  U M 0  mgz  z0   mgz0  z   mgh ,

where h  z  z0 is the difference of heights M 0 and M. B) work of elastic force lo is the length of nonstretched spring. Fx  cx, Fупр  {cx,0,0} M1

x1

M0

x0

A   (cx)dx  c  xdx 

c 2 ( x0  x12 ) 2

Fупр is potential (work does not depend on the type of trajectory). 174

3. Dinamics of the mass point and the system

C) work of friction force Fтр  fN (M1 )

(M1 )

(M 0 )

(M 0 )

A    Fds    fNds

If Fтр  const , then A   Fтр s , where s is the length of the arc of the curve M 0 M 1 which the mass point is moving along (work depends only on s  Fтр is nonpotential force). Questions 1. What is the work of force? 2. Derive the work of force on a finite displacement. 3. How can be defined the work of potential force? 4. What is the work of force of gravity? 5. What is the work of elastic force and friction force?

3.6. Work-Energy theorem for the mass point and mechanical system. Energy integral Work-Energy theorem for the mass point

Assume that the mass point of mass m moves along some curved path under action of force F . The basic law of dynamics is:

mw  F Multiply both sides of equation on the differential of positionvector dr  v dt of point of application of force F :

mw  dr  F  dr

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Note, that w  dv , dr  v dt , where v is the velocity of considt dered point relative to the fixed axis. Rewrite the left side as: m w  dr  m

dv  v dt  mv dv dt

1 As v 2  v 2 and v  d v  d (v 2 ) , we find: 2 mw  dr  m  v dv 

 mv 2 1 d (mv 2 )  d  2  2

  

Consequently our equation will take the form:  mv 2 d   2

   F dr 

(1)

The expression on the left-hand side is called the kinetic energy mv 2 of a mass point and it is denoted by T , T  . 2 2 Kinetic energy T  mv is a measure of the mechanical motion

2

of a mass point. The expression F dr  d ' A on the right-hand side is called the elementary work of the force applied to the mass point. Equation (1) expresses the work-energy theorem for a mass point in differential form: differential of kinetic energy of a mass point is equal to the elementary work of force acting on this point. Dividing both sides of (1) by dt we will get: d  mv 2  dt  2

 dr or d  mv 2    F   Fv  dt dt  2   176

3. Dinamics of the mass point and the system

The value N 

dA  F v is called power of force. dt

Theorem (also in differential form) sounds differently: the time derivative of kinetic energy equal to the power of the force acting on the mass point. Suppose that the mass point moving under the action of applied force F , has the velocity vO in the position M O and velocity v1 in

the position M 1 . Let’s integrate both sides of equation (1) along the arc of the trajectory from the point M O to the point M 1 , then we will get: ( M1 ) mv12 mvO2 (2)    F dr 2 2 (MO ) This equation expresses the theorem of change of kinetic energy in the integral form: change of kinetic energy of a mass point on a finite part of the trajectory is equal to the work of force acting on this point on the same part of the trajectory. Work-Energy theorem for mechanical system

Theorem of change of kinetic energy for a mass point (1) is easily generalized to the case of a mechanical system of mass points. Suppose that equation (1) is written for the k -th point of the mechanical system.  m v2  d  k k   Fk( e ) drk  Fk(i ) drk ,  2 

where Fk( e ) drk is the elementary work of the resultant of all external forces applied to the k-th mass point, Fk(i ) drk is the elementary work of the resultant of all internal forces applied to the k-th mass point. Summing the equations for each mass point we will get: n n mk v k2   Fk( e ) drk   Fk( i ) drk 2 k 1 k 1 k 1 n

d

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Textbook on Theoretical Mechanics

mk v k2 is the kinetic energy of mechanical system. 2 Then the equation can be rewritten in the T 

form: (e) (i )   dT  d A  d A The theorem of change of kinetic energy of mechanical system in the differential form: differential of kinetic energy of mechanical system is equal to the sum of all the elementary work of all external and internal forces acting on the system. Assume that the integrals can be calculated then we can write:

T1  T0  A ( e )  A ( i ) This is the theorem in integral form: change of kinetic energy of mechanical system in its finite displacement from one position to another is equal to the sum of work of all internal and external forces acting on the system, on this displacement. Any problems of dynamics can be solved with the help of this theorem if: 1). acting forces are constant or depend only on the distance; 2). if it is given F , S , v0 , v1 . If there is a force that depends on the velocity, this problem can not be solved with the help of this theorem (it is impossible to calculate mv and A ). In this case it is necessary to use a method of integrating the differential equations of motion. If the force F  F ( x, y, z ) and F dr  dU is the full differential of some function then U is called the force function and function V  U is called the potential energy. In this case the theorem can be written in the form:  mv 2 d   2

   F dr  dU . 

2 2 Integrating we will get  mv   U  h or mv  V  h .  2  2  

178

3. Dinamics of the mass point and the system

This is the law of conservation of mechanical energy: sum of kinetic and potential energy is the constant value. Questions 1. What is the kinetic energy? 2. Formulation of work-energy theorem for a mass point in differential form. 3. Formulation of work-energy theorem for a mass point in integral form. 4. Formulation of work-energy theorem for a mechanical system. 5. Formulate the law of conservation of mechanical energy.

PRACTICE 3.36. A nail is being driven into the wall that maintain resistance of 700 N. Nail is going deep into the wall on the length l=0.15 cm by each fall of the hammer. Determine the mass of hammer if it has the velocity v=1.25 m/sec when it falls. Answer: 1.344 kg. 3.37. Write the expression for potential energy of elastic spring that is sagging on a length of 1 cm due to the load of 4kN. Suppose that sag increases proportionally to the load. Answer: V  20 x 2  c 3.38. Determine the velocity v0 that must be imposed to the body on the surface of the Earth so that it lifted on the height equal to the Earth radius. Take into account only the gravitation force that is proportional to the square of distance from the center of the Earth. Answer: 7.9km/sec 3.39. The ball connected with the spring moves from the position A inside the tube. Ball separates from the spring when it pass the path h0. Define the velocity of the ball in the positions B,C. Mass of the mass point is equal to 0.5 kg, v A  0 m / sec , R=0.5 m, f=0.2, time of passing the path BD is equal to 0.2 sec,   45 ,   30 , h0  50cm , c=0.8N/cm.

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Textbook on Theoretical Mechanics

Answer: vB  2.95 m / sec, vC  2.4m / sec . 3.40. Hammer of mass 0.80 kg has the velocity of 1.5m/sec in the moment of strike against the nail. The hammer drives the nail into the beam on the depth of 5.00 mm. What should be the mass of the load applied to the nail so that the nail was driven in the beam on the same depth. Answer: 18.8 kg. 3.41. Calculate the kinetic energy of crank-slide mechanism if the mass of a crank m1, length of the crank r, mass of the slider m2, length of connecting rod l. Neglect the mass of the connecting rod. Consider the crank as homogeneous bar. The angular velocity of rotation of the crank ω.

Answer:

      r  1 1 T   m1  m2 sin   2 3  2l      

    sin 2  .   r  2 r l   sin 2     l   2

2

2

3.42. A planetary mechanism that is located in a horizontal plane is set in motion by crane OA that connect three equal wheels I, II and III. Wheel I is fixed, the crane rotates with angular velocity ω. The mass of each wheel is equal to M1, radius of each wheel is equal to r, mass of the crane is M2. Calculate the kinetic energy of the train assuming the wheels as homogeneous discs and the crane as homogeneous bar. What is the work of couple of forces applied to the wheel III? 180

3. Dinamics of the mass point and the system

2 2 Answer: T = r ω (33 M 1+ 8 M 2) .

3

3.43. Load A of mass M1 ascends by means of cable thrown over the block C and reeled on a barrel B of radius r and mass M2. Torque proportional to the square of rotation angle φ of a barrel mrot =αφ 2 (α is a constant) is applied to the barrel from the time of switching on. Define the velocity of a load at the time when it ascends at a height h. Assume the mass of barrel B equally distributed along the rim. Block C is a solid disc of mass M3. Neglect the mass of a cable. System was in rest at the initial time.

Answer:





2

4h a h  3 M 1 g r

2



3 r 2 M 1  2 M 2  M 3  3

3.44. An epicyclic mechanism that is located in the horizontal plane is set in motion from the state of rest by means of constant torque L attached to a crank OA. Define the angular velocity of a crank depending on its rotative angle if the fixed wheel I is of radius, r1 moving wheel II is of radius r2 and mass M1 and crank OA is of mass M2. Assume the wheel II as homogeneous disc and crank as homogeneous bar.

Answer:  

2  r1 r 2

3 L 9M1 2M 2

181

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3.45. To the barrel of capstan with radius r1 and mass M1 is set rotational moment L. To the end of a cable reeled onto a barrel attached an axis C of a wheel with mass M2. Wheel rolls without sliding up against the inclined plane that is placed at an angle  to the horizon. What angular velocity gains the barrel if it makes n rotations? Barrel and wheel consider as homogeneous round cylinders. At initial time the system was in rest. Neglect he cable mass and friction.

Answer:  

L  M 2 gr1 sin  2 2n r1 M 1  3M 2

3.46. Load with weight P=20 kg rises up an inclined plane via rope placed angularly   30  to the plane. Inclined plane subtends the angle   30 with horizon. Rope tension S=15kg. At initial position the load was in rest. Find the magnitude of movement of load along the inclined plane at time when load has velocity v =2mps . Load coefficient of sliding friction f equals to 0,2. Answer: l=3,54 m 3.47. Heavy body descends without initial velocity along the inclined plane making an angle 30° with the horizon. Friction coefficient is equal to 0.1. What velocity will have the body when it pass 2 meters from the beginning of motion. Answer: 4.02 m/sec

182

3. Dinamics of the mass point and the system

3.48. Mass point of mass 3kg moved along the horizontal line to the left with the velocity 5m/sec. Constant force directed to the right was applied to the mass point. Force stopped to act in 30 sec and velocity of the mass point was equal to 55 m/sec. Find this force and the value of work of this force. Answer: F=6N, A=4.5 joule 3.49. Determine the kinetic energy of the system consisting of two wheels connected by means of the coupler AB and rod О1О2 if the axis of the wheels move with the velocity v0. Mass of each wheel equals M1. Couple AB and connecting rod О1О2 has the equal mass M2. Mass of the wheels is uniformly distributed along its rims. O1 A  O2 B  r / 2 where r is the radius of the wheel. The wheels roll along the straight-line rail without sliding.

2 Answer: T  v0 16M 1  M 2 (9  4 sin  ) .

8

3.50. The reel is set in motion by means of a belt drive connecting the pulley II situated on the shaft of the reel with the pulley I situated on the shaft of the engine. Constant torque m is applied to the pulley I of mass M1 and radius r. Mass of the pulley II equals M2, its radius is R. Mass of the barrel of the reel is M3, its radius is r, mass of the lifting load is M4. The reel is set in motion from the state of rest. Find the velocity of the load at the moment when it lifts at the height h. Mass of the belt and rod and friction in bearings can be neglected. Consider the pulleys and reel as homogeneous round cylinders.

183

Teextbook on Theoreticaal Mechanics

Answer: v  2

h(mR / r 2  M 4 g ) M 1 ( R / r ) 2  M 2 ( R / r ) 2  M 3  2M 4

3.51. Nonstretchiing thread is fasteened to the load A of mass M1 and is i threw over the block D of mass M2 and reeled on n the side face of cy ylindrical roller B of mass M3. W When load A mov ves down the inclined plane making g an angle α withh the horizon blo ock D rotates and roller r B rolls up without w sliding along the inclined plane p making an angle a β with the horizon. Determiine the velocity of load A in depeendence on the paassed way s if thhe system was in rest at initial mom ment. Consider thee block D and rooller B as homogeeneous round cylin nders. Neglect thee friction force andd mass of the threead.

Answer: v  2 2 gs

2M1 sin   M 3 sin  . 8M 1  4M 2  3M 3

184

3. Dinamics of the mass point and the system

3.52. Crank OO1 of cyclic mechanism situated in horizontal plane rotates with the constant angular velocity 0 . Engine was switched off at some moment and mechanism stopped under action of constant moment of friction forces M fr . Determine the time of braking τ and rotation angle  of the crank if its mass equals M 1 , M 2 – mass of the satellite roller, R and r are radii of the big and small wheels. Consider crank as homogeneous thin rod and satellite roller as homogeneous disc.

Answer: 

rJ 1 rJ 2 0 ,   0 RMfr 2 RM fr

where M 3  J   1  M 2 R  r 2 . 3 2  

SELF STUDY OF THE STUDENT

1. Load with weight P=20kg rises up an inclined plane via rope  placed angularly   30 to the plane. Inclined plane subtends the angle   30 with horizon. Rope tension S=15kg. At initial position the load was in rest. Find the magnitude of movement of load along the inclined plane at time when load has velocity v =2mps . Load coefficient of sliding friction f equals to 0,2. Answer: l=3,54 m 185

Textbook on Theoretical Mechanics

2. Loads A and B get set in motion by 2 blocks: a raveling block K and fixed block L. As a result of a push imposed to the load A it started to descent with velocity v 0 . At what distance should load A descent in order that its velocity increased twice? Loads A and B are equal in weight. Blocks K and L consider as homogeneous round disks with equal weight Q. Coefficient of sliding friction of the load B against horizontal plane is f. Neglect the rope mass.

Answer: s 

3v 02 (10 P  7Q ) 4 g (Q  P (1  2 f ))

3. Load with weight P executes free oscillations on a spring suspended to a ceiling. Spring coefficient of elasticity equals c. To the load being in a position of static equilibrium was imparted velocity v0 via a push. Define the load velocity v depending on its 186

3. Dinamics of the mass point and the system

shift from a position of static equilibrium and the load vibration amplitude. 2 Answer: v  v0 

v cg 2 x ,a 0 P k

4. A load A with weight P1 is suspended to a homogeneous nonstretching cable with length L and weight Q. The cable is thrown over the block B that is rotating around O axis that is perpendicular to the picture plane. The second end of a cable is attached to the axis of a roller C that is driving without sliding along the fixed horizontal plane. Block B and roller C – homogeneous round cylinders with radius r and weight P2 each. Coefficient of rolling friction of roller C against the horizontal plane equals f k . At the initial time when system was in rest a part of a cable with length l hung over the block B. Define the velocity of a load A depending its vertical shift h.

Answer: v 

2 gh[ P1 

f Q h (r  l  )  k P2 ] L 2 r P1  Q  2 P2

5. Rectangular plate ABCD with sides a and b and weight P rotates around vertical axis z with initial angular velocity  0 . Each element of the plate experiences by this air resistance which direction is perpendicular to the plate plane and the magnitude is directly proportional to element plane and to the square of its velocity v, 187

Textbook on Theoretical Mechanics

coefficient of proportionality equals μ. How many revolutions makes the plate until the time when its angular velocity becomes twice less than initial.

