Simple pendulum

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 particle ${\displaystyle m}$ with the square of its distance ${\displaystyle r}$ to the pivot, that is

$${\displaystyle I=mr^{2}.}$$

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

$${\displaystyle {\begin{aligned}{\boldsymbol {\tau }}&=\mathbf {r} \times \mathbf {F} =\mathbf {r} \times (m{\boldsymbol {\alpha }}\times \mathbf {r} )\\&=m\left(\left(\mathbf {r} \cdot \mathbf {r} \right){\boldsymbol {\alpha }}-\left(\mathbf {r} \cdot {\boldsymbol {\alpha }}\right)\mathbf {r} \right)\\&=mr^{2}{\boldsymbol {\alpha }}=I\alpha \mathbf {\hat {k}} ,\end{aligned}}}$$

where ${\displaystyle \mathbf {\hat {k}} }$ is a unit vector perpendicular to the plane of the pendulum. (The second to last step uses the vector triple product expansion with the perpendicularity of ${\displaystyle {\boldsymbol {\alpha }}}$ and ${\displaystyle \mathbf {r} }$.) The quantity ${\displaystyle I=mr^{2}}$ is the moment of inertia of this single mass around the pivot point.

The quantity ${\displaystyle I=mr^{2}}$ also appears in the angular momentum of a simple pendulum, which is calculated from the velocity ${\displaystyle \mathbf {v} ={\boldsymbol {\omega }}\times \mathbf {r} }$ of the pendulum mass around the pivot, where ${\displaystyle {\boldsymbol {\omega }}}$ is the angular velocity of the mass about the pivot point. This angular momentum is given by

$${\displaystyle {\begin{aligned}\mathbf {L} &=\mathbf {r} \times \mathbf {p} =\mathbf {r} \times \left(m{\boldsymbol {\omega }}\times \mathbf {r} \right)\\&=m\left(\left(\mathbf {r} \cdot \mathbf {r} \right){\boldsymbol {\omega }}-\left(\mathbf {r} \cdot {\boldsymbol {\omega }}\right)\mathbf {r} \right)\\&=mr^{2}{\boldsymbol {\omega }}=I\omega \mathbf {\hat {k}} ,\end{aligned}}}$$

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 yield

$${\displaystyle E_{\text{K}}={\frac {1}{2}}m\mathbf {v} \cdot \mathbf {v} ={\frac {1}{2}}\left(mr^{2}\right)\omega ^{2}={\frac {1}{2}}I\omega ^{2}.}$$

This shows that the quantity ${\displaystyle I=mr^{2}}$ 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 values ${\displaystyle mr^{2}}$ for all of the elements of mass in the body.

Compound pendulums

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

A compound pendulum is a body formed from an assembly of particles 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} The natural frequency (${\displaystyle \omega _{\text{n}}}$) of a compound pendulum depends on its moment of inertia, ${\displaystyle I_{P}}$,

$${\displaystyle \omega _{\text{n}}={\sqrt {\frac {mgr}{I_{P}}}},}$$

where ${\displaystyle m}$ is the mass of the object, ${\displaystyle g}$ is local acceleration of gravity, and ${\displaystyle r}$ 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 point ${\displaystyle P}$ 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 (${\displaystyle t}$), to obtain

$${\displaystyle I_{P}={\frac {mgr}{\omega _{\text{n}}^{2}}}={\frac {mgrt^{2}}{4\pi ^{2}}},}$$

where ${\displaystyle t}$ is the period (duration) of oscillation (usually averaged over multiple periods).

Center of oscillation

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

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

or

$${\displaystyle L={\frac {g}{\omega _{\text{n}}^{2}}}={\frac {I_{P}}{mr}}.}$$

The seconds pendulum, which provides the "tick" and "tock" of a grandfather clock, takes one second to swing from side-to-side. This is a period of two seconds, or a natural frequency of ${\displaystyle \pi \ \mathrm {rad/s} }$ for the pendulum. In this case, the distance to the center of oscillation, ${\displaystyle L}$, can be computed to be

$${\displaystyle L={\frac {g}{\omega _{\text{n}}^{2}}}\approx {\frac {9.81\ \mathrm {m/s^{2}} }{(3.14\ \mathrm {rad/s} )^{2}}}\approx 0.99\ \mathrm {m} .}$$

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 a gravimeter.

