The equation of the dynamics of rotational motion of the body. Saveliev I.V. Course of General Physics, Volume I. An unproven and unrefuted hypothesis is called an open problem
Rigid body around a fixed axis.
angular momentum solid body during rotational movement around the z-axis is calculated as
Then the equation of the dynamics of rotational motion will take the form:
If the body is rigid, then, therefore, taking into account the fact that (angular acceleration), we obtain the expression
This is the equation of the dynamics of rotational motion of a rigid body around a fixed axis:
the angular acceleration of the rotational motion of a rigid body around a fixed axis is directly proportional to the magnitude of the moment of external forces about this axis.
Comment. By analogy with Newton's second law, in which acceleration is determined by force, the equation for the dynamics of rotational motion of a rigid body gives a relationship between angular acceleration and moment of force. In this sense, the moment of inertia of the body plays the role measures of inertia during rotational motion.
Examples of calculation of moments of inertia.
1) Moment of inertia of a thin ring (straight thin-walled cylinder) of mass m and radius R about the z-axis perpendicular to the plane of the ring passing through the center of the ring
2) Moment of inertia of a disk (solid cylinder) of mass m and radius R about the z axis, perpendicular to the disk plane passing through the center of the disk (solid cylinder).
Select a thin cylinder with radius r and thickness dr.
The mass of this cylinder 
, .
3) Moment of inertia of a thin rod about the z-axis, which is the perpendicular bisector. Rod mass m, length L.
Let us single out at a distance x from the axis a small part of the rod of length dx.
The mass of this part and . So
.
4) Moment of inertia of a thin-walled ball about any axis of symmetry z. Ball mass m, radius R.
On the surface of the sphere, we single out an annular segment for which the z-axis is the axis of symmetry. The segment rests on a small central angle dj, the position of the segment is determined by the angle j measured from the equatorial plane perpendicular to the z axis.
Then the radius of the ring,
its mass 
, That's why
or
5) The moment of inertia of a solid ball about any axis of symmetry z. The mass of the ball is m, the radius of the ball is R.
Let us imagine a ball as a set of thin-walled spheres of variable radius nested into each other r and thickness dr. The mass of one such sphere 
.
The moment of inertia of such a sphere is .
.
Huygens-Steiner theorem
How are the moments of inertia of a rigid body related to two parallel axes?
Consider two parallel axes z 1 and z 2 . We introduce two coordinate systems so that their x and y axes are parallel to each other, and the second coordinate system was obtained by parallel transfer from the first to a vector perpendicular to the axes z 1 and z 2 . Then the distance between the axes will be equal.
In this case, the coordinates of any i- and small particles of the body are related by the relations
The square of the distance from this point to the first z-axis is 1:
and up to the second axis z 2 .
We calculate the moment of inertia about the second axis:
In this equality
The moment of inertia of the body about the axis z 1,
We learn that 
and 
(where x 1C and y 1C - coordinates of the center of mass of the body in the 1st coordinate system) and we get
Assuming that the z-axis 1 passes through the center of mass of the body, then x 1C = 0 and y 1C = 0, so in this case the expression is simplified:
This expression is called the Huygens-Steiner theorem: the moment of inertia of a rigid body about an arbitrary axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through the center of mass of the body and the square of the distance between the axes multiplied by the mass of the body.
Example. The moment of inertia of the rod about the axis passing through the edge of the rod, perpendicular to it, is equal to the sum of the moment of inertia about the median axis and the mass, multiplied by the square of half the length of the rod:
.
Example. Consider the movement of loads on a weightless inextensible thread thrown over a block (disk). Weights m 1 and m 2 (m 1< m 2), масса блока m. Трения в оси блока нет. Нить не скользит по блоку. Силами сопротивления в воздухе пренебрегаем. Найти ускорение грузов. Радиус блока R.
Decision. We fix a frame of reference in which the axis of the block is fixed. We assume that this frame of reference is inertial. The z-axis of the coordinate system in this reference system will be directed along the rotation axis of the block (“away from us”).
“Mentally” we break the system into parts and find the forces between the parts of the system in accordance with the second and third laws of Newton.
At the same time, we take into account that the thread is weightless (the mass of any part of the thread is zero), therefore, if a piece of thread moves under the action of (tensile) forces, then from Newton's second law
The work during rotation of the body goes to increase its kinetic energy. Because , then or .
Considering that , we get . Therefore, the moment of force
acting on the body is equal to the product of the moment of inertia of the body and the angular acceleration. If the axis of rotation coincides with the free axis (see 7.7), then the vector equality holds
This equality is the basic equation of the dynamics of the rotational motion of a rigid body about the fixed axis.
Example 4.5.1. A thin rod of length and mass rotates around a fixed axis with angular acceleration. The axis of rotation is perpendicular to the rod and passes through its middle. Determine the moment of force acting on the rod.
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According to the basic equation of the dynamics of rotational motion, the torque is related to the angular acceleration by the following relationship: ; where is the moment of inertia of the rod about the axis of rotation. Because the axis of rotation passes through the center of mass of the rod, then .
Therefore, the moment of force acting on the rod is .
Answer : .
Example 4.5.2. The shaft in the form of a solid cylinder is mounted on a horizontal axis with a mass. An inextensible cord is wound around the cylinder, to the free end of which a weight of mass is suspended. With what acceleration will the weight fall if it is left to itself?
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Let's make a drawing (Fig. 4.5.1). The load descends with acceleration. It is affected by the forces of gravity and the tension of the cord. The shaft rotates counterclockwise with angular acceleration. The force of gravity acts on the shaft, the reaction force from the axis on which the shaft rests, and the reaction force from the side of the cord. The torque is created only by force, because. line of action of forces andpasses through the axis of rotation (the arm of these forces is equal to 0).
The basic equation of the dynamics of the translational movement of the load has the form:
. Projected onto the Oy axis: .
The basic equation of the dynamics of the rotational motion of the shaft has the form: .
If the force acting on the body creates a moment that promotes rotation in a given direction, then its moment is considered positive (the direction of the vector of the moment of force coincides with the direction of angular acceleration), if it interferes, the moment is considered negative (the directions and are opposite). Therefore, in scalar form (in projection onto the direction of angular acceleration) the basic equation of the dynamics of rotational motion will have the form: .
Considering that the axis of rotation passes through the center of mass of the cylindrical shaft perpendicular to the plane of its base, where the radius of the base of the cylinder, and the torque (the arm of force is equal to the radius of the base of the cylinder), then.
According to Newton's third law (the cord is inextensible), therefore . The tangential acceleration of the points lying on the shaft rim is related to its angular acceleration by the relation: . Any point of the cord on which the load is suspended moves with the same acceleration. Therefore, whence . Substituting into equation (1), we obtain: and.
Answer:.
Example 4.5.3. A thin flexible thread is thrown through a block in the form of a disk having mass , to the ends of which weights of masses and are suspended. With what acceleration will the loads move if they are left to themselves? Ignore friction.
Decision:
Let's make a drawing (Fig. 4.5.2). The first weight will move progressively upwards with acceleration , the second will fall with the same acceleration. The equations of translational motion of loads in vector form have the form .
Projected onto the axis direction:
, where .
According to the basic equation of the dynamics of rotational motion. When the masses move, the disk rotates rapidly clockwise, therefore, the force contributes to the rotation, and the force inhibits the rotation. Therefore, in scalar form (in projection onto the direction of angular acceleration), since the shoulder of forces is equal to the radius of the disk.
Considering that the moment of inertia of the disk , and the linear acceleration of the loads is equal to
tangential acceleration of the disk rim points associated with angular acceleration
wearing , then , from where .. In scalar form (projected onto the direction of angular acceleration)
Answer: .
Recall that elementary workdAstrengthFcalled the scalar product of the forceFfor an infinitesimal displacementdl:where is the angle between the direction of the force and the direction of movement.
Note that the normal component of the force F n(as opposed to tangential F τ ) and the support reaction force N work is not done, since they are perpendicular to the direction of movement.
Element dl=rd at small angles of rotation d (r is the radius vector of the body element). Then the work of this force is written as follows:
. (19)
The expression Fr cos is the moment of force (the product of the force F and the arm p=r cos):
(20)
Then the work is
. (21)
This work is spent on changing the kinetic energy of rotation:
. (22)
If I=const, then after differentiating the right side we get:
or, since 
, (23)
where 
- angular acceleration.
Expression (23) is the equation of the dynamics of the rotational motion of a rigid body relative to a fixed axis, which is better to represent from the point of view of cause-and-effect relationships as:
. (24)
The angular acceleration of a body is determined by the algebraic sum of the moments of external forces about the axis of rotation divided by the moment of inertia of the body about this axis.
Let's compare the main quantities and equations that determine the rotation of the body around a fixed axis and its translational motion (see table 1):
Table 1
|
translational movement |
rotational movement |
|
Moment of inertia I |
|
|
Speed |
Angular speed |
|
Acceleration |
Angular acceleration |
|
Force |
Moment of power |
|
Basic equation of dynamics: |
Basic equation of dynamics: |
|
Work |
Work |
|
Kinetic energy |
Kinetic energy |
or 
The dynamics of the translational motion of a rigid body is completely determined by the force and mass as a measure of their inertia. During the rotational motion of a rigid body, the dynamics of motion is determined not by the force as such, but by its moment, the inertia is not by the mass, but by its distribution relative to the axis of rotation. The body does not acquire angular acceleration if a force is applied, but its moment will be zero.
Methodology for performing work
circuit diagram laboratory setup shown in Fig.6. It consists of a disk of mass m d , four rods of mass m 2 fixed on it, and four weights of mass m 1 located symmetrically on the rods. A thread is wound around the disk, from which a weight m is suspended.
According to Newton's second law, we compose the equation of the translational motion of the load m without taking into account the forces of friction:
|
|
(25)or in scalar form, i.e. in projections on the direction of movement:
. (26)
, (27)
where T is the tension force of the thread. According to the basic equation of the dynamics of rotational motion (24), the moment of force T, under the influence of which the system of bodies m d , m 1, m 2 performs rotational motion, is equal to the product of the moment of inertia I of this system and its angular acceleration :
or 
, (28)
where R is the arm of this force equal to the disk radius.
Let us express the tension force of the thread from (28):
(29)
and equate the right sides of (27) and (29):
. (30)
Linear acceleration is related to the angular following relation a=R, therefore:
. (31)
Whence the acceleration of the load m without taking into account the friction forces in the block is:
. (32)
Consider the dynamics of the system's motion, taking into account the friction forces that act in the system. They occur between the rod on which the disk is fixed and the fixed part of the installation (inside the bearings), as well as between the moving part of the installation and the air. We will take into account all these friction forces using the moment of friction forces.
With considering friction moment the rotation dynamics equation is written as follows:
, (33)
where a' is linear acceleration under the action of friction forces, Mtr is the moment of friction forces.
Subtracting equation (33) from equation (28), we get:
,
. (34)
Acceleration without taking into account the friction force (a) can be calculated using formula (32). The acceleration of the weight, taking into account the forces of friction, can be calculated from the formula for uniformly accelerated motion, by measuring the distance traveled S and the time t:
. (35)
Knowing the values of accelerations (a and a’), formula (34) can be used to determine the moment of friction forces. For calculations, it is necessary to know the value of the moment of inertia of the system of rotating bodies, which will be equal to the sum of the moments of inertia of the disk, rods and loads.
The moment of inertia of the disk according to (14) is equal to:
. (36)
The moment of inertia of each of the rods (Fig. 6) relative to the O axis according to (16) and the Steiner theorem is:
where a c =l/2+R, R is the distance from the center of mass of the rod to the axis of rotation О; l is the length of the rod; I oc - its moment of inertia about the axis passing through the center of mass.
Similarly, the moments of inertia of the loads are calculated:
, (38)
where h is the distance from the center of mass of the load to the axis of rotation О; d is the length of the load; I 0 r is the moment of inertia of the load about the axis passing through its center of mass. Adding the moments of inertia of all bodies, we obtain a formula for calculating the moment of inertia of the entire system.
A rigid body can be represented as a collection of material points. When the body rotates, all these points have the same angular velocities and accelerations. Using the results of § 7.6, it is relatively easy to obtain the equation of motion of a rigid body as it rotates about a fixed axis.
Motion equation
To derive the basic equation for the dynamics of rotational motion, one can proceed as follows. Mentally divide the body into separate, sufficiently small elements that could be considered as material points (Fig. 7.33). Write equation (7.6.13) for each element, and add all these equations term by term. In this case, the internal forces acting between the individual elements will not be included in the equation of motion of the body. The sum of their moments as a result of adding the equations will be equal to zero, since, according to Newton's third law, the interaction forces are equal in absolute value and directed along one straight line in opposite directions. Considering further that during the rotation of a rigid body, all its points make the same angular displacements with the same speeds and accelerations, we can thus obtain the equation of rotational motion of the entire body.
However, the derivation of this equation is rather cumbersome, so we will not dwell on it. Moreover, this equation has the same form as equation (7.6.13) for a material point moving in a circle:
O"
O"
(7.7.1)
d(J In this equation JI
on the body relative to the axis of rotation.
Equation (7.7.1) is read as follows: the time derivative of the angular momentum is equal to the total moment of external forces.
It should be borne in mind that JITO rotation of a body around an axis can only be caused by forces Ft lying in a plane perpendicular to the axis of rotation (Fig. 7.34). The forces Fk directed parallel to the axis of rotation, obviously, can only cause the body to move along the axis. The moment of each force Fl is equal to the product of the modulus of this force and the arm d, taken with a plus or minus sign, i.e., the length of the segment of the perpendicular dropped from the point C of the axis to the line of action of the force Ft:
Mi = ±Ftd. (7.7.2)
The moment of force rotating the body around a given axis counterclockwise is considered positive, and clockwise - negative.
moment of inertia of the body
The formula (7.7.1) includes the moment of inertia of the body J. The moment of inertia of the body J is equal to the sum of the moments of inertia AJ - individual small elements into which the whole body can be divided:
(7.7.3)
і
Since the moment of inertia of a material point
AJ^Amtf, (7.7.4)
where Ampi is the mass of the body element, and r, is its distance to the axis of rotation (see Fig. 7.33), then
J = J A mtrf . (7.7.5)
385
13-Myakishev, 10 cells.
The moment of inertia of a body depends not only on the mass of the body, but also on the nature of the distribution of this mass. The more stretched
Rice. 7.35
body along the axis of rotation, the smaller its moment of inertia, since the closer to the axis of rotation are the individual elements of the body. It is also obvious that by changing the axis of rotation of the body, we thereby change its moment of inertia. For solids, the moment of inertia about a given axis is a constant. Therefore, a change in the angular momentum can occur only due to a change in the angular velocity. Accordingly, equation (7.7.1) can be written as:
jft = M. (7.7.6)
This equation is read as follows: the product of the moment of inertia of the body about the axis of rotation and the angular acceleration of the body is equal to the sum of the moments (relative to the same axis) of all external forces applied to the body.
Equation (7.7.6) shows that when a body rotates, the moment of inertia plays the role of mass, the moment of force plays the role of force, and the angular acceleration plays the role of linear acceleration during the movement of a material point or center of mass.
It is easy to verify that the angular acceleration is really determined by the moment of force, i.e., by the force and the shoulder, and not just by the force. So, it is possible to spin a bicycle wheel to the same angular velocity with the same force (for example, by finger force) much faster if the force is applied to the wheel rim (this creates a greater moment), and not to the spokes near the hub (Fig. .7.35).
In order to make sure that the angular acceleration is determined precisely by the moment of inertia, and not by the mass of the body, you need to have at your disposal a body whose shape can be easily changed without changing the mass. A bicycle wheel is not suitable here. But you can use your own body. Try twisting on your heel by pushing off the floor with your other foot. If you press your arms to your chest at the same time, then the angular velocity will be greater than if you spread your arms to the sides. The effect will be especially noticeable if you take a thick book in both hands.
Moments of inertia of the hoop and cylinder
Finding the moment of inertia of a body of an arbitrary asymmetric shape is quite difficult. It is easier to measure it empirically than to calculate it.
We restrict ourselves to calculating the moment of inertia of a thin hoop rotating around an axis passing through its center. If the mass of the wheel is concentrated mainly in its rim (as, for example, in a bicycle wheel), then such a wheel can be approximately considered as a hoop, neglecting the mass of the spokes and the bushing.
Let's break the hoop into N identical elements. If m is the mass of the entire hoop, then the mass of each element is Dmi = ^. Thickness
we will consider the hoop to be much smaller than its radius (Fig. 7.36). If the number of elements is chosen sufficiently large, then each element can be considered as a material point. Therefore, the moment of inertia of an arbitrary element with number i will be equal to:
D Jt = Dt; D2. (7.7.7)
Substituting expression (7.7.7) into formula (7.7.5) for the total moment of inertia, we obtain:
N
(7.7.8)
J= D^D miR2 = mR2.
Rice. 7.36
Here we took into account that the distance R for all elements is the same and that the sum
masses of elements is equal to the mass t about-
I
stream.
13*
387
A very simple result was obtained: the moment of inertia of the hoop is equal to the product of its mass and the square of the radius. The moment of inertia of a hoop of a given mass is greater, the greater its radius. Formula (7.7.8) also determines the moment of inertia
hollow thin-walled cylinder during its rotation around the axis of symmetry.
Calculating the moment of inertia of a solid homogeneous cylinder with mass m and radius R about its axis of symmetry is a more complex problem. We present only the calculation result: (7.7.9)
J =\mR2. Therefore, if we compare the moments of inertia of two cylinders of the same size and mass, one of which is hollow and the other solid, then the moment of inertia of the second cylinder will be half as much. This is due to the fact that the mass of a solid cylinder is, on average, closer to the axis of rotation.
We got acquainted with the equation of rotational motion of a rigid body. In form, it is similar to the equation for the translational motion of a rigid body. The definition of new physical quantities characterizing a solid body is given: the moment of inertia and the moment of impulse.
The basic equation of the dynamics of rotational motion - section Mechanics, The unproven and unrefuted hypothesis is called an open problem According to Equation (5.8) Newton's Second Law For Rotational Motion...
This expression is called the basic equation of the dynamics of rotational motion and is formulated as follows: the change in the angular momentum of a rigid body is equal to the momentum momentum of all external forces acting on this body.
Angular moment (kinetic moment, angular momentum, orbital momentum, angular momentum) characterizes the amount of rotational motion. A quantity that depends on how much mass is rotating, how it is distributed about the axis of rotation, and how fast the rotation occurs.
Comment: angular momentum about a point is a pseudovector, and angular momentum about an axis is a scalar quantity.
It should be noted that rotation here is understood in a broad sense, not only as a regular rotation around an axis. For example, even when rectilinear motion body past an arbitrary imaginary point, it also has an angular momentum. The angular momentum plays the greatest role in describing the actual rotational motion.
The angular momentum of a closed system is conserved.
Law of conservation of momentum(law of conservation of angular momentum) - the vector sum of all angular momenta about any axis for a closed system remains constant in the case of equilibrium of the system. In accordance with this, the angular momentum of a closed system relative to any fixed point does not change with time.
The law of conservation of angular momentum is a manifestation of the isotropy of space.
Where does the law of conservation of angular momentum apply? Who among us does not admire the beauty of the movements of figure skaters on the ice, their rapid rotations and equally rapid transitions to slow gliding, the most difficult somersaults of gymnasts or jumpers on a trampoline! This amazing craftsmanship is based on the same effect, which is a consequence of the law of conservation of angular momentum. Spreading his arms to the sides and winding his free leg, the skater informs himself of a slow rotation around the vertical axis (see Fig. 1). Having sharply "grouped", it reduces the moment of inertia and receives an increment in angular velocity.
If the axis of rotation of the body is free (for example, if the body falls freely), then the conservation of angular momentum does not mean that the direction of the angular velocity is conserved in the inertial frame of reference. With rare exceptions, the instantaneous axis of rotation is said to precess around the direction of the angular momentum of the body. This is manifested in the tumbling of the body when falling. However, the bodies have the so-called main axes of inertia, which coincide with the axes of symmetry of these bodies. The rotation around them is stable, the vectors of angular velocity and angular momentum coincide in direction, and no tumbling occurs.
If you carefully observe the work of the juggler, you will notice that, throwing up objects, he gives them rotation. Only in this case, maces, plates, hats are returned to his hands in the same position that they were given. Rifled weapons give better aim and greater range than smoothbore weapons. An artillery shell fired from a gun rotates around its longitudinal axis, and therefore its flight is stable.
Fig.2. fig.3.
The well-known top, or gyroscope, behaves in the same way (Fig. 2). In mechanics, a gyroscope is any massive homogeneous body rotating around an axis of symmetry with a large angular velocity. Usually the axis of rotation is chosen so that the moment of inertia about this axis is maximum. Then the rotation is most stable.
To create a free gyroscope in technology, a cardan suspension is used (Fig. 3). It consists of two annular clips that fit one into the other and can rotate relative to each other. Intersection point of all three axes 00, O"O" and O"0" coincides with the position of the center of mass of the gyroscope WITH. In such a suspension, the gyroscope can rotate around any of the three mutually perpendicular axes, while the center of mass relative to the suspension will be at rest.
While the gyroscope is stationary, it can be rotated around any axis without much effort. If the gyroscope is brought into rapid rotation about the axis 00 and after that try to rotate the gimbal, then the axis of the gyroscope tends to keep its direction unchanged. The reason for such stability of rotation is connected with the law of conservation of angular momentum. Since the moment of external forces is small, it is not able to noticeably change the angular momentum of the gyroscope. The axis of rotation of the gyroscope, with the direction of which the angular momentum vector almost coincides, does not deviate far from its position, but only trembles, remaining in place.
This property of the gyroscope finds wide practical application. The pilot, for example, must always know the position of the true earth's vertical in relation to the position of the aircraft in this moment. An ordinary plumb line is not suitable for this purpose: with accelerated movement, it deviates from the vertical. Fast-rotating gyroscopes on gimbals are used. If the axis of rotation of the gyroscope is set so that it coincides with the earth's vertical, then no matter how the aircraft changes its position in space, the axis will retain the direction of the vertical. Such a device is called a gyrohorizon.
If the gyroscope is in a rotating system, then its axis is set parallel to the axis of rotation of the system. In terrestrial conditions, this manifests itself in the fact that the axis of the gyroscope is eventually set parallel to the axis of rotation of the Earth, indicating the north-south direction. In marine navigation, such a gyroscopic compass is an absolutely indispensable instrument.
This, at first glance, strange behavior of the gyroscope is also in full agreement with the equation of moments and the law of conservation of angular momentum.
The law of conservation of angular momentum is, along with the laws of conservation of energy and momentum, one of the most important fundamental laws of nature and, generally speaking, is not derived from Newton's laws. Only in a particular case, when the motion along the circles of particles or material points, the totality of which forms a rigid body, is considered, such an approach is possible. Like other conservation laws, according to Noether's theorem, it is associated with a certain kind of symmetry.
End of work -
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