Kinetic energy during rotation. Kinetic energy during rotational motion

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 velocity of the i-th point , Ri is the distance to the axis of rotation. Hence,

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 a physical quantity determined by a 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 a chaotic, chaotic movement, are not bound or rather weakly bound 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 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 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 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)

The physical quantity determined by the 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 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.

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 velocity of the elementary mass m i– v i = w R i, where R i– distance of mass m i from the axis of rotation. Therefore, the 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 about the axis of rotation.

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

2. Let the body now revolves about some axis, and axis moves progressively, remaining parallel to itself.

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

Speed i-that elementary mass of the body is equal to , where is the speed of some point "O" of the body; – radius-vector that determines the position of the elementary mass in relation to the 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 into account this remark, we can write that , where is the distance of the mass from the axis of rotation. In the second term, we make a cyclic permutation of the factors, after which we obtain

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

The sum of elementary masses is the mass of the body "m". The expression is equal to the product of the body mass and the radius vector of the body's center of inertia (by definition of the center of inertia). Finally, - the moment of inertia of the body about the axis passing through the point "O". Therefore, one 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 the point "C", we get:

(4.6)

Thus, the kinetic energy of a body during plane motion is composed of the energy of translational motion with 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.

The work of external forces during the rotational motion of a rigid body.

Find the work done by the forces when the body rotates around the fixed 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 make in time dt job:

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

where , - respectively, the moments of the internal and external forces relative to the point "O".

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

The sum of the moments of internal forces is equal to 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 multiplied vectors and the projection of the second one onto the direction of the first, taking into account that , (the directions of the Z axis and coincide), we get

but w dt=d j, i.e. the angle through which the body rotates in time dt. So

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, the internal forces do no work, and the work of external forces is determined by the formula .

The work done over a finite time interval is found by integrating

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

, i.e. .

Those. the work of 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 and the direction and angle of rotation.

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

;

Hence,

. (4.7)

On one's own:

Elastic forces;

Hooke's law.

LECTURE 7

Hydrodynamics

Lines and tubes of current.

Hydrodynamics studies the motion of liquids, but its laws apply to the motion of gases as well. In a stationary fluid flow, the velocity of its particles at each point in space is a quantity independent of time and a function of the coordinates. In a stationary flow, the trajectories of fluid particles form a streamline. The set of streamlines forms a stream tube (Fig. 5.1). We assume that the liquid is incompressible, then the volume of liquid flowing through the sections S 1 and S 2 will be the same. In a second, a volume of fluid equal to

, (5.1)

where and are fluid velocities in cross 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 velocity is inversely proportional to the cross section of the current tube.

Bernoulli equation.

We will consider an ideal incompressible fluid in which there is no internal friction (viscosity). Let us single out a thin tube of current in a stationary flowing liquid (Fig. 5.2) with cross sections S1 and S2 perpendicular to the streamlines. in section 1 in a short time t particles move a distance l 1, and in the section 2 - at a distance l 2. Through both sections in time t equal small volumes of liquid will pass V= V 1 = V 2 and carry a lot of liquid m=rV, where r is the density of the liquid. In general, the change in the mechanical energy of the entire liquid in the current tube between sections S1 and S2, which happened during the time t, can be replaced by the change in volume energy V, which 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 - velocity of fluid particles in sections S1 and S2 respectively; g- acceleration of gravity; h1 and h2- heights of the center of the sections.

In an ideal fluid, there are no friction losses, so the energy increment DE must be equal to the work done by the 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 the terms with the same indices to one part of the equality, we obtain

. (5.4)

Tube sections S1 and S2 were taken arbitrarily, so it can be argued that the expression is valid in any section of the current tube

. (5.5)

Equation (5.5) is called the Bernoulli 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.

Forces of internal friction.

Viscosity is inherent in a real liquid, which manifests itself in the fact that any movement of liquid and gas spontaneously stops in the absence of the causes that caused it. Let us consider an experiment in which a liquid layer is located above a fixed surface, and a plate floating on it with a surface moves from above it with a speed S(Fig. 5.3). Experience shows that in order to move the 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 force equal to it in magnitude and oppositely directed, 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 direction of the vectors F tr and v o. If we examine the velocity of fluid 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) takes the form

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

where h- dynamic viscosity coefficient. Value dv/dz called the velocity gradient. It shows how fast the speed changes in the direction of the axis z. At dv/dz= const velocity gradient is numerically equal to velocity change v when it changes z per unit. We 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 liquid layer of unit area at a velocity gradient equal to one. 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.

Mechanics.

Question #1

Reference system. Inertial reference systems. Galileo-Einstein's principle of relativity.

reference system- 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 (ISO)- a system in which a freely moving body is at rest or uniform rectilinear motion.

Galileo-Einstein's principle of relativity- All phenomena of nature in any inertial frame of reference occur in the same way and have the same mathematical form. In other words, all ISOs are equal.

Question #2

The equation of motion. Types of motion of a rigid body. The main task of kinematics.

Equations of motion of a material point:

- kinematic equation of motion

Types of motion of a rigid body:

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 dependences of the velocity V= V(t) and the coordinates (or radius vector) r = r(t) of a material point from the known time dependence of its acceleration a = a(t) and the known initial conditions V 0 and r 0 .

Question #7

Pulse (Number of movement) is a vector physical quantity that characterizes the measure of the mechanical movement of the body. In classical mechanics, the momentum of a body is equal to the product of the mass m this point to its speed v, the direction of the momentum coincides with the direction of the velocity vector:

In theoretical mechanics generalized momentum is the partial derivative of the Lagrangian of the system with respect to the generalized velocity

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

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

The independence of the Lagrangian of a closed system from its position in space follows from the property homogeneity of space: for a well-isolated system, its behavior does not depend on where in space we place it. By Noether's theorem this homogeneity implies the conservation of some physical quantity. This quantity is called the impulse (ordinary, not generalized).

In classical mechanics, complete momentum system of material points is called a vector quantity equal to the sum of the products of the masses of material points at their speed:

accordingly, the quantity is called the momentum of one material point. It is a vector quantity directed in the same direction as the particle's velocity. The unit of momentum in the International System of 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 sum over them, as a result we get:

The momentum of a system that is not affected by any external forces (or they are compensated), preserved in time:

The conservation of momentum in this case follows from Newton's second and third laws: having written Newton's second law for each of the material points that make up the system and summing it over all the material points that make up the system, by virtue of Newton's third law we obtain the 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, the three-dimensional momentum is not a relativistically invariant quantity, since it depends on the frame of reference. A more meaningful value will be a four-dimensional momentum, which for one material point is defined as

In practice, the following relationships between the mass, momentum, and energy of a particle are often used:

In principle, for a system of non-interacting material points, their 4-momenta are summed. However, for interacting particles in relativistic mechanics, one should take into account the momenta not only of the particles that make up the system, but also the momentum of the field of interaction 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. It is characterized by the distribution of masses in the body: the moment of inertia is equal to the sum of the products of elementary masses and the square of their distances to the base set

Axial moment of inertia

Axial moments of inertia of some bodies.

The moment of inertia of a mechanical system relative to a fixed axis ("axial moment of inertia") is called the value J a equal to the sum of the products of the masses of all n material points of the system into 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 volume element of the body 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

Formula derivation

dm and moments of inertia DJ i. Then

Thin-walled cylinder (ring, hoop)

Formula derivation

The moment of inertia of a body is equal to the sum of the moments of inertia of its constituent parts. Dividing a thin-walled cylinder into elements with a 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 converted to the form

Steiner's theorem

Moment of inertia of a rigid body relative to any axis depends not only on the mass, shape and dimensions of the body, but also on the position of the body with respect to this axis. According to the Steiner theorem (Huygens-Steiner theorem), moment of inertia body J relative to an arbitrary axis is equal to the sum moment of inertia this body Jc relative to the axis passing through the center of mass of the body parallel to the considered axis, and the product of the body mass m per square distance d between axles:

If is the moment of inertia of the body about an axis passing through the center of mass of the body, then the moment of inertia about a parallel axis located at a distance from it is equal to

where is the total mass of the body.

For example, the moment of inertia of a rod about an axis passing through its end is:

Rotational energy

Kinetic energy of rotational motion- the energy of the body associated with its rotation.

The main kinematic characteristics of the rotational motion of a body are its angular velocity (ω) and angular acceleration. The main dynamic characteristics of rotational motion are the angular momentum about the rotation axis z:

Kz = Izω

and kinetic energy

where I z is the moment of inertia of the body about 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 are the principal components of the angular velocity.

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

, where I is the inertia tensor.

Question #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 about the axis of rotation, and how fast the rotation occurs.

It should be noted that rotation here is understood in a broad sense, not only as a regular rotation around an axis. For example, even with a rectilinear motion of a body past an arbitrary imaginary point that does not lie on the line of motion, it also has an angular momentum. Perhaps the greatest role is played by the angular momentum in describing the actual rotational motion. However, it is extremely important for a much wider class of problems (especially if the problem has central or axial symmetry, but not only in these cases).

Law of conservation of momentum(law of conservation of angular momentum) - the vector sum of all angular momenta about 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 with respect to any non-time derivative of the angular momentum is the moment of force:

Thus, the requirement of system closure 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 met).

Mathematically, the law of conservation of angular momentum follows from the isotropy of space, that is, from the invariance of space with respect to rotation through an arbitrary angle. When rotating through an arbitrary infinitesimal angle , the radius vector of the particle with the number will change by , and the velocities - . The Lagrange function of the system will not change during such a rotation, due to the isotropy of space. So

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 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 one to calculate the root-mean-square angular velocity of molecules.

