User:Prof McCarthy/momentofinertia

Calculating moment of inertia

Four objects racing down a plane while rolling without slipping. From back to front: spherical shell (red), solid sphere (orange), cylindrical ring (green) and solid cylinder (blue). The time for each object to reach the finishing line depends on their moment of inertia. (Details, Animated GIF version)

The moment of inertia of a body is calculated by summing mr² for every particle in the body about a given rotation axis. In order 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.

Consider the kinetic energy of an assembly of N masses m_i that lie at the distances r_i from the pivot point P, which is the sum of the kinetic energy of the individual masses,

$E_{\text{K}}=\sum _{i=1}^{N}{\frac {1}{2}}\,m_{i}\mathbf {v} _{i}\cdot \mathbf {v} _{i}=\sum _{i=1}^{N}{\frac {1}{2}}\,m_{i}(\omega r_{i})^{2}={\frac {1}{2}}\,\omega ^{2}\sum _{i=1}^{N}m_{i}r_{i}^{2}.$

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 ρ(r) is the mass density at each point r. The integration is evaluated over the volume V of the body.

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. A beam along the z-axis has stresses in the cross-section in the x-y plane that are calculated using the second moment of this area around either the x-axis or y-axis depending on the load. The moment of inertia of mass distributed along a body with the shape of this cross-section is the second moment of this area about the z-axis weighted by its density. The second moment of area around an axis perpendicular to the area is called the polar moment of the area, and is the sum of the second moments about the x and y axes.

Example calculation of moment of inertia

Main article: List of moments of inertia

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 centers of mass.

• The moment of inertia of a thin rod with constant cross-section s and density ρ and with length l about a perpendicular axis through its center of mass is determined by integration. Align the x-axis with the rod and locate the origin its center of mass at the center 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 an inertia of a thin disc of constant thickness s, radius R, and density ρ about an axis through its center and perpendicular to its face (parallel to its axis of rotational symmetry) is determined by integration. 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 center 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 center of mass. This is determined by summing the moment of inertias of the thin discs that form the sphere. 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,

$I_{C,{\text{ball}}}=\int _{-R}^{R}{\frac {\pi \rho }{2}}r(z)^{4}dz=\int _{-R}^{R}{\frac {\pi \rho }{2}}(R^{2}-z^{2})^{2}dz={\frac {\pi \rho }{2}}(R^{4}z-2R^{2}z^{3}/3+z^{5}/5){\bigg |}_{-R}^{R}$

$=\pi \rho (1-2/3+1/5)R^{5}={\frac {2}{5}}mR^{2},$

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 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 positions relative to a reference point R, and absolute velocities v_i

$\Delta \mathbf {r} _{i}=\mathbf {r} _{i}-\mathbf {R} ,\quad \mathbf {v} _{i}={\boldsymbol {\omega }}\times (\mathbf {r} _{i}-\mathbf {R} )+\mathbf {V} ,$

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 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 = k × e_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.

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

A figure skater can reduce her moment of inertia by pulling in her arms, allowing her to spin faster due to conservation of angular momentum.

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 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

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 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

$E_{\text{K}}={\frac {1}{2}}I_{C}\omega ^{2}+{\frac {1}{2}}M\mathbf {V} \cdot \mathbf {V} .$

The moment of inertia I_C 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, 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,

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

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 e_i from the reference point R to a point r_i and the unit vectors t_i = k × e_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

${\boldsymbol {\tau }}=\sum _{i=1}^{N}(m_{i}\Delta r_{i}\mathbf {e} _{i})\times (\alpha (\Delta r_{i}\mathbf {t} _{i})-\omega ^{2}(\Delta r_{i}\mathbf {e} _{i})+\mathbf {A} )=(\sum _{i=1}^{N}m_{i}\Delta r_{i}^{2})\alpha {\vec {k}}+(\sum _{i=1}^{N}m_{i}\Delta r_{i}\mathbf {e} _{i})\times \mathbf {A} ,$

where e_i × e_i = 0, and e_i × t_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

${\boldsymbol {\tau }}=I_{C}\alpha {\vec {k}}.$

The parameter I_C 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 P_i, i = 1,..., n be located at the coordinates r_i with velocities v_i 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 ω is the angular velocity of the system, and V_R is the velocity of R.

Angular momentum

If the reference point R in the assembly, or body, is chosen as the center 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 V_R sum to zero by definition of the center of mass.

In order to define 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

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

Now construct the skew-symmetric matrix [Δr_i]= [r_i-C] obtained from the relative position vector Δr_i=r_i - 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 [I_C ] 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 center of mass C.

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 with velocities v_i, 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 Δr_i= r_i-C is the position vector of a particle relative to the center 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 center of mass. Introduce the skew-symmetric matrix [Δr_i] so the kinetic energy becomes

$E_{\text{K}}={\frac {1}{2}}{\boldsymbol {\omega }}\cdot (-\sum _{i=1}^{n}m_{i}[\Delta r_{i}]^{2}){\boldsymbol {\omega }}+{\frac {1}{2}}(\sum _{i=1}^{n}m_{i})\mathbf {V} _{C}^{2}.$

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 [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,

${\boldsymbol {\tau }}=\sum _{i=1}^{n}(\mathbf {r_{i}} -\mathbf {R} )\times (m_{i}\mathbf {a} _{i}),$

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,

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

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

${\boldsymbol {\tau }}=(-\sum _{i=1}^{n}m_{i}[\Delta r_{i}]^{2})\times {\boldsymbol {\alpha }}+{\boldsymbol {\omega }}\times (-\sum _{i=1}^{n}m_{i}[\Delta r_{i}]^{2}){\boldsymbol {\omega }}.$

This calculation uses the identity

$\Delta \mathbf {r} _{i}\times ({\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}))+{\boldsymbol {\omega }}\times (\Delta \mathbf {r} _{i}\times (\Delta \mathbf {r} _{i}\times {\boldsymbol {\omega }}))=0,$

obtained from the Jacobi identity for the triple cross product.

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

${\boldsymbol {\tau }}=[I_{C}]{\boldsymbol {\alpha }}+{\boldsymbol {\omega }}\times [I_{C}]{\boldsymbol {\omega }},$

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

Parallel axis theorem

Main article: Parallel axis theorem

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 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

$[I_{R}]=-\sum _{i=1}^{n}m_{i}[r_{i}-R]^{2}.$

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,

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

Expand this equation to obtain

$[I_{R}]=(-\sum _{i=1}^{n}m_{i}[r_{i}-C]^{2})+(\sum _{i=1}^{n}m_{i}[r_{i}-C])[d]+[d](\sum _{i=1}^{n}m_{i}[r_{i}-C])+(-\sum _{i=1}^{n}m_{i})[d][d].$

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.

Note on the minus sign: By using the skew symmetric matrix of position vectors relative to the reference point, the inertia matrix of each particle has the form -m[r]², which is similar to the mr² that appears in planar movement. However, to make this to work out correctly a minus sign is needed. This minus sign can be absorbed into the term m[r]^T[r], if desired, by using the skew-symmetry property of [r].

The inertia matrix and the scalar moment of inertia around an arbitrary axis

The relationship between the inertia matrix of a rigid body and the scalar moment of inertia of the same body about a specified axis is important and rarely presented in detail. The following calculation expands the derivation presented by Kane and Levinson.

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

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

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.

In order to relate this scalar moment of inertia to the inertia matrix of the body, 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

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

The simplification of this equation uses the identity

$(\mathbf {S} \times (\mathbf {S} \times (\Delta \mathbf {r} _{i})))\cdot \mathbf {S} \times (\mathbf {S} \times (\Delta \mathbf {r} _{i}))=(\mathbf {S} \times (\mathbf {S} \times (\Delta \mathbf {r} _{i})))\times \mathbf {S} \cdot (\mathbf {S} \times (\Delta \mathbf {r} _{i})),$

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

$-(\Delta \mathbf {r} _{i})\times \mathbf {S} \cdot (\mathbf {S} \times (\Delta \mathbf {r} _{i})=-\mathbf {S} \cdot [\Delta r_{i}][\Delta r_{i}]\mathbf {S} ,$

where [Δr_i] is the skew symmetric matrix obtained from the vector Δr=r_i-R.

Thus, the moment of inertia around the line L through R in the direction S is obtained from the calculation

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

or

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

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

This shows that the inertia matrix can be used to calculate the moment of inertia of a body around any specified rotation axis in the body.and we also know that inertia can provide some moment of force

The inertia tensor

The inertia matrix is often described as the inertia tensor, which consists of the same moments of inertia and products of inertia about the three coordinate axes. The inertia tensor is constructed from the nine component tensors, (the symbol $\otimes$ is the tensor product)

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

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

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

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 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 inertia tensor is symmetric because I_ij= I_ji.

The inertia tensor can be used in the same way as the inertia matrix to compute the scalar moment of inertia about an arbitrary axis in the direction n,

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

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.

Identities for a skew-symmetric matrix

In order to compute moment of inertia of a mass around an axis, the perpendicular vector from the mass to the axis is needed. If the axis L is defined by the unit vector S through the reference point R, then the perpendicular vector from the line L to the point r is given by

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

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. Recall that skew-symmetric matrix [S] is constructed so that [S]y=S x y. The matrix [I-SS^T] in this equation subtracts the component of Δr=r-R that is parallel to S.

The previous sections show that in computing the moment of inertia matrix this operator yields a similar operator using the components of the vector Δr that is

$[I|\Delta \mathbf {r} |^{2}-\Delta \mathbf {r} \Delta \mathbf {r} ^{T}].$

It is helpful to keep the following identities in mind In order to compare the equations that define the inertia tensor and the inertia matrix.

Let [R] be the skew symmetric matrix associated with the position vector R=(x, y, z), then the product in the inertia matrix becomes

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

This can be viewed as another way of computing the perpendicular distance from an axis to a point, because the matrix formed by the outer product [R R^T] yields 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 identity 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.

The inertia matrix in different 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 eigendecomposition 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 Poinsot's ellipsoid. Let [Λ] be the inertia matrix relative to the center of mass aligned with the principal 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 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 the relationship presented above, between the inertia matrix and the scalar moment of inertia I_n around an axis in the direction n, yields

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

Thus, the magnitude of a point x in the direction n on the inertia ellipsoid is

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