Answer:  

4 P ln 2 3 gba

2

therefore  

2 P ln 2 3gba 2

6. Rectangular plate with sides a and b may rotate without friction around a vertical axis AB that passes through its center and parallel to he side b. There is a sheave C with radius r on the end of the axis. The sheave is reeled on with a flexible non-stretching thread and the other end of the thread is thrown over the block D to which the load with weight P is tied and that causes the plate to rotate. Neglecting masses of sheave and block find the acceleration of the load if weigh of the plate equals Q and there are no tractive resistance. dv

12r 2 P

Answer: w  dt  Qa 2  12r 2 P g  const 188

3. Dinamics of the mass point and the system

7. Epicyclic mechanism that lies in a horizontal plane is set in motion from the state of rest by a constant rotational moment L attached to a crank OA. Define the angular velocity of the crank depending on its rotation angle if a fixed wheel I has the radius r1, the moving wheel II has he radius r2 and the mass M1, and he crank OA has the mass M2. Consider wheel II as a homogeneous disk, and crank as a homogeneous bar.

Answer:  

2 r1  r2

3L 9M 1  2M 2

8. Wheel A rolls without sliding against the inclined plane OK and lifting wheel B via non-stretching cable. Wheel B rolls without sliding against the inclined plane ON. Cable is thrown over the block C that rotates around fixed horizontal axis O. Find the velocity of wheel A axis when its shift parallel to the line OK has distance s. At initial time the system was in rest. Both wheels consider as homogeneous disks with same mass and radius. Neglect the cable mass.

Answer: v  2

1 gs (sin   sin  ) 7

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9. To the barrel of capstan with radius r1 and mass M1 is set rotational moment L. To the end of a cable reeled onto a barrel attached an axis C of a wheel with mass M2. Wheel rolls without sliding up against the inclined plane that is placed at an angle α to the horizon. What angular velocity gains the barrel if it makes n rotations? Barrel and wheel consider as homogeneous round cylinders. At initial time the system was in rest. Neglect he cable mass and friction. L  M 2 gr1 sin  2 Answer:   r 2n M  3M 1

1

2

10. Homogeneous thread with length L that lies on a smooth horizontal table moves under he influence of gravity of another part that hangs from the table. Define the time interval T at the end of which the thread leaves the table if it is known that at initial time length of hanging part equals l and initial velocity equals to zero. Answer: T 

L L  L2  l 2 ln( ) g l

3.7. Rectilinear motion of a mass point. Harmonic oscillations of the mass point. Parameters of oscillations. Oscillations in a resistant medium Rectilinear oscillation of mass point

Consider the various cases of rectilinear oscillatory motion of a mass point around its equilibrium position. Oscillation of a mass point takes place, if the force F acts on a mass point M deviated from the position of rest O. The force F that tends to return this point in this position of rest is called the restoring force. 190

3. Dinamics of the mass point and the system

Consider the most simple case, when the restoring force is proportional to the deviation of a mass point from its position of rest. Assume that the body with the weight G lying on a smooth horizontal plane in the position O is connected to a rigid spring. Another end of a spring is fixed. Gravity force G and reaction force N attached to the body are balanced. Elastic force of stretched spring F1 will act on the body if the body is deviated in the position

M 1 . Elastic force tends to return the body to its position of rest O. F1  c  OM 1 where OM1 is the extension of the spring, c is the constant coefficient of proportionality (i.e., the elastic force of a spring). In the position M 2 elastic force of a spring F2 also tends to return the body in the position O. Thus, the elastic force is always directed to O. Oscillations can occur under the influence of the restoring forces varying according to other laws. There are four main cases of oscillation of a mass point

1. Free oscillations take place under action of restoring force. 2. Damped oscillations take place under action of restoring and resistance force. 3. Forced oscillations take place under action of restoring force and disturbing force with periodical character. 4. Forced oscillations with account of resistance take place under action of restoring force, disturbing force and resistance force. Free oscillations of a mass point

Let’s choose the axis x and origin of coordinate O in the position where the point M can be in rest. If the point M is led out from the state of rest than only the restoring force F is acting on it along the x axis: F  c  OM  с x , 191

Textbook on Theoretical Mechanics

where с is a stiffness coefficient of spring, determined as its elastic force when deformation of spring equal to 1. As F is directed to the point O in any position, its projection on the axis x always has the sign that is opposite to the sign of coordinate x, i.e. Fx  cx . (1) Let’s deduce the differential equation of motion of point M under action of force F :

mx  cx  x  Denote

c x0 m

c  k 2 then: m x  k 2 x  0

(2)

This equation is called the differential equation of free oscillations of a mass point. Let’s deduce the characteristic equation in order to integrate the equation (2).

2  k 2  0, 1, 2  ik General solution has the form:

x  C1 cos kt  C2 sin kt

(3)

Let’s find the velocity of the mass point in order to define the values of C1 ,C 2 :

x  kC1 sin kt  kC2 cos kt 192

(4)

3. Dinamics of the mass point and the system

Assume that the mass point has the coordinate x 0 and velocity

x 0 at the initial moment of time t  0 . Then, substituting this values in (3), (4) we will get:

C1  x0 ,

x 0  kC2 ,  C2 

x 0 k

Substituting in equation (3) we will get the equation of motion of the point M:

x  x0 cos kt 

x 0 sin kt k

(5)

This solution can be written in another form. Let’s introduce the new constants a and  instead of C1 and C 2 : C1  a sin  , C2  a cos  .

Substituting these values in (3) we will get:

x  a sin( kt   )

(6)

This equation is called the equation of harmonic vibration of a mass point. Thus, harmonic vibrations are free oscillations of a mass point under action of linear restoring force. Amplitude a and initial phase  of free oscillation are determined with the help of initial conditions of motion. Equation determining the velocity of a point has the form:

x  ak cos(kt   )

(7)

Substituting the values of x 0 , x0 and t 0  0 in (6), (7) we will

get: x0  a sin  , x 0  ak cos  . From here we will find a and  : 193

Teextbook on Theoreticaal Mechanics

 x  a  x  0   k  kx tg g  0 x 0 2 0

2

(8)

As each value off the tangent of anngle corresponds to t two angles [0, 2 ] then it is neccessary to define sin s  or cos  :

sin  

сos 

x0  a

x 0  ak

x0 2 0 2

x x  k 2 0

kx0



x 0 x 2 k x02  02 k

k x  x 20 2 2 0



x 0 k 2 x02  x 02

(9б)

(9в)

Angular frequenccy and period of free oscillations is i determined with h the help of formu ulas: c m

(10)

m 2  2 k c

(11)

k

T

Angular frequenccy is also called the frequency off free oscillationss or natural freq quency. As can bbe seen from (10), (11), the naturral frequency an nd the period off free oscillationss of a point depeends only on the mass of this poiint and the coeffficient c characteerizing the restorring force, and ddo not depend on o the initial cond ditions of motion. 194

3. Dinamics of the mass point and the system

Damped oscillations of a mass point

Mass point executing the oscillations in the real world is influenced by resistance to motion (friction, air resistance, etc.). It means that in addition to the restoring force acting on the point resisting force directed in the opposite direction of motion of the point also acts on the mass point. Air resistance at low velocities of the bodies is considered to be proportional to the velocity, and at higher velocities it is taken proportional to the square of the velocity of the moving body. Let’s consider the oscillations of a mass point M under action of restoring force F and air resistance force R proportional to the velocity of the point. Assume that x axis is directed along the path of mass point, origin O of coordinate system coincide with the position of rest of mass point. F , R are the forces, acting on the mass point:

Fx  cx, R  v If v  1 , то R   , that is coefficient of proportionality  is equal to the resistance force when velocity is equal to 1. R is directed opposite to the vector of velocity v , i.e. R  v , and as a projection to x – axis:

Rx  x Deduce the differential equations of motion of a mass point:

mw  F mx  Fx  R x  mx  cx  x  x 

 m

x 

c x  0. m

195

Textbook on Theoretical Mechanics

Assuming that:   2n, c  k 2 , we will get: m m

x  2nx  k 2 x  0

(12)

