Moment of Inertia

Motivation for Concept

It is clear that some objects 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. Thus, mass is one factor that plays into moment of inertia.

Moment of inertia is not the same as mass, however, because it depends quadratically on size as well.  This can be seen in a straightforward experiment. Find two boards that have the same weight but different lengths - for example a 1" X 2" board that is 12' long and a 2" by 4" board that is 3' long.  Grab each by the center in each hand and rotate them.  It will require dramatically more effort to rotate the longer board - 16 times as much, in fact.  Note that the mass has not changed in this exercise, only the distance between the mass and the axis of rotation.

Mathematical Definition

Rigid Body Simplification

For an introductory course, it is sufficient to consider the definition of the moment of inertia of a rigid body executing pure rotation (no translation relative to the axis) about a single axis of rotation. 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.

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:

{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. Everyday Definition

It is clear that some obejcts are more difficult to set into rotation or to stop from rotating than others.  Take, for example, the cases of 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

{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}

where N_f,i is the number of forces acting on the ith particle.

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:

\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

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rewrite

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this

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using

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the

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definition

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of

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the

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angular

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acceleration

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and

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the

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torque

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:

}\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}

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

\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}

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,

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:

 _I_ as:

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

Uses of the Moment of Inertia

Role in Rotational Analog of Newton's 2nd Law

The work of the previous section allows us to write:

{latex}

h2. Uses of the Moment of Inertia

h4. Role in Rotational Analog of Newton's 2nd Law

This expression lets us equate the sum of torques to:

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

This

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is

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the

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rotational

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analog

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of

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Newton's

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2nd

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Law,

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with

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the

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torque

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taking

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the

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place

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of

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the

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force

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,

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the

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angular

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acceleration

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taking

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the

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place

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of

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the

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(linear)

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acceleration

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and

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the

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moment

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of

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inertia

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taking

...

the

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place

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of

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the

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mass

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.

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Role

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in

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Angular

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Momentum

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Under

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the

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assumption

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we

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discussed

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at

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the

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beginning

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of

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the

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derivation

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above

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,

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the

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moment

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of

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inertia

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is

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a

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constant.

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Thus,

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using

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the

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definition

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of

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angular

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acceleration

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,

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we

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can

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write:

{

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

where,

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in

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the

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absence

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of

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a

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net

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torque,

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the

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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}

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

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:

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

We

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then

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multiply

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by

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the

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moment

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of

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inertia

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to

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find:

}\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

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the

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similarity

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to

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the

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Work-Kinetic Energy

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Theorem

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,

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and

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noting

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that

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each

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side

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has

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the

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units

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of

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Joules,

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a

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likely

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definition

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of

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rotational

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kinetic

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energy

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is:

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

The

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consistency

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of

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this

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definition

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with

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the

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principle

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of

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conservation

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of

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energy

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can

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be

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seen

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in

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example

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problems

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like:

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Summary of Analogies Between Mass and Moment of Inertia

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.

\begin{large}\[\sum \vec{F}= m\vec{a}\]\end{large}

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

\begin{large}\[\vec{p}= m\vec{v}\]\end{large}

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

\begin{large}\[K = \frac{1}{2}mv^{2}\]\end{large}

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

Calculating Moment of Inertia

Integrals in Cylindrical Coordinates

For continuous objects, the summation in our definition 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:

\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}

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

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

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

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

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.

Integrals in Rectangular Coordinates

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

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

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

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

Summary of Commonly Used Moments of Inertia

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.

\begin{Large}\[MR^{2}\]\end{Large}

\begin{Large}\[\frac{1}{2}MR^{2}\]\end{Large}

\begin{Large}\[\frac{1}{12}ML^{2}\]\end{Large}

\begin{Large}\[\frac{1}{3}ML^{2}\]\end{Large}

\begin{Large}\[\frac{1}{12}M(L^{2}+W^{2})\]\end{Large}

\begin{Large}\[\frac{2}{5}MR^{2}\]\end{Large}

\begin{Large}\[\frac{2}{3}MR^{2}\]\end{Large}