Mechanical energy. Kinetic energy during rotational motion

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 the movement of a given body and the coordinate system associated with it are described.

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 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 fully satisfies the 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: the moment of inertia is equal to the sum of the 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 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. Rotational energy 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

« 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 at 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. Consequently, for rotational motion 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.

The angular momentum of a rigid body is 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 duration 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 at a constant speed, is equal to

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.

Tasks

1. Determine how many times the effective mass is greater than the gravitating mass of a train weighing 4000 tons, if the mass of the wheels is 15% of the mass of the train. Consider the wheels to be discs with a diameter of 1.02 m. How will the answer change if the diameter of the wheels is half as large?

2. Determine the acceleration with which a wheel pair weighing 1200 kg rolls down a hill with a slope of 0.08. Consider wheels to be disks. Rolling resistance coefficient 0.004. Determine the adhesion force between wheels and rails.

3. Determine the acceleration with which a wheel pair weighing 1400 kg rolls up a hill with a slope of 0.05. Resistance coefficient 0.002. What should the coefficient of adhesion be so that the wheels do not slip? Consider wheels to be disks.

4. Determine with what acceleration a car weighing 40 tons rolls down a hill with a slope of 0.020, if it has eight wheels weighing 1200 kg and a diameter of 1.02 m. Determine the adhesion force of the wheels to the rails. Resistance coefficient 0.003.

5. Determine the pressure force of the brake pads on the tires if a train weighing 4000 tons brakes with an acceleration of 0.3 m/s 2 . The moment of inertia of one wheel pair is 600 kg m 2, the number of axles is 400, the sliding friction coefficient of the pad is 0.18, and the rolling resistance coefficient is 0.004.

6. Determine the braking force acting on a four-axle car weighing 60 tons on the braking platform of a hump if the speed on a track of 30 m decreased from 2 m/s to 1.5 m/s. The moment of inertia of one wheel pair is 500 kg m 2.

7. The locomotive’s speed gauge showed an increase in train speed within one minute from 10 m/s to 60 m/s. It is likely that the drive wheel pair has slipped. Determine the moment of forces acting on the armature of the electric motor. The moment of inertia of the wheelset is 600 kg m 2, the armature is 120 kg m 2. Gear ratio gear transmission 4.2. The pressure force on the rails is 200 kN, the sliding friction coefficient of the wheels on the rail is 0.10.

11. KINETIC ENERGY OF ROTATIONAL

MOVEMENTS

Let us derive the formula for the kinetic energy of rotational motion. Let the body rotate with angular velocity ω relative to a fixed axis. Any small particle of a body undergoes translational motion in a circle with a speed where r i – distance to the axis of rotation, radius of the orbit. Particle kinetic energy masses m i equal to . The total kinetic energy of a system of particles is equal to the sum of their kinetic energies. Let us sum up the formulas for the kinetic energy of particles of the body and take out half the square of the angular velocity, which is the same for all particles, as the sum sign. The sum of the products of the particle masses by the squares of their distances to the axis of rotation is the moment of inertia of the body relative to the axis of rotation . So, the kinetic energy of a body rotating relative to a fixed axis is equal to half the product of the moment of inertia of the body relative to the axis and the square of the angular velocity of rotation:

With the help of rotating bodies, mechanical energy can be stored. Such bodies are called flywheels. Usually these are bodies of revolution. The use of flywheels in the pottery wheel has been known since ancient times. In internal combustion engines, during the power stroke, the piston imparts mechanical energy to the flywheel, which then performs work on rotating the engine shaft for three subsequent strokes. In dies and presses, the flywheel is driven by a relatively low-power electric motor and accumulates mechanical energy for almost full turn and at a short moment of impact it gives it over to stamping work.

There are numerous attempts to use rotating flywheels to drive vehicles: cars, buses. They are called mahomobiles, gyromobiles. Many such experimental machines were created. It would be promising to use flywheels to accumulate energy during braking of electric trains in order to use the accumulated energy during subsequent acceleration. Flywheel energy storage is known to be used on New York City subway trains.

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

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

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

. (5.30)

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

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

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

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

In accordance with the basic equation of dynamics for rotational motion

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

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

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

If , then the formula simplifies

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

Gyroscope

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

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

1) Stability.

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

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

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

. (5.34)

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

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

2). Gyroscopic effect.

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

3). Precession of the gyroscope.

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

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

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

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

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

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

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

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

. (5.35)

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

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

Examples of application of the laws of dynamics

During rotational movement

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

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

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

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

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

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

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

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

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

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

According to the law of conservation of angular momentum

And .

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

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

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

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

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

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

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

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

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

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

Solving the resulting equation after substitution, we find

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

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

Having carried out the necessary transformations, we obtain

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

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

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

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

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

Solving these equations, we get

Then we find T 1 and T 2.

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

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

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

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

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

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

The kinetic energy of an elementary mass is equal to:

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

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

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

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

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

(4.6)

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

Work of external forces during rotational motion of a rigid body.

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

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

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

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

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

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

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

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

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

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

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

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

, i.e. .

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

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

;

Hence,

. (4.7)

On one's own:

Elastic forces;

Hooke's law.

LECTURE 7

Hydrodynamics

Current lines and tubes.

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

, (5.1)

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

Bernoulli's equation.

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

, (5.2)

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

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

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

. (5.4)

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

. (5.5)

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

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

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

Internal friction forces.

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

, (5.7)

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

v(z) = = (v 0 /d)·z.

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

formula (5.7) will take the form

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

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