Sectional areas moment calculated thus^[21]

1. Square: ${\displaystyle I_{xx}=I_{yy}={\frac {b^{4}}{12}}}$
   
2. Rectangular: ${\displaystyle I_{xx}={\frac {bh^{3}}{12}}}$ and; ${\displaystyle I_{yy}={\frac {hb^{3}}{12}}}$
   
3. Triangular: ${\displaystyle I_{xx}={\frac {bh^{3}}{36}}}$
   
4. Circular: ${\displaystyle I_{xx}=I_{yy}={\frac {1}{4}}{\pi }r^{4}={\frac {1}{64}}{\pi }d^{4}}$

Point mass

Four objects with identical masses and radii racing down a plane while rolling without slipping.

From back to front:

•   spherical shell,
•   solid sphere,
•   cylindrical ring, and
•   solid cylinder.

The time for each object to reach the finishing line depends on their moment of inertia. (OGV version)

The moment of inertia about an axis of a body is calculated by summing ${\displaystyle mr^{2}}$ for every particle in the body, where ${\displaystyle r}$ 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 of ${\displaystyle N}$ masses ${\displaystyle m_{i}}$ that lie at the distances ${\displaystyle r_{i}}$ from the pivot point ${\displaystyle P}$, 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}

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

This shows that the moment of inertia of the body is the sum of each of the ${\displaystyle mr^{2}}$ terms, that is

$${\displaystyle I_{P}=\sum _{i=1}^{N}m_{i}r_{i}^{2}.}$$

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:

$${\displaystyle I_{P}=\sum _{i}m_{i}r_{i}^{2}}$$

Another expression replaces the summation with an integral,

$${\displaystyle I_{P}=\iiint _{Q}\rho (x,y,z)\left\|\mathbf {r} \right\|^{2}dV}$$

Here, the function ${\displaystyle \rho }$ gives the mass density at each point ${\displaystyle (x,y,z)}$, ${\displaystyle \mathbf {r} }$ is a vector perpendicular to the axis of rotation and extending from a point on the rotation axis to a point ${\displaystyle (x,y,z)}$ in the solid, and the integration is evaluated over the volume ${\displaystyle V}$ of the body ${\displaystyle Q}$. 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 the second 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 the ${\displaystyle z}$-axis perpendicular to the cross-section, weighted by its density. This is also called the polar moment of the area, and is the sum of the second moments about the ${\displaystyle x}$- and ${\displaystyle y}$-axes.^[24] The stresses in a beam are calculated using the second moment of the cross-sectional area around either the ${\displaystyle x}$-axis or ${\displaystyle y}$-axis depending on the load.

Examples

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.^[23]

• The moment of inertia of a thin rod with constant cross-section ${\displaystyle s}$ and density ${\displaystyle \rho }$ and with length ${\displaystyle \ell }$ about a perpendicular axis through its center of mass is determined by integration.^[23]: 1301 Align the ${\displaystyle x}$-axis with the rod and locate the origin its center of mass at the center of the rod, then
  $${\displaystyle I_{C,{\text{rod}}}=\iiint _{Q}\rho \,x^{2}\,dV=\int _{-{\frac {\ell }{2}}}^{\frac {\ell }{2}}\rho \,x^{2}s\,dx=\left.\rho s{\frac {x^{3}}{3}}\right|_{-{\frac {\ell }{2}}}^{\frac {\ell }{2}}={\frac {\rho s}{3}}\left({\frac {\ell ^{3}}{8}}+{\frac {\ell ^{3}}{8}}\right)={\frac {m\ell ^{2}}{12}},}$$

  
  where ${\displaystyle m=\rho s\ell }$ is the mass of the rod.
• The moment of inertia of a thin disc of constant thickness ${\displaystyle s}$, radius ${\displaystyle R}$, and density ${\displaystyle \rho }$ about an axis through its center and perpendicular to its face (parallel to its axis of rotational symmetry) is determined by integration.^[23]: 1301 Align the ${\displaystyle z}$-axis with the axis of the disc and define a volume element as ${\displaystyle dV=sr\,dr\,d\theta }$, then
  $${\displaystyle I_{C,{\text{disc}}}=\iiint _{Q}\rho \,r^{2}\,dV=\int _{0}^{2\pi }\int _{0}^{R}\rho r^{2}sr\,dr\,d\theta =2\pi \rho s{\frac {R^{4}}{4}}={\frac {1}{2}}mR^{2},}$$

  
  where ${\displaystyle m=\pi R^{2}\rho s}$ 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 ${\displaystyle P}$ as,
  $${\displaystyle I_{P}=I_{C,{\text{rod}}}+M_{\text{rod}}\left({\frac {L}{2}}\right)^{2}+I_{C,{\text{disc}}}+M_{\text{disc}}(L+R)^{2},}$$

  
  where ${\displaystyle L}$ 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 inertia 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 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 ball is defined by the equation^[23]: 1301

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

then the square of the radius ${\displaystyle r}$ of the disc at the cross-section ${\displaystyle z}$ along the ${\displaystyle z}$-axis is

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

Therefore, the moment of inertia of the ball is the sum of the moments of inertia of the discs along the ${\displaystyle z}$-axis,

$${\displaystyle {\begin{aligned}I_{C,{\text{ball}}}&=\int _{-R}^{R}{\tfrac {1}{2}}\pi \rho r(z)^{4}\,dz=\int _{-R}^{R}{\tfrac {1}{2}}\pi \rho \left(R^{2}-z^{2}\right)^{2}\,dz\\[1ex]&={\tfrac {1}{2}}\pi \rho \left[R^{4}z-{\tfrac {2}{3}}R^{2}z^{3}+{\tfrac {1}{5}}z^{5}\right]_{-R}^{R}\\[1ex]&=\pi \rho \left(1-{\tfrac {2}{3}}+{\tfrac {1}{5}}\right)R^{5}\\[1ex]&={\tfrac {2}{5}}mR^{2},\end{aligned}}}$$

where ${\textstyle m={\frac {4}{3}}\pi R^{3}\rho }$ is the mass of the sphere.

Rigid body

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 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 ${\displaystyle \mathbf {\hat {k}} }$ parallel to this plane. In this case, the moment of inertia of the mass in this system is a scalar known as 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.^{[15][18][25][26]}

If a system of ${\displaystyle n}$ particles, ${\displaystyle P_{i},i=1,\dots ,n}$, are assembled into a rigid body, then the momentum of the system can be written in terms of positions relative to a reference point ${\displaystyle \mathbf {R} }$, and absolute velocities ${\displaystyle \mathbf {v} _{i}}$:

$${\displaystyle {\begin{aligned}\Delta \mathbf {r} _{i}&=\mathbf {r} _{i}-\mathbf {R} ,\\\mathbf {v} _{i}&={\boldsymbol {\omega }}\times \left(\mathbf {r} _{i}-\mathbf {R} \right)+\mathbf {V} ={\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}+\mathbf {V} ,\end{aligned}}}$$

where ${\displaystyle {\boldsymbol {\omega }}}$ is the angular velocity of the system and ${\displaystyle \mathbf {V} }$ is the velocity of ${\displaystyle \mathbf {R} }$.

For planar movement the angular velocity vector is directed along the unit vector ${\displaystyle \mathbf {k} }$ which is perpendicular to the plane of movement. Introduce the unit vectors ${\displaystyle \mathbf {e} _{i}}$ from the reference point ${\displaystyle \mathbf {R} }$ to a point ${\displaystyle \mathbf {r} _{i}}$, and the unit vector ${\displaystyle \mathbf {\hat {t}} _{i}=\mathbf {\hat {k}} \times \mathbf {\hat {e}} _{i}}$, so

$${\displaystyle {\begin{aligned}\mathbf {\hat {e}} _{i}&={\frac {\Delta \mathbf {r} _{i}}{\Delta r_{i}}},\quad \mathbf {\hat {k}} ={\frac {\boldsymbol {\omega }}{\omega }},\quad \mathbf {\hat {t}} _{i}=\mathbf {\hat {k}} \times \mathbf {\hat {e}} _{i},\\\mathbf {v} _{i}&={\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}+\mathbf {V} =\omega \mathbf {\hat {k}} \times \Delta r_{i}\mathbf {\hat {e}} _{i}+\mathbf {V} =\omega \,\Delta r_{i}\mathbf {\hat {t}} _{i}+\mathbf {V} \end{aligned}}}$$

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

The angular momentum vector for the planar movement of a rigid system of particles is given by^[15][18]

$${\displaystyle {\begin{aligned}\mathbf {L} &=\sum _{i=1}^{n}m_{i}\Delta \mathbf {r} _{i}\times \mathbf {v} _{i}\\&=\sum _{i=1}^{n}m_{i}\,\Delta r_{i}\mathbf {\hat {e}} _{i}\times \left(\omega \,\Delta r_{i}\mathbf {\hat {t}} _{i}+\mathbf {V} \right)\\&=\left(\sum _{i=1}^{n}m_{i}\,\Delta r_{i}^{2}\right)\omega \mathbf {\hat {k}} +\left(\sum _{i=1}^{n}m_{i}\,\Delta r_{i}\mathbf {\hat {e}} _{i}\right)\times \mathbf {V} .\end{aligned}}}$$

