# Moment of inertia

{{#invoke:Hatnote|hatnote}} Template:Classical mechanics Samuel Dixon uses the moment of inertia of the long rod to help maintain balance as he crosses the Niagara river (1890).

Moment of inertia is the mass property of a rigid body that determines the torque needed for a desired angular acceleration about an axis of rotation. Moment of inertia depends on the shape of the body and may be different around different axes of rotation. A larger moment of inertia around a given axis requires more torque to increase the rotation, or to stop the rotation, of a body about that axis. Moment of inertia depends on the amount and distribution of its mass, and can be found through the sum of moments of inertia of the masses making up the whole object, under the same conditions. For example, if ma + mb = mc, then Ia + Ib = Ic. In classical mechanics, moment of inertia may also be called mass moment of inertia, rotational inertia, polar moment of inertia, or the angular mass.

For planar movement of a body, the trajectories of all of its points lie in parallel planes, and the rotation occurs only about an axis perpendicular to this plane. In this case, the body has a single moment of inertia, which is measured around this axis.

For spatial movement of a body, the moment of inertia is defined by its symmetric 3 × 3 inertia matrix. The inertia matrix is often described as a symmetric rank two tensor, having six independent components. The inertia matrix includes off-diagonal terms called products of inertia that couple torque around one axis to angular acceleration about another axis. Each body has a set of mutually perpendicular axes, called principal axes, for which the off-diagonal terms of the inertia matrix are zero, and a torque around a principal axis only affects the acceleration about that axis.

## Introduction

When a body is rotating around an axis, a torque must be applied to change its angular momentum. The amount of torque needed for any given change in angular momentum is proportional to the size of that change. Moment of inertia may be expressed in terms of kilogram-square metres (kg·m2) in SI units and pound-square feet (lbm·ft2) in imperial or US units.

In 1673 Christiaan Huygens introduced this parameter in his study of the oscillation of a body hanging from a pivot, known as a compound pendulum. The term moment of inertia was introduced by Leonhard Euler in his book Theoria motus corporum solidorum seu rigidorum in 1765, and it is incorporated into Euler's second law.

The natural frequency of oscillation of a compound pendulum is obtained from the ratio of the torque imposed by gravity on the mass of the pendulum to the resistance to acceleration defined by the moment of inertia. Comparison of this natural frequency to that of a simple pendulum consisting of a single point of mass provides a mathematical formulation for moment of inertia of an extended body.

Moment of inertia also appears in momentum, kinetic energy, and in Newton's laws of motion for a rigid body as a physical parameter that combines its shape and mass. There is an interesting difference in the way moment of inertia appears in planar and spatial movement. Planar movement has a single scalar that defines the moment of inertia, while for spatial movement the same calculations yield a 3 × 3 matrix of moments of inertia, called the inertia matrix or inertia tensor.

The moment of inertia of a rotating flywheel is used in a machine to resist variations in applied torque to smooth its rotational output. The moment of inertia of an airplane about its longitudinal, horizontal and vertical axes determines how steering forces on the control surfaces of its wings, elevators and tail affect the plane in roll, pitch and yaw.

## Definition

Moment of inertia I is defined as the ratio of the angular momentum L of a system to its angular velocity ω around a principal axis, that is

$I={\frac {L}{\omega }}.$ If the momentum of a system is constant, then as the moment of inertia gets smaller, the angular velocity must increase. This occurs when spinning figure skaters pulls in their outstretched arms or divers move from a straight position to a tuck position during a dive.

If the shape of the body does not change, then its moment of inertia appears in Newton's law of motion as the ratio of an applied torque τ on a body to the angular acceleration α around a principal axis, that is

$\tau =I\alpha .$ For a simple pendulum, this definition yields a formula for the moment of inertia I in terms of the mass m of the pendulum and its distance r from the pivot point as,

$I=mr^{2}.$ Thus, moment of inertia depends on both the mass m of a body and its geometry, or shape, as defined by the distance r to the axis of rotation.

This simple formula generalizes to define moment of inertia for an arbitrarily shaped body as the sum of all the elemental point masses dm each multiplied by the square of its perpendicular distance r to an axis S .

In general, given an object of mass m, an effective radius r′ can be defined for an arbitrary axis of interest, with such a value that its moment of inertia is

$I=m(r')^{2},$ regardless of any symmetry that could justify, or not, defining a radius for that object.

## Simple pendulum

Moment of inertia can be measured using a simple pendulum, because it is the resistance to the rotation caused by gravity. Mathematically, the moment of inertia of the pendulum is the ratio of the torque due to gravity about the pivot of a pendulum to its angular acceleration about that pivot point. For a simple pendulum this is found to be the product of the mass of the particle m with the square of its distance r to the pivot, that is

$I=mr^{2}.$ This can be shown as follows: The force of gravity on the mass of a simple pendulum generates a torque ${\boldsymbol {\tau }}=\mathbf {r} \times \mathbf {F}$ around the axis perpendicular to the plane of the pendulum movement. Here r is the distance vector perpendicular to and from the force to the torque axis. Here F is the tangential component of the net force on the mass. Associated with this torque is an angular acceleration, ${\boldsymbol {\alpha }}$ , of the string and mass around this axis. Since the mass is constrained to a circle the tangential acceleration of the mass is ${\mathbf {a}}={\boldsymbol {\alpha }}\times {\mathbf {r}}$ . Since $F=ma$ the torque equation becomes:

${\boldsymbol {\tau }}=\mathbf {r} \times \mathbf {F} =\mathbf {r} \times (m{\boldsymbol {\alpha }}\times {\mathbf {r}})=(mr^{2}){\boldsymbol {\alpha }}=I\alpha {\mathbf {e}},$ where e is a unit vector perpendicular to the plane of the pendulum. (The second to the last step occurs because of the BAC-CAB rule using the fact that ${\boldsymbol {\alpha }}$ is always perpendicular to r.) The quantity I = mr2 is the moment of inertia of this single mass around the pivot point.

The quantity I = mr2 also appears in the angular momentum of a simple pendulum, which is calculated from the velocity v = ω×r of the pendulum mass around the pivot, where ω is the angular velocity of the mass about the pivot point. This angular momentum is given by

$\mathbf {L} =\mathbf {r} \times (m\mathbf {v} )=(mr^{2}){\boldsymbol {\omega }}=I\omega {\mathbf {e}},$ using math similar to that used to derive the previous equation.

Similarly, the kinetic energy of the pendulum mass is defined by the velocity of the pendulum around the pivot to yield

$E_{\text{K}}={\frac {1}{2}}m\mathbf {v} \cdot \mathbf {v} ={\frac {1}{2}}(mr^{2})\omega ^{2}={\frac {1}{2}}I\omega ^{2}.$ This shows that the quantity I = mr2 is how mass combines with the shape of a body to define rotational inertia. The moment of inertia of an arbitrarily shaped body is the sum of the values Template:Mvar2 for all of the elements of mass in the body.

## Compound 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.

A compound pendulum is a body formed from an assembly of particles or continuous shapes that rotates rigidly around a pivot. Its moments of inertia is the sum the moments of inertia of each of the particles that is composed of.:395–396:51–53 The natural frequency ($\omega _{n}$ ) of a compound pendulum depends on its moment of inertia, $I_{P}$ ,

$\omega _{n}={\sqrt {\frac {mgr}{I_{P}}}},$ where $m$ is the mass of the object, $g$ is local acceleration of gravity, and $r$ is the distance from the pivot point to the centre of mass of the object. Measuring this frequency of oscillation over small angular displacements provides an effective way of measuring moment of inertia of a body.:516–517

Thus, to determine the moment of inertia of the body, simply suspend it from a convenient pivot point $_{P}$ so that it swings freely in a plane perpendicular to the direction of the desired moment of inertia, then measure its natural frequency or period of oscillation ($t$ ), to obtain

$I_{P}={\frac {mgr}{\omega _{n}^{2}}}={\frac {mgrt^{2}}{4\pi ^{2}}},$ where $t$ is the period (duration) of oscillation (usually averaged over multiple periods).

