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

Moments of inertia homogeneous bodies of the simplest form with respect to some axes of rotation

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

Often we hear expressions: “it is inert”, “move by inertia”, “moment of inertia”. In a figurative sense, the word "inertia" can be interpreted as a lack of initiative and action. We are interested in direct meaning.

What is inertia

By definition inertia in physics, it is the ability of bodies to maintain a state of rest or motion in the absence of external forces.

If everything is clear with the very concept of inertia on an intuitive level, then moment of inertia- a separate issue. Agree, it is difficult to imagine in the mind what it is. In this article, you will learn how to solve basic problems on the topic "Moment of inertia".

Determining the moment of inertia

It is known from the school curriculum that mass is a measure of the inertia of a body. If we push two carts of different masses, then it will be more difficult to stop the one that is heavier. That is, the greater the mass, the greater the external influence is necessary to change the motion of the body. Considered refers to the translational movement, when the cart from the example moves in a straight line.

By analogy with mass and translational motion, the moment of inertia is a measure of the inertia of a body during rotational motion around an axis.

Moment of inertia- a scalar physical quantity, a measure of the inertia of a body during rotation around an axis. Denoted by letter J and in the system SI measured in kilograms multiplied by a square meter.

How to calculate the moment of inertia? There is a general formula by which the moment of inertia of any body is calculated in physics. If the body is broken into infinitely small pieces of mass dm , then the moment of inertia will be equal to the sum of the products of these elementary masses and the square of the distance to the axis of rotation.

This is the general formula for the moment of inertia in physics. For a material point of mass m , rotating about an axis at a distance r from it, this formula takes the form:

Steiner's theorem

What does the moment of inertia depend on? From the mass, the position of the axis of rotation, the shape and size of the body.

The Huygens-Steiner theorem is a very important theorem that is often used in solving problems.

By the way! For our readers there is now a 10% discount on

The Huygens-Steiner theorem states:

The moment of inertia of a body about an arbitrary axis is equal to the sum of the moment of inertia of the body about an axis passing through the center of mass parallel to an arbitrary axis and the product of the body's mass times the square of the distance between the axes.

For those who do not want to constantly integrate when solving problems of finding the moment of inertia, here is a figure showing the moments of inertia of some homogeneous bodies that are often found in problems:

An example of solving the problem of finding the moment of inertia

Let's consider two examples. The first task is to find the moment of inertia. The second task is to use the Huygens-Steiner theorem.

Problem 1. Find the moment of inertia of a homogeneous disk of mass m and radius R. The axis of rotation passes through the center of the disk.

Solution:

Let us divide the disk into infinitely thin rings, the radius of which varies from 0 before R and consider one such ring. Let its radius be r, and the mass dm. Then the moment of inertia of the ring:

The mass of the ring can be represented as:

Here dz is the height of the ring. Substitute the mass into the formula for the moment of inertia and integrate:

The result was a formula for the moment of inertia of an absolute thin disk or cylinder.

Problem 2. Let there again be a disk of mass m and radius R. Now we need to find the moment of inertia of the disk about the axis passing through the middle of one of its radii.

Solution:

The moment of inertia of the disk about the axis passing through the center of mass is known from the previous problem. We apply the Steiner theorem and find:

By the way, in our blog you can find other useful materials on physics and.

We hope that you will find something useful in the article. If there are difficulties in the process of calculating the inertia tensor, do not forget about the student service. Our experts will advise on any issue and help solve the problem in a matter of minutes.

Application. Moment of inertia and its calculation.

Let the rigid body rotate around the Z axis (Figure 6). It can be represented as a system of different material points m i , unchanged over time, each of which moves along a circle with a radius r i lying in a plane perpendicular to the Z axis. The angular velocities of all material points are the same. The moment of inertia of the body about the Z axis is the value:

where - the moment of inertia of a separate material point about the OZ axis. From the definition it follows that the moment of inertia is additive quantity, i.e., the moment of inertia of a body consisting of separate parts is equal to the sum of the moments of inertia of the parts.

Figure 6

Obviously, [ I] = kg × m 2. The importance of the concept of moment of inertia is expressed in three formulas:

; ; .

The first of them expresses the angular momentum of a body that rotates around a fixed axis Z (it is useful to compare this formula with the expression for the momentum of a body P = mVc, where Vc is the speed of the center of mass). The second formula is called the basic equation of the dynamics of the rotational motion of a body around a fixed axis, i.e., in other words, Newton's second law for rotational motion (compare with the law of motion of the center of mass: ). The third formula expresses the kinetic energy of a body rotating around a fixed axis (compare with the expression for the kinetic energy of a particle ). Comparison of the formulas allows us to conclude that the moment of inertia in rotational motion plays a role similar to mass in the sense that the greater the moment of inertia of the body, the less angular acceleration it acquires, all other things being equal (the body, figuratively speaking, is more difficult to spin). In reality, the calculation of the moments of inertia is reduced to the calculation of the triple integral and can be performed only for a limited number of symmetrical bodies and only for the axes of symmetry. The number of axes around which the body can rotate is infinitely large. Among all the axes, one stands out that passes through a wonderful point of the body - center of gravity (a point, to describe the movement of which it is enough to imagine that the entire mass of the system is concentrated at the center of mass and a force equal to the sum of all forces is applied to this point). But there are also infinitely many axes passing through the center of mass. It turns out that for any rigid body of arbitrary shape, there are three mutually perpendicular axes C x, C y, C z, called axes of free rotation , which have a remarkable property: if the body is twisted around any of these axes and thrown up, then during the subsequent movement of the body, the axis will remain parallel to itself, i.e. will not tumble. Twisting around any other axis does not have this property. The value of the moments of inertia of typical bodies about the indicated axes is given below. If the axis passes through the center of mass, but makes angles a, b, g with the axes C x, C y, C z accordingly, the moment of inertia about such an axis is equal to

I c = I cx cos 2 a + I cy cos 2 b + I cz cos 2 g (*)

Consider briefly the calculation of the moment of inertia for the simplest bodies.