Use the center of mass ${\displaystyle \mathbf {C} }$ as the reference point so

$${\displaystyle {\begin{aligned}\Delta r_{i}\mathbf {\hat {e}} _{i}&=\mathbf {r} _{i}-\mathbf {C} ,\\\sum _{i=1}^{n}m_{i}\,\Delta r_{i}\mathbf {\hat {e}} _{i}&=0,\end{aligned}}}$$

and define the moment of inertia relative to the center of mass ${\displaystyle I_{\mathbf {C} }}$ as

$${\displaystyle I_{\mathbf {C} }=\sum _{i}m_{i}\,\Delta r_{i}^{2},}$$

then the equation for angular momentum simplifies to^[23]: 1028

$${\displaystyle \mathbf {L} =I_{\mathbf {C} }\omega \mathbf {\hat {k}} .}$$

The moment of inertia ${\displaystyle I_{\mathbf {C} }}$ 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. Specifically, it is the second 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

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]

$${\displaystyle {\begin{aligned}E_{\text{K}}&={\frac {1}{2}}\sum _{i=1}^{n}m_{i}\mathbf {v} _{i}\cdot \mathbf {v} _{i},\\&={\frac {1}{2}}\sum _{i=1}^{n}m_{i}\left(\omega \,\Delta r_{i}\mathbf {\hat {t}} _{i}+\mathbf {V} \right)\cdot \left(\omega \,\Delta r_{i}\mathbf {\hat {t}} _{i}+\mathbf {V} \right),\\&={\frac {1}{2}}\omega ^{2}\left(\sum _{i=1}^{n}m_{i}\,\Delta r_{i}^{2}\mathbf {\hat {t}} _{i}\cdot \mathbf {\hat {t}} _{i}\right)+\omega \mathbf {V} \cdot \left(\sum _{i=1}^{n}m_{i}\,\Delta r_{i}\mathbf {\hat {t}} _{i}\right)+{\frac {1}{2}}\left(\sum _{i=1}^{n}m_{i}\right)\mathbf {V} \cdot \mathbf {V} .\end{aligned}}}$$

Let the reference point be the center of mass ${\displaystyle \mathbf {C} }$ of the system so the second term becomes zero, and introduce the moment of inertia ${\displaystyle I_{\mathbf {C} }}$ so the kinetic energy is given by^[23]: 1084

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

The moment of inertia ${\displaystyle I_{\mathbf {C} }}$ is the polar moment of inertia of the body.

Newton's laws

A 1920s 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 ${\displaystyle n}$ particles, ${\displaystyle P_{i},i=1,\dots ,n}$, can be written in terms of a resultant force and torque at a reference point ${\displaystyle \mathbf {R} }$, to yield^[15][18]

$${\displaystyle {\begin{aligned}\mathbf {F} &=\sum _{i=1}^{n}m_{i}\mathbf {A} _{i},\\{\boldsymbol {\tau }}&=\sum _{i=1}^{n}\Delta \mathbf {r} _{i}\times m_{i}\mathbf {A} _{i},\end{aligned}}}$$

where ${\displaystyle \mathbf {r} _{i}}$ denotes the trajectory of each particle.

The kinematics of a rigid body yields the formula for the acceleration of the particle ${\displaystyle P_{i}}$ in terms of the position ${\displaystyle \mathbf {R} }$ and acceleration ${\displaystyle \mathbf {A} }$ of the reference particle as well as the angular velocity vector ${\displaystyle {\boldsymbol {\omega }}}$ and angular acceleration vector ${\displaystyle {\boldsymbol {\alpha }}}$ of the rigid system of particles as,

$${\displaystyle \mathbf {A} _{i}={\boldsymbol {\alpha }}\times \Delta \mathbf {r} _{i}+{\boldsymbol {\omega }}\times {\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}+\mathbf {A} .}$$

For systems that are constrained to planar movement, the angular velocity and angular acceleration vectors are directed along ${\displaystyle \mathbf {\hat {k}} }$ perpendicular to the plane of movement, which simplifies this acceleration equation. In this case, the acceleration vectors can be simplified by introducing the unit vectors ${\displaystyle \mathbf {\hat {e}} _{i}}$ from the reference point ${\displaystyle \mathbf {R} }$ to a point ${\displaystyle \mathbf {r} _{i}}$ and the unit vectors ${\displaystyle \mathbf {\hat {t}} _{i}=\mathbf {\hat {k}} \times \mathbf {\hat {e}} _{i}}$, so

$${\displaystyle {\begin{aligned}\mathbf {A} _{i}&=\alpha \mathbf {\hat {k}} \times \Delta r_{i}\mathbf {\hat {e}} _{i}-\omega \mathbf {\hat {k}} \times \omega \mathbf {\hat {k}} \times \Delta r_{i}\mathbf {\hat {e}} _{i}+\mathbf {A} \\&=\alpha \Delta r_{i}\mathbf {\hat {t}} _{i}-\omega ^{2}\Delta r_{i}\mathbf {\hat {e}} _{i}+\mathbf {A} .\end{aligned}}}$$

