Kinematic energy of rotational motion. Kinetic energy of a rotating body

The main dynamic characteristics of rotational motion are the angular momentum about the rotation axis z:

and kinetic energy

In the general case, 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 forward movement, 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 one to calculate the root mean square angular velocity molecules.

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« Physics - Grade 10 "

Why does the skater stretch along the axis of rotation to increase the angular velocity of rotation.
Should a helicopter rotate when its propeller 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 must 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 speed in absolute value, but in the opposite direction (Fig. 6.9).

But the disk is moving, the angular velocity of rotation of all particles is the same. However, it is clear that the farther the particle is from the axis of rotation, the greater its momentum. Therefore, for rotational motion it is necessary to introduce one more characteristic, similar to an impulse, - the 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):

The linear and angular velocities are related by v = ωr, then

All points of a rigid matter move relative to a fixed axis of rotation with the same angular velocity. A rigid body can be represented as a set 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:

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

The basic equation of the dynamics of rotational motion in impulsive form.

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

In this way,

∆L = M∆t. (6.4)

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

Law of conservation of angular momentum:

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

∆L=0, L=const.

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

The spinning skater spreads his arms out to the sides, thereby increasing the moment of inertia in order to decrease the angular velocity of rotation.

The law of conservation of angular momentum can be demonstrated using the following experiment, called the "experiment with the Zhukovsky bench." A person stands on a bench with a vertical axis of rotation passing through its center. The man holds dumbbells in his hands. If the bench is made to rotate, then a person can change the speed of rotation by pressing the dumbbells to his chest or lowering his arms, and then spreading them apart. Spreading his arms, he increases the moment of inertia, and the angular velocity of rotation decreases (Fig. 6.11, a), lowering his hands, he reduces the moment of inertia, and the angular velocity 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 must 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 axis of rotation, if external forces do not act on this axis. In the 19th century gyroscopes were used by navigators to navigate the 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 brackets, as we already know, is the moment of inertia of the rigid body. Finally, the formula for the kinetic energy of a rigid body with 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 motions. So, 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 of the mechanics of the translational motion of a material point with similar formulas for the rotational motion of a rigid body.

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

The total kinetic energy of the rotating solid 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, so the elementary work external forces is equal to the increment of 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 .

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 i-th speed points , 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 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 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, making chaotic, chaotic movement, are not connected or rather weakly connected by interaction forces, because of which 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 of the 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 respects, 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)

Physical quantity determined 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 weight liquid (gas) displaced by the body: FA=ρgV, where ρ is the density of the liquid, V is the volume of the body immersed in the liquid.

Consider an absolutely rigid body rotating about a fixed axis. Let's mentally break this body into infinitely small pieces with infinitely small sizes and masses. m v t., t 3 ,... at distances R v R 0 , R 3 ,... from the axis. Kinetic energy of a rotating body we find as the sum of the kinetic energies of its small parts:

- moment of inertia rigid body relative to the given axis 00,. From a comparison of the formulas for the kinetic energy of translational and rotational motions, it is obvious that moment of inertia in rotational motion is analogous to mass in translational motion. Formula (4.14) is convenient for calculating the moment of inertia of systems consisting of individual material points. To calculate the moment of inertia of solid bodies, using the definition of the integral, you can convert it to the form

It is easy to see that the moment of inertia depends on the choice of axis and changes with its parallel translation and rotation. Let us find the values of the moments of inertia for some homogeneous bodies.

From formula (4.14) it is obvious that moment of inertia of a material point equals

where t - point mass; R- distance to the axis of rotation.

It is easy to calculate the moment of inertia for hollow thin-walled cylinder(or a special case of a cylinder with a small height - thin ring) radius R about the axis of symmetry. The distance to the axis of rotation of all points for such a body is the same, equal to the radius and can be taken out from under the sign of the sum (4.14):

Rice. 4.5

solid cylinder(or special case low height cylinder disk) radius R to calculate the moment of inertia about the axis of symmetry requires the calculation of the integral (4.15). It can be understood in advance that the mass in this case, on average, is concentrated somewhat closer to the axis than in the case of a hollow cylinder, and the formula will be similar to (4.17), but a coefficient less than one will appear in it. Let's find this coefficient. Let a solid cylinder have density p and height A. Let us divide it into hollow cylinders (thin cylindrical surfaces) with thickness dr(Fig. 4.5 shows a projection perpendicular to the axis of symmetry). The volume of such a hollow cylinder of radius r equal to area surface times thickness: dV = 2nrhdr, weight: dm=2nphrdr, and the moment of inertia in accordance with formula (4.17): dj=

= r 2 dm = 2lr/?g Wr. The total moment of inertia of a solid cylinder is obtained by integrating (summing) the moments of inertia of hollow cylinders:

Similarly searched moment of inertia of a thin rod length L and the masses t, if the axis of rotation is perpendicular to the rod and passes through its middle. Let's break this

Taking into account the fact that the mass of a solid cylinder is related to the density by the formula t = nR 2 hp, we finally have moment of inertia of a solid cylinder:

Rice. 4.6

rod in accordance with fig. 4.6 pieces thick dl. The mass of such a piece is dm = mdl/L, and the moment of inertia in accordance with formula (4.6): dj = l 2 dm = l 2 mdl/L. The total moment of inertia of a thin rod is obtained by integrating (summing) the moments of inertia of the pieces:

Taking the elementary integral gives the moment of inertia of a thin rod of length L and the masses t

Rice. 4.7

The integral is taken somewhat more complicated when searching moment of inertia of a homogeneous ball radius R and mass /77 with respect to the axis of symmetry. Let a solid ball have density p. Let's break it down as shown in Fig. 4.7 for hollow thin cylinders thickness dr, whose symmetry axis coincides with the axis of rotation of the ball. The volume of such a hollow cylinder of radius G is equal to the surface area multiplied by the thickness:

where is the height of the cylinder h found using the Pythagorean theorem: