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Moment of inertia

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(Redirected fromMoments of inertia)
Scalar measure of the rotational inertia with respect to a fixed axis of rotation
For the quantity also known as the "area moment of inertia", seeSecond moment of area.
Moment of inertia
Flywheels have large moments of inertia to smooth out changes in rates of rotational motion.
Common symbols
I
SI unitkg⋅m2
Other units
lb·ft·s2
Derivations from
other quantities
I=Lω{\displaystyle I={\frac {L}{\omega }}}
DimensionML2
Part of a series on
Classical mechanics
F=dpdt{\displaystyle {\textbf {F}}={\frac {d\mathbf {p} }{dt}}}
Tightrope walkers use the moment of inertia of a long rod for balance as they walk the rope. Samuel Dixon crossing theNiagara River in 1890.
To improve their maneuverability, combat aircraft are designed to minimize moments of inertia, while civil aircraft often are not.

Themoment of inertia, otherwise known as themass moment of inertia,angular/rotational mass,second moment of mass, or most accurately,rotational inertia, of arigid body is defined relatively to a rotational axis. It is the ratio between thetorque applied and the resultingangular acceleration about that axis.[1]: 279 [2]: 261  It plays the same role in rotational motion asmass does in linear motion. A body's moment of inertia about a particular axis depends both on the mass and its distribution relative to the axis, increasing with mass and distance from the axis.

It is anextensive (additive) property: for apoint mass the moment of inertia is simply the mass times the square of theperpendicular distance to the axis of rotation. The moment of inertia of a rigid composite system is the sum of the moments of inertia of its component subsystems (all taken about the same axis). Its simplest definition is the secondmoment of mass with respect to distance from anaxis.

For bodies forced to rotate in a plane, only their moment of inertia about an axis perpendicular to the plane, ascalar value, matters. For bodies free to rotate in all three dimensions, their moments can be described by asymmetric 3-by-3 matrix, with a set of mutually perpendicularprincipal axes for which this matrix isdiagonal and torques around the axes act independently of each other.

Introduction

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When a body is free to rotate around an axis,torque must be applied to change itsangular momentum. The amount of torque needed to cause any givenangular acceleration (the rate of change inangular velocity) is proportional to the moment of inertia of the body. Moments of inertia may be expressed in units of kilogram metre squared (kg·m2) inSI units and pound-foot-second squared (lbf·ft·s2) inimperial orUS units.

The moment of inertia plays the role in rotational kinetics thatmass (inertia) plays in linear kinetics—both characterize the resistance of a body to changes in its motion. The moment of inertia depends on how mass is distributed around an axis of rotation, and will vary depending on the chosen axis. For a point-like mass, the moment of inertia about some axis is given bymr2{\displaystyle mr^{2}}, wherer{\displaystyle r} is the distance of the point from the axis, andm{\displaystyle m} is the mass. For an extended rigid body, the moment of inertia is just the sum of all the small pieces of mass multiplied by the square of their distances from the axis in rotation. For an extended body of a regular shape and uniform density, this summation sometimes produces a simple expression that depends on the dimensions, shape and total mass of the object.

In 1673,Christiaan Huygens introduced this parameter in his study of the oscillation of a body hanging from a pivot, known as acompound pendulum.[3] The termmoment of inertia ("momentum inertiae" inLatin) was introduced byLeonhard Euler in his bookTheoria motus corporum solidorum seu rigidorum in 1765,[3][4] and it is incorporated intoEuler'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.[5][6]

The moment of inertia also appears inmomentum,kinetic energy, and inNewton'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.[7][8]

The moment of inertia of a rotatingflywheel 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 determine how steering forces on the control surfaces of its wings, elevators and rudder(s) affect the plane's motions in roll, pitch and yaw.

Definition

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Themoment of inertia is defined as the product of mass of section and the square of the distance between the reference axis and thecentroid of the section.

Spinning figure skaters can reduce their moment of inertia by pulling in their arms, allowing them to spin faster due toconservation of angular momentum.
Video of rotating chair experiment, illustrating moment of inertia. When the spinning professor pulls his arms, his moment of inertia decreases; to conserve angular momentum, his angular velocity increases.

The moment of inertiaI is also defined as the ratio of the netangular momentumL of a system to itsangular velocityω around a principal axis,[9][10] that isI=Lω.{\displaystyle I={\frac {L}{\omega }}.}

If the angular momentum of a system is constant, then as the moment of inertia gets smaller, the angular velocity must increase. This occurs when spinningfigure skaters pull in their outstretched arms ordivers curl their bodies into atuck position during a dive, to spin faster.[9][10][11][12][13][14]

If the shape of the body does not change, then its moment of inertia appears inNewton's law of motion as the ratio of anapplied torqueτ on a body to theangular accelerationα around a principal axis, that is[1]: 279 [2]: 261, eq.9-19 τ=Iα.{\displaystyle \tau =I\alpha .}

For asimple pendulum, this definition yields a formula for the moment of inertiaI in terms of the massm of the pendulum and its distancer from the pivot point as,I=mr2.{\displaystyle I=mr^{2}.}

Thus, the moment of inertia of the pendulum depends on both the massm of a body and its geometry, or shape, as defined by the distancer 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 massesdm each multiplied by the square of its perpendicular distancer to an axisk. An arbitrary object's moment of inertia thus depends on the spatial distribution of its mass.

In general, given an object of massm, an effective radiusk can be defined, dependent on a particular axis of rotation, with such a value that its moment of inertia around the axis isI=mk2,{\displaystyle I=mk^{2},}wherek is known as theradius of gyration around the axis.

Examples

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See also:List of moments of inertia

Simple pendulum

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Mathematically, the moment of inertia of a simple 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 particlem{\displaystyle m} with the square of its distancer{\displaystyle r} to the pivot, that isI=mr2.{\displaystyle I=mr^{2}.}

This can be shown as follows:

The force of gravity on the mass of a simple pendulum generates a torqueτ=r×F{\displaystyle {\boldsymbol {\tau }}=\mathbf {r} \times \mathbf {F} } around the axis perpendicular to the plane of the pendulum movement. Herer{\displaystyle \mathbf {r} } is the distance vector from the torque axis to the pendulum center of mass, andF{\displaystyle \mathbf {F} } is the net force on the mass. Associated with this torque is anangular acceleration,α{\displaystyle {\boldsymbol {\alpha }}}, of the string and mass around this axis. Since the mass is constrained to a circle the tangential acceleration of the mass isa=α×r{\displaystyle \mathbf {a} ={\boldsymbol {\alpha }}\times \mathbf {r} }. SinceF=ma{\displaystyle \mathbf {F} =m\mathbf {a} } the torque equation becomes:τ=r×F=r×(mα×r)=m((r⋅r)α−(r⋅α)r)=mr2α=Iαk^,{\displaystyle {\begin{aligned}{\boldsymbol {\tau }}&=\mathbf {r} \times \mathbf {F} =\mathbf {r} \times (m{\boldsymbol {\alpha }}\times \mathbf {r} )\\&=m\left(\left(\mathbf {r} \cdot \mathbf {r} \right){\boldsymbol {\alpha }}-\left(\mathbf {r} \cdot {\boldsymbol {\alpha }}\right)\mathbf {r} \right)\\&=mr^{2}{\boldsymbol {\alpha }}=I\alpha \mathbf {\hat {k}} ,\end{aligned}}}wherek^{\displaystyle \mathbf {\hat {k}} } is a unit vector perpendicular to the plane of the pendulum. (The second to last step uses thevector triple product expansion with the perpendicularity ofα{\displaystyle {\boldsymbol {\alpha }}} andr{\displaystyle \mathbf {r} }.) The quantityI=mr2{\displaystyle I=mr^{2}} is themoment of inertia of this single mass around the pivot point.

The quantityI=mr2{\displaystyle I=mr^{2}} also appears in theangular momentum of a simple pendulum, which is calculated from the velocityv=ω×r{\displaystyle \mathbf {v} ={\boldsymbol {\omega }}\times \mathbf {r} } of the pendulum mass around the pivot, whereω{\displaystyle {\boldsymbol {\omega }}} is theangular velocity of the mass about the pivot point. This angular momentum is given byL=r×p=r×(mω×r)=m((r⋅r)ω−(r⋅ω)r)=mr2ω=Iωk^,{\displaystyle {\begin{aligned}\mathbf {L} &=\mathbf {r} \times \mathbf {p} =\mathbf {r} \times \left(m{\boldsymbol {\omega }}\times \mathbf {r} \right)\\&=m\left(\left(\mathbf {r} \cdot \mathbf {r} \right){\boldsymbol {\omega }}-\left(\mathbf {r} \cdot {\boldsymbol {\omega }}\right)\mathbf {r} \right)\\&=mr^{2}{\boldsymbol {\omega }}=I\omega \mathbf {\hat {k}} ,\end{aligned}}}using a similar derivation to the previous equation.

Similarly, the kinetic energy of the pendulum mass is defined by the velocity of the pendulum around the pivot to yieldEK=12mv⋅v=12(mr2)ω2=12Iω2.{\displaystyle E_{\text{K}}={\frac {1}{2}}m\mathbf {v} \cdot \mathbf {v} ={\frac {1}{2}}\left(mr^{2}\right)\omega ^{2}={\frac {1}{2}}I\omega ^{2}.}

This shows that the quantityI=mr2{\displaystyle I=mr^{2}} 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 valuesmr2{\displaystyle mr^{2}} for all of the elements of mass in the body.

Compound pendulums

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Pendulums used in Mendenhallgravimeter 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.

Acompound pendulum is a body formed from an assembly of particles of continuous shape that rotates rigidly around a pivot. Its moment of inertia is the sum of the moments of inertia of each of the particles that it is composed of.[15][16]: 395–396 [17]: 51–53  Thenaturalfrequency (ωn{\displaystyle \omega _{\text{n}}}) of a compound pendulum depends on its moment of inertia,IP{\displaystyle I_{P}},ωn=mgrIP,{\displaystyle \omega _{\text{n}}={\sqrt {\frac {mgr}{I_{P}}}},}wherem{\displaystyle m} is the mass of the object,g{\displaystyle g} is local acceleration of gravity, andr{\displaystyle r} is the distance from the pivot point to the center 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.[18]: 516–517 

Thus, to determine the moment of inertia of the body, simply suspend it from a convenient pivot pointP{\displaystyle 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{\displaystyle t}), to obtainIP=mgrωn2=mgrt24π2,{\displaystyle I_{P}={\frac {mgr}{\omega _{\text{n}}^{2}}}={\frac {mgrt^{2}}{4\pi ^{2}}},}wheret{\displaystyle t} is the period (duration) of oscillation (usually averaged over multiple periods).

Center of oscillation

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A simple pendulum that has the same natural frequency as a compound pendulum defines the lengthL{\displaystyle L} from the pivot to a point called thecenter of oscillation of the compound pendulum. This point also corresponds to thecenter of percussion. The lengthL{\displaystyle L} is determined from the formula,ωn=gL=mgrIP,{\displaystyle \omega _{\text{n}}={\sqrt {\frac {g}{L}}}={\sqrt {\frac {mgr}{I_{P}}}},}orL=gωn2=IPmr.{\displaystyle L={\frac {g}{\omega _{\text{n}}^{2}}}={\frac {I_{P}}{mr}}.}

Theseconds 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 ofπ rad/s{\displaystyle \pi \ \mathrm {rad/s} } for the pendulum. In this case, the distance to the center of oscillation,L{\displaystyle L}, can be computed to beL=gωn2≈9.81 m/s2(3.14 rad/s)2≈0.99 m.{\displaystyle L={\frac {g}{\omega _{\text{n}}^{2}}}\approx {\frac {9.81\ \mathrm {m/s^{2}} }{(3.14\ \mathrm {rad/s} )^{2}}}\approx 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 a compound pendulum that uses this property to measure the local acceleration of gravity, and is called agravimeter.

Measuring moment of inertia

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The moment of inertia of a complex system such as a vehicle or airplane around its vertical axis can be measured by suspending the system from three points to form a trifilarpendulum. A trifilar pendulum is a platform supported by three wires designed to oscillate in torsion around its vertical centroidal axis.[19] The period of oscillation of the trifilar pendulum yields the moment of inertia of the system.[20]

Moment of inertia of area

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Moment of inertia of area is also known as thesecond moment of area and its physical meaning is completely different from the mass moment of inertia.

These calculations are commonly used in civil engineering for structural design ofbeams and columns. Cross-sectional areas calculated for vertical moment of the x-axisIxx{\displaystyle I_{xx}} and horizontal moment of the y-axisIyy{\displaystyle I_{yy}}.

