Kinetic energy of rotation

Lecture 3. Dynamics of a rigid body

Lecture plan

3.1. Moment of power.

3.2. Basic equations of rotational motion. Moment of inertia.

3.3. Kinetic energy of rotation.

3.4. moment of impulse. Law of conservation of angular momentum.

3.5. Analogy between translational and rotational motion.

Moment of power

Consider the motion of a rigid body around a fixed axis. Let a rigid body have a fixed axis of rotation ОО ( fig.3.1) and an arbitrary force is applied to it.

Rice. 3.1

We decompose the force into two components of the force, the force lies in the plane of rotation, and the force is parallel to the axis of rotation. Then we decompose the force into two components: – acting along the radius vector and – perpendicular to it.

Not any force applied to a body will rotate it. Forces and create pressure on the bearings, but do not rotate it.

The force may or may not bring the body out of balance, depending on where in the radius vector it is applied. Therefore, the concept of the moment of force about the axis is introduced. Moment of force relative to the axis of rotation is called the vector product of the radius vector and the force.

The vector is directed along the axis of rotation and is determined by the cross product rule or the right screw rule, or the gimlet rule.

Modulus of moment of force

where α is the angle between the vectors and .

From Fig.3.1. it's clear that .

r0- the shortest distance from the axis of rotation to the line of action of the force and is called the shoulder of the force. Then the moment of force can be written

M = F r 0 . (3.3)

From fig. 3.1.

where F is the projection of the vector onto the direction perpendicular to the vector radius vector . In this case, the moment of force is

. (3.4)

If several forces act on the body, then the resulting moment of force is equal to the vector sum of the moments separate forces, but since all the moments are directed along the axis, they can be replaced by an algebraic sum. The moment will be considered positive if it rotates the body clockwise and negative if counterclockwise. If all moments of forces are equal to zero (), the body will be in equilibrium.

The concept of a moment of force can be demonstrated using a "whimsical coil". The spool of thread is pulled by the free end of the thread ( rice. 3.2).

Rice. 3.2

Depending on the direction of the thread tension, the coil rolls in one direction or another. If you pull at an angle α , then the moment of force about the axis O(perpendicular to the figure) rotates the coil counterclockwise and it rolls back. In case of tension at an angle β the torque is counterclockwise and the coil rolls forward.

Using the equilibrium condition (), you can design simple mechanisms that are "converters" of force, i.e. By applying less force, you can lift and move loads of different weights. Leverage, wheelbarrows, blocks of various kinds, which are widely used in construction, are based on this principle. To comply with the equilibrium condition in construction cranes to compensate for the moment of force caused by the weight of the load, there is always a system of counterweights that creates a moment of force of the opposite sign.

3.2. Basic rotational equation
movement. Moment of inertia

Consider an absolutely rigid body rotating around a fixed axis OO(fig.3.3). Let's mentally divide this body into elements with masses Δ m 1, Δ m2, …, Δ m n. During rotation, these elements will describe circles with radii r1,r2 , …,rn. Forces act on each element F1,F2 , …,F n. Rotation of a body around an axis OO occurs under the influence of the total moment of forces M.

M \u003d M 1 + M 2 + ... + M n (3.4)

where M 1 = F 1 r 1, M 2 = F 2 r 2, ..., M n = F n r n

According to Newton's second law, each force F, acting on an element of mass D m, causes acceleration of the given element a, i.e.

F i = D m i a i (3.5)

Substituting the corresponding values ​​into (3.4), we obtain

Rice. 3.3

Knowing the relationship between linear angular acceleration ε () and that the angular acceleration is the same for all elements, formula (3.6) will look like

M = (3.7)

=I (3.8)

I is the moment of inertia of the body about the fixed axis.

Then we will get

M = I ε (3.9)

Or in vector form

(3.10)

This equation is the basic equation for the dynamics of rotational motion. It is similar in form to Equation II of Newton's law. From (3.10) the moment of inertia is

Thus, the moment of inertia of a given body is the ratio of the moment of force to the angular acceleration caused by it. From (3.11) it can be seen that the moment of inertia is a measure of the body's inertia with respect to rotational motion. The moment of inertia plays the same role as mass in translational motion. SI unit [ I] = kg m 2. From formula (3.7) it follows that the moment of inertia characterizes the distribution of the masses of the particles of the body relative to the axis of rotation.

So, the moment of inertia of an element of mass ∆m moving along a circle of radius r is equal to

I = r2 D m (3.12)

I= (3.13)

In the case of a continuous mass distribution, the sum can be replaced by the integral

I= ∫ r 2 dm (3.14)

where integration is performed over the entire body mass.

This shows that the moment of inertia of the body depends on the mass and its distribution relative to the axis of rotation. This can be demonstrated experimentally fig.3.4).

Rice. 3.4

Two round cylinders, one hollow (for example, metal), the other solid (wooden) with the same lengths, radii and masses, begin to roll down simultaneously. A hollow cylinder with a large moment of inertia will lag behind a solid one.

You can calculate the moment of inertia if you know the mass m and its distribution relative to the axis of rotation. The simplest case is a ring, when all elements of the mass are located equally from the axis of rotation ( rice. 3.5):

I= (3.15)

Rice. 3.5

Let us give expressions for the moments of inertia of different symmetrical bodies with mass m.

1. Moment of inertia rings, hollow thin-walled cylinder about the axis of rotation coinciding with the axis of symmetry.

, (3.16)

r is the radius of the ring or cylinder

2. For a solid cylinder and disk, the moment of inertia about the axis of symmetry

(3.17)

3. The moment of inertia of the ball about the axis passing through the center

(3.18)

r- ball radius

4. The moment of inertia of a thin rod of long l relative to an axis perpendicular to the rod and passing through its middle

(3.19)

l- the length of the rod.

If the axis of rotation does not pass through the center of mass, then the moment of inertia of the body about this axis is determined by Steiner's theorem.

(3.20)

According to this theorem, the moment of inertia about an arbitrary axis О'O' ( ) is equal to the moment of inertia about a parallel axis passing through the center of mass of the body ( ) plus the product of body mass times the square of the distance a between axles ( rice. 3.6).

Rice. 3.6

Kinetic energy of rotation

Consider the rotation of an absolutely rigid body around a fixed axis OO with an angular velocity ω (rice. 3.7). Let's split the rigid body into n elementary masses ∆ m i. Each element of the mass rotates on a circle of radius r i with linear speed (). Kinetic energy is the sum of the kinetic energies of individual elements.

(3.21)

Rice. 3.7

Recall from (3.13) that is the moment of inertia about the OO axis.

Thus, the kinetic energy of a rotating body

E k \u003d (3.22)

We have considered the kinetic energy of rotation around a fixed axis. If the body is involved in two movements: in translational and rotational motions, then the kinetic energy of the body is the sum of the kinetic energy of translational motion and the kinetic energy of rotation.

For example, a ball of mass m rolling; the center of mass of the ball moves forward with a speed u (rice. 3.8).

Rice. 3.8

The total kinetic energy of the ball will be equal to

(3.23)

3.4. moment of impulse. conservation law
angular momentum

Physical quantity equal to the product of the moment of inertia I to angular speed ω , is called the angular momentum (moment of momentum) L about the axis of rotation.

– angular momentum is a vector quantity and coincides in direction with the direction of the angular velocity .

Differentiating equation (3.24) with respect to time, we obtain

where, M is the total moment of external forces. In an isolated system, there is no moment of external forces ( M=0) and

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 a body is the sum of the moments of inertia of all material points bodies:

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 a significant amount of 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,

In this way,

Rotation energy

Consequently,

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.

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

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

The sum on the right side of this ratio is the moment of inertia of body 1 about the axis of rotation. Thus, the kinetic energy of a body rotating around a fixed axis is

Let an internal force and an external force act on the mass (see Fig. 41.1). According to (20.5), these forces will do work during the time

Carrying out a cyclic permutation of factors in mixed products of vectors (see (2.34)), we obtain:

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

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

The sum of moments of internal forces is equal to zero (see (29.12)). Therefore, denoting the total moment of external forces through N, we arrive at the expression

(we used formula (2.21)).

Finally, taking into account that there is an angle through which the body rotates in time, we get:

The sign of the work depends on the sign, i.e., on the sign of the projection of the vector N onto the direction of the vector

So, when the body rotates, the internal forces do not perform work, while the work of external forces is determined by formula (41.4).

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

According to the equation (38.8) so, replacing through we will come to the formula (41.4).

Table 41.1

In table. 41.1, the formulas of the mechanics of rotational motions are compared with similar formulas of the mechanics of translational motion (the mechanics of a point). From this comparison it is easy to conclude that in all cases the role of mass is played by the moment of inertia, the role of force is the moment of force, the role of momentum is played by the moment of momentum, etc.

Formula. (41.1) we obtained for the case when the body rotates around a fixed axis fixed in the body. Now let's assume that the body rotates arbitrarily about a fixed point coinciding with its center of mass.

Let us rigidly connect the Cartesian coordinate system with the body, the origin of which will be placed at the center of mass of the body. i-th speed elementary mass is Therefore, for the kinetic energy of the body, we can write the expression

where is the angle between the vectors Replacing a through and taking into account what we get:

We write the scalar products in terms of the projections of vectors on the axes of the coordinate system associated with the body:

Finally, by combining the terms with the same products of the components of the angular velocity and taking these products out of the signs of the sums, we get: so that formula (41.7) takes the form (compare with (41.1)). When an arbitrary body rotates around one of the main axes of inertia, say the axes and formula (41.7) goes into (41.10.

In this way. the kinetic energy of a rotating body is equal to half the product of the moment of inertia and the square of the angular velocity in three cases: 1) for a body rotating around a fixed axis; 2) for a body rotating around one of the main axes of inertia; 3) for a ball top. In other cases, the kinetic energy is determined by the more complex formulas (41.5) or (41.7).

Tasks

1. Determine how many times the effective mass is greater than the gravitating mass of a train with a mass of 4000 tons, if the mass of the wheels is 15% of the mass of the train. Consider the wheels as disks with a diameter of 1.02 m. How will the answer change if the diameter of the wheels is half that?

2. Determine the acceleration with which a wheel pair of mass 1200 kg rolls down a hill with a slope of 0.08. Consider wheels as disks. Rolling resistance coefficient 0.004. Determine the adhesion force of the wheels to the rails.

3. Determine the acceleration with which a wheel pair with a mass of 1400 kg rolls up a hill with a slope of 0.05. Drag coefficient 0.002. What should be the coefficient of adhesion so that the wheels do not slip. Consider wheels as disks.

4. Determine the acceleration with which a wagon weighing 40 tons rolls down a hill with a slope of 0.020 if it has eight wheels weighing 1200 kg and a diameter of 1.02 m. Determine the force of adhesion of the wheels to the rails. Drag coefficient 0.003.

5. Determine the pressure force of the brake shoes on the tires, if a train weighing 4000 tons slows down with an acceleration of 0.3 m/s 2 . The moment of inertia of one wheelset is 600 kg m 2 , the number of axles is 400, the sliding friction coefficient of the block is 0.18, the rolling resistance coefficient is 0.004.

6. Determine the braking force acting on a four-axle wagon weighing 60 tons on the brake pad hump if the speed on the path of 30 m decreased from 2 m/s to 1.5 m/s. The moment of inertia of one wheelset is 500 kg m 2 .

7. The speedometer of the locomotive showed an increase in the speed of the train within one minute from 10 m/s to 60 m/s. Probably, there was a slipping of the leading wheelset. Determine the moment of forces acting on the armature of the electric motor. Moment of inertia of wheelset 600 kg m 2 , anchors 120 kg m 2 . Gear ratio gear 4.2. The pressure force on the rails is 200 kN, the sliding friction coefficient of the wheels along the rail is 0.10.

11. KINETIC ENERGY OF THE ROTATOR

MOVEMENTS

We derive the formula for the kinetic energy of rotational motion. Let the body rotate with angular velocity ω about the fixed axis. Any small particle of the body performs translational motion in a circle with a speed , where r i - distance to the axis of rotation, radius of the orbit. Kinetic energy of a particle masses m i is equal to . The total kinetic energy of a system of particles is equal to the sum of their kinetic energies. Let us sum up the formulas for the kinetic energy of the particles of the body and take out the sign of the sum of half the square of the angular velocity, which is the same for all particles, . The sum of the products of the masses of particles and the squares of their distances to the axis of rotation is the moment of inertia of the body about the axis of rotation . So, the kinetic energy of a body rotating about a fixed axis is equal to half the product of the moment of inertia of the body about the axis and the square of the angular velocity of rotation:

Rotating bodies can store mechanical energy. Such bodies are called flywheels. Usually these are bodies of revolution. The use of flywheels in the potter's wheel has been known since antiquity. In internal combustion engines, during the working stroke, the piston imparts mechanical energy to the flywheel, which then performs work on the rotation of the engine shaft for the next three cycles. In stamps and presses, the flywheel is driven by a relatively low-power electric motor, accumulates mechanical energy for almost a full revolution and, in a short moment of impact, gives it to the work of stamping.

Numerous attempts are known to use rotating flywheels to drive Vehicle: cars, buses. They are called mahomobiles, gyro carriers. Many such experimental machines were created. It would be promising to use flywheels for energy storage during braking of electric trains in order to use the accumulated energy during subsequent acceleration. Flywheel energy storage is known to be used on New York City subway trains.

The main dynamic characteristics of rotational motion 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:

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.

see also

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