Energy of rotational motion. Kinetic energy and work during rotational motion Kinematic energy of rotational motion

1. Consider the rotation of the body around motionless Z axis. Let's divide the whole body into a set elementary masses m i. Linear speed of elementary mass m i– v i = w R i, where R i– mass distance m i from the axis of rotation. Therefore, kinetic energy i th elementary mass will be equal to . Total kinetic energy of the body: , here is the moment of inertia of the body relative to the axis of rotation.

Thus, the kinetic energy of a body rotating about a fixed axis is equal to:

2. Now let the body rotates relative to some axis, and itself axis moves progressively, remaining parallel to itself.

FOR EXAMPLE: A ball rolling without sliding makes a rotational motion, and its center of gravity, through which the axis of rotation passes (point “O”) moves translationally (Fig. 4.17).

Speed i-that elementary body mass is equal to , where is the speed of some point “O” of the body; – radius vector that determines the position of the elementary mass relative to point “O”.

The kinetic energy of an elementary mass is equal to:

NOTE: the vector product coincides in direction with the vector and has a modulus equal to (Fig. 4.18).

Taking this remark into account, we can write that , where is the distance of the mass from the axis of rotation. In the second term we make a cyclic rearrangement of the factors, after which we get

To obtain the total kinetic energy of the body, we sum this expression over all elementary masses, taking the constant factors beyond the sign of the sum. We get

The sum of elementary masses is the mass of the body “m”. The expression is equal to the product of the mass of the body by the radius vector of the center of inertia of the body (by definition of the center of inertia). Finally, the moment of inertia of the body relative to the axis passing through point “O”. Therefore we can write

If we take the center of inertia of the body “C” as the point “O”, the radius vector will be equal to zero and the second term will disappear. Then, denoting through – the speed of the center of inertia, and through – the moment of inertia of the body relative to the axis passing through point “C”, we obtain:

(4.6)

Thus, the kinetic energy of a body in plane motion is composed of the energy of translational motion at a speed equal to the speed of the center of inertia, and the energy of rotation around an axis passing through the center of inertia of the body.

Work of external forces during rotational motion solid.

Let's find the work done by the forces when the body rotates around the stationary Z axis.

Let an internal force and an external force act on the mass (the resulting force lies in a plane perpendicular to the axis of rotation) (Fig. 4.19). These forces perform in time dt job:

Having carried out in mixed works vectors cyclic permutation of factors, we find:

where , are, respectively, the moments of internal and external forces relative to point “O”.

Summing over all elementary masses, we obtain the elementary work done on the body in time dt:

The sum of the moments of internal forces is zero. Then, denoting the total moment of external forces through , we arrive at the expression:

It is known that the scalar product of two vectors is a scalar equal to the product of the modulus of one of the vectors being multiplied by the projection of the second to the direction of the first, taking into account that , (the directions of the Z axis coincide), we obtain

but w dt=d j, i.e. the angle through which a body turns in time dt. That's why

The sign of the work depends on the sign of M z, i.e. from the sign of the projection of the vector onto the direction of the vector.

So, when a body rotates, internal forces do no work, and the work of external forces is determined by the formula .

Work done in a finite period of time is found by integration

If the projection of the resulting moment of external forces onto the direction remains constant, then it can be taken out of the integral sign:

, i.e. .

Those. the work done by an external force during rotational motion of a body is equal to the product of the projection of the moment of the external force on the direction and angle of rotation.

On the other hand, the work of an external force acting on a body goes to increase the kinetic energy of the body (or is equal to the change in the kinetic energy of the rotating body). Let's show this:

;

Hence,

. (4.7)

On one's own:

Elastic forces;

Hooke's law.

LECTURE 7

Hydrodynamics

Current lines and tubes.

Hydrodynamics studies the movement of liquids, but its laws also apply to the movement of gases. In a stationary fluid flow, the speed of its particles at each point in space is a quantity independent of time and is a function of coordinates. In a steady flow, the trajectories of fluid particles form a streamline. The combination of current lines forms a current tube (Fig. 5.1). We assume that the fluid is incompressible, then the volume of fluid flowing through the sections S 1 and S 2 will be the same. In a second, a volume of liquid will pass through these sections equal to

, (5.1)

where and are the fluid velocities in sections S 1 and S 2 , and the vectors and are defined as and , where and are the normals to the sections S 1 and S 2. Equation (5.1) is called the jet continuity equation. It follows from this that the fluid speed is inversely proportional to the cross-section of the current tube.

Bernoulli's equation.

We will consider an ideal incompressible fluid in which internal friction(viscosity) is absent. Let us select a thin current tube in a stationary flowing liquid (Fig. 5.2) with sections S 1 And S 2, perpendicular to the streamlines. In cross section 1 in a short time t particles will move a distance l 1, and in section 2 - at a distance l 2. Through both sections in time t equal small volumes of liquid will pass through V= V 1 = V 2 and transfer a lot of liquid m=rV, Where r- liquid density. Overall change mechanical energy of all liquid in the flow tube between sections S 1 And S 2 that happened during t, can be replaced by changing the volume energy V that occurred when it moved from section 1 to section 2. With such a movement, the kinetic and potential energy of this volume will change, and the total change in its energy

, (5.2)

where v 1 and v 2 - velocities of fluid particles in sections S 1 And S 2 respectively; g- acceleration of gravity; h 1 And h 2- height of the center of the sections.

IN ideal liquid There are no friction losses, so the energy gain is DE must be equal to the work done by pressure forces on the allocated volume. In the absence of friction forces, this work:

Equating the right-hand sides of equalities (5.2) and (5.3) and transferring terms with the same indices to one side of the equality, we obtain

. (5.4)

Tube sections S 1 And S 2 were taken arbitrarily, therefore it can be argued that in any section of the current tube the expression is valid

. (5.5)

Equation (5.5) is called Bernoulli's equation. For a horizontal streamline h = const and equality (5.4) takes the form

r /2 + p 1 = r /2 + p2 , (5.6)

those. the pressure is less at those points where the speed is greater.

Internal friction forces.

A real liquid is characterized by viscosity, which manifests itself in the fact that any movement of liquid and gas spontaneously stops in the absence of the reasons that caused it. Let us consider an experiment in which a layer of liquid is located above a stationary surface, and on top of it moves at a speed of , a plate floating on it with a surface S(Fig. 5.3). Experience shows that in order to move a plate at a constant speed, it is necessary to act on it with a force. Since the plate does not receive acceleration, it means that the action of this force is balanced by another, equal in magnitude and oppositely directed force, which is the friction force . Newton showed that the force of friction

, (5.7)

Where d- thickness of the liquid layer, h - viscosity coefficient or coefficient of friction of the liquid, the minus sign takes into account the different directions of the vectors F tr And v o. If you examine the speed of liquid particles in different places of the layer, it turns out that it changes according to a linear law (Fig. 5.3):

v(z) = = (v 0 /d)·z.

Differentiating this equality, we get dv/dz= v 0 /d. With this in mind

formula (5.7) will take the form

F tr=- h(dv/dz)S , (5.8)

Where h- dynamic viscosity coefficient. Magnitude dv/dz called the velocity gradient. It shows how quickly the speed changes in the direction of the axis z. At dv/dz= const velocity gradient is numerically equal to the change in velocity v when it changes z per unit. Let us put numerically in formula (5.8) dv/dz =-1 and S= 1, we get h = F. this implies physical meaning h: the viscosity coefficient is numerically equal to the force that acts on a layer of liquid of unit area with a velocity gradient equal to unity. The SI unit of viscosity is called the pascal second (denoted Pa s). In the CGS system, the unit of viscosity is 1 poise (P), with 1 Pa s = 10P.

Let us first consider a rigid body rotating around a fixed axis OZ with angular velocity ω (Fig. 5.6). Let's break the body into elementary masses. The linear speed of the elementary mass is equal to , where is its distance from the axis of rotation. Kinetic energy i-that elementary mass will be equal to

The kinetic energy of the whole body is composed of the kinetic energies of its parts, therefore

Taking into account that the sum on the right side of this relation represents the moment of inertia of the body relative to the axis of rotation, we finally obtain

. (5.30)

Formulas for the kinetic energy of a rotating body (5.30) are similar to the corresponding formulas for the kinetic energy of translational motion of a body. They are obtained from the latter by a formal replacement .

In the general case, the motion of a rigid body can be represented as a sum of motions - translational at a speed equal to the speed of the body's center of mass, and rotation at an angular velocity around an instantaneous axis passing through the center of mass. In this case, the expression for the kinetic energy of the body takes the form

Let us now find the work done by the moment of external forces during the rotation of a rigid body. Elementary work of external forces in time dt will be equal to the change in kinetic energy of the body

Taking the differential from the kinetic energy of rotational motion, we find its increment

In accordance with the basic equation of dynamics for rotational motion

Taking into account these relations, we reduce the expression of elementary work to the form

where is the projection of the resulting moment of external forces on the direction of the axis of rotation OZ, is the angle of rotation of the body over the considered period of time.

Integrating (5.31), we obtain a formula for the work of external forces acting on a rotating body

If , then the formula simplifies

Thus, the work of external forces during rotation of a rigid body relative to a fixed axis is determined by the action of the projection of the moment of these forces onto this axis.

Gyroscope

A gyroscope is a rapidly rotating symmetrical body, the axis of rotation of which can change its direction in space. So that the axis of the gyroscope can rotate freely in space, the gyroscope is placed in the so-called gimbal suspension (Fig. 5.13). The gyroscope flywheel rotates in the inner ring around the axis C 1 C 2 passing through its center of gravity. The inner ring, in turn, can rotate in the outer ring around the axis B 1 B 2, perpendicular to C 1 C 2. Finally, the outer race can rotate freely in the bearings of the strut around the axis A 1 A 2, perpendicular to the axes C 1 C 2 and B 1 B 2. All three axes intersect at some fixed point O, called the center of the suspension or the fulcrum of the gyroscope. The gyroscope in a gimbal has three degrees of freedom and, therefore, can make any rotation around the center of the gimbal. If the center of the gyroscope's suspension coincides with its center of gravity, then the resulting moment of gravity of all parts of the gyroscope relative to the center of the suspension is zero. Such a gyroscope is called balanced.

Let us now consider the most important properties gyroscope, which have found wide application in various fields.

1) Stability.

For any rotation of the counterbalanced gyroscope, its axis of rotation remains unchanged in direction relative to the laboratory reference system. This is due to the fact that the moment of all external forces, equal to the moment of the friction forces, is very small and practically does not cause a change in the angular momentum of the gyroscope, i.e.

Since the angular momentum is directed along the axis of rotation of the gyroscope, its orientation must remain unchanged.

If the external force acts for a short time, then the integral that determines the increment in angular momentum will be small

. (5.34)

This means that under short-term influences of even large forces, the movement of a balanced gyroscope changes little. The gyroscope seems to resist any attempts to change the magnitude and direction of its angular momentum. This is due to the remarkable stability that the movement of the gyroscope acquires after it is brought into rapid rotation. This property of the gyroscope is widely used to automatically control the movement of aircraft, ships, missiles and other devices.

If the gyroscope is acted upon for a long time by a moment of external forces that is constant in direction, then the axis of the gyroscope is ultimately set in the direction of the moment of the external forces. This phenomenon is used in the gyrocompass. This device is a gyroscope, the axis of which can be freely rotated in a horizontal plane. Due to daily rotation Earth and the action of the moment of centrifugal forces, the axis of the gyroscope rotates so that the angle between and becomes minimal (Fig. 5.14). This corresponds to the position of the gyroscope axis in the meridian plane.

2). Gyroscopic effect.

If a pair of forces and is applied to a rotating gyroscope, tending to rotate it about an axis perpendicular to the axis of rotation, then it will begin to rotate around a third axis, perpendicular to the first two (Fig. 5.15). This unusual behavior of the gyroscope is called the gyroscopic effect. It is explained by the fact that the moment of the pair of forces is directed along the O 1 O 1 axis and the change in the vector by magnitude over time will have the same direction. As a result, the new vector will rotate relative to the O 2 O 2 axis. Thus, the behavior of the gyroscope, unnatural at first glance, fully corresponds to the laws of the dynamics of rotational motion

3). Precession of the gyroscope.

The precession of a gyroscope is the cone-shaped movement of its axis. It occurs in the case when the moment of external forces, remaining constant in magnitude, rotates simultaneously with the axis of the gyroscope, forming a right angle with it all the time. To demonstrate precession, a bicycle wheel with an extended axle set into rapid rotation can be used (Fig. 5.16).

If the wheel is suspended by the extended end of the axle, its axle will begin to precess around the vertical axis under the influence of its own weight. A rapidly rotating top can also serve as a demonstration of precession.

Let's find out the reasons for the precession of the gyroscope. Let's consider an unbalanced gyroscope, the axis of which can freely rotate around a certain point O (Fig. 5.16). The moment of gravity applied to the gyroscope is equal in magnitude

where is the mass of the gyroscope, is the distance from point O to the center of mass of the gyroscope, is the angle formed by the axis of the gyroscope with the vertical. The vector is directed perpendicular to the vertical plane passing through the axis of the gyroscope.

Under the influence of this moment, the angular momentum of the gyroscope (its origin is placed at point O) will receive an increment in time, and the vertical plane passing through the axis of the gyroscope will rotate by an angle. The vector is always perpendicular to , therefore, without changing in magnitude, the vector changes only in direction. However, after a while mutual arrangement vectors and will be the same as at the initial moment. As a result, the gyroscope axis will continuously rotate around the vertical, describing a cone. This movement is called precession.

Let us determine the angular velocity of precession. According to Fig. 5.16, the angle of rotation of the plane passing through the axis of the cone and the axis of the gyroscope is equal to

where is the angular momentum of the gyroscope, and is its increment over time.

Dividing by , taking into account the noted relations and transformations, we obtain the angular velocity of precession

. (5.35)

For gyroscopes used in technology, the angular velocity of precession is millions of times less than the rotation speed of the gyroscope.

In conclusion, we note that the phenomenon of precession is also observed in atoms due to the orbital motion of electrons.

Examples of application of the laws of dynamics

During rotational movement

1. Let's consider some examples on the law of conservation of angular momentum, which can be implemented using a Zhukovsky bench. In the simplest case, the Zhukovsky bench is a disk-shaped platform (chair), which can rotate freely around a vertical axis on ball bearings (Fig. 5.17). The demonstrator sits or stands on the bench, after which it is brought into rotation. Due to the fact that the friction forces due to the use of bearings are very small, the angular momentum of the system consisting of a bench and a demonstrator relative to the axis of rotation cannot change over time if the system is left to its own devices. If the demonstrator holds heavy dumbbells in his hands and spreads his arms to the sides, then he will increase the moment of inertia of the system, and therefore the angular velocity of rotation must decrease so that the angular momentum remains unchanged.

According to the law of conservation of angular momentum, we create an equation for this case

where is the moment of inertia of the person and the bench, and is the moment of inertia of the dumbbells in the first and second positions, and is the angular velocities of the system.

The angular speed of rotation of the system when raising dumbbells to the side will be equal to

The work done by a person when moving dumbbells can be determined through the change in the kinetic energy of the system

2. Let us give another experiment with the Zhukovsky bench. The demonstrator sits or stands on a bench and is handed a rapidly rotating wheel with a vertically directed axis (Fig. 5.18). The demonstrator then turns the wheel 180 0 . In this case, the change in the angular momentum of the wheel is entirely transferred to the bench and the demonstrator. As a result, the bench, together with the demonstrator, begins to rotate with an angular velocity determined on the basis of the law of conservation of angular momentum.

The angular momentum of the system in the initial state is determined only by the angular momentum of the wheel and is equal to

where is the moment of inertia of the wheel, and is the angular velocity of its rotation.

After turning the wheel through an angle of 180 0, the angular momentum of the system will be determined by the sum of the angular momentum of the bench with the person and the angular momentum of the wheel. Taking into account the fact that the angular momentum vector of the wheel has changed its direction to the opposite, and its projection onto the vertical axis has become negative, we obtain

where is the moment of inertia of the “person-platform” system, and is the angular velocity of rotation of the bench with the person.

According to the law of conservation of angular momentum

And .

As a result, we find the speed of rotation of the bench

3. Thin rod of mass m and length l rotates with an angular velocity ω=10 s -1 in the horizontal plane around a vertical axis passing through the middle of the rod. Continuing to rotate in the same plane, the rod moves so that the axis of rotation now passes through the end of the rod. Find the angular velocity in the second case.

In this problem, due to the fact that the distribution of the mass of the rod relative to the axis of rotation changes, the moment of inertia of the rod also changes. In accordance with the law of conservation of angular momentum of an isolated system, we have

Here is the moment of inertia of the rod relative to the axis passing through the middle of the rod; is the moment of inertia of the rod relative to the axis passing through its end and found by Steiner’s theorem.

Substituting these expressions into the law of conservation of angular momentum, we obtain

4. Rod length L=1.5 m and mass m 1=10 kg hingedly suspended from the upper end. A bullet with a mass of m 2=10 g, flying horizontally at a speed of =500 m/s, and gets stuck in the rod. At what angle will the rod deflect after the impact?

Let's imagine in Fig. 5.19. system of interacting bodies “rod-bullet”. The moments of external forces (gravity, axle reaction) at the moment of impact are equal to zero, so we can use the law of conservation of angular momentum

The angular momentum of the system before impact is equal to the angular momentum of the bullet relative to the suspension point

The angular momentum of the system after an inelastic impact is determined by the formula

where is the moment of inertia of the rod relative to the suspension point, is the moment of inertia of the bullet, is the angular velocity of the rod with the bullet immediately after the impact.

Solving the resulting equation after substitution, we find

Let us now use the law of conservation of mechanical energy. Let us equate the kinetic energy of the rod after a bullet hits it with its potential energy at the highest point of its rise:

where is the elevation height of the center of mass of this system.

Having carried out the necessary transformations, we obtain

The angle of deflection of the rod is related to the ratio

Having carried out the calculations, we get =0.1p=18 0 .

5. Determine the acceleration of the bodies and the tension of the thread on the Atwood machine, assuming that (Fig. 5.20). The moment of inertia of the block relative to the axis of rotation is equal to I, block radius r. Neglect the mass of the thread.

Let's arrange all the forces acting on the loads and the block, and draw up dynamic equations for them

If there is no slipping of the thread along the block, then the linear and angular acceleration are related to each other by the relation

Solving these equations, we get

Then we find T 1 and T 2.

6. A thread is attached to the pulley of the Oberbeck cross (Fig. 5.21), from which a load weighing M= 0.5 kg. Determine how long it takes for a load to fall from a height h=1 m to bottom position. Pulley radius r=3 cm. Four weights weighing m=250 g each at a distance R= 30 cm from its axis. The moment of inertia of the cross and the pulley itself is neglected in comparison with the moment of inertia of the loads.

The kinetic energy of a rotating body is equal to the sum of the kinetic energies of all particles of the body:

The mass of a particle, its linear (circumferential) speed, proportional to the distance of this particle from the axis of rotation. Substituting into this expression and taking the angular velocity o common for all particles out of the sum sign, we find:

This formula for the kinetic energy of a rotating body can be brought to a form similar to the expression for the kinetic energy of translational motion if we introduce the value of the so-called moment of inertia of the body. The moment of inertia of a material point is the product of the mass of the point and the square of its distance from the axis of rotation. The moment of inertia of a body is the sum of the moments of inertia of all material points of the body:

So, the kinetic energy of a rotating body is determined by the following formula:

Formula (2) differs from the formula that determines the kinetic energy of a body in translational motion in that instead of the mass of the body it includes the moment of inertia I and instead of the speed the group velocity

The large kinetic energy of a rotating flywheel is used in technology to maintain uniform running of the machine under suddenly changing loads. Initially, in order to bring a flywheel with a large moment of inertia into rotation, a significant amount of work is required from the machine, but when a large load is suddenly turned on, the machine does not stop and does the work using the reserve of kinetic energy of the flywheel.

Especially massive flywheels are used in rolling mills driven by an electric motor. Here is a description of one of these wheels: “The wheel has a diameter of 3.5 m and weighs. At a normal speed of 600 rpm, the reserve of kinetic energy of the wheel is such that at the moment of rolling the wheel gives the mill a power of 20,000 liters. With. Friction in the bearings is reduced to a minimum by the tale under pressure, and in order to avoid the harmful effects of centrifugal forces of inertia, the wheel is balanced so that a load placed on the circumference of the wheel brings it out of rest."

Let us present (without performing calculations) the values of the moments of inertia of some bodies (it is assumed that each of these bodies has the same density in all its areas).

The moment of inertia of a thin ring relative to an axis passing through its center and perpendicular to its plane (Fig. 55):

The moment of inertia of a circular disk (or cylinder) about an axis passing through its center and perpendicular to its plane (polar moment of inertia of the disk; Fig. 56):

The moment of inertia of a thin round disk relative to an axis coinciding with its diameter (equatorial moment of inertia of the disk; Fig. 57):

The moment of inertia of the ball relative to the axis passing through the center of the ball:

Moment of inertia of a thin spherical layer of radius about an axis passing through the center:

The moment of inertia of a thick spherical layer (a hollow ball having an outer surface radius and a cavity radius) about an axis passing through the center:

The moments of inertia of bodies are calculated using integral calculus. To give an idea of the progress of such calculations, let us find the moment of inertia of the rod relative to the axis perpendicular to it (Fig. 58). Let there be a cross section of the rod, density. Let us select an elementary small part of the rod, which has a length and is located at a distance x from the axis of rotation. Then its mass Since it is at a distance x from the axis of rotation, its moment of inertia is integrated over the range from zero to I:

Moment of inertia rectangular parallelepiped relative to the axis of symmetry (Fig. 59)

Moment of inertia of the ring torus (Fig. 60)

Let us consider how the rotational energy of a body rolling (without sliding) along a plane is related to the energy of the translational motion of this body,

The energy of translational motion of a rolling body is equal to , where is the mass of the body and the speed of translational motion. Let denote the angular velocity of rotation of a rolling body and the radius of the body. It is easy to understand that the speed of translational motion of a body rolling without sliding is equal to the peripheral speed of the body at the points of contact of the body with the plane (during the time when the body makes one revolution, the center of gravity of the body moves a distance, therefore,

Thus,

Rotation energy

hence,

Substituting here the above values of the moments of inertia, we find that:

a) the energy of the rotational motion of a rolling hoop is equal to the energy of its translational motion;

b) the rotational energy of a rolling homogeneous disk is equal to half the energy of translational motion;

c) the rotational energy of a rolling homogeneous ball is the energy of translational motion.

Dependence of the moment of inertia on the position of the axis of rotation. Let the rod (Fig. 61) with the center of gravity at point C rotate with angular velocity (o around the axis O, perpendicular to the plane of the drawing. Let us assume that over a certain period of time it has moved from position A B to and the center of gravity has described an arc. This movement of the rod can be considered as if the rod first translationally (i.e., remaining parallel to itself) moved to position and then rotated around C to position Let us denote (the distance of the center of gravity from the axis of rotation) by a, and the angle by When the rod moves from position A B to position, the movement of each of its particles is the same as the movement of the center of gravity, i.e., it is equal to or around an axis passing through O can be decomposed into two parts.

Lecture 3. Rigid body dynamics

Lecture outline

3.1. Moment of power.

3.2. Basic equations of rotational motion. Moment of inertia.

3.3. Kinetic energy of rotation.

3.4. Moment of impulse. Law of conservation of angular momentum.

3.5. Analogy between translational and rotational motion.

Moment of power

Let us consider the motion of a rigid body around a fixed axis. Let the rigid body have a fixed axis of rotation OO ( Fig.3.1) and an arbitrary force is applied to it.

Rice. 3.1

Let us decompose the force into two components of force, the force lies in the plane of rotation, and the force is parallel to the axis of rotation. Then we will decompose the force into two components: – acting along the radius vector and – perpendicular to it.

Not every force applied to a body will rotate it. The forces create pressure on the bearings, but do not rotate it.

A force may or may not throw a body out of balance, depending on where in the radius vector it is applied. Therefore, the concept of moment of force about an axis is introduced. A moment of power relative to the axis of rotation is called the vector product of the radius vector and the force.

The vector is directed along the axis of rotation and is determined by the cross product rule or the right screw rule or the gimlet rule.

Modulus of moment of force

where α is the angle between the vectors and .

From Fig. 3.1. it's clear that .

r 0 – shortest distance from the axis of rotation to the line of action of the force and is called the shoulder of the force. Then the moment of force can be written

M = F r 0 . (3.3)

From Fig. 3.1.

Where F– projection of the vector onto the direction perpendicular to the radius vector. In this case, the moment of force is equal to

. (3.4)

If several forces act on a body, then the resulting moment of force is equal to the vector sum of the moments of the individual forces, but since all moments are directed along the axis, they can be replaced by an algebraic sum. The moment will be considered positive if it rotates the body clockwise and negative if it rotates counterclockwise. If all moments of forces () are equal to zero, the body will be in equilibrium.

The concept of torque can be demonstrated using a "capricious coil". The spool of thread is pulled by the free end of the thread ( rice. 3.2).

Rice. 3.2

Depending on the direction of the thread tension, the spool rolls in one direction or another. If pulled at an angle α , then the moment of force about the axis ABOUT(perpendicular to the figure) rotates the coil counterclockwise and it rolls back. In case of tension at an angle β the torque is directed counterclockwise and the reel rolls forward.

Using the equilibrium condition (), we can construct simple mechanisms, which are “transformers” of force, i.e. By applying less force, you can lift and move loads of different weights. Levers, wheelbarrows, and various types of blocks, which are widely used in construction, are based on this principle. To maintain the equilibrium condition in construction cranes to compensate for the moment of force caused by the weight of the load, there is always a system of counterweights that creates a moment of force of the opposite sign.

3.2. Basic equation of rotation
movements. Moment of inertia

Consider an absolutely rigid body rotating around a fixed axis OO(Fig.3.3). Let us mentally divide this body into elements with masses Δ m 1, Δ m 2, …, Δ m n. When rotated, these elements will describe circles with radii r 1,r 2 , …,r n. Forces act on each element accordingly F 1,F 2 , …,Fn. Rotation of a body around an axis OO occurs under the influence of the full torque M.

M = M 1 + M 2 + … + M n (3.4)

Where M 1 = F 1 r 1, M 2 = F 2 r 2, ..., M n = F n r n

According to Newton's II law, every force F, acting on an element of mass D m, causes acceleration of this element a, i.e.

F i = D m i a i (3.5)

Substituting the corresponding values into (3.4), we obtain

Rice. 3.3

Knowing the relationship between linear angular acceleration ε () and that the angular acceleration is the same for all elements, formula (3.6) will have the form

M = (3.7)

=I (3.8)

I– moment of inertia of the body relative to the fixed axis.

Then we will get

M = I ε (3.9)

Or in vector form

(3.10)

This equation is the basic equation for the dynamics of rotational motion. It is similar in form to equation II of Newton's law. From (3.10) the moment of inertia is equal to

Thus, the moment of inertia of a given body is the ratio of the moment of force to the angular acceleration it causes. From (3.11) it is clear that the moment of inertia is a measure of the inertia of a body with respect to rotational motion. The moment of inertia plays the same role as mass in translational motion. SI unit [ I] = kg m 2. From formula (3.7) it follows that the moment of inertia characterizes the distribution of masses of body particles relative to the axis of rotation.

