Action of centrifugal force. Rotational movement of the body
Rice. 4.8
At each moment of time, the stone would have to move in a straight line tangential to the circle. However, it is connected to the rotation axis by a rope. The rope stretches, an elastic force appears acting on the stone, directed along the rope to the center of rotation. This is the centripetal force (when the Earth rotates around its axis, the force of gravity acts as a centripetal force).
But since then
| (4.5.2) | |||
| (4.5.3) |
The centripetal force arose as a result of the action of the stone on the rope, i.e. is the force applied to the body - inertia force of the second kind. It is fictitious - it does not exist.
The force applied to the connection and directed radially from the center is called centrifugal.
Remember that centripetal force is applied to the rotating body, and centrifugal force is applied to the connection.
The force of gravitational attraction is directed towards the center of the Earth.
The ground reaction force (normal pressure) is directed perpendicular to the surface of motion.
From the point of view of an observer associated with a non-inertial frame of reference, he is not approaching the center, although he sees that F the cs is in effect (this can be judged by the readings of the spring dynamometer). Consequently, from the point of view of the observer in a non-inertial system there is a force that balances F cs, equal to it in magnitude and opposite in direction:
Because a n= ω 2 R(here ω is the angular speed of rotation of the stone, and υ is linear), then
| F tsb = mω 2 R. | (4.5.4) |
All of us (and physical instruments too) are on the Earth, rotating around an axis, therefore, in a non-inertial system (Figure 4.9).
Rice. 4.9
Let us assume that rigid body A (Fig. 1.19, a) can rotate around some fixed axis. In order to cause rotation of a body (to change its angular velocity), an external influence is necessary. However, a force whose direction passes through the axis of rotation, or a force parallel to the axis, cannot change the angular velocity of bodies.
Therefore, from the external force applied to the body, it is necessary to isolate the components that do not cause rotation. Rotation can only be caused by a force (rotational force) lying in a plane perpendicular to the axis of rotation and directed tangentially to the circle described by the point of its application.
Note that when the body rotates, the components do not perform work, since the point of application of these forces moves perpendicular to their directions. The work is performed only by the rotational force; it is the projection of the force acting on the body onto the direction of movement of the point of application of this force.
Let us determine the amount of work performed by the rotating force if its point of application moves along a circle of radius by (Fig. 1.19, b). Let us assume that the magnitude of the force remains constant. Then
The product of a rotating force and a radius is the moment of the rotating force, or the torque acting on a given body, and is denoted by (recall that the moment of a given force relative to any axis is the product of this force by its arm, i.e. by the length of the perpendicular, carried out from the specified
axis to the direction of the force). Thus, in formula (2.8)
therefore, the work done by the torque is equal to the product of this moment and the angle of rotation of the body:
If the torque (force or its arm) changes over time, then the work done is determined as the sum:
The torque of the rotating force is represented as a vector coinciding with the axis of rotation; the positive orientation of this vector is chosen in the direction in which the right screw rotated by this moment would move.
The torque applied to the body imparts to it some angular acceleration according to the directions of the vectors we have chosen; they are oriented along the axis of rotation in the same direction. The relationship between the magnitude of the torque and the magnitude of the angular acceleration imparted by it can be established in two ways:
a) we can use the fact that the work of the driving force is equal to the change in the kinetic energy of the body to which this force is applied: For a rotating body, according to formulas (2.9) and (2.4), we have
Here we assume that the moment of inertia of the body does not change during rotation. Dividing this equation by and reducing by we get
b) you can take advantage of the fact that the moment of the rotating force is equal to the sum of the moments of forces that impart tangential accelerations to the individual components of the body; these forces are equal and their moments are
Let us replace tangential accelerations with angular acceleration, which is the same for all particles of a rotating body (if the body is not deformed during rotation): Then
Formula (2.12) expresses the basic law of the dynamics of rotational motion of solid (non-deformable) bodies, for which
the angular acceleration acquired by a body under the influence of a given torque is directly proportional to the magnitude of this moment and inversely proportional to the moment of inertia of the body relative to the axis of rotation:
In vector form, this law is written as
If a body is deformed during rotation, then its moment of inertia relative to the axis of rotation will change. Let us mentally imagine a rotating body consisting of many elementary (point) parts; then the deformation of the entire body will mean a change in the distances from these parts of the body to the axis of rotation. However, a change in the distance of a given angular velocity of rotation co will be accompanied by a change in the linear speed of movement of this particle, and therefore its kinetic energy. Thus, at a constant angular velocity of rotation of the body, a change in distances (hence, a change in the moment of inertia of the body) will be accompanied by a change in the kinetic energy of rotation of the entire body.
From formula (2.4), if we assume variables, we can obtain
The first term shows the change in the kinetic energy of a rotating body, which occurred only due to a change in the angular velocity of rotation (at a given moment of inertia of the body), and the second term shows the change in kinetic energy, which occurred only due to a change in the moment of inertia of the body (at a given angular velocity of rotation).
However, when the distance from a point body to the axis of rotation changes, the internal forces connecting this body with the axis of rotation will do work: negative if the body moves away, and positive if the body approaches the axis of rotation; this work can be calculated if we assume that the force connecting the particle to the axis of rotation is numerically equal to the centripetal force:
For the entire body, consisting of many particles with masses, we obtain
In the general case, when an external torque acts on a body, the change in kinetic energy must be equated to the sum of two works: external torque and internal forces. With accelerated rotation, the values will have positive signs, - negative
sign (since the particles of the body move away from the axis of rotation); Then
Substituting here the value from expression (2.15) and replacing with we get
or after reduction
This is a general form of the basic law of mechanics for bodies rotating about a fixed axis; it is also applicable for deforming bodies. When formula (2.16) transforms into formula (2.14).
Note that for deforming bodies, a change in the angular velocity of rotation is possible even in the absence of an external torque. Indeed, when - from formula (2.16) we obtain:
In this case, the angular velocity of rotation changes only due to a change in the moment of inertia of the body caused by internal forces.
3.1. Question. Is it possible to rotate “by inertia”? How does linear inertia differ from rotational inertia?
Answer. At first glance, rotation demonstrates the properties of inertia even more clearly than linear motion. A flywheel rotating in a vacuum on a magnetic suspension can move for years, since external influences on it are minimized.
