Kinetic energy of rotation. Rotational energy

« Physics - 10th grade"

Why does a skater stretch along the axis of rotation to increase the angular velocity of rotation?
Should a helicopter rotate when its rotor rotates?

The questions asked suggest that if external forces do not act on the body or their action is compensated and one part of the body begins to rotate in one direction, then the other part should rotate in the other direction, just as when fuel is ejected from a rocket, the rocket itself moves in the opposite direction.

Moment of impulse.

If we consider a rotating disk, it becomes obvious that the total momentum of the disk is zero, since any particle of the body corresponds to a particle moving with an equal velocity, but in the opposite direction (Fig. 6.9).

But the disk is moving, the angular velocity of rotation of all particles is the same. However, it is clear that the further a particle is from the axis of rotation, the greater its momentum. Therefore, for rotational movement it is necessary to introduce another characteristic similar to impulse - angular momentum.

The angular momentum of a particle moving in a circle is the product of the particle’s momentum and the distance from it to the axis of rotation (Fig. 6.10):

Linear and angular velocities are related by the relation v = ωr, then

All points of a solid object move relative to a fixed axis of rotation with the same angular velocity. A solid body can be represented as a collection of material points.

Momentum solid equal to the product of the moment of inertia and the angular velocity of rotation:

Angular momentum is a vector quantity; according to formula (6.3), angular momentum is directed in the same way as the angular velocity.

The basic equation for the dynamics of rotational motion in pulse form.

The angular acceleration of a body is equal to the change in angular velocity divided by the period of time during which this change occurred: Substitute this expression into the basic equation of the dynamics of rotational motion hence I(ω 2 - ω 1) = MΔt, or IΔω = MΔt.

Thus,

ΔL = MΔt. (6.4)

The change in angular momentum is equal to the product of the total moment of forces acting on a body or system and the time of action of these forces.

Law of conservation of angular momentum:

If the total moment of forces acting on a body or system of bodies having a fixed axis of rotation is equal to zero, then the change in angular momentum is also zero, i.e., the angular momentum of the system remains constant.

ΔL = 0, L = const.

The change in the momentum of the system is equal to the total momentum of the forces acting on the system.

A rotating skater spreads his arms out to the sides, thereby increasing the moment of inertia to reduce the angular velocity of rotation.

The law of conservation of angular momentum can be demonstrated using the following experiment, called the “Zhukovsky bench experiment.” A person stands on a bench that has a vertical axis of rotation passing through its center. A man holds dumbbells in his hands. If the bench is made to rotate, the person can change the speed of rotation by pressing the dumbbells to the chest or lowering the arms and then raising them. By spreading his arms, he increases the moment of inertia, and the angular speed of rotation decreases (Fig. 6.11, a), lowering his arms, he reduces the moment of inertia, and the angular speed of rotation of the bench increases (Fig. 6.11, b).

A person can also make a bench rotate by walking along its edge. In this case, the bench will rotate in the opposite direction, since the total angular momentum should remain equal to zero.

The principle of operation of devices called gyroscopes is based on the law of conservation of angular momentum. The main property of a gyroscope is the preservation of the direction of the rotation axis if external forces do not act on this axis. In the 19th century Gyroscopes were used by sailors for orientation at sea.

Kinetic energy of a rotating rigid body.

The kinetic energy of a rotating solid body is equal to the sum of the kinetic energies of its individual particles. Let us divide the body into small elements, each of which can be considered a material point. Then the kinetic energy of the body is equal to the sum of the kinetic energies of the material points of which it consists:

The angular velocity of rotation of all points of the body is the same, therefore,

The value in parentheses, as we already know, is the moment of inertia of the rigid body. Finally, the formula for the kinetic energy of a rigid body having a fixed axis of rotation has the form

In the general case of motion of a rigid body, when the axis of rotation is free, its kinetic energy is equal to the sum of the energies of translational and rotational motion. Thus, the kinetic energy of a wheel, the mass of which is concentrated in the rim, rolling along the road with constant speed, is equal

The table compares the formulas for the mechanics of translational motion of a material point with similar formulas for the rotational motion of a rigid body.

The main dynamic characteristics of rotational motion - angular momentum relative to the axis of rotation z:

and kinetic energy

In general, the energy during rotation with angular velocity is found by the formula:

, where is the inertia tensor.