This yields the resultant torque on the system as

$${\displaystyle {\begin{aligned}{\boldsymbol {\tau }}&=\sum _{i=1}^{n}m_{i}\,\Delta r_{i}\mathbf {\hat {e}} _{i}\times \left(\alpha \Delta r_{i}\mathbf {\hat {t}} _{i}-\omega ^{2}\Delta r_{i}\mathbf {\hat {e}} _{i}+\mathbf {A} \right)\\&=\left(\sum _{i=1}^{n}m_{i}\,\Delta r_{i}^{2}\right)\alpha \mathbf {\hat {k}} +\left(\sum _{i=1}^{n}m_{i}\,\Delta r_{i}\mathbf {\hat {e}} _{i}\right)\times \mathbf {A} ,\end{aligned}}}$$

where ${\displaystyle \mathbf {\hat {e}} _{i}\times \mathbf {\hat {e}} _{i}=\mathbf {0} }$, and ${\displaystyle \mathbf {\hat {e}} _{i}\times \mathbf {\hat {t}} _{i}=\mathbf {\hat {k}} }$ is the unit vector perpendicular to the plane for all of the particles ${\displaystyle P_{i}}$.

Use the center of mass ${\displaystyle \mathbf {C} }$ as the reference point and define the moment of inertia relative to the center of mass ${\displaystyle I_{\mathbf {C} }}$, then the equation for the resultant torque simplifies to^[23]: 1029

$${\displaystyle {\boldsymbol {\tau }}=I_{\mathbf {C} }\alpha \mathbf {\hat {k}} .}$$

Angular momentum

Note that the cross product can be equivalently written as matrix multiplication by combining the first operand and the operator into a skew-symmetric matrix, ${\displaystyle \left[\mathbf {b} \right]}$, constructed from the components of ${\displaystyle \mathbf {b} =(b_{x},b_{y},b_{z})}$:

$${\displaystyle {\begin{aligned}\mathbf {b} \times \mathbf {y} &\equiv \left[\mathbf {b} \right]\mathbf {y} \\\left[\mathbf {b} \right]&\equiv {\begin{bmatrix}0&-b_{z}&b_{y}\\b_{z}&0&-b_{x}\\-b_{y}&b_{x}&0\end{bmatrix}}.\end{aligned}}}$$

The inertia matrix is constructed by considering the angular momentum, with the reference point ${\displaystyle \mathbf {R} }$ of the body chosen to be the center of mass ${\displaystyle \mathbf {C} }$:^[4][7]

$${\displaystyle {\begin{aligned}\mathbf {L} &=\sum _{i=1}^{n}m_{i}\,\Delta \mathbf {r} _{i}\times \mathbf {v} _{i}\\&=\sum _{i=1}^{n}m_{i}\,\Delta \mathbf {r} _{i}\times \left({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}+\mathbf {V} _{\mathbf {R} }\right)\\&=\left(-\sum _{i=1}^{n}m_{i}\,\Delta \mathbf {r} _{i}\times \left(\Delta \mathbf {r} _{i}\times {\boldsymbol {\omega }}\right)\right)+\left(\sum _{i=1}^{n}m_{i}\,\Delta \mathbf {r} _{i}\times \mathbf {V} _{\mathbf {R} }\right),\end{aligned}}}$$

where the terms containing ${\displaystyle \mathbf {V_{R}} }$ (${\displaystyle =\mathbf {C} }$) sum to zero by the definition of center of mass.

Then, the skew-symmetric matrix ${\displaystyle [\Delta \mathbf {r} _{i}]}$ obtained from the relative position vector ${\displaystyle \Delta \mathbf {r} _{i}=\mathbf {r} _{i}-\mathbf {C} }$, can be used to define,

$${\displaystyle \mathbf {L} =\left(-\sum _{i=1}^{n}m_{i}\left[\Delta \mathbf {r} _{i}\right]^{2}\right){\boldsymbol {\omega }}=\mathbf {I} _{\mathbf {C} }{\boldsymbol {\omega }},}$$

where ${\displaystyle \mathbf {I_{C}} }$ defined by

$${\displaystyle \mathbf {I} _{\mathbf {C} }=-\sum _{i=1}^{n}m_{i}\left[\Delta \mathbf {r} _{i}\right]^{2},}$$

is the symmetric inertia matrix of the rigid system of particles measured relative to the center of mass ${\displaystyle \mathbf {C} }$.

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 ${\displaystyle n}$ particles ${\displaystyle P_{i},i=1,\dots ,n}$ be located at the coordinates ${\displaystyle \mathbf {r} _{i}}$ with velocities ${\displaystyle \mathbf {v} _{i}}$, then the kinetic energy is^[4][7]

$${\displaystyle E_{\text{K}}={\frac {1}{2}}\sum _{i=1}^{n}m_{i}\mathbf {v} _{i}\cdot \mathbf {v} _{i}={\frac {1}{2}}\sum _{i=1}^{n}m_{i}\left({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}+\mathbf {V} _{\mathbf {C} }\right)\cdot \left({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}+\mathbf {V} _{\mathbf {C} }\right),}$$

where ${\displaystyle \Delta \mathbf {r} _{i}=\mathbf {r} _{i}-\mathbf {C} }$ is the position vector of a particle relative to the center of mass.

This equation expands to yield three terms

$${\displaystyle E_{\text{K}}={\frac {1}{2}}\left(\sum _{i=1}^{n}m_{i}\left({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}\right)\cdot \left({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}\right)\right)+\left(\sum _{i=1}^{n}m_{i}\mathbf {V} _{\mathbf {C} }\cdot \left({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}\right)\right)+{\frac {1}{2}}\left(\sum _{i=1}^{n}m_{i}\mathbf {V} _{\mathbf {C} }\cdot \mathbf {V} _{\mathbf {C} }\right).}$$

Since the center of mass is defined by ${\displaystyle \sum _{i=1}^{n}m_{i}\Delta \mathbf {r} _{i}=0}$ , the second term in this equation is zero. Introduce the skew-symmetric matrix ${\displaystyle [\Delta \mathbf {r} _{i}]}$ so the kinetic energy becomes

$${\displaystyle {\begin{aligned}E_{\text{K}}&={\frac {1}{2}}\left(\sum _{i=1}^{n}m_{i}\left(\left[\Delta \mathbf {r} _{i}\right]{\boldsymbol {\omega }}\right)\cdot \left(\left[\Delta \mathbf {r} _{i}\right]{\boldsymbol {\omega }}\right)\right)+{\frac {1}{2}}\left(\sum _{i=1}^{n}m_{i}\right)\mathbf {V} _{\mathbf {C} }\cdot \mathbf {V} _{\mathbf {C} }\\&={\frac {1}{2}}\left(\sum _{i=1}^{n}m_{i}\left({\boldsymbol {\omega }}^{\mathsf {T}}\left[\Delta \mathbf {r} _{i}\right]^{\mathsf {T}}\left[\Delta \mathbf {r} _{i}\right]{\boldsymbol {\omega }}\right)\right)+{\frac {1}{2}}\left(\sum _{i=1}^{n}m_{i}\right)\mathbf {V} _{\mathbf {C} }\cdot \mathbf {V} _{\mathbf {C} }\\&={\frac {1}{2}}{\boldsymbol {\omega }}\cdot \left(-\sum _{i=1}^{n}m_{i}\left[\Delta \mathbf {r} _{i}\right]^{2}\right){\boldsymbol {\omega }}+{\frac {1}{2}}\left(\sum _{i=1}^{n}m_{i}\right)\mathbf {V} _{\mathbf {C} }\cdot \mathbf {V} _{\mathbf {C} }.\end{aligned}}}$$

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

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

where ${\displaystyle \mathbf {I_{C}} }$ is the inertia matrix relative to the center of mass and ${\displaystyle M}$ 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,^[4][7]

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

where ${\displaystyle \mathbf {a} _{i}}$ is the acceleration of the particle ${\displaystyle P_{i}}$. The kinematics of a rigid body yields the formula for the acceleration of the particle ${\displaystyle P_{i}}$ in terms of the position ${\displaystyle \mathbf {R} }$ and acceleration ${\displaystyle \mathbf {A} _{\mathbf {R} }}$ of the reference point, as well as the angular velocity vector ${\displaystyle {\boldsymbol {\omega }}}$ and angular acceleration vector ${\displaystyle {\boldsymbol {\alpha }}}$ of the rigid system as,