Newton, explaining the law of inertia he discovered, gives the following explanation: “A top, the parts of which, due to mutual adhesion, distract each other from rectilinear motion, does not cease to rotate uniformly, since this rotation is not slowed down by air resistance.” This phrase from Newton makes you think seriously about the question posed.
However, strictly speaking, movement by inertia can only be uniform and rectilinear. This means that there cannot be rotation due to inertia in the Newtonian mechanics we accept. But a solid massive body remains in a state of rest or uniform rotation until it is brought out of this state by a moment of external forces. Therefore, in fact, the phenomenon of inertia also takes place here, although different from the classical case. What is common and what is the difference between rotational inertia and inertia during rectilinear motion?
The inertia of a massive point (body) depends only on its mass. Mass is a measure of the inertia of a body during translational, including linear, motion. This means that with such a movement, inertia is not affected by the distribution of masses in the body, and this body can be safely taken as a material (massive) point. The mass of this point is equal to the mass of the body, and the point is located at the center of mass or center of inertia of the body. If you rotate a rod with massive weights mounted on it around the vertical axis Z (Fig. 6), you will notice that as long as the weights are near the center, it is easy to untwist the rod. But if the weights are moved apart, it will become more difficult to untwist the rod, although its mass has not changed.
Rice. 6. Scheme of changing the moment of inertia of the body.
Therefore, the inertia of a body during rotation depends not only on mass, but to a greater extent on the distribution of this mass relative to the axis of rotation. A measure of the inertia of a body during rotation is the axial moment of inertia I, equal to the sum of the products of masses T all particles of the body by the squares of their distances h from the axis of rotation:
The axial moment of inertia plays the same role during rotational motion as mass does during translational (rectilinear) motion, and thus it is a measure of the inertia (inertia) of a body during rotational motion.
As we know, the law of inertia establishes the equivalence of relative rest and uniform rectilinear motion - motion by inertia. It is impossible to determine by any mechanical experiment whether a given body is at rest or moves uniformly and in a straight line. This is not the case in rotational motion. For example, it is not at all indifferent whether the top is at rest or rotates uniformly with a constant angular velocity. As noted by A. Yu. Ishlinsky, the angular velocity of a solid body is a quantity that characterizes its physical state. Angular velocity can be measured, for example, by determining the elastic deformations of a body, without any information about the position of the body in relation to the “absolute” coordinate system. Therefore, the term “absolute angular velocity of a body”, in contrast to “absolute velocity of a point,” should be used in the literal sense (without quotation marks).
Thus, mechanical phenomena in a stationary and rotating system will proceed differently, not to mention the fact that if the body is twisted strongly enough, it will be torn apart due to the stresses that have arisen in it.
Another difference is that rectilinear uniform motion and rest are equivalent, and rotation, even with a constant angular velocity, can be clearly distinguished not only from rest, but also from rotation with a different angular velocity.
Here it is appropriate to mention the views of the Austrian physicist Ernst Mach (1838–1916), who had a great influence on the formation of Einstein’s equivalence principle. Mach, by “selecting” the appropriate coordinate system, sought to give the laws of mechanics such a form that they would not depend on rotation. What would happen if he succeeded? Let's place a rapidly rotating observer on a stationary flywheel. Then we can say that, relative to the observer, the flywheel rotates quickly, perhaps even faster than its strength allows. But the flywheel will not rupture, although it seems to the observer that enormous stress is acting on it. And the rotating observer himself may suffer, since during rotation it is in him that mechanical stresses arise.
3.2. Question. Is it possible to formulate the laws of rotational inertia in a similar way to Newton's first law?
Answer. You can take the liberty to formulate the “law” of inertia of rotational motion in the image and likeness of Newton’s first law: “An absolutely rigid body isolated from external moments will maintain a state of rest or uniform rotation around a fixed axis until the external moments applied to this body will force him to change this state.”
Why an absolutely solid body and not just any body? Because the moment of inertia of a non-rigid body will change due to forced deformations during rotation, and this is equivalent to a change in the mass of a point for Newton’s first law.
In the case of rotational motion, if the moment of inertia is not constant, it will be necessary to take as a constant not the angular velocity, but the product of the angular velocity ω and the moment of inertia / - the so-called kinetic moment TO. In this case, the “law” of rotational inertia will take a more general form: “A body isolated from external moments will keep the vector of its kinetic moment constant.” If the body rotates around a fixed axis: “A body isolated from external moments about the axis of rotation will maintain a constant kinetic moment about this axis.” These laws, however, in a slightly different formulation, are called the laws of conservation of angular momentum.
3.3. Question. The Earth and Moon rotate around a common center of mass. Do centrifugal forces act on these celestial bodies?
Answer. The idea that when material points and bodies rotate around an axis or a fixed point, centrifugal (i.e., directed from the center of rotation) forces must act on them is a common misconception.
For example, both the Earth and the Moon are affected by gravitational forces directed towards each other, and therefore towards the center of rotation (Fig. 7). There are no forces directed from the center here at all. In order for bodies moving by inertia, i.e., uniformly and rectilinearly, to turn away from this path and begin to move along curves, they must be affected by centripetal forces, i.e., directed towards the center of rotation, forces. These are the forces of gravity.
Rice. 7. Diagram of forces acting on the Earth-Moon system.
If the point rotates A, tied to a support ABOUT on a flexible weightless connection - thread (Fig. 8, A), then, neglecting the force of gravity (let’s say the experiment is carried out in weightlessness), we can say that the centripetal force also acts on this point Fts. On the thread itself, as a connection, from the side of the point A there is a reaction directed from the center R1 = Fc, and from the support side ABOUT - force R2 = Fc(Fig. 8, b). On support ABOUT force acts Fc, directed from the center. A balanced system of forces acts on the thread, which cannot influence the movement of the point A.
Rice. 8. Forces acting on bodies in a rotating system: A - forces acting on a point rotating in a circle A and support ABOUT; b – forces acting on the connection.
In some textbooks, for example, for schools with in-depth study of physics, it is specifically emphasized that “centrifugal forces of inertia do not act on all bodies on the surface of the Earth.” This formulation means that centrifugal forces exist and act on some bodies. Of course, this is not true.
3.4. Question. Why, when a body rotates rapidly, does it experience mechanical stress and may even collapse, since no other body is in contact with it, no force fields act on it, etc.?