In thermodynamics

By exactly the same reasoning as in the case of translational motion, equipartition implies that at thermal equilibrium the average rotational energy of each particle of a monatomic gas is: (3/2)k B T. Similarly, the equipartition theorem allows us to calculate the root mean square angular velocity of molecules.

§ 3.6 Work of external forces during rotation of a rigid body

When a rigid body rotates, its potential energy does not change, therefore the elementary work of external forces is equal to the increment in the kinetic energy of the body:

dA = dE or

Taking into account that Jβ = M, ωdr = dφ, we have α of the body at a finite angle φ is equal to

(3.25)

When a rigid body rotates around a fixed axis, the work of external forces is determined by the action of the moment of these forces relative to this axis. If the moment of forces relative to the axis is zero, then these forces do not produce work.

Examples of problem solving

Example 2.1. Flywheel massm=5kg and radiusr= 0.2 m rotates around a horizontal axis with frequencyν 0 =720 min -1 and when braking it stops behindt=20 s. Find the braking torque and the number of revolutions before stopping.

To determine the braking torque, we apply the basic equation of the dynamics of rotational motion

where I=mr 2 – moment of inertia of the disk; Δω =ω - ω 0, and ω =0 is the final angular velocity, ω 0 =2πν 0 is the initial. M is the braking moment of forces acting on the disk.

Knowing all the quantities, you can determine the braking torque

Mr 2 2πν 0 = МΔt (1)

(2)

From the kinematics of rotational motion, the angle of rotation during the rotation of the disk before stopping can be determined by the formula

(3)

where β is the angular acceleration.

According to the conditions of the problem: ω =ω 0 – βΔt, since ω=0, ω 0 = βΔt

Then expression (2) can be written as:

Example 2.2. Two flywheels in the form of disks of identical radii and masses were spun up to a rotation speedn= 480 rpm and left to our own devices. Under the influence of the friction forces of the shafts on the bearings, the first stopped throught=80 s, and the second one didN= 240 rpm to stop. Which flywheel had a greater moment of friction between the shafts and bearings and by how many times?

We will find the moment of forces of the thorn M 1 of the first flywheel using the basic equation of the dynamics of rotational motion

M 1 Δt = Iω 2 - Iω 1

where Δt is the time of action of the moment of friction forces, I=mr 2 is the moment of inertia of the flywheel, ω 1 = 2πν and ω 2 = 0 – the initial and final angular velocities of the flywheels

Then

The moment of friction forces M 2 of the second flywheel will be expressed through the connection between the work A of the friction forces and the change in its kinetic energy ΔE k:

where Δφ = 2πN is the angle of rotation, N is the number of revolutions of the flywheel.

Then where from

ABOUT the ratio will be equal

The frictional moment of the second flywheel is 1.33 times greater.

Example 2.3. Mass of a homogeneous solid disk m, mass of loads m 1 and m 2 (Fig. 15). There is no slipping or friction of the thread in the cylinder axis. Find the acceleration of the loads and the ratio of the thread tensionsin the process of movement.

There is no slipping of the thread, therefore, when m 1 and m 2 make translational motion, the cylinder will rotate about the axis passing through point O. Let us assume for definiteness that m 2 > m 1.

Then the load m 2 is lowered and the cylinder rotates clockwise. Let us write down the equations of motion of the bodies included in the system

The first two equations are written for bodies with masses m 1 and m 2 undergoing translational motion, and the third equation is written for a rotating cylinder. In the third equation on the left is the total moment of forces acting on the cylinder (the moment of force T 1 is taken with a minus sign, since the force T 1 tends to rotate the cylinder counterclockwise). On the right I is the moment of inertia of the cylinder relative to the O axis, which is equal to

where R is the radius of the cylinder; β is the angular acceleration of the cylinder.

Since there is no thread slippage, then
. Taking into account the expressions for I and β, we obtain:

Adding the equations of the system, we arrive at the equation

From here we find the acceleration a cargo

From the resulting equation it is clear that the thread tensions will be the same, i.e. =1 if the mass of the cylinder is much less than the mass of the loads.

Example 2.4. A hollow ball with mass m = 0.5 kg has an outer radius R = 0.08 m and an inner radius r = 0.06 m. The ball rotates around an axis passing through its center. At a certain moment, a force begins to act on the ball, as a result of which the angle of rotation of the ball changes according to the law
. Determine the moment of the applied force.

We solve the problem using the basic equation of the dynamics of rotational motion
. The main difficulty is to determine the moment of inertia of a hollow ball, and we find the angular acceleration β as
. The moment of inertia I of a hollow ball is equal to the difference between the moments of inertia of a ball of radius R and a ball of radius r:

where ρ is the density of the ball material. Finding the density by knowing the mass of a hollow ball

From here we determine the density of the ball material

For the moment of force M we obtain the following expression:

Example 2.5. A thin rod with a mass of 300 g and a length of 50 cm rotates with an angular velocity of 10 s -1 in a horizontal plane around a vertical axis passing through the middle of the rod. Find the angular velocity if, during rotation in the same plane, the rod moves so that the axis of rotation passes through the end of the rod.

We use the law of conservation of angular momentum

(1)

(J i is the moment of inertia of the rod relative to the axis of rotation).

For an isolated system of bodies, the vector sum of angular momentum remains constant. 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 (1):

J 0 ω 1 = J 2 ω 2 . (2)

It is known that the moment of inertia of the rod relative to the axis passing through the center of mass and perpendicular to the rod is equal to

J 0 = mℓ 2 /12. (3)

According to Steiner's theorem

J =J 0 +m A 2

(J is the moment of inertia of the rod relative to an arbitrary axis of rotation; J 0 is the moment of inertia relative to a parallel axis passing through the center of mass; A- distance from the center of mass to the selected axis of rotation).

Let us find the moment of inertia about the axis passing through its end and perpendicular to the rod:

J 2 =J 0 +m A 2, J 2 = mℓ 2 /12 +m(ℓ/2) 2 = mℓ 2 /3. (4)

Let's substitute formulas (3) and (4) into (2):

mℓ 2 ω 1 /12 = mℓ 2 ω 2 /3

ω 2 = ω 1 /4 ω 2 =10s-1/4=2.5s -1

Example 2.6 . Man of massm=60kg, standing on the edge of a platform with mass M=120kg, rotating by inertia around a fixed vertical axis with frequency ν 1 =12min -1 , moves to its center. Considering the platform to be a round homogeneous disk and the person to be a point mass, determine with what frequency ν 2 the platform will then rotate.

Given: m=60kg, M=120kg, ν 1 =12min -1 = 0.2s -1 .

Find:ν 1

Solution: According to the conditions of the problem, the platform with the person rotates by inertia, i.e. the resulting moment of all forces applied to the rotating system is zero. Therefore, for the “platform-person” system the law of conservation of angular momentum is satisfied

I 1 ω 1 = I 2 ω 2

Where
- moment of inertia of the system when a person stands on the edge of the platform (take into account that the moment of inertia of the platform is equal to (R – radius n
platform), the moment of inertia of a person at the edge of the platform is mR 2).

- moment of inertia of the system when a person stands in the center of the platform (take into account that the moment of a person standing in the center of the platform is zero). Angular velocity ω 1 = 2π ν 1 and ω 1 = 2π ν 2.

Substituting the written expressions into formula (1), we obtain

where does the desired rotation speed come from?

Answer: ν 2 =24min -1.

Kinetic energy is an additive quantity. Therefore, the kinetic energy of a body moving in an arbitrary manner is equal to the sum of the kinetic energies of all n material points into which this body can be mentally divided:

If a body rotates around a stationary axis z with angular velocity, then the linear speed i-th point , Ri – distance to the axis of rotation. Hence,

By comparison, we can see that the moment of inertia of body I is a measure of inertia during rotational motion, just as mass m is a measure of inertia during translational motion.

In the general case, the motion of a rigid body can be represented as the sum of two motions - translational with speed vc and rotational with angular speed ω around the instantaneous axis passing through the center of inertia. Then the total kinetic energy of this body

Here Ic is the moment of inertia about the instantaneous axis of rotation passing through the center of inertia.

The basic law of the dynamics of rotational motion.

Dynamics of rotational motion

The basic law of the dynamics of rotational motion:

or M=Je, where M is the moment of force M=[ r · F ] , J - moment of inertia is the moment of momentum of a body.

if M(external)=0 - the law of conservation of angular momentum. - kinetic energy of a rotating body.

work in rotational motion.

Law of conservation of angular momentum.

The angular momentum (momentum of motion) of a material point A relative to a fixed point O is called physical quantity, defined by the vector product:

where r is the radius vector drawn from point O to point A, p=mv is the momentum of the material point (Fig. 1); L is a pseudo-vector, the direction of which coincides with the direction of translational motion of the right propeller as it rotates from r to r.

Modulus of the angular momentum vector

where α is the angle between vectors r and p, l is the arm of vector p relative to point O.

The angular momentum relative to a fixed axis z is the scalar quantity Lz, equal to the projection onto this axis of the angular momentum vector defined relative to an arbitrary point O of this axis. The angular momentum Lz does not depend on the position of point O on the z axis.

When an absolutely rigid body rotates around a fixed axis z, each point of the body moves along a circle of constant radius ri with a speed vi. The velocity vi and momentum mivi are perpendicular to this radius, i.e. the radius is an arm of the vector mivi. This means that we can write that the angular momentum of an individual particle is equal to

and is directed along the axis in the direction determined by the right screw rule.

The momentum of a solid body relative to an axis is the sum of the angular momentum of individual particles:

Using the formula vi = ωri, we get

Thus, the angular momentum of a rigid body relative to an axis is equal to the moment of inertia of the body relative to the same axis, multiplied by the angular velocity. Let us differentiate equation (2) with respect to time:

This formula is another form of the equation for the dynamics of the rotational motion of a rigid body relative to a fixed axis: the derivative of the angular momentum of a rigid body relative to the axis is equal to the moment of force relative to the same axis.

It can be shown that there is a vector equality

In a closed system, the moment of external forces M = 0 and from where

Expression (4) represents the law of conservation of angular momentum: the angular momentum of a closed-loop system is conserved, that is, it does not change over time.

The law of conservation of angular momentum, as well as the law of conservation of energy, is a fundamental law of nature. It is associated with the property of symmetry of space - its isotropy, i.e. with the invariance of physical laws with respect to the choice of the direction of the coordinate axes of the reference system (relative to the rotation of a closed system in space at any angle).

Here we will demonstrate the law of conservation of angular momentum using a Zhukovsky bench. A person sitting on a bench rotating around a vertical axis and holding dumbbells in outstretched arms (Fig. 2) is rotated by an external mechanism with an angular velocity ω1. If a person presses the dumbbells to his body, the moment of inertia of the system will decrease. But the moment of external forces is zero, the angular momentum of the system is conserved and the angular velocity of rotation ω2 increases. Similarly, during an overhead jump, a gymnast presses his arms and legs towards his body in order to reduce his moment of inertia and thereby increase the angular velocity of rotation.

Pressure in liquid and gas.

Gas molecules, performing a chaotic, chaotic movement, are not connected or are rather weakly connected by interaction forces, which is why they move almost freely and, as a result of collisions, scatter in all directions, while filling the entire volume provided to them, i.e. the volume of gas is determined by the volume container occupied by gas.

And the liquid, having a certain volume, takes the shape of the vessel in which it is enclosed. But unlike gases in liquids, the average distance between molecules remains constant on average, so the liquid has a practically unchanged volume.

The properties of liquids and gases are very different in many ways, but in several mechanical phenomena their properties are determined by the same parameters and identical equations. For this reason, hydroaeromechanics is a branch of mechanics that studies the equilibrium and movement of gases and liquids, the interaction between them and between the solid bodies flowing around them, i.e. a unified approach to the study of liquids and gases is applied.

In mechanics, liquids and gases are considered with a high degree of accuracy as solid, continuously distributed in the part of space they occupy. For gases, density depends significantly on pressure. It has been established from experience. that the compressibility of liquid and gas can often be neglected and it is advisable to use unified concept- incompressibility of a liquid - a liquid with the same density everywhere, which does not change over time.

We place a thin plate at rest, as a result, parts of the liquid located along different sides from the plate, will act on each of its elements ΔS with forces ΔF, which will be equal in magnitude and directed perpendicular to the platform ΔS, regardless of the orientation of the platform, otherwise the presence of tangential forces would cause the fluid particles to move (Fig. 1)

A physical quantity determined by the normal force acting on the part of a liquid (or gas) per unit area is called pressure p/ of the liquid (or gas): p=ΔF/ΔS.

The unit of pressure is pascal (Pa): 1 Pa is equal to the pressure created by a force of 1 N, which is uniformly distributed over a surface normal to it with an area of 1 m2 (1 Pa = 1 N/m2).

Pressure in the equilibrium of liquids (gases) obeys Pascal's law: the pressure in any place of a liquid at rest is the same in all directions, and the pressure is equally transmitted throughout the entire volume occupied by the liquid at rest.