The moment of inertia of the body about its centre of mass, $I_{C}$ , is then calculated using the parallel axis theorem to be

$I_{C}=I_{P}-mr^{2},$ Moment of inertia of a body is often defined in terms of its radius of gyration, which is the radius of a ring of equal mass around the centre of mass of a body that has the same moment of inertia. The radius of gyration $K$ is calculated from the body's moment of inertia $I_{C}$ and mass $m$ as the length,:1296–1297

$K={\sqrt {\frac {I_{C}}{m}}}.$ ### Centre of oscillation

A simple pendulum that has the same natural frequency as a compound pendulum defines the length $L$ from the pivot to a point called the centre of oscillation of the compound pendulum. This point also corresponds to the centre of percussion. The length $L$ is determined from the formula,

$\omega _{n}={\sqrt {\frac {g}{L}}}={\sqrt {\frac {mgr}{I_{P}}}},$ or

$L={\frac {g}{\omega _{n}^{2}}}={\frac {I_{P}}{mr}}.$ The seconds pendulum, which provides the "tick" and "tock" of a grandfather clock, takes one second to swing from side-to-side. This is a period of two seconds, or a natural frequency of π radians/second for the pendulum. In this case, the distance to the center of oscillation, $L$ , can be computed to be

$L={\frac {g}{\omega _{n}^{2}}}={\frac {9.81\ \mathrm {m/s^{2}} }{(3.14\ \mathrm {rad/s} )^{2}}}=0.99\ \mathrm {m} .$ Notice that the distance to the center of oscillation of the seconds pendulum must be adjusted to accommodate different values for the local acceleration of gravity. Kater's pendulum is is a compound pendulum that uses this property to measure the local acceleration of gravity, and is called a gravimeter.

## Measuring moment of inertia

The moment of inertia of complex systems such as a vehicle or airplane around its vertical axis can be measured by suspending the system from three points to form a trifilar pendulum. A trifilar pendulum is a platform supported by three wires designed to oscillate in torsion around its vertical centroidal axis. The period of oscillation of the trifilar pendulum yields the moment of inertia of the system.

## Calculating moment of inertia about an axis

The moment of inertia about an axis of a body is calculated by summing mr2 for every particle in the body, where r is the perpendicular distance to the specified axis. To see how moment of inertia arises in the study of the movement of an extended body, it is convenient to consider a rigid assembly of point masses. (This equation can be used for axes that are not principal axes provided that it is understood that this does not fully describe the moment of inertia.)

Consider the kinetic energy of an assembly of Template:Mvar masses Template:Mvar that lie at the distances Template:Mvar from the pivot point P, which is the nearest point on the axis of rotation. It is the sum of the kinetic energy of the individual masses,:516–517:1084–1085 :1296–1300

$E_{\text{K}}=\sum _{i=1}^{N}{\frac {1}{2}}\,m_{i}\mathbf {v} _{i}\cdot \mathbf {v} _{i}=\sum _{i=1}^{N}{\frac {1}{2}}\,m_{i}(\omega r_{i})^{2}={\frac {1}{2}}\,\omega ^{2}\sum _{i=1}^{N}m_{i}r_{i}^{2}.$ This shows that the moment of inertia of the body is the sum of each of the mr2 terms, that is

$I_{P}=\sum _{i=1}^{N}m_{i}r_{i}^{2}.$ Thus, moment of inertia is a physical property that combines the mass and distribution of the particles around the rotation axis. Notice that rotation about different axes of the same body yield different moments of inertia.

The moment of inertia of a continuous body rotating about a specified axis is calculated in the same way, with the summation replaced by the integral,

$I_{P}=\int _{V}\rho (\mathbf {r} )\,\mathbf {r} ^{2}\,dV.$ Again r is the radius vector to a point in the body from the specified axis through the pivot P, and Template:Mvar(r) is the mass density at each point r. The integration is evaluated over the volume Template:Mvar of the body. The moment of inertia of a flat surface is similar with the mass density being replaced by its areal mass density with the integral evaluated over its area.

Note on second moment of area: The moment of inertia of a body moving in a plane and the second moment of area of a beam's cross-section are often confused. The moment of inertia of body with the shape of the cross-section is the second moment of this area about the z-axis perpendicular to the cross-section, weighted by its density. This is also called the polar moment of the area, and is the sum of the second moments about the x and y axes. The stresses in a beam are calculated using the second moment of the cross-sectional area around either the x-axis or y-axis depending on the load.

### Example calculation of moment of inertia

{{#invoke:main|main}}

The moment of inertia of a compound pendulum constructed from a thin disc mounted at the end of a thin rod that oscillates around a pivot at the other end of the rod, begins with the calculation of the moment of inertia of the thin rod and thin disc about their respective centres of mass.

• The moment of inertia of a thin rod with constant cross-section Template:Mvar and density ρ and with length Template:Mvar about a perpendicular axis through its centre of mass is determined by integration.:1301 Align the x-axis with the rod and locate the origin its centre of mass at the centre of the rod, then
$I_{C,{\text{rod}}}=\int \rho \,x^{2}dV=\int _{-\ell /2}^{\ell /2}\rho \,x^{2}sdx=\rho s{\frac {x^{3}}{3}}{\bigg |}_{-\ell /2}^{\ell /2}={\frac {\rho s}{3}}(\ell ^{3}/8+\ell ^{3}/8)={\frac {1}{12}}\,m\ell ^{2},$ where m = ρsℓ is the mass of the rod.

• The moment of inertia of a thin disc of constant thickness Template:Mvar, radius Template:Mvar, and density Template:Mvar about an axis through its centre and perpendicular to its face (parallel to its axis of rotational symmetry) is determined by integration.:1301 Align the z-axis with the axis of the disc and define a volume element as dV = sr drdθ, then
$I_{C,{\text{disc}}}=\int \rho r^{2}dV=\int _{0}^{2\pi }\int _{0}^{R}\rho r^{2}(srdrd\theta )=2\pi \rho s{\frac {R^{4}}{4}}={\frac {1}{2}}mR^{2},$ where m = πR2ρs is its mass.

• The moment of inertia of the compound pendulum is now obtained by adding the moment of inertia of the rod and the disc around the pivot point P as,
$I_{P}=I_{C,{\text{rod}}}+M_{\text{rod}}(L/2)^{2}+I_{C,{\text{disc}}}+M_{\text{disc}}(L+R)^{2},$ where L is the length of the pendulum. Notice that the parallel axis theorem is used to shift the moment of inertia from the centre of mass to the pivot point of the pendulum.

A list of moments of inertia formulas for standard body shapes provides a way to obtain the moment of inertial of a complex body as an assembly of simpler shaped bodies. The parallel axis theorem is used to shift the reference point of the individual bodies to the reference point of the assembly.

As one more example, consider the moment of inertia of a solid sphere of constant density about an axis through its centre of mass. This is determined by summing the moments of inertia of the thin discs that form the sphere. If the surface of the ball is defined by the equation:1301

$x^{2}+y^{2}+z^{2}=R^{2},$ 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 moments of inertia of the discs along the z-axis,

$I_{C,{\text{ball}}}=\int _{-R}^{R}{\frac {\pi \rho }{2}}r(z)^{4}dz=\int _{-R}^{R}{\frac {\pi \rho }{2}}(R^{2}-z^{2})^{2}dz={\frac {\pi \rho }{2}}(R^{4}z-2R^{2}z^{3}/3+z^{5}/5){\bigg |}_{-R}^{R}$ $=\pi \rho (1-2/3+1/5)R^{5}={\frac {2}{5}}mR^{2},$ where m = (4/3)πR3ρ is the mass of the ball.

## Moment of inertia in planar movement of a rigid body

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 Template:Mvar particles, Pi, i = 1,...,n, are assembled into a rigid body, then the momentum of the system can be written in terms of positions relative to a reference point R, and absolute velocities vi

$\Delta \mathbf {r} _{i}=\mathbf {r} _{i}-\mathbf {R} ,\quad \mathbf {v} _{i}={\boldsymbol {\omega }}\times (\mathbf {r} _{i}-\mathbf {R} )+\mathbf {V} ,$ where ω is the angular velocity of the system and V is the velocity of R.

For planar movement the angular velocity vector is directed along the unit vector Template:Mvar which is perpendicular to the plane of movement. Introduce the unit vectors ei from the reference point R to a point ri , and the unit vector ti = k × ei 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.

Note on the cross product: When a body moves parallel to a ground plane, the trajectories of all the points in the body lie in planes parallel to this ground plane. This means that any rotation that the body undergoes must be around an axis perpendicular to this plane. Planar movement is often presented as projected onto this ground plane so that the axis of rotation appears as a point. In this case, the angular velocity and angular acceleration of the body are scalars and the fact that they are vectors along the rotation axis is ignored. This is usually preferred for introductions to the topic. But in the case of moment of inertia, the combination of mass and geometry benefits from the geometric properties of the cross product. For this reason, in this section on planar movement the angular velocity and accelerations of the body are vectors perpendicular to the ground plane, and the cross product operations are the same as used for the study of spatial rigid body movement.

### Angular momentum in planar movement

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

{\begin{aligned}\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} .\\\end{aligned}} Use the centre 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 centre of mass IC as

$I_{C}=\sum _{i=1}^{n}m_{i}\Delta r_{i}^{2},$ then the equation for angular momentum simplifies to:1028

$\mathbf {L} =I_{C}\omega {\vec {k}}.$ The moment of inertia Template:Mvar about an axis perpendicular to the movement of the rigid system and through the centre 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 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 (International Library of Technology, 1906).

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

$E_{\text{K}}={\frac {1}{2}}\sum _{i=1}^{n}m_{i}\mathbf {v} _{i}\cdot \mathbf {v} _{i}={\frac {1}{2}}\sum _{i=1}^{n}m_{i}(\omega \Delta r_{i}\mathbf {t} _{i}+\mathbf {V} )\cdot (\omega \Delta r_{i}\mathbf {t} _{i}+\mathbf {V} ).$ This equation expands to yield three terms

$E_{\text{K}}={\frac {1}{2}}\omega ^{2}\sum _{i=1}^{n}m_{i}\Delta r_{i}^{2}(\mathbf {t} _{i}\cdot \mathbf {t} _{i})+\omega \mathbf {V} \cdot (\sum _{i=1}^{n}m_{i}\Delta r_{i}\mathbf {t} _{i})+{\frac {1}{2}}(\sum _{i=1}^{n}m_{i})\mathbf {V} \cdot \mathbf {V} .$ Let the reference point be the centre of mass C of the system so the second term becomes zero, and introduce the moment of inertia IC so the kinetic energy is given by:1084

$E_{\text{K}}={\frac {1}{2}}I_{C}\omega ^{2}+{\frac {1}{2}}M\mathbf {V} \cdot \mathbf {V} .$ The moment of inertia IC is the polar moment of inertia of the body.

### 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, Pi, i = 1,..., N, can be written in terms of a resultant force and torque at a reference point R, to yield

$\mathbf {F} =\sum _{i=1}^{N}m_{i}\mathbf {A} _{i},\quad {\boldsymbol {\tau }}=\sum _{i=1}^{N}(\mathbf {r} _{i}-\mathbf {R} )\times (m_{i}\mathbf {A} _{i}),$ where ri denotes the trajectory of each particle.

The kinematics of a rigid body yields the formula for the acceleration of the particle Template:Mvar in terms of the position R and acceleration A of the reference particle as well as the angular velocity vector Template:Mvar and angular acceleration vector Template:Mvar of the rigid system of particles as,

$\mathbf {A} _{i}={\boldsymbol {\alpha }}\times (\mathbf {r} _{i}-\mathbf {R} )+{\boldsymbol {\omega }}\times {\boldsymbol {\omega }}\times (\mathbf {r} _{i}-\mathbf {R} )+\mathbf {A} .$ 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 ei from the reference point R to a point ri and the unit vectors ti = k × ei , 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

${\boldsymbol {\tau }}=\sum _{i=1}^{N}(m_{i}\Delta r_{i}\mathbf {e} _{i})\times (\alpha (\Delta r_{i}\mathbf {t} _{i})-\omega ^{2}(\Delta r_{i}\mathbf {e} _{i})+\mathbf {A} )=(\sum _{i=1}^{N}m_{i}\Delta r_{i}^{2})\alpha {\vec {k}}+(\sum _{i=1}^{N}m_{i}\Delta r_{i}\mathbf {e} _{i})\times \mathbf {A} ,$ where ei × ei = 0, and ei × ti = k is the unit vector perpendicular to the plane for all of the particles Template:Mvar.

Use the centre of mass C as the reference point and define the moment of inertia relative to the centre of mass Template:Mvar, then the equation for the resultant torque simplifies to:1029

${\boldsymbol {\tau }}=I_{C}\alpha {\vec {k}}.$ The parameter Template:Mvar is the polar moment of inertia of the moving body.

## The inertia matrix for spatial movement of a rigid body

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.

An important application of the inertia matrix and Newton's laws of motion is the analysis of a spinning top. This is discussed in the article on gyroscopic precession. A more detailed presentation can be found in the article on Euler's equations of motion.

Let the system of particles Pi, i = 1,..., n be located at the coordinates ri with velocities vi relative to a fixed reference frame. For a (possibly moving) reference point R, the relative positions are

$\Delta \mathbf {r} _{i}=\mathbf {r} _{i}-\mathbf {R}$ and the (absolute) velocities are

$\mathbf {v} _{i}={\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}+\mathbf {V} _{R}$ where Template:Mvar is the angular velocity of the system, and VR is the velocity of R.

### Angular momentum

If the reference point R in the assembly, or body, is chosen as the centre of mass C, then its angular momentum takes the form,

$\mathbf {L} =\sum _{i=1}^{n}m_{i}\Delta \mathbf {r} _{i}\times \mathbf {v} _{i}=\sum _{i=1}^{n}m_{i}\Delta \mathbf {r} _{i}\times ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}),$ where the terms containing VR sum to zero by definition of the centre of mass.

To define the inertia matrix, introduce the skew-symmetric matrix [[[:Template:Mvar]]] constructed from a vector b that performs the cross product operation, such that

$[B]\mathbf {y} =\mathbf {b} \times \mathbf {y} .$ This matrix [[[:Template:Mvar]]] has the components of b = Template:Mvar as its elements, in the form

$[B]={\begin{bmatrix}0&-b_{z}&b_{y}\\b_{z}&0&-b_{x}\\-b_{y}&b_{x}&0\end{bmatrix}}.$ Now construct the skew-symmetric matrix [Δri]= [[[:Template:Mvar]]] obtained from the relative position vector Δri=ri - C, and use this skew-symmetric matrix to define,

$\mathbf {L} =(-\sum _{i=1}^{n}m_{i}[\Delta r_{i}]^{2}){\boldsymbol {\omega }}=[I_{C}]{\boldsymbol {\omega }},$ where [[[:Template:Mvar]]] defined by

$[I_{C}]=-\sum _{i=1}^{n}m_{i}[\Delta r_{i}]^{2},$ is the inertia matrix of the rigid system of particles measured relative to the centre of mass C.

### Kinetic energy

The kinetic energy of a rigid system of particles can be formulated in terms of the centre of mass and a matrix of mass moments of inertia of the system. Let the system of particles Pi, i = 1,...,n be located at the coordinates ri with velocities vi, then the kinetic energy is

$E_{\text{K}}={\frac {1}{2}}\sum _{i=1}^{n}m_{i}\mathbf {v} _{i}\cdot \mathbf {v} _{i}={\frac {1}{2}}\sum _{i=1}^{n}m_{i}({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}+\mathbf {V} _{C})\cdot ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}+\mathbf {V} _{C}),$ where Δri= ri-C is the position vector of a particle relative to the centre of mass.

This equation expands to yield three terms

$E_{\text{K}}={\frac {1}{2}}\sum _{i=1}^{n}m_{i}({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})\cdot ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}))+\sum _{i=1}^{n}m_{i}\mathbf {V} _{C}\cdot ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}))+{\frac {1}{2}}\sum _{i=1}^{n}m_{i}\mathbf {V} _{C}\cdot \mathbf {V} _{C}.$ The second term in this equation is zero because C is the centre of mass. Introduce the skew-symmetric matrix [Δri] so the kinetic energy becomes

$E_{\text{K}}={\frac {1}{2}}\sum _{i=1}^{n}m_{i}([\Delta r_{i}]{\boldsymbol {\omega }})\cdot ([\Delta r_{i}]{\boldsymbol {\omega }})+{\frac {1}{2}}(\sum _{i=1}^{n}m_{i})\mathbf {V} _{C}\cdot \mathbf {V} _{C}.$ $E_{\text{K}}={\frac {1}{2}}\sum _{i=1}^{n}m_{i}({\boldsymbol {\omega }}^{T}[\Delta r_{i}]^{T}[\Delta r_{i}]{\boldsymbol {\omega }})+{\frac {1}{2}}(\sum _{i=1}^{n}m_{i})\mathbf {V} _{C}\cdot \mathbf {V} _{C}.$ $E_{\text{K}}={\frac {1}{2}}{\boldsymbol {\omega }}\cdot (-\sum _{i=1}^{n}m_{i}[\Delta r_{i}]^{2}){\boldsymbol {\omega }}+{\frac {1}{2}}(\sum _{i=1}^{n}m_{i})\mathbf {V} _{C}\cdot \mathbf {V} _{C}.$ Thus, the kinetic energy of the rigid system of particles is given by

$E_{\text{K}}={\frac {1}{2}}{\boldsymbol {\omega }}\cdot [I_{C}]{\boldsymbol {\omega }}+{\frac {1}{2}}M\mathbf {V} _{C}^{2}.$ where [IC] is the inertia matrix relative to the centre 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,

${\boldsymbol {\tau }}=\sum _{i=1}^{n}(\mathbf {r_{i}} -\mathbf {R} )\times (m_{i}\mathbf {a} _{i}),$ where ai is the acceleration of the particle Pi. The kinematics of a rigid body yields the formula for the acceleration of the particle Pi 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,

$\mathbf {a} _{i}={\boldsymbol {\alpha }}\times (\mathbf {r} _{i}-\mathbf {R} )+{\boldsymbol {\omega }}\times {\boldsymbol {\omega }}\times (\mathbf {r} _{i}-\mathbf {R} )+\mathbf {A} _{R}.$ Use the centre of mass C as the reference point, and introduce the skew-symmetric matrix [Δri]=[ri-C] to represent the cross product (ri - C)x, to obtain

${\boldsymbol {\tau }}=\left(-\sum _{i=1}^{n}m_{i}[\Delta r_{i}]^{2}\right){\boldsymbol {\alpha }}+{\boldsymbol {\omega }}\times \left(-\sum _{i=1}^{n}m_{i}[\Delta r_{i}]^{2}\right){\boldsymbol {\omega }}$ The calculation uses the identity

$\Delta \mathbf {r} _{i}\times ({\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}))+{\boldsymbol {\omega }}\times (({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})\times \Delta \mathbf {r} _{i})=0,$ obtained from the Jacobi identity for the triple cross product as shown in the proof below:

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

${\boldsymbol {\tau }}=[I_{C}]{\boldsymbol {\alpha }}+{\boldsymbol {\omega }}\times [I_{C}]{\boldsymbol {\omega }},$ where [IC] is the inertia matrix relative to the centre of mass.

### Parallel axis theorem

{{#invoke:main|main}} The inertia matrix of a body depends on the choice of the reference point. There is a useful relationship between the inertia matrix relative to the centre of mass C and the inertia matrix relative to another point R. This relationship is called the parallel axis theorem.

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

$[I_{R}]=-\sum _{i=1}^{n}m_{i}[r_{i}-R]^{2}.$ Let C be the centre 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 centre of mass C to the reference point R. Use this equation to compute the inertia matrix,

$[I_{R}]=-\sum _{i=1}^{n}m_{i}[r_{i}-C-d]^{2}.$ Expand this equation to obtain