Height (h) and breadth (b) are the linear measures, except for circles, which are effectively half-breadth derived,r{\displaystyle r}

Sectional areas moment calculated thus

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Source:[21]

  1. Square:Ixx=Iyy=b412{\displaystyle I_{xx}=I_{yy}={\frac {b^{4}}{12}}}
  2. Rectangular:Ixx=bh312{\displaystyle I_{xx}={\frac {bh^{3}}{12}}} and;Iyy=hb312{\displaystyle I_{yy}={\frac {hb^{3}}{12}}}
  3. Triangular:Ixx=bh336{\displaystyle I_{xx}={\frac {bh^{3}}{36}}}
  4. Circular:Ixx=Iyy=14πr4=164πd4{\displaystyle I_{xx}=I_{yy}={\frac {1}{4}}{\pi }r^{4}={\frac {1}{64}}{\pi }d^{4}}

Motion in a fixed plane

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Point mass

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Four objects with identical masses and radii rolling down a plane without slipping.
From back to front:
  •   spherical shell,
  •   solid sphere,
  •   cylindrical ring, and
  •   solid cylinder.
The time to reach the finishing line is longer for objects with a greater moment of inertia. (OGV version)

The moment of inertia about an axis of a body is calculated by summingmr2{\displaystyle mr^{2}} for every particle in the body, wherer{\displaystyle 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.[22])

Consider the kinetic energy of an assembly ofN{\displaystyle N} massesmi{\displaystyle m_{i}} that lie at the distancesri{\displaystyle r_{i}} from the pivot pointP{\displaystyle P}, which is the nearest point on the axis of rotation. It is the sum of the kinetic energy of the individual masses,[18]: 516–517 [23]: 1084–1085 [23]: 1296–1300 EK=∑i=1N12mivi⋅vi=∑i=1N12mi(ωri)2=12ω2∑i=1Nmiri2.{\displaystyle 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}\left(\omega r_{i}\right)^{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 themr2{\displaystyle mr^{2}} terms, that isIP=∑i=1Nmiri2.{\displaystyle 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, except with infinitely many point particles. Thus the limits of summation are removed, and the sum is written as follows:IP=∑imiri2{\displaystyle I_{P}=\sum _{i}m_{i}r_{i}^{2}}

Another expression replaces the summation with anintegral,IP=∭Qρ(x,y,z)‖r‖2dV{\displaystyle I_{P}=\iiint _{Q}\rho (x,y,z)\left\|\mathbf {r} \right\|^{2}dV}

Here, thefunctionρ{\displaystyle \rho } gives the mass density at each point(x,y,z){\displaystyle (x,y,z)},r{\displaystyle \mathbf {r} } is a vector perpendicular to the axis of rotation and extending from a point on the rotation axis to a point(x,y,z){\displaystyle (x,y,z)} in the solid, and the integration is evaluated over the volumeV{\displaystyle V} of the bodyQ{\displaystyle Q}. 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 thesecond moment of area of a beam's cross-section are often confused. The moment of inertia of a body with the shape of the cross-section is the second moment of this area about thez{\displaystyle z}-axis perpendicular to the cross-section, weighted by its density. This is also called thepolar moment of the area, and is the sum of the second moments about thex{\displaystyle x}- andy{\displaystyle y}-axes.[24] The stresses in abeam are calculated using the second moment of the cross-sectional area around either thex{\displaystyle x}-axis ory{\displaystyle y}-axis depending on the load.

Examples

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Main article:List of moments of inertia

The moment of inertia of acompound 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.[23]

Alist of moments of inertia formulas for standard body shapes provides a way to obtain the moment of inertia of a complex body as an assembly of simpler shaped bodies. Theparallel 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 moments of inertia of the thin discs that can form the sphere whose centers are along the axis chosen for consideration. If the surface of the sphere is defined by the equation[23]: 1301 x2+y2+z2=R2,{\displaystyle x^{2}+y^{2}+z^{2}=R^{2},}

then the square of the radiusr{\displaystyle r} of the disc at the cross-sectionz{\displaystyle z} along thez{\displaystyle z}-axis isr(z)2=x2+y2=R2−z2.{\displaystyle r(z)^{2}=x^{2}+y^{2}=R^{2}-z^{2}.}

Therefore, the moment of inertia of the sphere is the sum of the moments of inertia of the discs along thez{\displaystyle z}-axis,IC,sphere=∫−RR12πρr(z)4dz=∫−RR12πρ(R2−z2)2dz=12πρ[R4z−23R2z3+15z5]−RR=πρ(1−23+15)R5=25mR2,{\displaystyle {\begin{aligned}I_{C,{\text{sphere}}}&=\int _{-R}^{R}{\tfrac {1}{2}}\pi \rho r(z)^{4}\,dz=\int _{-R}^{R}{\tfrac {1}{2}}\pi \rho \left(R^{2}-z^{2}\right)^{2}\,dz\\[1ex]&={\tfrac {1}{2}}\pi \rho \left[R^{4}z-{\tfrac {2}{3}}R^{2}z^{3}+{\tfrac {1}{5}}z^{5}\right]_{-R}^{R}\\[1ex]&=\pi \rho \left(1-{\tfrac {2}{3}}+{\tfrac {1}{5}}\right)R^{5}\\[1ex]&={\tfrac {2}{5}}mR^{2},\end{aligned}}}wherem=43πR3ρ{\textstyle m={\frac {4}{3}}\pi R^{3}\rho } is the mass of the sphere.

Rigid body

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The cylinders with higher moment of inertia roll down a slope with a smaller acceleration, as more of their potential energy needs to be converted into the rotational kinetic energy.

If amechanical system is constrained to move parallel to a fixed plane, then the rotation of a body in the system occurs around an axisk^{\displaystyle \mathbf {\hat {k}} } parallel to this plane. In this case, the moment of inertia of the mass in this system is a scalar known as thepolar 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.[15][18][25][26]

If a system ofn{\displaystyle n} particles,Pi,i=1,…,n{\displaystyle P_{i},i=1,\dots ,n}, are assembled into a rigid body, then the momentum of the system can be written in terms of positions relative to a reference pointR{\displaystyle \mathbf {R} }, and absolute velocitiesvi{\displaystyle \mathbf {v} _{i}}:Δri=ri−R,vi=ω×(ri−R)+V=ω×Δri+V,{\displaystyle {\begin{aligned}\Delta \mathbf {r} _{i}&=\mathbf {r} _{i}-\mathbf {R} ,\\\mathbf {v} _{i}&={\boldsymbol {\omega }}\times \left(\mathbf {r} _{i}-\mathbf {R} \right)+\mathbf {V} ={\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}+\mathbf {V} ,\end{aligned}}}whereω{\displaystyle {\boldsymbol {\omega }}} is the angular velocity of the system andV{\displaystyle \mathbf {V} } is the velocity ofR{\displaystyle \mathbf {R} }.

For planar movement the angular velocity vector is directed along the unit vectork{\displaystyle \mathbf {k} } which is perpendicular to the plane of movement. Introduce the unit vectorsei{\displaystyle \mathbf {e} _{i}} from the reference pointR{\displaystyle \mathbf {R} } to a pointri{\displaystyle \mathbf {r} _{i}}, and the unit vectort^i=k^×e^i{\displaystyle \mathbf {\hat {t}} _{i}=\mathbf {\hat {k}} \times \mathbf {\hat {e}} _{i}}, soe^i=ΔriΔri,k^=ωω,t^i=k^×e^i,vi=ω×Δri+V=ωk^×Δrie^i+V=ωΔrit^i+V{\displaystyle {\begin{aligned}\mathbf {\hat {e}} _{i}&={\frac {\Delta \mathbf {r} _{i}}{\Delta r_{i}}},\quad \mathbf {\hat {k}} ={\frac {\boldsymbol {\omega }}{\omega }},\quad \mathbf {\hat {t}} _{i}=\mathbf {\hat {k}} \times \mathbf {\hat {e}} _{i},\\\mathbf {v} _{i}&={\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}+\mathbf {V} =\omega \mathbf {\hat {k}} \times \Delta r_{i}\mathbf {\hat {e}} _{i}+\mathbf {V} =\omega \,\Delta r_{i}\mathbf {\hat {t}} _{i}+\mathbf {V} \end{aligned}}}

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

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The angular momentum vector for the planar movement of a rigid system of particles is given by[15][18]L=∑i=1nmiΔri×vi=∑i=1nmiΔrie^i×(ωΔrit^i+V)=(∑i=1nmiΔri2)ωk^+(∑i=1nmiΔrie^i)×V.{\displaystyle {\begin{aligned}\mathbf {L} &=\sum _{i=1}^{n}m_{i}\Delta \mathbf {r} _{i}\times \mathbf {v} _{i}\\&=\sum _{i=1}^{n}m_{i}\,\Delta r_{i}\mathbf {\hat {e}} _{i}\times \left(\omega \,\Delta r_{i}\mathbf {\hat {t}} _{i}+\mathbf {V} \right)\\&=\left(\sum _{i=1}^{n}m_{i}\,\Delta r_{i}^{2}\right)\omega \mathbf {\hat {k}} +\left(\sum _{i=1}^{n}m_{i}\,\Delta r_{i}\mathbf {\hat {e}} _{i}\right)\times \mathbf {V} .\end{aligned}}}

Use thecenter of massC{\displaystyle \mathbf {C} } as the reference point soΔrie^i=ri−C,∑i=1nmiΔrie^i=0,{\displaystyle {\begin{aligned}\Delta r_{i}\mathbf {\hat {e}} _{i}&=\mathbf {r} _{i}-\mathbf {C} ,\\\sum _{i=1}^{n}m_{i}\,\Delta r_{i}\mathbf {\hat {e}} _{i}&=0,\end{aligned}}}

and define the moment of inertia relative to the center of massIC{\displaystyle I_{\mathbf {C} }} asIC=∑imiΔri2,{\displaystyle I_{\mathbf {C} }=\sum _{i}m_{i}\,\Delta r_{i}^{2},}

then the equation for angular momentum simplifies to[23]: 1028 L=ICωk^.{\displaystyle \mathbf {L} =I_{\mathbf {C} }\omega \mathbf {\hat {k}} .}

The moment of inertiaIC{\displaystyle I_{\mathbf {C} }} about an axis perpendicular to the movement of the rigid system and through the center of mass is known as thepolar moment of inertia. Specifically, it is thesecond moment of mass with respect to the orthogonal distance from an axis (or pole).

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. A figure skater is not, however, a rigid body.

Kinetic energy

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

The kinetic energy of a rigid system of particles moving in the plane is given by[15][18]EK=12∑i=1nmivi⋅vi,=12∑i=1nmi(ωΔrit^i+V)⋅(ωΔrit^i+V),=12ω2(∑i=1nmiΔri2t^i⋅t^i)+ωV⋅(∑i=1nmiΔrit^i)+12(∑i=1nmi)V⋅V.{\displaystyle {\begin{aligned}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}\left(\omega \,\Delta r_{i}\mathbf {\hat {t}} _{i}+\mathbf {V} \right)\cdot \left(\omega \,\Delta r_{i}\mathbf {\hat {t}} _{i}+\mathbf {V} \right),\\&={\frac {1}{2}}\omega ^{2}\left(\sum _{i=1}^{n}m_{i}\,\Delta r_{i}^{2}\mathbf {\hat {t}} _{i}\cdot \mathbf {\hat {t}} _{i}\right)+\omega \mathbf {V} \cdot \left(\sum _{i=1}^{n}m_{i}\,\Delta r_{i}\mathbf {\hat {t}} _{i}\right)+{\frac {1}{2}}\left(\sum _{i=1}^{n}m_{i}\right)\mathbf {V} \cdot \mathbf {V} .\end{aligned}}}

Let the reference point be the center of massC{\displaystyle \mathbf {C} } of the system so the second term becomes zero, and introduce the moment of inertiaIC{\displaystyle I_{\mathbf {C} }} so the kinetic energy is given by[23]: 1084 EK=12ICω2+12MV⋅V.{\displaystyle E_{\text{K}}={\frac {1}{2}}I_{\mathbf {C} }\omega ^{2}+{\frac {1}{2}}M\mathbf {V} \cdot \mathbf {V} .}

The moment of inertiaIC{\displaystyle I_{\mathbf {C} }} is thepolar moment of inertia of the body.

Newton's laws

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

Newton's laws for a rigid system ofn{\displaystyle n} particles,Pi,i=1,…,n{\displaystyle P_{i},i=1,\dots ,n}, can be written in terms of aresultant force and torque at a reference pointR{\displaystyle \mathbf {R} }, to yield[15][18]F=∑i=1nmiAi,τ=∑i=1nΔri×miAi,{\displaystyle {\begin{aligned}\mathbf {F} &=\sum _{i=1}^{n}m_{i}\mathbf {A} _{i},\\{\boldsymbol {\tau }}&=\sum _{i=1}^{n}\Delta \mathbf {r} _{i}\times m_{i}\mathbf {A} _{i},\end{aligned}}}whereri{\displaystyle \mathbf {r} _{i}} denotes the trajectory of each particle.