Let us study the influence of the weight of the liquid on the pressure distribution inside a stationary incompressible liquid. When a fluid is in equilibrium, the pressure along any horizontal line is always the same, otherwise there would be no equilibrium. This means that the free surface of a liquid at rest is always horizontal (we do not take into account the attraction of the liquid by the walls of the vessel). If a fluid is incompressible, then the density of the fluid does not depend on pressure. Then, with a cross section S of the liquid column, its height h and density ρ, the weight P=ρgSh, while the pressure on the lower base: p=P/S=ρgSh/S=ρgh, (1)

that is, pressure varies linearly with altitude. The pressure ρgh is called hydrostatic pressure.

According to formula (1), the pressure force on the lower layers of the liquid will be greater than on the upper layers, therefore, a body immersed in a liquid is acted upon by a force determined by Archimedes’ law: a body immersed in a liquid (gas) is acted upon by a directed force from the side of this liquid upward buoyant force, equal to weight liquid (gas) displaced by the body: FA = ρgV, where ρ is the density of the liquid, V is the volume of the body immersed in the liquid.

Mechanics.

Question No. 1

Reference system. Inertial reference systems. The principle of relativity of Galileo - Einstein.

Frame of reference- this is a set of bodies in relation to which movement is described given body and the associated coordinate system.

Inertial reference system (IRS) is a system in which a freely moving body is in a state of rest or uniform rectilinear motion.

Galileo-Einstein principle of relativity- All natural phenomena in any inertial frame of reference occur in the same way and have the same mathematical form. In other words, all ISOs are equal.

Question No. 2

Equation of motion. Types of motion of a rigid body. The main task of kinematics.

Equations of motion of a material point:

- kinematic equation of motion

Types of rigid body motion:

1) Translational motion - any straight line drawn in the body moves parallel to itself.

2) Rotational movement - any point of the body moves in a circle.

φ = φ(t)

The main task of kinematics- this is obtaining the time dependence of the velocity V = V(t) and the coordinates (or radius vector) r = r(t) of a material point from the known time dependence of its acceleration a = a(t) and the known initial conditions V 0 and r 0 .

Question No. 7

Pulse (Quantity of movement) is a vector physical quantity characterizing the measure of mechanical motion of a body. In classical mechanics, the momentum of a body is equal to the product of mass m this point by its speed v, the direction of the impulse coincides with the direction of the velocity vector:

In theoretical mechanics generalized impulse is the partial derivative of the Lagrangian of the system with respect to the generalized velocity

If the Lagrangian of the system does not depend on some generalized coordinates, then due to Lagrange equations .

For a free particle, the Lagrange function has the form: , hence:

The independence of the Lagrangian of a closed system from its position in space follows from the property homogeneity of space: for a well-isolated system, its behavior does not depend on where in space we place it. By Noether's theorem From this homogeneity follows the conservation of some physical quantity. This quantity is called impulse (ordinary, not generalized).

In classical mechanics, complete impulse system of material points is called a vector quantity equal to the sum of the products of the masses of material points and their speed:

accordingly, the quantity is called the momentum of one material point. This is a vector quantity directed in the same direction as the particle velocity. The International System of Units (SI) unit of impulse is kilogram-meter per second(kg m/s)

If we are dealing with a body of finite size, to determine its momentum it is necessary to break the body into small parts, which can be considered material points and summed over them, as a result we get:

The impulse of a system that is not affected by any external forces (or they are compensated) saved in time:

Conservation of momentum in this case follows from Newton’s second and third laws: by writing Newton’s second law for each of the material points composing the system and summing over all the material points composing the system, by virtue of Newton’s third law we obtain equality (*).

In relativistic mechanics, the three-dimensional momentum of a system of non-interacting material points is the quantity

Where m i- weight i th material point.

For a closed system of non-interacting material points, this value is preserved. However, three-dimensional momentum is not a relativistically invariant quantity, since it depends on the reference frame. A more meaningful quantity will be the four-dimensional momentum, which for one material point is defined as

In practice, the following relationships between mass, momentum and energy of a particle are often used:

In principle, for a system of non-interacting material points, their 4-moments are summed. However, for interacting particles in relativistic mechanics, it is necessary to take into account not only the momentum of the particles that make up the system, but also the momentum of the interaction field between them. Therefore, a much more meaningful quantity in relativistic mechanics is the energy-momentum tensor, which in to the fullest satisfies conservation laws.

