Moment of inertia passing through the center of mass. Application
Moment of force and moment of inertia
In the dynamics of the translational motion of a material point, in addition to kinematic characteristics, the concepts of force and mass were introduced. When studying the dynamics of rotational motion, physical quantities are introduced - torque and moment of inertia, the physical meaning of which will be discussed below.
Let some body under the action of a force applied at a point BUT, comes into rotation around the axis OO "(Figure 5.1).
Figure 5.1 - To the conclusion of the concept of moment of force
The force acts in a plane perpendicular to the axis. Perpendicular R, dropped from the point O(lying on the axis) to the direction of the force, called shoulder of strength. The product of the force on the shoulder determines the modulus moment of force relative to the point O:
(5.1)
Moment of power is a vector determined by the vector product of the radius-vector of the force application point and the force vector:
(5.2)
Unit of moment of force - newton meter(H . m). The direction of the vector of the moment of force is found using right screw rules.
A measure of the inertia of bodies in translational motion is the mass. The inertia of bodies during rotational motion depends not only on the mass, but also on its distribution in space relative to the axis of rotation. The measure of inertia during rotational motion is a quantity called moment of inertia of the body about the axis of rotation.
Moment of inertia of a material point relative to the axis of rotation - the product of the mass of this point by the square of the distance from the axis:
moment of inertia of the body about the axis of rotation - the sum of the moments of inertia of the material points that make up this body:
(5.4)
In the general case, if the body is solid and is a collection of points with small masses dm, the moment of inertia is determined by integration:
, (5.5)
where r- distance from the axis of rotation to the element of mass d m.
If the body is homogeneous and its density ρ = m/V, then the moment of inertia of the body
(5.6)
The moment of inertia of a body depends on which axis it rotates and how the mass of the body is distributed throughout the volume.
The moment of inertia of bodies that have the correct geometric shape and a uniform distribution of mass over volume is most simply determined.
Moment of inertia of a homogeneous rod about the axis passing through the center of inertia and perpendicular to the rod,
Moment of inertia of a homogeneous cylinder about an axis perpendicular to its base and passing through the center of inertia,
(5.8)
Moment of inertia of a thin-walled cylinder or hoop about an axis perpendicular to the plane of its base and passing through its center,
Moment of inertia of the ball relative to diameter
(5.10)
Let us determine the moment of inertia of the disk about the axis passing through the center of inertia and perpendicular to the plane of rotation. Let the mass of the disk be m, and its radius is R.
The area of the ring (Figure 5.2) enclosed between r and , is equal to .
Figure 5.2 - To the conclusion of the moment of inertia of the disk
Disk area. With a constant ring thickness,
from where or 
.
Then the moment of inertia of the disk,
For clarity, Figure 5.3 shows homogeneous solids of various shapes and indicates the moments of inertia of these bodies about the axis passing through the center of mass.
Figure 5.3 - Moments of inertia I C some homogeneous solids.
Steiner's theorem
The above formulas for the moments of inertia of bodies are given under the condition that the axis of rotation passes through the center of inertia. To determine the moments of inertia of a body about an arbitrary axis, one should use Steiner's theorem : the moment of inertia of the body about an arbitrary axis of rotation is equal to the sum of the moment of inertia J 0 about the axis parallel to the given one and passing through the center of inertia of the body, and the value md 2:
(5.12)
where m- body mass, d- distance from the center of mass to the selected axis of rotation. Unit of moment of inertia - kilogram-meter squared (kg . m 2).
So, the moment of inertia of a homogeneous rod of length l with respect to the axis passing through its end, according to the Steiner theorem is equal to
Consider now the problem determination of the moment of inertia various bodies. General formula for finding the moment of inertia object relative to the z-axis has the form
In other words, you need to add all the masses, multiplying each of them by the square of its distance from the axis (x 2 i + y 2 i). Note that this is true even for a three-dimensional body, even though the distance has such a "two-dimensional appearance". However, in most cases we will restrict ourselves to two-dimensional bodies.
As a simple example, consider a rod rotating about an axis passing through its end and perpendicular to it (Fig. 19.3). We now need to sum all the masses multiplied by the squares of the distance x (in this case, all y are zero). By sum, of course, I mean the integral of x 2 multiplied by the "elements" of the mass. If we divide the rod into pieces of length dx, then the corresponding element of mass will be proportional to dx, and if dx were the length of the entire rod, then its mass would be equal to M. Therefore
The dimension of the moment of inertia is always equal to the mass times the square of the length, so the only significant value that we have calculated is the factor 1/3.
