User:Prof McCarthy/moment of inertia

Examples

Main article: List of moments of inertia

Diatomic molecule: The moment of inertia of a diatomic molecule about an axis through its center of mass is calculated by first locating the center of mass. For convenience let the x-axis be aligned with the axis of the molecule and place m₁ at the origin and m₂ at the distance d, then from the definition of the center of mass,

$m_{1}(0-r)+m_{2}(d-r)=0,\quad r={\frac {m_{2}d}{m_{1}+m_{2}}},$

where r locates the center of mass.

The moment of inertia of this molecule about the axis through the center of mass is given by

$I=m_{1}r^{2}+m_{2}(d-r)^{2}=m_{1}({\frac {m_{2}d}{m_{1}+m_{2}}})^{2}+m_{2}({\frac {m_{1}d}{m_{1}+m_{2}}})^{2}$

Thin rod: The moment of inertia of a thin rod of mass of constant cross-section s, density ρ and length ℓ about a perpendicular axis through its center of mass is determined by integration. Align the x-axis with the rod and located the origin its center of mass at the center of the rod, then

$I=\int \rho \,x^{2}dV=\int _{-\ell /2}^{\ell /2}\rho \,x^{2}sdx=\rho s{\frac {x^{3}}{3}}{\bigg |}_{-\ell /2}^{\ell /2}={\frac {\rho s}{3}}(\ell ^{3}/8+\ell ^{3}/8)={\frac {1}{12}}\,m\ell ^{2},$

where m=ρsℓ is the mass of the rod.

Thin disc: The moment of a inertia of a thin disc of constant thickness s and density ρ about an axis through its center is determined by integration. Align the z-axis with the axis of the disc and define a volume element as dV=srdrdθ, then

$\int \rho r^{2}dV=\int _{0}^{R}\int _{0}^{2\pi }\rho r^{2}(srdrd\theta )=2\pi \rho s{\frac {R^{4}}{4}}={\frac {1}{2}}mR^{2},$

where where R is the radius of the disc m=2πR²ρs is its mass.

Solid ball: The moment of inertia of a solid ball of constant density ρ about an axis through its center is determined by integration of the moment of inertia of thin discs along it axis. If the surface of the ball is defined by the equation

$x^{2}+y^{2}+z^{2}=R^{2},$

then the radius r of the disc at the cross-section z along the z-axis is

$r(z)^{2}=x^{2}+y^{2}=R^{2}-z^{2}.$

Therefore, the moment of inertia of the ball is the sum of the moment of inertias of the discs along the z-axis,

$I=\int _{-R}^{R}{\frac {\pi \rho }{2}}r(z)^{4}dz=\int _{-R}^{R}{\frac {\pi \rho }{2}}(R^{2}-z^{2})^{2}dz={\frac {\pi \rho }{2}}(R^{4}z-2R^{2}z^{3}/3+z^{5}/5){\bigg |}_{-R}^{R}=\pi \rho (1-2/3+1/5)R^{5}={\frac {2}{5}}mR^{2},$

where m=(4/3)πR³ρ is the mass of the ball.

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 + z². Thus, in order to compute the moment of inertia I_z, we need to evaluate the integral ∭(x² + y² + z²) 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

$I={\frac {1}{3}}(I_{x}+I_{y}+I_{z})={\frac {1}{3}}\iiint \rho \cdot (y^{2}+z^{2}+x^{2}+z^{2}+x^{2}+y^{2})\,\mathrm {d} V={\frac {2}{3}}\,\rho \int r^{2}\,\mathrm {d} V,$

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,

$I={\frac {2}{3}}\,\rho \int _{0}^{R}\!\!4\pi r^{4}\,\mathrm {d} r={\frac {2}{3}}\,\rho \cdot 4\pi {\frac {R^{5}}{5}}={\frac {m}{{\frac {4}{3}}\pi R^{3}}}\cdot {\frac {8\pi R^{5}}{15}}={\frac {2}{5}}\,mR^{2}.$

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.

Measuring moment of inertia

The moment of inertia of a part around an axis can be determined experimentally by forming a compound pendulum. Suspend the part from a convenient pivot point P so that it swings freely in a plane perpendicular to the direction of the desired moment of inertia. Locate the distance r from the pivot point P to the center of mass C of the part. The equations of motion of a compound pendulum show that the natural frequency of oscillation of the part as it swings from the pivot point is given by

$\omega ={\sqrt {\frac {mgr}{I_{P}}}}.$

Thus, if the period of oscillation is measured to be τ=2π/ω, then the moment of inertia about the pivot point is given by

$I_{P}=mgr({\frac {\tau }{2\pi }})^{2}.$

The moment of inertia of the part about the center of mass I_C is then determined using the parallel axis theorem,

$I_{C}=I_{P}+mr^{2}.$

Overview

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 rotating masses with large moment of inertia known as flywheels to maintain rotational velocity and resist small variations in applied torque.

The moment of inertia of an object measures the distribution of mass around an axis. It depends not only on the total mass of the object, but also 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 contain the same mass, one as large as bicycle wheel and another one that is half that size. The larger wheel has four times the moment of inertia though it is only twice the diameter.

Moment of inertia around a fixed axis is a scalar, however for the spatial movement of a body's rotation can occur around the three coordinate axes, which defines a matrix of scalars called the inertia matrix, also the inertia tensor.

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.

Scalar moment of inertia for a pendulum

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 pendulum 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 is defined by the angular acceleration α of the pendulum, therefore

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

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 using the angular velocity ω of the pendulum to yield

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

The angular momentum of the pendulum mass is given by

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

This shows that the quantity I=mr² plays the same role for rotational movement, as mass 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.

Definition

For a rigid object of $N$ point masses $m_{k}$, the moment of inertia tensor (with respect to the origin) has components given by

$\mathbf {I} ={\begin{bmatrix}I_{11}&I_{12}&I_{13}\\I_{21}&I_{22}&I_{23}\\I_{31}&I_{32}&I_{33}\end{bmatrix}}={\begin{bmatrix}\sum _{k=1}^{N}m_{k}(y_{k}^{2}+z_{k}^{2})&-\sum _{k=1}^{N}m_{k}x_{k}y_{k}&-\sum _{k=1}^{N}m_{k}x_{k}z_{k}\\-\sum _{k=1}^{N}m_{k}x_{k}y_{k}&\sum _{k=1}^{N}m_{k}(x_{k}^{2}+z_{k}^{2})&-\sum _{k=1}^{N}m_{k}y_{k}z_{k}\\-\sum _{k=1}^{N}m_{k}x_{k}z_{k}&-\sum _{k=1}^{N}m_{k}y_{k}z_{k}&\sum _{k=1}^{N}m_{k}(x_{k}^{2}+y_{k}^{2})\end{bmatrix}}.$

$I_{22}=I_{yy}\ {\stackrel {\mathrm {def} }{=}}\ \sum _{k=1}^{N}m_{k}(x_{k}^{2}+z_{k}^{2}),\,\!$

$I_{33}=I_{zz}\ {\stackrel {\mathrm {def} }{=}}\ \sum _{k=1}^{N}m_{k}(x_{k}^{2}+y_{k}^{2}),\,\!$

$I_{12}=I_{xy}\ {\stackrel {\mathrm {def} }{=}}\ -\sum _{k=1}^{N}m_{k}x_{k}y_{k},\,\!$

$I_{13}=I_{xz}\ {\stackrel {\mathrm {def} }{=}}\ -\sum _{k=1}^{N}m_{k}x_{k}z_{k},\,\!$

$I_{23}=I_{yz}\ {\stackrel {\mathrm {def} }{=}}\ -\sum _{k=1}^{N}m_{k}y_{k}z_{k},\,\!$

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.

Momentum for planar movement

Spinning figure skater pulling in her arms, reducing her moment of inertia, to rotate faster

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

$\mathbf {L} =\sum _{i=1}^{n}m_{i}(\mathbf {r} _{i}-\mathbf {R} )\times \mathbf {v} _{i}=\sum _{i=1}^{n}m_{i}\Delta r_{i}\mathbf {e} _{i}\times (\omega \Delta r_{i}\mathbf {t} _{i}+\mathbf {V} )=(\sum _{i=1}^{n}m_{i}\Delta r_{i}^{2})\omega {\vec {k}}+(\sum _{i=1}^{n}m_{i}\Delta r_{i}\mathbf {e} _{i})\times \mathbf {V} .$

Use the center of mass as the reference point R and define the moment of inertia relative to the center of mass I_R so

$\sum _{i=1}^{n}m_{i}\Delta r_{i}\mathbf {e} _{i}=0,\quad I_{R}=\sum _{i=1}^{n}m_{i}\Delta r_{i}^{2},$

then the equation for angular momentum simplifies to

$\mathbf {L} =I_{R}\omega {\vec {k}}.$

The moment of inertia I_R 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.

Kinetic energy for planar movement

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 R be the center of mass of the system so the second term becomes zero, and introduce the moment of inertia I_R so the kinetic energy is given by

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

Newton's laws for planar movement

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.

For planar movement 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 as the reference point R and define the moment of inertia relative to the center of mass I_R so

$\sum _{i=1}^{N}m_{i}\Delta r_{i}\mathbf {e} _{i}=0,\quad I_{R}=\sum _{i=1}^{N}m_{i}\Delta r_{i}^{2},$

then the equation for the resultant torque simplifies to

$\mathbf {T} =I_{R}\alpha {\vec {k}}.$

The moment of inertia I_R 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.

Copied for moment of inertia

Consider a massless rigid rod of length l with a point mass m at one end and rotating about the other end. Suppose the rod rotates at a constant rate so that the mass moves at speed v. Then the kinetic energy T of the mass is:

$T={\frac {1}{2}}\,mv^{2}$

Using v = lω, where ω is the angular velocity, one obtains:

$T={\frac {1}{2}}\,m(\omega l)^{2}$

which can be rearranged to give:

$T={\frac {1}{2}}\,ml^{2}\omega ^{2}$

This equation resembles the original expression for the kinetic energy, but in the place of the linear velocity v is the angular velocity ω, and instead of the mass m is ml². The quantity ml² can therefore be seen as an analogue of mass for rotational motion; in other words, it is a measure of rotational inertia.

Scalar moment of inertia for many bodies

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

$T=\sum _{i=1}^{N}{\frac {1}{2}}\,m_{i}v_{i}^{2}=\sum _{i=1}^{N}{\frac {1}{2}}\,m_{i}(\omega r_{i})^{2}={\frac {1}{2}}\,\omega ^{2}{\Big (}\textstyle \sum _{i=1}^{N}m_{i}r_{i}^{2}{\Big )}.$

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:

$I=\int _{V}\rho (\mathbf {r} )\,d(\mathbf {r} )^{2}\,\mathrm {d} V\!(\mathbf {r} ),$

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

Moment of inertia about a point

In discussions of the virial theorem, a different definition of moment of inertia is introduced, often qualified as moment of inertia about a point (as opposed to an axis). The defining equations look the same as the ones above, but $r_{i}$ and $d(\mathbf {r} )$ in these cases are understood to be the distance to the origin.

Inertia matrix

Angular momentum

The angular momentum 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. Select the center of mass as the 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. The angular momentum vector relative to the center of mass R is

$\mathbf {L} =\sum _{i=1}^{n}m_{i}(\mathbf {r} _{i}-\mathbf {R} )\times (\omega \times (\mathbf {r} _{i}-\mathbf {R} )).$

In order to simplify this equation and obtain a formula for 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

$[B]={\begin{bmatrix}0&-b_{z}&b_{y}\\b_{z}&0&-b_{x}\\-b_{y}&b_{x}&0\end{bmatrix}}.$

Introduce the skew-symmetric matrix [r_i-R] constructed from the relative position vector r_i - R to obtain the angular momentum as

$\mathbf {L} =(-\sum _{i=1}^{n}m_{i}[r_{i}-R][r_{i}-R])\omega =[I_{R}]\omega ,$

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

$T=\sum _{i=1}^{n}m_{i}\mathbf {v} _{i}\cdot \mathbf {v} _{i}=\sum _{i=1}^{n}m_{i}(\omega \times (\mathbf {r} _{i}-\mathbf {R} )+\mathbf {V} )\cdot (\omega \times (\mathbf {r} _{i}-\mathbf {R} )+\mathbf {V} ).$

This equation expands to yield three terms

$T=\sum _{i=1}^{n}m_{i}(\omega \times (\mathbf {r} _{i}-\mathbf {R} ))\cdot (\omega \times (\mathbf {r} _{i}-\mathbf {R} ))+2\sum _{i=1}^{n}m_{i}\mathbf {V} \cdot (\omega \times (\mathbf {r} _{i}-\mathbf {R} ))+\sum _{i=1}^{n}m_{i}\mathbf {V} \cdot \mathbf {V} .$

Let R be the center of mass of the system and introduce the skew-symmetric matrix [r_i-R] so the kinetic energy of the rigid system of particles becomes

$T=\omega \cdot (-\sum _{i=1}^{n}m_{i}[r_{i}-R][r_{i}-R])\omega +(\sum _{i=1}^{n}m_{i})\mathbf {V} \cdot \mathbf {V} ,$

or

$T=\omega \cdot [I_{R}]\omega +M\mathbf {V} \cdot \mathbf {V} ,$

where [I_R] is the inertia matrix and M is the total mass of the system.