Question #8

Moment of inertia- a scalar physical quantity, a measure of the inertia of a body in rotational motion around an axis, just as the mass of a body is a measure of its inertia in translational motion. Characterized by the distribution of masses in the body: moment of inertia equal to the sum products of elementary masses by the square of their distances to the base set

Axial moment of inertia

Axial moments of inertia of some bodies.

Moment of inertia of a mechanical system relative to a fixed axis (“axial moment of inertia”) is the quantity J a, equal to the sum of the products of the masses of all n material points of the system by the squares of their distances to the axis:

m i- weight i th point,
r i- distance from i th point to the axis.

Axial moment of inertia body J a is a measure of the inertia of a body in rotational motion around an axis, just as the mass of a body is a measure of its inertia in translational motion.

dm = ρ dV- mass of a small element of body volume dV,
ρ - density,
r- distance from element dV to axis a.

If the body is homogeneous, that is, its density is the same everywhere, then

Derivation of the formula

dm and moments of inertia dJ i. Then

Thin-walled cylinder (ring, hoop)

Derivation of the formula

The moment of inertia of a body is equal to the sum of the moments of inertia of its constituent parts. Divide a thin-walled cylinder into elements with mass dm and moments of inertia dJ i. Then

Since all elements of a thin-walled cylinder are at the same distance from the axis of rotation, formula (1) is transformed into the form

Steiner's theorem

Moment of inertia of a solid body relative to any axis depends not only on the mass, shape and size of the body, but also on the position of the body relative to this axis. According to Steiner's theorem (Huygens-Steiner theorem), moment of inertia body J relative to an arbitrary axis is equal to the sum moment of inertia this body J c relative to an axis passing through the center of mass of the body parallel to the axis under consideration, and the product of the body mass m per square of distance d between axes:

If is the moment of inertia of a body relative to an axis passing through the center of mass of the body, then the moment of inertia relative to a parallel axis located at a distance from it is equal to

where is the total body mass.

For example, the moment of inertia of a rod relative to an axis passing through its end is equal to:

Rotational energy

Kinetic energy of rotational motion- the energy of a body associated with its rotation.

The main kinematic characteristics of the rotational motion of a body are its angular velocity (ω) and angular acceleration. The main dynamic characteristics of rotational motion - angular momentum relative to the axis of rotation z:

K z = I zω

and kinetic energy

where I z is the moment of inertia of the body relative to the axis of rotation.

A similar example can be found when considering a rotating molecule with principal axes of inertia I 1, I 2 And I 3. The rotational energy of such a molecule is given by the expression

Where ω 1, ω 2, And ω 3- the main components of angular velocity.

In general, the energy during rotation with angular velocity is found by the formula:

, Where I- inertia tensor.

Question No. 9

Moment of impulse (angular momentum, angular momentum, orbital momentum, angular momentum) characterizes the amount of rotational motion. A quantity that depends on how much mass is rotating, how it is distributed relative to the axis of rotation, and at what speed the rotation occurs.

It should be noted that rotation here is understood in a broad sense, not only as regular rotation around an axis. For example, even with straight motion body past an arbitrary imaginary point not lying on the line of motion, it also has angular momentum. Perhaps the greatest role is played by angular momentum in describing the actual rotational motion. However, it is extremely important for a much wider class of problems (especially if the problem has central or axial symmetry, but not only in these cases).

Law of conservation of angular momentum(law of conservation of angular momentum) - the vector sum of all angular momentum relative to any axis for a closed system remains constant in the case of equilibrium of the system. In accordance with this, the angular momentum of a closed system relative to any non-derivative of the angular momentum with respect to time is the moment of force:

Thus, the requirement that the system be closed can be weakened to the requirement that the main (total) moment of external forces be equal to zero:

where is the moment of one of the forces applied to the system of particles. (But of course, if there are no external forces at all, this requirement is also satisfied).

Mathematically, the law of conservation of angular momentum follows from the isotropy of space, that is, from the invariance of space with respect to rotation through an arbitrary angle. When rotated by an arbitrary infinitesimal angle, the radius vector of the particle with number will change by , and the speed - . The Lagrange function of the system will not change with such a rotation, due to the isotropy of space. That's why

Kinetic energy of rotation. Rotational energy

In thermodynamics

see also

See what “Energy of rotational motion” is in other dictionaries:

§ 3.6 Work of external forces during rotation of a rigid body

Examples of problem solving