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 any 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 sign of the sum, we find:

This formula for the kinetic energy of a rotating body can be reduced 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 the 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 body mass, the moment of inertia I enters here and instead of the velocity, the group velocity

The large kinetic energy of a rotating flywheel is used in technology to maintain the uniformity of the machine under a suddenly changing load. At first, to bring the flywheel with a large moment of inertia into rotation, the machine requires a significant amount of work, but when a large load is suddenly turned on, the machine does not stop and does work due to the flywheel's kinetic energy reserve.

Particularly 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 kinetic energy of the wheel is such that at the time of rolling the wheel gives the mill a power of 20,000 liters. With. Friction in the bearings is kept to a minimum by a fairy tale under pressure, and in order to avoid the harmful effect of centrifugal inertia forces, the wheel is balanced so that the load placed on the circumference of the wheel brings it out of rest.

We 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 sections).

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

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

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

The moment of inertia of the ball about 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 calculation of the moments of inertia of bodies is carried out using integral calculus. To give an idea of ​​the course of such calculations, we find the moment of inertia of the rod relative to the axis perpendicular to it (Fig. 58). Let there be a section of the rod, density. We single out an elementarily 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, then its moment of inertia We integrate from zero to I:

Moment of inertia of a rectangular parallelepiped about the axis of symmetry (Fig. 59)

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

Let us consider how the energy of rotation of a body rolling (without sliding) along the plane is connected with the energy of the translational motion of this body,

The energy of the translational motion of a rolling body is , where is the mass of the body and the velocity of the translational motion. Let denote the angular velocity of rotation of the rolling body and the radius of the body. It is easy to understand that the speed of the translational motion of a body rolling without sliding is equal to the circumferential 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,

In this way,

Rotation energy

Consequently,

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

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

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

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

The 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 an angular velocity (o around the axis O, perpendicular to the plane of the drawing. Suppose that over a certain period of time it moved from position A B to and the center of gravity described an arc. This movement rod can be considered as if the rod first translationally (that is, 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 And In position, the displacement of each of its particles is the same as the displacement of the center of gravity, i.e. it is equal to or To obtain the actual movement of the rod, we can assume that both of these movements are performed simultaneously. about the axis passing through O can be decomposed into two parts.

Let us determine the kinetic energy of a rigid body rotating around a fixed axis. Let's divide this body into n material points. Each point moves with a linear speed υ i =ωr i , then the kinetic energy of the point

or

The total kinetic energy of a rotating rigid body is equal to the sum of the kinetic energies of all its material points:

(3.22)

(J - moment of inertia of the body about the axis of rotation)

If the trajectories of all points lie in parallel planes (like a cylinder rolling down an inclined plane, each point moves in its own plane fig), this is flat motion. According to Euler's principle, plane motion can always be decomposed in an infinite number of ways into translational and rotational motion. If the ball falls or slides along an inclined plane, it only moves forward; when the ball rolls, it also rotates.

If a body performs translational and rotational motions at the same time, then its total kinetic energy is equal to

(3.23)

From a comparison of the formulas of kinetic energy for translational and rotational motions, it can be seen that the measure of inertia during rotational motion is the moment of inertia of the body.

§ 3.6 The work of external forces during the 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

Considering that Jβ = M, ωdr = dφ, we have α of the body at a finite angle φ equals

(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 about a given axis. If the moment of forces about the axis is equal to 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 the horizontal axis with a frequencyν 0 =720 min -1 and stops when brakingt=20 s. Find the braking torque and the number of revolutions before stopping.

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

where I=mr 2 is the moment of inertia of the disk; Δω \u003d ω - ω 0, and ω \u003d 0 is the final angular velocity, ω 0 \u003d 2πν 0 is the initial one. M is the braking moment of the forces acting on the disk.

Knowing all the quantities, it is possible to determine the braking torque

Mr 2 2πν 0 = МΔt (1)

(2)

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

(3)

where β is the angular acceleration.

According to the condition 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 the same radii and masses were spun up to the speed of rotationn= 480 rpm and left to themselves. Under the action of the friction forces of the shafts on the bearings, the first one stopped aftert\u003d 80 s, and the second didN= 240 revolutions to stop. In which flywheel, the moment of the friction forces of the shafts on the bearings was greater and how many times.

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

M 1 Δt \u003d Iω 2 - Iω 1

where Δt is the time of action of the moment of friction forces, I \u003d mr 2 - the moment of inertia of the flywheel, ω 1 \u003d 2πν and ω 2 \u003d 0 are the initial and final angular velocities of the flywheels

Then

The moment of friction forces M 2 of the second flywheel is expressed through the relationship 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

O ratio will be

The friction torque of the second flywheel is 1.33 times greater.