Thekinematics of a rigid body yields the formula for the acceleration of the particlePi{\displaystyle P_{i}} in terms of the positionR{\displaystyle \mathbf {R} } and accelerationA{\displaystyle \mathbf {A} } of the reference particle as well as the angular velocity vectorω{\displaystyle {\boldsymbol {\omega }}} and angular acceleration vectorα{\displaystyle {\boldsymbol {\alpha }}} of the rigid system of particles as,Ai=α×Δri+ω×ω×Δri+A.{\displaystyle \mathbf {A} _{i}={\boldsymbol {\alpha }}\times \Delta \mathbf {r} _{i}+{\boldsymbol {\omega }}\times {\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}+\mathbf {A} .}

For systems that are constrained to planar movement, the angular velocity and angular acceleration vectors are directed alongk^{\displaystyle \mathbf {\hat {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 vectorse^i{\displaystyle \mathbf {\hat {e}} _{i}} from the reference pointR{\displaystyle \mathbf {R} } to a pointri{\displaystyle \mathbf {r} _{i}} and the unit vectorst^i=k^×e^i{\displaystyle \mathbf {\hat {t}} _{i}=\mathbf {\hat {k}} \times \mathbf {\hat {e}} _{i}}, soAi=αk^×Δrie^i−ωk^×ωk^×Δrie^i+A=αΔrit^i−ω2Δrie^i+A.{\displaystyle {\begin{aligned}\mathbf {A} _{i}&=\alpha \mathbf {\hat {k}} \times \Delta r_{i}\mathbf {\hat {e}} _{i}-\omega \mathbf {\hat {k}} \times \omega \mathbf {\hat {k}} \times \Delta r_{i}\mathbf {\hat {e}} _{i}+\mathbf {A} \\&=\alpha \Delta r_{i}\mathbf {\hat {t}} _{i}-\omega ^{2}\Delta r_{i}\mathbf {\hat {e}} _{i}+\mathbf {A} .\end{aligned}}}

This yields the resultant torque on the system asτ=∑i=1nmiΔrie^i×(αΔrit^i−ω2Δrie^i+A)=(∑i=1nmiΔri2)αk^+(∑i=1nmiΔrie^i)×A,{\displaystyle {\begin{aligned}{\boldsymbol {\tau }}&=\sum _{i=1}^{n}m_{i}\,\Delta r_{i}\mathbf {\hat {e}} _{i}\times \left(\alpha \Delta r_{i}\mathbf {\hat {t}} _{i}-\omega ^{2}\Delta r_{i}\mathbf {\hat {e}} _{i}+\mathbf {A} \right)\\&=\left(\sum _{i=1}^{n}m_{i}\,\Delta r_{i}^{2}\right)\alpha \mathbf {\hat {k}} +\left(\sum _{i=1}^{n}m_{i}\,\Delta r_{i}\mathbf {\hat {e}} _{i}\right)\times \mathbf {A} ,\end{aligned}}}

wheree^i×e^i=0{\displaystyle \mathbf {\hat {e}} _{i}\times \mathbf {\hat {e}} _{i}=\mathbf {0} }, ande^i×t^i=k^{\displaystyle \mathbf {\hat {e}} _{i}\times \mathbf {\hat {t}} _{i}=\mathbf {\hat {k}} } is the unit vector perpendicular to the plane for all of the particlesPi{\displaystyle P_{i}}.

Use thecenter of massC{\displaystyle \mathbf {C} } as the reference point and define the moment of inertia relative to the center of massIC{\displaystyle I_{\mathbf {C} }}, then the equation for the resultant torque simplifies to[23]: 1029 τ=ICαk^.{\displaystyle {\boldsymbol {\tau }}=I_{\mathbf {C} }\alpha \mathbf {\hat {k}} .}

Motion in space of a rigid body, and the inertia matrix

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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.[5][6][7][8][27]

For analysis of a spinning top, seePrecession § Classical (Newtonian), andEuler's equations (rigid body dynamics).

Let the system ofn{\displaystyle n} particles,Pi,i=1,…,n{\displaystyle P_{i},i=1,\dots ,n} be located at the coordinatesri{\displaystyle \mathbf {r} _{i}} with velocitiesvi{\displaystyle \mathbf {v} _{i}} relative to a fixed reference frame. For a (possibly moving) reference pointR{\displaystyle \mathbf {R} }, the relative positions areΔri=ri−R{\displaystyle \Delta \mathbf {r} _{i}=\mathbf {r} _{i}-\mathbf {R} }and the (absolute) velocities arevi=ω×Δri+VR{\displaystyle \mathbf {v} _{i}={\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}+\mathbf {V} _{\mathbf {R} }}whereω{\displaystyle {\boldsymbol {\omega }}} is the angular velocity of the system, andVR{\displaystyle \mathbf {V_{R}} } is the velocity ofR{\displaystyle \mathbf {R} }.

Angular momentum

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Note that thecross product can be equivalently written as matrix multiplication by combining the first operand and the operator into a skew-symmetric matrix,[b]{\displaystyle \left[\mathbf {b} \right]}, constructed from the components ofb=(bx,by,bz){\displaystyle \mathbf {b} =(b_{x},b_{y},b_{z})}:b×y≡[b]y[b]≡[0−bzbybz0−bx−bybx0].{\displaystyle {\begin{aligned}\mathbf {b} \times \mathbf {y} &\equiv \left[\mathbf {b} \right]\mathbf {y} \\\left[\mathbf {b} \right]&\equiv {\begin{bmatrix}0&-b_{z}&b_{y}\\b_{z}&0&-b_{x}\\-b_{y}&b_{x}&0\end{bmatrix}}.\end{aligned}}}

The inertia matrix is constructed by considering the angular momentum, with the reference pointR{\displaystyle \mathbf {R} } of the body chosen to be the center of massC{\displaystyle \mathbf {C} }:[5][8]L=∑i=1nmiΔri×vi=∑i=1nmiΔri×(ω×Δri+VR)=(−∑i=1nmiΔri×(Δri×ω))+(∑i=1nmiΔri×VR),{\displaystyle {\begin{aligned}\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 \left({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}+\mathbf {V} _{\mathbf {R} }\right)\\&=\left(-\sum _{i=1}^{n}m_{i}\,\Delta \mathbf {r} _{i}\times \left(\Delta \mathbf {r} _{i}\times {\boldsymbol {\omega }}\right)\right)+\left(\sum _{i=1}^{n}m_{i}\,\Delta \mathbf {r} _{i}\times \mathbf {V} _{\mathbf {R} }\right),\end{aligned}}}where the terms containingVR{\displaystyle \mathbf {V_{R}} } (=C{\displaystyle =\mathbf {C} }) sum to zero by the definition ofcenter of mass.

Then, the skew-symmetric matrix[Δri]{\displaystyle [\Delta \mathbf {r} _{i}]} obtained from the relative position vectorΔri=ri−C{\displaystyle \Delta \mathbf {r} _{i}=\mathbf {r} _{i}-\mathbf {C} }, can be used to define,L=(−∑i=1nmi[Δri]2)ω=ICω,{\displaystyle \mathbf {L} =\left(-\sum _{i=1}^{n}m_{i}\left[\Delta \mathbf {r} _{i}\right]^{2}\right){\boldsymbol {\omega }}=\mathbf {I} _{\mathbf {C} }{\boldsymbol {\omega }},}whereIC{\displaystyle \mathbf {I_{C}} } defined byIC=−∑i=1nmi[Δri]2,{\displaystyle \mathbf {I} _{\mathbf {C} }=-\sum _{i=1}^{n}m_{i}\left[\Delta \mathbf {r} _{i}\right]^{2},}is the symmetric inertia matrix of the rigid system of particles measured relative to the center of massC{\displaystyle \mathbf {C} }.

Kinetic energy

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The kinetic energy of a rigid system of particles can be formulated in terms of thecenter of mass and a matrix of mass moments of inertia of the system. Let the system ofn{\displaystyle n} particlesPi,i=1,…,n{\displaystyle P_{i},i=1,\dots ,n} be located at the coordinatesri{\displaystyle \mathbf {r} _{i}} with velocitiesvi{\displaystyle \mathbf {v} _{i}}, then the kinetic energy is[5][8]EK=12∑i=1nmivi⋅vi=12∑i=1nmi(ω×Δri+VC)⋅(ω×Δri+VC),{\displaystyle 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}\left({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}+\mathbf {V} _{\mathbf {C} }\right)\cdot \left({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}+\mathbf {V} _{\mathbf {C} }\right),}whereΔri=ri−C{\displaystyle \Delta \mathbf {r} _{i}=\mathbf {r} _{i}-\mathbf {C} } is the position vector of a particle relative to the center of mass.

This equation expands to yield three termsEK=12(∑i=1nmi(ω×Δri)⋅(ω×Δri))+(∑i=1nmiVC⋅(ω×Δri))+12(∑i=1nmiVC⋅VC).{\displaystyle E_{\text{K}}={\frac {1}{2}}\left(\sum _{i=1}^{n}m_{i}\left({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}\right)\cdot \left({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}\right)\right)+\left(\sum _{i=1}^{n}m_{i}\mathbf {V} _{\mathbf {C} }\cdot \left({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}\right)\right)+{\frac {1}{2}}\left(\sum _{i=1}^{n}m_{i}\mathbf {V} _{\mathbf {C} }\cdot \mathbf {V} _{\mathbf {C} }\right).}

Since the center of mass is defined by∑i=1nmiΔri=0{\displaystyle \sum _{i=1}^{n}m_{i}\Delta \mathbf {r} _{i}=0}, the second term in this equation is zero. Introduce the skew-symmetric matrix[Δri]{\displaystyle [\Delta \mathbf {r} _{i}]} so the kinetic energy becomesEK=12(∑i=1nmi([Δri]ω)⋅([Δri]ω))+12(∑i=1nmi)VC⋅VC=12(∑i=1nmi(ωT[Δri]T[Δri]ω))+12(∑i=1nmi)VC⋅VC=12ω⋅(−∑i=1nmi[Δri]2)ω+12(∑i=1nmi)VC⋅VC.{\displaystyle {\begin{aligned}E_{\text{K}}&={\frac {1}{2}}\left(\sum _{i=1}^{n}m_{i}\left(\left[\Delta \mathbf {r} _{i}\right]{\boldsymbol {\omega }}\right)\cdot \left(\left[\Delta \mathbf {r} _{i}\right]{\boldsymbol {\omega }}\right)\right)+{\frac {1}{2}}\left(\sum _{i=1}^{n}m_{i}\right)\mathbf {V} _{\mathbf {C} }\cdot \mathbf {V} _{\mathbf {C} }\\&={\frac {1}{2}}\left(\sum _{i=1}^{n}m_{i}\left({\boldsymbol {\omega }}^{\mathsf {T}}\left[\Delta \mathbf {r} _{i}\right]^{\mathsf {T}}\left[\Delta \mathbf {r} _{i}\right]{\boldsymbol {\omega }}\right)\right)+{\frac {1}{2}}\left(\sum _{i=1}^{n}m_{i}\right)\mathbf {V} _{\mathbf {C} }\cdot \mathbf {V} _{\mathbf {C} }\\&={\frac {1}{2}}{\boldsymbol {\omega }}\cdot \left(-\sum _{i=1}^{n}m_{i}\left[\Delta \mathbf {r} _{i}\right]^{2}\right){\boldsymbol {\omega }}+{\frac {1}{2}}\left(\sum _{i=1}^{n}m_{i}\right)\mathbf {V} _{\mathbf {C} }\cdot \mathbf {V} _{\mathbf {C} }.\end{aligned}}}

Thus, the kinetic energy of the rigid system of particles is given byEK=12ω⋅ICω+12MVC2.{\displaystyle E_{\text{K}}={\frac {1}{2}}{\boldsymbol {\omega }}\cdot \mathbf {I} _{\mathbf {C} }{\boldsymbol {\omega }}+{\frac {1}{2}}M\mathbf {V} _{\mathbf {C} }^{2}.}whereIC{\displaystyle \mathbf {I_{C}} } is the inertia matrix relative to the center of mass andM{\displaystyle M} is the total mass.