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,

$\mathbf {T} =\sum _{i=1}^{N}(\mathbf {R} _{i}-\mathbf {R} )\times (m_{i}\mathbf {A} _{i}).$

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 R as the reference point, and introduce the skew-symmetric matrix [R_i-R] to represent the (R_i - R)x in order to obtain

$\mathbf {T} =(-\sum _{i=1}^{N}m_{i}[R_{i}-R][R_{i}-R])\times \alpha +\omega \times (-\sum _{i=1}m_{i}[R_{i}-R][R_{i}-R])\omega .$

This calculation uses the identity

$(\mathbf {R} _{i}-\mathbf {R} )\times (\omega \times (\omega \times (\mathbf {R} _{i}-\mathbf {R} )))=\omega \times ((\mathbf {R} _{i}-\mathbf {R} )\times ((\mathbf {R} _{i}-\mathbf {R} )\times \omega ))).$

Thus, the resultant torque on the rigid system of particles is given by

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

where [I_R] is the inertia matrix.

Identities for a skew-symmetric matrix

In order to compare formulations of the inertia matrix in terms of a skew-symmetric matrix 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

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

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

$-[R]^{2}=|\mathbf {R} |^{2}[E_{3}]-[\mathbf {R} \mathbf {R} ^{T}]={\begin{bmatrix}x^{2}+y^{2}+z^{2}&0&0\\0&x^{2}+y^{2}+z^{2}&0\\0&0&x^{2}+y^{2}+z^{2}\end{bmatrix}}-{\begin{bmatrix}x^{2}&xy&xz\\yx&y^{2}&yz\\zx&zy&z^{2}\end{bmatrix}},$

where [E₃] 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.

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 R and the inertia matrix relative to another point S. This relationship is called the parallel axis theorem.

Consider the inertia matrix [I_S] obtained for a rigid system of particles measured relative to a reference point S, given by

$[I_{S}]=-\sum _{i=1}^{n}m_{i}[r_{i}-S][r_{i}-S].$

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

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

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

$[I_{S}]=-\sum _{i=1}^{n}m_{i}[r_{i}-R+d][r_{i}-R+d].$

Substitute the identity

$([r_{i}-R]+[-d])([r_{i}-R]+[d])=[r_{i}-R][r_{i}-R]+[r_{i}-R][d]+[d][r_{i}-R]+[d][d],$

into the inertia matrix to obtain

$[I_{S}]=(-\sum _{i=1}^{n}m_{i}[r_{i}-R][r_{i}-R])+(-\sum _{i=1}^{n}m_{i}[r_{i}-R])[d]+[d](-\sum _{i=1}^{n}m_{i}[r_{i}-R])+(-\sum _{i=1}^{n}m_{i})[d][d])$

The first term is the inertia matrix [I_R] relative to the center of mass. The second and third terms are zero by definition of the center of mass R. 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,

$[I_{S}]=[I_{R}]+M[d]^{2},$

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

Rigid system of particles

If a system of N particles, P_i, i=1,...,N, are assembled into a rigid body, then Newton's second law can be applied to each of the particles in the body. If F_i is the external force applied to particle P_i with mass m_i, then

$\mathbf {F} _{i}+\sum _{j=1}^{N}\mathbf {F} _{ij}=m_{i}\mathbf {A} _{i},\quad i=1,\ldots ,N,$

where F_ij is the internal force of particle P_j acting on particle P_i that maintains the constant distance between these particles.

An important simplification to these force equations is obtained by introducing the resultant force and torque that acts on the rigid system. This resultant force and torque is obtained by choosing one of the particles in the system as a reference point, R, where each of the external forces are applied with the addition of an associated torque. The resultant force F and torque T are given by the formulas,

$\mathbf {F} =\sum _{i=1}^{N}\mathbf {F} _{i},\quad \mathbf {T} =\sum _{i=1}^{N}(\mathbf {R} _{i}-\mathbf {R} )\times \mathbf {F} _{i},$

where R_i is the vector that defines the position of particle to P_i.

Newton's second law for a particle combines with these formulas for the resultant force and torque 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 the internal forces F_ij cancel in pairs. 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,

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

Substitute the particle accelerations into the resultant force to obtain

$\mathbf {F} =\alpha \times (\sum _{i=1}^{N}m_{i}(\mathbf {R} _{i}-\mathbf {R} ))+\omega \times \omega \times (\sum _{i=1}^{N}m_{i}(\mathbf {R} _{i}-\mathbf {R} ))+(\sum _{i=1}^{N}m_{i})\mathbf {A} .),$

If the reference point R is chosen to be the center of mass, this becomes

$\mathbf {F} =M\mathbf {A} .$

Substitute the particle accelerations into the resultant torque equation to obtain,

$\mathbf {T} =\sum _{i=1}^{N}(\mathbf {R} _{i}-\mathbf {R} )\times (m_{i}(\alpha \times (\mathbf {R} _{i}-\mathbf {R} )+\omega \times \omega \times (\mathbf {R} _{i}-\mathbf {R} )+\mathbf {A} )).$

Rearrange the terms to obtain,

$\mathbf {T} =-(\sum _{i=1}^{N}(\mathbf {R} _{i}-\mathbf {R} )\times (m_{i}(\mathbf {R} _{i}-\mathbf {R} ))\times \alpha +\sum _{i=1}m_{i}(\mathbf {R} _{i}-\mathbf {R} )\times \omega \times \omega \times (\mathbf {R} _{i}-\mathbf {R} )+(\sum _{i=1}m_{i}(\mathbf {R} _{i}-\mathbf {R} ))\times \mathbf {A} .$

Because R is the center of mass, the last term becomes zero. Notice that the order of the cross products in this equation are from right to left. Rearrange second term using the identify,

$(\mathbf {R} _{i}-\mathbf {R} )\times (\omega \times (\omega \times (\mathbf {R} _{i}-\mathbf {R} )))=\omega \times ((\mathbf {R} _{i}-\mathbf {R} )\times ((\mathbf {R} _{i}-\mathbf {R} )\times \omega ))).$

In order to simplify this equation and obtain a formula for the mass moment of inertia, introduce the skew symmetric matrix [B] associated with 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

$[B]={\begin{bmatrix}0&-b_{z}&b_{y}\\-b_{z}&0&-b_{z}\\-b_{y}&b_{x}&0\end{bmatrix}}.$

introduce the skew-symmetric matrix [R_i-R] constructed from the vector R_i - R to formulate the resultant torque as

$\mathbf {T} =(-\sum _{i=1}^{N}m_{i}[R_{i}-R][R_{i}-R])\times \alpha +\omega \times (-\sum _{i=1}m_{i}[R_{i}-R][R_{i}-R])\omega$

The matrix [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.

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 to all 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 the axis S defined by the unit vector S through the reference point. The moment of inertia of the system around the line L=R+tS is computed by determining the perpendicular vector from the axis L 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

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

This equation simplifies to

$|\Delta \mathbf {r} _{i}|^{2}=\mathbf {S} \cdot [r_{i}-R][r_{i}-R]\mathbf {S} .$

This is simplification 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

$I_{L}=\sum _{i=1}^{N}m_{i}|\Delta \mathbf {r} _{i}|^{2}=\sum _{i=1}^{N}m_{i}\mathbf {S} \cdot [r_{i}-R][r_{i}-R]\mathbf {S} ,$

or

$I_{L}=\mathbf {S} \cdot (\sum _{i=1}^{N}m_{i}[r_{i}-R][r_{i}-R])\mathbf {S} =\mathbf {S} \cdot [I_{R}]\mathbf {S} ,$

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

Body frame inertia matrix

The use of the inertia matrix in Newton's second law assumes its components are computed relative to axes parallel 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 constants.

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.

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.

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 reference 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

$\mathbf {I} =\sum _{k=1}^{N}m_{k}((\mathbf {r} _{k}\cdot \mathbf {r} _{k})\mathbf {E} -\mathbf {r} _{k}\otimes \mathbf {r} _{k}),$

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.

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.

Inertia ellipsoid

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

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

or

$I_{1}x^{2}+I_{2}y^{2}+I_{3}z^{2}=1,$

defines an ellipsoid in the body frame. Write this ellipsoid 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 its semi-principle diameters 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,

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

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

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