And what will be the moment of inertia I if the axis of rotation passes through the middle of the rod? To find it, we again need to take the integral, but already in the range from -1/2L to +1/2L. Note, however, one feature of this case. Such a rod with an axis passing through the center can be thought of as two rods with an axis passing through the end, each having a mass of M/2 and a length of L/2. The moments of inertia of two such rods are equal to each other and are calculated by formula (19.5). Therefore, the moment of inertia of the entire rod is
Thus, the rod is much easier to twist at the middle than at the end.
It is possible, of course, to continue the calculation of the moments of inertia of other bodies of interest to us. But since such calculations require a lot of experience in calculating integrals (which is very important in itself), they, as such, are of little interest to us. However, there are some very interesting and useful theorems here. Let there be some body and we want to know it moment of inertia about some axis. This means that we want to find its inertia when rotating around this axis. If we move the body by the rod supporting its center of mass so that it does not turn during rotation around the axis (in this case, no moments of inertia forces act on it, so the body will not turn when we start moving it), then for in order to turn it, you need exactly the same force as if all the mass were concentrated in the center of mass and the moment of inertia would simply be equal to I 1 = MR 2 c.m. , where R c.m is the distance from the center of mass to the axis of rotation. However, this formula is, of course, incorrect. It does not give the correct moment of inertia of the body. After all, in reality, when turning, the body rotates. Not only the center of mass is spinning (which would give the value I 1), the body itself must also rotate relative to the center of mass. Thus, to the moment of inertia I 1 you need to add I c - the moment of inertia about the center of mass. The correct answer is that the moment of inertia about any axis is
This theorem is called the parallel axis translation theorem. It is proved very easily. The moment of inertia about any axis is equal to the sum of the masses multiplied by the sum of the squares of x and y, i.e. I \u003d Σm i (x 2 i + y 2 i). We will now focus our attention on x, but the same can be said for y. Let the x-coordinate be the distance of a given particular point from the origin; let's see, however, how things change if we measure the distance x` from the center of mass instead of x from the origin. To find out, we must write
x i = x` i + X c.m.
Squaring this expression, we find
x 2 i = x` 2 i + 2X c.m. x` i + X 2 c.m.
What happens if you multiply it by m i and sum over all r? Taking the constants out of the summation sign, we find
I x = Σm i x` 2 i + 2X c.m. Σm i x` i + X2 c.m. Σm i
The third sum is easy to calculate; it's just MX 2 ts.m. . The second term consists of two factors, one of which is Σm i x` i ; it is equal to the x`-coordinate of the center of mass. But this must be zero, because x` is measured from the center of mass, and in this coordinate system, the average position of all particles, weighted by their masses, is zero. The first term, obviously, is a part of x from I c. Thus, we arrive at formula (19.7).
Let's check formula (19.7) with one example. Let's just check if it will be applicable for the rod. We have already found that the moment of inertia of the rod relative to its end must be equal to ML 2 /3. And the center of mass of the rod, of course, is at a distance of L/2. So we should get that ML 2 /3=ML 2 /12+M(L/2) 2 . Since one fourth + one twelfth = one third, we did not make any blunder.
By the way, to find the moment of inertia (19.5), it is not at all necessary to calculate the integral. One can simply assume that it is equal to the value of ML 2 multiplied by some unknown coefficient γ. After that, one can use the reasoning about two halves and obtain the coefficient 1/4γ for the moment of inertia (19.6). Using now the parallel axis translation theorem, we prove that γ=1/4γ + 1/4, whence γ=1/3. You can always find some detour!
When applying the parallel axis theorem, it is important to remember that the I axis must be parallel to the axis about which we want to calculate the moment of inertia.
It is perhaps worth mentioning one more property, which is often very useful in finding the moment of inertia of some types of bodies. It consists in the following: if we have a flat figure and a triple of coordinate axes with the origin located in this plane and the z-axis directed perpendicular to it, then the moment of inertia of this figure about the z-axis is equal to the sum of the moments of inertia about the x and y axes . It is proved quite simply. notice, that
The moment of inertia of a homogeneous rectangular plate, for example, with mass M, width ω and length L about an axis perpendicular to it and passing through its center, is simply
since the moment of inertia about an axis lying in the plane of the plate and parallel to its length is equal to Mω 2 /12, i.e. exactly the same as for a rod of length ω, and the moment of inertia about another axis in the same plane is equal to ML 2 / 12, the same as for a rod of length L.
So, let's list the properties of the moment of inertia about a given axis, which we will call the z-axis:
1. The moment of inertia is
2. If an object consists of several parts, and the moment of inertia of each of them is known, then the total moment of inertia is equal to the sum of the moments of inertia of these parts.
3. The moment of inertia about any given axis is equal to the moment of inertia about a parallel axis through the center of mass, plus the product of the total mass times the square of the distance of that axis from the center of mass.
4. The moment of inertia of a flat figure about an axis perpendicular to its plane is equal to the sum of the moments of inertia about any two other mutually perpendicular axes lying in the plane of the figure and intersecting with the perpendicular axis.
In table. 19.1 shows the moments of inertia of some elementary figures that have a uniform mass density, and in table. 19.2 - moments of inertia of some figures, which can be obtained from table. 19.1 using the properties listed above.
| Parameter name | Meaning |
| Article subject: | Moment of inertia |
| Rubric (thematic category) | Mechanics |
Consider a material point with mass m, which is at a distance r from the fixed axis (Fig. 26). The moment of inertia J of a material point about an axis is usually called a scalar physical quantity equal to the product of the mass m and the square of the distance r to this axis:
J = mr 2(75)
The moment of inertia of the system of N material points will be equal to the sum of the moments of inertia of individual points
(76)
To the definition of the moment of inertia of a point
If the mass is distributed in space continuously, then summation is replaced by integration. The body is divided into elementary volumes dv, each of which has a mass dm. The result is the following expression:
(77)
For a body homogeneous in volume, the density ρ is constant, and having written the elementary mass in the form
dm = ρdv, we transform formula (70) as follows:
(78)
The dimension of the moment of inertia is kg * m 2.
The moment of inertia of a body is a measure of the inertia of a body in rotational motion, just as the mass of a body is a measure of its inertia in translational motion.
The moment of inertia is a measure of the inertial properties of a rigid body during rotational motion, depending on the distribution of mass relative to the axis of rotation. In other words, the moment of inertia depends on the mass, shape, dimensions of the body and the position of the axis of rotation.
Any body, regardless of whether it is rotating or at rest, has a moment of inertia about any axis, just as a body has mass, regardless of whether it is moving or at rest. Like mass, the moment of inertia is an additive quantity.
In some cases, the theoretical calculation of the moment of inertia is quite simple. Below are the moments of inertia of some solid bodies of regular geometric shape about an axis passing through the center of gravity.
Moment of inertia of an infinitely flat disk of radius R about an axis perpendicular to the disk plane:
Moment of inertia of a ball of radius R:
Moment of inertia of a rod with a length L relative to the axis passing through the middle of the rod perpendicular to it:
Moment of inertia of an infinitely thin hoop of radius R about an axis perpendicular to its plane:
The moment of inertia of a body about an arbitrary axis is calculated using the Steiner theorem:
The moment of inertia of a body about an arbitrary axis is equal to the sum of the moment of inertia about an axis passing through the center of mass parallel to the given one, and the product of the body's mass times the square of the distance between the axes.
Using the Steiner theorem, we calculate the moment of inertia of a rod with length L about the axis passing through the end perpendicular to it (Fig. 27).
To the calculation of the moment of inertia of the rod
According to the Steiner theorem, the moment of inertia of the rod about the O′O′ axis is equal to the moment of inertia about the OO axis plus md 2. From here we get:
Obviously: the moment of inertia is not the same with respect to different axes, and therefore, when solving problems on the dynamics of rotational motion, the moment of inertia of the body relative to the axis of interest to us each time has to be searched separately. So, for example, when designing technical devices containing rotating parts (in railway transport, in aircraft construction, electrical engineering, etc.), knowledge of the values of the moments of inertia of these parts is required. With a complex shape of the body, the theoretical calculation of its moment of inertia can be difficult to perform. In these cases, it is preferable to measure the moment of inertia of a non-standard part empirically.
Moment of force F relative to point O
Moment of inertia - concept and types. Classification and features of the category "Moment of inertia" 2017, 2018.
Fig.35 Let us draw arbitrary axes Cx"y"z" through the center of mass C of the body, and through any point O on the axis Cx" - the axes Oxyz, such that Oy½½Сy", Oz½½Cz" (Fig. 35). We denote the distance between the axes Cz "and Oz by d. Then, as can be seen from the figure, for any point of the body or, a. Substituting ... .
The moment of inertia of a body is a quantity that determines its inertia in rotational motion. In the dynamics of translational motion, the inertia of a body is completely characterized by its mass. The influence of the body's own properties on the dynamics of rotational motion turns out to be more complex, ... .
Lecture 3. Forces. Mass, momentum of a material point and a mechanical system. Dynamics of translational motion in inertial reference systems. The law of change in the momentum of a mechanical system. Law of conservation of momentum. Dynamics studies the motion of bodies, taking into account the causes, ... .
Let us analyze the formula for the moment of inertia of a rigid body. The moment of inertia depends on 1) the mass of the body, 2) the shape and dimensions of the body, 3) the position of the axis of rotation relative to the body (Fig. 2). 2a Fig.2b So, the moment of inertia is a measure of the body's inertia during rotational motion,... .
The moment of inertia about any axis is equal to the moment of inertia about the central axis parallel to the given one, plus the product of the area of \u200b\u200bthe figure and the square of the distance between the axes. From the formula it can be seen that the moment of inertia about the central axis is less than the moment ...
Moment of inertia
To calculate the moment of inertia, we must mentally divide the body into sufficiently small elements, the points of which can be considered as lying at the same distance from the axis of rotation, then find the product of the mass of each element by the square of its distance from the axis, and, finally, sum up all the resulting products. Obviously, this is a very labor intensive task. For counting
moments of inertia of bodies of regular geometric shape, in some cases, the methods of integral calculus can be used.
Finding the finite sum of the moments of inertia of the elements of the body will be replaced by the summation of an infinitely large number of moments of inertia calculated for infinitely small elements:
lim i = 1 ∞ ΣΔm i r i 2 = ∫r 2 dm. (at ∆m → 0).
Let us calculate the moment of inertia of a homogeneous disk or a solid cylinder with a height h about its axis of symmetry
Let us divide the disk into elements in the form of thin concentric rings with centers on the axis of its symmetry. The resulting rings have an inner diameter r and external r + dr, and the height h. Because dr<< r
, then we can assume that the distance of all points of the ring from the axis is r.
For each individual ring, the moment of inertia
i = ΣΔmr 2 = r 2 ΣΔm,
where ΣΔm is the mass of the entire ring.
Ring volume 2prhdr. If the density of the disk material ρ
, then the mass of the ring
ρ2prhdr.
Ring moment of inertia
i = 2πρhr 3dr.
To calculate the moment of inertia of the entire disk, it is necessary to sum the moments of inertia of the rings from the center of the disk ( r = 0) to its edge ( r = R), i.e. calculate the integral:
I = 2πρh 0 R ∫r 3dr,
or
I = (1/2)πρhR 4.
But the mass of the disk m = ρπhR 2, Consequently,
I = (1/2)mR 2.
We present (without calculation) the moments of inertia for some bodies of regular geometric shape, made of homogeneous materials 
1.
The moment of inertia of a thin ring about an axis passing through its center perpendicular to its plane (or a thin-walled hollow cylinder about its axis of symmetry):
I = mR 2.
2.
Moment of inertia of a thick-walled cylinder about the axis of symmetry:
I = (1/2)m(R 1 2 − R 2 2)
where R1− internal and R2− outer radii.
3.
The moment of inertia of the disk about an axis coinciding with one of its diameters:
I = (1/4)mR 2.
4.
The moment of inertia of a solid cylinder about an axis perpendicular to the generatrix and passing through its middle:
I \u003d m (R 2 / 4 + h 2 / 12)
where R− radius of the base of the cylinder, h is the height of the cylinder.
5.
The moment of inertia of a thin rod about an axis passing through its middle:
I = (1/12) ml 2,
where l is the length of the rod.
6.
The moment of inertia of a thin rod about an axis passing through one of its ends:
I = (1/3) ml 2
7. The moment of inertia of the ball about an axis coinciding with one of its diameters:
I = (2/5)mR 2.
If the moment of inertia of a body about an axis passing through its center of mass is known, then the moment of inertia about any other axis parallel to the first can be found on the basis of the so-called Huygens-Steiner theorem.
moment of inertia of the body I relative to any axis is equal to the moment of inertia of the body I s about an axis parallel to the given one and passing through the center of mass of the body, plus the mass of the body m times the square of the distance l between axles:
I \u003d I c + ml 2.
As an example, we calculate the moment of inertia of a ball of radius R and weight m suspended on a thread of length l, relative to the axis passing through the suspension point O. The mass of the thread is small compared to the mass of the ball. Since the moment of inertia of the ball about the axis passing through the center of mass Ic = (2/5)mR 2, and the distance
between axles ( l + R), then the moment of inertia about the axis passing through the suspension point:
I = (2/5)mR 2 + m(l + R) 2.
Dimension of the moment of inertia:
[I] = [m] × = ML 2.
Relative to a fixed axis ("axial moment of inertia") is called the value J a equal to the sum of the products of the masses of all n material points of the system into the squares of their distances to the axis:
- m i- weight i-th point,
- r i- distance from i-th point to the axis.
Axial moment of inertia body J a is a measure of the inertia of a body in rotational motion around an axis, just as the mass of a body is a measure of its inertia in translational motion.
If the body is homogeneous, that is, its density is the same everywhere, then
Huygens-Steiner theorem
Moment of inertia of a solid body relative to any axis depends not only on the mass, shape and size of the body, but also on the position of the body with respect to this axis. According to the Steiner theorem (Huygens-Steiner theorem), moment of inertia body J relative to an arbitrary axis is equal to the sum moment of inertia this body Jc relative to the axis passing through the center of mass of the body parallel to the considered axis, and the product of the body mass m per square distance d between axles:
where is the total mass of the body.
For example, the moment of inertia of a rod about an axis passing through its end is:
Axial moments of inertia of some bodies
| Body | Description | Axis position a | Moment of inertia J a |
|---|---|---|---|
 |
Material point of mass m | On distance r from a point, fixed | |
 |
Hollow thin-walled cylinder or ring of radius r and the masses m | Cylinder axis | |
 |
Solid cylinder or disk radius r and the masses m | Cylinder axis | |
 |
Hollow thick-walled mass cylinder m with outer radius r2 and inner radius r1 | Cylinder axis | |
 |
Solid cylinder length l, radius r and the masses m | ||
 |
Hollow thin-walled cylinder (ring) length l, radius r and the masses m | The axis is perpendicular to the cylinder and passes through its center of mass | |
 |
Straight thin rod length l and the masses m | The axis is perpendicular to the rod and passes through its center of mass | |
 |
Straight thin rod length l and the masses m | The axis is perpendicular to the rod and passes through its end | |
 |
Thin-walled sphere of radius r and the masses m | The axis passes through the center of the sphere | |
 |
ball radius r and the masses m | The axis passes through the center of the ball | |
 |
Cone radius r and the masses m | cone axis | |
| Isosceles triangle with height h, base a and weight m | The axis is perpendicular to the plane of the triangle and passes through the vertex | ||
| Right triangle with side a and weight m | The axis is perpendicular to the plane of the triangle and passes through the center of mass | ||
| Square with side a and weight m | The axis is perpendicular to the plane of the square and passes through the center of mass |
Derivation of formulas
Thin-walled cylinder (ring, hoop)
Formula derivation
The moment of inertia of a body is equal to the sum of the moments of inertia of its constituent parts. Dividing a thin-walled cylinder into elements with a mass dm and moments of inertia DJ i. Then
Since all elements of a thin-walled cylinder are at the same distance from the axis of rotation, formula (1) is converted to the form
Thick-walled cylinder (ring, hoop)
Formula derivation
Let there be a homogeneous ring with outer radius R, inner radius R 1, thick h and density ρ. Let's break it into thin rings with a thickness dr. Mass and moment of inertia of a thin ring of radius r will be
We find the moment of inertia of a thick ring as an integral
Since the volume and mass of the ring are equal
we obtain the final formula for the moment of inertia of the ring
Homogeneous disk (solid cylinder)
Formula derivation
Considering the cylinder (disk) as a ring with zero inner radius ( R 1 = 0), we obtain the formula for the moment of inertia of the cylinder (disk):
solid cone
Formula derivation
Divide the cone into thin discs of thickness dh, perpendicular to the axis of the cone. The radius of such a disk is
where R is the radius of the base of the cone, H is the height of the cone, h is the distance from the top of the cone to the disk. The mass and moment of inertia of such a disk will be
Integrating, we get
Solid uniform ball
Formula derivation
Divide the ball into thin disks dh, perpendicular to the axis of rotation. The radius of such a disk located at a height h from the center of the sphere, we find by the formula
The mass and moment of inertia of such a disk will be
We find the moment of inertia of the sphere by integrating:
thin-walled sphere
Formula derivation
For the derivation, we use the formula for the moment of inertia of a homogeneous ball of radius R:
Let us calculate how much the moment of inertia of the ball will change if, at a constant density ρ, its radius increases by an infinitesimal value dR.
Thin rod (axis passes through the center)
Formula derivation
Divide the rod into small fragments of length dr. The mass and moment of inertia of such a fragment is
Integrating, we get
Thin rod (the axis goes through the end)
Formula derivation
When moving the axis of rotation from the middle of the rod to its end, the center of gravity of the rod moves relative to the axis by a distance l/2. According to the Steiner theorem, the new moment of inertia will be equal to
Dimensionless moments of inertia of planets and their satellites
Of great importance for studies of the internal structure of planets and their satellites are their dimensionless moments of inertia. Dimensionless moment of inertia of a body of radius r and the masses m is equal to the ratio of its moment of inertia about the axis of rotation to the moment of inertia of a material point of the same mass relative to a fixed axis of rotation located at a distance r(equal to mr 2). This value reflects the distribution of mass in depth. One of the methods for measuring it in planets and satellites is to determine the Doppler shift of the radio signal transmitted by the AMS flying around a given planet or satellite. For a thin-walled sphere, the dimensionless moment of inertia is equal to 2/3 (~0.67), for a homogeneous ball - 0.4, and in general the smaller, the greater the mass of the body is concentrated at its center. For example, the Moon has a dimensionless moment of inertia close to 0.4 (equal to 0.391), so it is assumed that it is relatively homogeneous, its density changes little with depth. The dimensionless moment of inertia of the Earth is less than that of a homogeneous ball (equal to 0.335), which is an argument in favor of the existence of a dense core in it.
centrifugal moment of inertia
The centrifugal moments of inertia of a body with respect to the axes of a rectangular Cartesian coordinate system are the following quantities:
where x, y and z- coordinates of a small element of the body with volume dV, density ρ and weight dm.
The OX axis is called main axis of inertia of the body if the centrifugal moments of inertia Jxy and Jxz are simultaneously zero. Three main axes of inertia can be drawn through each point of the body. These axes are mutually perpendicular to each other. Moments of inertia of the body relative to the three main axes of inertia drawn at an arbitrary point O bodies are called main moments of inertia of the body.
The principal axes of inertia passing through the center of mass of the body are called main central axes of inertia of the body, and the moments of inertia about these axes are its main central moments of inertia. The axis of symmetry of a homogeneous body is always one of its main central axes of inertia.
Geometric moment of inertia
Geometric moment of inertia - geometric characteristic of the section of the view
where is the distance from the central axis to any elementary area relative to the neutral axis.
The geometric moment of inertia is not related to the movement of the material, it only reflects the degree of rigidity of the section. It is used to calculate the radius of gyration, beam deflection, section selection of beams, columns, etc.
The SI unit of measurement is m 4 . In construction calculations, literature and assortments of rolled metal, in particular, it is indicated in cm 4.
From it the section modulus is expressed:
.| Geometric moments of inertia of some figures | |
|---|---|
| Rectangle Height and Width: | |
| Rectangular box section with height and width along the outer contours and , and along the inner and respectively | |
| Circle diameter | |
Central moment of inertia
Central moment of inertia(or the moment of inertia about the point O) is the quantity
The central moment of inertia can be expressed in terms of the main axial or centrifugal moments of inertia: .
Tensor of inertia and ellipsoid of inertia
The moment of inertia of a body about an arbitrary axis passing through the center of mass and having a direction given by a unit vector can be represented as a quadratic (bilinear) form:
(1),where is the inertia tensor. The inertia tensor matrix is symmetrical, has dimensions, and consists of centrifugal moment components:
.
By choosing an appropriate coordinate system, the matrix of the inertia tensor can be reduced to a diagonal form. To do this, you need to solve the eigenvalue problem for the tensor matrix:
,
where is the orthogonal transition matrix to the own basis of the inertia tensor. In its own basis, the coordinate axes are directed along the principal axes of the inertia tensor and also coincide with the principal semiaxes of the inertia tensor ellipsoid. The magnitudes are the principal moments of inertia. Expression (1) in its own coordinate system has the form:
where does the equation come from
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