Resultant torque

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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,[5][8]τ=∑i=1n(ri−R)×miai,{\displaystyle {\boldsymbol {\tau }}=\sum _{i=1}^{n}\left(\mathbf {r_{i}} -\mathbf {R} \right)\times m_{i}\mathbf {a} _{i},}whereai{\displaystyle \mathbf {a} _{i}} is the acceleration of the particlePi{\displaystyle P_{i}}. Thekinematics of a rigid body yields the formula for the acceleration of the particlePi{\displaystyle P_{i}} in terms of the positionR{\displaystyle \mathbf {R} } and accelerationAR{\displaystyle \mathbf {A} _{\mathbf {R} }} of the reference point, as well as the angular velocity vectorω{\displaystyle {\boldsymbol {\omega }}} and angular acceleration vectorα{\displaystyle {\boldsymbol {\alpha }}} of the rigid system as,ai=α×(ri−R)+ω×(ω×(ri−R))+AR.{\displaystyle \mathbf {a} _{i}={\boldsymbol {\alpha }}\times \left(\mathbf {r} _{i}-\mathbf {R} \right)+{\boldsymbol {\omega }}\times \left({\boldsymbol {\omega }}\times \left(\mathbf {r} _{i}-\mathbf {R} \right)\right)+\mathbf {A} _{\mathbf {R} }.}

Use the center of massC{\displaystyle \mathbf {C} } as the reference point, and introduce the skew-symmetric matrix[Δri]=[ri−C]{\displaystyle \left[\Delta \mathbf {r} _{i}\right]=\left[\mathbf {r} _{i}-\mathbf {C} \right]} to represent the cross product(ri−C)×{\displaystyle (\mathbf {r} _{i}-\mathbf {C} )\times }, to obtainτ=(−∑i=1nmi[Δri]2)α+ω×(−∑i=1nmi[Δri]2)ω{\displaystyle {\boldsymbol {\tau }}=\left(-\sum _{i=1}^{n}m_{i}\left[\Delta \mathbf {r} _{i}\right]^{2}\right){\boldsymbol {\alpha }}+{\boldsymbol {\omega }}\times \left(-\sum _{i=1}^{n}m_{i}\left[\Delta \mathbf {r} _{i}\right]^{2}\right){\boldsymbol {\omega }}}

The calculation uses the identityΔri×(ω×(ω×Δri))+ω×((ω×Δri)×Δri)=0,{\displaystyle \Delta \mathbf {r} _{i}\times \left({\boldsymbol {\omega }}\times \left({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}\right)\right)+{\boldsymbol {\omega }}\times \left(\left({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}\right)\times \Delta \mathbf {r} _{i}\right)=0,}obtained from theJacobi identity for the triplecross product as shown in the proof below:

Proof

τ=∑i=1n(ri−R)×(miai)=∑i=1nΔri×(miai)=∑i=1nmi[Δri×ai]… cross-product scalar multiplication=∑i=1nmi[Δri×(atangential,i+acentripetal,i+AR)]=∑i=1nmi[Δri×(atangential,i+acentripetal,i+0)]{\displaystyle {\begin{aligned}{\boldsymbol {\tau }}&=\sum _{i=1}^{n}(\mathbf {r_{i}} -\mathbf {R} )\times (m_{i}\mathbf {a} _{i})\\&=\sum _{i=1}^{n}\Delta \mathbf {r} _{i}\times (m_{i}\mathbf {a} _{i})\\&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times \mathbf {a} _{i}]\;\ldots {\text{ cross-product scalar multiplication}}\\&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times (\mathbf {a} _{{\text{tangential}},i}+\mathbf {a} _{{\text{centripetal}},i}+\mathbf {A} _{\mathbf {R} })]\\&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times (\mathbf {a} _{{\text{tangential}},i}+\mathbf {a} _{{\text{centripetal}},i}+0)]\\\end{aligned}}}In the last statement,AR=0{\displaystyle \mathbf {A} _{\mathbf {R} }=0} becauseR{\displaystyle \mathbf {R} } is either at rest or moving at a constant velocity but not accelerated, or the origin of the fixed (world) coordinate reference system is placed at the center of massC{\displaystyle \mathbf {C} }. And distributing the cross product over the sum, we getτ=∑i=1nmi[Δri×atangential,i+Δri×acentripetal,i]τ=∑i=1nmi[Δri×(α×Δri)+Δri×(ω×vtangential,i)]τ=∑i=1nmi[Δri×(α×Δri)+Δri×(ω×(ω×Δri))]{\displaystyle {\begin{aligned}{\boldsymbol {\tau }}&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times \mathbf {a} _{{\text{tangential}},i}+\Delta \mathbf {r} _{i}\times \mathbf {a} _{{\text{centripetal}},i}]\\{\boldsymbol {\tau }}&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times ({\boldsymbol {\alpha }}\times \Delta \mathbf {r} _{i})+\Delta \mathbf {r} _{i}\times ({\boldsymbol {\omega }}\times \mathbf {v} _{{\text{tangential}},i})]\\{\boldsymbol {\tau }}&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times ({\boldsymbol {\alpha }}\times \Delta \mathbf {r} _{i})+\Delta \mathbf {r} _{i}\times ({\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}))]\end{aligned}}}

Then, the followingJacobi identity is used on the last term:0=Δri×(ω×(ω×Δri))+ω×((ω×Δri)×Δri)+(ω×Δri)×(Δri×ω)=Δri×(ω×(ω×Δri))+ω×((ω×Δri)×Δri)+(ω×Δri)×−(ω×Δri)… cross-product anticommutativity=Δri×(ω×(ω×Δri))+ω×((ω×Δri)×Δri)+−[(ω×Δri)×(ω×Δri)]… cross-product scalar multiplication=Δri×(ω×(ω×Δri))+ω×((ω×Δri)×Δri)+−[0]… self cross-product0=Δri×(ω×(ω×Δri))+ω×((ω×Δri)×Δri){\displaystyle {\begin{aligned}0&=\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})+({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})\times (\Delta \mathbf {r} _{i}\times {\boldsymbol {\omega }})\\&=\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})+({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})\times -({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})\;\ldots {\text{ cross-product anticommutativity}}\\&=\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})+-[({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})\times ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})]\;\ldots {\text{ cross-product scalar multiplication}}\\&=\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]\;\ldots {\text{ self cross-product}}\\0&=\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})\end{aligned}}}

The result of applyingJacobi identity can then be continued as follows:Δri×(ω×(ω×Δri))=−[ω×((ω×Δri)×Δri)]=−[(ω×Δri)(ω⋅Δri)−Δri(ω⋅(ω×Δri))]… vector triple product=−[(ω×Δri)(ω⋅Δri)−Δri(Δri⋅(ω×ω))]… scalar triple product=−[(ω×Δri)(ω⋅Δri)−Δri(Δri⋅(0))]… self cross-product=−[(ω×Δri)(ω⋅Δri)]=−[ω×(Δri(ω⋅Δri))]… cross-product scalar multiplication=ω×−(Δri(ω⋅Δri))… cross-product scalar multiplicationΔri×(ω×(ω×Δri))=ω×−(Δri(Δri⋅ω))… dot-product commutativity{\displaystyle {\begin{aligned}\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})]\\&=-[({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})({\boldsymbol {\omega }}\cdot \Delta \mathbf {r} _{i})-\Delta \mathbf {r} _{i}({\boldsymbol {\omega }}\cdot ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}))]\;\ldots {\text{ vector triple product}}\\&=-[({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})({\boldsymbol {\omega }}\cdot \Delta \mathbf {r} _{i})-\Delta \mathbf {r} _{i}(\Delta \mathbf {r} _{i}\cdot ({\boldsymbol {\omega }}\times {\boldsymbol {\omega }}))]\;\ldots {\text{ scalar triple product}}\\&=-[({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})({\boldsymbol {\omega }}\cdot \Delta \mathbf {r} _{i})-\Delta \mathbf {r} _{i}(\Delta \mathbf {r} _{i}\cdot (0))]\;\ldots {\text{ self cross-product}}\\&=-[({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})({\boldsymbol {\omega }}\cdot \Delta \mathbf {r} _{i})]\\&=-[{\boldsymbol {\omega }}\times (\Delta \mathbf {r} _{i}({\boldsymbol {\omega }}\cdot \Delta \mathbf {r} _{i}))]\;\ldots {\text{ cross-product scalar multiplication}}\\&={\boldsymbol {\omega }}\times -(\Delta \mathbf {r} _{i}({\boldsymbol {\omega }}\cdot \Delta \mathbf {r} _{i}))\;\ldots {\text{ cross-product scalar multiplication}}\\\Delta \mathbf {r} _{i}\times ({\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}))&={\boldsymbol {\omega }}\times -(\Delta \mathbf {r} _{i}(\Delta \mathbf {r} _{i}\cdot {\boldsymbol {\omega }}))\;\ldots {\text{ dot-product commutativity}}\\\end{aligned}}}

The final result can then be substituted to the main proof as follows:τ=∑i=1nmi[Δri×(α×Δri)+Δri×(ω×(ω×Δri))]=∑i=1nmi[Δri×(α×Δri)+ω×−(Δri(Δri⋅ω))]=∑i=1nmi[Δri×(α×Δri)+ω×{0−Δri(Δri⋅ω)}]=∑i=1nmi[Δri×(α×Δri)+ω×{[ω(Δri⋅Δri)−ω(Δri⋅Δri)]−Δri(Δri⋅ω)}]…ω(Δri⋅Δri)−ω(Δri⋅Δri)=0=∑i=1nmi[Δri×(α×Δri)+ω×{[ω(Δri⋅Δri)−Δri(Δri⋅ω)]−ω(Δri⋅Δri)}]… addition associativity{\displaystyle {\begin{aligned}{\boldsymbol {\tau }}&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times ({\boldsymbol {\alpha }}\times \Delta \mathbf {r} _{i})+\Delta \mathbf {r} _{i}\times ({\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}))]\\&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times ({\boldsymbol {\alpha }}\times \Delta \mathbf {r} _{i})+{\boldsymbol {\omega }}\times -(\Delta \mathbf {r} _{i}(\Delta \mathbf {r} _{i}\cdot {\boldsymbol {\omega }}))]\\&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times ({\boldsymbol {\alpha }}\times \Delta \mathbf {r} _{i})+{\boldsymbol {\omega }}\times \{0-\Delta \mathbf {r} _{i}(\Delta \mathbf {r} _{i}\cdot {\boldsymbol {\omega }})\}]\\&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times ({\boldsymbol {\alpha }}\times \Delta \mathbf {r} _{i})+{\boldsymbol {\omega }}\times \{[{\boldsymbol {\omega }}(\Delta \mathbf {r} _{i}\cdot \Delta \mathbf {r} _{i})-{\boldsymbol {\omega }}(\Delta \mathbf {r} _{i}\cdot \Delta \mathbf {r} _{i})]-\Delta \mathbf {r} _{i}(\Delta \mathbf {r} _{i}\cdot {\boldsymbol {\omega }})\}]\;\ldots \;{\boldsymbol {\omega }}(\Delta \mathbf {r} _{i}\cdot \Delta \mathbf {r} _{i})-{\boldsymbol {\omega }}(\Delta \mathbf {r} _{i}\cdot \Delta \mathbf {r} _{i})=0\\&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times ({\boldsymbol {\alpha }}\times \Delta \mathbf {r} _{i})+{\boldsymbol {\omega }}\times \{[{\boldsymbol {\omega }}(\Delta \mathbf {r} _{i}\cdot \Delta \mathbf {r} _{i})-\Delta \mathbf {r} _{i}(\Delta \mathbf {r} _{i}\cdot {\boldsymbol {\omega }})]-{\boldsymbol {\omega }}(\Delta \mathbf {r} _{i}\cdot \Delta \mathbf {r} _{i})\}]\;\ldots {\text{ addition associativity}}\\\end{aligned}}}τ=∑i=1nmi[Δri×(α×Δri)+ω×{ω(Δri⋅Δri)−Δri(Δri⋅ω)}−ω×ω(Δri⋅Δri)]… cross-product distributivity over addition=∑i=1nmi[Δri×(α×Δri)+ω×{ω(Δri⋅Δri)−Δri(Δri⋅ω)}−(Δri⋅Δri)(ω×ω)]… cross-product scalar multiplication=∑i=1nmi[Δri×(α×Δri)+ω×{ω(Δri⋅Δri)−Δri(Δri⋅ω)}−(Δri⋅Δri)(0)]… self cross-product=∑i=1nmi[Δri×(α×Δri)+ω×{ω(Δri⋅Δri)−Δri(Δri⋅ω)}]=∑i=1nmi[Δri×(α×Δri)+ω×{Δri×(ω×Δri)}]… vector triple product=∑i=1nmi[Δri×−(Δri×α)+ω×{Δri×−(Δri×ω)}]… cross-product anticommutativity=−∑i=1nmi[Δri×(Δri×α)+ω×{Δri×(Δri×ω)}]… cross-product scalar multiplication=−∑i=1nmi[Δri×(Δri×α)]+−∑i=1nmi[ω×{Δri×(Δri×ω)}]… summation distributivityτ=−∑i=1nmi[Δri×(Δri×α)]+ω×−∑i=1nmi[Δri×(Δri×ω)]…ω is not characteristic of particle Pi{\displaystyle {\begin{aligned}{\boldsymbol {\tau }}&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times ({\boldsymbol {\alpha }}\times \Delta \mathbf {r} _{i})+{\boldsymbol {\omega }}\times \{{\boldsymbol {\omega }}(\Delta \mathbf {r} _{i}\cdot \Delta \mathbf {r} _{i})-\Delta \mathbf {r} _{i}(\Delta \mathbf {r} _{i}\cdot {\boldsymbol {\omega }})\}-{\boldsymbol {\omega }}\times {\boldsymbol {\omega }}(\Delta \mathbf {r} _{i}\cdot \Delta \mathbf {r} _{i})]\;\ldots {\text{ cross-product distributivity over addition}}\\&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times ({\boldsymbol {\alpha }}\times \Delta \mathbf {r} _{i})+{\boldsymbol {\omega }}\times \{{\boldsymbol {\omega }}(\Delta \mathbf {r} _{i}\cdot \Delta \mathbf {r} _{i})-\Delta \mathbf {r} _{i}(\Delta \mathbf {r} _{i}\cdot {\boldsymbol {\omega }})\}-(\Delta \mathbf {r} _{i}\cdot \Delta \mathbf {r} _{i})({\boldsymbol {\omega }}\times {\boldsymbol {\omega }})]\;\ldots {\text{ cross-product scalar multiplication}}\\&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times ({\boldsymbol {\alpha }}\times \Delta \mathbf {r} _{i})+{\boldsymbol {\omega }}\times \{{\boldsymbol {\omega }}(\Delta \mathbf {r} _{i}\cdot \Delta \mathbf {r} _{i})-\Delta \mathbf {r} _{i}(\Delta \mathbf {r} _{i}\cdot {\boldsymbol {\omega }})\}-(\Delta \mathbf {r} _{i}\cdot \Delta \mathbf {r} _{i})(0)]\;\ldots {\text{ self cross-product}}\\&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times ({\boldsymbol {\alpha }}\times \Delta \mathbf {r} _{i})+{\boldsymbol {\omega }}\times \{{\boldsymbol {\omega }}(\Delta \mathbf {r} _{i}\cdot \Delta \mathbf {r} _{i})-\Delta \mathbf {r} _{i}(\Delta \mathbf {r} _{i}\cdot {\boldsymbol {\omega }})\}]\\&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times ({\boldsymbol {\alpha }}\times \Delta \mathbf {r} _{i})+{\boldsymbol {\omega }}\times \{\Delta \mathbf {r} _{i}\times ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})\}]\;\ldots {\text{ vector triple product}}\\&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times -(\Delta \mathbf {r} _{i}\times {\boldsymbol {\alpha }})+{\boldsymbol {\omega }}\times \{\Delta \mathbf {r} _{i}\times -(\Delta \mathbf {r} _{i}\times {\boldsymbol {\omega }})\}]\;\ldots {\text{ cross-product anticommutativity}}\\&=-\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times (\Delta \mathbf {r} _{i}\times {\boldsymbol {\alpha }})+{\boldsymbol {\omega }}\times \{\Delta \mathbf {r} _{i}\times (\Delta \mathbf {r} _{i}\times {\boldsymbol {\omega }})\}]\;\ldots {\text{ cross-product scalar multiplication}}\\&=-\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times (\Delta \mathbf {r} _{i}\times {\boldsymbol {\alpha }})]+-\sum _{i=1}^{n}m_{i}[{\boldsymbol {\omega }}\times \{\Delta \mathbf {r} _{i}\times (\Delta \mathbf {r} _{i}\times {\boldsymbol {\omega }})\}]\;\ldots {\text{ summation distributivity}}\\{\boldsymbol {\tau }}&=-\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times (\Delta \mathbf {r} _{i}\times {\boldsymbol {\alpha }})]+{\boldsymbol {\omega }}\times -\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times (\Delta \mathbf {r} _{i}\times {\boldsymbol {\omega }})]\;\ldots \;{\boldsymbol {\omega }}{\text{ is not characteristic of particle }}P_{i}\end{aligned}}}

Notice that for any vectoru{\displaystyle \mathbf {u} }, the following holds:−∑i=1nmi[Δri×(Δri×u)]=−∑i=1nmi([0−Δr3,iΔr2,iΔr3,i0−Δr1,i−Δr2,iΔr1,i0]([0−Δr3,iΔr2,iΔr3,i0−Δr1,i−Δr2,iΔr1,i0][u1u2u3]))… cross-product as matrix multiplication=−∑i=1nmi([0−Δr3,iΔr2,iΔr3,i0−Δr1,i−Δr2,iΔr1,i0][−Δr3,iu2+Δr2,iu3+Δr3,iu1−Δr1,iu3−Δr2,iu1+Δr1,iu2])=−∑i=1nmi[−Δr3,i(+Δr3,iu1−Δr1,iu3)+Δr2,i(−Δr2,iu1+Δr1,iu2)+Δr3,i(−Δr3,iu2+Δr2,iu3)−Δr1,i(−Δr2,iu1+Δr1,iu2)−Δr2,i(−Δr3,iu2+Δr2,iu3)+Δr1,i(+Δr3,iu1−Δr1,iu3)]=−∑i=1nmi[−Δr3,i2u1+Δr1,iΔr3,iu3−Δr2,i2u1+Δr1,iΔr2,iu2−Δr3,i2u2+Δr2,iΔr3,iu3+Δr2,iΔr1,iu1−Δr1,i2u2+Δr3,iΔr2,iu2−Δr2,i2u3+Δr3,iΔr1,iu1−Δr1,i2u3]=−∑i=1nmi[−(Δr2,i2+Δr3,i2)u1+Δr1,iΔr2,iu2+Δr1,iΔr3,iu3+Δr2,iΔr1,iu1−(Δr1,i2+Δr3,i2)u2+Δr2,iΔr3,iu3+Δr3,iΔr1,iu1+Δr3,iΔr2,iu2−(Δr1,i2+Δr2,i2)u3]=−∑i=1nmi[−(Δr2,i2+Δr3,i2)Δr1,iΔr2,iΔr1,iΔr3,iΔr2,iΔr1,i−(Δr1,i2+Δr3,i2)Δr2,iΔr3,iΔr3,iΔr1,iΔr3,iΔr2,i−(Δr1,i2+Δr2,i2)][u1u2u3]=−∑i=1nmi[Δri]2u−∑i=1nmi[Δri×(Δri×u)]=(−∑i=1nmi[Δri]2)u…u is not characteristic of Pi{\displaystyle {\begin{aligned}-\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times (\Delta \mathbf {r} _{i}\times \mathbf {u} )]&=-\sum _{i=1}^{n}m_{i}\left({\begin{bmatrix}0&-\Delta r_{3,i}&\Delta r_{2,i}\\\Delta r_{3,i}&0&-\Delta r_{1,i}\\-\Delta r_{2,i}&\Delta r_{1,i}&0\end{bmatrix}}\left({\begin{bmatrix}0&-\Delta r_{3,i}&\Delta r_{2,i}\\\Delta r_{3,i}&0&-\Delta r_{1,i}\\-\Delta r_{2,i}&\Delta r_{1,i}&0\end{bmatrix}}{\begin{bmatrix}u_{1}\\u_{2}\\u_{3}\end{bmatrix}}\right)\right)\;\ldots {\text{ cross-product as matrix multiplication}}\\[6pt]&=-\sum _{i=1}^{n}m_{i}\left({\begin{bmatrix}0&-\Delta r_{3,i}&\Delta r_{2,i}\\\Delta r_{3,i}&0&-\Delta r_{1,i}\\-\Delta r_{2,i}&\Delta r_{1,i}&0\end{bmatrix}}{\begin{bmatrix}-\Delta r_{3,i}\,u_{2}+\Delta r_{2,i}\,u_{3}\\+\Delta r_{3,i}\,u_{1}-\Delta r_{1,i}\,u_{3}\\-\Delta r_{2,i}\,u_{1}+\Delta r_{1,i}\,u_{2}\end{bmatrix}}\right)\\[6pt]&=-\sum _{i=1}^{n}m_{i}{\begin{bmatrix}-\Delta r_{3,i}(+\Delta r_{3,i}\,u_{1}-\Delta r_{1,i}\,u_{3})+\Delta r_{2,i}(-\Delta r_{2,i}\,u_{1}+\Delta r_{1,i}\,u_{2})\\+\Delta r_{3,i}(-\Delta r_{3,i}\,u_{2}+\Delta r_{2,i}\,u_{3})-\Delta r_{1,i}(-\Delta r_{2,i}\,u_{1}+\Delta r_{1,i}\,u_{2})\\-\Delta r_{2,i}(-\Delta r_{3,i}\,u_{2}+\Delta r_{2,i}\,u_{3})+\Delta r_{1,i}(+\Delta r_{3,i}\,u_{1}-\Delta r_{1,i}\,u_{3})\end{bmatrix}}\\[6pt]&=-\sum _{i=1}^{n}m_{i}{\begin{bmatrix}-\Delta r_{3,i}^{2}\,u_{1}+\Delta r_{1,i}\Delta r_{3,i}\,u_{3}-\Delta r_{2,i}^{2}\,u_{1}+\Delta r_{1,i}\Delta r_{2,i}\,u_{2}\\-\Delta r_{3,i}^{2}\,u_{2}+\Delta r_{2,i}\Delta r_{3,i}\,u_{3}+\Delta r_{2,i}\Delta r_{1,i}\,u_{1}-\Delta r_{1,i}^{2}\,u_{2}\\+\Delta r_{3,i}\Delta r_{2,i}\,u_{2}-\Delta r_{2,i}^{2}\,u_{3}+\Delta r_{3,i}\Delta r_{1,i}\,u_{1}-\Delta r_{1,i}^{2}\,u_{3}\end{bmatrix}}\\[6pt]&=-\sum _{i=1}^{n}m_{i}{\begin{bmatrix}-(\Delta r_{2,i}^{2}+\Delta r_{3,i}^{2})\,u_{1}+\Delta r_{1,i}\Delta r_{2,i}\,u_{2}+\Delta r_{1,i}\Delta r_{3,i}\,u_{3}\\+\Delta r_{2,i}\Delta r_{1,i}\,u_{1}-(\Delta r_{1,i}^{2}+\Delta r_{3,i}^{2})\,u_{2}+\Delta r_{2,i}\Delta r_{3,i}\,u_{3}\\+\Delta r_{3,i}\Delta r_{1,i}\,u_{1}+\Delta r_{3,i}\Delta r_{2,i}\,u_{2}-(\Delta r_{1,i}^{2}+\Delta r_{2,i}^{2})\,u_{3}\end{bmatrix}}\\[6pt]&=-\sum _{i=1}^{n}m_{i}{\begin{bmatrix}-(\Delta r_{2,i}^{2}+\Delta r_{3,i}^{2})&\Delta r_{1,i}\Delta r_{2,i}&\Delta r_{1,i}\Delta r_{3,i}\\\Delta r_{2,i}\Delta r_{1,i}&-(\Delta r_{1,i}^{2}+\Delta r_{3,i}^{2})&\Delta r_{2,i}\Delta r_{3,i}\\\Delta r_{3,i}\Delta r_{1,i}&\Delta r_{3,i}\Delta r_{2,i}&-(\Delta r_{1,i}^{2}+\Delta r_{2,i}^{2})\end{bmatrix}}{\begin{bmatrix}u_{1}\\u_{2}\\u_{3}\end{bmatrix}}\\&=-\sum _{i=1}^{n}m_{i}[\Delta r_{i}]^{2}\mathbf {u} \\[6pt]-\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times (\Delta \mathbf {r} _{i}\times \mathbf {u} )]&=\left(-\sum _{i=1}^{n}m_{i}[\Delta r_{i}]^{2}\right)\mathbf {u} \;\ldots \;\mathbf {u} {\text{ is not characteristic of }}P_{i}\end{aligned}}}

Finally, the result is used to complete the main proof as follows:τ=−∑i=1nmi[Δri×(Δri×α)]+ω×−∑i=1nmiΔri×(Δri×ω)]=(−∑i=1nmi[Δri]2)α+ω×(−∑i=1nmi[Δri]2)ω{\displaystyle {\begin{aligned}{\boldsymbol {\tau }}&=-\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times (\Delta \mathbf {r} _{i}\times {\boldsymbol {\alpha }})]+{\boldsymbol {\omega }}\times -\sum _{i=1}^{n}m_{i}\Delta \mathbf {r} _{i}\times (\Delta \mathbf {r} _{i}\times {\boldsymbol {\omega }})]\\&=\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 }}\end{aligned}}}

Thus, the resultant torque on the rigid system of particles is given byτ=ICα+ω×ICω,{\displaystyle {\boldsymbol {\tau }}=\mathbf {I} _{\mathbf {C} }{\boldsymbol {\alpha }}+{\boldsymbol {\omega }}\times \mathbf {I} _{\mathbf {C} }{\boldsymbol {\omega }},}whereIC{\displaystyle \mathbf {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 massC{\displaystyle \mathbf {C} } and the inertia matrix relative to another pointR{\displaystyle \mathbf {R} }. This relationship is called the parallel axis theorem.[5][8]

Consider the inertia matrixIR{\displaystyle \mathbf {I_{R}} } obtained for a rigid system of particles measured relative to a reference pointR{\displaystyle \mathbf {R} }, given byIR=−∑i=1nmi[ri−R]2.{\displaystyle \mathbf {I} _{\mathbf {R} }=-\sum _{i=1}^{n}m_{i}\left[\mathbf {r} _{i}-\mathbf {R} \right]^{2}.}

LetC{\displaystyle \mathbf {C} } be the center of mass of the rigid system, thenR=(R−C)+C=d+C,{\displaystyle \mathbf {R} =(\mathbf {R} -\mathbf {C} )+\mathbf {C} =\mathbf {d} +\mathbf {C} ,}whered{\displaystyle \mathbf {d} } is the vector from the center of massC{\displaystyle \mathbf {C} } to the reference pointR{\displaystyle \mathbf {R} }. Use this equation to compute the inertia matrix,IR=−∑i=1nmi[ri−(C+d)]2=−∑i=1nmi[(ri−C)−d]2.{\displaystyle \mathbf {I} _{\mathbf {R} }=-\sum _{i=1}^{n}m_{i}[\mathbf {r} _{i}-\left(\mathbf {C} +\mathbf {d} \right)]^{2}=-\sum _{i=1}^{n}m_{i}[\left(\mathbf {r} _{i}-\mathbf {C} \right)-\mathbf {d} ]^{2}.}

Distribute over the cross product to obtainIR=−(∑i=1nmi[ri−C]2)+(∑i=1nmi[ri−C])[d]+[d](∑i=1nmi[ri−C])−(∑i=1nmi)[d]2.{\displaystyle \mathbf {I} _{\mathbf {R} }=-\left(\sum _{i=1}^{n}m_{i}[\mathbf {r} _{i}-\mathbf {C} ]^{2}\right)+\left(\sum _{i=1}^{n}m_{i}[\mathbf {r} _{i}-\mathbf {C} ]\right)[\mathbf {d} ]+[\mathbf {d} ]\left(\sum _{i=1}^{n}m_{i}[\mathbf {r} _{i}-\mathbf {C} ]\right)-\left(\sum _{i=1}^{n}m_{i}\right)[\mathbf {d} ]^{2}.}

The first term is the inertia matrixIC{\displaystyle \mathbf {I_{C}} } relative to the center of mass. The second and third terms are zero by definition of the center of massC{\displaystyle \mathbf {C} }. And the last term is the total mass of the system multiplied by the square of the skew-symmetric matrix[d]{\displaystyle [\mathbf {d} ]} constructed fromd{\displaystyle \mathbf {d} }.