This equation is a differential equation of motion of a mass point under action of restoring force and resistance force that is proportional to the velocity of the mass point. In order to integrate this equation we deduce the characteristic equation and find its roots:

2  2n  k 2  0 1, 2  n  n 2  k 2 А) The case of a low-resistance ( n  k ) The roots of characteristic equation are imaginary in this case ( D  0 ). Denote:

k 2  n 2  k1 , then 1, 2   n  ik 1 As it is known, the general solution of equation (12) will take the form:

x  e nt (C1 cos k1t  C2 sin k1t ) Let’s introduce the constants a and  instead of C1 and C 2 : C1  a sin  C 2  a cos 

Substituting these values C1 and C 2 we will get the equations of motion of a mass point in the form:

x  ae  nt sin(k1t   ) 196

(13)

3. Dinaamics of the mass poiint and the system

This equation describes the oscilllation as sin( k1t   ) is periodicall function. Mu ultiplier e  nt show ws that amplitude of oscillations decreeases with time. Such oscillationss are called damp ped. The values and a  are define ed from initial a cond ditions. Assume that the mass point has initial position x0 and initial velocity x 0 at the moment t  0 . Subbstituting the initiial conditions quation (13) and equation e for velocity: in eq



x  ae  nt sin(k1t   )  nae

 nt



t



sin(k1t   )  akk1e

 nt

cos(k1t   )

i.е. x  nx  ak1e  nt cos(k1t   ) We will get:

x0  a sin  , x  nx0 x0  nx0  ak1 cos  or 0  a cos  k1 From here we willl have:

(x0  nx0 ) 2 k1 x0 k1 tg  x0  nx0

a  x02 

197

,

Textbook on Theoretical Mechanics

As the function (13) is nonperiodic it is usually introduced the notion of period. Period of damped oscillation T * represent the time interval between two sequential paths of the mass point through the position of rest in one direction. T 

2 2  , k1 k 2  n2

k1  k 2  n 2 .

The value k1 is called the frequency of damped oscillations. B) The case of high resistance ( n  k ). The roots of characteristic equation are real, negative and distinct ( D  0 ):

1, 2  n  n 2  k 2 General solution of equation (12) has the form:

x  e nt (C1e

n2 k 2 t

 C2 e 

n2 k 2 t

)

(14)

Introduce the constants B1 and B2 instead of C1 and C 2 :

C1 

B1  B2 B  B2 , C2  1 2 2

Substituting this in equation (14) we will get: nt

x  e (B1 As

e

n2 k 2 t

e x  ex  chx, 2

 e 2

n2 k 2 t

 B2

e

n2 k 2 t

e x  ex  shx, then 2 198

 e 2

n2 k 2t

)

3. Dinaamics of the mass poiint and the system

x  e  ntt ( B1ch n 2  k 2 t  B2 sh n 2  k 2 t )

can be b reduced to ano other form assuminng that B1  ash , B2  ach , then x  ae  nt sh( n 2  k 2 t   )

(15)

This equation sho ows that considereed motion of masss point is not oscilllating, as shx is not a perioddic function. Masss point can perfo orm one of the mo otions depending on the initial cond ditions:

All of these graph hs correspond to tthe initial deflectiion of a mass poin nt on the value x0  0 from the possition of rest. C) The case of n  k . The roots of characteristic equationn are real, negativ ve and equal:

1  2  n General solution of equation (12) hhas the form:

x  e  nt (C1t  C 2 )

(15а)

The motion of a mass point is alsoo aperiodic. Consstants C1 and

C2 are a defined with help h of initial condditions.

199

Textbook on Theoretical Mechanics Questions 1. What is the definition of harmonic oscillations? 2. Write the equation of harmonic oscillations. 3. What is the definition of dumped oscillations? 4. What is the character of dumped oscillation in case of a low-resistance? 5. What is the character of dumped oscillation in case of a high-resistance?

3.8. Forced oscillations in a medium without resistance and in a resistant medium. Resonance Forced oscillations of a mass point

A mass point execute the forced oscillations when it is influenced by restoring force and periodic varying force that is called perturbing force. The most important case is when perturbing force Q varies in accordance with the harmonic law, i.e. Qx  H sin( pt   ) . Deduce the differential equation of motion of a mass point:

Fx  cx , mx  cx  H sin( pt   ) or x 

Denote

H c x  sin( pt   ) m m

c H  k2,  h , then m m x  k 2 x  h sin( pt   )

(16)

This is the differential equation of forced oscillations of a mass point. General solution of equation (16) is represented as a sum of general solution and particular solution of inhomogeneous equation: x  x   x  . Homogeneous equation: x  k 2 x  0 . 200

3. Dinamics of the mass point and the system 

General solution of this equation is: x  C1 cos kt  C 2 sin kt . Particular solution can be obtained in dependence of the type of inhomogeneity: (17) x   A sin( pt   ) .

x



Let’s define the constant A by substituting (17) in (16). As   Ap 2 sin( pt   ) , then after the substitution we will get:

 Ap 2 sin( pt   )  Ak 2 sin( pt   )  h sin( pt   ) This equation must be satisfied for any sin( pt   ) , therefore: A(k 2  p 2 )  h  A 

h . k  p2 2

Substituting the value A in (17) we will get the desired particular solution: h x   2 sin( pt   ) k  p2 Then the general solution of equation (16) has the form:

x  C1 cos kt  C 2 sin kt 

h sin( pt   ) k  p2 2

Or if x   a sin(kt   ) , then general solution is:

x  a sin( kt   ) 

h sin( pt   ) k  p2 2

This equation shows that the point M performs a complex oscillatory motion that is composed of two harmonic oscillations. 201

Textbook on Theoretical Mechanics

The first term defines the free oscillations, and the second term defines forced oscillations of a mass point. Thus, a mass point performs a complex oscillatory motion, which is the result of the superposition of free and forced oscillations when it is influenced by the restoring and perturbing forces. Constants C1 ,C 2 or a,  are defined from the initial conditions. Forced oscillations are characterized by the equation: x  

h sin( pt   ) k  p2 2

Frequency p and period   2 of forced oscillations coincide p

with the frequency and period of perturbing force variation. If p  k , then there is the case of forced oscillations with low frequency. If p  k , then there is the case of forced oscillations with high frequency. Amplitude of forced oscillations does not depend on the initial conditions. When the frequency of perturbing force increases the amplitude of forced oscillations tends to zero. Consequently, the effect of perturbing force with a very high frequency ( p  k ) almost does not disturb the natural oscillations. If p  k , then particular solution of considered type does not exist. Let’s find the particular solution in the form: x   Bt cos(kt   )

Find the derivatives x  : x   B cos( kt   )  Bkt sin( kt   ) ,

x   Bk sin(kt   )  Bk sin(kt   )  Bk 2 t cos(kt   ) . Substituting the values x  and x  in (16) and taking into account that p  k we will find B:  Bksin(kt  )  Bksin(kt  )  Bk2t cos(kt  )  k 2 Btcos(kt  )  hsin(kt  ) 202

3. Dinaamics of the mass poiint and the system

or  2Bk B sin(kt   )  h sin((kkt   )  B  

h 2k

General solution has the form: x  C1 cos kt  C 2 sin kt 

h t cos( kt   ) 2k

or x  C1 coos kt  C 2 sin kt 

h  t cos( kt    ) 2 2k

This equation shows s that ampplitude increasess with time indefinitely, i.e. reson nance takes place ((see the figure). Influence of resistance on fforced oscillation ns

Acting forces: resstoring force F , perturbing force Q , resistance fo orce R : Fx  cxx, Q x  H sin( pt   ), R x  x . Let’s deduce thee differential equuation of motion n of a mass poin nt M:

mx  cx  x  H sin( pt   )

or

x 

 m

x 

c H x  sin( pt   ) m m

Or introducing th he notations

 m

 2 n,

c H  k2,  h : m m

x  2nx  k 2 x  h sin( pt   )

