Kinetic energy around a circle formula. Kinetic energy and work during rotational motion

Main dynamic characteristics rotational movement- moment of impulse relative to the axis of rotation z:

and kinetic energy

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

, where is the inertia tensor.

In thermodynamics

Exactly by the same reasoning as in the case forward movement, equidistribution implies that when thermal equilibrium the average rotational energy of each particle of a monatomic gas: (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 oscillations 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. Ephron

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

1. Consider the rotation of the body around motionless axis Z. Let's 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:

COMMENT: 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 constant factors for the sum sign. 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 with a speed equal speed 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 get basic work, performed 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 scalar product two vectors is called 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 the body rotates internal forces no work is done, 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 the 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. Overall change mechanical energy of all liquid 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 this volume, and the complete 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 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 horizontal line current 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.

Real liquid inherent 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 to move a plate with constant speed it is necessary to act on it with 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 different direction vectors F tr And v o. If we examine the speed of liquid particles in different places of the layer, it turns out that it varies according to 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- coefficient dynamic viscosity . 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 changing z per unit. Let us put numerically in formula (5.8) dv/dz =-1 and S= 1, we get h = F. It follows physical meaning h: viscosity coefficient numerically equal to force, which acts on a layer of liquid of unit area with a velocity gradient, equal to one. The SI unit of viscosity is called the pascal second (denoted Pa s). In the system GHS unit viscosity is 1 poise (P), with 1 Pa s = 10P.

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 linear motion.

Galileo-Einstein principle of relativity- All natural phenomena at any time inertial system counts 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 movement solid. 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 dependence of its acceleration a = a(t) on the time and the known initial conditions V 0 and r 0 .

Question No. 7

Pulse (Quantity of movement) - vector physical quantity, characterizing the measure mechanical movement bodies. IN classical mechanics body impulse equal to the product masses 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 generalized speed

If the Lagrangian of the system does not depend on some generalized coordinates, then due to Lagrange equations .

For free particle the Lagrange function has the form: , hence:

Independence of the Lagrangian closed system from its position in space follows from the property homogeneity of space: for good 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 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 unit of impulse is in International system units (SI) 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: moment of inertia equal to the sum products of elementary masses by the square of their distances to base set

Axial moment of inertia

Axial moments of inertia of some bodies.

Moment of inertia of a 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 - gross weight bodies.

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.

Basic kinematic characteristics rotational motion of a body - 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. The rotational energy of 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 in a broad sense, not just 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 a 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 by 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

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

Theorem about the kinetic energy of a body performing complex movement, will have the same form as for its translational motion: .

Example 1. A body of mass m is tied 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 pull of the thread 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 general moment 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 .

The kinetic energy of the system, due to the conservation of its angular momentum, immediately after the thread jerks, 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 inclined plane with inclination angle α report the same initial speeds, directed upward along an 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). Works normal reaction and the adhesion friction forces (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.

Reaction horizontal plane. 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 . Law of conservation of energy for starting point and the separation point 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 . From where, taking into account we get . For a body sliding along the surface of a hemisphere without friction, k=0 and , , .

Answer: , , .

« 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 time period 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 fixed axis rotation is 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.” On a bench that has vertical axis rotation passing through its center, a person stands up. 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:

Angular velocity 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.

Kinetic energy around a circle formula. Kinetic energy and work during rotational motion

In thermodynamics

See also

See what “Energy of rotational motion” is in other dictionaries: