Theoretical mechanics: educational manual 9786010418127

Citation preview

AL-FARABI KAZAKH NATIONAL UNIVERSITY

Z. B. Rakisheva A. S. Sukhenko

THEORETICAL MECHANICS Educational manual

Almaty «Qazaq university» 2016

UDC 531.011 (076) LBC 22.21 R 17 Recommended for publication by the Academic Council of the Faculty of Mathematics and Mechanics and Editorial and Publishing Council of the Kazakh National University named after Al-Farabi (Protocol №2 dated 12.02.2016) Reviewers: Doctor of Physical and Mathematical Sciences, Professor A.N. Turekhodjayev 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

R 17

Rakisheva Z.B. Theoretical mechanics: educational manual / Z.B. Rakisheva, A.S. Sukhenko. – Almaty: Qazaq university, 2016. – 196 p. ISBN 978-601-04-1812-7 Educational manual was prepared on the base of a compulsory subject of theoretical mechanics, read by the authors for the students of specialty «Mechanics». It contains a lecture course of five modules, including kinematics, statics, dynamics of the mass point and the mechanical system, rigid body dynamics, analytical mechanics, as well as some of the tasks and exercises on these topics. It will be of interest for students of specialties «Mathematics», «Mathematical and Computer Modeling» and others, which curriculum includes the study of theoretical mechanics. Publishing in authorial release. Учебное пособие подготовлено на основе обязательного курса теоретической механики, читаемого авторами для студентов специальности «Механика». Содержит курс лекций из пяти модулей, включающих кинематику, статику, динамику материальной точки и механической системы, динамику твердого тела, аналитическую механику, а также некоторые задачи и упражнения по этим темам. Представляет интерес для студентов специальностей «Математика», «Математическое и компьютерное моделирование» и других, в учебный план которых входит изучение теоретической механики. Издается в авторской редакции.

UDC 531.011 (076) LBC 22.21 ISBN 978-601-04-1812-7

© Rakisheva Z.B., Sukhenko A.S., 2016 © Al-Farabi KazNU, 2016

Content

LECTURE COURSE ......................................................................6 Module I. KINEMATICS ...............................................................6 Lecture 1. The subject of mechanics. Models of material bodies studied in mechanics. Basic concepts and laws of mechanics. Kinematics of a point. Problems of kinematic. Methods of the point’s motion setting...............................................6 Lecture 2. The speed and acceleration. The decomposition of the velocity and acceleration on the radial and transversal components. Decomposition of acceleration of the axes of the natural trihedral .....................................................10 Lecture 3. Mechanical system. The number of the freedom degrees of the system and a rigid body. Basic movements of a rigid body. Translational motion of a rigid body. Acceleration and velocity in the translational motion .......................15 Lecture 4. Rotational motion of a rigid body about a fixed axis. Angular velocity and the angular acceleration of a rigid body ..........17 Lecture 5. Plane-parallel motion of a rigid body. Velocities of the points of a plane figure. Instantaneous center of velocity .......22 Lecture 6. Acceleration of the points of a plane figure. Instantaneous center of acceleration .................................................26 Lecture 7. Compound motion of a point. Basic concepts. Full and relative derivatives of the vector. Addition of velocities. Theorem on the addition of accelerations (Coriolis theorem) ............................................................................30 Lecture 8. Complex motion of a solid body. The problem statement. Addition of the translational velocities. Addition of the instant angular velocities. Addition of the instant angular and translational velocities .......................................36 Module II. STATICS ......................................................................40 Lecture 9. Statics. Basic definitions and axioms of statics. Constraints. Constraints. Constraint reactions. Axiom of constraints ...................40 3

Theoretical mechanics

Lecture 10. Center of gravity ...........................................................44 Lecture 11. Equilibrium conditions of arbitrary spatial system of forces ............................................................................................47 Lecture 12. Special cases of the equilibrium conditions (convergent system of forces, parallel forces, plane system of forces) ...........................................................................................51 Module III. DYNAMICS OF THE MASS POINT AND THE SYSTEM .......................................................................53 Lecture 13. The laws of Newton. Direct and inverse problems of dynamics. Motion equations .........................................53 Lecture 14. Basic dynamic variables. Properties of internal forces of the system...........................................................................58 Lecture 16. Angular momentum theorem of a mass point and mechanical system ............................................................72 Lecture 17. Work of force. Work of potential force ........................78 Lecture 18. Work-Energy theorem for the mass point and mechanical system. Energy integral ...........................................84 Lecture 19. Rectilinear motion of a mass point. Harmonic oscillations of the mass point. Parameters of oscillations. Oscillations in a resistant medium ....................................................87 Lecture 20. Forced oscillations in a medium without resistance and in a resistant medium. Resonance ..............................................96 Module IV. DYNAMICS OF THE SOLID BODY.......................103 Lecture 21. Mass geometry. Inertia moment. Inertia moments of the body relative to the axis and point. Centrifugal moment of inertia ............................................................................................103 Lecture 22. Theorem of Guigens-Shteiner. Inertia moments relative to the axes of the beam, coming from this point ..................107 Lecture 25. General formulation of the problem of heavy solid body with one fixed point. First integrals of equations of motion of the heavy solid body around the fixed point ................110 Lecture 26. Special cases of integration and its geometrical interpretation: the case of Euler-Puanso, the case of Lagrange-Puasson, the case of Kovalevskaya .................116 Module V. ANALYTICAL MECHANICS ...................................119 4

Content

Lecture 27. The notion of holonomic and nonholonomic systems. Actual and virtual displacement of the point. Conditions imposed by reactions on the variations of coordinates ....................................................................................127 Lecture 28. Virtual work of forces. Ideal constraints. Principle of virtual work. d'Alambert's principle for the point and system. General dynamic equation ..............................130 Lecture 29. Generalized coordinates. Generalized forces. Equations of motion in generalized coordinates (Lagrange equations of 2-nd kind) ...................................................133 Lecture 30. Expression of kinetic energy in generalized coordinates. Lagrange equations of 2-nd type for the system under action of potential force ..............................................136 PRACTICE ......................................................................................139 1. Velocity and acceleration of a mass point and methods of its decomposition ..........................................................................139 2. Rotational motion of a perfectly rigid body around the fixed axis. Angular velocity and angular acceleration of a rigid body ...................................................................................143 3. Velocities and accelerations of points of plane figure ...................146 4. Compound motion of a mass point. Theorem on velocity addition. Theorem on acceleration addition ......................................152 5. Center of gravity ...........................................................................158 6. Special cases of the equilibrium conditions ..................................161 7. Differential equations of motion. Direct and inverse problems of dynamics .......................................................................165 8. Theorems of dynamics ..................................................................169 Theorem of change of linear momentum (momentum theorem). Angular momentum theorem ............................................................169 9. Work – energy theorem .................................................................176 10. Oscillations. Free oscillations. Free oscillations in the medium with resistance. Forced oscillations. Resonance........182 11. Inertia moments...........................................................................186 12. Principle of virtual work. Lagrange equations of 2-nd kind ........190 REFERENCES ................................................................................194

5

Theoretical mechanics

Lecture 1. THE SUBJECT OF MECHANICS. MODELS OF MATERIAL BODIES STUDIED IN MECHANICS. BASIC CONCEPTS AND LAWS 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 is: 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. 6

Module I. Kinematics

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:  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. 7

Theoretical mechanics

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 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) 8

(1)

Module I. Kinematics

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. 2. Coordinate methods. 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, q2, 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 ) , q2  q2 (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 9

Theoretical mechanics

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. According to the character of the trajectory the motion of the point can be linear and curvilinear in the dependence on the reference frame. Lecture 2. THE VELOCITY AND ACCELERATION. THE DECOMPOSITION OF THE VELOCITY AND ACCELERATION ON THE RADIAL AND TRANSVERSAL COMPONENTS. DECOMPOSITION OF ACCELERATION OF THE AXES OF THE NATURAL TRIHEDRAL. 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: v

dr  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 10

Module I. Kinematics

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.

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 direction 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 v  r  r 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  d  0 , v  vr  v p  r 0  r p dt dt 11

Theoretical mechanics

where the first term is the radial component of the velocity an the second is the transversal component.

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 2  d dp 0 . w  2 r   p  r 2 p0  r dt dt dt dt dt dt dt dt dt

Let’s find the magnitude and direc0 tion of the vector dp . d p 0  p 0 is the dt differential of the unite vector that is perpendicular to the direction of the  vector itself and directs opposite to r 0  (direction of d p 0 is obtained by the rotation of p 0 at an angle 90° in the direction of positive count of a angle).  Besides, dp 0  d , therefore:

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

2   d 2 r  d    0  d 2 dr d   0 w   2  r   r  r 2  2 p dt dt   dt    dt  dt

12

Module I. Kinematics



 d 2r

2  d    0  r  r directs along the posi  2  dt    dt

Radial component wr   tion vector.



 d 2 dr d   0 2  p is perpendicular 2 dt dt   dt

Transversal componens w p  r to the position vector.

NATURAL TRIHEDRAL 1) Let’s draw a tangent in the point M and define the positive  direction with the unit vector  0 (in the direction of increasing the curvilinear coordinates S). Osculating plane is the end position of the plane passing through any three points of the curve when these points tend to M. IOW: Osculating plane is the end position of the plane passing through the tangent Mτ and the point M' when M'→M. 2) Let’s draw a plane through the point M and its plane will be perpendicular to the tangent  0 . This plane is called the normal plane of the trajectory at the point M. All lines lying in this plane is named normal, and the line of intersection of normal plane to the osculating plane is called the principal normal n 0 . 3) Straight line perpendicular to the tangent and the principal normal is called binormal. The unit vector b 0 . The positive direction   is selected so that  0 , b 0 , n 0 form a right-handed system. This system is called the natural axes. The plane passing through  ( 0 , n 0 ) is called osculating plane, through ( n 0 , b 0 ) – normal plane,  through ( b 0 ,  0 ) – rectifying plane. Trihedral with vertices at the point M is a natural trihedral, it moves along a trajectory with the point M. 13

Theoretical mechanics

CURVATURE OF A CURVE Let’s identify with  the angle of contingence between the 

tangent  0 and  0 ' , and arc length ММ '  S , than   k  is the aveср S

rage curvature on the segment ММ ' .  d lim   k is the curvature of the S 0

S

dS

curve. The value  

1 dS  is the radius of k d

curvature.

DECOMPOSITION OF THE ACCELERATION ON THE AXIS OF NATURAL TRIHEDRAL    v  v  0  v 0    dv dv  0 d 0 w    v dt dt dt

The firs term directs along the tangent. Let’s consider the second 0 term d v . The vector d 0 directs along the normal to n 0 . Its dt

magnitude equals:  sin  2  lim  ) . ( lim  lim lim t 0 t t 0 t t 0  t 0 t 2 0 2   d  v v d  d  dS 1 But v  n 0v  n 0   v , then dt   dt dS dt  2  dv  v  finally, w   0  n 0 dt   d 0 d  dt dt

  0

14

Module I. Kinematics

w 

v2 dv is the tangential component, wn  – is the normal  dt

component. Since wb  0 , than w  rectifying plane. Absolute magnitude of acceleration is: 2 2  dv   v  w  w2  wn2        dt    

2



The angle μ 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.

Lecture 3. MECHANICAL SYSTEM. THE NUMBER OF THE FREEDOM DEGREES OF THE SYSTEM AND A RIGID BODY. BASIC MOVEMENTS OF A RIGID BODY. TRANSLATIONAL MOTION OF A RIGID BODY. ACCELERATION AND VELOCITY IN THE TRANSLATIONAL MOTION 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, i.e.: x1 , y1 , z1 , x2 , y2 , z2 , ..., xn , yn , zn . 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. 15

Theoretical mechanics

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. Let the system be impose by the k- geometric constraints: f  x1 , y1 , ...., zn ; t   0, (   1, k )

Then from 3n - coordinates will be (3n-k) – independent coordinates, that is, if it is set any (3n-k) – coordinates, the rest of them will be determined from the constraint equations. These independent coordinates are called the coordinates of the system. Only in case of geometric constraints the number of coordinates of the system is called the number of degrees of freedom of the system. Mechanical system where the distance between any two points is constant is called absolutely rigid (solid) body. The basic motions of a rigid body. There are five types of motion of a rigid body: 1) translational motion; 2) rotational motion; 3) plane-parallel or plane motion; 4) spherical motion; 5) general case of rigid body motion. Translational and rotational motion are simple, basic motions of a rigid body.

TRANSLATIONAL MOTION OF A RIGID BODY If every line connecting two points of the body moves parallel to itself this motion is called translational motion. Theorem. All points of the solid body executing a translational motion describe identical and parallel trajectories and have 16

Module I. Kinematics

geometrically equal velocities and accelerations at each moment of time. Thus, the study of translational motion of a rigid body can be reduced to the study of the motion of a single point of this body, that is, to the problem of kinematics of the point.   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 ) z c  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. Lecture 4. ROTATIONAL MOTION OF A RIGID BODY ABOUT A FIXED AXIS. ANGULAR VELOCITY AND THE ANGULAR ACCELERATION OF A RIGID BODY 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 17

Theoretical mechanics

with their plane perpendicular to the axis of rotation, and with the centers lying on this axis in this case. Let's suppose that the rotational axis AB coincides with the axis z. To determine the position of a rotating body let's draw two planes through z-axis: Q – fixed plane and P – fixed and invariably associated plane with the body. The dihedral angle φ between the fixed plane Q and unsteady plane P is called the angle of rotation of the body φ. Angle φ assumed to be positive if from the end of positive z axis it is seen directed opposite clockwise from the fixed plane. Position of the plane P is defined with the magnitude and the sign of angle φ. Therefore the position of the body is defined with this angle. That is φ can be considered as the angular coordinate of the body. Angle of rotation changes with time, therefore it is the function of time:

   (t )

– the law of rotational motion. The body have one degree of freedom in case of rotational motion. If the angle φ becomes increment  for the time interval t the  value  

 is called mean angular velocity of the body for the t

given time interval. Proceeding to the limit we will become:  d    t 0 t dt

  lim

– angular velocity of rotation of the solid body. 18

Module I. Kinematics

Angular velocity changes with time in general case, i.e.    (t ) .  If the increment of ω for t equals to  , than the value   is t

the mean angular acceleration of the body for the given time interval. Proceeding to the limit we will become:   lim

t 0

 d      t dt

– angular acceleration of the body. 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:

  0 t If   const at all the time of motion, this motion is called a uniformly accelerated motion. The law of this rotation is:   0 0 t 

 t2 2

And angular velocity varies according to the law:   0   t

The angle of rotation is often expressed with number of revolutions N. Then the angle φ in radians, corresponding to N revolutions is defined as:   2 N 19

Theoretical mechanics

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    Cross product r  a is called the moment of vector a relative to the center:    i j k       momO a  r  a  x y z , momO a  r  a  sin( r , a ) ax a y az

The direction of the cross product is defined by the «right-hand screw rule».  As any cross product the moment momO a equals to the area of     parallelogram covering the vectors r and a , that is a h  r a sin( r , a ) . Moments relative to the axes:  y M x  momx a  ay

z  z , M y  mom y a  az az

x  x , M z  momz a  ax ax

y ay

VELOCITIES AND ACCELERATIONS OF THE POINTS OF ROTATING SOLID BODY Any point M of rotating solid body makes circular motion, that is:

v M  O0 M   20

Module I. Kinematics

 Then v  mom  , as O0 M is the arm of M



vector  relative to the point M. In accordance with the definition of the   moment: v coincides with mom M  by the   direction, that is v M  mom M  . If О is the arbitrary point of the axis, to  which a sliding vector  is applied and  r  OM then:         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       vy   x v r  0 0   v 0 x y z  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. 21

Theoretical mechanics

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 velocity  . dt

Lecture 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 PERFECTLY 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. Thus all points of the perpendicular M1M2 describe identical and parallel to each other trajectories and have geometrically equal 22

Module I. Kinematics

velocities and accelerations. That is, trajectories A1B1, A2B2, AB are identical and parallel and their velocities and accelerations equal to       v  v  v and w  w  w respectively at the moment. 1

2

1

2

Thus, the motion of each point of the plane figure in a fixed plane determines the motion of all the points of a rigid body, located on the perpendicular to the plane Q. It allows to reduce the study of plane motion of a rigid body to the study of the motion of a plane figure in its plane. As the position of a plane figure on the plane defined by the position of two points or a segment containing two points, the motion of a plane figure can be studied as a motion of rectilinear segment in this plane. Translational and rotational motion are the main types of motion of figure in its plane. When every line in the plane of the moving figure moves parallel to itself this motion is called translational motion of a plane figure. When one point of the moving figure, called the center of rotation, is fixed this motion is called rotational motion of the figure in its plane. Here all the points of the figure moves along the circles with centers at the center of rotation. And v М  OM   , w M  OM   2   4 , tg 

w   2 (Accelerawn 

tion deviates from the radius of rotation at an angle μ which is the same for all points). Theorem 1. Any displacement of a plane figure in its plane can be composed of translational motion and rotation about an arbitrary center (pole) (see the figure). Theorem 2. Every displacement of a plane figure in its plane that is not translational can be implemented by one turn around the appointed center, called the center or pole of finite rotation (see the figure). 23

Theoretical mechanics

These theorems inform about the displacement of the plane figure from one fixed position to another. But one can have the geometrical description of the plane figure motion. Any motion, including the motion of plane figure in its plane can be treated as a continuous sequence of elementary displacements, which can be represented in two ways. 1) According to Theorem 1 elementary displacement can be obtained by an infinitesimal translational displacement with arbitrary selected pole and turn at an infinitely small angle around this pole. This implyies, that any motion of a plane figure can be considered as a combination of translational motion determining by the motion of arbitrary selected pole and rotational motion around this pole. 2) According to Theorem 2, any elementary displacement of figure can be achieved by turning on only one infinitesimal angle around a specified center, called the instantaneous center of rotation. Therefore, any motion of a plane figure that is not translational can be considered as a continuous sequence of infinitesimal rotations about the instantaneous center of rotation (the position of the instantaneous center of rotation is changing continuously). Rotation around any pole or around the instantaneous center occurs with the same angular velocity for this moment t:  (independent of the   lim t 0 t choice of pole). ω is the angular velocity of the figure at the moment or the instantaneous angular velocity. Instantaneous center of rotation is the end position of the center of end rotation О when position П2 tends to position П1. That is, the figure moves from the given position to infinitely close adjacent position by elementary rotation around the point that is the instantaneous center of rotation. 24

Module I. Kinematics

Velocities of all the points А1, А2,…, Аn of plane figure will be perpendicular to the radius of rotation А1P, А2P,…, АnP at the moment. Velocity of a point of plane figure which coincides with the point P, will be equal to zero at the moment. This point of figure is called the instantaneous center of velocities. Knowing the position P, the direction of velocity of any point of plane figure can be determined at the given moment t. To determine the position of instantaneous center of velocity we must know the direction of velocity of any two points of the figure. Restoring the perpendiculars from these points we will get the instantaneous center of rotation P at the point of their intersection. Centroids. During the motion the position of instantaneous center of rotation of plane figure changes continuously both on the fixed plane and on a plane associated with a moving figure. The locus of the instantaneous centers of rotation on a fixed plane is a continuous curve, which is called a fixed centroid. The locus of the instantaneous centers of rotation on a moving plane, related to the moving figure is also a continuous curve, which is called the moving centroid. This implyies, that when the figure is moving the moving centroid is rolling without sliding on a fixed plane (the point of contact, which is the instantaneous center of velocity for the figure, has a velocity equal to zero).

VELOCITIES OF THE POINTS OF A PLANE FIGURE Let the plane figure moves relative to the origin of reference frame  .  Point А is the pole and  A is its position vector, point М is the  arbitrary point of the figure and  M its position vector. Then at any moment of time t:







M  A r 25

Theoretical mechanics

Differentiating with respect to time t  d A dr . As AM=const, then r  

 d M dt

dt

dt

changes only by the direction, therefore  dr   r . dt        Hence, v  v    r or v  v  v M

A

M

A

MA

Velocity of any point M is composed of velocity of arbitrary selected pole A which is common to all points of the figure (velocity    of translational motion of the body) and velocity vMA    r ,



which occurs due to the rotation of figure around the pole A. v MA directs perpendicularly to МА in the direction of rotation of figure.

v MA    AM Thus, knowing the velocity of any point A of plane figure and angular velocity of figure  , we can find the velocity of any point of the figure. Theorem. If it is known the velocity of any point of plane figure and direction of velocity of its other point, it is possible to determine the velocity of any point of a figure using the instantaneous center of rotation. Theorem. The projections of the velocities of the ends of invariable segment on its direction are equal. Lecture 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. 26

Module I. Kinematics

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

rot wMA cen wMA



 . 2

rot always corresponds to the direction of ε. The direction of w MA 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.  wn does not depend on the pole and is perpendicular to the velocity, that is, it is directed along the instantaneous radius to the  instantaneous center of velocity. wcen depends on the pole and is  always directed radially towards the pole. w is directed along the  velocity of the point or opposite to v , that is, it does not depend on  rot the pole, w depends on the choice of the pole and is perpendicular to the line AM.

INSTANTANEOUS CENTER OF ACCELERATION Theorem. At each moment of motion of plane figure in its plane (if   0,   0 ) there is a point of a plane figure whose acceleration equals to zero at this time. 27

Theoretical mechanics

Point Q, which have the acceleration equal to zero at this time, is called the instantaneous center of accelerations. If we assume the instantaneous center of accelerations as a pole then the acceleration of any point of plane figure can be determined as:    wM  wQ  wMQ

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

That is, accelerations of the points of plane figure are proportional to the distances of these points to the instantaneous center of acelerations at each moment of time. The angle between the acceleration of the point and the segment connecting it with the instantaneous center of acceleration is the same for all points of the figure and equal to μ. 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

28

Module I. Kinematics

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 wА  wQ  wQA  wQA  wА  wQA

2.   0,   0 . This is possible when it is the instant translational motion. Then tg         . This angle is measured from 2



 w

2

in the direction of  . Instantaneous center of acceleration lies at the point of intersection of perpendiculars to accelerations.

   rot  cen  rot wA  wQ  wQA  wQA  wQA

METHODS OF CALCULATING THE ANGULAR ACCELERATION 1. Rotation angle φ or angular velocity ω in dependence of time t is known, then: d d 2   2 dt dt 2. Angular velocity is found with the help of instantaneous center of velocity:  

vA . Differentiating in terms of time: AP



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

1

 A а) AP  const all the time of motion, therefore,   AP dt  AP wA .

29

Theoretical mechanics

b) AP  AP(t ) is a variable value all the time of motion. Than  can be found by forming the equation:

   rot  cen wM  wA  wMA  wMA  rot Here w . Hence, we find ε. MA    MA

Lecture 7. COMPOUND MOTION OF A POINT. BASIC CONCEPTS. FULL AND RELATIVE DERIVATIVES OF THE VECTOR. ADDITION OF VELOCITIES. THEOREM ON THE ADDITION OF ACCELERATIONS (CORIOLIS THEOREM) Motion of a point relative to two or several reference frames is called compound motion. For example, a boat crossing the river, a passenger moving in the carriage of the moving train, a passenger moving along the deck of the ship, etc. Suppose that the point is moving relative to a moving coordinate system Oxyz, which is moving relative to the main (fixed) coordinate system  . The motion, velocity   and acceleration of the point ( vr , wr ) considered in relation to the system Oxyz is called relative. The motion, velocity and acceleration of the point relative to the system  is called   absolute ( va , wa ). The motion of the moving coordinate system Oxyz relative to a fixed coordinate system  is called transportation. In transportation motion velocity and aceleration is considered relative to the point where the moving point is located at the moment. In other words transportation velocity and acceleration is the velocity and acceleration of the point which doesn't execute the relative motion. 30

Module I. Kinematics

For example, if a person walks along a radius of the rotating platform, the platform can be associated with a moving reference frame, and the surface of the earth with the fixed reference frame. Then motion of the platform is transportation, motion of a person relative to the platform is relative, motion of a person in relation to the Earth is the absolute motion. The main problem of studying the compound motion is to establish the relationship between relative, transportation and absolute velocities and accelerations of the point. Suppose that the point is moving relative to a moving frame Oxyz that moves in relation to the basic (fixed) coordinate system  . Then the motion, velocity, and acceleration of a point considered in relation to the system Oxyz are called relative, in relation to the system  – absolute (the motion is also called compound). The motion of the moving system Oxyz relative to the fixed coordinate system  is called transportation motion. Velocity and aceleration of a point where the moving point is located at the moment are called transporation.

ADDITION OF VELOCITIES Let's consider the compound motion of the point М. This point executes relative displacement (vector MM  ) along the trajectory АВ for a period of time t  t1  t . Curve АВ transferres to the curve A1B1 for the period of time t . At the same time the point M of the curve АВ where the point M located at the moment t executing the transportation displacement mm1  M M1 . As result the point M will come to the position M1 and execute the absolute displacement MM1 for the time t : 31

Theoretical mechanics

MM1  Mm1  m1M1 Dividing by t and proceeding to the limit: MM1 Mm1 mM  lim  lim 1 1 t 0 t t 0 t t 0 t lim

By definition, as when t  0 the curve A1B1  AB , then we have: MM  = v  v m1M 1 = rel r

lim t 0

t

lim

t 0

t

Mm1 = v  v tr t t 0 t lim

MM1  vabs  va t 0 t lim

      As result: va  vt  vr ( vr , vt , va directed along the tangent to the 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 bet  ween the directions of v t and vr .

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, 32

Module I. Kinematics

 executing a motion, regardless of Oxyz. Then vector r  OM  changes 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  lim dt t  0 t ~

~

dt

t  0 t

– absolute or total derivative and d r  lim r – relative or local derivative. Let's find the relationship between these derivatives.   From the definition of vr and v it follows that, a ~  dr vr  , dt

     dr va  , as va = v r + 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 ve    r (Euler formula for any point of rotating solid body).

 ~   ~  dr d r      r or, for arbitrary vector a : da  d 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 dax  da y  daz  di dj dk ( i j k )  (a x  a y  a z ) dt dt dt dt dt dt dt 33

Theoretical mechanics

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

~

dt

dt

Finally we have: da  d a    a .

ADDITION OF ACCELERATIONS Absolute acceleration of the point М equals to the derivative of absolute velocity of the point M:

va 

d d 0 dr ,   dt dt dt

r  xi  y j  z k ,  dva  wa  dt

Differentiating with respect to t, we will get:

        d 2i d2 j d 2k    di dj dk (1) wa  0  x 2  y 2  z 2  xi  yj  zk  2( x  y  z ) dt dt dt dt dt dt Let’s divide the terms of the right side of this equation into three groups. The first group comprises the terms containing derivatives 34

Module I. Kinematics

only on the relative coordinates x, y, z, but do not contain derivatives of the vectors, that is:

    wr  xi  yj  zk

The second group comprises the terms which only contain     derivatives of vectors  0 , i , j , k but do not contain the derivatives of the relative coordinates x, y, z, that is:     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 accelera tion of the point M. The third group of terms does not refer to wr or      to wt . Accounting that di =   i , dj =   j , dk =   k the third dt dt dt acceleration 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 35

Theoretical mechanics

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

(3)

Equation (3) represents the acceleration addition theorem: Coriolis theorem: Absolute acceleration of the point equals to the vector sum of transportation, relative and Coriolis acceleration.  From (2) it follows, that wc  2evr sin(e , vr ) .  The direction of the Coriolis acceleration wc is determined by the corkscrew rule. Let's consider the cases when Coriolis acceleration is equal to zero: 1. If transportation motion is translational then  e =0  w =0; c  2. If v e , i.e. relative motion of the point has a direction par  rallel to the axis of transportation rotation then wc =0; 3. Point don't move relative to the moving coordinate system then   v r =0  wc =0. Lecture 8. COMPLEX MOTION OF A SOLID BODY. THE PROBLEM STATEMENT. 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 36

Module I. Kinematics

  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1 is the velocity of instant translational motion of a  rigid body relative to the coordinate system Oxyz and v 2 is the 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 va = vr + v t in accordance with the theorem of addition of    velocities, or in our case we have v  v1  v 2. 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. For the case of n ° terms: n   v   vi . i 1 37

Theoretical mechanics

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 Oa . 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

ADDITION OF INSTANTANEOUS ANGULAR VELOCITY AND TRANSLATIONAL VELOCITY 1. The translational velocity is perpendicular to the instantaneous  axis of rotation. Instead of v we take a pair (  ' , ' ). Then 38

Module I. Kinematics

 and   ' are destroyed. It remains only a momentary rotation about another axis Bb with angular velocity  ' =  . Thus, when adding the instantaneous rotational motion and translational motion resulting motion represents the instantaneous rotation with same angular velocity, but around another axis displaced on a distance

v.  2. Translational velocity is parallel to the axis of rotation (the resulting motion is a permanent screw motion or instantaneous screw motion). d

2.1. Permanent screw motion.   v : Both motions are uniform. The resulting motion is called a permanent screw motion, the axis of rotation is called the axis of the screw. 2.2. If instant rotation with angular velocity  is added to  translational motion with the velocity v  then instantaneous screw motion will take place. The resulting motion is instantaneous screw motion, the axis of this screw is the instant screw axis. Position of the instantaneous screw axis changes with time. 3. Translational velocity forms an arbitrary angle with the instantaneous axis of rotation. Resulting motion will be the instant-screw motion.

39

Theoretical mechanics

Lecture 9. 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. 40

Module II. Statics

MOMENT ABOUT A POINT The magnitude of the moment of a force at a point O, is equal to the perpendicular distance from O to the line of action of F, multiplied by the magnitude of the force: M = Fd, where F ° the force applied, d ° the perpendicular distance from the axis to the line of action of the force. This perpendicular distance is called the moment arm.

The moment counts as positive if force tends to turn the arm counterclockwise. In vector format, the moment can be defined as the cross product between the radius vector, r (the vector from point O to the line of action), and the force vector, F: Mo  r  F

Force couple is a pair of forces, equal in magnitude, oppositely directed, and displaced by perpendicular distance or moment. The simplest kind of couple consists of two equal and opposite forces whose lines of action do not coincide. The forces have a turning effect or moment called a torque about an axis which is 41

Theoretical mechanics

normal to the plane of the forces. If the two forces are F and −F, then the magnitude of the torque is given by the following formula:

  Fd where 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. 42

Module II. Statics

2. The system of two equal by magnitude mutually opposing forces applied to any two points of a rigid body and directed by a 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 (S1) can be replaced with another system (S2) 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.

CONSTRAINTS. CONSTRAINT FORCE (REACTION) If each mass point of 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 unfree. Conditions that impose restrictions on a motion of the material system have the name of constraints. There are geometrical constraints, kinematic relations, stationary, nonstationary. Сonstraints are usually represented as various bodies that hinder (стесняет) 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 43

Theoretical mechanics

a)

c)

f)

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;

b)

d)

e)

g)

h)

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. Lecture 10. 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. 44

Module II. Statics

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. P dP  The value   lim is called a weight per unit of voluV  0 V dV  me of the body at this point, and the value   g represents the density of the body (weight of unit of volume) at this point. If the body is homogeneous, and γ 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   Pi   i Vi

As  i   i g i , then 45

Theoretical mechanics

rc 

  i g i Vi ri   i g i Vi

If the body is sufficiently small, it is possible to reduce g i , as acceleration of force of gravity for all the points of the body is the same. Then we will get:  i Vi ri , rc   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 ρ are continuous functions of coordinates, than the amounts represent the volume integral. If the body is homogeneous, than ρ 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. The projections of the static moment on the axis of a Cartesian coordinate system: xc  

xdV , V

yc  

46

ydV , V

zc  

zdV V

Module II. Statics

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

rc 

  l r .   l

If the line is homogeneous than rc   rdl . l Lecture 11. EQUILIBRIUM CONDITIONS OF ARBITRARY SPATIAL SYSTEM OF FORCES Lemma (main). Any force applied to the absolutely solid body at a given point A, is equivalent to the same force applied to another point B, and the pair of forces with the moment, which is equal to the moment of force applied at point A with respect to point B. Proof. Assume that the force FA is apllied at the point A. Let's apply two mutually opposing forces FB   FB at the point B. These forces are equal by magnitude to the force FA and parallel to it. Then 47

Theoretical mechanics

the forces FB and FA give a torque couple equal to the moment of this force FA with respect to point B, i.e. BA F A . Which was to be proved. 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

 (r  F )   M i 1

i

i

i 1

i

 M0 .

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: 48

Module II. Statics

M    (ri Fi ).

The position-vector of the point of application of one of the forces of the system, drawn from the new center is: ri  ri  OO  . Substituting this expression, 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 moment is changed by an amount equal to the moment of vector R relative to the new center of reduction. The second invariant is the quantity R M  R M cos(R, M ) or the projection of the vector M on the direction R . Proof: For 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 )  R M  R  (OO  R ) .

But R  (OO  R ) is equal to zero, as it have to equal multipliers R  (OO  R )  ОО( R  R )  ( R  R )  0, as. sin O o  0

 R   M   R  M  const is the second invariant.

Thus, the projection of M on the direction R is the constant value for a given system of forces and does not depend on the choice of the center of reduction. 49

Theoretical mechanics

POSSIBLE LOCATION OF THE VECTORS M AND R 1). R  M  0 а) R  0 ( R lies in the plane of the pair ( F , F  ), which is replaced by M ). The system of forces is reduced to one force, i.e. to the resultant force: R   Fi . 

M It is possible because of R  M  0  p  R  M 0  M   0. 2 R

R

b). R  0 , но M  0  M  M   const , 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. 50

Module II. Statics

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,

Ry  0,

Rz  0

 mom y Fi  0,

 momz Fi  0

Lecture 12. SPECIAL CASES OF THE EQUILIBRIUM CONDITIONS (CONVERGENT SYSTEM OF FORCES, PARALLEL FORCES, PLANE SYSTEM OF FORCES) 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,

Ry  0 ,  momz Fi  0 .

B) momA Fi  0,  momB Fi  0,  ( Fi )l  0i , where the straight line l must not be perpendicular to the line AB. 51

Theoretical mechanics

C)  momA Fi  0,

 momB Fi  0 ,  momC Fi  0 .

Points A,B,C must not lie on one straight line. 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 B) 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,  Zi  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,

Y  0 .

52

i

Module III. Dynamics of the mass point and the system

Lecture 13. 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. 2 As w  dv  d 2r , then equation (1) will take the form:

dt

dt

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 53

Theoretical mechanics

of free point in this coordinate system. Cartesian coordinate system or the natural trihedral axes are used most often. If the rectangular Cartesian coordinate system Oxyz is in the state of rest, then, projecting equation (2) on its axis, we will get:

y  Fy , 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); C) 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 position-vector 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  Fy (t; x, y, z; x, y , z) mz  Fz (t; x, y, z; x, y , z)

(4)

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 54

Module III. Dynamics of the mass point and the system

tangent (τ), 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:

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 (τ, 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 , Fy , 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. 55

Theoretical mechanics

Example. Let the mass point moves under the law:

x  a sin kt, y  b cos rt, z  0

(a)

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  Fy (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  56

(7)

Module III. Dynamics of the mass point and the system

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  z0 . 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 , y0 , 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.:

ck  f k (t , x0 , y0 , z 0 , x 0 , y 0 , z0 ), (k  1,6)

(9)

Replacing all c k 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: 57

Theoretical mechanics

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

y  y(t , x0 , y0 , 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.

Lecture 14. 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

58

Module III. Dynamics of the mass point and the system

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 )   mom0 Fk(i )  0 . k 1

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 a perfectly 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 : 59

Theoretical mechanics

rk  rk r 0 Let’s consider the position vector rc (the position vector of the center of mass in the new coordinate 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 coor60

Module III. Dynamics of the mass point and the system

dinates 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, 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 center n of mass of the system, it follows that Mrc   m r , as v  dr , we

 1

n

can write: Mvc   m v  Q ;  1

Thus, Q  Mvc , Q  Mvc 61

dt

Theoretical mechanics

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 Cxy z  with the origin in the center of mass is equal to zero, because vc  0 relative to these axes. The principle moment of linear momentum (angular momentum) of the system relative to the center O:

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 vc )   (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: Mv is any point with the mass mv, 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. 62

Module III. Dynamics of the mass point and the system

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: 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 Mv relative to the axis Cxy z  . Then the linear momentum will take the form:



G0  rc  Mv c  r  m v 

63



Theoretical mechanics

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. 

Hence: G0  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  , т.е.      c

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  G A  BA  Q

Thus, GB  GA  BA  Q 64

Module III. Dynamics of the mass point and the system

ANGULAR MOMENTUM OF THE ROTATING BODY RELATIVE TO THE AXIS OF ROTATION Assume that perfectly solid body rotates around axis x with angular velocity  . G X 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 G x   m v h , as v  h , 

then Gx   m h 2    m h2 . 

Value



 m h2  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: T

n 1 n 1 1 n 2 2   m ( v  v )  m v  m v  v  m v 2 .   c  2  c    c  2  1 2  1  1

65

Theoretical mechanics

Let's transform the right side: 1 n 1  m vc2  2 Mvc2 2  1 n

m v   1

n

c

 v  vc   m v  vc  m 

 1

,

dr d d  vc  m r  vc  Mrc  0 , dt dt dt

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 Perfectly rigid body is the body in which deformation is neglectted. 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 v c is the velocity of the center of mass and T

1 n 1 1  m v2  2  m vc2  2 Mvc2 . 2  1

Now we calculate the kinetic energy of a perfectly rigid body rotating about the axis Ox with angular velocity ω. It is known, that 66

Module III. Dynamics of the mass point and the system

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 Mvc2  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. Lecture 15. 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 ,

where F is the resultant of all forces, acting on the point. 67

(1)

Theoretical mechanics

As w  v , m  const is the weight, equation (1) can be rewritten: d (mv ) (2) F dt As it is known, a vector mv  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 )  Fdt

(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 v 0 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: 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.

68

Module III. Dynamics of the mass point and the system

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   mk v k is called the linear momentum of mechanical k 1

system. Summing the equations of motion of mass points of the mechanical system:  mwk   Fk(e)   Fk(i) , It is known that the sum of internal forces for mechanical system is equal to zero:  Fk(i)  0 Let’s transform the left side:

 mw   m k

k

dVk d dQ ,  ( mk Vk )  dt dt dt

69

Theoretical mechanics

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: dQ  0  Q  const  Q0 dt

Or, as Q  Mvc , 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: Qx  c1 , Q y  c2 , Qz  c3 , or x c  c1 , y c  c2 , zc  c3 70

Module III. Dynamics of the mass point and the system

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 difn

ficult to count Q   mk v k . Let’s find a different formula from the k 1

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

dr dr mk k  M c or MvC   dt dt k 1

n

m v k 1

k

k

,

where vC is the velocity of center of mass of mechanical 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: n dv d (mvC )   Fk( e ) or M C   Fk( e )  MwC   Fk( e ) dt dt k 1

where wC is the acceleration of center of mass of the system. 71

Theoretical mechanics

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. Lecture 16. 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 , dt

where 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

As d dr dv (r  mv )   mv  r  m , dt dt dt 72

(1)

Module III. Dynamics of the mass point and the system

dr  mv  v  mv  0 dt

Then we will get: d (r  mv )  r  F dt

(2)

vector K O  momO (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 y dK x dK z  momx F ,  mom y F ,  momz 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. 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, momO F  0  73

Theoretical mechanics

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 that dK z momz F  0 , тогда  0  K z  const . That is, if the moment of dt 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 dt 2

(  1, n )

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   









(1)

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: 74

Module III. Dynamics of the mass point and the system

r  F   0   i

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

Rewrite (1) in the form:



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



(2)

Denote the angular momentum relative to the center O as dr   G O    r  m   , dt    e  r  F  M 0e is the principle momentum of all external 





forces acting on the mechanical system. Then we can write (2) in the form: e dG0 (2’)  MO. dt 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. Obtain a theorem in integral form, if we integrate (2 ') from t1 up to t2: t2

G0  G02  G01   M 0e dt t1

75

(3)

Theoretical mechanics 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 G0 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:

Gox  c,1 Goy  c2 , Goz  c3 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: Gx  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: 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 , 76

(4)

Module III. Dynamics of the mass point and the system

Denote GO  G ,

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

dQ  OO   Fe . dt 

dG by formula and substitute it in the expression for dt

the theorem: dG   r  F e , dt

then we will get:





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





Or taking into account that r  OO  r , finally we will have: dG   v   Q   r  Fe , dt 





(5)

This equation expresses the theorem of change of angular momentum of the system relative to the center O  , moving relative to the main coordinate system with the velocity v  . 77

Theoretical mechanics

Lecture 17. 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 will dt dS dt dS  v

take the form:

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

where dS is the increment of arc constant (the algebraic value).    mv2  , i.e. the elemv dv  F dS or d   2   d ( A)  

mentary work is equal to F dS .



But F is the projection of F on the direction

of  , i.е. F  F  CosF , 

Then, the elementary work force d A on the displacement dS 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 78

Module III. Dynamics of the mass point and the system

  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.:   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 

then Fx 

U U U , dx  dy  dz x y z

U U U , Fy  , Fz  x y z

79

(1)

Theoretical mechanics

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 : grad (U ) 

U U U i j k  Fx i  Fy j  Fz k  U x y z    i j k x y z F  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 2 2 , as Fx   U   U  Fy  Fy  Fx  0 .  Fz Fy  0 y xy yx x x y  z  y  Fx Fz  0  x  z

Example: Gravity force field  P  Px , Py , Pz   0,0,mg

Let’s check up the conditions of potentiality: 80

Module III. Dynamics of the mass point and the system

 Px Py     0  0  0  y x y x   Py Pz        0    mg   0  0  0  z y z y   Pz Px        mg    0  0  0  0  x z x z   

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

  F Fy  rotF x  z   0 k y z   ,   Fx Fz rotF y    0 z z x   Fz Fy Fx  rotF z    0 x y 













conditions of potentiality can be written as:  rotF  0 



As 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  Fy 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 0 0

M 0M

M 0M

is equal 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 81

Theoretical mechanics

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  . 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 , 82

Module III. Dynamics of the mass point and the system

where h  z  z0 is the difference of heights M 0 and M .

B) work of elastic force l o is the length of nonstretched spring. Fx  cx, Fупр  {cx,0,0} A

M1



M0

x1

(cx )dx  c  xdx  x0

c 2 ( x0  x12 ) 2

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

C) work of friction force

Fтр  fN A

( M1 )



Fds  

( M1 )

 fNds

(M0 )

(M0 )

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). 83

Theoretical mechanics

Lecture 18. 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 Note, that w  dv , dr  v dt , where v is the velocity of considedt

red point relative to the fixed axis. Rewrite the left side as: mw  dr  m

dv  v dt  mv dv dt

As v 2  v 2 и v  d v  1 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    F dr d   2 

84

(1)

Module III. Dynamics of the mass point and the system

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 of

2

a mass point. The expression Fdr  d ' A on the right-hand side is called the elementary work of the force applied to the mass point. Equation (1) expresses the theorem of change of kinetic energy of 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   F dt 

or

d  mv 2     F v dt  2 

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: (M )

1 mv12 mvO2    F dr 2 2 (MO )

(2)

This equation expresses the theorem of change of kinetic energy in the integral form: change of kinetic energy of a mass point on a 85

Theoretical mechanics

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 vk2   Fk( e ) drk   Fk(i ) drk 2 k 1 k 1 k 1 n

d

mk v k2 is the kinetic energy of mechanical system. 2 Then the equation can be rewritten in the form: dT  d A(e)  d A(i ) 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: T 

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 86

Module III. Dynamics of the mass point and the system

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 Fdr  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

This is the law of conservation of mechanical energy: sum of kinetic and potential energy is the constant value.

Lecture 19. 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 87

Theoretical mechanics

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. 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  OM1 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 88

Module III. Dynamics of the mass point and the system

state of rest than only the restoring force F is acting on it along the x axis: F  c  OM  с x , 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. (1) Fx  cx . Let’s deduce the differential equation of motion of point M under action of force F : c mx  cx  x  x  0 m Denote 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 : (4) x  kC1 sin kt  kC2 cos kt 89

Theoretical mechanics

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 (5) x  x0 cos kt  0 sin kt k 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 β 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  : 90

Module III. Dynamics of the mass point and the system

 x  a  x02   0   k  kx tg   0 x 0

2

(8)

As each value of the tangent of angle corresponds to two angles [0, 2 ] then it is necessary to define sin  or cos  : sin  

сos 

x0  a

x 0  ak

x0 x02 



x k

2 0 2

x 0 k x02 

x k



2 0 2

kx0 k x  x 2 0 2 2 0

(9б)

x 0 k x  x 2

2 0

(9в) 2 0

Angular frequency and period of free oscillations is determined with the help of formulas: c (10) k m

2 m T  2 k c

(11)

Angular frequency is also called the frequency of free oscillations or natural frequency. As can be seen from (10), (11), the natural frequency and the period of free oscillations of a point depends only on the mass of this point and the coefficient c characterizing the restoring force, and do not depend on the initial conditions of motion. 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 91

Theoretical mechanics

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 



x 

c x  0. m

m  c Assuming that:  2n,  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: 92

Module III. Dynamics of the mass point and the system

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  ik1 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   )

(13)

This equation describes the oscillation as sin(k1t   ) is periodical function. Multiplier e  nt shows that amplitude of oscillations decreases with time. Such oscillations are called damped. The values a and β are defined from initial conditions. Assume that the mass point has initial position x 0 and initial velocity x 0 at the moment t  0 . Substituting the initial conditions in equation (13) and equation for velocity: 93

Theoretical mechanics





 x  aent sin(k1t   ) t  naent sin(k1t   )  ak1e nt cos( k1t   ) , i.е.

x  nx  ak1e  nt cos(k1t   )

We will get: x0  a sin  , x0  nx0  ak1 cos  or x 0  nx0  a cos  k1

From here we will have: a  x02  tg 

( x 0  nx0 ) 2 k1 x0 k1 x 0  nx0

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

94

 C2 e 

n 2 k 2 t

)

(14)

Module III. Dynamics of the mass point and the system

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: xe

 nt

( B1

e

n2 k 2 t

 e 2

n2 k 2 t

 B2

e

n2 k 2 t

 e 2

n 2  k 2t

)

As e x  ex  chx, 2

e x  ex  shx, 2

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

can be reduced to another form assuming that B1  ash , B2  ach , then (15) x  ae  nt sh( n 2  k 2 t   ) This equation shows that considered motion of mass point is not oscillating, as shx is not a periodic function. Mass point can perform one of the motions depending on the initial conditions:

All of these graphs correspond to the initial deflection of a mass point on the value x0  0 from the position of rest. 95

Theoretical mechanics

C. The case of n  k . The roots of characteristic equation are real, negative and equal: 1  2  n

General solution of equation (12) has the form: x  e  nt (C1t  C2 )

(15а)

The motion of a mass point is also aperiodic. Constants C1 and C 2 are defined with help of initial conditions.

Lecture 20. 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  c x  H sin( pt   ) m

m

Denote c  k 2 , H  h , then m

m

x  k 2 x  h sin( pt   ) 96

(16)

Module III. Dynamics of the mass point and the system

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 .

General solution of this equation is:

x   C1 cos kt  C2 sin kt . Particular solution can be obtained in dependence of the type of inhomogeneity: (17) x   A sin( pt   ) . Let’s define the constant A by substituting (17) in (16). As x    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: x  

h sin( pt   ) k  p2 2

97

Theoretical mechanics

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 2  p2

This equation shows that the point M performs a complex oscillatory motion that is composed of two harmonic oscillations. 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 2  p2

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. 98

Module III. Dynamics of the mass point and the system

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:

 Bk sin(kt   )  Bk sin(kt   )  Bk 2t cos(kt   )   k 2 Bt cos(kt   )  h sin(kt   ) or  2 Bk sin(kt   )  h sin(kt   )  B  

h 2k

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

or x  C1 cos kt  C 2 sin kt 

h t cos(kt   ) 2k

h  t cos(kt    ) 2k 2

This equation shows that amplitude increases with time indefinitely, i.e. resonance takes place (see the figure).

INFLUENCE OF RESISTANCE ON FORCED OSCILLATIONS Acting forces: restoring force F , perturbing force Q , resistance force R : Fx  cx, Qx  H sin( pt   ), Rx  x . 99

Theoretical mechanics

Let’s deduce the differential equation of motion of a mass point M:

mx  cx  x  H sin( pt   )

or x 

 m

x 

c H x  sin( pt   ) m m

Or introducing the notations   2n, c  k 2 , H  h : m

m

m

x  2nx  k 2 x  h sin( pt   )

(18)

This is the differential equation of forced oscillations with the resistance to motion. 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 p 2 sin( pt     )  2nAC p cos( pt     )   AC k 2 sin( pt     )  h sin( pt   )

Express: h sin( pt   )  h sin( pt       )   h sin( pt     )cos   h cos( pt     )sin  100

Module III. Dynamics of the mass point and the system

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, h

AC 

( k  p )  4n 2 p 2 2

2 2

tg 

2np k  p2 2

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

h ( k  p )  4n 2 p 2 2

2

sin( pt     )

Then the general solution will have the form: 1) when n  k h x  ae nt sin( k 2  n 2 t   )  sin( pt     ) ( k 2  p 2 )  4n 2 p 2 2) when n  k h

x  ae nt sh( n 2  k 2 t   ) 

( k  p )  4n 2 p 2 2

2

sin( pt     )

3) when n  k x  e nt (C1t  C2 ) 

h ( k  p )  4n 2 p 2 2

2

101

sin( pt     )

Theoretical mechanics

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.

Module IV. Dynamics of the solid body

Lecture 21. MASS GEOMETRY. INERTIA MOMENT. INERTIA MOMENTS OF THE BODY RELATIVE TO THE AXIS AND POINT. CENTRIFUGAL MOMENT OF INERTIA GENERAL FORMULAS FOR INERTIA MOMENTS Position of center of mass characterizes the distribution of mass of the system incompletely. Therefore one more characteristic of the mass distribution the moment of inertia of the system is introduced. Inertia moment is the sum of products of masses of all the points of the system on a homogeneous function of coordinates of these points:

 m x y  z  , i i

i

i

i

n       is the degree of the moment. Inertia moments of first

and second order are used in mechanics. First order inertia moments are called the static moments.  mi ri   mi xi i +  mi yi j +  mi zi k is the static moment of the system relative the center O,

m x , m y ,  m z i i

i

i

i i

are the static

moments relative to the coordinate planes. It is known that  mi ri  MrC . If the center O is coincide with the center of mass then the static moment vanishes (as the static moments relative to the planes passing through the center of mass). 103

Theoretical mechanics

 m x  Mx , С is the xC  0   m x  0 . i

i

C

i

origin of coordinate system, therefore,

i

Second order inertia moments. Scalar quantity equal to the sum of products of masses of all the points on the square of their distances from a given point O, axis L or plane is called the inertia moment of mechanical system of mass points with respect to a given point O, axis L or a plane. Therefore we denote inertia moments relative to the point O by J 0 , relative to the axis L by J L and relative to the plane P by J  : n

J 0   mk rk2 , k 1

n

J L   mk hk2 , k 1

n

J    mk d k2 k 1

where mk is the mass of k -th mass point of the system, rk , hk , d k are its distance to the given point O, axis L and plane P. Let's consider the system of mass points and Cartesian coordinate system. Denoting the distance of k-th point to the axis z by hk , the inertia moment about z axis will have the form: n

J z   mk hk2 , but h 2 k  xk2  y k2 , k 1

from here n

J z   mk ( xk2  y k2 ) . k 1

Inertia moments relative to the other axes are calculated similarly. Thus, inertia moments of the system with respect to the coordinate axes are: n

n

n

k 1

k 1

k 1

J xx   mk ( y k2  z k2 ) , J yy   mk ( xk2  z k2 ) , J zz   mk ( xk2  yk2 ) , (1) Inertia moments relative to the coordinate planes yOz, zOx, xOy are: 104

Module IV. Dynamics of the solid body n

J ( yz )   mk xk2 , k 1

n

n

k 1

k 1

J ( zx)   mk yk2 , J ( xy )   mk z k2

(2)

Centrifugal moment of inertia are: n

J yz   mk yk z k , k 1

n

J zx   mk zk xk , k 1

n

J xy   mk xk yk

(3)

k 1

Polar moment of inertia or inertia moment relative to the origin of coordinate system O is: n

2

n

k 1

k

k 1

J 0   mk r   mk ( xk2  y k2  z k2 )

(4)

Inertia moments of the system for solid body will take the form:

J xx   ( y 2  z 2 )dm,

J yy   ( z 2  x 2 )dm,

V

J zz   ( x 2  y 2 )dm

V

J ( yz )   x 2 dm,

J ( zx)   y 2 dm,

J ( xy )   z 2 dm

V

V

V

J yz   yzdm,

J zx   zxdm,

V

( 1 )

V

J xy   xydm

V

J 0   ( x 2  y 2  z 2 )dm,

(2’) (3’)

V

(4’)

V

Typically, inertia moments for continuous solid bodies are calculated with the help of integral calculus. It is easy if these bodies are homogeneous and have a regular geometric shape. If the bodies are complex and non-uniform shape, it is easier to calculate the inertia moments experimentally. Examples. Let's calculate the inertia moments of certain homogeneous symmetric bodies relative to the axes passing through the center of gravity of bodies and are the axes of symmetry. Axis passing through the center of gravity is called the central axis. 105

Theoretical mechanics

1) Inertia moment of homogeneous thin rod. Let's define the inertia moment relative to the axis C y that passes through the center of gravity and is perpendicular to the rod. l is the length of the rod, F is the area of section,  is the density. Mass of the rod is m  V  Fl . Let's split the rod into small elements of length xi . Then mass of the element is: mi  Fxi

J cy   mi xi2   Fx i2 xi

Proceeding to the limit of sum we will have:

J cy 

l 2

 

l 2

l 2

 Fx dx  F   2



x3 x dx  F  3 2

l 2



l 2 

 l 2

3 3 F   l   l           3  2   2  

F  2l 3 F  l 3  3 8 12

As Fl  m , then finally we will have J cy 

ml 2 . 12

2) Inertia moment of a thin homogeneous round plate (disc). Let's find the inertia moments relative to the axes C x , C y , lying on the plane of the disc and relative to the axis C z perpendicular to the plane of the disc. Assume that R is the radius of the disc, h is the thickness of the disc,  is the density. As it is known the mass is m  V  hR 2 . Let's split the disc into many elementary rings with radius ri and width ri . Mass of the ring is mi  h2ri ri . As the thickness is small, then we can assume that z i  0 for all points of the plate. Then: 106

Module IV. Dynamics of the solid body n

n

J cx   mi ( yi2  z i2 )   mi yi2 , J cy   mi ( xi2  z i2 )   mi xi2 , i 1

i 1

n

J cz   mi ( xi2  yi2 ) , i 1

 J cz  J cx  J cy , т.к.J cx  J cy ,  J cz  2 J cx n

Calculate J cz   mi ( xi2  yi2 )   mi ri 2 . i 1

As the distances ri to C z are equal for all points of the ring, then

mi can be the mass of all the ring. Then: J cz   2hri3 ri or R

R

0

0

J cz   2 hr 3 dr  2 h  r 3 dr 2 h

r4 R4  h 4 2

As hR 2  m , then we finally have: 2

J cz

mR . mR 2 and J cx  J cy   4 2

Lecture 22. THEOREM OF GUIGENS-SHTEINER. INERTIA MOMENTS RELATIVE TO THE AXES OF THE BEAM, COMING FROM THIS POINT Huygens-Steiner theorem. Inertia moment of the body relative to some axis is equal to the inertia moment of a body relative to the axis parallel to the given axis and passing through the center of mass of the body, summed with a product of mass of the body by the square of the distance between the axes. 107

Theoretical mechanics

J z  J z  Md 2 Proof: Consider the inertia moment relative to z axis: J z   mi hi2   mi ( xi2  yi2 )

(*)

Use the axis z  that is parallel to axis z and passing through the center of mass of the system С( xC , yC , z C ). Draw the lines

x || x, y || y  . Then the coordinates of the point М will be:

xi  xi  xC yi  y i  yC Substituting in (*): 2 2 J z   mi  xi  xC    yi  yC       i

  m  xi2  yi2    mi  xC2  yC2   2 mi ( xixC  yi yC ). i

The last term is equal to zero as there is the static moments relative to the planes passing through the center of mass. (You need to note that:  mi xi'  MxC' =0). i

Finally: J z  J z  Md 2 .

INERTIA MOMENT RELATIVE TO THE AXIS OF THE BEAM COMING FROM THE GIVEN POINT Let's find the inertia moment of the system relative to the axis l, which passes through the given point O and whose direction with 108

Module IV. Dynamics of the solid body

respect to the axis Oxyz is defined by the direction cosines α, β, γ. According to the definition of the axial moment of inertia we have: n

J l   mi hi2 . i 1

Assume that the origin of coordinate system located at the point О and consider OM i  r i  xi i  yi j  zi k . Then: 0

r i l  xi  yi  zi  ri cos i , wherе  i is the angle between the directions of vectors r i and l . 0

As hi  ri sin i , then n

n

i 1

i 1

J l   mi hi2  

mi ri 2 sin 2 i  n

  mi  ri 2  (ri cos i ) 2 

.

i 1

Taking into account that  2   2   2  1, we find: 2 J l   mi  xi2  yi2  zi2  2   2   2    xi   yi   zi      i 1 n

  mi  2 ( yi2  zi2 )   2 ( zi2  xi2 )   2 ( xi2  yi2 )  2 yi zi  2 zi xi  2 xi yi 

Taking into account the formulas (1) and (3) of last lecture we finally get: J l  J xx 2  J yy  2  J zz 2  2 J yz   2 J zx  2 J xy

109

Theoretical mechanics

Lecture 23. ROTATION OF SOLID BODY AROUND FIXED AXIS. DIFFERENTIAL EQUATIONS OF MOTION. AXLE PRESSURE. MOTION OF PERFECTLY RIGID BODY WITH ONE FIXED POINT. EULER KINEMATIC AND DYNAMIC EQUATIONS ROTATION OF A RIGID BODY ABOUT A FIXED AXIS The equation of motion. Assume that the rigid body rotates about a fixed axis under the action of active forces    F1 , F2,......,Fn . Let’s apply the theorem of change of angular momentum about a fixed axis (axis z) in order to derive the equations of motion: dGz   momz F i. dt

(1)

But we know that the following formula can be used to characterize the rotation about the fixed axis z: Gz  J zz  J zz

d dt

substituting this value in equation (1), we obtain the equation of motion of a rigid body rotating about a fixed axis in the form:

J zz

d 2   momz F i . dt 2

(2)

Integrating this equation, we find the angle φ as a function of time and determine the motion completely, because in this case the system has one degree of freedom and its position is characterized by 110

Module IV. Dynamics of the solid body

angle φ. If there are no external forces or their line of action crosses the axis of rotation z, then  momz F i  0, and we have: d    const , dt

d 2  0, dt 2

  0  t. i.e. it is the rotation with constant angular velocity. Pressure on axis. Rotation of a rigid body about a fixed axis can be considered as a motion of a rigid body when two its points A and B are fixed. Thus in accordance with the axiom of coupling we can say that the restraint which is carried out by two fixed points A and B is equivalent to two reactions, applied to these points. Let’s denote





these reactions as A and B . Denote the active external forces as    F1 , F2,......,Fn . Assume that the origin of coordinate system is located at point A and Az is the axis of rotation. Applying the theorem of change of momentum, we have: d  mi  i  A  B   F i . dt

(3)

Projecting both parts of equation (3) on the axes of coordinate system we will get the three equations: d  mi xi  Ax  Bx   Fix , dt d  mi yi  Ay  By   Fiy , dt

(4)

d  mi zi  Az  Bz   Fiz . dt 111

Theoretical mechanics

Applying the theorem of change of angular momentum relative to the center A we have:









d  r i  mi i  AB  B   r i  F i . dt

(5)

Projecting the both parts of equation (5) on the coordinate axes we will get three equations: d  mi ( yi zi  zi yi )   AB  By   ( yi Fiz  zi Fiy ), dt d  mi zi xi  xi zi   AB  Bx   zi Fix  xi Fiz , dt d  mi x, yi  yi xi   xi Fiy  yi Fix . dt



(6)



Expressing all the coordinate derivatives in (4) and (6) through  of the body and the angular velocity  and angular acceleration  taking into account, that: 2 x  r cos  x  r sin  *    y , x   x  y , y   y 2  x , y  r sin  y  r cos  *   x , z  0; z  const z  0,

We will get the following formulas by substituting the values x, x, y , y, z, z in the equations (4) and (6):   2  mi xi    mi yi  Ax  Bx   Fix ,

  2  mi yi   mi xi  Ay  By   Fiy , 0  Az  Bza   Fiz , 112

(7)

Module IV. Dynamics of the solid body

   mi zi xi   2  mi zi yi   AB  By   yi Fiz  zi Fiy ,

   mi zi yi   2  mi zi xi  AB  Bx   zi Fix  xi Fiz ,   mi xi2  yi2    xi Fiy  yi Fix .

Taking into account that:

m x i

i

 Mxc ,  mi yi  Myc ,  mi yi zi  J yz ,

 mzx i i

we will get:

j

 J zx ,  mi xi2  y2i   J zz,

 Mxc 2  Myc  Ax  Bx   Fxi ,

 Myc 2  Mxc  Ay  By   Fyi , 0  Az  Bz   Fzi ,

J yz  J zx   AB  By   momx Fi ,

(8)

2

ииииии  J zx 2  J yz  AB  Bx   momy Fi ,

J zz 

 mom F . z

i

The last equation of the system (7) or (8) does not contain reactions and, therefore, gives the equation of rotation of the rigid body about a fixed axis (2). Integrating this equation, we will find the angular velocity ω, and then the angle φ as a function of time t. Remai ning equations contain the projections of unknown reactions A and  B . As the quantity of unknown reactions is equal to six the system is  uncertain. And we can determine only the sum A + B from the third equation of the first group. CONDITIONS UNDER WHICH THE DYNAMIC REACTIONS ARE EQUAL TO STATIC REACTIONS

 = 0 in the left sides of first five equations (8). Assume that  =  Keeping the values of forces in the right sides of (8) we will get the 113

Theoretical mechanics

ordinary equilibrium equations of statics which will serve for definition of reactions that we will name as static reactions. In general these reactions can be variable. In contrast to static reactions we will call the reactions that take place when the solid body rotates as dynamic reactions. It appears a question about the conditions when static reactions are equal to dynamic reactions, i.e. the conditions when rotation doesn’t cause the additional pressure on the axis of rotation. Generally dynamic reactions may differ from static reactions. This is evident from the fact that if in the case of equilibrium all active external forces were equal to zero, then the reaction would also be equal to zero. But dynamic reactions will not be equal to zero in the case of absence of external forces as the left parts of equations (8) are not equal to zero. Hence it is evident, that in order to get the pressure on axis caused by rotation of a rigid body, it is necessary to make the external active forces equal to zero in these equations. Let’s find the conditions under which the rotation of a rigid body does not cause the additional reactions, i.e. conditions when dynamic reactions are equal to static reactions. For this purpose it is necessary and sufficient that the left sides of equations (8), except the last, were equal to zero:

xc 2  yc  0,

J yz 2  J zx  0,

xc  yc 2  0

J yz  J zx 2  0.

(9)

and

As the determinants are not equal to zero during the rotation, equations (9) can take place only when:

xc  yc  0,

J yz  J zx  0,

Combining the two results, we can say that there are no additional pressures when the body rotates, i.e. dynamic pressures equal to 114

Module IV. Dynamics of the solid body

static pressures only when the axis of rotation is one of the principal central axes of inertia. Example. Given: Isosceles right triangle OO1A rotates around the vertical axis. OO1=a (Fig. 2). What should be the angular velocity of rotation so that the lateral pressure on the lower support was equal to zero? Consider the triangle as thin, uniform plate. Solution: Let’s use equations (8) to solve the problem. In this case:

xC  0, J yz 

 yzdm 

yC  a / 3, h  a, J xz  0, a z a  2m  m 3 1   z ydy dz  z dz  ma2 . 2   2   a 0 0 a 0 4 

Further

Rx  Ry  0, Rz  mg, M x   1 mga, M y  M z  0. 3

Taking into account that Fx  Fy  0, (as it was given) we will get the equations (8) in the form: 1 1  ma  F1x ,  ma 2  F1 y , 0  mg  Fz  F1z , 3 3 1 2 1 2 2 1   0. ma    mga  aF1 y ,  ma   aF1x , J z 4 4 3     const

Answer:   2 g / a.

115

Theoretical mechanics

Lecture 24. MOTION OF PERFECTLY RIGID BODY WITH ONE FIXED POINT. EULER KINEMATIC AND DYNAMIC EQUATIONS Assume that the rigid body is moving about a fixed point O. Consider the main coordinate system Оξηζ (fixed) and moving coordinate system Oxyz that is invariably associated with the body and moving with it relative to the fixed coordinate system Oξηζ. Theorem of change of the angular momentum relative to the main (fixed) system Oξηζ gives the equation: dG M0 dt

(1)

where M 0 is the principle moment of all external forces acting on a rigid body relative to a fixed point O. If we will project equation (1) on the fixed axes Oξηζ (where G=ω(J)) the equations of motion in the projections will be complicated. Euler made two simplifications in the process of deriving the dynamic equations of motion of a rigid body. First simplification is: projection both sides of equation (1) on the axis of the moving system. We know that d a is the derivative of some vector a relative to dt

~

the main coordinate system, d a is the local derivative (variation of dt a relative to the fixed coordinate system), then: ~ da d a  a dt dt

116

Module IV. Dynamics of the solid body

Applying this formula to the vector G we will get: ~ dG d G   G , dt dt

and equation (1) has the form: ~ dG  G  M 0 . dt

(2)

When projecting both sides of equation (2) on the axis of moving coordinate system Оxyz we don’t take into account the sign of local derivative as the local derivative is considered in accordance to the fixed coordinate system, then we will get: dG x   qG x  rG y  M x  dt  dG y   rGx  pG z  M y  dt  dG z   pG y  qG x  M z  dt 

(3)

where Gx, Gy и Gz are defined with the help of formulas: G x  Ap  Fq  Er ,   G y   Fp  Bq  Dr   G z   Ep  Dr  Cr 

(4)

Here the components of tensor of inertia A, B, C, D, E, F are constants. The second simplification of Euler is: principal axes of inertia of the body about a fixed point is considered as the axes of moving coordinate system. Due to this fact the formulas (4) will simplify and can be written as: 117

Theoretical mechanics

G x  Ap ,  G y  Bq ,  G z  Cr , 

(5)

As result the equations (3) will take the form: dp   (C  B)qr  M x ,  dt  dq  B  ( A  C )rp  M y ,  dt  dr  C  ( B  A) pq  M z , dt  A

(6)

System (6) is the differential equation of motion of a rigid body about a fixed point that were derived by Euler (1765). These equations are called dynamic Euler equations. Adding kinematic equations to the equation (6) we will get the system of six differential nonlinear equations of first order relative to six unknown functions of time p, q, r, φ, ψ, θ. p   sin  sin    cos  ,  q   sin  cos    sin  ,   r   sin    , 

(7)

General integrals must contain six arbitrary constants, which can be determined if you set the initial position and initial angular velocity of the body, i.e. p0, q0, r0, φ0, ψ0, θ0. Excluding p, q, and r from equations (6) and (7) we can obtaine three differential equations of second order with respect to the three Euler angles φ, ψ, θ. If we take into account that the projections of principle moment of all external forces Mx, My, Mz are the functions of p, q, r, φ, ψ, θ, t, it becomes clear that it is very difficult to integrate the system of six 118

Module IV. Dynamics of the solid body

equations (6) and (7). Even a special case, namely the problem of a rigid body about a fixed point under the action of gravity force can not be solved in general form.

Lecture 25. GENERAL FORMULATION OF THE PROBLEM OF HEAVY SOLID BODY WITH ONE FIXED POINT. FIRST INTEGRALS OF EQUATIONS OF MOTION OF THE HEAVY SOLID BODY AROUND THE FIXED POINT GENERAL FORMULATION OF THE PROBLEM OF HEAVY SOLID BODY WITH ONE FIXED POINT Consider the motion of a body about a fixed point under the action of gravity force P. Axis Oζ of the main coordinate system Oξηζ connected with the Earth is directed vertically straight up. Oxyz is the moving coordinate system; r C ( xc , yc , zc ) is the position vector of the center of gravity C relative to the origin O; γ1, γ2, γ3 are the cosines of angles that made by vertical fixed axis Oζ with moving axes. Since the angular velocity ψ `is directed along the axis Oξ, then the direction cosines Oξ will be equal to multipliers by ψ` in kinematic Euler equations (7). Consequently, cos( , x)   1  sin  sin  ,   cos( , y )   2  sin  cos ,  cos( , z )   3  cos . 

(1)

It is evident that γ1, γ2, γ3 is the projections of unit vector ζ0 on the moving axes. Therefore: 119

Theoretical mechanics

ζ0 = γ1i+ γ2j+ γ3k, where i, j, k are the unit vectors of moving axes connected with the body. As the direction of gravity force or the direction of axis Оζ, is fixed relative to the Earth then: 0

d 0 dt

or using the symbol of local derivative: 0 ~ 0 0 d d       0. dt dt

Therefore, i ~ 0 0 d       p dt

j q

k r.

(2)

1  2  3

Projecting both sides of this equation on the moving axes we will get the equations derived by Poisson: d 1   r 2  q 3,  dt  d 2   r 3  q 1,  dt  d 3   r  1  q 2 ,  dt 

(3)

Let's write the dynamic Euler equations. Gravity force P is directed opposite to the axis ζ; therefore, P=-Pζ0 , 120

Module IV. Dynamics of the solid body

Principle moment of all external forces is equal to i 0

M 0  r c  P   Pr c    P  1 xc

j

k

2 3 . yc

zc

Thus, Euler equations for considered case will take the form: dp   (C  B)qr  P( 2 z c   3 yc ),  dt  dq  B  ( A  C )rp  P( 3 xc   1 z c ),  dt  dr  C  ( B  A) pq  P( 1 yc   2 xc ), dt  A

(4)

The system of equations (3) and (4) is a system of six nonlinear ordinary differential equations of first order with six unknown functions of time γ1, γ2, γ3, p, q, r. The quantities A , B, C , xc , yc , zc are constants . If γ1, γ2, γ3, p, q, r are obtained by integrating (3) and (4) as the functions of time then for obtaining the complete solution of the problem, i.e. for determining the Euler angles as a functions of time it is necessary only to find ψ(t) with the help of one quadrature from the kinematic equations. Thus, the main part of solution of the problem of motion of a heavy rigid body about a fixed point is to find the six integrals of the system (3) and (4). But it is easy to prove that the problem is reduced to finding only one integral. Indeed, let's represent the equations of the system in the canonical form: dp dq dr d 1 d 2 d 3       dt П Q R 1 2 3

where 121

(5)

Theoretical mechanics

1 P( 2 zc   3 yc )  (C  B)qr , 1  r 2  q 3 ;  A  1  Q  P( 3 xc   1 z c )  ( A  C )rp , 2  p 3  r 1 ;  B  1  R  P( 1 yc   2 xc )  ( B  A) pq , 3  q 1  p 2 ; C  П

(6)

Since the time t does not appear explicitly in any of the functions (6) we can integrate separately first five equations of the system instead of six equations (5): dp dq dr d 1 d 2 d 3      . П Q R 1 2 3

(5`)

If the system (`5) will be able to integrate, the time t can be found with the help of simple quadrature. The system (5 ') has already only five integrals. According to the theory of Jacobi's last multiplier for the canonical system of differential equations the factor M for the system (5 ') has a value of M = 1, since: П Q R 1 2 3      0 p q r  1  2  3

But if the last multiplier of the system is known, then it is enough to know only four integrals of the system (5 ') for reducing the problem to quadratures. One of these four integrals is obtained directly from simple geometric consideration. Indeed, γ1, γ2, γ3 are the direction cosines of the axis Oζ, or what is the same, the projection of the unit vector of this axis. Consequently, γ12+ γ22+ γ32=1 (7) This trivial integral can be obtained from equation (3). Multiplying them respectively by γ1, γ2, γ3 and adding term by term, we obtain 122

Module IV. Dynamics of the solid body

1

d d 1 d   2 2   3 3  0, dt dt dt

or

d 2 ( 1   22   32 )  0, dt that will give us the integral (8). Thus, it remains to find the three integrals. We obtain the first of them by applying the theorem of change of angular momentum relative to the axis Oζ. Since the moment of gravity force P relative to axis Oζ is equal to zero then: d G  M   0, and therefore Gζ =const. dt

But 0

G  G    ( Api  Bq j  Cr k )  ( 1 i   2 j   3 k )  Ap 1  Bq 2  Cr 3

consequently,

Ap 1  Bq 2  Cr 3  const

(8)

And this integral (8) can be obtained directly from the equations of motion. Indeed, multiplying equation (4), respectively by γ1, γ2, γ3 and adding them term by term, we obtain A 1

dp dq dr  B 2  C 3  ( B  C ) 1qr  (C  A) 2 rp  ( A  B) 3 pq. dt dt dt

This equation can be transformed to the form: d ( Ap 1  Bq 2  Cr 3 )  ( B  C ) 1qr  (C  A) 2 rp  ( A  B) 3 pq  dt d d d  Ap 1  Bq 2  Cr 3 dt dt dt 123

Theoretical mechanics

If we replace the derivatives of γ1, γ2, γ3 in the right side by their expressions from equations (3), it is easy to verify that the right-hand side is equal to zero, and we get: d ( Ap 1  Bq 2  Cr 3 )  0 dt

Integrating we will find the integral (8). Let's apply the theorem of change of kinetic energy for obtaining the second integral of three remaining integrals dT   Pd c ,

where ζс denotes the coordinate of center of gravity of the body relative to the main coordinate system. Integrating we will get:

T   P c  const. Taking into account that 0

 c  r c    ( xc i  yc j  zc k )  ( 1 i   2 j   3 k )   xc 1  yc 2  zc 3 , we will get the energy integral in the form: 1 ( Ap 2  Bq 2  Cr 2 )   P( xc 1  yc 2  zc 3 )  const . 2

(9)

Integral (9) is also found directly from the equations (4), if each of them multiply respectively by pdt, qdt, rdt and add the obtained expressions term by term. Then we have: 1 d ( Ap 2  Bq 2  Cr 2 )   Pxc (r 2  q 3 )  yc ( p 3  r 1 )  zc (q 1  p 2 )dt 2

124

Module IV. Dynamics of the solid body

or on the base of Poisson equations (3): 1 d ( Ap 2  Bq 2  Cr 2 )   P( xc d 1  yc d 2  zc d 3 ). 2

Hence, integrating and taking into account that xc, yc, zc are the constant values we will get the equation (9). It remains to find just one more integral. And this is the main difficulty of the problem. This third integral was obtained only for three particular assumptions about the moving body and conditions of motion. These three particular cases of motion are as follows: the case of Euler-Poinsot, the case of Lagrange-Poisson, the case of Kovalevskaya.

Lecture 26. SPECIAL CASES OF INTEGRATION AND ITS GEOMETRICAL INTERPRETATION: THE CASE OF EULER-PUANSO, THE CASE OF LAGRANGE-PUASSON, THE CASE OF KOVALEVSKAYA 1) The case of Euler-Poinsot. This is the case of moving by inertia when the resultant of the external forces passes through a fixed point O. Fixed point O coincides with the center of gravity C in the particular case of a heavy rigid body researched by Euler and Poinsot. There are no other external forces except the gravity force, but the latter is balanced by reaction of the support. Solid body can be have any shape. 2) The case of Lagrange - Poisson. In this case, the ellipsoid of inertia, built for the fixed point O is an ellipsoid of revolution, i.e., A = B ≠ C, and the center of gravity C lies on the axis of dynamic 125

Theoretical mechanics

symmetry (the rotation axis of the ellipsoid of inertia). Such a body rapidly rotating about the axis of symmetry is called symmetric gyroscope.

3) The case of Kovalevskaya. In this case, the ellipsoid of inertia for a fixed point is an elongated ellipsoid of revolution, and there is the relationship between the principal moments of inertia: A=B=2C, center of gravity lies in the equatorial plane of ellipsoid of inertia.

Module V. Analytical mechanics

Lecture 27. THE NOTION OF HOLONOMIC AND NONHOLONOMIC SYSTEMS. ACTUAL AND VIRTUAL DISPLACEMENT OF THE POINT. CONDITIONS IMPOSED BY REACTIONS ON THE VARIATIONS OF COORDINATES ACTUAL AND VIRTUAL DISPLACEMENT If geometric constraints only are imposed on mechanical system, it is called holonomic system. If differential nonintegrable constraints are imposed on mechanical system, it is called nonholonomic system. When we studied kinematics we considered displacements of the moving point for a period of time in order to determine its velocity or position at some moment of time, etc. We will call the displacements of moving point over a period of time in dependence on the law of its motion as actual or real. Thus, when the point moves according to the law:

r  r (t ) or in projections on the axes, x=x(t), y=y(t), z=z(t) its actual displacement is d r (dx, dy, dz ) for infinitely small period of time dt, where dx= x dt, dy= y dt, dz= z dt. 127

Theoretical mechanics

As we can see, coordinates of the point executing the actual displacement vary due to the fact that the argument t (time) changes by dt. Mathematically these changes are expressed by the differentials of coordinates. If the point is stationary (with respect to the considered system of reference), its actual displacement is equal to zero. As opposed to actual displacement that executes for some period time there is also a displacement which represent a set of all possible displacements that could be imparted to the point at the present time, they depend only on the position of the point at this moment and constraints. Any elementary displacement, which can be imparted to the point from its current position under condition of keeping the constraints is called virtual displacement. As opposed to actual displacement we will note the virtual displacement of the point as r , and its projections on the coordinate axes as x, y, z . Set of virtual displacements of all points of mechanical system is called virtual displacement of the system. CONDITIONS IMPOSED BY REACTIONS ON THE VARIATIONS OF COORDINATES Assume that stationary nonreleasing constraint is imposed on the point M: f ( x, y, z )  0 Let the point execute the virtual displacement, then the coordinates of new point М1 are: x  x, y  y, z  z . As the constraint is nonreleasing, new coordinates must satisfy the constraint equation, i.e. f ( x  x, y  y, z  z)  0 . Expanding it into Taylor series and taking into account the constraint equation and neglecting the terms with high order of smallness we will get the condition imposed by stationary nonreleasing constraint on variation of coordinates: 128

Module V. Analytical mechanics

f f f x  y  z  0 x y z

One can check that if rheonomic nonreleasing constraint is imposed on the point then the conditions imposed by this constraint on the variation of coordinates stay the same. Now we consider the actual displacement of the point. Assume that stationary nonreleasing constraint is imposed on the point M:

f ( x, y, z )  0 . Point M had coordinates x, y, z at a time t and coordinates x  dx, y  dy, z  dz in a period of time dt. As the constraint is nonreleasing, the new coordinates must satisfy the equation of constraints. Expanding it into Taylor series, we obtain the condition: f f f dx  dy  dz  0 x y z

Consequently, in case of stationary constraints the projection of actual displacement satisfy the same relation as the virtual, i.e. actual displacements belong to virtual displacements. If the contsraint is rheonomic, then:

f ( x  dx, y  dy, z  dz, t  dt )  f ( x, y, z, t )  f f f f dx  dy  dz  dt  ...  0 . x y z t

neglecting the terms with high order of smallness and taking into account the constraint equation we will get: f f f f dx  dy  dz  dt  0 x y z t

In this case projection of actual displacement satisfy another relation, that don't coincide with the relation for variations x, y, z , i.e. actual displacements don't belong to virtual displacements. 129

Theoretical mechanics

Lecture 28. VIRTUAL WORK OF FORCES. IDEAL CONSTRAINTS. PRINCIPLE OF VIRTUAL WORK. D'ALAMBERT'S PRINCIPLE FOR THE POINT AND SYSTEM. GENERAL DYNAMIC EQUATION PRINCIPLE OF VIRTUAL WORK The principle of virtual work: In order that the mechanical system of mass points with stationary and ideal constraints were in equilibrium it is necessary and sufficient that the sum of all elementary works of all acting forces for any virtual displacement of the system was equal to zero (for nonreleasing constraints) or was equal to zero or less than zero (for releasing constraints), i.e. n

F

 r k  0

(1)

 rk  0

(1`)

k

k 1

or n

F

k

k 1

This principle follows from the postulate of ideal constraints. According to this postulate sum of elementary works of these constraint reactions for any virtual displacement is equal to zero (if the constraint is nonreleasing) or equal to or greater than zero (if the constraint is releasing), i.e. n

n

 N k  rk  0

or

N k 1

k 1

k

 rk  0.

(2)

D'ALAMBERT'S PRINCIPLE FOR THE POINT AND SYSTEM Consider the mechanical system of n mass points. Mass point of mass mk. will move relative to the inertial coordinate system with 130

Module V. Analytical mechanics

some acceleration w k under action of external and internal forces Fke and Fki . Let's introduce the force of inertia for this point:

Fkи  mk wk , then we will get:

Fa  N  Fи  0

(3)

Doing this operation for every point of mechanical system we will come to the result expressing the d'Alambert's principle for mechanical system: If at any moment of time we apply to the mass point of mechanical system the force of inertia except the external and internal forces, this system will be balanced and we can use all the equations of statics. We can deduce all the theorems of dynamics from the principle of d'Alambert. According to the theory of statics we know, that geometric sum of balanced forces and its moments with respect to any center O are equal to zero. This is true for the forces acting on a solid body and any varying mechanical system. Then on the basis of the principle of d'Alembert it must be:

 (F

e k

 m (F 0

Denoting:

e k

 Fki  Fkи )  0



)  m0 ( Fki )  m0 ( Fkи )  0

R и   Fkи ,

M 0и   m0 ( Fkи )

и

Values R и and M 0 are the principle vector and moment of forces of inertia relative to the center O. Finally, taking into account that geometric sum of internal forces and moments is equal to zero, we will get:  Fke  R и  0,  m0 (Fke )  M 0и  0 131

Theoretical mechanics

Using the principle of d'Alambert simplify the solving of problems as these equations don't contain the internal forces.

GENERAL DYNAMIC EQUATION. EQUATION OF D'ALAMBERT-LAGRANGE Using the principle of d'Alembert we can reduce the equations of motion to the equation of equilibrium. This allows to use the principle of virtual work for solving the problems of dynamics. Consider the system of n – mass points with nonreleasing ideal constraints. Then it takes place the equation for each point of the system: F   m

d 2 r  N  0 dt 2

Let's impart the virtual displacement to the points of system and multiply each of these equations on  r . Summing we will get:

 

  d 2 r  F   m  N     r  0 2  dt  

For ideal constraints

N    r  

 

 0 , then:

 d 2 r  F   m  dt 2 

    r  0  

(4)

Or in projections:  d 2 x  F  m     x dt 2   

  d 2 y x   Fy  m dt 2  

  d 2 z y   Fz  m dt 2  

  z   0 .  

Equation (4) is called the equation of d'Alambert-Lagrange. It is the most common equation of mechanics. All the theorems of dyna132

Module V. Analytical mechanics

mics and equation of motion for mechanical system can be deduced from this equation.

Lecture 29. GENERALIZED COORDINATES. GENERALIZED FORCES. EQUATIONS OF MOTION IN GENERALIZED COORDINATES (LAGRANGE EQUATIONS OF 2-ND KIND) Consider the motion of mechanical system consisting of N material points with respect to the principal inertial frame. Assume that only holonomic constraints are imposed on the system: f ( x; t )  0 (  1, k )

Then the number of independent coordinates will n=3N-k and position of system can be defined by n-independent parameters of any dimension q1, q2, …, qn, that is called the generalized coordinates or lagrange coordinates. Express all rectangular coordinates x in terms of generalized coordinates: (1) x  x (q1 , q2 ,...qn ; t ) All holonomic constraints are taken into account in the equation (1). If the constraint equation don’t contain the time evidently then this constraint is called stationary. As n x x    qi (  1,3N ) , i 1 qi then substituting this expressions in the equation of d’Alambert-Lagrange, we get: 3N  d 2 x  n x X m  q  0 .

  1

 





 dt 2  i 1 qi 133

i

Theoretical mechanics

Remove the parentheses: 3N

n

 1

i 1

 X 

3N x d 2 x qi   m qi dt 2  1

n

x

 q q i 1

i

 0.

(2)

i

The first sum in the left side is the elementary work of all acting forces. 3N

3N

n

 1

 1

i 1

A   X  x  X  

n 3N n x x qi  qi  X    Qi qi qi qi i 1 i 1  1

(3)

where the value: 3N

Qi   X   1

x qi

is called the generalized force correspond to the coordinate qi. Transforming the second term (2): 3N

 m  1

d 2 x dt 2

3N n n  3N x d 2 x x d 2 x x  (4)   q  m  q  m    i  i    dt 2 q qi dt 2 qi i 1 qi  1 i 1 i 1   1 i  n

Note, that: n dx x x    q i   . dt t i 1 qi

Take partial derivative with respect to q i :  dx x .  q i dt qi

(5)

Now we will find: n  2 x  2 x d x   q j   dt qi qi t qi j 1 qi q j

134

 n x x   dx  q      j 1 q j  t  qi dt j 

(6)

Module V. Analytical mechanics

On the base of (5) and (6) transform the double force in (4):  3N d 2 x x  m     dt 2 q i 1   1 i n

   n   d 3N      m x  i 1  dt  1   

n   dx x d  3N qi      m   dt qi i 1  dt   1 

 x  3 N   m x qi   1 из( 5 ) 

 3N dx d  x    m    dt dt  qi   1

 qi  

 x  qi qi   из( 6 ) 

(7)

Introduce the expression for kinetic energy of the system: T

1 3N  m x2 , 2  1

As x  x(q, t ) , то x  x(q, q, t ) . Then its partial derivatives are: x T 3 N   m x  ; q i  1 q i

x T 3 N   m x  qi  1 qi

Therefore the double sum in (7) is transformed to the form: n  3N  d T T  . d 2 x x   m     dt 2 q qi    dt q  q qi i 1   1 i 1  i  i i  n

Returning to the equation (2) we will get the equation of d’Alambert-Lagrange in generalized coordinates: 

n

 Q i 1



i

 d T T   .    qi  0  dt  q  q i i  

(8)

As all variations qi are independent equation (8) takes place only in the case of: d T T (9)   Q , i  1, n dt q i

qi

135

i

Theoretical mechanics

This system is the system of ordinary differential equations of second order relative to the generalized coordinates qi. Equations (8) is called the Lagrange equations of second kind. Advantages: а) number of equations is equal to the number of degrees of freedom; b) there aren’t unknown constraint reactions in these equations. This equation is applied for studying the motion of holonomic systems. Lecture 30. EXPRESSION OF KINETIC ENERGY IN GENERALIZED COORDINATES. LAGRANGE EQUATIONS OF 2-ND TYPE FOR THE SYSTEM UNDER ACTION OF POTENTIAL FORCE. EXPRESSION OF KINETIC ENERGY IN GENERALIZED COORDINATES Kinetic energy of system in rectangular coordinates is: T

1 3N  m x2 2  1

(1)

n As x  x(q, t ) , then dx   x q i  x . Substituting it in (1)

dt

i 1

qi

t

we’ll get: 2

T

 n x x  x x 1 3N 1 n 3N m    qi       m   qi q j   2  1  q  t 2 qi q j i  1 i , j  1   1 i   n

3N

  m i 1  1

x x 1 3N  x  qi   m    qi t 2  1  t 

denote: 3N

aij   m  1

x x qi q j 136

aij  a ji ;

2

Module V. Analytical mechanics 3N

bi   m  1

3N x x ,  x  c   m    . qi t  t   1 2

Then T

n 1 n 1 aij q i q j   bi q i  c  T2  T1  T0 .  2 i , j 1 2 i 1

If the system is stationary, then:

T

1 n  aij qi q j . 2 i , j 1

In this case kinetic energy is the function of generalized velocities. LAGRANGE EQUATIONS OF 2-ND KIND FOR THE SYSTEM UNDER ACTION OF POTENTIAL FORCE If all active forces are potential, then: X 

Taking

into

account U  U (q1 , q2 ,...qn ; t ) . Then: 3N

Qi   X   1

U x

(  1,3N )

that

x  x (q1 , q2 ,...qn ; t ) ,

express

n x 3 N U x U ,   qi  1 x qi i 1 qi

i.e. generalized force is equal to the derivative of force function with respect to appropriate coordinate. In this case the Lagrange equation takes the form: 137

Theoretical mechanics

d T T U   . dt qi qi q i

(2)

Taking into account that U  U (q1 , q2 ,...qn ; t ) then

U 0 q i

and

equation (2) we can write in the form: d (T  U ) (T  U )  0 dt q i qi

Assuming that T  U  L(q, q, t ) we’ll get: d L L   0, dt q i qi

i  1, n

(3)

This is Lagrange equations of second kind for holonomic system under action of potential forces. Function L  T  U is called the Lagrange function or kinetic potential.

Practice

1. VELOCITY AND ACCELERATION OF A MASS POINT AND METHODS OF ITS DECOMPOSITION Class work 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 sin10t , y  3 cos 10t 4) x  2  3 cos 5t , y  4 sin 5t  1 Answer: 1) half-line 2x  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 ; x2 y2   1 with origin in x  0 y  3 ; 4) ellipse 25 9 ( x  2) 2 ( y  1) 2   1 with origin in x  1, y   1 . 9 16

3) ellipse

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 139

Theoretical mechanics

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 / sec 2 . 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  500 m / sec, w0  10 m / sec 2 , h  8 km, s  24 km,  0  41,67 km,   9 km . 1.5. Motion of material point is given with the equations: 1 x  v0t cos  0 , y  v0t sin  0  gt 2 . 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 

v02 2g

sin 2 0 , y 

v02 2g

g 2v02

cos  0 2

x2 ;

sin 2  0 .

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 2 , 0,25 2 . sec 2 sec sec 140

Practice

1.7. Find the trajectory of the point M of coupler link of the slidercrank mechanism if

1 r  l  60 cm, MB  l ,   4t 3

1 r  l  60 cm, MB  l ,   4t (t in seconds) and 3

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 A2 b a2

cos 3  1  3 sin 2  ,

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

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 .

141

Theoretical mechanics

Home work 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  0,5gt 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 142

Practice

distance from weapon at which the projectile will hit the mark situated at the sea level. 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 / sec 2 , (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 .

2. ROTATIONAL MOTION OF A PERFECTLY RIGID BODY AROUND THE FIXED AXIS. ANGULAR VELOCITY AND ANGULAR ACCELERATION OF A RIGID BODY Class work 2.1. 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 . 2.2. 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 / sec 2 . 143

Theoretical mechanics

2.3. 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 . 2.4. 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  15 sec . Answer: v  150 m / sec, wn  11250 m / sec 2 , w  10 m / sec 2. 2.5. 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 / sec 2 , w  200 1  400t 4 cm / sec 2. 2.6. 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 . 2.7. Airscrew rotating with angular velocity 40 rad / sec made 80 turns before its stop after turning off the engine. How much time 144

Practice

it was passed from the moment of turning off the engine before the stop? Assume rotation of the airscrew uniformly retarded. Answer: 8 sec. 2.8. 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. 2.9. 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? 3 Answer: 1)    / 16 sin t rad 4 2) in plumbness max 

3 2  rad / sec 2. 64

2.10. 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 / sec 2 . Answer: v  7,9 km / sec, T  1,4 h.

Home work 2.11. 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 145

Theoretical mechanics

proportional to the cube of time and angular velocity of disc equals   27 rad / sec when t=3 sec. Answer:   t 3 rad . 2.12. 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  5cm / sec 2. 2.13. 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. 2.14. 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 . Find the angular velocity and angular acceleration of balance wheel in 2 seconds after the moment when balance wheel pass the position of equilibrium. 2 Answer:   2 rad / sec,   0.

3. VELOCITIES AND ACCELERATIONS OF POINTS OF PLANE FIGURE Class work 3.1. In crank mechanism the length of the crank OA=40cm, length of link AB=2m; crank rotates uniformly with angular velocity 146

Practice

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

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

sion shows that the link rotates in direction opposite to cranks'. 3.2. Define the velocity of the piston E of pump drive mechanism in the position showed in a picture if OA=20cm, O1 B=O 1 D. Crank OA rotates uniformly with angular velocity of 2 rad/sec. Answer: 46,2 cm/sec. 3.3. 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 ω0=12radpsec. is perpendicular to the slider guide. Sizes given: OA = 10 cm, OD = 12 cm, AB = = 26 cm, EB = 12, DE= 12 3 сm . 147

Theoretical mechanics

Answer: OD  10 3 rad / sec,  DE 

10 3 rad / sec . 3

3.4. 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=20cm and the angular velocity of the link OA equals at the moment to 1 rad/sec. Answer: υ K=40cmpsec , ωDE =3,46radps 3.5. 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  2m / sec 2 , w2  3,16m / sec 2 , w3  6,32m / sec 2 , w4  5,83m / sec 2

3.6. 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 148

Practice

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. 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. 3.7. The bar OB rotates about the axis with constant angular velocity ω=2 c−1 and sets in motion the bar 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. x 2 y 2 Answer: υ D=53,66 cmpsec , ( ) + ( ) =1 12 36

O

3.8. 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 2π 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 π radpsec , III . ω 1= 10 π radpsec , II . ω1=π radpsec , IV . ω1=π 11

9

10 radpsec , III . ω1= π radpsec , II . ω1=π radpsec , IV . ω1=π radpsec 9 149

Theoretical mechanics

picture;

AD  OB

3.9. The load K that is connected via non-stretching thread with the coil L moves down vertically under the law x  t 2 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 OD=2 OC=0,2 m.

Answer: wA  20,9cm / sec 2 , wB  22,4cm / sec 2 , wD  20,1cm / sec 2 ,   10rad / sec,   20rad / sec

3.10. 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 3.11. 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 .

Home work 3.12. The wheel rolls along the plane making an angle 30 degrees with the horizon. Center O of the wheel moves in accordance with 150

Practice

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 . Find the velocity of the end A of the rod AO in the 3

6

moment of time t=1 sec. Answer: velocity equals 2,8m/sec and directed down parallel to the inclined plane. 3.13. 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 . 3.14. 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, v A  6m / sec, vB  4m / sec, vO  2m / sec,,vE  4,46m sec,   20rad / se

m / sec, vB  4m / sec, vO  2m / sec, vE  4,46m sec,   20rad / sec .

151

Theoretical mechanics

3.15. The bar OA of the four-bar 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=2OA=2a . Answer:   0,  

3 2 3 0 ,  B  a02 . 6 6

3.16. 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 .

4. COMPOUND MOTION OF A MASS POINT. THEOREM ON VELOCITY ADDITION. THEOREM ON ACCELERATION ADDITION Class work 4.1. 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 O angle of 40 . Define the absolute velocity of falling of raindrops neglecting the friction of drops against the glass. 152

Practice

Answer: v a 

ve tg 40 

 23.8 m / sec .

4.2. 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: vr  36 km / hour directed at north-east. 4.3. The balls of Watt centrifugal governor that rotates around the vertical axis with angular velocity ω=10radpsec 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 α1=α 2=α=30 O . Answer: v  3.06 m / sec . 4.4. 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. l tg Answer: v r  . cos  153

Theoretical mechanics

4.5. A truck on which an engine is placed moves horizontally to the right with constant acceleration w  0.4m / sec 2 . The 2 engine rotates under the law φ=1 /2 t . 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 / sec 2 , w2  0 m / sec 2 , w3  0,4 2 m / sec 2 , w4  0,8m / sec 2 .

4.6. A disc rotates about the axis that is perpendicular to the plane of disc clockwise uniformly accelerated at angular velocity of 1 radpsec2 ; 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, η connected with a disc at 2 3

time t  1 sec. 2 3 Answer: ωξ =10,95 mpsec ,ωη=−4,37 mpsec

4.7. 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. 2 Answer: υ=υr + h ω ,ω=ω h+ 2 ωυr 154

Practice

4.8. 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 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 . 4.9. 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 incli-ned to the radius OC at an angle 45°, OC = 0,5 m, ω=4 π radpsec , υr=2 mpsec . Answer: v  7,7 m / sec, v  1,414 m / sec, w  35,54 m / sec 2 , w  114,5 m / sec 2 .

4.10. 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 diagonal AC π under the law ξ =a cos t m , where m – 2 meters. Define the magnitude of absolute acceleration of the point at t=1sec and t=2sec.

155

Theoretical mechanics

Answer: wa  a 2 5 m / sec 2 , wa  0,44a 2 m / sec 2 . 4.11. 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 t2 cm . The disc rotates about the axis O1O2 at angular velocity ω=2 t radpsec . Radius OM makes with the axis O1O2 an angle 60°. Define the magnitude of absolute acceleration of the point M at time t=1sec. 2 Answer: wм  35,55 сm / sec . 4.12. 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 / sec 2 .

Home work 4.13. 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 υ=15 mpsec , the angle between 156

Practice

absolute velocity and radius is α = 60°, entrance radius R=2m, angular velocity of the wheel is π radpsec 

Answer: vr  10,106 m / sec, (vr , R)  4150' . 4.14. A stone A 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, O1A = 0,3 m, distance between the axis O1 of the rackwheel E and center B of the link rolling O1B = 0,7 m. Define the angular velocity of the link at moments when segment O1A 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 , ωII =ωIV =0 , ωIII =1,5 radpsec . 4.15. 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. Answer: u2 u2 u2 w 1=r ω 2− , w3 =3 r ω2+ , w 2=w4 =2 r ω 2+ r r r 4.16. 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: 157

Theoretical mechanics

1) transportation and relative motion of one direction; 2) transportation and relative motion are opposite in direction. Answer: 1) w1  r 2  u 2 / r  2u, w2  w4  (u 2 / r  2u   2 r ) 2  4 2 r 2 , w3  3r 2  u 2 / r  2u

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

5. CENTER OF GRAVITY Class work 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 hole is r2 . Center of this hole is situated at the distance

r1 from the center of the disc. 2

Answer: xc  

r1r22 2(r12  r22 )

2. Determine the coordinates of the center of gravity of the figure shown in the picture. Answer: xc  1,6а 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. 158

.

Practice

Answer: yc 

ad 2  bh 2  bd 2 2(ad  bh  bd )

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

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. Answer: xc  3,2cm, yc  9,6cm, zc  6cm 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

7. Let’s consider the homogeneous tetrahedron ABCDEF which is truncated parallel to its base. Area of ABC is equal 159

Theoretical mechanics

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. Answer: z 

3(a( a  b )  3b) 4(a  ab  b)

Home work 8. Find the center of gravity of the contour shown in the picture. Answer: xc  9cm 9. Determine the position of the center of gravity of the ring shown in the picture. Answer: xc  yc  1,38cm 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 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 160

Practice

6. SPECIAL CASES OF THE EQUILIBRIUM CONDITIONS. CONVERGENT AND PARALLEL FORCES. Class work 4.1. 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 T1 and T2 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 4.2. 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 4.3. 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 4.4. 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 161

Theoretical mechanics

4.5. 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. 4.6. 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: RA  2kN , RB  4kN 4.7. 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 4.9. 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 4.10. 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 162

Practice

4.11. 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. Answer: X A  2kN , YA  4,32kN , YB  7,78kN

4.12. 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 , YA  4,2kN , X B  15,6kN

4.13. Determine the reactions of supports A,B,C and the joint D of the composite beam showed in the picture. Answer:

X A  2.8kN ,YA  4.4kN ,YB  22.2kN ,YC  5kN , X D  0,YD  5kN kN ,YA  4.4kN ,YB  22.2kN ,YC  5kN , X D  0,YD  5kN

Home work 4.14 A homogeneous ball of weight 20N is held on the smooth inclined plane with the cable that is connected to the springbalance fixed over the plane; registration of 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

163

Theoretical mechanics

pressure Q of the ball on the plane. Neglect the weight of the springbalance. Answer: α=60°, Q=17,3 N 4.15. 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. 4.16. 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. 4.17. 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=1kN, Q=2kN, q=2kN/m, a=0,8m. Answer: RA  1,5kN , RB  2,1kN . 4.18. 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 164

Practice

4.19. 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 4.20. 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 4.21. 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

7. DIFFERENTIAL EQUATIONS OF MOTION. DIRECT AND INVERSE PROBLEMS OF DYNAMICS Class work 1.1. 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: ωmin=4,494 radps 165

Theoretical mechanics

1.2. Sporting airplane of mass 2000kg flies horizontally with acceleration 5 m / sec 2 and now has velocity of 200m/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. 1.3. Define the motion of heavy ball along the imaginary channel passing through the Earth center if we assume that the 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 / sec 2 . Answer: The distance of ball from Earth center changes under a g t , v  7,9  103 m / sec, T  1266,4 sec  21,1 min law x  R cos R 1.4. The airplane A flies at a height of 4000m over the ground with a horizontal velocity of 140m/sec. At what distance x 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: x=4000 m 1.5. 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. 166

Practice

1.6 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 v 0=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. 1.7. From a gun that is in the point O was fired a shot at angle a to the horizon at initial velocity v 0 . 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 v 1 a second shell should be fired in order that it hit the first shell if velocity v 0 and point A lie in one vertical plane. Neglect the air resistance. Answer: v 1=v 0 sin a 1.8. 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 v 0 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. g  kv 0 g (1  e kt ) Answer: H  t  k k2 1.9. 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. 167

Theoretical mechanics

1.10 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. Answer: α1=30 o 18 ' , α2=59 o 42'

Home work 1.11. A weight with mass 0.2kg is hung to an end of a thread with length 1m. In the result of a push the weight got horizontal velocity of 5m/sec. Find tension of the thread right after the push. Answer: 6,96N 1.12. A point with mass m starts moving from the state of rest from the position x 0=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 2 cube of a distance: Q x =c 2 mx . 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

1.13. A plane A flies over the ground at a height h with a horizontal velocity v 1 . 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 v 0 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. Answer: 1. v 20≥v 21+ 2gh; 2. cosα=v 1 /v 0 1.14. 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 168

Practice

1.15. 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 1.16. Stone is thrown vertically up at v 0 . 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 axis of its velocity, k 2 is the constant coefficient. X axis is directed vertically up. Answer: H 

1 2k 2

ln 4

g  k 2 v02 4 g  k 2 v02

1.17. The greatest horizontal range of a missile equals to L. Equations of motion of missile are: x  v0t cos  , y  vot sin   gt 2 / 2 . Define its horizontal range l at throwing angle α = 30° and height h of trajectory in that case. Neglect the air resistance. Answer: l  L 3 / 2 , h= L 8

8. THEOREMS OF DYNAMICS THEOREM OF CHANGE OF LINEAR MOMENTUM (MOMENTUM THEOREM). ANGULAR MOMENTUM THEOREM Table of moments of inertia № 1 1

Name 2 Thin linear rod

Scheme 3

Inertia moment 4 Jz 

169

Ml 2 3

Theoretical mechanics

2

Ring (circle)

3

Thin round disc

J z  Mr 2

Jx  Jy  M Jz  M

4

Round cylinder

5

Sphere

6

Parallelepiped

7

Cone

r2 2

Jx  Jy  M ( Jz  M

r2 4

r2 l2  ) 4 3

r2 2

Jx  Jy  Jz 

2 Mr 2 5

1 m(b 2  c 2 ) 12 1 J y  m( a 2  c 2 ) 12 1 J z  m(b 2  a 2 ) 12 Jx 

Jz 

170

3 Mr 2 10

Practice

Class work 1. 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 2. 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. 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=0.00095sec . Define the mean magnitude of the gases pressure that throw the bullet if the area of bore section is S  150mm 2 . Answer: mean pressure is P  9.12 10 4 N / mm 2 4. 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  500m / sec,  0  60 , v1  200m / sec , mass of shell is 100kg. Answer: Projections of resultant impulse: S x  5000N  sec, S x  43300N  sec.

5. 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 12,5 m/sec in 50 sec after the entry to a rise. Find the thrust of a diesel locomotive. Answer: 23120N. 171

Theoretical mechanics

6. 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? m

Answer:  ln 2 . 7. 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. 8. An acrobat performing a somersault and pushing off from the ground imposes himself at the initial time an angular velocity 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 kgm 2 . 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. 9. 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 172

Practice

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 )

10. 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 disc motion 2

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. 3r Answer:    0 cos kt, T  2 2g 11. 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 173

Theoretical mechanics

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.

Home work 12. 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  2N  sec, F  1N . 13. 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 2

gR

that equal to 1 kg in a distance of 1 m assume equal to m N where m – mass of the Earth. It follows from Kepler laws that the attractive 2 3 4π a m force of Earth by the Sun equals where r – is a distance 2 2 T r between Earth and Sun. Answer: M  1,966  1030 kg . 14. 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 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. 174

Practice

Answer:   0e

k  t J .

15. 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 ω 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

16. Barrel A of mass M 1 and radius r is set in rotation by means of the load C of mass M 2 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  . 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. Answer:  

M 2 gr



1  e   where   r  t

2

2 . ( M 1  2M 2 )

17. 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 through the mass center C perpendicular to the plane of solid 175

Theoretical mechanics

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 .

18 Define whit what angular velocity ω 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 2 where α is constant. The moment of inertia of the tree mcz   relative to z axis coinciding with an axis around which the falling tree rotates, is equal to J. Answer:  



 2Mghl  (e J  2 ) 2 2 J  4 J

9. WORK – ENERGY THEOREM Class work 6.1. 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 6.2. 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 6.3. 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 176

Practice

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 6.4 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. Answer: vB  2.95 m / sec, vC  2.4m / sec 6.5. 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 6.6. 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 2  l   sin     l   2

2

177

2

Theoretical mechanics

6.7. 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? 2 2 Answer: T = r ω (33 M 1+ 8 M 2) 3 6.8. Lifting mechanism of a crab is shown at the picture. 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:  



4h a h  3 M 1 g r 2

2



3 r 2 M 1  2 M 2  M 3  3

6.9. 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 178

Practice

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 3L .  9 r1 r 2 M 1  2 M 2

6.10. 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. 

 Answer:   2 2n L  M 2 gr1 sin  (e J  2  ) r1 M 1  3M 2 J

6.11. Load with weight P=20kg rises up an inclined plane via rope placed angularly α = 30° to the plane. Inclined plane subtends the angle β = 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

Home work 6.13. Heavy body descends without initial velocity along the inclined plane making an angle 30° with the horizon. Friction coeffi179

Theoretical mechanics

cient is equal to 0.1. What velocity will have the body when it pass 2 meters from the beginning of motion. Answer: 4.02m/sec 6.14. 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 6.15. 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 M 1 . Couple AB and connecting rod О1О2 has the equal mass M 2 . 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. Answer: T 

v02 16M1  M 2 (9  4 sin  ) 8

6.16. 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 M 1 and radius r. Mass of the pulley II equals M 2 , its radius is R. Mass of the barrel of the reel is M 3 , its radius is r, mass of the lifting load is M 4 . The reel is set in motion from the state of rest. Find the velocity of the 180

Practice

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. Answer: v  2

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

6.17. Nonstretching thread is fastened to the load A of mass M 1 and is threw over the block D of mass M 2 and reeled on the side face of cylindrical roller B of mass M 3 . When load A moves down the inclined plane making an angle  with the horizon block D rotates and roller B rolls up without sliding along the inclined plane making an angle  with the horizon. Determine the velocity of load A in dependence on the passed way s if the system was in rest at initial moment. Consider the block D and roller B as homogeneous round cylinders. Neglect the friction force and mass of the thread. 2M1 sin   M 3 sin  Answer: v  2 2 gs 8M 1  4M 2  3M 3

6.18. 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  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. 181

Theoretical mechanics

Answer:  

rJ 1 rJ 0 ,   02 RMfr 2 RM fr

 M1 3   M 2 R  r 2 . 2  3 

where J  

10. OSCILLATIONS. FREE OSCILLATIONS. FREE OSCILLATIONS IN THE MEDIUM WITH RESISTANCE. FORCED OSCILLATIONS. RESONANCE Classwork 7.1. 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 с1  4кг/см , с 2  5кг/см . 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  0,21sec 7.2. 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 c1, c2 are equal.

√

m ; c=c 1+ c 2 ; the location of the load is (c 1+ c2 ) such that a 1 /a 2=c2 /c1

Answer: T =2π

182

Practice

7.3. 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. m(c1  c2  c3 ) Answer: c  c1 (c2  c3 ) , T  2 c1 (c2  c3 ) c1  c2  c3 7.4. Two loads of mass m1 = 0.5 kg, m2 = 0.8 kg, 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  1sec 7.5. The load D of mass 4 kg passing the distance s=0.1m without initial velocity along the inclined plane (   30 ) struck against the springs with the series connection ( c1  48N / 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 cos 1.25t  1.25 sin 1.25t

183

Theoretical mechanics

7.6. 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 v0  4сm/sec. Answer: x  7,2e8t sin(6t  0,59) 7.7. 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 7.8. 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 7.9. 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 184

Practice

law of the point motion if at the initial time x  x0  0 , v  v0  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 )

Home work 7.10. 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

 6,7 cm

2

2  2 2  1  p   4 n p  k 2  k2 k2 

7.11. 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 = 2cm, n = 7 rad/sec. 185

Theoretical mechanics

Answer: x=4 sin 7 t cm . 7.12. 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 Answer: x  e  26t (29e  24t  5e 24t ) 6

7.13. 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   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 ) 11. INERTIA MOMENTS Class work 4.1. Determine the moment of inertia J x , J y , J z of the thin homogeneous disc of mass m and radius r with respect to axes X,Y,Z passing through its center of mass. 186

Practice 2

Answer: J x  J y  MR , J z  MR 4

2

2

4.2. Determine the moment of inertia J x , J y , J z of the solid cylinder of mass m, radius r and height h with respect to axes X,Y,Z passing through its center of mass. 2 2 2 Answer: J x  J y  MR  Mh , J z  MR

4

12

2

4.3. Concentric aperture of radius r is drilled in the thin homogeneous round disc of radius R. Determine the moment of inertia of this disc of mass M relative to the axis z passing through it’s center of mass perpendicular to the plane of the disc. Answer: J z 

M 2 (R  r 2 ) 2

4.4. Determine the moment of inertia of an axle relative to the axis of rotation z. Weight of an axle is 100 kg, radius of an axle is 10 cm. Flywheel of a mass 1 kg and radius of 1 m is stuck to the axle. Axle can be considered as the homogeneous cylinder. Answer: J z  102,05 kg m 2 4.5. Homogeneous round disc of a mass M is stuck to the axis z perpendicular to it’s plane. Radius of the disc is r, eccentricity OC=a, where C is the center of mass of the disc. Determine the moments of inertia J x , J y , J z relative to the axis X,Y,Z and centrifugal moment of inertia J xy , J xz , J yz of the disc. Axis of coordinates are shown at the picture. 187

Theoretical mechanics

Answer: Jx 

Mr 2 r2 r2 , J y  M (  a 2 ), J z  M (  a 2 ), J xy  J xz  J yz  0 4 4 2

4.6. Homogeneous equilateral triangular plate has a mass M and a height of a side l. Calculate the moment of inertia of the plate relative to z axis passing through the vertex of the plate perpendicularly to its plane. Answer: I z  3 Ml 2 16 4.7. The pendulum consists of a homogeneous pin-type rod AB with mass M 1 to the end of which the homogeneous disk with mass M 2 is fastened. The length of the rod is 4r, where r is a radius of disk. Calculate the moment of inertia of a pendulum relative to its axis of suspension, perpendicular to the plane of a pendulum and situated at the distance r from the end of a rod. Answer: J 

4M1  99M 2 2 r 6

4.8. Homogeneous round disk of mass M and radius r is attached to an axis AB situated at the distance ОС  r / 2 from the center of mass center C. Calculate the axial moment of inertia and centrifugal moment of inertia. Answer: Jx 

3 Mr 2 Mr 2 Mr 2 Mr 2 , J y  ,Jy  , Jz  , J xy  J xz  J yz  0 4 4 4 2

4.9. Calculate the moment of inertia J of a homogeneous rod with mass m and length L relative to the axis that is perpendicular to the 188

Practice

rod and passing through its center. Cross-sizes of the rod should be neglected. Answer: J C  1 mL 2 12

4.10. Calculate the moment of inertia J of a homogeneous rod with mass m and length L relative to the axis that is perpendicular to the rod and passing through its end. Crosssizes of the rod should be neglected. 1 2 Answer: J C  mL 3 Home work 4.11. A pin-type homogeneous rod AB with the length of 2l and the mass of M is fastened at center O to the vertical axis forming with it an angle  . Calculate the moments of inertia of the rod J x , J y and the centrifugal moment of inertia J xy . Axes of coordinates are shown in a picture. Answer: J x 

Ml 2 Ml 2 Ml 2 cos 2  , J y  sin 2  , J xy  sin 2 3 3 6

4.12. Determine the moment of inertia J x , J y , J z of the solid parallelepiped of mass m, radius r and height h with respect to axes X,Y,Z passing through its center of mass.

189

Theoretical mechanics

Answer: J x 

1 1 1 m(b 2  c 2 ), J y  m(a 2  c 2 ), J z  m(b 2  a 2 ) 12 12 12

4.13. Determine the moment inertia J x , J y of a homogeneous rectangular plate of mass m shown at the picture. 4 3

4 3

Answer: J x  Ma 2 , J y  Mb 2 4.14. Homogeneous round disc of a mass M is stuck to the axis z perpendicular to it’s plane. Radius of the disc is r, eccentricity OC=a, where C is the center of mass of the disc. Determine the moments of inertia of the disc relative to the axis z1 lying in the vertical plane xz and making an angle  with the axis z. Answer: J z1 

Mr 2 r2 sin 2   M (  a 2 ) cos 2  4 2

12. PRINCIPLE OF VIRTUAL WORK. LAGRANGE EQUATIONS OF 2-ND KIND 4.9. The bush 2 is put on the link AB of the jointed parallelogram OABC. Vertical rod 3 is fastened to the bush. Define the moment M of the pair of forces in the case of equilibrium of the mechanism if the force F=400N and the length of the crank OA=0.2m. Answer: M = 40Nm

190

Practice

4.10. Define the module of the moment M of the couple of forces that should be attached to the pulley 3 for the even lift of the load 1 of weight 900N. Pulleys' radii are R=2r=40cm. Answer: M = 90Nm 1.3. Horizontal force F=50N is applied to the rocker of the hinged parallelogram OABC. Define the magnitude of the moment M of couple of forces that should apply to the crank OA of length 10 cm in order to balance the mechanism. Answer: M=4,33 Nm 1.4. The force F=100N acts on the wedge 3. Define the force with which the pusher 2 press the component 1 to the reference plane in the position of equilibrium if   11 . Answer: F=514N 2.1. Tappet K of mass M 1 is in the state rest of the smooth horizontal plane. It supports the rod AB of mass M 2 which is situated in the vertical direction. System is in rest under action of force F applied to the tappet K horizontally to the right. Define the magnitude of force F if the lateral surface of the tappet make an angle  with the horizon. Answer: F  M 2 tg 2.2. Two beams BC and CD are connected in C by means of the joint and fastened to the vertical column AB fixed in the section A by means of cylindrical hinge D. Also they are connected with the floor by means of cylindrical joint. Horizontal forces P1 and P2 are 191

Theoretical mechanics

applied to the beams. Define the horizontal component of reaction in the section A. Dimensions are shown in the picture. 2.4. A polyspast consists of fixed block A and of the n moving blocks. In the case of equilibrium define the relation of mass M of the lifting load to the force P that is attached to the end of the cable that goes down from the fixed block A. n

Answer: Mg/ P=2

2.5 The loads A and B of equal mass fastened to the end of nonstretching thread. Thread goes parallel to the horizontal plane from the load A and rounds the fixed block C, embraces moving block D, then rounds fixed block E where the load B is tied to the other end of the thread. Load K of mass M is suspended to the axis of the moving block D. Define the mass M1 of each load A and B and a coefficient of sliding friction f of the load A against the horizontal plane if the system of loads is in rest. Neglect the mass of the thread. Answer: M 1=M /2 ; f =1 1. Determine the angular acceleration of the crank OA of planetary gear situated in the vertical plane. Tooth-wheel 2 is set in motion by the help of the crank OA to which the couple of forces with rotation moment m0 is applied. m is the mass of the crank, m2 is the mass of the wheel 2, r2 is the radius of the wheel 2, r1 is the radius of the fixed wheel 1. Consider the wheel 2 as the homogeneous round disc and the crank OA as the thin homogeneous rod. Neglect the forces of motion resistance. 192

Practice

Lagrange equations of 2-nd kind

Answer:   3

2m0  g (m  2m2 )( r1  r2 ) cos  2m  9m2

1. Masses of the bodies 1,2,3 are: m1  m , m2  2m , m3  3m . External mo-

ments M 1 , M 2 are given. Work out the equations of motion of the system in generalized coordinates 1 ,  2 . All dimensions are shown in the picture. 4. Masses of the bodies 1,2,3,4 are: m1  2m , m2  m , m3  m , m4  2m . Exter-

nal moment M is given. Work out the equations of motion of the system in generalized coordinates  ,  . All dimensions are shown in the picture. 8. Masses of the bodies 1,2,3 are: m1  2m , m2  2m , m3  m . External for-

ces P1 , P2 are given. Coefficient of sliding friction for the bodies 1,2,3 is f. Work out the equations of motion of the system in

193

Theoretical mechanics

generalized coordinates x1 , x2 . All dimensions are shown in the picture. 10. Masses of the bodies 1,2 are: m1  m , m2  3m . External moment M is given. Work out the equations of motion of the system in generalized coordinates x ,  . All dimensions are shown in the picture.

Module I. Kinematics

1. Bukhgoltsz N.N. Basic course of theoretical mechanics. P.1. – St-P.: Lan, 2009. – 480 p.; P.2. – St-P.: Lan, 2009. – 336 p. 2. Gross D., Hauger W., Schroder J., Wall W.A., Rajapakse N. Engineering Mechanics. P.1 Springer, Verlag. – 2012. – 301 p. 3. Gross D., Hauger W., Schroder J., Wall W.A., Govindjee S. Engineering Mechanics. P.3 Springer, Verlag. – 2011. – 359 p. 4. Markeyev A.P. Theoretical mechanics. – M.-Ijevsk: «Regulyarnaya I haoticheskaya dinamika». – 2001. – 592 p. 5. Meschersky I.V. Collection of problems on theoretical mechanics. – St-P.: Lan’. – 2010. – 448 p. 6. Collection of the tasks for course works on theoretical mechanics. Ed. Yablonsky. – M.: KnoRus. – 2011. – 392 p. 7. Collection of short tasks on theoretical mechanics. Ed. Kepe O. – St-P.: Lan, 2009. – 368 p. 8. Targ S.M. Short course of theoretical mechanics. – М.: Vysshaya shkola. – 2010. – 416 p. 9. Golubev Yu.F. Principles of theoretical mechanics. – М.: MSU. – 2000. – 719 p. 10. Nesterova L.F., Rakisheva Z.B. Kinematics of the plane motion of the solid body. – Almaty: Kazakh university. – 2001. – 84 p.

195

Еducational issue

Rakisheva Zaure Bayanovna Sukhenko Anna Sergeevna THEORETICAL MECHANICS Educational manual Typesetting and cover design G. Кaliyeva Cover design used photos from sites www.mechanics-a-level-link-image.com

IB No. 9291

Signed for publishing 31.03.2016. Format 60x84 1/16. Offset paper. Digital printing. Volume 12,25 printer’s sheet. 100 copies. Order No. 888. Publishing house «Qazakh university» Al-Farabi Qazaq National University KazNU, 71 Al-Farabi, 050040, Almaty Printed in the printing office of the «Qazaq university» publishing house.