The angular momentum of the body relative to the fixed axis of rotation

Definition

angular momentum- vector physical quantity characterizing the momentum, numerically equal to the vector product
Angular moment about a point is a pseudovector, and angular momentum about the axis is a pseudoscalar.

The angular momentum of a closed system is conserved.
This quantity is called the angular momentum about the axis.

Law of conservation of angular momentum(law of conservation of angular momentum) is one of the fundamental laws of conservation. Expressed mathematically in terms of the vector sum of all angular momenta about a chosen axis for a closed system of bodies and remains constant until the system is affected external forces. In accordance with this, the angular momentum of a closed system in any coordinate system does not change with time. Simplified: if the system is in equilibrium.

Let's first define isotropy to advance further in the study.

Isotropy is one of the key properties of space in classical mechanics. The space is called isotropic if the rotation of the reference system by an arbitrary angle does not lead to a change in the measurement results.

The law of conservation of angular momentum is a manifestation of the isotropy of space with respect to rotation.
The law of conservation of angular momentum is a fundamental law of nature. The validity of this law is determined by the property of the symmetry of space - its isotropy, i.e. with invariance physical laws regarding the choice of the direction of the coordinate axes of the reference system.

Example

The validity of the law of conservation of angular momentum relative to a fixed axis of rotation can be demonstrated by experiment with the Zhukovsky bench. The Zhukovsky bench is a horizontal platform that freely rotates without friction around a fixed vertical axis. A person standing or sitting on a bench holds gymnastic dumbbells in outstretched hands and is rotated along with the bench around an axis at an angular velocity ω1. By bringing the dumbbells closer to himself, a person reduces the moment of inertia of the system, and since the moment of external forces is zero, the angular momentum of the system is preserved and the angular velocity of its rotation ω2 increases.

Similarly to the moment of force, the moment of impulse (moment of momentum) of a material point is determined

Similarly to the moment of force, the moment of impulse (moment of momentum) of a material point is determined. The angular momentum relative to the point O is equal to

The angular momentum about the z-axis is the component Lz along this axis of angular momentum L relative to the point O lying on the axis (Fig. 97):

where R is the component of the radius vector r perpendicular to the z axis, and p τ is the component of the vector p perpendicular to the plane passing through the z axis and the point m.

Let us find out what determines the change in the angular momentum with time. To do this, we differentiate (37.1) with respect to time t, using the product differentiation rule:

(3 7.5 )

The first term is equal to zero, since it is a vector product of vectors of the same direction. Indeed, the vector equal to the vector speed v and, therefore, coincides in direction with the vector p=mv. According to Newton's second law, the vector is equal to the force f acting on the body [see. (22.3)]. Therefore, expression (37.5) can be written as follows:

(3 7.6 )

where M is the moment of forces applied to the material point, taken relative to the same point O, relative to which the angular momentum L is taken.

From relation (37.6) it follows that if the resulting moment of forces acting on a material point relative to any point O is equal to zero, then the angular momentum of the material point, taken relative to the same point O, will remain constant.

Taking the components along the z axis from the vectors included in the formula (37.6), we obtain the expression:

(3 7.7 )

Formula (37.6) is similar to formula (22.3). From a comparison of these formulas, it follows that, just as the time derivative of momentum is equal to the force acting on a material point, the time derivative of the moment of momentum is equal to the moment of force.

Let's look at a few examples.

Example 1. Let the material point m move along the dotted line in Fig.96. Since the motion is rectilinear, the momentum of the material point changes only in absolute value, and

where f is the modulus of the force [in this case, f has the same direction as p (see Fig. 96), so that].

The arm t remains unchanged. Hence,

which is consistent with formula (37.6) (in this case, L changes only in absolute value, and it increases, therefore ).

Example 2. A material point of mass m moves along a circle of radius R (Fig. 98).

The angular momentum of a material point relative to the center of the circle O is equal in absolute value:

L=mυR

(3 7.8 )

The vector L is perpendicular to the plane of the circle, and the direction of motion of the point and the vector L form a right-handed system.

Since the arm equal to R remains constant, the angular momentum can only be changed by changing the velocity modulus. With a uniform motion of a material point along a circle, the angular momentum remains constant both in magnitude and in direction. It is easy to see that in this case the moment of the force acting on the material point is equal to zero.

Example 3. Consider the motion of a material point in the central field of forces (see § 26). In accordance with (37.6), the angular momentum of a material point, taken relative to the center of forces, must remain constant in magnitude and direction (the moment of the central force relative to the center is zero). The radius vector r drawn from the center of forces to the point m and the vector L are perpendicular to each other. Therefore, the vector r remains all the time in the same plane, perpendicular to the direction L. Consequently, the movement of a material point in the central field of forces will occur along a curve lying in a plane passing through the center of forces.

Depending on the sign of the central forces (that is, whether they are attractive or repulsive forces), as well as on the initial conditions, the trajectory is a hyperbola, parabola, or ellipse (in particular, a circle). For example, the Earth moves in an elliptical orbit, in one of the focuses of which the Sun is placed.

Law of conservation of angular momentum. Consider a system of N material points. Just as it was done in §23, we divide the forces acting on points into internal and external. The resulting moment of internal forces acting on i-th material point, we denote by the symbol , the resulting moment of external forces acting on the same point, by the symbol M i . Then equation (37.6) for i-th material points will look like:

(i=1, 2,…, N)

This expression is a set of N equations that differ from each other by the values ​​of the index i . Adding these equations, we get:

is called the angular momentum of the system of material points.

The sum of the moments of internal forces [the first of the sums on the right side of formula (37.9)], as shown at the end of §36, is equal to zero. Therefore, denoting the total moment of external forces by the symbol M, we can write that

(3 7.11 )

[the symbols L and M in this formula have a different meaning than the same symbols in formula (37.6)].

For a closed system of material points, M=0, as a result of which the total angular momentum L does not depend on time. Thus, we have come to the law of conservation of angular momentum: the angular momentum of a closed system of material points remains constant.

Note that the angular momentum remains constant for a system subjected to external influences, provided that the total moment of external forces acting on the bodies of the system is equal to zero.

Taking from the vectors on the left and right sides of equation (37.11), their components along the z axis, we arrive at the relation:

(3 7.12 )

It may happen that the resulting moment of external forces relative to the point O is different from zero (M≠0), but the component M z of the vector M in some direction z is equal to zero. Then, according to (37.12), the component L z of the angular momentum of the system along the z axis will be preserved.

According to formula (2.1 1)

where is the projection onto the z-axis of the vector , and L z is the projection onto the z-axis of the vector L . Multiply both sides of the equality by the ort e z z axis and, taking into account that e z does not depend on t, we introduce it on the right side under the derivative sign. As a result, we get:

But the product of e z times the projection of the vector on the z-axis gives the z-component of that vector (see footnote on page 132). Hence,

where is the component along the axis z vector .

Angular moment refers to the fundamental, fundamental laws of nature. It is directly related to the symmetry properties of the space of the physical world in which we all live. Thanks to the law of its conservation, the angular momentum determines the physical laws that are familiar to us for the movement of material bodies in space. This value characterizes the amount of translational or rotary motion.

The angular momentum, also called "kinetic", "angular" and "orbital", is an important characteristic that depends on the mass of a material body, the features of its distribution relative to the imaginary axis of circulation and the speed of movement. Here it should be clarified that in mechanics rotation has a broader interpretation. Even past some point arbitrarily lying in space can be considered rotational, taking it as an imaginary axis.

The angular momentum and the laws of its conservation were formulated by Rene Descartes in relation to a progressively moving system. True, he did not mention the conservation of type. Only a century later, Leonhard Euler, and then another Swiss scientist, physicist and mathematician, when studying the rotation of a material system around a fixed central axis, concluded that this law also applies to this type of movement in space.

Further studies fully confirmed that in the absence of external influence the sum of the product of the mass of all points by the total speed of the system and the distance to the center of rotation remains unchanged. Somewhat later, the French scientist Patrick Darcy expressed these terms in terms of the areas swept by the radius vectors over the same period of time. This made it possible to connect the angular momentum of a material point with some well-known postulates of celestial mechanics and, in particular, with the most important position on the motion of the planets

angular momentum solid body- the third dynamic variable, to which the provisions of the fundamental conservation law are applicable. It says that regardless of the nature and in the absence of external influence, a given value in an isolated material system will always remain unchanged. This physical indicator can undergo any changes only if there is a non-zero moment of the acting forces.

From this law it also follows that if M = 0, any change in the distance between the body (system of material points) and the central axis of rotation will certainly cause an increase or decrease in the speed of its circulation around the center. For example, a gymnast performing somersaults in order to make several turns in the air initially rolls her body into a ball. And ballerinas or figure skaters, spinning in a pirouette, spread their arms to the sides if they want to slow down the movement, and, conversely, press them to the body when they try to spin at a faster speed. Thus, the fundamental laws of nature are used in sports and art.

Angular moment relative to the fixed axis z called a scalar Lz, equal to the projection onto this axis of the angular momentum vector, defined relative to an arbitrary point 0 of this axis. The value of the angular momentum Lz does not depend on the position of point 0 on the axis z.
When an absolutely rigid body rotates around a fixed axis, each individual point of the body moves along a circle of constant radius r i at some speed v i. Speed v i and momentum m i v i are perpendicular to this radius, i.e. the radius is the arm of the vector m i v i. Therefore, it can be written that the angular momentum of an individual point about the axis z equals

The moment of momentum of a rigid body about the axis is the sum of the moments of momentum of its individual points:

Considering the relationship between linear and angular velocities ( v i = ωr i), we obtain the following expression for the angular momentum of the body relative to the fixed axis:

Those. the angular momentum of a rigid body about an axis is equal to the product of the moment of inertia of the body about the same axis and the angular velocity.
Differentiating expression (4.12) with respect to time, we obtain:

(4.13)

This is another form of the equation of the dynamics of the rotational motion of a rigid body relative to a fixed axis: the rate of change of the angular momentum of the body relative to the fixed axis of rotation is equal to the resulting moment relative to this axis of all external forces acting on the body.
Law of conservation of angular momentum follows from the basic equation of the dynamics of the rotational motion of a body fixed at a fixed point (Equation 4.8), and is as follows:
if the resulting moment of external forces relative to a fixed point is identically equal to zero, then the angular momentum of the body relative to this point does not change over time.
Indeed, if M= 0, then dL / dt= 0 , whence

(4.14)

In other words, the angular momentum of a closed system does not change over time.
From the basic law of the dynamics of a body rotating around a fixed axis z(Equation 4.13), follows law of conservation of angular momentum of a body about an axis:
if the moment of external forces relative to the fixed axis of rotation of the body is identically equal to zero, then the angular momentum of the body relative to this axis does not change in the process of motion, i.e. If Mz= 0, then dLz / dt= 0, whence

The law of conservation of angular momentum is a fundamental law of nature. The validity of this law is determined by the property of the 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.
The validity of the law of conservation of angular momentum relative to a fixed axis of rotation can be demonstrated by experiment with the Zhukovsky bench. The Zhukovsky bench is a horizontal platform that freely rotates without friction around a fixed vertical axis OO 1. A person standing or sitting on a bench holds gymnastic dumbbells in outstretched hands and is rotated together with the bench around the axis OO 1 with an angular velocity ω 1. By bringing the dumbbells closer to himself, a person reduces the moment of inertia of the system, and since the moment of external forces is zero, the angular momentum of the system is preserved and the angular velocity of its rotation ω 2 increases. Then, according to the law of conservation of angular momentum relative to the axis OO 1, we can write:

Where J0- the moment of inertia of the person and the bench; 2 Mr 1 2 and 2 Mr 2 2- moments of inertia of dumbbells in the first and second positions; m- weight of one dumbbell; r1, r2- the distance from the dumbbells to the axis of the OO 1.
The change in the moment of inertia of the system is associated with a change in its kinetic energy:

Using the expression for ω 2 obtained from (4.16)

after transformations we get:

This change in the kinetic energy of the system is numerically equal to the work done by a person when moving dumbbells.
In table. 4.2 mapped main physical quantities and equations that determine the rotation of the body around a fixed axis and its translational motion.

