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{excerpt}An object's moment of inertia is a measure of the effort required to change that object's rotational velocity about a specified axis of rotation. {excerpt}

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h2. Motivation of Concept
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It is clear that some obejcts are more difficult to set into rotation or to stop from rotating than others.  Consider four very different objects that are often rotated: a CD, a bicycle wheel, a merry-go-round in a park, and a carousel at an amusement park.  Rotating a CD about its natural axis is trivial (simply brush it with your finger), and stopping its rotation is similarly trivial.  Rotating a bicycle wheel is fairly easy (a push with your hand) and stopping its rotation is similarly straightforward.  Rotating a park merry-go-round requires some effort (a full push with your legs) and stopping it takes some thought if you wish to avoid injury.  Starting an amusement park carousel requires a large motor and stopping it requires sturdy brakes.  These objects have distinctly different moments of inertia.  Of course, they also have very different [masses|mass].  Thus, mass is one factor that plays into moment of inertia.

Moment of inertia is _not_ the same as mass, however, as can be seen in a straightforward experiment.  Find a desk chair that swivels fairly easily and grab a pair of dumbbells or other objects with significant mass.  Sit on the chair holding the dumbbells at your chest and swivel back and forth a few times.  Get a sense of the effort your feet exert to start and stop your motion.  Next, hold the dumbbells out to your sides at your full arms' length.  Repeat the experiment and note the effort required in the new configuration.  Note that your mass (plus the chair and dumbbells) has not changed in this exercise, only the position of the mass has changed.

h2. Mathematical Definition
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{anchor:der}
h4. An Important Assumption

For an introductory course, it is sufficient to consider the definition of the moment of inertia of a [rigid body] executing pure rotation (no tranlation relative to the axis) about an [axis of rotation] that maintains a fixed distance from that [rigid body's|rigid body] center of mass.  The importance of this statement is that every point in the body will *maintain a fixed distance from the [axis of rotation]*.  This condition is specified so that the moment of inertia of the body remains constant.  

h4. Body as Sum of Point Particles

Under this condition, we can quickly derive the form and the utility of the moment of inertia by considering the [rigid body] to be a collection of _N_~p~ [point particles].  Each of the _N_~p~ point particles (of mass _m_~i~ where i runs from 1 to _N_~p~) will obey [Newton's 2nd Law|Newton's Second Law]:

{latex}\begin{large}\[ \sum_{j=1}^{N_{\rm f,i}} \vec{F}_{i,j} = m_{i}\vec{a}_{i} \] \end{large}{latex}

where _N_~f,i~ is the number of forces acting on the ith particle.

h4. Cross Product with Radius

Taking the cross product of each side of this equation with respect to the radial distance from the axis of rotation:

{latex}\begin{large}\[ \sum_{j=1}^{N_{\rm f,i}} \vec{r}_{i} \times \vec{F}_{i,j} = m_{i} \vec{r}_{i}\times \vec{a}_{i} \]\end{large}{latex}

We can rewrite this using the definition of the [angular acceleration] and the [torque]:

{latex}\begin{large} \[ \sum_{j=1}^{N_{\rm f,i}} \tau_{i,j} = m_{i} r_{i}^{2} \alpha \] \end{large}{latex}

{note}Note that for a [rigid body] that is undergoing pure rotation about a certain axis (recall our assumption), all particles will have the same angular acceleration.{note}

Implementing a sum over the particles that make up the body then gives:

{latex}\begin{large}\[ \sum_{i=1}^{N_{\rm p}} \sum_{j=1}^{N_{\rm f,i}} \tau_{i,j} = \alpha \sum_{i=1}^{N_{\rm p}} m_{i}r_{i}^{2}\]\end{large}{latex}

{anchor:sum}
h4. Moment of Inertia as Sum

The left side of this equation is simply the sum of all torques acting on the body.  On the right side, we define the moment of inertia, _I_ as:

{latex}\begin{large} \[ I = \sum_{i=1}^{N_{\rm p}} m_{i}r_{i}^{2} \] \end{large}{latex}

h2. Uses of the Moment of Inertia
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h4. Role in Rotational Analog of Newton's 2nd Law

The work of the previous section allows us to write:

{latex}\begin{large} \[ \sum \tau = I\alpha \]\end{large}{latex}

This is the rotational analog of Newton's 2nd Law, with the [torque] taking the place of the [force], the [angular acceleration] taking the place of the (linear) [acceleration] and the [moment of inertia] taking the place of the [mass].

h4. Role in Angular Momentum

Under the assumption we discussed at the beginning of the [derivation above|#der], the moment of inertia is a constant.  Thus, using the definition of [angular acceleration], we can write:

{latex}\begin{large} \[ \sum \tau = \frac{d(I\omega)}{dt} = \frac{dL}{dt}\]\end{large}{latex}

where, in the absence of a net torque, the quantity:

{latex}\begin{large}\[ L = I\omega \] \end{large}{latex}

is [conserved].  By analogy with the linear case, we refer to _L_ as the *angular momentum* of the rigid body about the specified axis.

h4. Role in Rotational Kinetic Energy

We can similarly define a quantity analogous to the translational kinetic energy.  We start with a relationship from [angular kinematics]:

{latex} \begin{large}\[ \omega_{f}^{2} = \omega_{i}^{2} + 2\alpha(\theta_{f}-\theta_{i}) \] \end{large}{latex}

We then multiply by the moment of inertia to find:

{latex}\begin{large} \[ \frac{1}{2}I\omega_{f}^{2} - \frac{1}{2}I\omega_{i}^{2} = I \alpha(\theta_{f} - \theta_{i}) = \Delta\theta \sum \tau \] \end{large}{latex}

Noting the similarity to the [Work-Energy Theorem], and noting that each side has the units of Joules, a likely definition of rotational kinetic energy is:

{latex} \begin{large}\[ K_{\rm rot} = \frac{1}{2} I\omega^{2} \] \end{large}{latex}

The consistency of this definition with the [principle of conservation of energy] can be seen in example problems like:

{contentbylabel:energy_conservation,rotational_energy|operator=AND|maxResults=20|showSpace=false|excerpt=true}

h2. Summary of Analogies Between Mass and Moment of Inertia
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This table presents a list of formulas in which moment of inertia plays a role in the angular formula analogous to that of [mass] in the linear formula.

{table:border=1|cellspacing=0|cellpadding=4}{tr}{th:align=center|bgcolor=#F2F2F2}Description{th}{th:align=center|bgcolor=#F2F2F2}Linear Formula{th}{th:align=center|bgcolor=#F2F2F2}Angular Formula{th}{tr}
{tr}{td}Newton's 2nd Law / Angular Version{td}{td:align=center}{latex}\begin{large}\[\sum \vec{F} = m\vec{a}\]\end{large}{latex}{td}{td:align=center}{latex}\begin{large}\[\sum \tau = I\alpha\]\end{large}{latex}{td}{tr}
{tr}{td}Momentum / Angular Momentum{td}{td:align=center}{latex}\begin{large}\[\vec{p}= m\vec{v}\]\end{large}{latex}{td}{td:align=center}{latex}\begin{large}\[L = I\omega\]\end{large}{latex}{td}{tr}
{tr}{td}Kinetic Energy / Rotational Kinetic Energy{td}{td:align=center}{latex}\begin{large}\[K = \frac{1}{2}mv^{2}\]\end{large}{latex}{td}{td:align=center}{latex}\begin{large}\[K_{\rm rot} = \frac{1}{2}I\omega^{2}\]\end{large}{latex}{td}{tr}
{table}

h2. Calculating Moment of Inertia
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h4. Integrals in Cylindrical Coordinates

For _continuous_ objects, the summation in our [definition|#sum] of the moment of inertia must be converted to an integral. Because the definition involves the radial distance from a specific axis, the integrals are often best performed in cylindrical coordinates with the z-axis of the coordinate system identified with the axis of rotation.  In this case, the sum can be converted to the following integral:

{latex}\begin{large}\[ I = \sum_{i=1}^{N} r_{i}^{2} m_{i} \rightarrow  \int\int\int \:r^{2}\: \rho(r,\theta,z)\:r\:dr\:d\theta:dz \]\end{large}{latex}


where ρ is the density of the object and the quantity:

{latex}\begin{large} \[ \rho(r,\theta,z)\: r\: dr\: d\theta\: dz = \rho dV = dm \] \end{large}{latex}

is the differential mass.  Thus, the mass of the object can be expressed:

{latex}\begin{large} \[ M = \int\int\int \:\rho(r,\theta,z)\:r\:dr\:d\theta\:dz \] \end{large}{latex}

The total mass of the object is usually calculated to allow the moment of inertia to be expressed in a form not involving the density.  

Examples illustrating the use of these coordinates are:

{contentbylabel:moment_of_inertia,cylindrical_coordinates|operator=AND|showSpace=false|excerpt=true|maxResults=20}

h4. Integrals in Rectangular Coordinates

It is preferable in certain cases to perform the integral in rectangular coordinates instead, where the mass differential is:

{latex}\begin{large}\[ dm = \rho(x,y,z) dx dy dz \] \end{large}{latex}

and the axis of rotation is again usually identified with the coordinate z-axis, giving:

{latex}\begin{large}\[ I = \int \int \int \:(x^{2}+y^{2})\:\rho(x,y,z)\:dx\:dy\:dz \]\end{large}{latex}

Examples illustrating the use of these coordinates are:

{contentbylabel:moment_of_inertia,rectangular_coordinates|operator=AND|showSpace=false|excerpt=true|maxResults=20}

h2. Summary of Commonly Used Moments of Inertia
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Certain objects have simple forms for their moments of inertia.  The most commonly referenced such objects are summarized in the table below.  Note that the moments reported are only valid about the axis shown (the vertical line in all figures and in some cases shown as an "x" when more than one point of view is provided).  In each case, the object has a total mass M and is assumed to be of uniform density.

{table:border=1|cellspacing=0|cellpadding=4}{tr}{th:align=center|bgcolor=#F2F2F2}Description{th}{th:align=center|bgcolor=#F2F2F2}Illustration{th}{th:align=center|bgcolor=#F2F2F2}Moment of Inertia{th}{tr}
{tr}{th}Thin Hoop or Hollow Cylinder of Radius _R_{th}{td}!hollowcylinder.png|height=150!{td}{td}{latex}\begin{Large}\[MR^{2}\]\end{Large}{latex}{td}{tr}
{tr}{th}Disc or Solid Cylinder of Radius _R_{th}{td}!solidcylinder.png|height=150!{td}{td}{latex}\begin{Large}\[\frac{1}{2}MR^{2}\]\end{Large}{latex}{td}{tr}
{tr}{th}Thin Rod or Plane of Length _L_ (rotated about center){th}{td}!rodaboutcenter.png|height=150!{td}{td}{latex}\begin{Large}\[\frac{1}{12}ML^{2}\]\end{Large}{latex}{td}{tr}
{tr}{th}Thin Rod or Plane of Length _L_ (rotated about edge){th}{td}!rodaboutedge.png|height=150!{td}{td}{latex}\begin{Large}\[\frac{1}{3}ML^{2}\]\end{Large}{latex}{td}{tr}
{tr}{th}[Solid Block of Length _L_ and Width _W_|Moment of Inertia of a Block]{th}{td}!block.png|height=150!{td}{td}{latex}\begin{Large}\[\frac{1}{12}M(L^{2}+W^{2})\]\end{Large}{latex}{td}{tr}
{tr}{th}[Solid Sphere of Radius _R_|Moment of Inertia of a Solid Sphere]{th}{td}!solidsphere.png|height=150!{td}{td}{latex}\begin{Large}\[\frac{2}{5}MR^{2}\]\end{Large}{latex}{td}{tr}
{tr}{th}Thin Hollow Spherical Shell of Radius _R_{th}{td}!hollowsphere.png|height=150!{td}{td}{latex}\begin{Large}\[\frac{2}{3}MR^{2}\]\end{Large}{latex}{td}{tr}
{table}




h2. The Parallel Axis Theorem
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h4. Statement of the Theorem

The [parallel axis theorem] states that if the moment of inertia of a rigid body about an axis *passing through the body's center of mass* is _I_~cm~ then the moment of inertia of the body about any parallel axis can be found by evaluating the sum:

{latex}\begin{large}\[ I_{||} = I_{cm} + Md^{2} \] \end{large}{latex}

where _d_ is the (shortest) distance between the original center of mass axis and the new parallel axis.


h4. Complex Objects as a Sum of Simple Constituents 

The principle utility of the parallel axis theorem is in quickly finding the moment of inertia of complicated objects.  For example, suppose we were asked to find the moment of inertia of an object created by screwing two hollow spheres of radius _R_ and mass _M_~s~ to the end of a thin rod of length _L_ and mass _M_~r~.  If the object is rotated about the center of the rod, then the total moment of inertia is found by adding the contributions from the rod to that from the spheres.  From the table above, we can see that the rod contributes:

{latex}\begin{large}\[ I_{r} = \frac{1}{12}M_{r}L^{2}\]\end{large}{latex}

Since the centers of the spheres are a distance _L/2_+_R_ away from the axis of rotation of the composite object, they each contribute:

{latex}\begin{large}\[ I_{s} = \frac{2}{3}M_{s}R^{2} + M_{s}\left(\frac{L}{2}+R\right)^{2} \] \end{large}{latex}

so the total moment of inertia is:

{latex}\begin{large}\[ I_{\rm obj}= I_{r}+2I_{s} = \frac{1}{12}M_{r}L^{2} + \frac{4}{3}M_{s}R^{2} + 2M_{s}\left(\frac{L}{2}+R\right)^{2}\]\end{large}{latex}