Editing Moment of inertia (section)

== Motion in a fixed plane ==
=== Point mass ===
[[File:Rolling Racers - Moment of inertia.gif|thumb|right|Four objects with identical masses and radii rolling down a plane without slipping. {{paragraph}}From back to front: {{unbulleted list
 | {{color box|red}} spherical shell,
 | {{color box|orange}} solid sphere,
 | {{color box|green}} cylindrical ring, and
 | {{color box|blue}} solid cylinder.
}}{{paragraph}} The time to reach the finishing line is longer for objects with a greater moment of inertia. ([[:File:Rolling Racers - Moment of inertia.ogv|OGV version]])]]

The moment of inertia about an axis of a body is calculated by summing <math>mr^2</math> for every particle in the body, where <math>r</math> 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.<ref>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.</ref>)

Consider the kinetic energy of an assembly of <math>N</math> masses <math>m_i</math> that lie at the distances <math>r_i</math> from the pivot point <math>P</math>, which is the nearest point on the axis of rotation. It is the sum of the kinetic energy of the individual masses,<ref name="Uicker">
{{cite book
 |first1=John J. |last1=Uicker
 |first2=Gordon R. |last2=Pennock
 |first3=Joseph E. |last3=Shigley
 |title=Theory of Machines and Mechanisms
 |edition=4th
 |publisher=Oxford University Press
 |year=2010
 |isbn=978-0195371239
}}</ref>{{rp|pp=516–517}}<ref name="Beer"/>{{rp|pp=1084–1085}}<ref name="Beer">{{cite book|author=Ferdinand P. Beer | author2=E. Russell Johnston, Jr.|author3= Phillip J. Cornwell| title=Vector mechanics for engineers: Dynamics|year=2010| publisher=McGraw-Hill | location=Boston| isbn=978-0077295493 | edition=9th}}</ref>{{rp|pp=1296–1300}}
<math display="block">
  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 =
  \frac12\, \omega^2 \sum_{i=1}^N m_i r_i^2.
</math>

This shows that the moment of inertia of the body is the sum of each of the <math>mr^2</math> terms, that is
<math display="block">I_P = \sum_{i=1}^N m_i r_i^2.</math>

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:
<math display="block">I_P = \sum_i m_i r_i^2</math>

Another expression replaces the summation with an [[multiple integral|integral]],
<math display="block">I_P = \iiint_{Q} \rho(x, y, z) \left\|\mathbf{r}\right\|^2 dV</math>

Here, the [[function (mathematics)|function]] <math>\rho</math> gives the mass density at each point <math>(x, y, z)</math>, <math>\mathbf{r}</math> is a vector perpendicular to the axis of rotation and extending from a point on the rotation axis to a point <math>(x, y, z)</math> in the solid, and the integration is evaluated over the volume <math>V</math> of the body <math>Q</math>. 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 <math>z</math>-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 <math>x</math>- and <math>y</math>-axes.<ref>Walter D. Pilkey, [https://books.google.com/books?id=4hEsqvplmFMC&q=%22polar+moment+of+inertia%22&pg=PA437 Analysis and Design of Elastic Beams: Computational Methods], John Wiley, 2002.</ref> The stresses in a [[beam (structure)|beam]] are calculated using the second moment of the cross-sectional area around either the <math>x</math>-axis or <math>y</math>-axis depending on the load.

==== Examples ====

{{main|List of moments of inertia}}
[[File:Moment of inertia rod center.svg|thumb|right]]

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.<ref name="Beer"/>
* The moment of inertia of a '''thin rod''' with constant cross-section <math>s</math> and density <math>\rho</math> and with length <math>\ell</math> about a perpendicular axis through its center of mass is determined by integration.<ref name="Beer"/>{{rp|p=1301}}  Align the <math>x</math>-axis with the rod and locate the origin its center of mass at the center of the rod, then <math display="block">
  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},
</math> where <math>m = \rho s \ell</math> is the mass of the rod.
* The moment of inertia of a '''thin disc''' of constant thickness <math>s</math>, radius <math>R</math>, and density <math>\rho</math> about an axis through its center and perpendicular to its face (parallel to its axis of [[rotational symmetry]]) is determined by integration.<ref name="Beer"/>{{rp|p=1301}}{{Failed verification|date=June 2019|reason=page 1301 is the index of the book. I assume someone made a mistake with the page number}}  Align the <math>z</math>-axis with the axis of the disc and define a volume element as <math>dV = sr \, dr\, d\theta</math>, then <math display="block">
  I_{C, \text{disc}} = \iiint_Q \rho \, r^2\, dV =
  \int_0^{2\pi} \int_0^R \rho r^2 s r\, dr\, d\theta =
  2\pi \rho s \frac{R^4}{4} =
  \frac{1}{2}mR^2,
</math> where <math>m = \pi R^2 \rho s</math> 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 <math>P</math> as, <math display="block"> 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,</math> where <math>L</math> 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.

[[File:Moment of inertia solid sphere.svg|right|thumb]]
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<ref name="Beer"/>{{rp|p=1301}}
<math display="block"> x^2 + y^2 + z^2 = R^2,</math>

then the square of the radius <math>r</math> of the disc at the cross-section <math>z</math> along the <math>z</math>-axis is
<math display="block">r(z)^2 = x^2 + y^2 = R^2 - z^2.</math>

Therefore, the moment of inertia of the sphere is the sum of the moments of inertia of the discs along the <math>z</math>-axis,
<math display="block">\begin{align}
  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^4z - \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{align}</math>
where <math display="inline">m = \frac{4}{3}\pi R^3 \rho</math> is the mass of the sphere.

=== Rigid body ===
[[File:RollingVsInertia.gif|400px|thumb|right|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 <math>\mathbf{\hat{k}}</math> 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.<ref name="B-Paul"/><ref name="Uicker"/><ref name="Goldstein">{{cite book |last=Goldstein |first=H. |year=1980 |title=Classical Mechanics |edition=2nd |publisher=Addison-Wesley |isbn=0-201-02918-9}}</ref><ref>L. D. Landau and E. M. Lifshitz, [https://archive.org/details/Mechanics_541 Mechanics], Vol 1. 2nd Ed., Pergamon Press, 1969.</ref>

If a system of <math>n</math> particles, <math>P_i, i = 1, \dots, n</math>, are assembled into a rigid body, then the momentum of the system can be written in terms of positions relative to a reference point <math>\mathbf{R}</math>, and absolute velocities <math>\mathbf{v}_i</math>:
<math display="block">\begin{align}
  \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{align}</math>
where <math>\boldsymbol{\omega}</math> is the angular velocity of the system and <math>\mathbf{V}</math> is the velocity of <math>\mathbf{R}</math>.

For planar movement the angular velocity vector is directed along the unit vector <math>\mathbf{k}</math> which is perpendicular to the plane of movement. Introduce the unit vectors <math>\mathbf{e}_i</math> from the reference point <math>\mathbf{R}</math> to a point <math>\mathbf{r}_i</math>, and the unit vector <math>\mathbf{\hat{t}}_i = \mathbf{\hat{k}} \times \mathbf{\hat{e}}_i</math>, so
<math display="block">\begin{align}
  \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{align}</math>

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<ref name="B-Paul"/><ref name="Uicker"/>
<math display="block">\begin{align}
  \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{align}</math>

Use the [[center of mass]] <math>\mathbf{C}</math> as the reference point so
<math display="block">\begin{align}
    \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{align}</math>

and define the moment of inertia relative to the center of mass <math>I_\mathbf{C}</math> as
<math display="block">I_\mathbf{C} = \sum_{i} m_i\,\Delta r_i^2,</math>

then the equation for angular momentum simplifies to<ref name="Beer"/>{{rp|p=1028}}
<math display="block">\mathbf{L} = I_\mathbf{C} \omega \mathbf{\hat{k}}.</math>

The moment of inertia <math>I_\mathbf{C}</math> 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 [[moment (physics)|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 ====
[[File:Lever shear flywheel.jpg|thumb|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<ref name="B-Paul"/><ref name="Uicker"/>
<math display="block">\begin{align}
  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{align}</math>

Let the reference point be the center of mass <math>\mathbf{C}</math> of the system so the second term becomes zero, and introduce the moment of inertia <math>I_\mathbf{C}</math> so the kinetic energy is given by<ref name="Beer"/>{{rp|p=1084}}
<math display="block">E_\text{K} = \frac{1}{2} I_\mathbf{C} \omega^2 + \frac{1}{2} M\mathbf{V} \cdot \mathbf{V}.</math>

The moment of inertia <math>I_\mathbf{C}</math> is the ''polar moment of inertia'' of the body.

==== Newton's laws ====
[[File:Johndeered.jpg|thumb|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 <math>n</math> particles, <math>P_i, i = 1, \dots, n</math>, can be written in terms of a [[resultant force]] and torque at a reference point <math>\mathbf{R}</math>, to yield<ref name="B-Paul"/><ref name="Uicker"/>
<math display="block">\begin{align}
       \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{align}</math>
where <math>\mathbf{r}_i</math> denotes the trajectory of each particle.

The [[kinematics]] of a rigid body yields the formula for the acceleration of the particle <math>P_i</math> in terms of the position <math>\mathbf{R}</math> and acceleration <math>\mathbf{A}</math> of the reference particle as well as the angular velocity vector <math>\boldsymbol{\omega}</math> and angular acceleration vector <math>\boldsymbol{\alpha}</math> of the rigid system of particles as,
<math display="block">
  \mathbf{A}_i =
  \boldsymbol\alpha \times \Delta\mathbf{r}_i + \boldsymbol{\omega} \times \boldsymbol{\omega} \times \Delta\mathbf{r}_i + \mathbf{A}.
</math>

For systems that are constrained to planar movement, the angular velocity and angular acceleration vectors are directed along <math>\mathbf{\hat{k}}</math> 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 <math>\mathbf{\hat{e}}_i</math> from the reference point <math>\mathbf{R}</math> to a point <math>\mathbf{r}_i</math> and the unit vectors <math>\mathbf{\hat{t}}_i = \mathbf{\hat{k}} \times \mathbf{\hat{e}}_i</math>, so
<math display="block">\begin{align}
  \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{align}</math>

This yields the resultant torque on the system as
<math display="block">\begin{align}
  \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{align}</math>

where <math>\mathbf{\hat{e}}_i \times \mathbf{\hat{e}}_i = \mathbf{0}</math>, and <math>\mathbf{\hat{e}}_i \times \mathbf{\hat{t}}_i = \mathbf{\hat{k}}</math> is the unit vector perpendicular to the plane for all of the particles <math>P_i</math>.

Use the [[center of mass]] <math>\mathbf{C}</math> as the reference point and define the moment of inertia relative to the center of mass <math>I_\mathbf{C}</math>, then the equation for the resultant torque simplifies to<ref name="Beer"/>{{rp|p=1029}}
<math display="block">\boldsymbol{\tau} = I_\mathbf{C}\alpha\mathbf{\hat{k}}.</math>