§13. Steiner's theorem on the moment of inertia about an arbitrary axis
body m per square distance d between axles:
J = J c + m d 2 , (\displaystyle J=J_(c)+md^(2),)Where m- total body weight.
For example, the moment of inertia of a rod about an axis passing through its end is:
J \u003d J c + m d 2 \u003d 1 12 m l 2 + m (l 2) 2 \u003d 1 3 m l 2. (\displaystyle J=J_(c)+md^(2)=(\frac (1)(12))ml^(2)+m\left((\frac (l)(2))\right)^ (2)=(\frac (1)(3))ml^(2).)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 | m r 2 (\displaystyle mr^(2)) | |
| Solid cylinder or disk radius r and the masses m | Cylinder axis | 1 2 m r 2 (\displaystyle (\frac (1)(2))mr^(2)) | |
| Hollow thick-walled mass cylinder m with outer radius r 2 and inner radius r 1 | Cylinder axis | m r 2 2 + r 1 2 2 (\displaystyle m(\frac (r_(2)^(2)+r_(1)^(2))(2))) | |
| Solid cylinder length l, radius r and the masses m | 1 4 m ⋅ r 2 + 1 12 m ⋅ l 2 (\displaystyle (1 \over 4)m\cdot r^(2)+(1 \over 12)m\cdot l^(2)) | ||
| 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 | 1 2 m ⋅ r 2 + 1 12 m ⋅ l 2 (\displaystyle (1 \over 2)m\cdot r^(2)+(1 \over 12)m\cdot l^(2)) | |
| Straight thin rod length l and the masses m | The axis is perpendicular to the rod and passes through its center of mass | 1 12 m l 2 (\displaystyle (\frac (1)(12))ml^(2)) | |
| Straight thin rod length l and the masses m | The axis is perpendicular to the rod and passes through its end | 1 3 m l 2 (\displaystyle (\frac (1)(3))ml^(2)) | |
| Thin-walled sphere of radius r and the masses m | The axis passes through the center of the sphere | 2 3 m r 2 (\displaystyle (\frac (2)(3))mr^(2)) | |
| ball radius r and the masses m | The axis passes through the center of the ball | 2 5 m r 2 (\displaystyle (\frac (2)(5))mr^(2)) | |
| Cone radius r and the masses m | cone axis | 3 10 m r 2 (\displaystyle (\frac (3)(10))mr^(2)) | |
| 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 | 1 24 m (a 2 + 12 h 2) (\displaystyle (\frac (1)(24))m(a^(2)+12h^(2))) | |
| 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 | 1 12 m a 2 (\displaystyle (\frac (1)(12))ma^(2)) | |
| Square with side a and weight m | The axis is perpendicular to the plane of the square and passes through the center of mass | 1 6 m a 2 (\displaystyle (\frac (1)(6))ma^(2)) | |
| Rectangle with sides a And b and weight m | The axis is perpendicular to the plane of the rectangle and passes through the center of mass | 1 12 m (a 2 + b 2) (\displaystyle (\frac (1)(12))m(a^(2)+b^(2))) | |
| Regular n-gon of radius r and weight m | The axis is perpendicular to the plane and passes through the center of mass | m r 2 6 [ 1 + 2 cos (π / n) 2 ] (\displaystyle (\frac (mr^(2))(6))\left) | |
| Torus (hollow) with guide circle radius R, the radius of the generating circle r and weight m | The axis is perpendicular to the plane of the guide circle of the torus and passes through the center of mass | I = m (3 4 r 2 + R 2) (\displaystyle I=m\left((\frac (3)(4))\,r^(2)+R^(2)\right)) |
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. Let us divide the thin-walled cylinder into elements with mass dm and moments of inertia DJ i. Then
J = ∑ d J i = ∑ R i 2 d m . (1) . (\displaystyle J=\sum dJ_(i)=\sum R_(i)^(2)dm.\qquad (1).)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
J = ∑ R 2 d m = R 2 ∑ d m = m R 2 . (\displaystyle J=\sum R^(2)dm=R^(2)\sum dm=mR^(2).)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
d m = ρ d V = ρ ⋅ 2 π r h d r ; d J = r 2 d m = 2 π ρ h r 3 d r . (\displaystyle dm=\rho dV=\rho \cdot 2\pi rhdr;\qquad dJ=r^(2)dm=2\pi \rho hr^(3)dr.)We find the moment of inertia of a thick ring as an integral
J = ∫ R 1 R d J = 2 π ρ h ∫ R 1 R r 3 d r = (\displaystyle J=\int _(R_(1))^(R)dJ=2\pi \rho h\int _ (R_(1))^(R)r^(3)dr=) = 2 π ρ h r 4 4 | R 1 R = 1 2 π ρ h (R 4 − R 1 4) = 1 2 π ρ h (R 2 − R 1 2) (R 2 + R 1 2) . (\displaystyle =2\pi \rho h\left.(\frac (r^(4))(4))\right|_(R_(1))^(R)=(\frac (1)(2 ))\pi \rho h\left(R^(4)-R_(1)^(4)\right)=(\frac (1)(2))\pi \rho h\left(R^(2 )-R_(1)^(2)\right)\left(R^(2)+R_(1)^(2)\right).)Since the volume and mass of the ring are equal
V = π (R 2 − R 1 2) h ; m = ρ V = π ρ (R 2 − R 1 2) h , (\displaystyle V=\pi \left(R^(2)-R_(1)^(2)\right)h;\qquad m= \rho V=\pi \rho \left(R^(2)-R_(1)^(2)\right)h,)we obtain the final formula for the moment of inertia of the ring
J = 1 2 m (R 2 + R 1 2) . (\displaystyle J=(\frac (1)(2))m\left(R^(2)+R_(1)^(2)\right).)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):
J = 1 2 m R 2 . (\displaystyle J=(\frac (1)(2))mR^(2).)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
r = R h H , (\displaystyle r=(\frac (Rh)(H)),)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
d J = 1 2 r 2 d m = 1 2 π ρ r 4 d h = 1 2 π ρ (R h H) 4 d h ; (\displaystyle dJ=(\frac (1)(2))r^(2)dm=(\frac (1)(2))\pi \rho r^(4)dh=(\frac (1)( 2))\pi \rho \left((\frac (Rh)(H))\right)^(4)dh;)Integrating, we get
J = ∫ 0 H d J = 1 2 π ρ (R H) 4 ∫ 0 H h 4 d h = 1 2 π ρ (R H) 4 h 5 5 | 0 H == 1 10 π ρ R 4 H = (ρ ⋅ 1 3 π R 2 H) 3 10 R 2 = 3 10 m R 2 . (\displaystyle (\begin(aligned)J=\int _(0)^(H)dJ=(\frac (1)(2))\pi \rho \left((\frac (R)(H)) \right)^(4)\int _(0)^(H)h^(4)dh=(\frac (1)(2))\pi \rho \left((\frac (R)(H) )\right)^(4)\left.(\frac (h^(5))(5))\right|_(0)^(H)==(\frac (1)(10))\pi \rho R^(4)H=\left(\rho \cdot (\frac (1)(3))\pi R^(2)H\right)(\frac (3)(10))R^( 2)=(\frac (3)(10))mR^(2).\end(aligned)))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
r = R 2 − h 2 . (\displaystyle r=(\sqrt (R^(2)-h^(2))).)The mass and moment of inertia of such a disk will be
d m = ρ d V = ρ ⋅ π r 2 d h ; (\displaystyle dm=\rho dV=\rho \cdot \pi r^(2)dh;) d J = 1 2 r 2 d m = 1 2 π ρ r 4 d h = 1 2 π ρ (R 2 − h 2) 2 d h = 1 2 π ρ (R 4 − 2 R 2 h 2 + h 4) d h . (\displaystyle dJ=(\frac (1)(2))r^(2)dm=(\frac (1)(2))\pi \rho r^(4)dh=(\frac (1)( 2))\pi \rho \left(R^(2)-h^(2)\right)^(2)dh=(\frac (1)(2))\pi \rho \left(R^( 4)-2R^(2)h^(2)+h^(4)\right)dh.)The moment of inertia of the ball is found by integration:
J = ∫ − R R d J = 2 ∫ 0 R d J = π ρ ∫ 0 R (R 4 − 2 R 2 h 2 + h 4) d h = = π ρ (R 4 h − 2 3 R 2 h 3 + 1 5 h 5) | 0 R = π ρ (R 5 − 2 3 R 5 + 1 5 R 5) = 8 15 π ρ R 5 = = (4 3 π R 3 ρ) ⋅ 2 5 R 2 = 2 5 m R 2 . (\displaystyle (\begin(aligned)J&=\int _(-R)^(R)dJ=2\int _(0)^(R)dJ=\pi \rho \int _(0)^(R )\left(R^(4)-2R^(2)h^(2)+h^(4)\right)dh=\\&=\pi \rho \left.\left(R^(4) h-(\frac (2)(3))R^(2)h^(3)+(\frac (1)(5))h^(5)\right)\right|_(0)^( R)=\pi \rho \left(R^(5)-(\frac (2)(3))R^(5)+(\frac (1)(5))R^(5)\right) =(\frac (8)(15))\pi \rho R^(5)=\\&=\left((\frac (4)(3))\pi R^(3)\rho \right) \cdot (\frac (2)(5))R^(2)=(\frac (2)(5))mR^(2).\end(aligned)))thin-walled sphere
Formula derivation
For the derivation, we use the formula for the moment of inertia of a homogeneous ball of radius R :
J 0 = 2 5 M R 2 = 8 15 π ρ R 5 . (\displaystyle J_(0)=(\frac (2)(5))MR^(2)=(\frac (8)(15))\pi \rho R^(5).)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 .