(18)

This is the differrential equation oof forced oscillatiions with the resisstance to motion. 203

Textbook on Theoretical Mechanics

Thus, we have the inhomogeneous equation of second order. General solution of this equation is: x  x   x  General solution of homogeneous equation x  has the form (13), (15) or (15а) in dependence on values of k and n . We will find the particular solution of this equation in the form:

x   AC sin( pt     )

(19)

Constants AC and  will be found by substituting (19) in (18).

x   AC p cos( pt     )

x   AC p 2 sin( pt     ) In (18) we will have:  AC p2 sin(pt     )  2nAC p cos(pt     )  AC k 2 sin(pt     )  hsin(pt   )

Express: hsin(pt  )  hsin(pt      )  hsin(pt    ) cos  hcos(pt    )sin Substituting this expression and grouping the terms we will get:

A (k C

2



 p 2 )  h cos sin( pt     )  (2nAC p  h sin  ) cos( pt     )  0

Equating the coefficients we will get: AC ( k 2  p 2 )  h cos  , 2npAC  h sin  ,

Therefore,

AC 

h ( k  p )  4n 2 p 2 2

2 2

204

3. Dinamics of the mass point and the system

2np k  p2

tg 

2

Let’s substitute this value ( AC ) in particular solution:

x  

h (k 2  p 2 )  4n 2 p 2

sin( pt     )

Then the general solution will have the form: 1). when n  k x  ae  nt sin( k 2  n 2 t   ) 

h ( k  p )  4n 2 p 2 2

2

sin( pt     )

2). when n  k h

x  ae nt sh( n 2  k 2 t   ) 

( k 2  p 2 )  4n 2 p 2

sin( pt     )

3). when n  k x  e  nt (C1t  C 2 ) 

h ( k  p )  4n 2 p 2 2

2

sin( pt     )

The values a,  or C1 ,C 2 are defined from the initial conditions. The motion is a superposition of forced oscillations on the damped oscillations (when n  k ) or the superposition of forced oscillations on the aperiodic motion (when n  k ). Presence of the factor e  nt causes the rapid damping of motion and the mass point will execute only the forced oscillation after a certain period of time. Questions 1. What is the definition of forced oscillations? 2. Write the equation of forced oscillations.

205

Textbook on Theoretical Mechanics 3. What is resonance? 4. When the resonance takes place? 5. What is the influence of resistance on forced oscillations?

PRACTICE 3.53. The load with the weight P = 9.8 kg lies on a smooth horizontal plane. It is connected on the left and on the right to the ends of the two horizontal springs with coefficients of elasticity c1 = 4 kg/cm, c2 = 5 kg/cm. Both of the springs are unstrained in the position of equilibrium of the load. Find the equation of motion and the oscillation period of the load if at the initial time it was moved from its equilibrium position to the right by 4 cm and it was given the initial velocity of 90 cm / sec. Answer: x  5 sin(30t  0,092) , T  0,21 sec . 3.54. Define the free period of a load of mass m joined to two parallel springs and the coefficient of stiffness of a spring that is equivalent to the given doubled spring if the load is located so that the elongations of both springs with the coefficients of rigidity c 1 , c 2 are equal.

Answer: T =2π

√

m ; c=c 1+ c 2 ; the location of the load is (c 1+ c2 )

such that a 1 /a 2=c2 /c1 . 3.55. The load P of mass m is suspended to the rod AB which is connected with the rod DE by the help of two springs with stiffness coefficients equal to c2 and c3 . The rod DE is fastened to the ceiling

in the point H by the spring with stiffness coefficient equal to c1 . Rods AB and DE stay horizontal during the oscillations. Define the stiffness coefficient of equivalent spring so that load P oscillated with the same frequency. Find the period of free oscillations of the load. The mass of rods can be neglected. 206

3. Dinamics of the mass point and the system

Answer: c  c1 (c2  c3 ) , T  2 m(c1  c2  c3 ) c1 (c2  c3 ) c1  c2  c3 3.56. Two loads of mass m1  0.5kg, m2  0.8kg are suspended to the spring with the stiffness coefficient c  19.6 N / m . The system was in the state of static equilibrium when the load m2 was removed. Find the equation of motion, circular frequency and period of oscillations of the load m1. Answer: x  0.4 cos 6.26t , k  2rad / sec, T  1 sec 3.57. The load D of mass 4 kg passing the distance s=0.1 m without initial velocity along the inclined plane (α = 30°) struck against the springs with the series connection (c1 = 48 N/cm, c2 = 24 N/cm). Find the law of motion of the load D from the moment of contact with the springs assuming that it is rigidly connected with them after the strike. Assume the position of rest of the load as the origin. Answer: x  3.2 cos1.25t  1.25 sin 1.25t 3.58. The load with the weight P = 98g suspended at the end of the spring moves in a fluid. Stiffness coefficient of the spring is c = 10g/sm. Resistance force is proportional to the velocity of the load: R  v , where   1,6g sec/сe . Find the equation of motion of the load if at the initial moment the load was moved from its position of equilibrium by 4 cm and it was given the initial velocity v 0  4сm/sec .

Answer: x  7,2e8t sin(6t  0,59) . 3.59. The disc suspended to the elastic cabel executes the torsional vibrations in the fluid. Inertia moment of the disc relative to the axis of the cabel is equal to J. Moment needed for twisting the cabel on 1 radian is equal to c. Resistance moment is equal to aS , where  is the angular velocity of the disc. Define the period of oscillations of the disc in the fluid. 4J Answer: T  . 4cJ   2 S 2

207

Textbook on Theoretical Mechanics

3.60. Disturbing force F  0.3 sin t is acting on the load of mass 0.1 kg suspended to the spring with stiffness coefficient c  0.5N / cm. Define the amplitude of forced oscillations in mm. Answer: 6.01 mm 3.61. The mass point of mass m = 50 kg is moving along a horizontal line, attracting to the fixed center O with the force F proportional to the distance of the point from the center. Proportionality constant c=200N/m. In addition disturbing force F   2 sin 2t , expressed in Newtons is acting on the point. Find the law of the point motion if at the initial time x  x 0  0 , v  v 0  1 cm/sec and the disturbing force coincides with the direction of the initial velocity at the beginning of motion. Answer: x  0,01(sin 2t  t cos 2t ) . 3.62. The body with the weight P = 49N, immersed in a fluid, is suspended on a spring, whose static extension is equal to 1 cm under the influence of weight of the body. Free end of the spring makes the vertical oscillations about a fixed point А0 in accordance with the law y A  AA0  0,05 sin 5t where y A is expressed in meters, t in seconds. Resistance force of the fluid during the motion is proportional to velocity of the load v and it is equal to 15,7N when v  1m/sec . Find the amplitude of the forced oscillations of the load.

Answer: b 

0,05 2

2  2 2  1  p   4 n p 2 2 2  k  k k 

 6,7 cm .