So, the moment of inertia of an element of mass ∆m moving in a circle of radius r is equal to

I = r 2 D m (3.12)

I= (3.13)

In the case of a continuous mass distribution, the sum can be replaced by the integral

I= ∫ r 2 dm (3.14)

where integration is performed over the entire body mass.

This shows that the moment of inertia of a body depends on the mass and its distribution relative to the axis of rotation. This can be demonstrated experimentally ( Fig.3.4).

Rice. 3.4

Two round cylinders, one hollow (for example, metal), the other solid (wooden) with the same lengths, radii and masses begin to roll simultaneously. A hollow cylinder, which has a large moment of inertia, will lag behind the solid one.

The moment of inertia can be calculated if the mass is known m and its distribution relative to the axis of rotation. The simplest case is a ring, when all elements of the mass are located equally from the axis of rotation ( rice. 3.5):

I = (3.15)

Rice. 3.5

Let us present expressions for the moments of inertia of various symmetrical bodies of mass m.

1. Moment of inertia rings, hollow thin-walled cylinder relative to the axis of rotation coinciding with the axis of symmetry.

, (3.16)

r– radius of the ring or cylinder

2. For a solid cylinder and disk, the moment of inertia about the axis of symmetry

(3.17)

3. Moment of inertia of the ball about an axis passing through the center

(3.18)

r– radius of the ball

4. Moment of inertia of a thin rod with a long length l relative to an axis perpendicular to the rod and passing through its middle

(3.19)

l– length of the rod.

If the axis of rotation does not pass through the center of mass, then the moment of inertia of the body relative to this axis is determined by Steiner’s theorem.

(3.20)

According to this theorem, the moment of inertia about an arbitrary axis O’O’ ( ) is equal to the moment of inertia about a parallel axis passing through the center of mass of the body ( ) plus the product of body mass times the square of the distance A between axes ( rice. 3.6).

Rice. 3.6

Kinetic energy of rotation

Let us consider the rotation of an absolutely rigid body around a fixed axis OO with angular velocity ω (rice. 3.7). Let's break the solid body into n elementary masses ∆ m i. Each element of mass rotates along a circle of radius r i with linear speed (). Kinetic energy consists of the kinetic energies of individual elements.

(3.21)

Rice. 3.7

Let us recall from (3.13) that – moment of inertia relative to the OO axis.

Thus, the kinetic energy of a rotating body

E k = (3.22)

We considered the kinetic energy of rotation around a fixed axis. If a body is involved in two movements: translational and rotational motion, then the kinetic energy of the body consists of the kinetic energy of translational motion and the kinetic energy of rotation.

For example, a ball of mass m rolls; the center of mass of the ball moves translationally at a speed u (rice. 3.8).

Rice. 3.8

The total kinetic energy of the ball will be equal to

(3.23)

3.4. Moment of impulse. Conservation Law
angular momentum

Physical quantity equal to the product of the moment of inertia I to angular velocity ω , is called angular momentum (angular momentum) L relative to the axis of rotation.

– angular momentum is a vector quantity and its direction coincides with the direction of angular velocity.

Differentiating equation (3.24) with respect to time, we obtain

Where, M– total moment of external forces. In an isolated system there is no torque of external forces ( M=0) and

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Two cases of transformation of the mechanical motion of a material point or system of points:

mechanical movement is transferred from one mechanical system to another as a mechanical movement;
mechanical motion turns into another form of motion of matter (into the form of potential energy, heat, electricity, etc.).

When the transformation of mechanical motion without its transition to another form of motion is considered, the measure of mechanical motion is the vector of momentum of a material point or mechanical system. The measure of the force in this case is the vector of the force impulse.

When mechanical motion turns into another form of motion of matter, the kinetic energy of a material point or mechanical system acts as a measure of mechanical motion. The measure of the action of force when transforming mechanical motion into another form of motion is the work of force

Kinetic energy

Kinetic energy is the body's ability to overcome an obstacle while moving.

Kinetic energy of a material point

The kinetic energy of a material point is a scalar quantity that is equal to half the product of the mass of the point and the square of its speed.

Kinetic energy:

characterizes both translational and rotational movements;
does not depend on the direction of movement of the points of the system and does not characterize changes in these directions;
characterizes the action of both internal and external forces.

Kinetic energy of a mechanical system

The kinetic energy of the system is equal to the sum of the kinetic energies of the bodies of the system. Kinetic energy depends on the type of motion of the bodies of the system.

Determination of the kinetic energy of a solid body at different types movements movements.

Kinetic energy of translational motion
During translational motion, the kinetic energy of the body is equal to T=m V 2 /2.

The measure of the inertia of a body during translational motion is mass.

Kinetic energy of rotational motion of a body

During the rotational motion of a body, kinetic energy is equal to half the product of the moment of inertia of the body relative to the axis of rotation and the square of its angular velocity.

A measure of the inertia of a body during rotational motion is the moment of inertia.

The kinetic energy of a body does not depend on the direction of rotation of the body.

Kinetic energy of plane-parallel motion of a body

With plane-parallel motion of a body, the kinetic energy is equal to

Work of force

The work of force characterizes the action of a force on a body during some movement and determines the change in the velocity modulus of a moving point.

Elementary work of force

The elementary work of a force is defined as a scalar quantity equal to the product of the projection of the force onto the tangent to the trajectory, directed in the direction of motion of the point, and the infinitesimal displacement of the point, directed along this tangent.

Work done by force on final displacement

The work done by a force on a final displacement is equal to the sum of its work on elementary sections.

The work of a force on a final displacement M 1 M 0 is equal to the integral of the elementary work along this displacement.

The work of a force on displacement M 1 M 2 is depicted by the area of the figure, limited by the abscissa axis, the curve and the ordinates corresponding to the points M 1 and M 0.

The unit of measurement for the work of force and kinetic energy in the SI system is 1 (J).

Theorems about the work of force

Theorem 1. The work done by the resultant force on a certain displacement is equal to the algebraic sum of the work done by the component forces on the same displacement.

Theorem 2. The work done by a constant force on the resulting displacement is equal to the algebraic sum of the work done by this force on the component displacements.

Power

Power is a quantity that determines the work done by a force per unit of time.

The unit of power measurement is 1W = 1 J/s.

Cases of determining the work of forces

Work of internal forces

The sum of the work done by the internal forces of a rigid body during any movement is zero.

Work of gravity

Work of elastic force

Work of friction force

Work of forces applied to a rotating body

The elementary work of forces applied to a rigid body rotating around a fixed axis is equal to the product of the main moment of external forces relative to the axis of rotation and the increment in the angle of rotation.

Rolling resistance

In the contact zone of the stationary cylinder and the plane, local deformation of contact compression occurs, the stress is distributed according to an elliptical law, and the line of action of the resultant N of these stresses coincides with the line of action of the load force on the cylinder Q. When the cylinder rolls, the load distribution becomes asymmetrical with a maximum shifted towards movement. The resultant N is displaced by the amount k - the arm of the rolling friction force, which is also called the rolling friction coefficient and has the dimension of length (cm)

Theorem on the change in kinetic energy of a material point

The change in the kinetic energy of a material point at a certain displacement is equal to the algebraic sum of all forces acting on the point at the same displacement.

Theorem on the change in kinetic energy of a mechanical system

The change in the kinetic energy of a mechanical system at a certain displacement is equal to the algebraic sum of the robot internal and external forces acting on material points systems at the same displacement.

Theorem on the change in kinetic energy of a solid body

The change in the kinetic energy of a rigid body (unchanged system) at a certain displacement is equal to the sum of the external forces acting on points of the system at the same displacement.

Efficiency

Forces acting in mechanisms

Forces and pairs of forces (moments) that are applied to a mechanism or machine can be divided into groups:

1. Driving forces and moments that perform positive work (applied to the driving links, for example, gas pressure on the piston in an internal combustion engine).

2. Forces and moments of resistance that perform negative work:

useful resistance (they perform the work required from the machine and are applied to the driven links, for example, the resistance of the load lifted by the machine),
resistance forces (for example, friction forces, air resistance, etc.).

3. Gravity forces and elastic forces of springs (both positive and negative work, while the work for a full cycle is zero).

4. Forces and moments applied to the body or stand from the outside (reaction of the foundation, etc.), which do not do work.

5. Interaction forces between links acting in kinematic pairs.

6. The inertial forces of the links, caused by the mass and movement of the links with acceleration, can perform positive, negative work and do not perform work.

Work of forces in mechanisms

When the machine operates at a steady state, its kinetic energy does not change and the sum of the work of the driving forces and resistance forces applied to it is zero.

The work expended in setting the machine in motion is expended in overcoming useful and harmful resistances.

Mechanism efficiency

The mechanical efficiency during steady motion is equal to the ratio of the useful work of the machine to the work expended on setting the machine in motion:

Machine elements can be connected in series, parallel and mixed.

Efficiency in series connection

When mechanisms are connected in series, the overall efficiency is less than the lowest efficiency of an individual mechanism.

Efficiency in parallel connection

When mechanisms are connected in parallel, the overall efficiency is greater than the lowest and less than the highest efficiency of an individual mechanism.

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