J = d J 0 d R d R = d d R (8 15 π ρ R 5) d R = = 8 3 π ρ R 4 d R = (ρ ⋅ 4 π R 2 d R) 2 3 R 2 = 2 3 m R 2 . (\displaystyle (\begin(aligned)J&=(\frac (dJ_(0))(dR))dR=(\frac (d)(dR))\left((\frac (8)(15))\ pi \rho R^(5)\right)dR=\\&=(\frac (8)(3))\pi \rho R^(4)dR=\left(\rho \cdot 4\pi R^ (2)dR\right)(\frac (2)(3))R^(2)=(\frac (2)(3))mR^(2).\end(aligned)))Thin rod (axis passes through the center)
Formula derivation
Let's split the rod into small fragments of length dr. The mass and moment of inertia of such a fragment is
d m = m d r l ; d J = r 2 d m = m r 2 d r l . (\displaystyle dm=(\frac (mdr)(l));\qquad dJ=r^(2)dm=(\frac (mr^(2)dr)(l)).)Integrating, we get
J = ∫ − l / 2 l / 2 d J = 2 ∫ 0 l / 2 d J = 2 m l ∫ 0 l / 2 r 2 d r = 2 m l r 3 3 | 0 l / 2 = 2 m l l 3 24 = 1 12 m l 2 . (\displaystyle J=\int _(-l/2)^(l/2)dJ=2\int _(0)^(l/2)dJ=(\frac (2m)(l))\int _ (0)^(l/2)r^(2)dr=(\frac (2m)(l))\left.(\frac (r^(3))(3))\right|_(0) ^(l/2)=(\frac (2m)(l))(\frac (l^(3))(24))=(\frac (1)(12))ml^(2).)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 ⁄2. According to the Steiner theorem, the new moment of inertia will be equal to
J \u003d J 0 + m r 2 \u003d J 0 + m (l 2) 2 \u003d 1 12 m l 2 + 1 4 m l 2 \u003d 1 3 m l 2. (\displaystyle J=J_(0)+mr^(2)=J_(0)+m\left((\frac (l)(2))\right)^(2)=(\frac (1)( 12))ml^(2)+(\frac (1)(4))ml^(2)=(\frac (1)(3))ml^(2).)Dimensionless moments of inertia of planets and 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 for 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:
J x y = ∫ (m) x y d m = ∫ (V) x y ρ d V , (\displaystyle J_(xy)=\int \limits _((m))xydm=\int \limits _((V))xy\ rho dV,) J x z = ∫ (m) x z d m = ∫ (V) x z ρ d V , (\displaystyle J_(xz)=\int \limits _((m))xzdm=\int \limits _((V))xz\ rho dV,) J y z = ∫ (m) y z d m = ∫ (V) y z ρ d V , (\displaystyle J_(yz)=\int \limits _((m))yzdm=\int \limits _((V))yz\ rho dV,)Where x , y And z- coordinates of a small element of the body with volume dV, density ρ and mass 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 this 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 principal central axes of inertia.
Geometric moments of inertia
Geometric moment of inertia of the volume
J V a = ∫ (V) r 2 d V , (\displaystyle J_(Va)=\int \limits _((V))r^(2)dV,)where, as before r- distance from element dV to axis a .
Geometric moment of inertia of area relative to the axis - the geometric characteristic of the body, expressed by the formula:
J S a = ∫ (S) r 2 d S , (\displaystyle J_(Sa)=\int \limits _((S))r^(2)dS,)where integration is performed over the surface S, A dS is an element of this surface.
Dimension J Sa- length to the fourth power ( d i m J S a = L 4 (\displaystyle \mathrm (dim) J_(Sa)=\mathrm (L^(4)) )), respectively, the SI unit is 4. In construction calculations, literature and assortments of rolled metal, it is often indicated in cm 4.
Through the geometric moment of inertia of the area, the moment of section resistance is expressed:
W = J S a r m a x . (\displaystyle W=(\frac (J_(Sa))(r_(max))).)Here rmax- the maximum distance from the surface to the axis.
| Geometric moments of inertia of the area of some figures | |
|---|---|
| Rectangle Height h (\displaystyle h) and width b (\displaystyle b): | J y = b h 3 12 (\displaystyle J_(y)=(\frac (bh^(3))(12)))
J z = h b 3 12 (\displaystyle J_(z)=(\frac (hb^(3))(12))) |
| Rectangular box section height and width along the outer contours H (\displaystyle H) And B (\displaystyle B), and for internal h (\displaystyle h) And b (\displaystyle b) respectively | J z = B H 3 12 − b h 3 12 = 1 12 (B H 3 − b h 3) (\displaystyle J_(z)=(\frac (BH^(3))(12))-(\frac (bh^( 3))(12))=(\frac (1)(12))(BH^(3)-bh^(3)))
J y = H B 3 12 − h b 3 12 = 1 12 (H B 3 − h b 3) (\displaystyle J_(y)=(\frac (HB^(3))(12))-(\frac (hb^( 3))(12))=(\frac (1)(12))(HB^(3)-hb^(3))) |
| Circle diameter d (\displaystyle d) | J y = J z = π d 4 64 (\displaystyle J_(y)=J_(z)=(\frac (\pi d^(4))(64))) |
Moment of inertia about a plane
The moment of inertia of a rigid body relative to a certain plane is called a scalar value equal to the sum of the products of the mass of each point of the body and the square of the distance from this point to the plane under consideration.
If through an arbitrary point O (\displaystyle O) draw coordinate axes x , y , z (\displaystyle x,y,z), then the moments of inertia about the coordinate planes x O y (\displaystyle xOy), y O z (\displaystyle yOz) And zO x (\displaystyle zOx) will be expressed by the formulas:
J x O y = ∑ i = 1 n m i z i 2 , (\displaystyle J_(xOy)=\sum _(i=1)^(n)m_(i)z_(i)^(2)\ ,) J y O z = ∑ i = 1 n m i x i 2 , (\displaystyle J_(yOz)=\sum _(i=1)^(n)m_(i)x_(i)^(2)\ ,) J z O x = ∑ i = 1 n m i y i 2 . (\displaystyle J_(zOx)=\sum _(i=1)^(n)m_(i)y_(i)^(2)\ .)In the case of a solid body, summation is replaced by integration.
Central moment of inertia
Central moment of inertia (moment of inertia about the point O, moment of inertia about the pole, polar moment of inertia) J O (\displaystyle J_(O)) is the value determined by the expression:
J a = ∫ (m) r 2 d m = ∫ (V) ρ r 2 d V , (\displaystyle J_(a)=\int \limits _((m))r^(2)dm=\int \limits _((V))\rho r^(2)dV,)The central moment of inertia can be expressed through the main axial moments of inertia, as well as through the moments of inertia about the planes:
J O = 1 2 (J x + J y + J z) , (\displaystyle J_(O)=(\frac (1)(2))\left(J_(x)+J_(y)+J_(z) \right),) J O = J x O y + J y O z + J x O z . (\displaystyle J_(O)=J_(xOy)+J_(yOz)+J_(xOz).)Tensor of inertia and ellipsoid of inertia
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 s → = ‖ s x , s y , s z ‖ T , | s → | = 1 (\displaystyle (\vec (s))=\left\Vert s_(x),s_(y),s_(z)\right\Vert ^(T),\left\vert (\vec (s) )\right\vert=1), can be represented as a quadratic (bilinear) form:
I s = s → T ⋅ J ^ ⋅ s → , (\displaystyle I_(s)=(\vec (s))^(T)\cdot (\hat (J))\cdot (\vec (s)) ,\qquad ) (1)where is the inertia tensor. The matrix of the inertia tensor is symmetrical, has dimensions 3 × 3 (\displaystyle 3\times 3) and consists of the components of centrifugal moments:
By choosing an appropriate coordinate system, the matrix of the inertia tensor can be reduced to a diagonal form. To do this, we need to solve the eigenvalue problem for the tensor matrix J ^ (\displaystyle (\hat(J))):
Where Q ^ (\displaystyle (\hat (Q)))- orthogonal matrix of transition to own basis of 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. Quantities J X , J Y , J Z (\displaystyle J_(X),J_(Y),J_(Z)) are the principal moments of inertia. Expression (1) in its own coordinate system has the form:
whence the equation of the ellipsoid in eigencoordinates is obtained. Dividing both sides of the equation by I s (\displaystyle I_(s))
and making the substitutions:
ξ = s x I s , η = s y I s , ζ = s z I s , (\displaystyle \xi =(s_(x) \over (\sqrt (I_(s)))),\eta =(s_(y ) \over (\sqrt (I_(s)))),\zeta =(s_(z) \over (\sqrt (I_(s)))),)we obtain the canonical form of the ellipsoid equation in coordinates ξ η ζ (\displaystyle \xi \eta \zeta ):
ξ 2 ⋅ J X + η 2 ⋅ J Y + ζ 2 ⋅ J Z = 1. (\displaystyle \xi ^(2)\cdot J_(X)+\eta ^(2)\cdot J_(Y)+\zeta ^( 2)\cdot J_(Z)=1.)The distance from the center of the ellipsoid to some of its points is related to the value of the moment of inertia of the body along a straight line passing through the center of the ellipsoid and this point.
Let there be a rigid body. Let's choose some line OO (Fig.6.1), which we will call the axis (the line OO can also be outside the body). Let us divide the body into elementary sections (material points) by masses 
, located at a distance from the axis 
respectively.
The moment of inertia of a material point about the axis (OO) is the product of the mass of a material point and the square of its distance to this axis:
. (6.1)
The moment of inertia (MI) of the body relative to the axis (OO) is the sum of the products of the masses of the elementary sections of the body and the square of their distance to the axis:
. (6.2)
As you can see, the moment of inertia of a body is an additive quantity - the moment of inertia of the whole body about a certain axis is equal to the sum of the moments of inertia of its individual parts about the same axis.
In this case
.
The moment of inertia is measured in kgm 2 . Because
, (6.3)
where
is the density of matter, 
- volume i- th section, then
,
or, passing to infinitesimal elements,
. (6.4)
Formula (6.4) is convenient to use to calculate the MI of homogeneous bodies of regular shape with respect to the axis of symmetry passing through the center of mass of the body. For example, for an MI cylinder with respect to an axis passing through the center of mass parallel to the generatrix, this formula gives
,
Where T- weight; R is the radius of the cylinder.
Steiner's theorem is of great help in calculating the MI of bodies with respect to some axes: MI of a body I relative to any axis is equal to the sum of the MI of this body I c relative to the axis passing through the center of mass of the body and parallel to the given one, and the product of the body mass by the square of the distance d between the specified axes:
. (6.5)
Moment of force about the axis
Let the force act on the body F. Let us assume for simplicity that the force F lies in a plane perpendicular to some straight line OO (Fig. 6.2, A), which we will call the axis (for example, this is the axis of rotation of the body). On fig. 6.2, A
A- point of application of force F,
- the point of intersection of the axis with the plane in which the force lies; r- radius vector defining the position of the point A relative to the point ABOUT";
O"B
= b
-
shoulder of strength. The shoulder of the force relative to the axis is the smallest distance from the axis to the straight line on which the force vector lies F(the length of the perpendicular drawn from the point 
to this line).
The moment of force about the axis is a vector quantity defined by the equality
. (6.6)
The modulus of this vector is . Sometimes, therefore, they say that the moment of force about the axis is the product of the force on its shoulder.
If strength F is directed arbitrarily, then it can be decomposed into two components; 
And 
(fig.6.2, b), i.e. 
+
, Where 
is a component directed parallel to the OO axis, and 
lies in a plane perpendicular to the axis. In this case, under the moment of force F relative to the OO axis understand the vector
. (6.7)
In accordance with expressions (6.6) and (6.7), the vector M directed along the axis (see Fig.6.2, A,b).
The angular momentum of the body about the axis of rotation
P 
let the body rotate around some axis OO with an angular velocity 
. Let's break this body mentally into elementary sections with masses 
, which are located from the axis, respectively, at distances 
and rotate in circles, having linear velocities 
It is known that the value is 
- there is momentum i-plot. angular momentum i-a section (material point) relative to the axis of rotation is called a vector (more precisely, a pseudovector)
, (6.8)
Where r i is the radius vector that defines the position i- area relative to the axis.
The angular momentum of the whole body about the axis of rotation is called the vector
(6.9)
whose module 
.
In accordance with expressions (6.8) and (6.9), the vectors
And 
directed along the axis of rotation (Fig. 6.3). It is easy to show that the angular momentum of the body L relative to the axis of rotation and the moment of inertia I of this body relative to the same axis are related by the relation
. (6.10)
The moment of inertia of a body (system) about a given axis Oz (or the axial moment of inertia) is a scalar value that is different from the sum of the products of the masses of all points of the body (system) and the squares of their distances from this axis:
It follows from the definition that the moment of inertia of a body (or system) about any axis is a positive quantity and not equal to zero.
Later it will be shown that the axial moment of inertia plays the same role during the rotational motion of the body as the mass during translational motion, i.e., that the axial moment of inertia is a measure of the inertia of the body during rotational motion.
According to formula (2), the moment of inertia of a body is equal to the sum of the moments of inertia of all its parts about the same axis. For one material point located at a distance h from the axis, . The unit of measurement of the moment of inertia in SI will be 1 kg (in the MKGSS system -).
To calculate the axial moments of inertia, the distances of points from the axes can be expressed in terms of the coordinates of these points (for example, the square of the distance from the Ox axis will be, etc.).
Then the moments of inertia about the axes will be determined by the formulas:
Often in the course of calculations, the concept of the radius of gyration is used. The radius of gyration of a body relative to an axis is a linear quantity determined by the equality
where M is the mass of the body. It follows from the definition that the radius of inertia is geometrically equal to the distance from the axis of the point at which the mass of the entire body must be concentrated so that the moment of inertia of this one point is equal to the moment of inertia of the entire body.
Knowing the radius of inertia, it is possible to find the moment of inertia of the body using formula (4) and vice versa.
Formulas (2) and (3) are valid both for a rigid body and for any system of material points. In the case of a solid body, dividing it into elementary parts, we find that in the limit the sum in equality (2) turns into an integral. As a result, given that where is the density and V is the volume, we get
The integral here extends to the entire volume V of the body, and the density and distance h depend on the coordinates of the points of the body. Similarly, formulas (3) for solid bodies will take the form
Formulas (5) and (5) are convenient to use when calculating the moments of inertia of homogeneous bodies of regular shape. In this case, the density will be constant and will go out from under the integral sign.
Let us find the moments of inertia of some homogeneous bodies.
1. A thin homogeneous rod of length l and mass M. Let us calculate its moment of inertia about the axis perpendicular to the rod and passing through its end A (Fig. 275). Let us direct the coordinate axis along AB. Then for any elementary segment of length d, the value is , and the mass is , where is the mass of a unit length of the rod. As a result, formula (5) gives
Replacing here its value, we finally find
2. A thin round homogeneous ring with radius R and mass M. Let's find its moment of inertia about the axis perpendicular to the plane of the ring and passing through its center C (Fig. 276).
Since all points of the ring are at a distance from the axis, formula (2) gives
Therefore, for the ring
Obviously, the same result will be obtained for the moment of inertia of a thin cylindrical shell with mass M and radius R about its axis.
3. Round homogeneous plate or cylinder with radius R and mass M. Let us calculate the moment of inertia of the round plate about the axis perpendicular to the plate and passing through its center (see Fig. 276). To do this, we select an elementary ring with a radius and width (Fig. 277, a). The area of this ring is , and the mass is where is the mass per unit area of the plate. Then, according to formula (7), for the selected elementary ring it will be and for the entire plate
As noted above, three figures are among the simple flat figures: a rectangle, a triangle and a circle. These figures are considered simple because the position of the center of gravity of these figures is known in advance. All other figures can be composed of these simple figures and are considered complex. Let us calculate the axial moments of inertia of simple figures about their central axes.
1. Rectangle. Consider a section of a rectangular profile with dimensions (Fig. 4.6). Select a section element by two infinitely close sections at a distance 
from the central axis 
.
Calculate the moment of inertia of a rectangular section about the axis:
.
(4.10)
Moment of inertia of a rectangular section about an axis 
find similarly. The output is not shown here.
.
(4.11)
And 
is zero, since the axes 
And 
are the axes of symmetry and hence the principal axes.
2. Isosceles triangle. Consider a section of a triangular profile with dimensions 
(Fig.4.7). Select a section element by two infinitely close sections at a distance 
from the central axis 
. The center of gravity of a triangle is at a distance 
from the base. The triangle is assumed to be isosceles, so that the axis 
section is the axis of symmetry.
Calculate the moment of inertia of the section about the axis 
:
.
(4.12)
the value 
we define from the similarity of triangles:
; where 
.
Substituting expressions for 
in (4.12) and integrating, we get:
.
(4.13)
Moment of inertia for an isosceles triangle about an axis 
is found in the same way and is equal to:
(4.14)
Centrifugal moment of inertia about the axes 
And 
is zero because the axis 
is the axis of symmetry of the section.
3. Circle. Consider a section of a circular profile with a diameter 
(Fig.4.8). Let us select the element of the section by two infinitely close concentric circles located at a distance 
from the center of gravity of the circle 
.
Let's calculate the polar moment of inertia of the circle using the expression (4.5):
.
(4.15)
Using the invariance condition for the sum of the axial moments of inertia about two mutually perpendicular axes (4.6) and taking into account that for a circle, due to symmetry 
, we determine the value of the axial moments of inertia:
.
(4.16)
.
(4.17)
Centrifugal moment of inertia about the axes 
And 
is zero, since the axes 
And 
are the axes of symmetry of the section.
4.4. Relationships between moments of inertia about parallel axes
When calculating the moments of inertia for complex figures, one rule should be remembered: the values \u200b\u200bof the moments of inertia can be added, if they are calculated with respect to the same axis. For complex figures, most often the centers of gravity of individual simple figures and the entire figure do not coincide. The central axes for separate simple figures and the whole figure do not coincide, respectively. In this regard, there are methods for bringing the moments of inertia to one axis, for example, the central axis of the entire figure. This may be due to the parallel translation of the axes of inertia and additional calculations.
Consider the definition of moments of inertia about parallel axes of inertia, shown in Fig.4.9.
Let the axial and centrifugal moments of inertia shown in Figure 4.9. figures about arbitrarily chosen axes 
And 
with the origin at the point 
known. It is required to calculate the axial and centrifugal moments of inertia of the figure relative to arbitrary parallel axes 
And 
with the origin at the point 
. axes 
And 
carried out at distances 
And 
respectively from the axes 
And 
.
Let us use the expressions for the axial moments of inertia (4.4) and for the centrifugal moment of inertia (4.7). Substitute in these expressions instead of the current coordinates 
And 
element with infinitely small coordinate area 
And 
in the new coordinate system. We get:
Analyzing the obtained expressions, we come to the conclusion that when calculating the moments of inertia relative to parallel axes to the moments of inertia calculated relative to the initial axes of inertia, it is necessary to add additions in the form of additional terms, which can be much larger than the values for the moments of inertia relative to the initial axes. Therefore, these additional terms should not be neglected in any case.
The considered case is the most general case of parallel transfer of axes, when arbitrary axes of inertia were taken as initial ones. In most calculations, there are special cases of determining the moments of inertia.
First special case. The reference axes are the central axes of inertia of the figure. Then, using the main property for the static moment of the area, it is possible to exclude from the equations (4.18)-(4.20) the members of the equations, which include the static moment of the area of the figure. As a result, we get:
.
(4.21)
.
(4.22)
.
(4.23)
Here the axes 
And 
- central axis of inertia.
Second special case. The reference axes are the principal axes of inertia. Then, given that the centrifugal moment of inertia is zero relative to the principal axes of inertia, we get:
.
(4.24)
.
(4.25)
.
(4.26)
Here the axes 
And 
- Principal axes of inertia.
Let's use the obtained expressions and consider several examples of calculating the moments of inertia for plane figures.
Example 4.2. Determine the axial moments of inertia of the figure shown in fig. 4.10, relative to the central axes 
And 
.
In the previous example 4.1, for the figure shown in Fig. 4.10, the position of the center of gravity C was determined. The coordinate of the center of gravity was plotted from the axis 
and made 
. Let's calculate the distances 
And 
between axles 
And 
and axes 
And 
. These distances were respectively 
And 
. Since the original axes 
And 
are the central axes for simple figures in the form of rectangles, to determine the moment of inertia of the figure about the axis 
we use the derivations for the first particular case, in particular, formula (4.21).
Moment of inertia about the axis 
obtained by adding the moments of inertia of simple figures about the same axis, since the axis 
is a common central axis for simple figures and for the whole figure.
cm 4.
Centrifugal moment of inertia about the axes 
And 
is zero, since the axis of inertia 
is the main axis (axis of symmetry of the figure).
Example 4.3. What is the size b(in cm)
the figure shown in Fig. 4.11, if the moment of inertia of the figure about the axis 
equal to 1000 cm 4?
We express the moment of inertia about the axis 
through an unknown section size 
, using formula (4.21), taking into account that the distance between the axes 
And 
equals 7cm:
cm 4. (A)
Solving expression (a) with respect to section size 
, we get:
cm.
Example 4.4. Which of the figures shown in Fig. 4.12 has a greater moment of inertia about the axis 
if both shapes have the same area 
cm 2?
1. We express the areas of the figures in terms of their sizes and determine:
a) section diameter for a circular section:
cm 2; Where 
cm.
b) the size of the side of the square:
; Where 
cm.
2. Calculate the moment of inertia for a circular section:
cm 4.
3. Calculate the moment of inertia for a square section:
cm 4.
Comparing the results obtained, we come to the conclusion that the largest moment of inertia will have a square section in comparison with a round section with the same area.
Example 4.5. Determine the polar moment of inertia (in cm 4) of a rectangular section relative to its center of gravity, if the section width 
cm, section height 
cm.
1. Find the moments of inertia of the section relative to the horizontal 
and vertical 
central axes of inertia:
cm 4; 
cm 4.
2. Determine the polar moment of inertia of the section as the sum of the axial moments of inertia:
cm 4.
Example 4.6. Determine the moment of inertia of the triangular shape shown in Fig. 4.13, relative to the central axis 
, if the moment of inertia of the figure about the axis 
equal to 2400 cm 4.
Moment of inertia of a triangular section about the main axis of inertia 
will be less than the moment of inertia about the axis 
by the amount 
. Therefore, when 
see the moment of inertia of the section about the axis 
find in the following way.
DEFINITION
The measure of the inertia of a rotating body is moment of inertia(J) relative to the axis around which the rotation occurs.
This is a scalar (in the general case, tensor) physical quantity, which is equal to the product of the masses of material points () into which the body under consideration should be partitioned, by the squares of distances () from them to the axis of rotation:
where r is a function of the position of a material point in space; - body density; - the volume of the body element.
For a homogeneous body, expression (2) can be represented as:
The moment of inertia in the international system of units is measured in:
The value of J is included in the basic laws that describe the rotation of a rigid body.
In general, the magnitude of the moment of inertia depends on the direction of the axis of rotation, and since the vector usually changes its direction relative to the body in the process of motion, the moment of inertia should be considered as a function of time. An exception is the moment of inertia of a body rotating around a fixed axis. In this case, the moment of inertia remains constant.
Steiner's theorem
The Steiner theorem makes it possible to calculate the moment of inertia of a body about an arbitrary axis of rotation, when the moment of inertia of the body under consideration is known with respect to the axis passing through the center of mass of this body and these axes are parallel. In mathematical form, the Steiner theorem is represented as:
where is the moment of inertia of the body about the axis of rotation passing through the center of mass of the body; m is the mass of the considered body; a is the distance between the axles. Be sure to remember that the axes must be parallel. From expression (4) it follows that:
Some expressions for calculating the moments of inertia of a body
When rotating around an axis, a material point has a moment of inertia equal to:
where m is the mass of the point; r is the distance from the point to the axis of rotation.
For a homogeneous thin rod of mass m and length l J relative to the axis passing through its center of mass (the axis is perpendicular to the rod), is equal to:
A thin ring, with a mass rotating about an axis that passes through its center, perpendicular to the plane of the ring, then the moment of inertia is calculated as:
where R is the radius of the ring.
A round homogeneous disk of radius R and mass m has J relative to the axis passing through its center and perpendicular to the disk plane, equal to:
For a uniform ball
where m is the mass of the ball; R is the radius of the ball. The ball rotates about an axis that passes through its center.
If the axes of rotation are the axes of a rectangular Cartesian coordinate system, then for a continuous body the moments of inertia can be calculated as:
where are the coordinates of an infinitely small element of the body.
Examples of problem solving
EXAMPLE 1
| Exercise | Two balls, which can be considered as point ones, are held together by a thin weightless rod. Rod length l. What is the moment of inertia of this system, with respect to the axis that runs perpendicular to the rod through the center of mass. The point masses are the same and equal to m. |
| Solution | Let's find the moment of inertia of one ball () relative to an axis located at a distance from it:
The moment of inertia of the second ball will be equal to: The total moment of inertia of the system is equal to the sum:
|
| Answer |
EXAMPLE 2
| Exercise | What is the moment of inertia of the physical pendulum about the axis that passes through the point O (Fig. 1)? The axis is perpendicular to the plane of the figure. Consider that a physical pendulum consists of a thin rod of length l with mass m and a disk of mass . The disc is attached to the lower end of the rod and has a radius equal to |
| Solution | The moment of inertia of our pendulum (J) will be equal to the sum of the moment of inertia of the rod () rotating about the axis passing through the point O and the disk () rotating around the same axis: |