3.63. The weight M is suspended on a spring AB the upper end of which makes harmonic oscillations along the vertical straight line with amplitude a and frequency n so that O1 C=a sin nt cm . Define the forced oscillation of the weight M if the mass of weight is 400 g, the spring lengthens at 1m when the force of 39,2 N acts on it, a = 2 cm, n = 7 rad/sec. 208

3. Dinamics of the mass point and the system

Answer: x=4 sin 7 t cm . 3.64. The load with the weight P = 98g suspended at the end of the spring moves in a fluid. Stiffness coefficient of the spring is c = 10g/sm. Resistance force is proportional to the velocity of the load: R  v , where   5,2g sec/сm . Find the equation of motion of the load if at the initial moment the load was moved from its position of equilibrium by 4 cm and it was given the initial velocity v0  240сm/sec . 1 6

 26t  24t  5e 24t ) . Answer: x  e (29e

3.65. The load with the weight P = 98 g suspended at the end of the spring moves in a fluid. Stiffness coefficient of the spring is c = 10 g/sm. Resistance force is proportional to the velocity of the load: R  v , where   2g sec/сm . Find the equation of motion of the load if at the initial moment the load was moved from its position of equilibrium by 4 cm without initial velocity. Answer: x  4e 10t (1  10t ) . 209

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3.9. Motion of mass point under the action of central forces. Law of areas. Binet formulas Motion of mass point under the action of central forces

Let us consider the motion of mass point under the action of central forces (the line of action continually passes through the fixed point О taken as the origin of coordinates). Let us consider the repulsive force to be positive, and the attractive force to be negative. Since for the central force mom0 F  0 , then in accordance with the theorem on change in angular-momentum there is the first integral which is called the area integral:

r v  c

(1)

Projections of vector c on the axis of Cartesian coordinate system are evaluated by formulas: c x  yz  y z , c y  zx  zx , c z  xy  xy .

(2)

If c x  c y  c z  0 , then, it is obvious that the point will move along the straight line, passing through the center О. But if at least one of the values (2) is nonzero, then vector r at all motion time lays in fixed plane, which is perpendicular to vector c . The equation of this plane is:

cx x  c y y  cz z  0 . Thus, orbit of a point moving under the action of central force is a plane curve. Let us find out the geometric sense of an area integral. Let us choose the system of coordinates O~x ~y~z in such a way that the plane 210

3. Dinamics of the mass point and the system

of point motion coincided with the plane O~x ~y , let us introduce polar coordinates of the point r and  : ~ x  r cos ,

~ y  r sin  .

Then c ~x  c ~y  0, c~z  ~x ~y  ~x ~y and we get the polar form of the area integral: r2

d  c ~z . dt

(3)

Now let the angle  for time dt change to value d , radius r – to dr , then the area of the curvilinear triangular, circumscribed by radius for time dt will be (within the accuracy of infinitesimals of higher order): 1 1 S  r  r  r 2  2 2

If we divide both parts of this equation by t and switch to the limit, we will get: dS 1 2 d . (4)  r dt

Derivative

2

dt

dS in mechanics is called sector velocity. Then for it dt

considering (3) and (4) we have an expression: dS 1  c~z . dt 2 211

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Thus, geometric sense of the area integral is that sector velocity of a point is constant. Binet formula

Let us redesignate the polar angle through φ. It is known that the velocity of a point in polar coordinates is: 2

 dr   d  v2  r 2   r 2   dt   dt 

2

Let us assign dt from the area integral: r2

r 2 d d  C  dt  C dt

and substitute into expression for velocity: 2

2

2 (d ) 2 C 2  dr  C 2 C 2  dr  C   v2  r 2   4  r 2 4  2 r (d ) 2  d  r 4 r  dt  r

Let us made a change:

1 dr d 1 dr d 1  r 2 or   2  , d  r  d d  r  r d

then for the square of velocity we will get:  d  1  2 1    d  1  2 1 1   dr  2 1 1  v 2  C 2   4  2   C 2 r 4    4  2   C 2     2  r  r   d  r  r    d  r  r  d  r

or  d  1  2 1  v 2  C 2      2   d  r  r 

212

3. Dinamics of the mass point and the system

This formula is called Binet's first formula for velocity of a point. The formula lets us define the velocity of a mass point moving in central field when its trajectory r  r ( ) and sector velocity are known. Let us make use of work-energy theorem: Fdr  d

mv 2 2

Let us divide by d and substitute v 2 from Binet's first formula: F

2 2 1  mC2 d  d  1  1  dr d m 2  d  1   C     2       2   2 d  d  r  r  d d 2  d  r  r 



2 mC2  d  1  d 2  1   1  1  1  dr  2  d 1 d 21 dr      2    2 2 mC r          2 2   2  d  r  d  r  d   d  r  d  r  r  r  d 

2  d  1  d 2  1  1 d  1   1  1 2 d  1  d  mC2    2       mC   2     d  r  d  r  r   d  r  d  r  r d  r 

Let us substitute

F (r 2 )

dr d 1  r 2   to the left side and get: d d  r 

2 d 1  1  1 2 d  1  d    mC   2      d  r  d  r   d   r  r  2 mC  d 2  1  1  F   2  2   . r  d  r  r 

This is Binet's second formula which is used for defining the central force, when the trajectory and the sector velocity of a point are known. Binet formulas let us solve an inverse problem as well – finding the trajectory of a point by given central force. Then the problem comes to the integration of second-order differential equation. 213

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Example. A mass point of mass m circumscribes a circle of radius R . What should be the central force if its center is on the circle? For the polar axis let us take the diameter of a circle ОА. Then the path equation will be

r  2R cos  or

Let us calculate derivatives from

1 1  r 2 R cos  1 : r

d 1 1 d 1 1 sin  (cos  ) 1  (1)(cos  ) 2 ( sin  )  ;   d  r  2 R d 2R 2 R cos 2  d 2  1  1  cos  cos 2   2 cos  ( sin  ) sin   1  sin 2  1      2 cos 3   cos   2   4 d  r  2 R  cos  2 R   

Let us substitute these values into the Binet formula for force: F 

mC2 r2

 d 2  1  1 mC2  1 2 sin2  1 1      2     2 2  3    2 2 cos 2 cos r r R R R    d 4 R cos cos     



mC2  2(1  cos2 ) 2  mC2  2 2 2  mC2    3 5    3 2  3   3 2 3 cos  8R cos   cos  8R cos   cos  cos cos  4R cos 



8mC2 R 2 8mC2 R 2  5 5 32R cos  r5

That is the point is acted upon by central attractive force which is inversely proportional to the fifth order of the distance of a point from the attractive center. The value of force depends upon law of motion of a point along the trajectory. Let us assume that in the outermost point of the trajectory (r  2 R ) velocity equals v0 , then C  2 Rv0 and for F we will get the expression: F 

8m4 R 2 v02 R 2 32mv02 R 4  5 r r5 .

214

3. Dinamics of the mass point and the system Questions 1. What is the central force? 2. What is the main property of central force? 3. What is the orbit of a point moving under the action of central force? 4. What is the sector velocity? 5. What can be defined using the first Binet formula? 6. What can be defined using the second Binet formula?

3.10. Planetary motion problem. Kepler's laws. Derivation of Newton’s law of gravitaion from Kepler's laws Planetary motion problem

Planetary motion laws were opened by German astronomer Johannes Kepler (1571-1640). He was banished from Germany and he worked in Prague with famous Tycho Brahe (1546-1601). Processing numerous observations of the planet Mars by Tycho Brahe Kepler determined the planetary motion laws. The orbit of a planet (and comet) is an ellipse with the Sun at one of the two foci. Areas circumscribed by radius vector of planets relative to the Sun proportionate to planetary motion times. The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, i.e.

T 2 : a 3  const . Kepler's laws gave quite clear notion of planetary motion and showed that the planet world is a harmonious system controlled by a single force associated with Sun. However, Kepler could not determine the law of gravity action against the Sun because the basic laws of mechanics were still unknown. Newton was the first person who determined the force acting on planets. Newton carried out his reasonings using a complicated geometric method. We shall use Binet formulas for derivation of the law of gravity.

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Derivation of Newton's law from Kepler's laws

From Kepler's second and first laws follows that the force acting on planets is central and its center is the Sun. Actually, from the area law xy  yx  C we have xy  yx  0 . By defining the acceleration from the equation of motion mx  X ,

my  Y ,

mz  Z we will get

that the moment of force relative to the origin of coordinates equals to zero, consequently, the force is central. Kepler's first law determines the orbit and enables to determine the force with the help of Binet formulas. Let us write down the ellipse polar equation: 1 1  e cos  , (1)  r p where e 

a2  b2 – eccentricity of ellipse, a and b – semimajor a 2

and semiminor axes of ellipse, p  b – focal parameter. a

Let us substitute (1) to the Binet's second formula: F 

mC 2  d 2  1  1  mC 2   e cos  1  e cos   mC 2 .           p p r 2  d 2  r  r  r2  pr 2 

Thus, the central force acting on a planet is attractive (as there is a « minus» sign) and inversely proportional to the square of a planet distance from Sun. A value С of doubled sector velocity is determined from the planet motion law. Let us express the force acting on a planet as: F 

where  

m , r2

2

C . p

216

3. Dinamics of the mass point and the system

Let us show that the acting force is in direct proportion to the mass of the planets. For that in beforehand it is necessary to show that the value μ is equal for all planets. It can be expressed as: 

C 2 C 2a .  2 p b

For the rotation period the radius vector will cover the whole area (πab) of an ellipse, consequently: C dS  T   T  ab  C  2ab dt 2 T ,

then 

4 2 a 2 b 2 a a3  4 2 2 2 2 T b T

According to Kepler's third law this ratio is constant for all planets, consequently, μ is constant for all planets of solar system. Assuming that μ = k2M, where M – mass of the Sun, then the force acting on planet is: mM F  k 2 2 . r Law of universal gravitation: two bodies gravitate with force which is in direct proportion to the product of their masses and inversely proportional to the square of distance between them. Value 2 k is also constant for all planets of Solar system; it is called gravitation constant: γ= k 2 . The derived law of gravitation of the bodies is true not only for planets, but also for all bodies in whole. Questions 1. What is the formulation of first Kepler’s law? 2. What is the formulation of second Kepler’s law? 3. What is the formulation of third Kepler’s law? 4. Formulate the law of universal gravitation. 5. How can be defined the force using first Kepler’s law?

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3.11. Motion of a non-free mass point. Concept of a constraint. Motion of a mass point over given curve. Motion of a mass point over given surface. Geodetic line Motion of a non-free mass point

Formulation of a problem. A mass point is called non-free if it cannot hold an arbitrary position in space. Conditions restraining the freedom of motion of a point are called constraints. Constraints imposed on a point may hold it on a certain curve or surface. The essential distinction of a non-free mass point from a free mass point is that the non-free mass point is affected not only by active forces but also by constraint forces when it performs the motion. If a constraint is ideal (friction-free) then the constrained force will be directed on a normal to the curve or the surface where the point has to stay by virtue of imposed constraints. The value of this force is unknown in advance and depends upon both the acting active forces and the law of a point motion. Determination the law of a mass point motion and the forces of imposed constraints knowing the acting active forces and initial conditions is the primal problem of dynamics for a non-free mass point. Differential equations of motion of a point over a given curve in projections on Cartesian coordinate axes

Let us assume that the constraint is rheonomic, that is the curve, on which a point has to move, may change with time and is given by equations f1 x, y, z, t   0 f 2 x, y, z, t   0

Differential equation of motion of a point in vector form 218

3. Dinamics of the mass point and the system

m

d 2r  F  N , dt 2

(1)

N – reaction force of a curve directed on one of normals. If considering the given curve as intersection of two surfaces f1 and f 2 , then the reaction force N may be considered as a sum of forces N1 and N 2 of these surfaces directed on normals to the respective surfaces, that is

N = N1 + N 2

(2)

It is obvious that N1 and N 2 lie in a normal plane of a given curve. As the direction of a normal to a surface f  0 coincides with the vector direction grad f = 0, then

N1 = 1 grad f1 , N 2   2 grad f 2

(3)

where 1 and 2 – are multipliers to be determined. Taking into account the equalities (2) and (3) we will get an equation of motion of a point (1) in the form m

d 2r  F  1 grad f 1 +  2 grad f 2 dt 2

From the equation (3) it follows that 1 

2

N1 , gradf1

2

2

 f1   f1   f1          1 f1 .  x   y   z 

where gradf1 = 

219

(4)

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Similarly for λ2 , then 1 

N1 N2 , 2  1 f1 2 f2 2

2

(5)

2

 f   f   f          gradf is called first  x   y   z 

The value f1  

differential parameter of function f (according to Lame). From (5) it is seen that multipliers λ1 and λ2 are equal to the normal reaction forces divided by first differential parameter of a constraint. In projections on coordinate axes an equation of motion (4) gives: f1 f   2 2  x x  f1 f 2   2 my  Fy  1  y y  f f  mz  Fz  1 1   2 2  z z 

mx  Fx  1

(6)

This is a system of simultaneous differential equations of second order; let us add equations of constraint f1x, y, z, t   0 , f 2 x, y, z, t   0 to them and we get a system of equations from which we can define x, y, z, λ1 and λ2 as a function t. Natural equations of motion of a mass point over a given curve

When a given curve AB on which a point moves is fixed (stationary constraint), it is convenient to use equations of motion in projections on axis of natural trihedron: τ – directed towards positive count of distance S; n – towards the concavity of trajectory; b – binormal perpendicular to τ , n. 220

3. Dinamics of the mass point and the system

Let the active force that acts on a point is equal to F and constraint force equals to N ; if constraint is ideal, then force N is normal towards the curve, that is lies in plane nb. Then the motion equation mw  F  N in projections to  , n, b gives: m

d 2s dv  F or m 2  F dt dt mv 2



 Fn  N n

0  Fb  N b

(7а)

(7б) (7в)

Equations (7) are called natural equations of motion of a point on a given fixed smooth curve. First of them does not contain an unknown constraint force and it is used for determination of a law of motion of a point; equations (7b) and (7c) determine the constraint force that depends on force F and velocity. Thus, using equations (7), we may determine the law of motion of non-free mass point without finding the constraint force, what cannot be done using a system (6). Example. A mathematical pendulum moves in a vertical plane xy. Constraint equations are of the form:

f1 x, y, z   x 2  y 2  z 2  l 2  0 – equation of sphere f 2 ( x, y, z )    Z  0 – equation of plane xy.

Intersection of these surfaces gives a circle in the plane xy. Equations of motion will be: 221

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f1 f   2 2  x x  f 2  f1  2 my  0  1  y y  f f  mz  0  1 1   2 2  z z  mx  mg  1

 f 2 f1   2x   0 x  x  f1  f 2   2 y  0 y  y   f 2  f1  2z   1 z  , z  

m x  mg  2  1 x  mx  mg  21 x    m y  2  1 y y  21 y m    0  m z  2  1 z   2   0  2  (1)     my m y 2  0, 1   (1)  mx  mg  21 x  xy  xy  gy 2y , 2y

Substituting x  l cos  , y  l sin  .  x  l sin   2  x  l cos   l sin     y  l cos   y  l sin  2  l cos  

 l cos 

2





(2)



 l sin  l sin   l cos   l sin  2  l cos  

 sin  cos  2  sin 2   sin  cos  2  cos 2    

g sin   0 . l

Find   y , y , and then find 1 . Initial conditions: t  0 :   0,    0   

d g  2 g  sin   0  cos   c  0 d l 2 l

 

d d  2 g c 0  d dt 2 l.

222

g l2

g l sin  l2

3. Dinamics of the mass point and the system

Work-energy theorem for a non-free mass point

If we apply constraint forces to a non-free mass point except the active force F , then the point can be considered as free, and we can apply to it all the theorems that are true to the free point. According to the work-energy theorem we have:  mv 2    F dr  1 grad f1 dr  2 grad f 2 dr d   2 

As grad fdr 

f f f dx  dy  dz x y z ,

And form constrain equations we have f f f f dx  dy  dz  dt  0 x y z t

f x, y, z, t   0 

f

  grad fdr  t dt .

Thus, theorem for rheonomous constraint is of the form:  mv 2  f f   F dr  1 1 dt  2 2 dt d  t t  2 

(8)

for stationary constraint: f 1 f 2   0, t t

then  mv 2 d   2

   F dr 

(9)

Consequently, in the case of stationary ideal constraint the constraint forces will not be a part of elementary work expression, and the work-energy theorem keeps the same form that is for a free point. 223

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The reason is that by stationary ideal constraints the actual displacement dr will always be perpendicular to the force N , that is why the elementary work of the force will be equal to zero. From (9) we will get: mv 2 mv02 a   AMoM 2 2

(10)



a where AMoM – work on displacement MoM of active force F . If the active force that acts on a point is potential the elementary work will equal to: F dr  dU   dV ,

where V ( x, y , z ) – potential energy of a point. Then from (9) we will get the energy integral: mv 2  V ( x, y , z )  h 2

(11)

where h – initial energy of a point. Thus, energy integral exists for non-free motion, if the acting force is potential and the constraint is ideal and stationary. Trajectory of a point moving along the surface will be, obviously, a curve lying on that surface. The problem of determining the movement of a point along the surface is solved analogically to the motion of a point on a given curve. Geodetic line on a surface is a line in each point of which main normal coincides with normal to the surface. Geodetic line on a surface plays for that surface a part of a straight line: if the mass point has to stay on the surface, then in the absence of other external forces it moves on a surface along the geodetic line; an elastic thread stretched on a surface takes the form of geodetic line; a line of shortest distance between two points on a surface is geodetic. 224

3. Dinamics of the mass point and the system Questions 1. What is the non-free mass point? 2. What is the primal problem of dynamics for a non-free mass point? 3. Write the differential equations of motion of a point over a given curve. 4. Write the natural equations of motion of a point over a given curve. 5. Formulate the work-energy theorem for a non-free mass point

3.12. Relative motion and equlibrium of a mass point. Equations of relative motion. Inertial forces of transportation motion, Coriolis inertial force Relative motion of a mass point

Any motion of a mass point or a body is considered relative to some reference system. Before now we were considering the motion relative to inertial reference system. Inertial reference system is a system where fundamental laws of dynamics are true. Mass point moves inertially (rectilinearly and uniformly) relative to this system if it’s not influenced by any forces. Inertial system is considered as relatively motionless and the motion relative to it as absolute. We will consider motion relative to reference system which moves arbitrarily relative to inertial reference system. That kind of motion of a point is called relative. If motion of this reference system relative to inertial reference system is not translational, uniform and rectilinear then the system is noninertial. The fundamental law of dynamics, particularly the law of inertia do not take place in noninertial reference system. Reference system related to the Earth is also noninertial. However only delicate experiments (observation of deflection of falling to the east bodies, rotation of pendulum oscillation) can determine that geocentric reference system is noninertial. In most cases reference system related to the Earth can be determined as inertial. Noninertiality is much distinctly revealed in reference systems related to acceleratedly moving technical objects: lift going up with acceleration, artificial satellite or space ship taking off from the Earth. If we relate reference system to the ship, automobile or 225

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airplane moving along the curved trajectory, then the noninertiality becomes so significant that the basic dynamic equation will be found wrong then numerous conclusions derived from it will turn to be also wrong. In order to extend all dynamic equations to noninertial reference sysytems the consistent inertial forces should be introduced. Differential equations of relative motion

Let us consider a mass point М which is influenced by the force F . Let us set up differential equations of motion of that point relative to the reference sysytem Axyz which arbitrarily moves relative to inertial reference system Bx1 y1z1 .

Wa  Wr  We  Wc , Wc  2 e   r ,

e – angular velocity of reference system A relative to the reference system B. Equation of motion relative to B:

mWa  F . Replacing Wa we will get:

mWr  mWe  mWc  F or

mWr  F   mWe    mWc 

Let us designate: J e  mWe , J c   mWc . Let us call the values J e and J c as transportation inertial force and Coriolis inertial force correspondingly. Then we will get: 226

3. Dinamics of the mass point and the system

mW r  F  J e  J c

an equation of relative motion of a point (relative to the reference system А). These equations can be set up as well as equations of absolute motion if we add transportation and Coriolis inertial forces to the acting forces interacting with other bodies. In Cartesian reference system: mx  Fx  J ex  J cx  my  Fy  J ey  J cy   mz  Fz  J ez  J cz

Introduction of inertial forces J e and J c when studying the relative motion let us to set up the equations of motion of a point in noninertial reference system in the same form that had the point M for inertial system. That is with the help of forces J e and J c we take into account the influence of motion of moving reference system onto relative motion of a point. Normal forces with which we deal in inertial reference system appear in noninertial coordinate system. Transportation and Coriolis inertial forces cause relative acceleration. They can deform the body and even destroy it, they perform work etc. But it should be noted that unlike of normal forces, for example, gravity force the value and the direction of which depend on the character of bodies interaction and do not depend on choice of the reference system, inertial forces J e and J c are determined by the choice of noninertial coordinate system. Moreover, we cannot indicate the bodies within the Solar system (to which related the heliocentric system) in result of interaction with appear inertial forces.

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Methods of determination of inertial forces

a) In order to find We it is necessary to know the motion of moving coordinate system. Formula of transportation acceleration has the form: We  W0           

 and  – angular velocity and angular acceleration of moving coordinate system; W0 – acceleration of its origin;  – position vector of a point in a moving coordinate system. In all cases of calculation of We and J e we should consider We as acceleration of a point fixed in a moving coordinates system («to freeze» a point in a moving trihedron). b) Wc  2e   r . If moving coordinates system performs translational motion then J c  0 . Equations of relative rest of a point

A point relative to reference system А is in rest then r  0 . Wr  0  Wc  0 and we get an equation of relative rest in the form of: F  Je  0 . That is, the equations of relative rest are set up in the same way as equations for inertial reference system if we add to active forces the transportation and Corilolis inertial forces. The difference is in conditions of equilibrium: in inertial reference system F  0 means that a point may be either in the state of rest or uniform rectilinear motion. In noninertial reference system F  J e  0 determines only a condition of relative rest of a point. And if the point performs uniform and rectilinear relative motion then the forces acting on it will satisfy the equation: 228

3. Dinamics of the mass point and the system

F  Je  Jc  0,

where J c  0 if moving coordinates system does not perform translational motion or  and  are not parallel. Work-energy theorem for relative motion

All general theorems obtained for inertial reference system take place in relative motion too, if only we add J e and J c to the forces of interaction with other bodies acting on a point. d mv   F dt – in inertial reference system,

d mv   F dt  J e dt  J c dt – in noninertial reference system

 mv 2    F dr  J e dr  J c dr . d  2     2  2   

After proving that J c dr  0 we get: d  mv   F dr  J dr . e Theorem: in case of relative motion the differential from kinetic energy of a point equals to elementary work of applied forces of interaction with other bodies added to elementary work of transportation inertial force. Questions 1. What is relative motion of a mass point? 2. Write the differential equations of relative motion of a mass point. 3. How can be determined the inertial forces? 4. Write the equations of relative rest of a mass point. 5. Formulate the work-energy theorem for relative motion.

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3..13. Apparent weeight of a body. D Deflection of bod dies falling on the Earth from the vertical. W Work-energy theo orem for relative motiion. Relative mottion and equilibrrium Rellative rest and rellative motion closse to the surface of o the Earth

The Earth is a non-inertial n system m as it rotates around its axis and moves non-rectiilinearly about thhe Sun. The lastt motion for interrvals of t