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

There is no slippage of the thread, therefore, when m 1 and m 2 will make translational motion, the cylinder will rotate about the axis passing through the point O. Let's 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 performing translational motion, and the third equation is 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 turn the cylinder counterclockwise). On the right, I is the moment of inertia of the cylinder about the axis O, 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 slip,
. Taking into account the expressions for I and β, we get:

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

From here we find the acceleration a cargo

It can be seen from the resulting equation 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 weights.

Example 2.4. A hollow ball with mass m = 0.5 kg has an outer radius R = 0.08m and an inner radius r = 0.06m. 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 the hollow ball, and the angular acceleration β is found 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. We find the density, knowing the mass of a hollow ball

From here we determine the density of the material of the ball

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 - moment of inertia of the rod relative to the axis of rotation).

For an isolated system of bodies, the vector sum of the 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 about the axis passing through the center of mass and perpendicular to the rod is equal to

J 0 \u003d mℓ 2 / 12. (3)

According to the Steiner theorem

J = J 0 +m a 2

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

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

J 2 \u003d J 0 +m a 2 , J 2 = mℓ 2 /12 +m(ℓ/2) 2 = mℓ 2 /3. (four)

Let us substitute formulas (3) and (4) into (2):

mℓ 2 ω 1 /12 = mℓ 2 ω 2 /3

ω 2 \u003d ω 1 /4 ω 2 \u003d 10s-1/4 \u003d 2.5s -1

Example 2.6 . mass manm= 60 kg, standing on the edge of the platform with mass M = 120 kg, rotating by inertia around a fixed vertical axis with a frequency ν 1 =12min -1 , goes to its center. Considering the platform as a round homogeneous disk, and the person as 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: v 1

Solution: According to the condition 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-man” system, the law of conservation of momentum is fulfilled

I 1 ω 1 = I 2 ω 2

where
- the moment of inertia of the system when a person is standing on the edge of the platform (we took into account that the moment of inertia of the platform is equal to (R is the radius p
platform), the moment of inertia of a person at the edge of the platform is mR 2).

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

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

whence the desired rotational speed

Answer: v 2 =24 min -1 .

Let's start by considering the rotation of the body around a fixed axis, which we will call the z-axis (Fig. 41.1). The linear speed of the elementary mass is where is the distance of the mass from the axis. Therefore, for the kinetic energy of an elementary mass, the expression is obtained

The kinetic energy of a body is composed of the kinetic energies of its parts:

The sum on the right side of this ratio is the moment of inertia of body 1 about the axis of rotation. Thus, the kinetic energy of a body rotating around a fixed axis is

Let an internal force and an external force act on the mass (see Fig. 41.1). According to (20.5), these forces will do work during the time

Carrying out a cyclic permutation of factors in mixed products of vectors (see (2.34)), we obtain:

where N is the moment of the internal force relative to the point O, N is the analogous moment of the external force.

Summing expression (41.2) over all elementary masses, we obtain the elementary work performed on the body during the time dt:

The sum of moments of internal forces is equal to zero (see (29.12)). Therefore, denoting the total moment of external forces through N, we arrive at the expression

(we used formula (2.21)).

Finally, taking into account that there is an angle through which the body rotates in time, we get:

The sign of the work depends on the sign, i.e., on the sign of the projection of the vector N onto the direction of the vector

So, when the body rotates, the internal forces do not perform work, while the work of external forces is determined by formula (41.4).

Formula (41.4) can be arrived at by using the fact that the work done by all the forces applied to the body goes to increase its kinetic energy (see (19.11)). Taking the differential of both sides of equality (41.1), we arrive at the relation

According to the equation (38.8) so, replacing through we will come to the formula (41.4).

Table 41.1

In table. 41.1, the formulas of the mechanics of rotational motions are compared with similar formulas of the mechanics of translational motion (the mechanics of a point). From this comparison it is easy to conclude that in all cases the role of mass is played by the moment of inertia, the role of force is the moment of force, the role of momentum is played by the moment of momentum, etc.

Formula. (41.1) we obtained for the case when the body rotates around a fixed axis fixed in the body. Now let's assume that the body rotates arbitrarily about a fixed point coinciding with its center of mass.

Let us rigidly connect the Cartesian coordinate system with the body, the origin of which will be placed at the center of mass of the body. The speed of the i-th elementary mass is Therefore, for the kinetic energy of the body, we can write the expression

where is the angle between the vectors Replacing a through and taking into account what we get:

We write the scalar products in terms of the projections of vectors on the axes of the coordinate system associated with the body:

Finally, by combining the terms with the same products of the components of the angular velocity and taking these products out of the signs of the sums, we get: so that formula (41.7) takes the form (compare with (41.1)). When an arbitrary body rotates around one of the main axes of inertia, say the axes and formula (41.7) goes into (41.10.

