Из полученной формулы следует, что

Из полученной формулы следует, что работа силы упругости положительна, если начальная деформация больше конечной по модулю, и отрицательна в противном случае. Она зависит только от начального и конечного положений груза М.
7.4.3. Работа силы, приложенной к твердому телу, вращающемуся во-круг неподвижной оси
Пусть к точке М твердого тела, вращающегося вокруг неподвижной оси Oz c угловой скоростью , приложена сила (рис. 7.10). Элементарная работа этой силы:
. (7.18)
Введем радиус-вектор точки М относительно точки О, лежащей на оси вращения, тогда скорость точки М можно определить по формуле Эйлера
.
Преобразуем формулу (7.18), используя свойство смешанного произ-ведения векторов
.
Но так как произведение

– это момент силы относительно точки О,

.
Учтем, что
–
момент силы относительно оси z, а – элементарный угол поворота тела, потому
. (7.19)
Таким образом, элементарная работа равна произведению момента силы от-носительно оси вращения на элементарный угол поворота тела. Работа силы на конечном перемещении тела
. (7.20)
Если , то
, (7.21)
т.е. работа равна произведению момента силы относительно оси вращения на угол поворота тела.
7.4.4. Работа постоянной силы
Рассмотрим перемещение точки приложения постоянной силы по ду-ге (рис. 7.11). Радиус-векторы и определяют начало и конец траекто-рии относительно неподвижной точки О.
Для определения работы силы используем формулу (7.13), где поло-жим ,
.
Учитывая, что , окончательно получим
, (7.21)
т.е. для постоянной силы работа вычисляется так же, как и при прямолиней-ном перемещении точки ее приложения вдоль хорды, стягивающей дугу .
7.4.5. Работа внутренних сил твердого тела
Рассмотрим точки и твердого тела, которые действуют друг на друга с силами и (рис. 7.12), причем по закону равенства действия и про-тиводействия .
Пусть точки и получили элементарные перемещения и . Вы-числим сумму элементарных работ

,
где – скорости точек и .

В соответствии с теоремой о проекциях скоростей двух точек твердого тела и . Так как каждой внутренней силе соответствует дру-гая сила, равная ей по модулю и противоположная по направлению, сумма элементарных работ всех внутренних сил равна нулю. Конечное перемеще-ние тела складывается из его элементарных перемещений и поэтому сумма работ всех внутренних сил твердого тела на любом его перемещении равна нулю
.
7.4.6. Работа сил сопротивления качению
Рассмотрим цилиндрический каток, который движется по горизонталь-ной плоскости без скольжения (рис. 7.13). Возникновение сопротивления ка-чению связано с деформациями соприкасающихся поверхностей, из-за чего линия действия нормальной составляющей равнодействующей реактивных сил оказывается смещенной в сторону движения катка на некоторое расстоя-ние δ от линии действия силы тяжести . Силы ( , ) образуют пару сил со-противления качению. Момент этой пары называют моментом сопротивле-ния качению. Его модуль равен произведению модуля нормальной реакции на плечо пары δ, называемое коэффициентом трения качения
. (7.22)
Этот коэффициент измеряется в единицах длины.

Пусть центр катка получил элементарное перемещение , тогда каток по-вернулся на угол и элементарная работа пары сил сопротивления ка-чению
. (7.23)
При конечном перемещении центра катка на расстояние работа сил сопро-тивления качению
. (7.24)
Элементарная работа горизонтальной составляющей равнодействую-щей реактивных сил (силы сцепления ), которую можно считать прило-женной в мгновенном центре скоростей катка Р, равна нулю, так как скорость этой точки и ее элементарное перемещение . Поскольку полная работа равна сумме элементарных работ, приходим к следующему выводу: сила сцепления в рассмотренном случае качения без скольжения работу не совершает.

7.5. Теорема об изменении кинетической энергии материальной точки
Рассмотрим материальную точку, движущуюся под действием системы сил , и запишем для нее основное уравнение динамики (1.2)
.
Умножим скалярно обе части этого равенства на вектор элементарного пере-мещения точки
. (7.25)
Преобразуем левую часть полученного уравнения
.
Учитывая, что в правой части уравнения (7.25) произведения представ-ляют собой элементарные работы сил , получим
.
Разделим это равенство на величину и с учетом (7.14) запишем
. (7.26)
Соотношение (7.26) позволяет сформулировать теорему об изменении кинетической энергии в дифференциальной форме. Производная по времени от кинетической энергии материальной точки равна сумме мощностей всех действующих на нее сил.
Интегрируя равенство (7.26), получим
, (7.27)
где

Результаты (английский) 1:

The formula should be received, that the work force of elasticity is positive, if the initial deformation more ultimate modulo, and negative otherwise. It depends only on the initial and final provisions cargo m. 7.4.3. the work of the force applied to a solid body, rotating in a circle of fixed axis Let the point m is solid, rotating around a fixed axis Oz (c) angular velocity, applied force (Figure 7.10). Elementary work of this force: . (7.18)We introduce the radius-vector of the point m of a point on the axis of rotation, then the speed point of m can be identified by Euler's formula .Convert formula (7.18), using the mixed performance of reference vectors .But as the product of is the torque about the point o .Take into account that –moment of force about an axis z-elementary rotation angle of the body, because . (7.19)Thus, the basic work equals to the product of the moment of force-' axis of rotation on the elementary body rotation angle. The work force on a target moving body . (7.20)If, then (7.21)i.e. work equals to the product of the moment of force about an axis of rotation by an angle of rotation of the body. 7.4.4. the work of the permanent force Consider moving the point of application of the permanent force do-Gyo (fig. 7.11). RADIUS-vectors and define the beginning and end of the traekto-history of relatively fixed point of.To determine the work force use formula (7.13), where Polo-lifter .Given that, finally get (7.21)i.e. for permanent work force is calculated the same way as when Mr. prjamolinej point moved it along the chord, stjagivajushhej arc. 7.4.5. the work of the internal forces of a rigid body Consider the point and a rigid body that Act on each other with the force and (Figure 7.12), and according to the law of equality actions and counter.Let point and got the basic move and. You-chislim sum of elementary works ,where is speed and points. In accordance with the theorem on projections of the speeds of the two points of a rigid body and. Because each inner strength matches the other guy force equal modulo and opposite in direction, the amount of the basic works of all internal forces is zero. The target of bringing the body develops from its basic movements and therefore the amount of works by all the internal forces of a rigid body on any moving it is zero . 7.4.6. the work force of rolling resistance Consider a cylindrical roller moves on the horizontal plane-without slipping (Figure 7.13). The emergence of resistance to Ka for deformities associated with surfaces, leaving a line of actions a normal component of the resultant reaction forces turns out to be biased in the direction of movement of the rink for a an ocular micrometer of δ from the effect of gravity. Forces () comprise a couple of forces from-rolling resistance. The moment this pair is called the point of rolling-resistance. It is the product of the module module normal reactions to leverage pair δ, called friction bearings . (7.22)This ratio is measured in units of length. Let the center of the rink got elementary move, then skating rink-returned to the corner and elementary work couples resistance forces Ka-clude . (7.23)The eventual displacement of the center of the rink on distance work force re-rolling tivlenija . (7.24) Elementary jobs ravnodejstvuju future horizontal component reactive forces (adhesion), which can be considered as annex immediate policy Center rink r speeds, is zero, because the speed of this point and its basic movement. Because the total work is the sum of the basic works, come to the following conclusion: in this case the adhesive force of rolling without slipping does not work.7.5. changing the theorem of kinetic energy of a material point Let's take a look at the material point moving under the action of the force system and write for her Dynamics equation (1.2) .Multiply both parts of the scalar by a vector that equality of basic re-location of points . (7.25)Convert the left part of the resulting equation .Given that on the right side of equation (7.25) works submitted reflected a basic work force get .Divide this by the amount of equity and given (7.14) write . (7.26) Ratio (7.26) permits to formulate the theorem of kinetic energy change in differential form. Derived from the kinetic energy of a material point is equal to the sum of the capacities of all the forces acting on it. Integrating equality (7.26), we obtain (7.27)where

Результаты (английский) 2:

From this formula it follows that the work of the elastic force is positive, if the initial deformation of the ultimate greater in magnitude and negative otherwise. It depends only on the initial and final position of the load M.
7.4.3. Work force applied to a solid body, rotating in a circle-fixed axis
Let the point M of the solid body rotating around a fixed axis Oz c angular velocity, force is applied (Fig. 7.10). Elementary work of this force:
. (7.18)
We introduce the radius vector of the point M with respect to the point O lies on the axis of rotation, then the velocity of the point M can be determined from the Euler equation
.
We transform equation (7.18), using the properties of the mixed pro-reference vectors
.
However, since the product

- it is a moment of force with respect to point O

.
We take into account that
-
the moment of force about the z-axis, and - the elementary angle of rotation of the body, because
. (7.19)
Thus, the elementary work is the product of the torque from the axis of rotation, relatively to the elementary angle of rotation of the body. The work force in the final movement of the body
. (7.20)
If, then
, (7.21)
ie, work is the product of a moment of force about the rotational axis by an angle of rotation of the body.
7.4.4. Work constant force
Consider moving the point of application of the constant force on the do-re (Fig. 7.11). Radius vectors and define the beginning and end of the trajectory of relativity-a fixed point O.
To determine the work force we use the formula (7.13), where polo-press,
.
Considering that, we finally obtain
(7.21),
ie, for permanent work force is calculated in the same manner as in rectilinear Mr. move its point of application along the chord subtending the arc.
7.4.5. The work of internal rigid body forces
consider the point and rigid body, which act on each other with forces and (Fig. 7.12), and according to the law of equality of action and pro-counteracting.
Let the points and got the basic movement and. You are the number as a sum of elementary work

,
where -. Points and speed

in accordance with the theorem on projections of the velocities of two points and a solid body. Since each inner strength corresponds Dru-Gaya force equal to it in magnitude and opposite in direction, the sum of all the elementary work of the internal forces is zero. End-displacements of the body is made up of its basic movements and, therefore, the sum of all the work of the internal forces of a rigid body in any of his move is zero
.
7.4.6. Job rolling resistance forces
consider a cylindrical roller, which moves in a horizontal-plane without sliding (Fig. 7.13). The emergence of resistance ka-cheniyu due to the deformation of the contacting surfaces, which is why the line of action of the normal component of the resultant reactive force is shifted in the direction of motion of the roller on some of δ-distances from the line of gravity. Force (,) form a pair of forces from the rolling-resistance. The moment the pair is called the point of resis--rolling. His unit is equal to the product of the modulus of the normal response to a pair of shoulder δ, called rolling friction
. (7.22)
This ratio is measured in units of length.

Let the roller center has received an elementary movement, then the rink is back on the angle and the elementary work of a pair of resistance ka-cheniyu forces
. (7.23)
In the final movement of the roller center distance work force accompanied resistivity-rolling
. (7.24)
The elementary work of the horizontal component of the resultant-ing reaction forces (grip strength), which can be considered APPENDIX-zhennoy in the instant center of the rink P velocity is zero, since the velocity of this point and its elementary displacement. Since the total work is the sum of the elementary works, we arrive at the following conclusion: the adhesive force in the present case of rolling without slipping does not work.

7.5. The theorem of change of kinetic energy of a material point
Consider a material point moving under the influence of a system of forces, and write for her the fundamental equation of dynamics (1.2)
.
We take the inner sides of this equation by elementary vector re-displaced point
. (7.25)
We transform the left-hand side of this equation
.
Given that the right-hand side of equation (7.25) works appeared-lyayut an elementary work forces, we obtain
.
We divide this equality by value and taking into account (7.14), we write
. (7.26)
Equation (7.26) allows us to formulate the theorem on change of kinetic energy in differential form. The time derivative of the kinetic energy of the material point is equal to the sum of the capacities of all the forces acting on it.
Integrating the equality (7.26), we obtain
(7.27)
where

Результаты (английский) 3:

from the obtained formula should be that the work force of the positive, if the initial deformation more ultimate module, and negative otherwise. it depends only on the initial and final provisions of the goods.7.4.3. the work force is applied to the solid body, вращающемуся in terms of fixed axisget to the point, the rigid body rotating around a fixed axis, the angular speed of the oz, accompanied by strong (figure 1). 7.10). the basic work of the force.. (7.18)inject the radius vector, on the point of point lying on the axis of rotation, then the speed of the point m can be defined by euler's formula.convert formula (7.18), using the characteristic of mixed production of vectors.but as the workis a moment force on the point of.take into consideration that-moment of force about the axis z, and is the angle of body, because. (7.19)thus, the basic work is the product of the moment of force of the носительно axis of rotation in the rotation body. the work force in the final movement of the body. (7.20)if,(7.21)i.e. work is the product of the moment of force about the axis of rotation of the rotation body.7.4.4.. the constant forceconsider the movement of constant force application points on du gi (figure 1). 7.11). radius vectors and define the beginning and end of траекто - рии relatively stationary point of.to determine the work force, use the formula (7.13), where polo - press..given that, finally get(7.21)i.e. for constant force work is calculated as the прямолиней by displacement along the хорды in its applications, стягивающей arc.7.4.5. the internal forces of rigid bodyconsider the point and a solid body, which act on each other with the forces and (figure 1). 7.12), and the equality of action and the тиводействия.let's point and obtained the basic movements and. you числим amount basic works,where is the speed of the points and.in accordance with the theorem of rigid body and projections; the two points. as each inner strength is consistent with the other gaia power equal to the module and the other in the direction of the basic work of internal forces is zero. the final перемеще body consists of the basic movements, and therefore the amount of work of all the internal forces of rigid body in any of its displacement is zero.7.4.6. the force resistanceconsider the cylindrical roller, which moves in the horizontal plane without sliding, and the (figure 2). 7.13). the emergence of resistance to this drug is associated with деформациями shared surfaces, which line of the normal component of the равнодействующей reactive force is displaced toward the movement of the roller at a расстоя of garden of the line of action of the force of gravity. force (,) form a pair.