Answer. Indeed, if an experiment on the rotation of, say, a metal ring is carried out in weightlessness and in a vacuum, then no other body, not even air, will interact with this body. This ring can be accelerated by a rotating electromagnetic field (for example, arising in the stator of an asynchronous electric motor), especially if the ring is made of steel. After acceleration is completed, does it rotate freely at angular velocity? the ring will have kinetic energy E:
and will be stretched by mechanical stress?:
Where I– axial moment of inertia of the ring;
? – density of the ring material;
v – linear speed of the ring.
What causes this tension? We saw above that the connection is a thread (see Fig. 8, a, b) there are tensile forces caused by the point A, rotating around a support ABOUT. After all, it is the connection that acts on the point A centripetal force Fc, constantly turns it off the natural straight path. In this case, the mass (point A) and the connection (weightless thread) are clearly distinguished. But if the point A eliminate, instead of a thread, take a massive body - a rod or chain - and rotate it around a point ABOUT, then the picture will become more complicated.
In such cases, when the connection itself has mass, it is convenient to imagine it in the form of a weightless connection (thread) loaded with individual massive points (Fig. 9).
Rice. 9. Weightless connection - a thread loaded with point masses.
If the number of points is small, the centripetal forces acting on these points are easy to determine: at point 1 it is Fts1, At point 2 – the sum of two forces (Ft1+ Fts2), and at point 3 it is maximum - the sum of three forces (Ft1+ Ft2 + Ft3). From here it is easy to move on to the case when the mass is distributed uniformly along the bond length.
So it is with a rotating ring - if you imagine that it is replaced by a polygon of weightless threads with weights placed at the vertices of the corners T(Fig. 10, a), then by selecting one of the loads (Fig. 10, b), we can determine the forces Fst, acting on the load (their reactions act on the thread):
Where Fts = m?2R or mv2/R, which follows from formula (2.4).
Having distributed the loads T evenly along the thread, we obtain a massive ring with a density of ?, which has bond strength (Fig. 11). For simplicity of calculations, we discard the lower half of the ring and denote it by F tensile forces acting on its side on the upper half-ring. Considering that the center of mass of the upper semiring C is located at a distance 2R/? up from center ABOUT, the normal acceleration of this center of mass is:
We write Newton's second law in projection onto the direction of normal acceleration:
Considering what voltage? = F/S, Where S – cross-sectional area of the ring, mass of the semi-ring M= ??R.S. and that linear speed v= ?R, we write taking into account (3.6):
Thus, we obtain formula (3.3).
Consequently, the rotating ring will stretch with force F and stress? even without contact with any other body. In a similar way, stresses arise in rotating bodies of any configuration, for example, in moving flexible massive closed connections - belts, chains, as well as flywheels - accumulators of kinetic energy.
Rice. 10. Schematic representation of a rotating ring: A - a closed rotating polygon with point masses placed at the vertices of the corners; b – forces acting on a single load.
Rice. 11. Scheme for determining stresses in a rotating ring.
3.5. Question. How to accumulate the greatest kinetic energy in a rotating flywheel?
Answer. Kinetic energy of a rotating thin ring of mass T, as for a rectilinearly moving mass, it is proportional to the square of its linear (circumferential) speed:
Indeed, in both cases the mass T moves at the same speed v. The only difference is that in the case of rectilinear motion, no stresses arise in the moving body, but when the ring rotates (as well as a belt, chain, any flat massive closed connection), stresses arise in it that do not depend on the radius of the ring and are determined by the formula ( 3.3). Consequently, in a rectilinearly moving mass it is possible to increase the speed and kinetic energy infinitely (within the framework of classical mechanics). In a rotating mass, in this case a ring, we are strictly limited by the strength of the material, and both the kinetic energy and stress in the material are proportional to the square of the peripheral speed.
What if it is not a ring, but a body of a different shape? Will it be possible to accumulate greater kinetic energy with the same strength of the material? To analyze this issue, it is most convenient to express energy and strength through specific indicators - specific energy intensity e = E/t and specific strength x = ?/?. Then for a flywheel in the form of a rotating ring:
For flywheels of other shapes, the coefficient k will take different values. For example, for a disk with a very small central hole it will be 0.3; for a disk without a hole at all - 0.6. The best flywheel shape for storing kinetic energy is a disk of equal strength. For example, the disks of steam and gas turbines have this shape - thick in the center and thin at the periphery.
3.6. Question. Is it possible to create an energy-intensive flywheel with a variable moment of inertia?
Answer. The device shown in Fig. 6, in principle, allows both the accumulation of kinetic energy and the change in the moment of inertia. But due to low strength, such a design will have negligible specific energy intensity. If you make a flywheel from rubber, then during rotation its moment of inertia will increase, the greater the angular speed of the flywheel. In this case, the potential energy accumulated during rubber stretching will be added to the kinetic energy.
But the interest is not in flywheels with a “passive” change in the moment of inertia, but in those in which this indicator can be changed forcibly. Why might this be needed?
With a constant angular momentum of the flywheel, the moment of inertia can be increased by decreasing the angular velocity and vice versa. An example is a man with dumbbells in his hands on the so-called Zhukovsky platform - a disk mounted on a stand on bearings (Fig. 12, a, b).
Rice. 12. Man on the platform (bench) of Zhukovsky: A– with arms spread to the side and a large moment of inertia; b– with hands shifted to the center and minimal moment of inertia
If a person, standing on this platform with his arms spread to the sides, rotates (Fig. 12, a), then by bringing his hands with dumbbells to the center (Fig. 12, b), he reduces his moment of inertia, thereby significantly increasing the angular velocity . Flywheels with an adjustable variable moment of inertia could provide almost any angular speed required by the working part of the machine, for example, the wheels of a car.
3.7. Question. What consequences can result from replacing an inertial frame of reference with a non-inertial one, for example, a rotating one?
Answer. Each relative motion of a body in a rotating reference frame can be associated with the motion of exactly the same body relative to an inertial coordinate system. But for such a correspondence it is necessary to reproduce not only those real forces that acted on the original body, but also to add new forces corresponding to the Euler forces of inertia in the relative motion of the original body. Euler inertia forces are defined here as real forces acting on a body, under the assumption that a moving reference frame is conventionally taken to be a stationary one. For example, if we take a turning bus as a stationary one, then we will have to consider the centrifugal forces acting on the turn as real.
Thus, if we connect the moving coordinate system with the Earth, then the acceleration of a point on the Earth in the “absolute” system - real acceleration - will be the vector sum of three accelerations: relative, portable and Coriolis (named after the 19th century French mechanic Gustav Coriolis), which occurs when the moving coordinate system rotates. It is with this Coriolis acceleration and the corresponding Coriolis force that “miracles” begin to happen, similar to those that happen with d'Alembert's inertial forces. They begin to be considered as really existing, corresponding actions are attributed to them, etc.
Here we must firmly remember that both the transfer and Coriolis forces of inertia are unreal forces, they depend only on the choice of the coordinate system and do not reflect the interactions of a given point with other points. These forces do not have a reaction, which, according to Newton’s third law, every force must have. The forces of inertia, whatever they may be, are always unreal; and you cannot believe it, even if the textbook says that they “act” on something (see question 3.3). These forces, in the figurative expression of the famous physicist Richard Feynman, are “pseudo-forces.”
3.8. Question. Is it possible to define Euler inertial forces not formally, but based on the physical essence of the phenomena?
Answer. It's possible, although it will take some imagination. Let's consider an auxiliary body, completely identical to the main one. Let this auxiliary body perform exactly the same movements in relation to an arbitrarily chosen “absolute” coordinate system as the main body makes in relation to the selected non-inertial coordinate system. Thus, the same physical forces act on all points of the auxiliary body as on the main body. However, in order for the movement of the auxiliary body relative to the “absolute” coordinate system to exactly repeat the movement of the main body relative to the non-inertial coordinate system, it is necessary to apply additional forces to the auxiliary system, in addition to all the physical forces of the main system. Since motion is considered in relation to the “absolute”, inertial frame of reference, these can only be physical forces. Obviously, they correspond exactly to the Euler inertial forces.
Thus, the Euler forces of inertia are equal to those physical forces that should be added to the original physical forces in order to accurately reproduce the relative motion of any body as absolute motion, i.e. in an inertial reference frame.
3.9. Question. If Coriolis inertial forces are unreal, how can they cause erosion of river banks? What is the gyroscopic effect?
Answer. The erosion of river banks can be qualitatively explained without the use of a moving frame of reference, Euler inertia forces and other assumptions.
It is known that the right banks of rivers flowing in the Northern Hemisphere are washed away. Let's look at the Earth from above from its North Pole. Let us imagine for simplicity that the river, starting at the equator, flows directly north, crosses the North Pole and also ends at the equator, but on the other side. The water in a river at the equator has the same speed in the direction from west to east as its banks (not the flow of the river, but the speed of the water together with the banks and with the Earth). With the daily rotation of the Earth, this is about 0.5 km/s. As you approach the pole, the speed of the shores decreases, and at the pole itself it is zero. But the water in the river “does not want” to reduce its speed - it obeys the law of inertia. And this speed is directed in the direction of the Earth’s rotation - from west to east. So the water begins to “press” on the eastern bank of the river, which turns out to be on the right side of the flow. Having reached the pole, the water in the river will completely lose its speed in the “lateral” direction, since the pole is a stationary point on Earth. But the river continues to flow now to the south, and its banks rotate again from west to east with an ever-increasing speed as it approaches the equator. The western bank begins to “press” on the water in the river, accelerating it from west to east, and the water, according to Newton’s third law, “presses” on this bank, which happens to be on the right side of the flow.
In the Southern Hemisphere, the opposite happens. If you look at the Earth from the South Pole, it rotates in a different direction. Anyone with a globe can check this out. Here's Baer's law, named after the Russian naturalist Karl Baer (1792–1876), who noticed this feature of rivers.
And here it’s not far from explaining the gyroscopic effect in general. Let's continue our river further and use it to describe a vicious circle on the surface of the Earth. At the same time, we note that the entire northern part of the river, located in the Northern Hemisphere, will tend to the right, and the entire southern part - to the left. That's all the explanation of the gyroscopic effect, which is considered perhaps the most difficult in theoretical mechanics!
So, our river is a huge ring or flywheel, rotating in the same direction as the flow of the river. If you turn this flywheel in the direction of the Earth’s rotation, then its entire northern part will deviate to the right, and the southern part will deviate to the left (Fig. 13). In other words, the flywheel will rotate so that its rotation coincides with the direction of rotation of the Earth! This is a qualitative manifestation of the gyroscopic effect.
Rice. 13. Scheme of rotation of a flywheel “wrapped” around the Earth.
3.10. Question. The gyroscopic effect is said to keep the bike from falling over. Is it so?
The gyroscopic effect is the occurrence of a moment when an attempt is made to forcibly rotate the axis of a rotating body. But we have not yet determined the magnitude of the gyroscopic moment. When turning the axis of a bicycle wheel, this moment is equal to the product of the moment of inertia of the wheel and the angular speeds of its rotation and rotation of the axis (forced precession). For simplicity, we decide that the mass of the wheel is 2 kg, its radius is 0.25 m and, therefore, the moment of inertia, approximately equal to the product of the mass by the square of the radius, is equal to 0.125 kg? m2. A cyclist calmly maneuvers already at a speed of 1 m/s, and the wheel rotates at an angular speed of 4 rad/s. The angular speed of rotation of the wheel axis is 20 times less and is approximately 0.2 rad/s. As a result, we obtain a gyroscopic moment equal to 0.1 N?m. This is the same as hanging a 1 kg weight on the end of a nail protruding only 1 cm from the wall. It is unlikely that such an insignificant moment can change anything in the movement of the bicycle.
At the same time, a riding cyclist, having turned only 10 cm from the straight line, if he does not lean towards the turn, will create a tipping moment equal to his weight plus approximately half the weight of the bicycle, multiplied by 0.1 m, which reaches about 100 N?m. This moment is a thousand times greater than the gyroscopic moment! In this way, leaning towards the center of the turn, the cyclist maintains stability.
By the way, if we are talking about special “monorail” vehicles that maintain balance precisely thanks to a massive and rapidly rotating flywheel, then the gyroscopic effect really helps here. By producing forced precession (rotation of the axis) of the flywheel with a large kinetic moment, we cause huge gyroscopic moments that hold multi-ton machines in a vertical position. For example, with a moment of inertia of the flywheel of 100 kg? m2 (this is approximately a wheel from a railway passenger car), an angular velocity of 600 rad/s and the same forced precession as before of 0.2 rad/s, the gyroscopic moment will be equal to 12 kN? m, which is equivalent to a 1.2 t load suspended on a 1 m arm. Such a large moment can not only stabilize a heavy vehicle, but also destroy the rapidly rotating flywheel bearings. Therefore, the possibility of the occurrence of gyroscopic moments must always be taken into account when calculating bearings.
3.11. Question. If you fire a cannon vertically upward, will the shell fall back into the cannon barrel?
Answer. This problem haunted the mechanics of the 19th century. Of course, the projectile will fall back into the barrel if everything happens in an absolute frame of reference. But in real life, that is, on a rotating Earth, everything will be different. Usually this problem is considered with the transition to a rotating reference frame, which greatly complicates it, at least in mathematical terms. Let's try here to consider only the qualitative side of this problem in the inertial frame of reference.
Suppose, at the latitude of Moscow, a massive point falls in a vacuum from a tower 100 m high. The Earth rotates from west to east, and at the moment of its fall this point had a circumferential speed greater than the surface of the Earth, since it was further away from its center. While falling, the point retains its peripheral speed, and it will come into contact with the Earth, moving towards the excess speed, i.e. to the east. Calculations show that this displacement is small - only 1.2 cm.
Now let's shoot a point projectile vertically upward. At the moment of the shot - on the surface of the Earth - the peripheral speed of the point is less than at altitude. Therefore, rising upward, the point will deviate to the west. The point will spend a particularly long time in the upper zone of its flight, since the vertical speed there is low, and therefore the path traveled to the west will be quite long. On the way back, the point will also deviate to the west, although now it’s slower and slower. Thus, it will fall west of the cannon muzzle.
By the way, by tilting the cannon barrel a little to the east, you can, in principle, ensure that the projectile, when falling, touches the cannon muzzle again; but in reality, especially taking into account the influence of the atmosphere, this is impossible to do - this task is purely theoretical.
Of course, the entire calculation could be carried out accurately, and without the use of fictitious Coriolis forces. But most mechanical experts believe that by placing our gun in a relative rotating coordinate system and introducing fictitious Coriolis forces, the calculation can be performed shorter and simpler. Even if this is so, then we would not lose the main thing - the feeling of the reality of what is happening, which plays an important role in physics!
The simplest motion of a rigid body is rotation around a fixed axis: the body is mounted on an axis, the position of which in space is fixed by bearings. The position of the body is determined by one parameter - the angle of rotation cf. The rate of change of this angle with time ω = d(p/d/ is called the angular velocity of rotation of the body. All points of the body move in circles with a speed v = south, where G- distance from the point to the axis of rotation.
Let's break the body into small elements, At.- mass of the i-th element, g. - distance from it to the axis. The speed of this element is v, = c or.. We have (see formula (2.43)):
Here Fxt- tangential external force acting on the element, AF.- tangential internal force. Let's multiply equation (3.102) by g p Let's express the speed of the element through the angular velocity and sum the resulting equation over all elements. We get
The sum on the left side of this equality
called moment of inertia of the body relative to a given axis, the first sum on the right side
called the moment of external forces relative to a given axis.
Note. The contribution at this moment comes only from the tangential components of external forces, i.e., the projection of forces onto the tangent to the circle at the point of application of the force. This means that forces directed along the perpendicular to the axis or parallel to the axis do not contribute to the moment.
The second sum on the right side of (3.103) is equal to zero (internal forces do not affect the rotation of the body around its axis). Thus, we get equation of motion of a rigid body around a given axis:
The quantity e = ~ is called angular acceleration.
Note. Equation (3.106) is scalar. However, the signs of the quantities included in the equation should be taken into account. This is done as follows: we set (arbitrarily) the positive direction of the rotation angle; moments of forces rotating a body in a positive direction are written with a plus sign, in the opposite direction - with a minus sign.
Problem 3.24. Homogeneous disk with radius R can rotate around a horizontal axis passing through its center. Wound on the disk
thread at the end of which a force is applied F. The thread unwinds from the disk under the influence of this force (Fig. 3.5). Determine the length of the thread unwound from the disk at time /.
Solution. If the disk rotates through an angle d
Rdtp. From here
a piece of thread of length ds = /?d will fit
The problem comes down to finding the angle through which the disk will rotate in time /. Let's turn to equation (3.106). The tension force of the thread acts on the disk at the point where the thread leaves the disk. If the mass of the thread is
zero, this force is equal to the force F. This force is tangential, and its moment relative to the axis of rotation is equal to A/= FR. The equation becomes
note. Equation (3.106) in its mathematical structure is identical to the second law for the one-dimensional motion of a particle (a mathematician would say that the equations are identical up to notation), therefore the methods for solving this equation (up to notation) are the same as in section 2.2.8.
since the initial angular velocity is zero. Further,
We found the rotation angle of the disk as a function of time. This was an example of uniformly accelerated rotational motion.
The axis of rotation of the disk coincides with one of the main axes, so
/ = /, = mR 72.
Problem 3.25. The disk from the previous problem rotates by inertia, when t= 0 its angular velocity is equal to co(0). The disk is acted upon by a moment of frictional forces (about the air), proportional to the speed: M= eye. What will be the speed of the disk at time /?
Solution. We write:
Thus,
(It is useful to compare this with the solution to Problem 2.30.)
Problem 3.26. How many revolutions will the disk from the previous problem make by the time P
Solution. The problem, obviously, is that the time for one revolution varies. Speed n(t)= (ср(г) - ф(0))/2я, and it comes down to finding the angle of rotation in time t. We write:
which solves the problem. Check: for small /, expanding the exponential, we obtain
we obtain, assuming / -»: Df = co(0)-.
Comments. The solution for the angular velocity obtained in Problem 3.25 does not entirely correspond to reality: according to this solution, the angular velocity tends to zero asymptotically, but, obviously, the disk will actually stop after a finite period of time. This means that the accepted law for the moment of friction forces is violated at a sufficiently low angular velocity. However, the result for the full rotation angle is reasonable (why?)
Let's return to equation (3.106). Let's multiply this equation by co = dtp/d t. We get
Fr.i1F t ds., but this is the sum of the work done by all external forces when the body rotates through an angle dtp. From the law of conservation of energy (see formula (3.30)) it follows that the expression in parentheses on the left side of (3.107) is the kinetic energy of a rotating rigid body (since the distances between the particles of the rigid body do not change, the internal potential energy of the rigid body is constant and the internal forces of work do not commit). From formula (3.107) we obtain
The change in kinetic energy of a rotating body is equal to the work of external forces. This is a special case of the law of conservation of energy. In this case, the kinetic energy of a rotating rigid body around a fixed axis, equal to
work of external forces
Problem 3.27. To the disk from problem 3.24, rotating with an angular
speed with, press with force F brake pad. How many revolutions will the disk make before stopping? Friction coefficient between disc and pad To.
Solution. By analogy with the solution to Problem 3.25, one could find the entire kinematics of the disk motion, but the answer to the question posed can be given immediately based on formula (3.108). The disk is acted upon by a frictional force /^ (tangential force!) with a moment M= - kFR. There are no other moments. We have:
Problem 3.28. A rotating flywheel is an example of a mechanical energy storage device. Estimate to what angular velocity you need to spin a disk with a radius R= 0.3 m and a mass of 100 kg, so that due to this energy the car can travel 20 km.
Solution. We assume that a car with an engine power of 80 hp. s., or 60 kW, covers this distance in 20 minutes. The engine does work A = Nt. If work is done due to the energy of the flywheel, then
Substituting the numbers, we get
or 900 rps. (Cars with such an energy source have been practically tested.)
Let's return to equation (3.106) again. As we have already seen, this equation allows us to determine the entire kinematics of body motion around a fixed axis. The question arises: does this equation allow us to answer all the questions associated with such a movement, and in what relation does this equation stand with the equations
The answer to the first question is negative. The equation takes into account only the moments of forces rotating the body around an axis (lying in a plane orthogonal to the axis so that their lines of action do not pass through the axis). The equation does not allow us to determine the forces acting on the axis.
As for the answer to the second question, we emphasize once again that the motion of a rigid body is determined by the laws
Here V- speed of the center of mass; ?„ - intrinsic angular momentum (relative to the center of mass); defined by formula (3.86), M 0- moment of forces relative to the center of mass. The kinetic energy of the body is determined by formula (3.99). This fundamental (always fair) laws. Let us apply these formulas to the case under consideration.
Let us choose the origin of coordinates at some point on the rotation axis.
Let R- radius vector of the center of mass of the body and l - unit vector along the axis of rotation, coinciding in direction with the vector
angular velocity. We have: co = lo>, |k| = |wxl| = aso, where A - distance from the axis of rotation to the center of mass.
Body kinetic energy
(The last equality was obtained based on formula (3.109).) If A - 0 (axis passes through the center of mass),
where / 0 - moment of inertia of a body about an axis passing through the center of mass.
Magnitude am I? 0 is the projection of the body’s own angular momentum onto the axis of rotation. From formula (3.113) we obtain
Returning to formula (3.112), taking into account (3.114) we will have
From here we find the connection between the moments of inertia about a given axis and the axis parallel to it, passing through the center of mass:
(the so-called Steiner theorem).
Let's take a look at the angular momentum. Let's choose the origin of coordinates on the rotation axis at the point of intersection of the axis with the plane in which the center of mass rotates (this is not necessary, but it makes the analysis easier). We have:
The first term on the right side (orbital momentum) gives the vector directed along the rotation axis: pta 2 a>. Intrinsic angular momentum 
or, given that co = soy,
The basis vectors rotate with the body, so, for example, cl/ _ r
mer, - = co x /, therefore
(it was taken into account that th is a constant vector and dn/dt= 0). This result means that the components of the vector d are constant, and this, in turn,
turn, means (according to formula (3.117)) that the vector Z 0 rotates with the body and changes with time, even if the angular velocity of rotation of the body is constant (the vector Z 0 describes a conical surface, the axis of which is specified by the vector d). From formula (3.117) we obtain
(Recall that for any vector “frozen” into a body, ^ = c oh ha.)
The first of equations (3.111) will take the form
(origin at the center of the circle |l| = A, along which the center of mass moves), the second -
These two equations determine the forces and moments of force acting on a body rotating about a fixed axis. If a body rotates with a constant angular velocity and no forces, except those from the axis, act on it, formulas (3.119) and (3.120) determine the force and moment of forces acting on the body from the axis, and taken with the opposite sign - from sides of the body to the axis. The first of the equations gives the “centrifugal force” directed perpendicular to the axis. If the axis passes through the center of mass, this force is zero. The second takes the form
We see that the vector of the moment of force is directed perpendicular to the plane in which the axis of rotation and the moment of momentum lie, and rotates along with the body. This moment tends to rotate the axis in a plane orthogonal to the moment of force, and must be compensated by forces in the bearings holding the axis. This moment will vanish if the axis of rotation and the angular momentum vector are parallel, and this is only possible if the axis of rotation is parallel to one of the main axes of the inertia tensor. In technology, the problem of balancing rapidly rotating flywheels is very important.
Turning to formula (3.116), we write
Multiplying this equality scalarly by the vector H and taking into account formulas (3.114), (3.115), we obtain
Thus, the quantity appearing on the left side of equality (3.106) is projection of the total angular momentum onto the axis of rotation of the body. Then the right side of this equality is the projection of the complete
moment of force on the axis of rotation: M= I M(this could be verified directly). Thus, equation (3.106) derived at the beginning of this paragraph is simply a consequence of the fundamental equation
conclusions
The kinematics of rotation of a rigid body around a fixed axis is determined by formula (3.106). The moment of inertia about the axis is determined by formula (3.105) and is related in a complex way to the inertia tensor. The moment of forces relative to the axis (formula (3.105)) is the projection of the moment of forces onto the axis of rotation. The kinetic energy of a body rotating around a fixed axis is determined by formula (3.109), which is a consequence of the general formula (3.9). The forces acting on the axis can be found from formulas (3.111).
Note. It is necessary to distinguish between the concepts of “moments of impulse and forces relative to the axis” and simply “moments...”. The first are scalar quantities, the second are vector. To determine the former, you need to specify an axis, for the latter, a point.
Problem 3.29. A rigid body can rotate around a horizontal axis that does not pass through the center of mass. Moment of inertia of the body relative to the axis /, distance from the axis to the center of mass /, body mass T. The body is deviated from its equilibrium position by an angle 
Solution. Axis X- horizontal, axis at- vertically downward, the center of mass moves in the plane xOu, the axis of rotation passes through the origin, R- radius vector of the center of mass, R and axis u. The body is subject to forces: from the F axis and the force of gravity. When the body is deflected by an angle M = -wg/sincp (here / is the distance from the axis to the center of mass). Equation (3.106) becomes
This equation, up to notation, is identical to equation (2.149) and can be solved in exactly the same way (do this). For small deflection angles we get
This is a harmonic oscillation.
Problem 3.30. The pendulum is a disk with radius /?, mass T on a weightless rod of length /. The plane of the disk is in the plane of the pendulum's swing. How will such a pendulum move at small angles of deflection?
Solution. Formula (3.123) gives the answer, but it is necessary to determine the moment of inertia of this system relative to the axis of rotation. The axis of rotation is parallel to one of the main axes of the disk with a moment of inertia about this axis /. = - tYa 2. This value will be equal to the moment of inertia of the system / 0 relative to the axis passing through the center of mass. We find the moment of inertia of the pendulum using Steiner’s theorem: / = / n + ta 2 = tYa 2 /2 + m(l + I / 2) This value must be substituted into formula (3.123). Instead of / in this formula you need to substitute / + I/ 2.
Problem 3.31. Will the result of the previous problem change if the disk is rotated so that its plane is perpendicular to the plane of the pendulum's swing?
Answer. Will change. In this case, the axis of rotation of the system is parallel to the other main axis of the disk with a smaller (half) moment of inertia.
Problem 3.32. How will the solution to problem 3.30 change if we take into account the mass of the rod /i?
Solution. The moment of inertia about an axis is an additive quantity: the moment of inertia of a body is equal to the sum of the moments of inertia of its parts. Therefore, to the moment of inertia of the disk found in Problem 3.30, it is necessary to add the moment of inertia of the rod relative to the axis passing through its end perpendicular to the rod. This axis is parallel to one of the main axes of the rod with a moment / 2 = / 3 = ml 2/12. Using Steiner's theorem, we find that the moment about the end of the rod will be equal to ml 2 /3.
Problem 3.33. In the conditions of problem 3.29, determine the force acting on the axis of the pendulum.
Solution. There are two external forces acting on the pendulum: Fx from the axis and gravity mg. We have (for notation see Problem 3.29):
Equation (3.119) becomes
The solution to Problem 3.29 shows that - = 1 -sintp.
It comes down to finding the angular velocity. Let's turn to the law of conservation of energy. It is obvious that in the absence of friction forces, which we do not take into account, the mechanical energy of the system is conserved: W k + W n= const. Potential energy W n - it's energy
pendulum in the field of gravity. We have: R= /7 sintp + jl costp,
The law of conservation of energy gives the equation
(on the right side is the initial energy of the system). From here
We found the angular velocity as a function of the position of the pendulum. Returning to formula (3.124) for force, we obtain
This is the force acting on the axis of the pendulum. We see that the horizontal component of the force is non-zero, but equal to zero in the equilibrium position. The vertical component is maximum at the equilibrium position. It is obvious that a mathematical pendulum (a material point at the end of a weightless rod) is a special case of the considered system. Assuming /= ml 2, we get the result for a mathematical pendulum.
Problem 3.34. TO leaning against a vertical wall is a board of length 1/2 and mass T. At a moment in time t= 0 the board starts to fall. Find the force acting on the supporting end of the board.
the mass of the board moves in the plane xOy, R= / -jsin
radius vector of the center of mass (
- - g dtp g
- (O = -k- =-xo. External forces act on the board: F to the bottom at
end and gravity mg at the center of mass. Equation (3.119) becomes
The angular velocity, as in the previous problem, will be found from the law of conservation of energy:
(It was taken into account that the moment of inertia of the board, like a thin rod,
t t 1 g
equals I = -^-.)
To determine the angular acceleration, it is necessary to refer to equation (3.106). The center of mass of the board, to which the force of gravity is applied, moves in a circle, the tangential component of the force of gravity is equal to /ngsin
axes M =-^-sincp, there are no other points. Thus, Equation (3.126) becomes
where m is the unit tangent vector to the trajectory of the center of mass. Substituting this into formula (3.127) and solving the resulting equation for force, we obtain
This is the force acting on the bottom end of the board. Horizontal component of force at
then it begins to decrease, and when
This means that the board then loses contact with the wall, and at large angles the solution is incorrect. (If the bottom end of the board were hinged, the solution would be correct at any angle.) Moreover, numerical analysis shows that the vertical component of the force at an angle
Vladimir.erashov.rf
First, we formulate the unified law of inertia, which applies to all bodies and all types of motion:
The subsequent kinematic state of the body differs from the previous one only if, in the period between states, a new external force or moment of force begins to act on the body and it differs only in the magnitude of the body’s response to this effect.
With this law we are not opening any new pages in the kinematics of bodies; it was obtained on the basis of Newton’s laws, but with complex motion of a body it helps to simplify the task of describing this motion. We proceed from the fact that in the previous kinematic state, no matter what forces act on the body, it has already responded to the action of these forces and will continue to move according to the acquired laws. For example, in the initial state the acceleration acts on the body A , the body under the influence of this acceleration acquired speed v, but the acceleration continues to act until the subsequent state. This means that the body will increase the speed between states by the amount at. If some additional acceleration appears between the states, then it is enough to superimpose its effect on the previous result obtained, that is, to take advantage of the independence of the action of the forces. The main thread of the unified law of inertia is that if there is no change in the acting forces between states, then there are no changes in the laws of body motion, as in life, the next day is strung together with the previous one. If yesterday you didn’t have a penny of money in your soul, then today you will wake up without a penny of money. If yesterday you went on a long sea voyage on a cruise ship, then today you will wake up on a cruise ship. If your shirt is clean, it means someone washed it. Not a speck of dust or a hair will fall off of you by itself, there must be a reason for this (read some kind of force). If before rotation the main axis of inertia of the body was perpendicular to the surface of the Earth and was at rest relative to this surface, then after spinning the body it will be at rest relative to the Earth, as before (in case of stable rotation, in case of unstable rotation a specific force acts on the body). Changes in the state of the body can occur, but only under the influence of a specific force or moment of force and nothing else.
To make it easier to understand the action of the formulated law, and even try to derive practical benefit from this law, let’s consider a specific example - this is our rotating Earth and the bodies on its surface.
The first, let us clarify, acts on the Earth Newton's law of universal gravitation , so it is round, like a ball.
Second, the Earth is subject to centrifugal acceleration from rotation; under the influence of this acceleration, the Earth has acquired the shape of a geoid of rotation. Let us clarify that the property of the Earth-geoid is that at any point on the Earth’s surface any body remains motionless (even if it is capable of moving freely) due to the fact that the resulting force acting on the body from the forces of gravity and the centrifugal force of inertia is directed perpendicular to the surface and is balanced by the reaction of this surface (property of the geoid). Due to the geoid of rotation, even the ocean on the surface of the Earth came to an equilibrium state and became immobile relative to the surface, hence the geoid.
Let's return to the body on the surface of the Earth; no one is stopping us from assuming that a block in the shape of a rectangular parallelepiped lies on the surface of the Earth. The main axis of inertia of this block passes through the fulcrum on the surface and is perpendicular to the surface. Note that relative to the Earth the block lies motionless, but relative to the stars it, together with the Earth, makes one revolution per day.
Let us highlight for the readers that, relative to the stars, the bar is a rotating body with one revolution per day, the main axis of inertia of this bar is perpendicular to the surface of the Earth and motionless relative to the Earth. Let's spin the block to high speeds relative to its main axis of inertia. Will the axis of the block remain perpendicular and motionless relative to the Earth? Or, as is commonly believed, will it acquire movement (rotation) in relation to the Earth and, relative to the stars, change its state from rotating with one revolution per day to a stationary state?
According to the unified law of inertia, after spinning, the block must maintain a stationary state of the axis of rotation (the main axis of inertia) relative to the Earth, and relative to the stars it must still rotate at an angular speed of one revolution per day. This is motivated by the fact that when spinning a block, if the block is balanced relative to the axis of rotation, the same forces will act on the center of mass of the block as in the previous state (before spinning). Consequently, the subsequent state of the block (after spinning) is identical to the previous state (before spinning) and the block must retain all the properties of the previous state and not receive any changes, including the axis of rotation of the block must remain stationary and perpendicular to the surface of the Earth.
If someone does not like the combined law of inertia and does not agree with the conclusions according to the combined law, then the behavior of the block (rotating body) after spinning to maintain its original state, we explain by the fact that spinning did not add any new forces to the centimeter of mass of the block, and that’s all the parameters of the movement of the center of mass of the block in space remained the same.
In general, under the conditions of the Earth, the following forces act on a body, whether rotating or not:
1. The force of gravity of the Earth.
2. The force of inertia.
3. Ground reaction.
There are no other forces in nature; there is also Coriolis acceleration, but it is a derivative of the forces of inertia (it is not an independent force) and appears only when there is a movement of the body relative to the surface of the Earth. The Coriolis acceleration itself cannot transfer a body from a stationary state relative to the Earth to a moving one, there is no movement relative to the earth, and there is no Coriolis acceleration.
Bodies rotating rapidly relative to the main axis of inertia are called gyroscopes. Gyroscopes have a number of unique properties. Let's consider these properties too. It is generally accepted that the main property of a gyroscope is that they always keep the position of the axis of rotation relative to the stars fixed.
Our theory makes a significant clarification of this property of the gyroscope. In inertial reference systems, this property of the gyroscope is strictly observed, here we agree with the accepted theory, but in non-inertial reference systems, in particular those associated with the surface of the rotating Earth, this property does not act differently; the axis of the gyroscope, if the rotation is stable, retains its original position and relative to the stars, and relative to the Earth. But since in the initial position the axis of the gyroscope rotated relative to the stars, it will continue to rotate relative to the stars at the same speed, and relative to the Earth, as it was motionless, it will remain motionless. The state of the body is inert, the movement of the axis is inert, and not the direction towards anything.
The conclusion is unusual at first (inertia of thinking), that additional comments are required. Take the simple wolf meme to (yule). Let's launch the top to. Let's assume that the friction forces at the base of the top's axis are minimal, and it can maintain rotation for a relatively long time. According to our theory, the axis of rotation of the top remains stationary and perpendicular to the surface of the Earth, therefore, nothing prevents the top from stable and long-term rotation. In life, a top cannot be completely isolated from external forces; some external forces, let’s call them random, still act on the axis of the top and deflect it from its vertical position. Further, the weight force deviates from the fulcrum, a moment of force arises, to which the wolf responds with precession.
If the axis of rotation of the top, as is commonly believed, must remain stationary relative to the stars, then it cannot maintain a vertical position relative to the surface of the Earth for a long time; it will tilt from east to west at a speed of one revolution per day (12 degrees per hour). The axis of rotation of such a top will deviate from the vertical by about one degree within five minutes of rotation. If earlier, with the vertical position of the axis of rotation, the force of gravity acting on the center of mass lay on the axis of rotation and passed through the fulcrum and did not cause any movement of the center of mass, then when the axis of rotation is tilted, an overturning moment should occur. Moreover, the overturning moment circulates not only in direction, but also in magnitude. It is maximum in the lower position of the center of mass and minimum in the upper position. Thus, this moment should not cause precession of the top, but its nutation. This contradicts the results of the spinning top experiments. In a top, the main movement is precession, and nutation appears only at the very end of rotation, when the rotation is already close to random.
There are such units in industry as centrifuges. Due to the very high number of revolutions, these units are very sensitive to external forces. If their axis of rotation remained stationary relative to the stars, but tilted relative to the surface of the Earth, then these units would fly apart and fly apart, but they work. Consequently, our version of the interpretation of the behavior of rotating bodies in a non-inertial coordinate system is valid, and not the generally accepted one. Which was accepted based on experience, and not on theoretical grounds. This means that they did not properly understand the experimental material; they took what was not as a postulate.
Conclusion
The unified law of inertia operates in all reference systems, both inertial and non-inertial. Based on this law, the erroneous idea about the existing first law of the gyroscope was revealed, according to which the axis of rotation of the gyroscope must always be stationary relative to the stars. It has been established that gyroscopes behave this way only in inertial frames of reference; in non-inertial frames it is necessary to use not this rule, but the unified law of inertia.
July 12, 2018