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

Use the center of mass ${\displaystyle \mathbf {C} }$ as the reference point, and introduce the skew-symmetric matrix ${\displaystyle \left[\Delta \mathbf {r} _{i}\right]=\left[\mathbf {r} _{i}-\mathbf {C} \right]}$ to represent the cross product ${\displaystyle (\mathbf {r} _{i}-\mathbf {C} )\times }$, to obtain

$${\displaystyle {\boldsymbol {\tau }}=\left(-\sum _{i=1}^{n}m_{i}\left[\Delta \mathbf {r} _{i}\right]^{2}\right){\boldsymbol {\alpha }}+{\boldsymbol {\omega }}\times \left(-\sum _{i=1}^{n}m_{i}\left[\Delta \mathbf {r} _{i}\right]^{2}\right){\boldsymbol {\omega }}}$$

The calculation uses the identity

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

obtained from the Jacobi identity for the triple cross product as shown in the proof below:

Proof

$${\displaystyle {\begin{aligned}{\boldsymbol {\tau }}&=\sum _{i=1}^{n}(\mathbf {r_{i}} -\mathbf {R} )\times (m_{i}\mathbf {a} _{i})\\&=\sum _{i=1}^{n}\Delta \mathbf {r} _{i}\times (m_{i}\mathbf {a} _{i})\\&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times \mathbf {a} _{i}]\;\ldots {\text{ cross-product scalar multiplication}}\\&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times (\mathbf {a} _{{\text{tangential}},i}+\mathbf {a} _{{\text{centripetal}},i}+\mathbf {A} _{\mathbf {R} })]\\&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times (\mathbf {a} _{{\text{tangential}},i}+\mathbf {a} _{{\text{centripetal}},i}+0)]\\\end{aligned}}}$$

In the last statement, ${\displaystyle \mathbf {A} _{\mathbf {R} }=0}$ because ${\displaystyle \mathbf {R} }$ 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 mass ${\displaystyle \mathbf {C} }$. And distributing the cross product over the sum, we get

$${\displaystyle {\begin{aligned}{\boldsymbol {\tau }}&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times \mathbf {a} _{{\text{tangential}},i}+\Delta \mathbf {r} _{i}\times \mathbf {a} _{{\text{centripetal}},i}]\\{\boldsymbol {\tau }}&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times ({\boldsymbol {\alpha }}\times \Delta \mathbf {r} _{i})+\Delta \mathbf {r} _{i}\times ({\boldsymbol {\omega }}\times \mathbf {v} _{{\text{tangential}},i})]\\{\boldsymbol {\tau }}&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times ({\boldsymbol {\alpha }}\times \Delta \mathbf {r} _{i})+\Delta \mathbf {r} _{i}\times ({\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}))]\end{aligned}}}$$

Then, the following Jacobi identity is used on the last term:

$${\displaystyle {\begin{aligned}0&=\Delta \mathbf {r} _{i}\times ({\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}))+{\boldsymbol {\omega }}\times (({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})\times \Delta \mathbf {r} _{i})+({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})\times (\Delta \mathbf {r} _{i}\times {\boldsymbol {\omega }})\\&=\Delta \mathbf {r} _{i}\times ({\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}))+{\boldsymbol {\omega }}\times (({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})\times \Delta \mathbf {r} _{i})+({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})\times -({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})\;\ldots {\text{ cross-product anticommutativity}}\\&=\Delta \mathbf {r} _{i}\times ({\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}))+{\boldsymbol {\omega }}\times (({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})\times \Delta \mathbf {r} _{i})+-[({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})\times ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})]\;\ldots {\text{ cross-product scalar multiplication}}\\&=\Delta \mathbf {r} _{i}\times ({\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}))+{\boldsymbol {\omega }}\times (({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})\times \Delta \mathbf {r} _{i})+-[0]\;\ldots {\text{ self cross-product}}\\0&=\Delta \mathbf {r} _{i}\times ({\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}))+{\boldsymbol {\omega }}\times (({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})\times \Delta \mathbf {r} _{i})\end{aligned}}}$$

The result of applying Jacobi identity can then be continued as follows:

