Mechanical energy called the ability of a body or system of bodies to do work. There are two types of mechanical energy: kinetic and potential energy.

Kinetic energy of translational motion

Kinetic called energy due to the movement of a body. It is measured by the work done by the resultant force to accelerate a body from rest to a given speed.

Let the body have mass m begins to move under the action of a resultant force. Then the elementary work dA equal to dA = F· dl· cos. In this case, the direction of force and displacement coincide. Therefore= 0, cos = 1 and dl=  · dt, Where  - the speed with which the body is moving at a given time. This force imparts acceleration to the body
According to Newton's second law F = ma =
That's why
and full work A on a way l is equal to:
According to the definition, W k = A, That's why

(6)

From formula (6) it follows that the value of kinetic energy depends on the choice of reference system, since the velocities of bodies in different reference systems are different.

Kinetic energy of rotational motion

Let a body with a moment of inertia I z rotates about the axis z with some angular velocity. Then from formula (6), using the analogy between translational and rotational movements, we obtain:

(7)

Kinetic energy theorem

Let the body have mass T moves forward. Under the influence of various forces applied to it, the speed of the body varies from before
Then work A of these forces is equal

(8)

Where W k 1 and W k 2 - kinetic energy of the body in the initial and final states. Relationship (8) is called the theorem on kinetic energy. Its wording: the work done by all forces acting on a body is equal to the change in its kinetic energy. If a body simultaneously participates in translational and rotational movements, for example, rolling, then its kinetic energy is equal to the sum of the kinetic energy during these movements.

Conservative and non-conservative forces

If some force acts on a body at every point in space, then the totality of these forces is called force field or field . There are two types of fields - potential and non-potential (or vortex). In potential fields, bodies placed in them are subject to forces that depend only on the coordinates of the bodies. These forces are called conservative or potential . They have a remarkable property: the work of conservative forces does not depend on the path of transfer of the body and is determined only by its initial and final position. It follows that when a body moves along a closed path (Fig. 1), no work is done. Indeed, work A along the entire path is equal to the amount of work A 1B2 made on the way 1B2, and work A 2C1 on the way 2C1, i.e. A = A 1B2+ A 2C1. But work A 2C1 = – A 1C2, since the movement occurs in the opposite direction and A 1B2 = A 1C2. Then A = A 1B2 – A 1C2 = 0, which is what needed to be proven. The equality of work along a closed path to zero can be written in the form

(9)

The symbol "" on the integral means that the integration is performed along a closed curve of length l. Equality (9) is the mathematical definition of conservative forces.

In the macrocosm there are only three types of potential forces: gravitational, elastic and electrostatic forces. Non-conservative forces include friction forces called dissipative . In this case, the direction of the force And always opposite. Therefore, the work of these forces along any path is negative, as a result of which the body continuously loses kinetic energy.

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 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 a cyclic rearrangement of factors in mixed products of vectors, 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 there is no internal friction (viscosity). 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 an ideal fluid there are no friction losses, so the energy increase 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.

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Two cases of transformation of the mechanical motion of a material point or system of points:

1. mechanical motion is transferred from one mechanical system to another as mechanical motion;
2. mechanical motion turns into another form of motion of matter (into the form of potential energy, heat, electricity, etc.).

When the transformation of mechanical motion without its transition to another form of motion is considered, the measure of mechanical motion is the vector of momentum of a material point or mechanical system. The measure of the force in this case is the vector of the force impulse.

When mechanical motion turns into another form of motion of matter, the kinetic energy of a material point or mechanical system acts as a measure of mechanical motion. The measure of the action of force when transforming mechanical motion into another form of motion is the work of force

Kinetic energy

Kinetic energy is the body's ability to overcome an obstacle while moving.

Kinetic energy of a material point

The kinetic energy of a material point is a scalar quantity that is equal to half the product of the mass of the point and the square of its speed.

Kinetic energy:

• characterizes both translational and rotational movements;
• does not depend on the direction of movement of the points of the system and does not characterize changes in these directions;
• characterizes the action of both internal and external forces.

Kinetic energy of a mechanical system

The kinetic energy of the system is equal to the sum of the kinetic energies of the bodies of the system. Kinetic energy depends on the type of motion of the bodies of the system.

Determination of the kinetic energy of a solid body for different types of motion.

Kinetic energy of translational motion
During translational motion, the kinetic energy of the body is equal to T=m V 2 /2.

The measure of the inertia of a body during translational motion is mass.

Kinetic energy of rotational motion of a body

During the rotational motion of a body, kinetic energy is equal to half the product of the moment of inertia of the body relative to the axis of rotation and the square of its angular velocity.

A measure of the inertia of a body during rotational motion is the moment of inertia.

The kinetic energy of a body does not depend on the direction of rotation of the body.

Kinetic energy of plane-parallel motion of a body

With plane-parallel motion of a body, the kinetic energy is equal to

Work of force

The work of force characterizes the action of a force on a body during some movement and determines the change in the velocity modulus of the moving point.

Elementary work of force

The elementary work of a force is defined as a scalar quantity equal to the product of the projection of the force onto the tangent to the trajectory, directed in the direction of motion of the point, and the infinitesimal displacement of the point, directed along this tangent.

Work done by force on final displacement

The work done by a force on a final displacement is equal to the sum of its work on elementary sections.

The work of a force on a final displacement M 1 M 0 is equal to the integral of the elementary work along this displacement.

The work of a force on displacement M 1 M 2 is depicted by the area of ​​the figure, limited by the abscissa axis, the curve and the ordinates corresponding to the points M 1 and M 0.

The unit of measurement for the work of force and kinetic energy in the SI system is 1 (J).

Theorems about the work of force

Theorem 1. The work done by the resultant force on a certain displacement is equal to the algebraic sum of the work done by the component forces on the same displacement.

Theorem 2. The work done by a constant force on the resulting displacement is equal to the algebraic sum of the work done by this force on the component displacements.

Power

Power is a quantity that determines the work done by a force per unit of time.

The unit of power measurement is 1W = 1 J/s.

Cases of determining the work of forces

Work of internal forces

The sum of the work done by the internal forces of a rigid body during any movement is zero.

Work of gravity

Work of elastic force

Work of friction force

Work of forces applied to a rotating body

The elementary work of forces applied to a rigid body rotating around a fixed axis is equal to the product of the main moment of external forces relative to the axis of rotation and the increment in the angle of rotation.

Rolling resistance

In the contact zone of the stationary cylinder and the plane, local deformation of contact compression occurs, the stress is distributed according to an elliptical law, and the line of action of the resultant N of these stresses coincides with the line of action of the load force on the cylinder Q. When the cylinder rolls, the load distribution becomes asymmetrical with a maximum shifted towards movement. The resultant N is displaced by the amount k - the arm of the rolling friction force, which is also called the rolling friction coefficient and has the dimension of length (cm)

Theorem on the change in kinetic energy of a material point

The change in the kinetic energy of a material point at a certain displacement is equal to the algebraic sum of all forces acting on the point at the same displacement.

Theorem on the change in kinetic energy of a mechanical system

The change in the kinetic energy of a mechanical system at a certain displacement is equal to the algebraic sum of the internal and external forces acting on the material points of the system at the same displacement.

Theorem on the change in kinetic energy of a solid body

The change in the kinetic energy of a rigid body (unchanged system) at a certain displacement is equal to the sum of the external forces acting on points of the system at the same displacement.

Efficiency

Forces acting in mechanisms

Forces and pairs of forces (moments) that are applied to a mechanism or machine can be divided into groups:

1. Driving forces and moments that perform positive work (applied to the driving links, for example, gas pressure on the piston in an internal combustion engine).

2. Forces and moments of resistance that perform negative work:

• useful resistance (they perform the work required from the machine and are applied to the driven links, for example, the resistance of the load lifted by the machine),
• resistance forces (for example, friction forces, air resistance, etc.).

3. Gravity forces and elastic forces of springs (both positive and negative work, while the work for a full cycle is zero).

4. Forces and moments applied to the body or stand from the outside (reaction of the foundation, etc.), which do not do work.

5. Interaction forces between links acting in kinematic pairs.

6. The inertial forces of the links, caused by the mass and movement of the links with acceleration, can perform positive, negative work and do not perform work.

Work of forces in mechanisms

When the machine operates at a steady state, its kinetic energy does not change and the sum of the work of the driving forces and resistance forces applied to it is zero.

The work expended in setting the machine in motion is expended in overcoming useful and harmful resistances.

Mechanism efficiency

The mechanical efficiency during steady motion is equal to the ratio of the useful work of the machine to the work expended on setting the machine in motion:

Machine elements can be connected in series, parallel and mixed.

Efficiency in series connection

When mechanisms are connected in series, the overall efficiency is less than the lowest efficiency of an individual mechanism.

Efficiency in parallel connection

When mechanisms are connected in parallel, the overall efficiency is greater than the lowest and less than the highest efficiency of an individual mechanism.

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Calculation example of a spur gear
An example of calculating a spur gear. The choice of material, calculation of permissible stresses, calculation of contact and bending strength have been carried out.

An example of solving a beam bending problem
In the example, diagrams of transverse forces and bending moments were constructed, a dangerous section was found and an I-beam was selected. The problem analyzed the construction of diagrams using differential dependencies and carried out a comparative analysis of various cross sections of the beam.

An example of solving a shaft torsion problem
The task is to test the strength of a steel shaft at a given diameter, material and allowable stress. During the solution, diagrams of torques, shear stresses and twist angles are constructed. The shaft's own weight is not taken into account

An example of solving a problem of tension-compression of a rod
The task is to test the strength of a steel bar at specified permissible stresses. During the solution, diagrams of longitudinal forces, normal stresses and displacements are constructed. The rod's own weight is not taken into account

Application of the theorem on conservation of kinetic energy
An example of solving a problem using the theorem on the conservation of kinetic energy of a mechanical system

> Rotational kinetic energy: work, energy and power

Explore kinetic energy of rotational motion– formulas. Read about moment of inertia, mechanical work, translational and rotational motion.

Caused by rotation of the body.

Learning Objective

• Express rotational kinetic energy based on angular velocity and moment of inertia, and relate it to total kinetic energy.

Main points

• Rotational kinetic energy is expressed as E rotation = 0.5 Iω 2 (where ω is the moment of inertia about the axis of rotation).
• Mechanical work – W = τθ.
• Instantaneous power of the angular accelerating body – P = τω.
• There is a close connection between the result for rotational energy and that retained by linear motion.

Terms

• Inertia is the property of a body to resist any change in its uniform motion.
• Torque is the rotating effect of a force, measured in newtons per meter.
• Angular velocity is a vector quantity that characterizes a body in circular motion. The magnitude is equal to the speed of the particle, and the direction is perpendicular to the plane.

Rotational kinetic energy is the kinetic energy created by the rotation of a body and is part of the total kinetic energy. If we want to analyze a specific case, then we will need the formula E rotation = 0.5 Iω 2 (I is the moment of inertia around the axis of rotation, ω is the angular velocity).

During rotation, mechanical work is applied, representing torque (τ) multiplied by the angle of rotation (θ): W = τθ.

Instantaneous power of an angular accelerating object: P = τω.

There is a close connection between the result for rotational energy and that retained by linear (translational) motion: E translational = 0.5 mv 2 .

In a rotating system, the moment of inertia resembles mass, and the angular velocity appears linear.

Let's look at the kinetic energy of our planet. The Earth makes one axial revolution in 23.93 hours at an angular velocity of 7.29 x 10 -5. Moment of inertia – 8.04 x 10 37 kg m 2. Therefore, the rotational kinetic energy is 2.148 × 10 29 J.

The rotation of the Earth is the clearest example of rotational kinetic energy

The kinetic energy of rotational motion can also be calculated using tidal force. The additional friction from the two large tidal waves creates energy that slows the planet's angular velocity. Angular momentum is conserved, so the process transfers angular momentum to the lunar orbital motion, increasing the distance from Earth and the orbital period.

Number of rotational kinematics
Angular acceleration
Rotational kinematics
Dynamics
Rotational kinetic energy
Conservation of Angular Momentum
Vector nature of rotational kinematics
Problem solving
Linear and rotational quantities
Energy saving

« 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 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 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.