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The kinetic energy of a rotating body is equal to the sum of the kinetic energies of all particles of the body:

The mass of any particle, its linear (circumferential) speed, proportional to the distance of this particle from the axis of rotation. Substituting into this expression and taking the angular velocity o common for all particles out of the sign of the sum, we find:

This formula for the kinetic energy of a rotating body can be reduced to a form similar to the expression for the kinetic energy of translational motion if we introduce the value of the so-called moment of inertia of the body. The moment of inertia of a material point is the product of the mass of the point and the square of its distance from the axis of rotation. The moment of inertia of the body is the sum of the moments of inertia of all material points of the body:

So, the kinetic energy of a rotating body is determined by the following formula:

Formula (2) differs from the formula that determines the kinetic energy of a body in translational motion in that instead of the body mass, the moment of inertia I enters here and instead of the velocity, the group velocity

The large kinetic energy of a rotating flywheel is used in technology to maintain the uniformity of the machine under a suddenly changing load. At first, to bring the flywheel with a large moment of inertia into rotation, the machine requires considerable work, but when a large load is suddenly turned on, the machine does not stop and does work due to the flywheel's kinetic energy reserve.

Particularly massive flywheels are used in rolling mills driven by an electric motor. Here is a description of one of these wheels: “The wheel has a diameter of 3.5 m and weighs At a normal speed of 600 rpm, the kinetic energy of the wheel is such that at the time of rolling the wheel gives the mill a power of 20,000 liters. with. Friction in the bearings is kept to a minimum by a fairy tale under pressure, and in order to avoid the harmful effect of centrifugal inertia forces, the wheel is balanced so that the load placed on the circumference of the wheel brings it out of rest.

We present (without performing calculations) the values of the moments of inertia of some bodies (it is assumed that each of these bodies has the same density in all its sections).

The moment of inertia of a thin ring about an axis passing through its center and perpendicular to its plane (Fig. 55):

The moment of inertia of a round disk (or cylinder) about an axis passing through its center and perpendicular to its plane (the polar moment of inertia of the disk; Fig. 56):

The moment of inertia of a thin round disk about an axis coinciding with its diameter (equatorial moment of inertia of the disk; Fig. 57):

The moment of inertia of the ball about the axis passing through the center of the ball:

Moment of inertia of a thin spherical layer of radius about an axis passing through the center:

The moment of inertia of a thick spherical layer (a hollow ball having an outer surface radius and a cavity radius) about an axis passing through the center:

The calculation of the moments of inertia of bodies is carried out using integral calculus. To give an idea of the course of such calculations, we find the moment of inertia of the rod relative to the axis perpendicular to it (Fig. 58). Let there be a section of the rod, density. We single out an elementarily small part of the rod, which has a length and is located at a distance x from the axis of rotation. Then its mass Since it is at a distance x from the axis of rotation, then its moment of inertia We integrate from zero to I:

Moment of inertia of a rectangular parallelepiped about the axis of symmetry (Fig. 59)

Moment of inertia of the annular torus (Fig. 60)

Let us consider how the energy of rotation of a body rolling (without sliding) along the plane is connected with the energy of the translational motion of this body,

The energy of the translational motion of a rolling body is , where is the mass of the body and the velocity of the translational motion. Let denote the angular velocity of rotation of the rolling body and the radius of the body. It is easy to understand that the speed of the translational motion of a body rolling without sliding is equal to the circumferential speed of the body at the points of contact of the body with the plane (during the time when the body makes one revolution, the center of gravity of the body moves a distance, therefore,

Thus,

Rotation energy

hence,

Substituting here the above values of the moments of inertia, we find that:

a) the energy of the rotational motion of the rolling hoop is equal to the energy of its translational motion;

b) the energy of rotation of a rolling homogeneous disk is equal to half the energy of translational motion;

c) the energy of rotation of a rolling homogeneous ball is the energy of translational motion.

The dependence of the moment of inertia on the position of the axis of rotation. Let the rod (Fig. 61) with the center of gravity at point C rotate with an angular velocity (o around the axis O, perpendicular to the plane of the drawing. Suppose that over a certain period of time it moved from position A B to and the center of gravity described an arc. This movement rod can be considered as if the rod first translationally (that is, remaining parallel to itself) moved to position and then rotated around C to position Let us denote (the distance of the center of gravity from the axis of rotation) by a, and the angle by When the rod moves from position And In position, the displacement of each of its particles is the same as the displacement of the center of gravity, i.e. it is equal to or To obtain the actual movement of the rod, we can assume that both of these movements are performed simultaneously. about the axis passing through O can be decomposed into two parts.

Kinetic energy during rotation. Kinetic energy during rotational motion

In thermodynamics

see also

See what "Energy of rotational motion" is in other dictionaries: