Moment of inertia

This article is about the moment of inertia of a rotating object, also termed the mass moment of inertia. For the moment of inertia dealing with the bending of a beam, also termed the area moment of inertia, see second moment of area.

In classical mechanics, moment of inertia is also called mass moment of inertia, rotational inertia, polar moment of inertia of mass, or the angular mass, (SI units kg·m², US units lb_m ft²). It is a property of a distribution of mass in space that measures its resistance to rotational acceleration about an axis. Newton's first law, which describes the inertia of a body in linear motion, can be extended to the inertia of a body rotating about an axis using the moment of inertia. That is, an object that is rotating at constant angular velocity will remain rotating unless acted upon by an external torque. In this way, the moment of inertia plays the same role in rotational dynamics as mass does in linear dynamics, describing the relationship between angular momentum and angular velocity, torque and angular acceleration. The symbols I and sometimes J are usually used to refer to the moment of inertia or polar moment of inertia.
The moment of the inertia force on a particle around an axis multiplies the mass of the particle by the square of its distance to the axis, and forms a parameter called the moment of inertia. The moments of inertia of individual particles sum to define the moment of inertia of a body rotating about an axis. For rigid bodies moving in a plane, such as a compound pendulum, the moment of inertia is a scalar, but for movement in three dimensions, such as a spinning top, the moment of inertia becomes a matrix, also called a tensor.
In 1673 Christiaan Huygens derived the equation for the center of oscillation and oscillation period of compound pendulums, thus making use of the general equation for the moment of inertia for the first time. The term moment of inertia was introduced by Leonhard Euler in his book Theoria motus corporum solidorum seu rigidorum in 1765. In this book, he discussed the moment of inertia and many related concepts, such as the principal axis of inertia.

Overview

A flywheel is a wheel with a large moment of inertia used to smooth out motion in machines. This example is in a Russian factory.

Moment of inertia appears in Newton's second law for the rotation of a rigid body, which states that the torque necessary to accelerate rotation is proportional to the moment of inertia of the body. Thus, the greater the moment of inertia the greater the torque needed for the same acceleration. Many systems use masses with large moment of inertia to maintain a rotational velocity and resist small variations in applied torque. For example, the long pole held by a tight-rope walker maintains a zero angular velocity resisting the small torque applied by the walker to maintain balance. Another example is the rotating mass of a flywheel which maintains a constant angular velocity resisting the torque variations in a machine.
The moment of inertia of an object is defined by the distribution of mass around an axis. It depends not only on the total mass of the object, but also on the square of the perpendicular distance from the axis to each each element of mass. This means the moment of inertia increases rapidly as masses are distributed more distant from the axis. For example, consider two wheels that have the same mass, one that is the size of a bicycle wheel and one that is half that size. The larger wheel has four times the moment of inertia even though it is only twice the diameter.
Moment of inertia around a fixed axis is a scalar, however the rotation of a body in space can occur around the three coordinate axes. In this case, the moments of inertia associated with the three coordinate axes define a matrix of scalars called the inertia matrix, also known as the inertia tensor.

Scalar moment of inertia of a simple pendulum

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Pendulums used in Mendenhall gravimeter apparatus, from 1897 scientific journal. The portable gravimeter developed in 1890 by Thomas C. Mendenhall, provided the most accurate relative measurements of the local gravitational field of the Earth.

Moment of inertia can be obtained by considering the movement of a mass at the end of a light-weight rod forming a simple pendulum, which can be studied using Newton's second law of motion. The weight of the mass is a force that accelerates it around the pivot point.
This weight also generates a torque T on the pendulum around the pivot point and the acceleration of the mass a=rα is defined by the angular acceleration α of the pendulum, therefore

$\mathbf{T}=(mr^2)\alpha = I\alpha$

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where r is the length of the pendulum. The quantity I=mr² is the moment of inertia of the pendulum mass around the pivot point.
In the same way, the kinetic energy of the pendulum mass is defined by its velocity v=rω using the angular velocity ω of the pendulum to yield

$T = \frac{1}{2}(mr^2)\omega^2= \frac{1}{2}I\omega^2.$

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The angular momentum of the pendulum mass is given by

$\mathbf{L} = (mr^2)\omega = I\omega.$

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This shows that the quantity I=mr² plays the same role for rotational movement, as mass does for translational movement. The moment of inertia of an arbitrarily shaped body is the sum of the values mr² for all of the elements of mass in the body.

Scalar moment of inertia of a rigid body

Consider a rigid body rotating with angular velocity ω around a certain axis. The body consists of N point masses m_i whose distances to the axis of rotation are denoted r_i. Each point mass will have the speed v_i = ωr_i, so that the total kinetic energy T of the body can be calculated as

$T = \sum_{i=1}^N \frac12\,m_i v_i^2 = \sum_{i=1}^N \frac12\,m_i (\omega r_i)^2 = \frac12\, \omega^2 \Big( \textstyle \sum_{i=1}^N m_i r_i^2 \Big).$

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In this expression the quantity in parentheses is called the moment of inertia of the body (with respect to the specified axis of rotation). It is a purely geometric characteristic of the object, as it depends only on its shape and the position of the rotation axis. The moment of inertia is usually denoted with the capital letter I:

$I = \sum_{i=1}^N m_i r_i^2\ .$

It is worth emphasizing that r_i here is the distance from a point to the axis of rotation, not to the origin. As such, the moment of inertia will be different when considering rotations about different axes.
Similarly, the moment of inertia of a continuous solid body rotating about a known axis can be calculated by replacing the summation with the integral:

$I = \int_V \rho(\mathbf{r})\,\mathbf{r}^2 \, dV,$

where r is the radius vector of a point within the body from the axis, and ρ(r) is the mass density at each point r. The integration is evaluated over the volume V of the body.

Moment of inertia theorems

Calculations for the moment of inertia (MOI) of a body are not easy in general. The process can be simplified in the following ways:

Choosing axes to take advantage of geometric symmetries
Physical homogeneity (i.e., uniform mass distribution) making the density function ρ(r) become constant (elementary calculations, or generally an approximation)
Use of the following theorems see below for details:

Theorem	Nomenclature	Equation
Superposition Principle for Moment Of Inertia (MOI) about any chosen Axis	I_net = resultant MOI (about any one axis)	$I_\mathrm{net} = \sum_j I_j\,\!$
Parallel axis theorem	M = total mass of body d = perpendicular distance from an axis through the Center Of Mass (COM) to another parallel axis I_COM = MOI about the axis through the COM I_d = MOI about the parallel axis	$I_d = I_\mathrm{com} + Md^2 \,\!$
Perpendicular axis theorem	i, j, k refer to MOI about any three mutually perpendicular axes. The sum of MOI about any two is greater than or equal to the third.	$I_k \leq I_i + I_j \,\!$

Properties

Superposition
The moment of inertia of the body is additive. That is, if a body can be decomposed (either physically or conceptually) into several constituent parts, then the moment of inertia of the whole body about a given axis is equal to the sum of moments of inertia of each part around the same axis.
Basing just on the dimensional analysis, the moment of inertia must take the form I = c·M·L², where M is the mass, L is the “size” of the body in the direction perpendicular to the axis of rotation, and c is a dimensionless inertial constant. Additionally, the length $\scriptstyle R\,=\,\sqrt{I/M}$ is called the radius of gyration of the body.
Perpendicular Axes
If I_x, I_y, I_z are moments of inertia around three perpendicular axes passing through the body’s center of mass, then each of them cannot be greater than the sum of two others: for example I_z ≤ I_x + I_y. Here the equality holds only if the body is flat and located in the Oxy coordinate plane.
Parallel Axes
If the object’s moment of inertia I_COM around a certain axis passing through the center of mass is known, then the parallel axis theorem or Huygens–Steiner theorem provides a convenient formula to compute the moment of inertia I_d of the same body around a different axis, which is parallel to the original and located at a distance d from it. The formula is only suitable when the initial and final axes are parallel. In order to compute the moment of inertia about an arbitrary axis, one has to use the object’s moment of inertia tensor.

Examples

Main article: List of moments of inertia

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Diatomic molecule, with atoms m₁ and m₂ at a distance d from each other, rotating around the axis which passes through the molecule’s center of mass and is perpendicular to the direction of the molecule.
The easiest way to calculate this molecule’s moment of inertia is to use the parallel axis theorem. If we consider rotation around the axis passing through the atom m₁, then the moment of inertia will be I₁ = m₁·0 + m₂·d² = m₂d². On the other hand, by the parallel axis theorem this moment is equal to I₁ = I + (m₁ + m₂)·a², where I is the moment of inertia around the axis passing through the center of mass, and a is the distance between the center of mass and the first atom. By the center of mass formula, this distance is equal to a = m₂ d / (m₁ + m₂). Thus,

$I = I_1 - (m_1+m_2)a^2 = m_2d^2 - \frac{m_2^2d^2}{m_1+m_2} = \frac{m_1m_2}{m_1+m_2}\,d^2 .$

Thin rod of mass m and length ℓ, rotating around the axis which passes through its center and is perpendicular to the rod.
Let Oz be the axis of rotation, and Ox the axis along the rod. If ρ is the density, and s the cross-section of the rod (so that m = ρℓs), then the volume element for the integral formula will be equal to dV = s·dx, where x changes from −½ℓ to ½ℓ. The moment of inertia can be found by computing the integral:

$I = \int_{-\ell/2}^{\ell/2} \rho\,x^2 s \mathrm{d}x = \rho s \,\frac{x^3}{3}\bigg|_{-\ell/2}^{\ell/2} = \frac{m}{s\ell} \cdot s \cdot 2\frac{\ell^3/8}{3} = \frac{1}{12}\, m\ell^2 .$

Solid ball of mass m and radius R, rotating around an axis which passes through the center.
Suppose Oz is the axis of rotation. The distance from point r = (x, y, z) towards the axis Oz is equal to d(r)² = x² + y². Thus, in order to compute the moment of inertia I_z, we need to evaluate the integral ∭(x² + y²) dV. The calculation considerably simplifies if we notice that by symmetry of the problem, the moments of inertia around all axes are equal: I_x = I_y = I_z. Then

$I = \frac13(I_x + I_y + I_z) = \frac13 \iiint \rho\cdot(y^2+z^2 + x^2+z^2 + x^2+y^2)\,\mathrm{d}V = \frac23\,\rho \int r^2 \,\mathrm{d}V ,$

where r² = x² + y² + z² is the distance from point r to the origin. This integral is easy to evaluate in the spherical coordinates, the volume element will be equal to dV = 4πr² dr, where r goes from 0 to R. Thus,

$I = \frac23\,\rho \int_0^R \!\!4\pi r^4\,\mathrm{d}r = \frac23\,\rho\cdot4\pi\frac{R^5}{5} = \frac{m}{\frac43\pi R^3}\cdot\frac{8\pi R^5}{15} = \frac25\,mR^2 .$

In mechanics of machines, when designing rotary parts like gears, pulleys, shafts, couplings etc., which are used to transmit torques, the moment of inertia has to be considered. The moment of inertia is given about an axis and it depends on the shape, density of a rotating element.
When considering mechanisms like gear trains, worm and wheel, where there are more than one rotating element, more than one axis of rotation, an equivalent moment of inertia for the system should be found. Practically when a geared system is enclosed, equivalent moment of inertia can be measured by measuring the angular acceleration for a known torque or theoretically it can be estimated when the masses and dimensions of the rotating elements and shafts are known. In this practical the equivalent moment of inertia of a worm and wheel system is measured using above mention methods.

Polar moment of inertia

If a mechanical system is constrained to move parallel to a fixed plane, then the rotation of a body in the system occurs around an axis k perpendicular to this plane. In this case, the moment of inertia of the mass in this system is a scalar known as the polar moment of inertia. The definition of the polar moment of inertia can be obtained by considering momentum, kinetic energy and Newton's laws for the planar movement of a rigid system of particles.
If a system of n particles, P_i, i=1,...,n, are assembled into a rigid body, then the momentum of the system can be written in terms of position and velocity relative to a reference point R,

$\mathbf{r}_i = (\mathbf{r}_i - \mathbf{R}) + \mathbf{R}, \quad \mathbf{v}_i = \omega\times(\mathbf{r}_i - \mathbf{R}) + \mathbf{V},$

where ω is the angular velocity of the system.
For planar movement the angular velocity vector is directed along the unit vector k which is perpendicular to the plane of movement. Introduce the unit vectors e_i from the reference point R to a point r_i, and the unit vector t_i=kxe_i so

$\Delta r_i\mathbf{e}_i = \mathbf{r}_i-\mathbf{R}, \quad \mathbf{v}_i = \omega \Delta r_i\mathbf{t}_i + \mathbf{V},\quad i=1,\dots, n.$

This defines the relative position vector and the velocity vector for the rigid system of the particles moving in a plane.

Angular momentum in planar movement

The angular momentum vector for the planar movement of a rigid system of particles is given by

$\mathbf{L} = \sum_{i=1}^n m_i (\mathbf{r}_i-\mathbf{R})\times\mathbf{v}_i = \sum_{i=1}^n m_i \Delta r_i\mathbf{e}_i \times(\omega \Delta r_i\mathbf{t}_i + \mathbf{V}) = (\sum_{i=1}^n m_i \Delta r_i^2)\omega \vec{k} + (\sum_{i=1}^n m_i \Delta r_i\mathbf{e}_i)\times\mathbf{V}.$

A figure skater can reduce her moment of inertia to spin faster

Use the center of mass C as the reference point so

$\Delta r_i\mathbf{e}_i = \mathbf{r}_i-\mathbf{C},\quad \sum_{i=1}^n m_i\Delta r_i\mathbf{e}_i=0,$

and define the moment of inertia relative to the center of mass I_C as

$I_C= \sum_{i=1}^n m_i\Delta r_i^2,$

then the equation for angular momentum simplifies to

$\mathbf{L} = I_C \omega \vec{k}.$

The moment of inertia I_C about an axis perpendicular to the movement of the rigid system and through the center of mass is known as the polar moment of inertia.
For a given amount of angular momentum, a decrease in the moment of inertia results in an increase in the angular velocity. Figure skaters can change their moment of inertia by pulling in their arms. Thus, the angular velocity achieved by a skater with outstretched arms results in a greater angular velocity when the arms are pulled in, because of the reduced moment of inertia.

Kinetic energy in planar movement

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This 1906 rotary shear uses the moment of inertia of two flywheels to store kinetic energy which when released is used to cut metal stock.

The kinetic energy of a rigid system of particles moving in the plane is given by

$T=\frac{1}{2}\sum_{i=1}^n m_i \mathbf{v}_i\cdot\mathbf{v}_i = \frac{1}{2}\sum_{i=1}^n m_i (\omega \Delta r_i\mathbf{t}_i + \mathbf{V})\cdot(\omega \Delta r_i\mathbf{t}_i + \mathbf{V}).$

This equation expands to yield three terms

$T= \frac{1}{2}\omega ^2\sum_{i=1}^n m_i \Delta r_i^2(\mathbf{t}_i\cdot\mathbf{t}_i) + \omega\mathbf{V}\cdot(\sum_{i=1}^n m_i \Delta r_i\mathbf{t}_i )+ \frac{1}{2}(\sum_{i=1}^n m_i) \mathbf{V}\cdot\mathbf{V}.$

Let the reference point be the center of mass C of the system so the second term becomes zero, and introduce the moment of inertia I_C so the kinetic energy is given by

$T=\frac{1}{2}I_C \omega^2 +\frac{1}{2}M\mathbf{V}\cdot\mathbf{V}.$

The moment of inertia I_C about an axis perpendicular to the movement of the rigid system and through the center of mass is known as the polar moment of inertia.

Newton's laws for planar movement

A 1920's John Deere tractor with the spoked flywheel on the engine. The large moment of inertia of the flywheel smooths the operation of the tractor

Newton's laws for a rigid system of N particles, P_i, i=1,...,N, can be written in terms of a resultant force and torque at a reference point R, to yield

$\mathbf{F} = \sum_{i=1}^N m_i\mathbf{A}_i,\quad \mathbf{T} = \sum_{i=1}^N (\mathbf{r}_i-\mathbf{R})\times (m_i\mathbf{A}_i),$

where r_i denotes the trajectory of each particle.
The kinematics of a rigid body yields the formula for the acceleration of the particle P_i in terms of the position R and acceleration A of the reference particle as well as the angular velocity vector ω and angular acceleration vector α of the rigid system of particles as,

$\mathbf{A}_i = \alpha\times(\mathbf{r}_i-\mathbf{R}) + \omega\times\omega\times(\mathbf{r}_i-\mathbf{R}) + \mathbf{A}.$

For systems that are constrained to planar movement, the angular velocity and angular acceleration vectors are directed along k perpendicular to the plane of movement, which simplifies this acceleration equation. In this case, the acceleration vectors can be simplified by introducing the unit vectors e_i from the reference point R to a point r_i and the unit vectors t_i=kxe_i, so

$\mathbf{A}_i = \alpha(\Delta r_i\mathbf{t}_{i}) - \omega^2(\Delta r_i\mathbf{e}_{i}) + \mathbf{A}.$

This yields the resultant torque on the system as

$\mathbf{T} =\sum_{i=1}^N (m_i\Delta r_i\mathbf{e}_i)\times (\alpha(\Delta r_i\mathbf{t}_{i}) - \omega^2(\Delta r_i\mathbf{e}_{i}) + \mathbf{A}) = (\sum_{i=1}^N m_i\Delta r_i^2)\alpha\vec{k} + (\sum_{i=1}^N m_i\Delta r_i\mathbf{e}_i)\times\mathbf{A},$

where e_ixe_i=0, and e_ixt_i=k is the unit vector perpendicular to the plane for all of the particles P_i.
Use the center of mass C as the reference point and define the moment of inertia relative to the center of mass I_C, then the equation for the resultant torque simplifies to

$\mathbf{T} = I_C\alpha\vec{k}.$

The moment of inertia I_C about an axis perpendicular to the movement of the rigid system and through the center of mass is known as the polar moment of inertia.

Moment of inertia matrix

The scalar moments of inertia appear as elements in a matrix when a system of particles is assembled into a rigid body that moves in three dimensional space. This inertia matrix appears in the calculation of the angular momentum, kinetic energy and resultant torque of the rigid system of particles. Let the system of particles P_i, i=1,...,n be located at the coordinates r_i and velocities v_i. Select a reference point R and compute the relative position and velocity vectors,

$\mathbf{r}_i = (\mathbf{r}_i - \mathbf{R}) + \mathbf{R}, \quad \mathbf{v}_i = \omega\times(\mathbf{r}_i - \mathbf{R}) + \mathbf{V},$

where ω is the angular velocity of the system.

Angular momentum

The angular momentum of a rigid system of particles measured relative to the center of mass R is

$\mathbf{L} = \sum_{i=1}^n m_i(\mathbf{r}_i-\mathbf{R})\times \mathbf{v}_i = \sum_{i=1}^n m_i (\mathbf{r}_i-\mathbf{R})\times(\omega\times(\mathbf{r}_i - \mathbf{R})).$

In this equation the terms containing V sum to zero by definition of the center of mass.
In order to use this formula for angular momentum to obtain the matrix of mass moment of inertias, also called the inertia matrix, introduce the skew-symmetric matrix [B] constructed from a vector b that performs the cross product operation, such that

$[B]\mathbf{y} =\mathbf{b}\times\mathbf{y}.$

This matrix [B] has the components of b=(b_x, b_y,b_z) as its elements, in the form

$[B] = \begin{bmatrix} 0 & -b_z & b_y \\ b_z & 0 & -b_x \\ -b_y & b_x & 0 \end{bmatrix}.$

For the cross products in the angular momentum formula, introduce the skew-symmetric matrix [r_i-R] constructed from the relative position vector r_i - R to obtain

$\mathbf{L} = (-\sum_{i=1}^n m_i [r_i-R][r_i-R])\omega = [I_R]\omega,$

where [I_R] defined by

$[I_R] = -\sum_{i=1}^n m_i[r_i-R][r_i-R],$

is the inertia matrix of the rigid system of particles.

Kinetic energy

The kinetic energy of a rigid system of particles can be formulated in terms of the center of mass and a matrix of mass moments of inertia of the system. Let the system of particles P_i, i=1,...,n be located at the coordinates r_i and velocities v_i, then the kinetic energy is

$T=\frac{1}{2}\sum_{i=1}^n m_i \mathbf{v}_i\cdot\mathbf{v}_i = \frac{1}{2}\sum_{i=1}^n m_i (\omega\times(\mathbf{r}_i - \mathbf{C}) + \mathbf{V})\cdot(\omega\times(\mathbf{r}_i - \mathbf{C}) + \mathbf{V}),$

where C is the center of mass. This equation expands to yield three terms

$T= \frac{1}{2}\sum_{i=1}^n m_i (\omega\times(\mathbf{r}_i - \mathbf{C})) \cdot(\omega\times(\mathbf{r}_i - \mathbf{C})) + \sum_{i=1}^n m_i \mathbf{V}\cdot(\omega\times(\mathbf{r}_i - \mathbf{C}) )+ \frac{1}{2}\sum_{i=1}^n m_i \mathbf{V}\cdot\mathbf{V}.$

The second term in this equation is zero because C is the center of mass. Introduce the skew-symmetric matrix [r_i-C] so the kinetic energy becomes

$T=\frac{1}{2}\omega\cdot(- \sum_{i=1}^n m_i [r_i - C][r_i - C]) \omega + \frac{1}{2}(\sum_{i=1}^n m_i) \mathbf{V}\cdot\mathbf{V}.$

Thus, the kinetic energy of the rigid system of particles is given by

$T=\frac{1}{2}\omega\cdot[I_C]\omega +\frac{1}{2}M\mathbf{V}\cdot\mathbf{V}.$

where [I_C] is the inertia matrix relative to the center of mass and M is the total mass.

Resultant torque

The inertia matrix appears in the application of Newton's second law to a rigid assembly of particles. The resultant torque on this system is,

$\mathbf{T} = \sum_{i=1}^n (\mathbf{r_i}-\mathbf{R})\times (m_i\mathbf{a}_i),$

where a_i is the acceleration of the particle P_i. The kinematics of a rigid body yields the formula for the acceleration of the particle P_i in terms of the position R and acceleration A of the reference point, as well as the angular velocity vector ω and angular acceleration vector α of the rigid system as,

$\mathbf{a}_i = \alpha\times(\mathbf{r}_i-\mathbf{R}) + \omega\times\omega\times(\mathbf{r}_i-\mathbf{R}) + \mathbf{A}.$

Use the center of mass C as the reference point, and introduce the skew-symmetric matrix [r_i-C] to represent the cross product (r_i - C)x, in order to obtain

$\mathbf{T} =( -\sum_{i=1}^n m_i[r_i-C][r_i-C])\times\alpha + \omega\times(- \sum_{i=1}^n m_i [r_i-C][r_i-C])\omega.$

This calculation uses the identity

$(\mathbf{r}_i-\mathbf{C})\times(\omega\times(\omega\times(\mathbf{r}_i-\mathbf{C}) )) + \omega\times((\mathbf{r}_i-\mathbf{C})\times((\mathbf{r}_i-\mathbf{C})\times\omega)))=0,$

obtained from the Jacobi identity for the triple cross product.
Thus, the resultant torque on the rigid system of particles is given by

$\mathbf{T} =[I_C]\alpha + \omega\times[I_C]\omega,$

where [I_C] is the inertia matrix relative to the center of mass.

Parallel axis theorem

The inertia matrix of a rigid system of particles depends on the choice of the reference point. There is a useful relationship between the inertia matrix relative to the center of mass C and the inertia matrix relative to another point R. This relationship is called the parallel axis theorem.
Consider the inertia matrix [I_R] obtained for a rigid system of particles measured relative to a reference point R, given by

$[I_R] = -\sum_{i=1}^n m_i[r_i-R][r_i-R].$

Let C be the center of mass of the rigid system, then

$\mathbf{R} = (\mathbf{R}-\mathbf{C}) + \mathbf{C} = \mathbf{d} + \mathbf{C},$

where d is the vector from the center of mass C to the reference point R. Use this equation to compute the inertia matrix,

$[I_R] = -\sum_{i=1}^n m_i[r_i - C - d][r_i - C - d].$

Expand this equation to obtain

$[I_R] = (-\sum_{i=1}^n m_i [r_i - C][r_i - C]) + (\sum_{i=1}^n m_i[r_i - C])[d] + [d](\sum_{i=1}^n m_i[r_i - C]) + (-\sum_{i=1}^n m_i)[d][d].$

The first term is the inertia matrix [I_C] relative to the center of mass. The second and third terms are zero by definition of the center of mass C. And the last term is the total mass of the system multiplied by the square of the skew-symmetric matrix [d] constructed from d.
The result is the parallel axis theorem,

$[I_R] = [I_C] - M[d]^2,$

where d is the vector from the center of mass C to the reference point R.

Moment of inertia around an arbitrary axis

The moment of inertia of a body around an arbitrary axis in space is a scalar that is computed as the sum of the distance squared from the axis to each of the mass elements. This scalar can be computed from the moment inertia matrix of the body using the unit vector along the axis.
Let a rigid assembly of rigid system of N particles, P_i, i=1,...,N, have coordinates r_i. Choose R as a reference point and compute the moment of inertia around an axis L defined by the unit vector S through the reference point R. The moment of inertia of the system around this line L=R+tS is computed by determining the perpendicular vector from this axis to the particle P_i given by

$\Delta\mathbf{r}_i = (\mathbf{r}_i-\mathbf{R}) - (\mathbf{S}\cdot(\mathbf{r}_i-\mathbf{R}))\mathbf{S} = [[I]-[\mathbf{S}\mathbf{S}^T]](\mathbf{r}_i-\mathbf{R}),$

where [I] is the identity matrix and [S S^T] is the outer product matrix formed from the unit vector S along the line L.
Introduce the skew-symmetric matrix [S] such that [S]y=S x y, then we have the identity

$-[S]^2 = [I]-[\mathbf{S}\mathbf{S}^T],$

which relies on the fact that S is a unit vector.
The magnitude squared of the perpendicular vector is

$|\Delta\mathbf{r}_i|^2 = (-[S]^2(\mathbf{r}_i-\mathbf{R})) \cdot (-[S]^2(\mathbf{r}_i-\mathbf{R})) = -\mathbf{S}\cdot[r_i-R][r_i-R]\mathbf{S}.$

The simplification of this equation uses the identity

$(\mathbf{S}\times(\mathbf{S}\times(\mathbf{r}_i-\mathbf{R}))) \cdot \mathbf{S}\times(\mathbf{S}\times(\mathbf{r}_i-\mathbf{R})) = (\mathbf{S}\times(\mathbf{S}\times(\mathbf{r}_i-\mathbf{R})))\times\mathbf{S}\cdot (\mathbf{S}\times(\mathbf{r}_i-\mathbf{R})),$

where the dot and the cross products have been interchanged. Expand the cross products to compute

$-(\mathbf{r}_i-\mathbf{R})\times\mathbf{S}\cdot(\mathbf{S}\times(\mathbf{r}_i-\mathbf{R})=-\mathbf{S}\cdot[r_i-R][r_i-R]\mathbf{S},$

where [r_i-R] is the skew symmetric matrix obtained from the vector r_i-R.
Thus, the moment of inertia around the line L through R in the direction S is given by the scalar

$I_L = \sum_{i=1}^N m_i |\Delta\mathbf{r}_i|^2= -\sum_{i=1}^N m_i \mathbf{S}\cdot[r_i-R][r_i-R]\mathbf{S},$

$I_L = \mathbf{S}\cdot(-\sum_{i=1}^N m_i [r_i-R][r_i-R])\mathbf{S}=\mathbf{S}\cdot[I_R]\mathbf{S},$

where [I_R] is the moment of inertia matrix of the system relative to the reference point R.

Moment of inertia tensor

The moment of inertia for a rigid body moving in space is a tensor formed from the scalars obtained from the moments of inertia and products of inertia about the three coordinate axes. The moment of inertia tensor is constructed from the nine component tensors,

$\mathbf{e}_i\otimes\mathbf{e}_j,\quad i,j=1,2,3,$

where e_i, i=1,2,3 are the three orthogonal unit vectors defining the inertial frame in which the body moves. Using this basis the inertia tensor is given by

$\mathbf{I}=\sum_{i=1}^3\sum_{j=1}^3 I_{ij}\mathbf{e}_i\otimes\mathbf{e}_j.$

This tensor is of degree two because the component tensors are each constructed from two basis vectors. In this form the inertia tensor is also called the inertia binor.
For a rigid system of particles P_k, k=1,...,N each of mass m_k with position coordinates r_k=(x_k, y_k, z_k), the inertia tensor is given by

$\mathbf{I} =\sum_{k=1}^Nm_k((\mathbf{r}_k\cdot\mathbf{r}_k)\mathbf{E}-\mathbf{r}_k\otimes\mathbf{r}_k),$

where E is the identity tensor

$\mathbf{E}= \mathbf{e}_1\otimes\mathbf{e}_1+ \mathbf{e}_2\otimes\mathbf{e}_2+\mathbf{e}_3\otimes\mathbf{e}_3.$

The moment of inertia tensor for a continuous body is given by

$\mathbf{I}=\int_V \rho(\mathbf{r}) \left( \left( \mathbf{r} \cdot \mathbf{r} \right) \mathbf{E} - \mathbf{r}\otimes \mathbf{r}\right)\, dV,$

where r defines the coordinates of a point in the body and ρ(r) is the mass density at that point. The integral is taken over the volume V of the body. The moment of inertia tensor is symmetric because I_ij= I_ji.
The inertia tensor defines the moment of inertia about an arbitrary axis defined by the unit vector n as the product,

$I_n = \mathbf{n}\cdot\mathbf{I}\cdot\mathbf{n},$

where the dot product is taken with the corresponding elements in the component tensors. A product of inertia term such as I₁₂ is obtained by the computation

$I_{12} = \mathbf{e}_1\cdot\mathbf{I}\cdot\mathbf{e}_2,$

and can be interpreted as the moment of inertia around the x-axis when the object rotates around the y-axis.
The components of tensors of degree two can be assembled into a matrix. For the inertia tensor this matrix is given by,

$[I] = \begin{bmatrix} I_{11} & I_{12} & I_{13} \\ I_{21} & I_{22} & I_{23} \\ I_{31} & I_{32} & I_{33} \end{bmatrix}=\begin{bmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{xy} & I_{yy} & I_{yz} \\ I_{xz} & I_{yz} & I_{zz} \end{bmatrix}.$

It is common in rigid body mechanics to use notation that explicitly identifies the x, y, and z axes, such as I_xx and I_xy, for the components of the inertia tensor.

Moment of inertia reference frames

The use of the inertia matrix in Newton's second law assumes its components are computed relative to axes parallel to the inertial frame and not relative to a body-fixed reference frame. This means that as the body moves the components of the inertia matrix change with time. In contrast, the components of the inertia matrix measured in a body-fixed frame are constant.

Body frame inertia matrix

Let the body frame inertia matrix relative to the center of mass be denoted [I_C^B], and define the orientation of the body frame relative to the inertial frame by the rotation matrix [A], such that,

$\mathbf{x}=[A]\mathbf{y},$

where vectors y in the body fixed coordinate frame have coordinates x in the inertial frame. Then, the inertia matrix of the body measured in the inertial frame is given by

$[I_C]=[A][I_C^B][A^T].$

Notice that [A] changes as the body moves, while [I_C^B] remains constant.

Principal axes

Measured in the body frame the inertia matrix is a constant real symmetric matrix. A real symmetric matrix has the eigendecomponsition into the product of a rotation matrix [Q] and a diagonal matrix [Λ], given by

$[I_C^B]=[Q][\Lambda][Q^T],$

where

$[\Lambda]= \begin{bmatrix} I_{1} & 0 & 0 \\ 0 & I_{2} & 0 \\ 0 & 0 & I_{3} \end{bmatrix}.$

The columns of the rotation matrix [Q] define the directions of the principal axes of the body, and the constants I₁, I₂ and I₃ are called the principal moments of inertia. This result was first shown by J. J. Sylvester (1852), and is a form of Sylvester's law of inertia.
For bodies with constant density an axis of rotational symmetry is a principal axis.

Inertia ellipsoid

An ellipsoid with the semi-principal diameters labeled a, b, and c.

The moment of inertia matrix in body-frame coordinates is a quadratic form that defines a surface in the body called the inertia ellipsoid. Let [Λ] be the inertia matrix relative to the center of mass aligned with the principle axes, then the surface

$\mathbf{x}^T[\Lambda]\mathbf{x}=1,$

$I_1x^2 + I_2y^2 + I_3z^2 =1,$

defines an ellipsoid in the body frame. Write this equation in the form,

$\frac{x^2}{(1/\sqrt{I_1})^2} + \frac{y^2}{(1/\sqrt{I_2})^2} + \frac{z^2}{(1/\sqrt{I_3})^2} = 1,$

to see that the semi-principal diameters of this ellipsoid are given by

$a=\frac{1}{\sqrt{I_1}}, \quad b=\frac{1}{\sqrt{I_2}}, \quad c=\frac{1}{\sqrt{I_3}}.$

Let a point x on this ellipsoid be defined in terms of its magnitude and direction, x=|x|n, where n is a unit vector, then from the definition of the inertia matrix,

$\mathbf{x}^T[\Lambda]\mathbf{x}=|\mathbf{x}|^2\mathbf{n}^T[\Lambda]\mathbf{n} = |\mathbf{x}|^2I_n = 1,$

where I_n is the moment of inertia of the body around the axis in the direction n. Thus, the magnitude of a point x in the direction n on the inertia ellipsoid is

$|\mathbf{x}| = \frac{1}{\sqrt{I_n}}.$

Parallel axis theorem

It is useful to note here that if the moment of inertia matrix or tensor relative to the center of mass, then it can be determined relative to any other reference point in the body using the parallel axis theorem. If [I_C^B] is the moment of inertia matrix in the body frame relative to the center of mass C, then the moment of inertia matrix [I_R^B] in the same frame but relative to a different point R is given by

$[I_R^B] = [I_C^B] - M[d]^2,$

where M is the mass of the body, and [d] is the skew-symmetric matrix obtained from the vector d= R-C.
The tensor form of the parallel axis theorem is given by

$\mathbf{I}_R^B = \mathbf{I}_C^B + M((\mathbf{d} \cdot \mathbf{d}) \mathbf{E} - \mathbf{d} \otimes \mathbf{d}).$

Identities for a skew-symmetric matrix

In order to compare formulations of the inertia matrix in terms of a product of skew-symmetric matrices and in terms of a tensor formulation, the following identities are useful.
Let [R] be the skew symmetric matrix associated with the position vector R=(x, y, z), then the product in the inertia matrix becomes

$-[R][R]= -\begin{bmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{bmatrix}^2 = \begin{bmatrix} y^2+z^2 & -xy & -xz \\ -y x & x^2+z^2 & -yz \\ -zx & -zy & x^2+y^2 \end{bmatrix}.$

This product can be computed using the matrix formed by the outer product [R R^T] using the identify

$-[R]^2 = |\mathbf{R}|^2[I] -[\mathbf{R}\mathbf{R}^T]= \begin{bmatrix} x^2+y^2+z^2 & 0 & 0 \\ 0& x^2+y^2+z^2 & 0 \\0& 0& x^2+y^2+z^2 \end{bmatrix}- \begin{bmatrix}x^2 & xy & xz \\ yx & y^2 & yz \\ zx & zy & z^2\end{bmatrix},$

where [I] is the 3x3 identify matrix.
Also notice, that

$|\mathbf{R}|^2 = \mathbf{R}\cdot\mathbf{R} =\operatorname{tr}[\mathbf{R}\mathbf{R}^T],$

where tr denotes the sum of the diagonal elements of the outer product matrix, known as its trace.

knowledge

Thursday 8 November 2012

Moment of inertia

Moment of inertia

This article is about the moment of inertia of a rotating object, also termed the mass moment of inertia. For the moment of inertia dealing with the bending of a beam, also termed the area moment of inertia, see second moment of area.

Overview

Scalar moment of inertia of a simple pendulum

Scalar moment of inertia of a rigid body

Moment of inertia theorems

Properties

Examples

Polar moment of inertia

Angular momentum in planar movement

Kinetic energy in planar movement

Newton's laws for planar movement

Moment of inertia matrix

Angular momentum

Kinetic energy

Resultant torque

Parallel axis theorem

Moment of inertia around an arbitrary axis

Moment of inertia tensor

Moment of inertia reference frames

Body frame inertia matrix

Principal axes

Inertia ellipsoid

Parallel axis theorem

Identities for a skew-symmetric matrix

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