Table 4.2

Task 1.A ball with a radius of 10 cm and a mass of 5 kg rotates around the axis of symmetry according to the law φ = A + Bt 2 + Ct 3, Where IN\u003d 2 rad / s 2, WITH\u003d -0.5 rad / s 3. Determine the moment of forces about the axis of rotation for the moment of time t= 3 s.
Given: R= 0.1 m; m= 5 kg; φ = A + Bt 2 + Ct 3 glad; IN\u003d 2 rad / s 2; WITH\u003d -0.5 rad / s 3; t= 3 s.
Find: Mz.
Solution
According to the equation of dynamics of rotational motion of a rigid body relative to a fixed axis

Answer: Mz= -0.1H*m.

Task 2. On a homogeneous solid cylindrical shaft with a radius of 20 cm, the moment of inertia of which is 0.15 kg * m 2, a light thread is wound, to the end of which a weight of 0.5 kg is attached. Before the drum began to rotate, the height of the load above the floor was 2.3 m (Fig. 4.7). Determine: a) the time of lowering the load to the floor; b) the force of the thread tension; c) the kinetic energy of the load at the moment of impact on the floor.
Given: R= 0.2 m; Jz\u003d 0.15 kg * m 2; m= 0.5 kg; h= 2.3 m.
Find: t, T, E k.

Solution
According to the law of conservation of energy

Answer: t= 2 s; T= 4.31 N; E k= 1.32 J.

Tasks for independent solution

1. A ball and a solid cylinder, made of the same material, of the same mass roll without slipping at the same speed. Determine how many times the kinetic energy of the ball is less than the kinetic energy of a solid cylinder.
2. A hollow thin-walled cylinder of mass 0.5 kg, rolling without slipping, hits a wall and rolls away from it. The speed of the cylinder before hitting the wall is 1.4 m/s, after hitting it is 1 m/s. Determine the amount of heat released during the impact.
3. A constant tangential force of 30 N is applied to the rim of a homogeneous solid disk with a mass of 10 kg, mounted on an axle. Determine the kinetic energy 4 s after the onset of the force.
4. The fan rotates at 600 rpm. After turning it off, it began to rotate uniformly and, having made 50 revolutions, stopped. The work of the braking forces is 31.4 J. Determine: a) the moment of the braking forces; b) the moment of inertia of the fan.
5. A constant tangential force of 100 N is applied to the rim of a homogeneous solid disk with a radius of 0.5 m. When the disk rotates, a friction moment of 2 N * m acts on it. Determine the mass of the disk if its angular acceleration is known to be constant and equal to 16 rad/s 2 .
6. A ball rolls down from an inclined plane making an angle of 30° with the horizontal. Neglecting friction, determine the time it takes for a ball to move down an inclined plane if it is known that its center of mass has decreased by 30 cm as it rolls down.
7. A light thread is wound on a homogeneous solid cylindrical shaft with a radius of 50 cm, to the end of which a weight of 6.4 kg is attached. The load, unwinding the thread, falls with an acceleration of 2 m / s 2. Determine: a) the moment of inertia of the shaft; b) the mass of the shaft.
8. A horizontal platform with a mass of 25 kg and a radius of 0.8 m rotates at a frequency of 18 rpm. A man stands in the center and holds weights in his outstretched hands. Considering the platform as a disk, determine the frequency of rotation of the platform if a person, lowering his hands, reduces his moment of inertia from 3.5 kg * m 2 to 1 kg * m 2.
9. A person weighing 60 kg, standing on the edge of a horizontal platform weighing 120 kg, rotating by inertia around a fixed vertical axis with a frequency of 10 rpm, moves to its center. Considering the platform as a round homogeneous disk, and the person as a point mass, determine with what frequency the platform will then rotate.
10. The platform, which has the form of a solid homogeneous disk, can rotate by inertia around a fixed vertical axis. On the edge of the platform stands a man whose mass is 3 times less than the mass of the platform. Determine how and how many times the angular velocity of rotation of the platform will change if the person moves closer to the center at a distance equal to half the radius of the platform.

Angular moment in classical mechanics

Relationship between momentum and moment

Definition

The angular momentum of a particle relative to some origin is determined by the vector product of its radius vector and momentum :

where is the radius-vector of the particle relative to the selected reference point, which is motionless in the given reference frame, is the momentum of the particle.

For several particles, angular momentum is defined as the (vector) sum of such terms:

where are the radius vector and the momentum of each particle in the system, the angular momentum of which is determined.

(In the limit, the number of particles can be infinite, for example, in the case of a solid body with a continuously distributed mass or a distributed system in general, this can be written as where is the momentum of an infinitely small point element of the system).

From the definition of angular momentum follows its additivity: both for a system of particles in particular, and for a system consisting of several subsystems, the following is true:

• Remark: in principle, the angular momentum can be calculated with respect to any reference point (the resulting different values ​​are related in an obvious way); however, most often (for convenience and definiteness) it is calculated relative to the center of mass or a fixed point of rotation of a rigid body, etc.).

Torque calculation

Since the angular momentum is determined by the cross product, it is a pseudovector perpendicular to both vectors and. However, in cases of rotation around a constant axis, it is convenient to consider not the angular momentum as a pseudovector, but its projection onto the rotation axis as a scalar, the sign of which depends on the direction of rotation. If such an axis is chosen, passing through the origin, to calculate the projection of the angular momentum on it, you can specify a number of recipes in accordance with the general rules for finding vector product two vectors.

where is the angle between and , determined so that the rotation from to is counterclockwise from the point of view of an observer located on the positive part of the axis of rotation. The direction of rotation is important in the calculation, as it determines the sign of the desired projection.

We write in the form , where is the component of the radius vector, parallel to the momentum vector, and - similarly, perpendicular to it. is, in fact, the distance from the axis of rotation to the vector, which is usually called the "shoulder". Similarly, you can divide the momentum vector into two components: parallel to the radius vector and perpendicular to it. Now, using the linearity of the vector product, as well as the property that the product of parallel vectors is zero, we can get two more expressions for .

Conservation of angular momentum

Symmetry in physics
transformation	Relevant invariance	Corresponding law conservation
↕ Broadcast time	…energy
⊠ , , and -symmetries	... parity
↔ Broadcast space	Uniformity space	…impulse
↺ Rotation of space	Isotropy space	… moment momentum
⇆ Lorentz group	Relativity Lorentz invariance	…4-pulses
~ Gauge transformation	Gauge invariance	... charge

Thus, the requirement of system closure 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 met).

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 rotating through an arbitrary infinitesimal angle , the radius vector of the particle with the number will change by , and the velocities - . The Lagrange function of the system will not change during such a rotation, due to the isotropy of space. That's why

Taking into account , where is the generalized momentum of the -th particle, each term in the sum from the last expression can be rewritten as

Now, using the mixed product property, we perform a cyclic permutation of vectors, as a result of which we obtain, taking out the common factor:

where, is the angular momentum of the system. In view of the arbitrariness of , it follows from the equality that .

On orbits, the angular momentum is distributed between the planet's own rotation and the angular momentum of its orbital motion:

Angular moment in electrodynamics

When describing the motion of a charged particle in an electromagnetic field, the canonical momentum is not invariant. As a consequence, the canonical angular momentum is also not invariant. Then we take the real momentum, which is also called the "kinetic momentum":

where is the electric charge, is the speed of light, is the vector potential. Thus, the (invariant) Hamiltonian of a charged mass particle in an electromagnetic field is:

where is the scalar potential. From this potential follows the Lorentz law. The invariant angular momentum or "kinetic angular momentum" is defined by:

Angular momentum in quantum mechanics

Moment operator

Calculation of angular momentum in nonrelativistic mechanics

If there is a material point with a mass moving at a speed and located at a point described by the radius vector, then the angular momentum is calculated by the formula:

where is the sign of the vector product.

To calculate the angular momentum of a body, it must be broken into infinitesimal pieces and vector sum their moments as moments of momentum of material points, that is, take the integral:

We can rewrite this in terms of density: