The kinetic energy of rotational motion depends on. Kinetic energy and work during rotational motion

The expression for the kinetic energy of a rotating body, taking into account that the linear speed of an arbitrary material point composing the body relative to the axis of rotation is equal, has the form

where is the moment of inertia of the body relative to the selected axis of rotation, its angular velocity relative to this axis, and the angular momentum of the body relative to the axis of rotation.

If a body undergoes translational rotational motion, then the calculation of kinetic energy depends on the choice of the pole with respect to which the body’s motion is described. The end result will be the same. So, if for a round body rolling at speed v without slipping with radius R and coefficient of inertia k, the pole is taken at its CM, at point C, then its moment of inertia is , and the angular velocity of rotation around the axis C is . Then the kinetic energy of the body is .

If the pole is taken at the point O of contact between the body and the surface through which the instantaneous axis of rotation of the body passes, then its moment of inertia relative to the axis O will become equal . Then the kinetic energy of the body, taking into account that the angular velocities of rotation of the body are the same relative to parallel axes and the body performs pure rotation around the O axis, will be equal to . The result is the same.

The theorem about the kinetic energy of a body performing a complex movement will have the same form as for its forward motion: .

Example 1. A body of mass m is attached to the end of a thread wound around a cylindrical block of radius R and mass M. The body is raised to a height h and released (Fig. 65). After an inelastic jerk of the thread, the body and the block immediately begin to move together. How much heat will be released during the jerk? What will be the acceleration of the body and the tension of the thread after the jerk? What will be the speed of the body and the distance traveled by it after the thread is jerked after time t?

Given: M, R, m, h, g, t. Find: Q -?,a - ?, T - ?,v -?, s - ?

Solution: Body speed before the thread jerks. After a jerk of the thread, the block and the body will go into rotational motion relative to the block axis O and will behave like bodies with moments of inertia relative to this axis equal to and . Their total moment of inertia about the axis of rotation.

Thread jerking is a fast process and during a jerk, the law of conservation of angular momentum of the block-body system takes place, which, due to the fact that the body and block immediately after the jerk begin to move together, has the form: . Where does the initial angular velocity of rotation of the block come from? , and the initial linear velocity of the body .

Kinetic energy of the system, due to the conservation of its angular momentum immediately after the thread is jerked, is equal to . The heat released during the jerk according to the law of conservation of energy

The dynamic equations of motion of the bodies of the system after a jerk of the thread do not depend on their initial speed. For a block it has the form or, and for the body. Adding these two equations, we get . Where does the acceleration of body motion come from? Thread tension

The kinematic equations of body motion after a jerk will have the form , where all parameters are known.

Answer: . .

Example 2. Two round bodies with inertia coefficients (hollow cylinder) and (ball) located at the base of an inclined plane with an angle of inclination α report identical initial velocities directed upward along the inclined plane. To what height and in what time will the bodies rise to this height? What are the accelerations of rising bodies? How many times do the heights, times and accelerations of the bodies rise differ? Bodies move along an inclined plane without slipping.

Given: . Find:

Solution: The body is acted upon by: gravity m g, inclined plane reaction N, and clutch friction force (Fig. 67). The work of normal reaction and adhesion friction force (there is no slipping and no heat is released at the point of adhesion of the body and the plane.) are equal to zero: , therefore, to describe the motion of bodies it is possible to use the law of conservation of energy: . Where .

We will find the times and accelerations of motion of bodies from kinematic equations . Where , . The ratio of heights, times and accelerations of lifting bodies:

Answer: , , , .

Example 3. A bullet of mass , flying at speed, strikes the center of a ball of mass M and radius R, attached to the end of a rod of mass m and length l, suspended at point O by its second end, and flies out of it with speed (Fig. 68). Find the angular velocity of rotation of the rod-ball system immediately after the impact and the angle of deflection of the rod after the bullet impact.

Given: . Find:

Solution: Moments of inertia of the rod and the ball relative to the suspension point O of the rod according to Steiner’s theorem: and . Total moment of inertia of the rod-ball system . The impact of a bullet is a fast process, and the law of conservation of angular momentum of the bullet-rod-ball system takes place (bodies after a collision enter into rotational motion): . Where does the angular velocity of motion of the rod-ball system immediately after impact come from?

Position of the CM of the rod-ball system relative to the suspension point O: . The law of conservation of energy for the CM of a system after an impact, taking into account the law of conservation of angular momentum of the system upon impact, has the form . Where does the height of the system's CM rise from after an impact? . The angle of deflection of the rod after impact is determined by the condition .

Answer: , , .

Example 4. A block is pressed with a force N to a round body of mass m and radius R, with a coefficient of inertia k, rotating with an angular velocity . How long will it take for the cylinder to stop and how much heat will be released when the pad rubs against the cylinder during this time? The coefficient of friction between the block and the cylinder is .

Given: Find:

Solution: The work done by the friction force before the body stops according to the theorem on kinetic energy is equal to . Heat released during rotation .

The equation of rotational motion of a body has the form . Where does the angular acceleration of its slow rotation come from? . The time it takes for a body to rotate until it stops.

Answer: , .

Example 5. A round body of mass m and radius R with inertia coefficient k is spun to an angular velocity counterclockwise and placed on a horizontal surface adjacent to a vertical wall (Fig. 70). How long will it take for the body to stop and how many revolutions will it make before stopping? What will be the amount of heat released when the body rubs against the surface during this time? The coefficient of friction of the body on the surface is equal to .

Given: . Find:

Solution: The heat released during the rotation of a body until it stops is equal to the work of friction forces, which can be found using the theorem on the kinetic energy of a body. We have.

Horizontal plane reaction. The friction forces acting on the body from the horizontal and vertical surfaces are equal: and .From the system of these two equations we obtain and .

Taking these relations into account, the equation of rotational motion of a body has the form (. Whence the angular acceleration of rotation of the body is equal to. Then the time of rotation of the body before it stops, and the number of revolutions it makes.

Answer: , , , .

Example 6. A round body with a coefficient of inertia k rolls without slipping from the top of a hemisphere of radius R standing on a horizontal surface (Fig. 71). At what height and at what speed will it break away from the hemisphere and at what speed will it fall onto a horizontal surface?

Given: k, g, R. Find:

Solution: Forces act on the body . Work and 0, (there is no slipping and heat is not released at the point of adhesion of the hemisphere and the ball) therefore, to describe the motion of a body it is possible to use the law of conservation of energy. Newton's second law for the CM of a body at the point of its separation from the hemisphere, taking into account that at this point has the form , from where . The law of conservation of energy for the initial point and the point of separation of the body has the form . Whence the height and speed of separation of the body from the hemisphere are equal, .

After the body is separated from the hemisphere, only its translational kinetic energy changes, therefore the law of conservation of energy for the points of separation and fall of the body to the ground has the form . Where, taking into account, we get . For a body sliding along the surface of a hemisphere without friction, k=0 and , , .

Answer: , , .

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. The gear ratio is 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 a 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 into rotation by a relatively low-power electric motor, accumulates mechanical energy during almost a full revolution and, at a short moment of impact, releases it to the 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.

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

and kinetic energy

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

, where is the inertia tensor.

In thermodynamics

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

- (French marées, German Gezeiten, English tides) periodic fluctuations in water level due to the attraction of the Moon and the Sun. General information. P. is most noticeable along the shores of the oceans. Immediately after low tide, the ocean level begins... ... encyclopedic Dictionary F. Brockhaus and I.A. Efron

Reefer vessel Ivory Tirupati initial stability is negative Stability ability ... Wikipedia

Reefer vessel Ivory Tirupati initial stability is negative Stability is the ability of a floating craft to withstand external forces that cause it to roll or trim and return to a state of equilibrium after the end of the disturbance... ... Wikipedia

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

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

The sum on the right side of this relationship represents the moment of inertia of body 1 relative to the axis of rotation. Thus, the kinetic energy of a body rotating around a fixed axis is equal to

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 in time

Having carried out in mixed works vectors cyclic permutation of factors (see (2.34)), we obtain:

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

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

The sum of the moments of internal forces is equal to zero (see (29.12)). Consequently, denoting the total moment of external forces by 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 over time, we obtain:

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 no work, but the work of external forces is determined by formula (41.4).

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

According to equation (38.8) so, replacing through we arrive at formula (41.4).

Table 41.1

In table 41.1 the formulas of the mechanics of rotational motion are compared with similar formulas of the mechanics of translational motion (point mechanics). 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 played by the moment of force, the role of momentum is played by angular momentum, etc.

Formula. (41.1) we obtained for the case when the body rotates around a stationary axis fixed in the body. Now let us assume that the body rotates in an arbitrary manner relative to a fixed point coinciding with its center of mass.

We will rigidly associate a Cartesian coordinate system with the body, the origin of which will be placed at the center of mass of the body. Speed i-th elementary mass is equal 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 that we get:

Let us write the scalar products through projections of vectors on the axes of the coordinate system associated with the body:

Finally, combining terms with identical products of angular velocity components and taking these products out of the signs of the sums, we obtain: so formula (41.7) takes the form (cf. (41.1)). When an arbitrary body rotates around one of the main axes of inertia, say the axis and, formula (41.7) becomes (41.10.

Thus. 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, kinetic energy is determined by more complex formulas (41.5) or (41.7).

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

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

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

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

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

The basic law of the dynamics of rotational motion.

Dynamics of rotational motion

The basic law of the dynamics of rotational motion:

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

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

work in rotational motion.

Law of conservation of angular momentum.

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

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

Modulus of the angular momentum vector

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

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

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

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

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

Using the formula vi = ωri, we get

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

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

It can be shown that there is a vector equality

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

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

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

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

Pressure in liquid and gas.

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

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

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

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

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

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

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

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

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

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

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

The kinetic energy of rotational motion depends on. Kinetic energy and work during rotational motion

In thermodynamics

see also

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