The result is the parallel axis theorem,IR=IC−M[d]2,{\displaystyle \mathbf {I} _{\mathbf {R} }=\mathbf {I} _{\mathbf {C} }-M[\mathbf {d} ]^{2},}whered{\displaystyle \mathbf {d} } is the vector from the center of massC{\displaystyle \mathbf {C} } to the reference pointR{\displaystyle \mathbf {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{\displaystyle -m\left[\mathbf {r} \right]^{2}}, which is similar to themr2{\displaystyle mr^{2}} 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 termm[r]T[r]{\displaystyle m\left[\mathbf {r} \right]^{\mathsf {T}}\left[\mathbf {r} \right]}, if desired, by using the skew-symmetry property of[r]{\displaystyle [\mathbf {r} ]}.

Scalar moment of inertia in a plane

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The scalar moment of inertia,IL{\displaystyle I_{L}}, of a body about a specified axis whose direction is specified by the unit vectork^{\displaystyle \mathbf {\hat {k}} } and passes through the body at a pointR{\displaystyle \mathbf {R} } is as follows:[8]IL=k^⋅(−∑i=1Nmi[Δri]2)k^=k^⋅IRk^=k^TIRk^,{\displaystyle I_{L}=\mathbf {\hat {k}} \cdot \left(-\sum _{i=1}^{N}m_{i}\left[\Delta \mathbf {r} _{i}\right]^{2}\right)\mathbf {\hat {k}} =\mathbf {\hat {k}} \cdot \mathbf {I} _{\mathbf {R} }\mathbf {\hat {k}} =\mathbf {\hat {k}} ^{\mathsf {T}}\mathbf {I} _{\mathbf {R} }\mathbf {\hat {k}} ,}whereIR{\displaystyle \mathbf {I_{R}} } is the moment of inertia matrix of the system relative to the reference pointR{\displaystyle \mathbf {R} }, and[Δri]{\displaystyle [\Delta \mathbf {r} _{i}]} is the skew symmetric matrix obtained from the vectorΔri=ri−R{\displaystyle \Delta \mathbf {r} _{i}=\mathbf {r} _{i}-\mathbf {R} }.

This is derived as follows. Let a rigid assembly ofn{\displaystyle n} particles,Pi,i=1,…,n{\displaystyle P_{i},i=1,\dots ,n}, have coordinatesri{\displaystyle \mathbf {r} _{i}}. ChooseR{\displaystyle \mathbf {R} } as a reference point and compute the moment of inertia around a line L defined by the unit vectork^{\displaystyle \mathbf {\hat {k}} } through the reference pointR{\displaystyle \mathbf {R} },L(t)=R+tk^{\displaystyle \mathbf {L} (t)=\mathbf {R} +t\mathbf {\hat {k}} }. The perpendicular vector from this line to the particlePi{\displaystyle P_{i}} is obtained fromΔri{\displaystyle \Delta \mathbf {r} _{i}} by removing the component that projects ontok^{\displaystyle \mathbf {\hat {k}} }.Δri⊥=Δri−(k^⋅Δri)k^=(E−k^k^T)Δri,{\displaystyle \Delta \mathbf {r} _{i}^{\perp }=\Delta \mathbf {r} _{i}-\left(\mathbf {\hat {k}} \cdot \Delta \mathbf {r} _{i}\right)\mathbf {\hat {k}} =\left(\mathbf {E} -\mathbf {\hat {k}} \mathbf {\hat {k}} ^{\mathsf {T}}\right)\Delta \mathbf {r} _{i},}whereE{\displaystyle \mathbf {E} } is the identity matrix, so as to avoid confusion with the inertia matrix, andk^k^T{\displaystyle \mathbf {\hat {k}} \mathbf {\hat {k}} ^{\mathsf {T}}} is the outer product matrix formed from the unit vectork^{\displaystyle \mathbf {\hat {k}} } along the lineL{\displaystyle L}.

To relate this scalar moment of inertia to the inertia matrix of the body, introduce the skew-symmetric matrix[k^]{\displaystyle \left[\mathbf {\hat {k}} \right]} such that[k^]y=k^×y{\displaystyle \left[\mathbf {\hat {k}} \right]\mathbf {y} =\mathbf {\hat {k}} \times \mathbf {y} }, then we have the identity−[k^]2≡|k^|2(E−k^k^T)=E−k^k^T,{\displaystyle -\left[\mathbf {\hat {k}} \right]^{2}\equiv \left|\mathbf {\hat {k}} \right|^{2}\left(\mathbf {E} -\mathbf {\hat {k}} \mathbf {\hat {k}} ^{\mathsf {T}}\right)=\mathbf {E} -\mathbf {\hat {k}} \mathbf {\hat {k}} ^{\mathsf {T}},}noting thatk^{\displaystyle \mathbf {\hat {k}} } is a unit vector.

The magnitude squared of the perpendicular vector is|Δri⊥|2=(−[k^]2Δri)⋅(−[k^]2Δri)=(k^×(k^×Δri))⋅(k^×(k^×Δri)){\displaystyle {\begin{aligned}\left|\Delta \mathbf {r} _{i}^{\perp }\right|^{2}&=\left(-\left[\mathbf {\hat {k}} \right]^{2}\Delta \mathbf {r} _{i}\right)\cdot \left(-\left[\mathbf {\hat {k}} \right]^{2}\Delta \mathbf {r} _{i}\right)\\&=\left(\mathbf {\hat {k}} \times \left(\mathbf {\hat {k}} \times \Delta \mathbf {r} _{i}\right)\right)\cdot \left(\mathbf {\hat {k}} \times \left(\mathbf {\hat {k}} \times \Delta \mathbf {r} _{i}\right)\right)\end{aligned}}}

The simplification of this equation uses the triple scalar product identity(k^×(k^×Δri))⋅(k^×(k^×Δri))≡((k^×(k^×Δri))×k^)⋅(k^×Δri),{\displaystyle \left(\mathbf {\hat {k}} \times \left(\mathbf {\hat {k}} \times \Delta \mathbf {r} _{i}\right)\right)\cdot \left(\mathbf {\hat {k}} \times \left(\mathbf {\hat {k}} \times \Delta \mathbf {r} _{i}\right)\right)\equiv \left(\left(\mathbf {\hat {k}} \times \left(\mathbf {\hat {k}} \times \Delta \mathbf {r} _{i}\right)\right)\times \mathbf {\hat {k}} \right)\cdot \left(\mathbf {\hat {k}} \times \Delta \mathbf {r} _{i}\right),}where the dot and the cross products have been interchanged. Exchanging products, and simplifying by noting thatΔri{\displaystyle \Delta \mathbf {r} _{i}} andk^{\displaystyle \mathbf {\hat {k}} } are orthogonal:(k^×(k^×Δri))⋅(k^×(k^×Δri))=((k^×(k^×Δri))×k^)⋅(k^×Δri)=(k^×Δri)⋅(−Δri×k^)=−k^⋅(Δri×Δri×k^)=−k^⋅[Δri]2k^.{\displaystyle {\begin{aligned}&\left(\mathbf {\hat {k}} \times \left(\mathbf {\hat {k}} \times \Delta \mathbf {r} _{i}\right)\right)\cdot \left(\mathbf {\hat {k}} \times \left(\mathbf {\hat {k}} \times \Delta \mathbf {r} _{i}\right)\right)\\={}&\left(\left(\mathbf {\hat {k}} \times \left(\mathbf {\hat {k}} \times \Delta \mathbf {r} _{i}\right)\right)\times \mathbf {\hat {k}} \right)\cdot \left(\mathbf {\hat {k}} \times \Delta \mathbf {r} _{i}\right)\\={}&\left(\mathbf {\hat {k}} \times \Delta \mathbf {r} _{i}\right)\cdot \left(-\Delta \mathbf {r} _{i}\times \mathbf {\hat {k}} \right)\\={}&-\mathbf {\hat {k}} \cdot \left(\Delta \mathbf {r} _{i}\times \Delta \mathbf {r} _{i}\times \mathbf {\hat {k}} \right)\\={}&-\mathbf {\hat {k}} \cdot \left[\Delta \mathbf {r} _{i}\right]^{2}\mathbf {\hat {k}} .\end{aligned}}}

Thus, the moment of inertia around the lineL{\displaystyle L} throughR{\displaystyle \mathbf {R} } in the directionk^{\displaystyle \mathbf {\hat {k}} } is obtained from the calculationIL=∑i=1Nmi|Δri⊥|2=−∑i=1Nmik^⋅[Δri]2k^=k^⋅(−∑i=1Nmi[Δri]2)k^=k^⋅IRk^=k^TIRk^,{\displaystyle {\begin{aligned}I_{L}&=\sum _{i=1}^{N}m_{i}\left|\Delta \mathbf {r} _{i}^{\perp }\right|^{2}\\&=-\sum _{i=1}^{N}m_{i}\mathbf {\hat {k}} \cdot \left[\Delta \mathbf {r} _{i}\right]^{2}\mathbf {\hat {k}} =\mathbf {\hat {k}} \cdot \left(-\sum _{i=1}^{N}m_{i}\left[\Delta \mathbf {r} _{i}\right]^{2}\right)\mathbf {\hat {k}} \\&=\mathbf {\hat {k}} \cdot \mathbf {I} _{\mathbf {R} }\mathbf {\hat {k}} =\mathbf {\hat {k}} ^{\mathsf {T}}\mathbf {I} _{\mathbf {R} }\mathbf {\hat {k}} ,\end{aligned}}}whereIR{\displaystyle \mathbf {I_{R}} } is the moment of inertia matrix of the system relative to the reference pointR{\displaystyle \mathbf {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.

Inertia tensor

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For the same object, different axes of rotation will have different moments of inertia about those axes. In general, the moments of inertia are not equal unless the object is symmetric about all axes. Themoment of inertiatensor is a convenient way to summarize all moments of inertia of an object with one quantity. It may be calculated with respect to any point in space, although for practical purposes the center of mass is most commonly used.

Definition

[edit]

For a rigid object ofN{\displaystyle N} point massesmk{\displaystyle m_{k}}, the moment of inertiatensor is given byI=[I11I12I13I21I22I23I31I32I33].{\displaystyle \mathbf {I} ={\begin{bmatrix}I_{11}&I_{12}&I_{13}\\I_{21}&I_{22}&I_{23}\\I_{31}&I_{32}&I_{33}\end{bmatrix}}.}

Its components are defined asIij =def ∑k=1Nmk(‖rk‖2δij−xi(k)xj(k)){\displaystyle I_{ij}\ {\stackrel {\mathrm {def} }{=}}\ \sum _{k=1}^{N}m_{k}\left(\left\|\mathbf {r} _{k}\right\|^{2}\delta _{ij}-x_{i}^{(k)}x_{j}^{(k)}\right)}where

Note that, by the definition,I{\displaystyle \mathbf {I} } is asymmetric tensor.

The diagonal elements are more succinctly written asIxx =def ∑k=1Nmk(yk2+zk2),Iyy =def ∑k=1Nmk(xk2+zk2),Izz =def ∑k=1Nmk(xk2+yk2),{\displaystyle {\begin{aligned}I_{xx}\ &{\stackrel {\mathrm {def} }{=}}\ \sum _{k=1}^{N}m_{k}\left(y_{k}^{2}+z_{k}^{2}\right),\\I_{yy}\ &{\stackrel {\mathrm {def} }{=}}\ \sum _{k=1}^{N}m_{k}\left(x_{k}^{2}+z_{k}^{2}\right),\\I_{zz}\ &{\stackrel {\mathrm {def} }{=}}\ \sum _{k=1}^{N}m_{k}\left(x_{k}^{2}+y_{k}^{2}\right),\end{aligned}}}while the off-diagonal elements, also called theproducts of inertia, areIxy=Iyx =def −∑k=1Nmkxkyk,Ixz=Izx =def −∑k=1Nmkxkzk,Iyz=Izy =def −∑k=1Nmkykzk.{\displaystyle {\begin{aligned}I_{xy}=I_{yx}\ &{\stackrel {\mathrm {def} }{=}}\ -\sum _{k=1}^{N}m_{k}x_{k}y_{k},\\I_{xz}=I_{zx}\ &{\stackrel {\mathrm {def} }{=}}\ -\sum _{k=1}^{N}m_{k}x_{k}z_{k},\\I_{yz}=I_{zy}\ &{\stackrel {\mathrm {def} }{=}}\ -\sum _{k=1}^{N}m_{k}y_{k}z_{k}.\end{aligned}}}

HereIxx{\displaystyle I_{xx}} denotes the moment of inertia around thex{\displaystyle x}-axis when the objects are rotated around the x-axis,Ixy{\displaystyle I_{xy}} denotes the moment of inertia around they{\displaystyle y}-axis when the objects are rotated around thex{\displaystyle x}-axis, and so on.

These quantities can be generalized to an object with distributed mass, described by a mass density function, in a similar fashion to the scalar moment of inertia. One then hasI=∭Vρ(x,y,z)(‖r‖2E3−r⊗r)dxdydz,{\displaystyle \mathbf {I} =\iiint _{V}\rho (x,y,z)\left(\|\mathbf {r} \|^{2}\mathbf {E} _{3}-\mathbf {r} \otimes \mathbf {r} \right)\,dx\,dy\,dz,}wherer⊗r{\displaystyle \mathbf {r} \otimes \mathbf {r} } is theirouter product,E3 is the 3×3identity matrix, andV is a region of space completely containing the object.

Alternatively it can also be written in terms of theangular momentum operator[r]x=r×x{\displaystyle [\mathbf {r} ]\mathbf {x} =\mathbf {r} \times \mathbf {x} }:I=∭Vρ(r)[r]T[r]dV=−∭Qρ(r)[r]2dV{\displaystyle \mathbf {I} =\iiint _{V}\rho (\mathbf {r} )[\mathbf {r} ]^{\textsf {T}}[\mathbf {r} ]\,dV=-\iiint _{Q}\rho (\mathbf {r} )[\mathbf {r} ]^{2}\,dV}

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 directionn{\displaystyle \mathbf {n} },In=n⋅I⋅n,{\displaystyle I_{n}=\mathbf {n} \cdot \mathbf {I} \cdot \mathbf {n} ,}where thedot product is taken with the corresponding elements in the component tensors. A product of inertia term such asI12{\displaystyle I_{12}} is obtained by the computationI12=e1⋅I⋅e2,{\displaystyle I_{12}=\mathbf {e} _{1}\cdot \mathbf {I} \cdot \mathbf {e} _{2},}and can be interpreted as the moment of inertia around thex{\displaystyle x}-axis when the object rotates around they{\displaystyle 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=[I11I12I13I21I22I23I31I32I33]=[IxxIxyIxzIyxIyyIyzIzxIzyIzz]=∑k=1N[mk(yk2+zk2)−mkxkyk−mkxkzk−mkxkykmk(xk2+zk2)−mkykzk−mkxkzk−mkykzkmk(xk2+yk2)].{\displaystyle {\begin{aligned}\mathbf {I} &={\begin{bmatrix}I_{11}&I_{12}&I_{13}\\[1.8ex]I_{21}&I_{22}&I_{23}\\[1.8ex]I_{31}&I_{32}&I_{33}\end{bmatrix}}={\begin{bmatrix}I_{xx}&I_{xy}&I_{xz}\\[1.8ex]I_{yx}&I_{yy}&I_{yz}\\[1.8ex]I_{zx}&I_{zy}&I_{zz}\end{bmatrix}}\\[2ex]&=\sum _{k=1}^{N}{\begin{bmatrix}m_{k}\left(y_{k}^{2}+z_{k}^{2}\right)&-m_{k}x_{k}y_{k}&-m_{k}x_{k}z_{k}\\[1ex]-m_{k}x_{k}y_{k}&m_{k}\left(x_{k}^{2}+z_{k}^{2}\right)&-m_{k}y_{k}z_{k}\\[1ex]-m_{k}x_{k}z_{k}&-m_{k}y_{k}z_{k}&m_{k}\left(x_{k}^{2}+y_{k}^{2}\right)\end{bmatrix}}.\end{aligned}}}

It is common in rigid body mechanics to use notation that explicitly identifies thex{\displaystyle x},y{\displaystyle y}, andz{\displaystyle z}-axes, such asIxx{\displaystyle I_{xx}} andIxy{\displaystyle I_{xy}}, for the components of the inertia tensor.

Alternate inertia convention

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There are some CAD and CAE applications such as SolidWorks, Unigraphics NX/Siemens NX and MSC Adams that use an alternate convention for the products of inertia. According to this convention, the minus sign is removed from the product of inertia formulas and instead inserted in the inertia matrix:Ixy=Iyx =def ∑k=1Nmkxkyk,Ixz=Izx =def ∑k=1Nmkxkzk,Iyz=Izy =def ∑k=1Nmkykzk,I=[I11I12I13I21I22I23I31I32I33]=[Ixx−Ixy−Ixz−IyxIyy−Iyz−Izx−IzyIzz]=∑k=1N[mk(yk2+zk2)−mkxkyk−mkxkzk−mkxkykmk(xk2+zk2)−mkykzk−mkxkzk−mkykzkmk(xk2+yk2)].{\displaystyle {\begin{aligned}I_{xy}=I_{yx}\ &{\stackrel {\mathrm {def} }{=}}\ \sum _{k=1}^{N}m_{k}x_{k}y_{k},\\I_{xz}=I_{zx}\ &{\stackrel {\mathrm {def} }{=}}\ \sum _{k=1}^{N}m_{k}x_{k}z_{k},\\I_{yz}=I_{zy}\ &{\stackrel {\mathrm {def} }{=}}\ \sum _{k=1}^{N}m_{k}y_{k}z_{k},\\[3pt]\mathbf {I} ={\begin{bmatrix}I_{11}&I_{12}&I_{13}\\[1.8ex]I_{21}&I_{22}&I_{23}\\[1.8ex]I_{31}&I_{32}&I_{33}\end{bmatrix}}&={\begin{bmatrix}I_{xx}&-I_{xy}&-I_{xz}\\[1.8ex]-I_{yx}&I_{yy}&-I_{yz}\\[1.8ex]-I_{zx}&-I_{zy}&I_{zz}\end{bmatrix}}\\[1ex]&=\sum _{k=1}^{N}{\begin{bmatrix}m_{k}\left(y_{k}^{2}+z_{k}^{2}\right)&-m_{k}x_{k}y_{k}&-m_{k}x_{k}z_{k}\\[1ex]-m_{k}x_{k}y_{k}&m_{k}\left(x_{k}^{2}+z_{k}^{2}\right)&-m_{k}y_{k}z_{k}\\[1ex]-m_{k}x_{k}z_{k}&-m_{k}y_{k}z_{k}&m_{k}\left(x_{k}^{2}+y_{k}^{2}\right)\end{bmatrix}}.\end{aligned}}}

Determine inertia convention (principal axes method)

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If one has the inertia data(Ixx,Iyy,Izz,Ixy,Ixz,Iyz){\displaystyle (I_{xx},I_{yy},I_{zz},I_{xy},I_{xz},I_{yz})} without knowing which inertia convention that has been used, it can be determined if one also has theprincipal axes. With the principal axes method, one makes inertia matrices from the following two assumptions:

  1. The standard inertia convention has been used(I12=Ixy,I13=Ixz,I23=Iyz){\displaystyle (I_{12}=I_{xy},I_{13}=I_{xz},I_{23}=I_{yz})}.
  2. The alternate inertia convention has been used(I12=−Ixy,I13=−Ixz,I23=−Iyz){\displaystyle (I_{12}=-I_{xy},I_{13}=-I_{xz},I_{23}=-I_{yz})}.

Next, one calculates the eigenvectors for the two matrices. The matrix whose eigenvectors are parallel to the principal axes corresponds to the inertia convention that has been used.

Derivation of the tensor components

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The distancer{\displaystyle r} of a particle atx{\displaystyle \mathbf {x} } from the axis of rotation passing through the origin in then^{\displaystyle \mathbf {\hat {n}} } direction is|x−(x⋅n^)n^|{\displaystyle \left|\mathbf {x} -\left(\mathbf {x} \cdot \mathbf {\hat {n}} \right)\mathbf {\hat {n}} \right|}, wheren^{\displaystyle \mathbf {\hat {n}} } is a unit vector. The moment of inertia on the axis isI=mr2=m(x−(x⋅n^)n^)⋅(x−(x⋅n^)n^)=m(x2−2x(x⋅n^)n^+(x⋅n^)2n^2)=m(x2−(x⋅n^)2).{\displaystyle I=mr^{2}=m\left(\mathbf {x} -\left(\mathbf {x} \cdot \mathbf {\hat {n}} \right)\mathbf {\hat {n}} \right)\cdot \left(\mathbf {x} -\left(\mathbf {x} \cdot \mathbf {\hat {n}} \right)\mathbf {\hat {n}} \right)=m\left(\mathbf {x} ^{2}-2\mathbf {x} \left(\mathbf {x} \cdot \mathbf {\hat {n}} \right)\mathbf {\hat {n}} +\left(\mathbf {x} \cdot \mathbf {\hat {n}} \right)^{2}\mathbf {\hat {n}} ^{2}\right)=m\left(\mathbf {x} ^{2}-\left(\mathbf {x} \cdot \mathbf {\hat {n}} \right)^{2}\right).}

Rewrite the equation usingmatrix transpose:I=m(xTx−n^TxxTn^)=m⋅n^T(xTx⋅E3−xxT)n^,{\displaystyle I=m\left(\mathbf {x} ^{\textsf {T}}\mathbf {x} -\mathbf {\hat {n}} ^{\textsf {T}}\mathbf {x} \mathbf {x} ^{\textsf {T}}\mathbf {\hat {n}} \right)=m\cdot \mathbf {\hat {n}} ^{\textsf {T}}\left(\mathbf {x} ^{\textsf {T}}\mathbf {x} \cdot \mathbf {E_{3}} -\mathbf {x} \mathbf {x} ^{\textsf {T}}\right)\mathbf {\hat {n}} ,}whereE3 is the 3×3identity matrix.

This leads to a tensor formula for the moment of inertiaI=m[n1n2n3][y2+z2−xy−xz−yxx2+z2−yz−zx−zyx2+y2][n1n2n3].{\displaystyle I=m{\begin{bmatrix}n_{1}&n_{2}&n_{3}\end{bmatrix}}{\begin{bmatrix}y^{2}+z^{2}&-xy&-xz\\[0.5ex]-yx&x^{2}+z^{2}&-yz\\[0.5ex]-zx&-zy&x^{2}+y^{2}\end{bmatrix}}{\begin{bmatrix}n_{1}\\[0.7ex]n_{2}\\[0.7ex]n_{3}\end{bmatrix}}.}

For multiple particles, we need only recall that the moment of inertia is additive in order to see that this formula is correct.

Inertia tensor of translation

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Main article:Parallel axis theorem § Tensor generalization

LetI0{\displaystyle \mathbf {I} _{0}} be the inertia tensor of a body calculated at itscenter of mass, andR{\displaystyle \mathbf {R} } be the displacement vector of the body. The inertia tensor of the translated body respect to its original center of mass is given by:I=I0+m[(R⋅R)E3−R⊗R]{\displaystyle \mathbf {I} =\mathbf {I} _{0}+m[(\mathbf {R} \cdot \mathbf {R} )\mathbf {E} _{3}-\mathbf {R} \otimes \mathbf {R} ]}wherem{\displaystyle m} is the body's mass,E3 is the 3 × 3 identity matrix, and⊗{\displaystyle \otimes } is theouter product.

Inertia tensor of rotation

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LetR{\displaystyle \mathbf {R} } be thematrix that represents a body's rotation. The inertia tensor of the rotated body is given by:[28]I=RI0RT{\displaystyle \mathbf {I} =\mathbf {R} \mathbf {I_{0}} \mathbf {R} ^{\textsf {T}}}

Inertia matrix in different reference frames

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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.[8][25] 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

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Let the body frame inertia matrix relative to the center of mass be denotedICB{\displaystyle \mathbf {I} _{\mathbf {C} }^{B}}, and define the orientation of the body frame relative to the inertial frame by the rotation matrixA{\displaystyle \mathbf {A} }, such that,x=Ay,{\displaystyle \mathbf {x} =\mathbf {A} \mathbf {y} ,}where vectorsy{\displaystyle \mathbf {y} } in the body fixed coordinate frame have coordinatesx{\displaystyle \mathbf {x} } in the inertial frame. Then, the inertia matrix of the body measured in the inertial frame is given byIC=AICBAT.{\displaystyle \mathbf {I} _{\mathbf {C} }=\mathbf {A} \mathbf {I} _{\mathbf {C} }^{B}\mathbf {A} ^{\mathsf {T}}.}

Notice thatA{\displaystyle \mathbf {A} } changes as the body moves, whileICB{\displaystyle \mathbf {I} _{\mathbf {C} }^{B}} remains constant.

Principal axes

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Measured in the body frame, the inertia matrix is a constant real symmetric matrix. A real symmetric matrix has theeigendecomposition into the product of a rotation matrixQ{\displaystyle \mathbf {Q} } and a diagonal matrixΛ{\displaystyle {\boldsymbol {\Lambda }}}, given byICB=QΛQT,{\displaystyle \mathbf {I} _{\mathbf {C} }^{B}=\mathbf {Q} {\boldsymbol {\Lambda }}\mathbf {Q} ^{\mathsf {T}},}whereΛ=[I1000I2000I3].{\displaystyle {\boldsymbol {\Lambda }}={\begin{bmatrix}I_{1}&0&0\\0&I_{2}&0\\0&0&I_{3}\end{bmatrix}}.}

The columns of the rotation matrixQ{\displaystyle \mathbf {Q} } define the directions of the principal axes of the body, and the constantsI1{\displaystyle I_{1}},I2{\displaystyle I_{2}}, andI3{\displaystyle I_{3}} are called theprincipal moments of inertia. This result was first shown byJ. J. Sylvester (1852), and is a form ofSylvester's law of inertia.[29][30] When the body has an axis of symmetry (sometimes called thefigure axis oraxis of figure) then the other two moments of inertia will be identical and any axis perpendicular to the axis of symmetry will be a principal axis.

A toytop is an example of a rotating rigid body, and the wordtop is used in the names of types of rigid bodies. When all principal moments of inertia are distinct, the principal axes throughcenter of mass are uniquely specified and the rigid body is called anasymmetric top. If two principal moments are the same, the rigid body is called asymmetric top and there is no unique choice for the two corresponding principal axes. If all three principal moments are the same, the rigid body is called aspherical top (although it need not be spherical) and any axis can be considered a principal axis, meaning that the moment of inertia is the same about any axis.

The principal axes are often aligned with the object's symmetry axes. If a rigid body has an axis of symmetry of orderm{\displaystyle m}, meaning it is symmetrical under rotations of360°/m about the given axis, that axis is a principal axis. Whenm>2{\displaystyle m>2}, the rigid body is a symmetric top. If a rigid body has at least two symmetry axes that are not parallel or perpendicular to each other, it is a spherical top, for example, a cube or any otherPlatonic solid.

Themotion ofvehicles is often described in terms ofyaw, pitch, and roll which usually correspond approximately to rotations about the three principal axes. If the vehicle has bilateral symmetry then one of the principal axes will correspond exactly to the transverse (pitch) axis.

A practical example of this mathematical phenomenon is the routine automotive task ofbalancing a tire, which basically means adjusting the distribution of mass of a car wheel such that its principal axis of inertia is aligned with the axle so the wheel does not wobble.

Rotating molecules are also classified as asymmetric, symmetric, or spherical tops, and the structure of theirrotational spectra is different for each type.

Ellipsoid

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An ellipsoid with the semi-principal diameters labelleda{\displaystyle a},b{\displaystyle b}, andc{\displaystyle c}.

The moment of inertia matrix in body-frame coordinates is a quadratic form that defines a surface in the body calledPoinsot's ellipsoid.[31] LetΛ{\displaystyle {\boldsymbol {\Lambda }}} be the inertia matrix relative to the center of mass aligned with the principal axes, then the surfacexTΛx=1,{\displaystyle \mathbf {x} ^{\mathsf {T}}{\boldsymbol {\Lambda }}\mathbf {x} =1,}orI1x2+I2y2+I3z2=1,{\displaystyle I_{1}x^{2}+I_{2}y^{2}+I_{3}z^{2}=1,}defines anellipsoid in the body frame. Write this equation in the form,(x1/I1)2+(y1/I2)2+(z1/I3)2=1,{\displaystyle \left({\frac {x}{1/{\sqrt {I_{1}}}}}\right)^{2}+\left({\frac {y}{1/{\sqrt {I_{2}}}}}\right)^{2}+\left({\frac {z}{1/{\sqrt {I_{3}}}}}\right)^{2}=1,}to see that the semi-principal diameters of this ellipsoid are given bya=1I1,b=1I2,c=1I3.{\displaystyle a={\frac {1}{\sqrt {I_{1}}}},\quad b={\frac {1}{\sqrt {I_{2}}}},\quad c={\frac {1}{\sqrt {I_{3}}}}.}

Let a pointx{\displaystyle \mathbf {x} } on this ellipsoid be defined in terms of its magnitude and direction,x=‖x‖n{\displaystyle \mathbf {x} =\|\mathbf {x} \|\mathbf {n} }, wheren{\displaystyle \mathbf {n} } is a unit vector. Then the relationship presented above, between the inertia matrix and the scalar moment of inertiaIn{\displaystyle I_{\mathbf {n} }} around an axis in the directionn{\displaystyle \mathbf {n} }, yieldsxTΛx=‖x‖2nTΛn=‖x‖2In=1.{\displaystyle \mathbf {x} ^{\mathsf {T}}{\boldsymbol {\Lambda }}\mathbf {x} =\|\mathbf {x} \|^{2}\mathbf {n} ^{\mathsf {T}}{\boldsymbol {\Lambda }}\mathbf {n} =\|\mathbf {x} \|^{2}I_{\mathbf {n} }=1.}

Thus, the magnitude of a pointx{\displaystyle \mathbf {x} } in the directionn{\displaystyle \mathbf {n} } on the inertia ellipsoid is‖x‖=1In.{\displaystyle \|\mathbf {x} \|={\frac {1}{\sqrt {I_{\mathbf {n} }}}}.}

See also

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References

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  1. ^abLerner, Lawrence S. (1996).Physics for Scientists and Engineers. Jones and Bartlett.ISBN 0867204796.
  2. ^abTipler, Paul A. (1999).Physics for Scientists and Engineers, Vol. 1: Mechanics, Oscillations and Waves, Thermodynamics. Macmillan.ISBN 1572594918.
  3. ^abMach, Ernst (1919).The Science of Mechanics. pp. 173–187. RetrievedNovember 21, 2014.
  4. ^Euler, Leonhard (1765).Theoria motus corporum solidorum seu rigidorum: Ex primis nostrae cognitionis principiis stabilita et ad omnes motus, qui in huiusmodi corpora cadere possunt, accommodata [The theory of motion of solid or rigid bodies: established from first principles of our knowledge and appropriate for all motions which can occur in such bodies.] (in Latin). Rostock and Greifswald (Germany): A. F. Röse. p. 166.ISBN 978-1-4297-4281-8.{{cite book}}:ISBN / Date incompatibility (help) From page 166:"Definitio 7. 422. Momentum inertiae corporis respectu eujuspiam axis est summa omnium productorum, quae oriuntur, si singula corporis elementa per quadrata distantiarum suarum ab axe multiplicentur." (Definition 7. 422. A body's moment of inertia with respect to any axis is the sum of all of the products, which arise, if the individual elements of the body are multiplied by the square of their distances from the axis.)
  5. ^abcdefMarion, JB; Thornton, ST (1995).Classical dynamics of particles & systems (4th ed.). Thomson.ISBN 0-03-097302-3.
  6. ^abSymon, KR (1971).Mechanics (3rd ed.). Addison-Wesley.ISBN 0-201-07392-7.
  7. ^abTenenbaum, RA (2004).Fundamentals of Applied Dynamics. Springer.ISBN 0-387-00887-X.
  8. ^abcdefghKane, T. R.; Levinson, D. A. (1985).Dynamics, Theory and Applications. New York: McGraw-Hill.
  9. ^abWinn, Will (2010).Introduction to Understandable Physics: Volume I - Mechanics. AuthorHouse. p. 10.10.ISBN 978-1449063337.
  10. ^abFullerton, Dan (2011).Honors Physics Essentials. Silly Beagle Productions. pp. 142–143.ISBN 978-0983563334.
  11. ^Wolfram, Stephen (2014)."Spinning Ice Skater".Wolfram Demonstrations Project. Mathematica, Inc. RetrievedSeptember 30, 2014.
  12. ^Hokin, Samuel (2014)."Figure Skating Spins".The Physics of Everyday Stuff. RetrievedSeptember 30, 2014.
  13. ^Breithaupt, Jim (2000).New Understanding Physics for Advanced Level. Nelson Thomas. p. 64.ISBN 0748743146.
  14. ^Crowell, Benjamin (2003).Conservation Laws. Light and Matter. pp. 107.ISBN 0970467028.ice skater conservation of angular momentum.
  15. ^abcdePaul, Burton (June 1979).Kinematics and Dynamics of Planar Machinery. Prentice Hall.ISBN 978-0135160626.
  16. ^Halliday, David; Resnick, Robert; Walker, Jearl (2005).Fundamentals of physics (7th ed.). Hoboken, NJ: Wiley.ISBN 9780471216438.
  17. ^French, A.P. (1971).Vibrations and waves. Boca Raton, FL: CRC Press.ISBN 9780748744473.
  18. ^abcdefUicker, John J.; Pennock, Gordon R.; Shigley, Joseph E. (2010).Theory of Machines and Mechanisms (4th ed.). Oxford University Press.ISBN 978-0195371239.
  19. ^C. Couch and J. Mayes,Trifilar Pendulum for MOI, Happresearch.com, 2016.
  20. ^Gracey, William, The experimental determination of the moments of inertia of airplanes by a simplified compound-pendulum method,NACA Technical Note No. 1629, 1948
  21. ^Morrow, H. W.; Kokernak, Robert (2011).Statics and Strengths of Materials (7 ed.). New Jersey: Prentice Hall. pp. 192–196.ISBN 978-0135034521.
  22. ^In that situation this moment of inertia only describes how a torque applied along that axis causes a rotation about that axis. But, torques not aligned along a principal axis will also cause rotations about other axes.
  23. ^abcdefghiFerdinand P. Beer; E. Russell Johnston, Jr.; Phillip J. Cornwell (2010).Vector mechanics for engineers: Dynamics (9th ed.). Boston: McGraw-Hill.ISBN 978-0077295493.
  24. ^Walter D. Pilkey,Analysis and Design of Elastic Beams: Computational Methods, John Wiley, 2002.
  25. ^abGoldstein, H. (1980).Classical Mechanics (2nd ed.). Addison-Wesley.ISBN 0-201-02918-9.
  26. ^L. D. Landau and E. M. Lifshitz,Mechanics, Vol 1. 2nd Ed., Pergamon Press, 1969.
  27. ^L. W. Tsai, Robot Analysis: The mechanics of serial and parallel manipulators, John-Wiley, NY, 1999.
  28. ^David, Baraff."Physically Based Modeling - Rigid Body Simulation"(PDF).Pixar Graphics Technologies.
  29. ^Sylvester, J J (1852)."A demonstration of the theorem that every homogeneous quadratic polynomial is reducible by real orthogonal substitutions to the form of a sum of positive and negative squares"(PDF).Philosophical Magazine. 4th Series.4 (23):138–142.doi:10.1080/14786445208647087. RetrievedJune 27, 2008.
  30. ^Norman, C.W. (1986).Undergraduate algebra.Oxford University Press. pp. 360–361.ISBN 0-19-853248-2.
  31. ^Mason, Matthew T. (2001).Mechanics of Robotics Manipulation. MIT Press.ISBN 978-0-262-13396-8. RetrievedNovember 21, 2014.

External links

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Wikimedia Commons has media related toMoments of inertia.
Linear/translational quantitiesAngular/rotational quantities
Dimensions1LL2Dimensions1θθ2
Ttime:t
s		absement:A
m s		Ttime:t
s
1distance:d,position:r,s,x,displacement
m		area:A
m2		1angle:θ,angular displacement:θ
rad		solid angle:Ω
rad2, sr
T−1frequency:f
s−1,Hz		speed:v,velocity:v
m s−1		kinematic viscosity:ν,
specific angular momentum: h
m2 s−1		T−1frequency:f,rotational speed:n,rotational velocity:n
s−1,Hz		angular speed:ω,angular velocity:ω
rad s−1
T−2acceleration:a
m s−2		T−2rotational acceleration
s−2		angular acceleration:α
rad s−2
T−3jerk:j
m s−3		T−3angular jerk:ζ
rad s−3
Mmass:m
kg		weighted position:M ⟨x⟩ = ∑mxmoment of inertia: I
kg m2		ML
MT−1Mass flow rate:m˙{\displaystyle {\dot {m}}}
kg s−1		momentum:p,impulse:J
kg m s−1,N s		action:𝒮,actergy:ℵ
kg m2 s−1,J s		MLT−1angular momentum:L,angular impulse:ΔL
kg m rad s−1
MT−2force:F,weight:Fg
kg m s−2,N		energy:E,work:W,Lagrangian:L
kg m2 s−2,J		MLT−2torque:τ,moment:M
kg m rad s−2,N m
MT−3yank:Y
kg m s−3, N s−1		power:P
kg m2 s−3, W		MLT−3rotatum:P
kg m rad s−3, N m s−1
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