In this way. the kinetic energy of a rotating body is equal to half the product of the moment of inertia and the square of the angular velocity in three cases: 1) for a body rotating around a fixed axis; 2) for a body rotating around one of the main axes of inertia; 3) for a ball top. In other cases, the kinetic energy is determined by the more complex formulas (41.5) or (41.7).

Consider first a rigid body rotating around a fixed axis OZ with an angular velocity ω (fig.5.6). Let's break the body into elementary masses. The linear velocity of an elementary mass is , 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 made up of the kinetic energies of its parts, therefore

.

Considering that the sum on the right side of this relation represents the moment of inertia of the body about the axis of rotation, we finally obtain

. (5.30)

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

In the general case, the motion of a rigid body can be represented as a sum of motions - translational with a speed equal to the speed of the center of mass of the body, and rotation with an angular velocity around the 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 the 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 for 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 for 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 is simplified

Thus, the work of external forces during the rotation of a rigid body about a fixed axis is determined by the action of the projection of the moment of these forces on a given 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 freely rotate in space, the gyroscope is placed in the so-called gimbal suspension (Fig. 5.13). The flywheel of the gyroscope rotates in the inner annular cage around the C 1 C 2 axis passing through its center of gravity. The inner cage, in turn, can rotate in the outer cage around the axis B 1 B 2 perpendicular to C 1 C 2 . Finally, the outer race can freely rotate in the strut bearings 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 suspension or the fulcrum of the gyroscope. The gyroscope in the gimbal has three degrees of freedom and, therefore, can make any rotation around the center of the gimbal. If the gyroscope's suspension center coincides with its center of gravity, then the resulting moment of gravity of all parts of the gyroscope relative to the suspension center is equal to zero. Such a gyroscope is called balanced.

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

1) Sustainability.

With any rotation of the balanced gyroscope rack, its axis of rotation remains the same direction with respect to the laboratory frame of reference. This is due to the fact that the moment of all external forces, equal to the moment of 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 an external force acts for a short time, then the integral that determines the increment of the 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, as it were, resists all attempts to change the magnitude and direction of its angular momentum. Connected with this is the remarkable stability that the motion of a gyroscope acquires after bringing it into rapid rotation. This property of the gyroscope is widely used to automatically control the movement of aircraft, ships, rockets and other vehicles.

If, however, the gyroscope is acted upon for a long time by a moment of external forces constant in direction, then the axis of the gyroscope is established, in the end, in the direction of the moment of external forces. This phenomenon is used in the gyrocompass. This device is a gyroscope, the axis of which can freely rotate in a horizontal plane. Due to the daily rotation of the 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 rotate around the 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 a pair of forces is directed along the O 1 O 1 axis and a change in the vector by a value over time will have the same direction. As a result, the new vector will rotate about the O 2 O 2 axis. Thus, the seemingly unnatural behavior of the gyroscope fully corresponds to the laws of the dynamics of rotational motion

3). Gyro precession.

The precession of a gyroscope is the conical movement of its axis. It occurs 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, brought into rapid rotation (Fig. 5.16), can serve.

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

Find out the reasons for the precession of the gyroscope. Consider an unbalanced gyroscope whose axis 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 the 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 action of this moment, the angular momentum of the gyroscope (its beginning 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 only changes in direction. In this case, after a while, the relative position of the vectors and will be the same as at the initial moment. As a result, the axis of the gyroscope 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 above 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 rotational 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 applying the laws of dynamics

When rotating

1. Consider some examples of the law of conservation of angular momentum, which can be implemented using the Zhukovsky bench. In the simplest case, the Zhukovsky bench is a disk-shaped platform (chair) that can freely rotate around a vertical axis on ball bearings (Fig. 5.17). The demonstrator sits or stands on the bench, after which it is brought into rotational motion. 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 the bench and the demonstrator, relative to the axis of rotation, cannot change in time if the system is left to itself. 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 compose 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 are the angular velocities of the system.

The angular velocity of rotation of the system when breeding dumbbells to the side will be equal to

.

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

2. Let's give one more experiment with Zhukovsky's bench. The demonstrator sits or stands on a bench and is given 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, comes into rotation 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, is the angular velocity of its rotation.

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

,

where is the moment of inertia of the "man-platform" system, 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 mass m and length l rotates with an angular velocity ω=10 s -1 in a 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 the angular momentum of an isolated system, we have

Here - the moment of inertia of the rod about the axis passing through the middle of the rod; - the moment of inertia of the rod about 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 weight m 1=10 kg is hinged at the upper end. A bullet hits the center of the rod with a mass m2=10 g, flying horizontally at a speed of =500 m/s, and gets stuck in the rod. Through what angle will the rod deviate after the impact?

Let's imagine in Fig. 5.19. system of interacting bodies "rod-bullet". The moments of external forces (gravity, axis 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 point of suspension, 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 the bullet hits it with its potential energy at the highest point of the ascent:

,

where is the height of the center of mass of the given system.

Having carried out the necessary transformations, we obtain

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

.

Having carried out the calculations, we obtain =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 about the axis of rotation is I, block radius r. Ignore the mass of the thread.

Let's arrange all the forces acting on the loads and the block, and compose the dynamics equations for them

If there is no slippage of the thread along the block, then the linear and angular acceleration are related 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), to which a load of mass M= 0.5 kg. Determine how long it takes for a load to fall from a height h=1 m to the bottom position. Pulley radius r\u003d 3 cm. Four weights of mass m=250g each at a distance R= 30 cm from its axis. Neglect the moment of inertia of the cross itself and the pulley compared to the moment of inertia of the weights.

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

If the body rotates around a fixed axis z with an angular velocity , then the linear velocity of the i-th point , Ri is the distance to the axis of rotation. Consequently,

Comparing and it can be seen that the moment of inertia of the body I is a measure of inertia during rotational motion, just as the 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 a speed vc and rotational with an angular velocity 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.

Rotational dynamics

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 the body.

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

rotational work.

Law of conservation of angular momentum.

The angular momentum (momentum) of a material point A relative to a fixed point O is a physical quantity determined by a 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 pseudovector, the direction of which coincides with the direction of the translational movement of the right screw during its rotation from r to p.

Momentum vector modulus

where α is the angle between the vectors r and p, l is the shoulder of the vector p with respect to the point O.

The angular momentum relative to the fixed axis z is the scalar value Lz, which is 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 the 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 the arm of the vector mivi . So we can write that the angular momentum of an individual particle is

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

The momentum of a rigid body relative to the axis is the sum of the momentum of the individual particles:

Using the formula vi = ωri, we get

Thus, the angular momentum of a rigid body about an axis is equal to the moment of inertia of the body about 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 of the dynamics of the rotational motion of a rigid body about a fixed axis: the derivative of the angular momentum of a rigid body about an axis is equal to the moment of forces about the same axis.

It can be shown that the vector equality holds

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

Expression (4) is the law of conservation of angular momentum: the angular momentum of a closed system is conserved, i.e., 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 symmetry property 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 (with respect to the rotation of a closed system in space by any angle).

Here we will demonstrate the law of conservation of angular momentum using the Zhukovsky bench. A person sitting on a bench, rotating around a vertical axis, and holding dumbbells in outstretched hands (Fig. 2), is rotated by an external mechanism with an angular velocity ω1. If a person presses the dumbbells to the body, then the moment of inertia of the system will decrease. But the moment of external forces is equal to zero, the angular momentum of the system is preserved and the angular velocity of rotation ω2 increases. Similarly, the gymnast, while jumping over his head, draws his arms and legs close to the 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 bound or rather weakly bound 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 vessel occupied by the gas.

And the liquid, having a certain volume, takes the form 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 an almost constant 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 continuous, continuously distributed in the part of space occupied by them. In gases, the density depends on pressure significantly. Established from experience. that the compressibility of a liquid and a gas can often be neglected and it is advisable to use a single concept - the incompressibility of a liquid - a liquid with the same density everywhere, which does not change over time.

We place it in a thin plate at rest, as a result, parts of the liquid located on opposite sides of the plate will act on each of its elements ΔS with forces ΔF, which will be equal in absolute value and directed perpendicular to the site ΔS, regardless of the orientation of the site, otherwise the presence of tangential forces would set the particles of the liquid in motion (Fig. 1)

The physical quantity determined by the normal force acting from the side of the liquid (or gas) per unit area is called the pressure p / 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 evenly distributed over a surface of 1 m2 normal to it (1 Pa = 1 N/m2).

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

Let us investigate the effect of the weight of a fluid on the distribution of pressure inside a stationary incompressible fluid. When a liquid 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 fluid at rest is always horizontal (we do not take into account the attraction of the fluid by the walls of the vessel). If a fluid is incompressible, then the density of the fluid is independent of pressure. Then, with the cross section S of the liquid column, its height h and density ρ, the weight is P=ρgSh, while the pressure on the lower base is: p=P/S=ρgSh/S=ρgh, (1)

i.e. pressure changes 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 ones, therefore, a force determined by the law of Archimedes acts on a body immersed in a liquid (gas): upward buoyant force equal to the weight of the 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.