1.The moment of inertia of a long thin homogeneous rod about an axis passing through the center of mass of the rod and perpendicular to it.

Let t - rod mass, l - its length.

,

Index " With» at the moment of inertia I c means that this is the moment of inertia about the axis passing through the point of the center of mass (the center of symmetry of the body), C(0,0,0).

2. Moment of inertia of a thin rectangular plate.

; ;

3. Moment of inertia of a rectangular parallelepiped.

, t. C(0,0,0)

4. Moment of inertia of a thin ring.

;

, t. C(0,0,0)

5. Moment of inertia of a thin disk.

Because of the symmetry

; ;

6. Moment of inertia of a solid cylinder.

;

Due to symmetry:

7. Moment of inertia of a solid ball.

, t. C(0,0,0)

8. Moment of inertia of a solid cone.

, t. C(0,0,0)

where R is the radius of the base, h is the height of the cone.

Recall that cos 2 a + cos 2 b + cos 2 g = 1. Finally, if the axis O does not pass through the center of mass, then the moment of inertia of the body can be calculated using the Huygens Steiner theorem

I o \u003d I c + md 2, (**)

where I o is the moment of inertia of the body about an arbitrary axis, I s- the moment of inertia about an axis parallel to it, passing through the center of mass,
m- body mass, d- the distance between the axles.

The procedure for calculating the moments of inertia for bodies of standard shape with respect to an arbitrary axis is as follows.

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.

Moment of inertia of a body about an axis and about a point. The moment of inertia of a material point about the axis is equal to the product of the mass of the point and the square of the distance of the point to the axis. To find the moment of inertia of a body (with a continuous distribution of matter) about the axis, it is necessary to mentally divide it into such small elements that each of them can be considered a material point of an infinitely small mass dm =  dV. Then the moment of inertia of the body about the axis is equal to the integral over the volume of the body:

where r– element distance dm to the axis.

The calculation of the moment of inertia of a body about an axis is often simplified if it is pre-calculated moment of inertia about a point . It is calculated by a formula similar to (1):

(2)

where r– element distance dm to the selected point (with respect to which the  ). Let this point be the origin of the coordinate system X, Y, Z(Fig. 1). Element distance squares dm to coordinate axes X, Y, Z and to the origin are equal respectively y 2 + z 2 , z 2 + x 2 , x 2 + y 2 , x 2 + y 2 + z 2 . Moments of inertia of the body about the axes X, Y, Z and relative to the origin

From these relations it follows that

In this way, the sum of the moments of inertia of the body about any three mutually perpendicular axes passing through one point is equal to twice the moment of inertia of the body about this point.

Moment of inertia of a thin ring. All elements of the ring dm(Fig. 2) are at the same distance equal to the radius of the ring R, from its axis of symmetry (Y-axis) and from its center. Moment of inertia of the ring about the Y axis

(4)

Moment of inertia of a thin disk. Let a thin homogeneous disk of mass m with a concentric hole (Fig. 3) has an inner and outer radius R 1 and R 2 . Let's mentally divide the disk into thin rings of radius r, thickness dr. The moment of inertia of such a ring about the axis Y(Fig. 3, it is perpendicular to the figure and not shown), in accordance with (4):

Disc moment of inertia:

(6)

In particular, setting in (6) R 1 = 0, R 2 = R, we obtain a formula for calculating the moment of inertia of a thin continuous homogeneous disk about its axis:

The moment of inertia of the disk about its axis of symmetry does not depend on the thickness of the disk. Therefore, formulas (6) and (7) can be used to calculate the moments of inertia of the corresponding cylinders relative to their axes of symmetry.

The moment of inertia of a thin disk relative to its center is also calculated by formula (6),  = J y , and the moments of inertia about the axes X and Z are equal to each other J x = J z. Therefore, according to (3): 2 J x + J y = 2 J y , J x = J y /2, or

(8)

moment of inertia of the cylinder. Let there be a hollow symmetric cylinder of mass m, length h, whose inner and outer radii are equal R 1 and R 2 . Find its moment of inertia about the axis Z, drawn through the center of mass perpendicular to the axis of the cylinder (Fig. 4). To do this, mentally divide it into disks of infinitesimal thickness dy. One of these disks, weighing dm = mdy/ h located at a distance y from the origin of coordinates, shown in Fig. 4. Its moment of inertia about the axis Z, in accordance with (8) and the Huygens – Steiner theorem

Moment of inertia of the whole cylinder

Moment of inertia of the cylinder about the axis Z (the axis of rotation of the pendulum) we find by the Huygens-Steiner theorem

where d is the distance from the center of mass of the cylinder to the axis Z . In Ref. 16 this moment of inertia is denoted as J c

(11)

LEAST SQUARE METHOD

Drawing experimental points and drawing a graph on them “by eye”, as well as determining the abscissa and ordinates of points from the graph, are not very accurate. It can be increased by using the analytical method. The mathematical rule for constructing a graph is to select such values ​​of the parameters "a" and "b" in a linear relationship of the form y = ax + b so that the sum of the squared deviations  at i (Fig. 5) of all experimental points from the graph line was the smallest ( least square method"), i.e. so that the value

(1)