$${\displaystyle {\begin{aligned}\Delta \mathbf {r} _{i}\times ({\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}))&=-[{\boldsymbol {\omega }}\times (({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})\times \Delta \mathbf {r} _{i})]\\&=-[({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})({\boldsymbol {\omega }}\cdot \Delta \mathbf {r} _{i})-\Delta \mathbf {r} _{i}({\boldsymbol {\omega }}\cdot ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}))]\;\ldots {\text{ vector triple product}}\\&=-[({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})({\boldsymbol {\omega }}\cdot \Delta \mathbf {r} _{i})-\Delta \mathbf {r} _{i}(\Delta \mathbf {r} _{i}\cdot ({\boldsymbol {\omega }}\times {\boldsymbol {\omega }}))]\;\ldots {\text{ scalar triple product}}\\&=-[({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})({\boldsymbol {\omega }}\cdot \Delta \mathbf {r} _{i})-\Delta \mathbf {r} _{i}(\Delta \mathbf {r} _{i}\cdot (0))]\;\ldots {\text{ self cross-product}}\\&=-[({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i})({\boldsymbol {\omega }}\cdot \Delta \mathbf {r} _{i})]\\&=-[{\boldsymbol {\omega }}\times (\Delta \mathbf {r} _{i}({\boldsymbol {\omega }}\cdot \Delta \mathbf {r} _{i}))]\;\ldots {\text{ cross-product scalar multiplication}}\\&={\boldsymbol {\omega }}\times -(\Delta \mathbf {r} _{i}({\boldsymbol {\omega }}\cdot \Delta \mathbf {r} _{i}))\;\ldots {\text{ cross-product scalar multiplication}}\\\Delta \mathbf {r} _{i}\times ({\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}))&={\boldsymbol {\omega }}\times -(\Delta \mathbf {r} _{i}(\Delta \mathbf {r} _{i}\cdot {\boldsymbol {\omega }}))\;\ldots {\text{ dot-product commutativity}}\\\end{aligned}}}$$

The final result can then be substituted to the main proof as follows:

$${\displaystyle {\begin{aligned}{\boldsymbol {\tau }}&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times ({\boldsymbol {\alpha }}\times \Delta \mathbf {r} _{i})+\Delta \mathbf {r} _{i}\times ({\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}))]\\&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times ({\boldsymbol {\alpha }}\times \Delta \mathbf {r} _{i})+{\boldsymbol {\omega }}\times -(\Delta \mathbf {r} _{i}(\Delta \mathbf {r} _{i}\cdot {\boldsymbol {\omega }}))]\\&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times ({\boldsymbol {\alpha }}\times \Delta \mathbf {r} _{i})+{\boldsymbol {\omega }}\times \{0-\Delta \mathbf {r} _{i}(\Delta \mathbf {r} _{i}\cdot {\boldsymbol {\omega }})\}]\\&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times ({\boldsymbol {\alpha }}\times \Delta \mathbf {r} _{i})+{\boldsymbol {\omega }}\times \{[{\boldsymbol {\omega }}(\Delta \mathbf {r} _{i}\cdot \Delta \mathbf {r} _{i})-{\boldsymbol {\omega }}(\Delta \mathbf {r} _{i}\cdot \Delta \mathbf {r} _{i})]-\Delta \mathbf {r} _{i}(\Delta \mathbf {r} _{i}\cdot {\boldsymbol {\omega }})\}]\;\ldots \;{\boldsymbol {\omega }}(\Delta \mathbf {r} _{i}\cdot \Delta \mathbf {r} _{i})-{\boldsymbol {\omega }}(\Delta \mathbf {r} _{i}\cdot \Delta \mathbf {r} _{i})=0\\&=\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times ({\boldsymbol {\alpha }}\times \Delta \mathbf {r} _{i})+{\boldsymbol {\omega }}\times \{[{\boldsymbol {\omega }}(\Delta \mathbf {r} _{i}\cdot \Delta \mathbf {r} _{i})-\Delta \mathbf {r} _{i}(\Delta \mathbf {r} _{i}\cdot {\boldsymbol {\omega }})]-{\boldsymbol {\omega }}(\Delta \mathbf {r} _{i}\cdot \Delta \mathbf {r} _{i})\}]\;\ldots {\text{ addition associativity}}\\\end{aligned}}}$$

$${\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 vector ${\displaystyle \mathbf {u} }$, the following holds:

$${\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:

$${\displaystyle {\begin{aligned}{\boldsymbol {\tau }}&=-\sum _{i=1}^{n}m_{i}[\Delta \mathbf {r} _{i}\times (\Delta \mathbf {r} _{i}\times {\boldsymbol {\alpha }})]+{\boldsymbol {\omega }}\times -\sum _{i=1}^{n}m_{i}\Delta \mathbf {r} _{i}\times (\Delta \mathbf {r} _{i}\times {\boldsymbol {\omega }})]\\&=\left(-\sum _{i=1}^{n}m_{i}[\Delta r_{i}]^{2}\right){\boldsymbol {\alpha }}+{\boldsymbol {\omega }}\times \left(-\sum _{i=1}^{n}m_{i}[\Delta r_{i}]^{2}\right){\boldsymbol {\omega }}\end{aligned}}}$$