Let us first consider a rigid body rotating around a fixed axis OZ with angular velocity ω (Fig. 5.6). Let's break the body into elementary masses. The linear speed of the elementary mass is equal to , where is its distance from the axis of rotation. Kinetic energy i-that elementary mass will be equal to
.
The kinetic energy of the whole body is composed of the kinetic energies of its parts, therefore
.
Taking into account that the sum on the right side of this relation represents the moment of inertia of the body relative to the axis of rotation, we finally obtain
. (5.30)
Formulas for the kinetic energy of a rotating body (5.30) are similar to the corresponding formulas for the kinetic energy of translational motion of a body. They are obtained from the latter by a formal replacement 
.
In the general case, the motion of a rigid body can be represented as a sum of motions - translational at a speed equal to the speed of the center of mass of the body, and rotation at an angular velocity around an instantaneous axis passing through the center of mass. In this case, the expression for the kinetic energy of the body takes the form
.
Let us now find the work done by the moment of external forces during the rotation of a rigid body. Elementary work of external forces in time dt will be equal to the change in kinetic energy of the body
Taking the differential from the kinetic energy of rotational motion, we find its increment
.
In accordance with the basic equation of dynamics for rotational motion
Taking into account these relations, we reduce the expression of elementary work to the form
where is the projection of the resulting moment of external forces on the direction of the axis of rotation OZ, is the angle of rotation of the body over the considered period of time.
Integrating (5.31), we obtain a formula for the work of external forces acting on a rotating body
If , then the formula simplifies
Thus, the work of external forces during rotation of a rigid body relative to a fixed axis is determined by the action of the projection of the moment of these forces onto this axis.
Gyroscope
A gyroscope is a rapidly rotating symmetrical body, the axis of rotation of which can change its direction in space. So that the axis of the gyroscope can rotate freely in space, the gyroscope is placed in the so-called gimbal suspension (Fig. 5.13). The gyroscope flywheel rotates in the inner ring around the axis C1C2 passing through its center of gravity. The inner ring, in turn, can rotate in the outer ring around the axis B 1 B 2, perpendicular to C 1 C 2. Finally, the outer race can rotate freely in the bearings of the strut around the axis A 1 A 2, perpendicular to the axes C 1 C 2 and B 1 B 2. All three axes intersect at some fixed point O, called the center of the suspension or the fulcrum of the gyroscope. The gyroscope in a gimbal has three degrees of freedom and, therefore, can make any rotation around the center of the gimbal. If the center of the gyroscope's suspension coincides with its center of gravity, then the resulting moment of gravity of all parts of the gyroscope relative to the center of the suspension is zero. Such a gyroscope is called balanced.
Let us now consider the most important properties of the gyroscope, which have found it widespread use in various fields.
1) Stability.
For any rotation of the counterbalanced gyroscope, its axis of rotation remains unchanged in direction relative to the laboratory reference system. This is due to the fact that the moment of all external forces, equal to the moment of the friction forces, is very small and practically does not cause a change in the angular momentum of the gyroscope, i.e.
Since the angular momentum is directed along the axis of rotation of the gyroscope, its orientation must remain unchanged.
If the external force acts for a short time, then the integral that determines the increment in angular momentum will be small
. (5.34)
This means that under short-term influences of even large forces, the movement of a balanced gyroscope changes little. The gyroscope seems to resist any attempts to change the magnitude and direction of its angular momentum. This is due to the remarkable stability that the movement of the gyroscope acquires after it is brought into rapid rotation. This property of the gyroscope is widely used to automatically control the movement of aircraft, ships, missiles and other devices.
If the gyroscope is acted upon for a long time by a moment of external forces that is constant in direction, then the axis of the gyroscope is ultimately set in the direction of the moment of the external forces. This phenomenon is used in the gyrocompass. This device is a gyroscope, the axis of which can be freely rotated in a horizontal plane. Due to the daily rotation of the Earth and the action of the moment of centrifugal forces, the axis of the gyroscope rotates so that the angle between and becomes minimal (Fig. 5.14). This corresponds to the position of the gyroscope axis in the meridian plane.
2). Gyroscopic effect.
If a pair of forces and is applied to a rotating gyroscope, tending to rotate it about an axis perpendicular to the axis of rotation, then it will begin to rotate around a third axis, perpendicular to the first two (Fig. 5.15). This unusual behavior of the gyroscope is called the gyroscopic effect. It is explained by the fact that the moment of the pair of forces is directed along the O 1 O 1 axis and the change in the vector by magnitude over time will have the same direction. As a result, the new vector will rotate relative to the O 2 O 2 axis. Thus, the behavior of the gyroscope, unnatural at first glance, fully corresponds to the laws of the dynamics of rotational motion
3). Precession of the gyroscope.
The precession of a gyroscope is the cone-shaped movement of its axis. It occurs in the case when the moment of external forces, remaining constant in magnitude, rotates simultaneously with the axis of the gyroscope, forming a right angle with it all the time. To demonstrate precession, a bicycle wheel with an extended axle set into rapid rotation can be used (Fig. 5.16).
If the wheel is suspended by the extended end of the axle, its axle will begin to precess around the vertical axis under the influence of its own weight. A rapidly rotating top can also serve as a demonstration of precession.
Let's find out the reasons for the precession of the gyroscope. Let's consider an unbalanced gyroscope, the axis of which can freely rotate around a certain point O (Fig. 5.16). The moment of gravity applied to the gyroscope is equal in magnitude
where is the mass of the gyroscope, is the distance from point O to the center of mass of the gyroscope, is the angle formed by the axis of the gyroscope with the vertical. The vector is directed perpendicular to the vertical plane passing through the axis of the gyroscope.
Under the influence of this moment, the angular momentum of the gyroscope (its origin is placed at point O) will receive an increment in time, and the vertical plane passing through the axis of the gyroscope will rotate by an angle. The vector is always perpendicular to , therefore, without changing in magnitude, the vector changes only in direction. Moreover, after a while, the relative position of the vectors will be the same as at the initial moment. As a result, the gyroscope axis will continuously rotate around the vertical, describing a cone. This movement is called precession.
Let us determine the angular velocity of precession. According to Fig. 5.16, the angle of rotation of the plane passing through the axis of the cone and the axis of the gyroscope is equal to
where is the angular momentum of the gyroscope, and is its increment over time.
Dividing by , taking into account the noted relations and transformations, we obtain the angular velocity of precession
. (5.35)
For gyroscopes used in technology, the angular velocity of precession is millions of times less than the rotation speed of the gyroscope.
In conclusion, we note that the phenomenon of precession is also observed in atoms due to the orbital motion of electrons.
Examples of application of the laws of dynamics
During rotational movement
1. Let's consider some examples on the law of conservation of angular momentum, which can be implemented using a Zhukovsky bench. In the simplest case, the Zhukovsky bench is a disk-shaped platform (chair), which can rotate freely around a vertical axis on ball bearings (Fig. 5.17). The demonstrator sits or stands on the bench, after which it is brought into rotation. Due to the fact that the friction forces due to the use of bearings are very small, the angular momentum of the system consisting of a bench and a demonstrator relative to the axis of rotation cannot change over time if the system is left to its own devices. If the demonstrator holds heavy dumbbells in his hands and spreads his arms to the sides, then he will increase the moment of inertia of the system, and therefore the angular velocity of rotation must decrease so that the angular momentum remains unchanged.
According to the law of conservation of angular momentum, we create an equation for this case
where is the moment of inertia of the person and the bench, and is the moment of inertia of the dumbbells in the first and second positions, and is the angular velocities of the system.
The angular speed of rotation of the system when raising dumbbells to the side will be equal to
.
The work done by a person when moving dumbbells can be determined through the change in the kinetic energy of the system
2. Let us give another experiment with the Zhukovsky bench. The demonstrator sits or stands on a bench and is handed a rapidly rotating wheel with a vertically directed axis (Fig. 5.18). The demonstrator then turns the wheel 180 0 . In this case, the change in the angular momentum of the wheel is entirely transferred to the bench and the demonstrator. As a result, the bench, together with the demonstrator, begins to rotate with an angular velocity determined on the basis of the law of conservation of angular momentum.
The angular momentum of the system in the initial state is determined only by the angular momentum of the wheel and is equal to
where is the moment of inertia of the wheel, is the angular velocity of its rotation.
After turning the wheel through an angle of 180 0, the angular momentum of the system will be determined by the sum of the angular momentum of the bench with the person and the angular momentum of the wheel. Taking into account the fact that the angular momentum vector of the wheel has changed its direction to the opposite, and its projection onto the vertical axis has become negative, we obtain
,
where is the moment of inertia of the “person-platform” system, and is the angular velocity of rotation of the bench with the person.
According to the law of conservation of angular momentum
And 
.
As a result, we find the speed of rotation of the bench
3. Thin rod of mass m and length l rotates with an angular velocity ω=10 s -1 in the horizontal plane around a vertical axis passing through the middle of the rod. Continuing to rotate in the same plane, the rod moves so that the axis of rotation now passes through the end of the rod. Find the angular velocity in the second case.
In this problem, due to the fact that the distribution of the mass of the rod relative to the axis of rotation changes, the moment of inertia of the rod also changes. In accordance with the law of conservation of angular momentum of an isolated system, we have
Here is the moment of inertia of the rod relative to the axis passing through the middle of the rod; 
is the moment of inertia of the rod relative to the axis passing through its end and found by Steiner’s theorem.
Substituting these expressions into the law of conservation of angular momentum, we obtain
,
.
4. Rod length L=1.5 m and mass m 1=10 kg hingedly suspended from the upper end. A bullet with a mass of m 2=10 g, flying horizontally at a speed of =500 m/s, and gets stuck in the rod. At what angle will the rod deflect after the impact?
Let's imagine in Fig. 5.19. system of interacting bodies “rod-bullet”. The moments of external forces (gravity, axle reaction) at the moment of impact are equal to zero, so we can use the law of conservation of angular momentum
The angular momentum of the system before impact is equal to the angular momentum of the bullet relative to the suspension point
The angular momentum of the system after an inelastic impact is determined by the formula
,
where is the moment of inertia of the rod relative to the suspension point, is the moment of inertia of the bullet, is the angular velocity of the rod with the bullet immediately after the impact.
Solving the resulting equation after substitution, we find
.
Let us now use the law of conservation of mechanical energy. Let us equate the kinetic energy of the rod after a bullet hits it with its potential energy at the highest point of its rise:
,
where is the elevation height of the center of mass of this system.
Having carried out the necessary transformations, we obtain
The angle of deflection of the rod is related to the ratio
.
Having carried out the calculations, we get =0.1p=18 0 .
5. Determine the acceleration of the bodies and the tension of the thread on the Atwood machine, assuming that (Fig. 5.20). The moment of inertia of the block relative to the axis of rotation is equal to I, block radius r. Neglect the mass of the thread.
Let's arrange all the forces acting on the loads and the block, and draw up dynamic equations for them
If there is no slipping of the thread along the block, then the linear and angular acceleration are related to each other by the relation
Solving these equations, we get
Then we find T 1 and T 2.
6. A thread is attached to the pulley of the Oberbeck cross (Fig. 5.21), from which a load weighing M= 0.5 kg. Determine how long it takes for a load to fall from a height h=1 m to bottom position. Pulley radius r=3 cm. Four weights weighing m=250 g each at a distance R= 30 cm from its axis. The moment of inertia of the cross and the pulley itself is neglected in comparison with the moment of inertia of the loads.
Let us determine the kinetic energy of a rigid body rotating around a fixed axis. Let's divide this body into n material points. Each point moves with linear speed υ i =ωr i , then the kinetic energy of the point
or 
The total kinetic energy of a rotating rigid body is equal to the sum of the kinetic energies of all its material points:
(3.22)
(J is the moment of inertia of the body relative to the axis of rotation)
If the trajectories of all points lie in parallel planes (like a cylinder rolling down an inclined plane, each point moves in its own plane), this flat movement.
According to Euler's principle, plane motion can always be decomposed into translational and rotational motion in countless ways. If a ball falls or slides along an inclined plane, it moves only translationally; when the ball rolls, it also rotates.
(3.23)
From a comparison of the formulas for kinetic energy for translational and rotational motions, it is clear that the measure of inertia during rotational motion is the moment of inertia of the body.
§ 3.6 Work of external forces during rotation of a rigid body
When a rigid body rotates, its potential energy does not change, therefore the elementary work of external forces is equal to the increment in the kinetic energy of the body:
dA = dE or 
Taking into account that Jβ = M, ωdr = dφ, we have α of the body at a finite angle φ is equal to
(3.25)
When a rigid body rotates around a fixed axis, the work of external forces is determined by the action of the moment of these forces relative to this axis. If the moment of forces relative to the axis is zero, then these forces do not produce work.
Examples of problem solving
Example 2.1. Flywheel massm=5kg and radiusr= 0.2 m rotates around a horizontal axis with frequencyν 0 =720 min -1 and when braking it stops behindt=20 s. Find the braking torque and the number of revolutions before stopping.
To determine the braking torque, we apply the basic equation of the dynamics of rotational motion
where I=mr 2 – moment of inertia of the disk; Δω =ω - ω 0, and ω =0 is the final angular velocity, ω 0 =2πν 0 is the initial. M is the braking moment of forces acting on the disk.
Knowing all the quantities, you can determine the braking torque
Mr 2 2πν 0 = МΔt (1)
(2)
From the kinematics of rotational motion, the angle of rotation during the rotation of the disk before stopping can be determined by the formula
(3)
where β is the angular acceleration.
According to the conditions of the problem: ω =ω 0 – βΔt, since ω=0, ω 0 = βΔt
Then expression (2) can be written as:
Example 2.2. Two flywheels in the form of disks of identical radii and masses were spun up to a rotation speedn= 480 rpm and left to our own devices. Under the influence of the friction forces of the shafts on the bearings, the first stopped throught=80 s, and the second one didN= 240 rpm to stop. Which flywheel had a greater moment of friction between the shafts and bearings and by how many times?
We will find the moment of forces of the thorn M 1 of the first flywheel using the basic equation of the dynamics of rotational motion
M 1 Δt = Iω 2 - Iω 1
where Δt is the time of action of the moment of friction forces, I=mr 2 is the moment of inertia of the flywheel, ω 1 = 2πν and ω 2 = 0 – the initial and final angular velocities of the flywheels
Then 
The moment of friction forces M 2 of the second flywheel will be expressed through the connection between the work A of the friction forces and the change in its kinetic energy ΔE k:
where Δφ = 2πN is the angle of rotation, N is the number of revolutions of the flywheel.
Then where from
ABOUT 
the ratio will be equal
The frictional moment of the second flywheel is 1.33 times greater.
Example 2.3.
Mass of a homogeneous solid disk m, mass of loads m 1
and m 2
(Fig. 15). There is no slipping or friction of the thread in the cylinder axis. Find the acceleration of the loads and the ratio of the thread tensions
in the process of movement.
There is no slipping of the thread, therefore, when m 1 and m 2 make translational motion, the cylinder will rotate about the axis passing through point O. Let us assume for definiteness that m 2 > m 1.
Then the load m 2 is lowered and the cylinder rotates clockwise. Let us write down the equations of motion of the bodies included in the system
The first two equations are written for bodies with masses m 1 and m 2 undergoing translational motion, and the third equation is written for a rotating cylinder. In the third equation on the left is the total moment of forces acting on the cylinder (the moment of force T 1 is taken with a minus sign, since the force T 1 tends to rotate the cylinder counterclockwise). On the right I is the moment of inertia of the cylinder relative to the O axis, which is equal to
where R is the radius of the cylinder; β is the angular acceleration of the cylinder.
Since there is no thread slippage, then 
. Taking into account the expressions for I and β, we obtain:
Adding the equations of the system, we arrive at the equation
From here we find the acceleration a cargo
From the resulting equation it is clear that the thread tensions will be the same, i.e. 
=1 if the mass of the cylinder is much less than the mass of the loads.
Example 2.4.
A hollow ball with mass m = 0.5 kg has an outer radius R = 0.08 m and an inner radius r = 0.06 m. The ball rotates around an axis passing through its center. At a certain moment, a force begins to act on the ball, as a result of which the angle of rotation of the ball changes according to the law
. Determine the moment of the applied force.
We solve the problem using the basic equation of the dynamics of rotational motion 
. The main difficulty is to determine the moment of inertia of a hollow ball, and we find the angular acceleration β as 
. The moment of inertia I of a hollow ball is equal to the difference between the moments of inertia of a ball of radius R and a ball of radius r:
where ρ is the density of the ball material. Finding the density by knowing the mass of a hollow ball
From here we determine the density of the ball material
For the moment of force M we obtain the following expression:
Example 2.5. A thin rod with a mass of 300 g and a length of 50 cm rotates with an angular velocity of 10 s -1 in a horizontal plane around a vertical axis passing through the middle of the rod. Find the angular velocity if, during rotation in the same plane, the rod moves so that the axis of rotation passes through the end of the rod.
We use the law of conservation of angular momentum
(1)
(J i is the moment of inertia of the rod relative to the axis of rotation).
For an isolated system of bodies, the vector sum of angular momentum remains constant. Due to the fact that the distribution of the mass of the rod relative to the axis of rotation changes, the moment of inertia of the rod also changes in accordance with (1):
J 0 ω 1 = J 2 ω 2 .
(2)
It is known that the moment of inertia of the rod relative to the axis passing through the center of mass and perpendicular to the rod is equal to
J 0 = mℓ 2 /12.
(3) According to Steiner's theorem 2
J =J 0 +m According to Steiner's theorem A
(J is the moment of inertia of the rod relative to an arbitrary axis of rotation; J 0 is the moment of inertia relative to a parallel axis passing through the center of mass;
- distance from the center of mass to the selected axis of rotation). According to Steiner's theorem Let's find the moment of inertia about the axis passing through its end and perpendicular to the rod:
J 2 =J 0 +m
2, J 2 = mℓ 2 /12 +m(ℓ/2) 2 = mℓ 2 /3.
(4)
Let's substitute formulas (3) and (4) into (2): mℓ 2 ω 1 /12 = mℓ 2 ω 2 /3mω 2 = ω 1 /4 ω 2 =10s-1/4=2.5s -1 1 Example 2.6 -1 . Man of mass 2 =60kg, standing on the edge of a platform with mass M=120kg, rotating by inertia around a fixed vertical axis with frequency ν
=12min, moves to its center. Considering the platform to be a round homogeneous disk and the person to be a point mass, determine with what frequency ν .
the platform will then rotate. Given:
m=60kg, M=120kg, ν 1 =12min -1 = 0.2s -1 Find:
ν 1
Solution: 
According to the conditions of the problem, the platform with the person rotates by inertia, i.e. the resulting moment of all forces applied to the rotating system is zero. Therefore, for the “platform-person” system the law of conservation of angular momentum is satisfied 
I 1 ω 1 = I 2 ω 2 
Where
- moment of inertia of the system when a person stands on the edge of the platform (take into account that the moment of inertia of the platform is equal to
(R – radius n
platform), the moment of inertia of a person at the edge of the platform is mR 2).
- moment of inertia of the system when a person stands in the center of the platform (take into account that the moment of a person standing in the center of the platform is zero). Angular velocity ω 1 = 2π ν 1 and ω 1 = 2π ν 2. Substituting the written expressions into formula (1), we obtain
where does the required rotation speed come from? Answer: ν 2 =24min -1. 1. Consider the rotation of the body around motionless 1. Consider the rotation of the body around axis Z. Let us divide the whole body into a set of elementary masses m 1. Consider the rotation of the body around i 1. Consider the rotation of the body around. Linear speed of elementary mass m 1. Consider the rotation of the body around– v i = w R i, where R 
– mass distance m 
from the axis of rotation. Therefore, kinetic energy
th elementary mass will be equal to
. Total kinetic energy of 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
| |
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 the 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.
Let us consider an absolutely rigid body rotating about a fixed axis. Let's mentally break this body into infinitesimal pieces with infinitely small sizes and masses m v t., t 3,... located at distances R v R 0 , R 3,... from the axis. Kinetic energy of a rotating body we find it as the sum of the kinetic energies of its small parts:
- moment of inertia of a rigid body relative to a given axis 00,. From a comparison of the formulas for the kinetic energy of translational and rotational motions, it is obvious that the moment of inertia in rotational motion is analogous to mass in translational motion. Formula (4.14) is convenient for calculating the moment of inertia of systems consisting of individual material points. To calculate the moment of inertia of solid bodies, using the definition of the integral, you can transform it to the form
It is easy to notice that the moment of inertia depends on the choice of the axis and changes with its parallel translation and rotation. Let's find the values of the moments of inertia for some homogeneous bodies.
From formula (4.14) it is obvious that moment of inertia of a material point equals
Where T - point mass; R- distance to the axis of rotation.
It is easy to calculate the moment of inertia for hollow thin-walled cylinder(or the special case of a cylinder with low height - thin ring) radius R relative to the axis of symmetry. The distance to the axis of rotation of all points for such a body is the same, equal to the radius and can be taken out from under the sum sign (4.14):
Rice. 4.5
Solid cylinder(or a special case of a cylinder with low height - disk) radius R to calculate the moment of inertia relative to the axis of symmetry requires calculating the integral (4.15). You can understand in advance that the mass in this case, on average, is concentrated somewhat closer to the axis than in the case of a hollow cylinder, and the formula will be similar to (4.17), but it will contain a coefficient less than unity. Let's find this coefficient. Let a solid cylinder have density p and height A. Let us divide it into hollow cylinders (thin cylindrical surfaces) of thickness dr(Figure 4.5 shows a projection perpendicular to the axis of symmetry). The volume of such a hollow cylinder of radius r is equal to the surface area multiplied by the thickness: dV = 2nrhdr, weight: dm = 2nphrdr, and the moment of inertia in accordance with formula (4.17): dj =
= r 2 dm = 2lr/?g Wr. The total moment of inertia of a solid cylinder is obtained by integrating (summing) the moments of inertia of hollow cylinders:
Search in the same way moment of inertia of a thin rod length L and masses T, if the axis of rotation is perpendicular to the rod and passes through its middle. Let's break this one down
Taking into account the fact that the mass of a solid cylinder is related to density by the formula t = nR 2 hp, we finally have moment of inertia of a solid cylinder:
Rice. 4.6
rod in accordance with fig. 4.6 pieces thick dl. The mass of such a piece is equal to dm = mdl/L, and the moment of inertia in accordance with formula (4.6): dj = l 2 dm = l 2 mdl/L. The total moment of inertia of a thin rod is obtained by integrating (summing) the moments of inertia of the pieces:
Taking the elementary integral gives the moment of inertia of a thin rod of length L and masses T
Rice. 4.7
It is somewhat more difficult to take the integral when searching moment of inertia of a homogeneous ball radius R and mass /77 relative to the axis of symmetry. Let a solid ball have density p. Let's break it down in accordance with Fig. 4.7 for hollow thin cylinders thick dr, the axis of symmetry of which coincides with the axis of rotation of the ball. The volume of such a hollow cylinder of radius G equal to the surface area multiplied by the thickness:
where is the height of the cylinder h found using the Pythagorean theorem:
Then it is easy to find the mass of the hollow cylinder:
as well as the moment of inertia in accordance with formula (4.15):
The total moment of inertia of a solid ball is obtained by integrating (summing) the moments of inertia of hollow cylinders:
Taking into account the fact that the mass of a solid ball is related to the density of the form-4.
loy T = -npR A y we finally have the moment of inertia about the axis
symmetry of a homogeneous ball of radius R masses T:
Since a rigid body is a special case of a system of material points, the kinetic energy of the body when rotating around a fixed Z axis will be equal to the sum of the kinetic energies of all its material points, that is
All material points of a rigid body rotate in this case in circles with radii and with the same angular velocities. The linear speed of each material point of a rigid body is equal to . The kinetic energy of a solid body will take the form
The sum on the right side of this expression, in accordance with (4.4), represents the moment of inertia of this body relative to a given axis of rotation. Therefore, the formula for calculating the kinetic energy of a rigid body rotating relative to a fixed axis will take its final form:
. (4.21)
It is taken into account here that
Calculating the kinetic energy of a rigid body in the case of arbitrary motion becomes much more complicated. Let us consider plane motion when the trajectories of all material points of the body lie in parallel planes. The speed of each material point of a rigid body, according to (1.44), can be represented in the form
,
where as the instantaneous axis of rotation we choose the axis passing through the center of inertia of the body perpendicular to the plane of the trajectory of any point of the body. In this case, in the last expression it represents the speed of the center of inertia of the body, the radii of the circles along which points of the body rotate with angular velocity around an axis passing through its center of inertia. Since with such a movement ^, the vector equal to lies in the plane of the point’s trajectory.
Based on the above, the kinetic energy of a body during its plane motion is equal to
.
By squaring the expression in parentheses and taking the constant quantities for all points of the body out of the sum sign, we obtain
It is taken into account here that ^.
Let's consider each term on the right side of the last expression separately. The first term, by obvious equality, is equal to
The second term is equal to zero, since the sum determines the radius vector of the center of inertia (3.5), which in this case lies on the axis of rotation. Taking into account (4.4), the last term will take the form . Now, finally, kinetic energy during arbitrary but plane motion of a rigid body can be represented as the sum of two terms:
, (4.23)
where the first term represents the kinetic energy of a material point with a mass equal to the mass of the body and moving with the speed of the center of mass of the body;
the second term represents the kinetic energy of a body rotating around an axis (moving at speed) passing through its center of inertia.
Conclusions: So, the kinetic energy of a rigid body during its rotation around a fixed axis can be calculated using one of the relations (4.21), and in the case of plane motion using (4.23).
Control questions.
4.4. In what cases does (4.23) transform into (4.21)?
4.5. What will the formula for the kinetic energy of a body look like when it moves in a plane if the instantaneous axis of rotation does not pass through the center of inertia? What is the meaning of the quantities included in the formula?
4.6. Show that the work done by internal forces during rotation of a rigid body is zero.
