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.
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Motivation of Concept
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. 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.
CD
Bike Wheel
Merry-Go-Round
Carousel
Photo courtesy Wikimedia Commons
by user Ubern00b.
Photo courtesy Wikimedia Commons,
by user Herr Kriss
Photo by Eric Hart, courtesy Flickr.
Photo courtesy Wikimedia Commons,
by user Benutzer:KMJ
Mathematical Definition
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. 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 Np [point particles]. Each of the Np point particles (of mass mi where i runs from 1 to Np) will obey Newton's 2nd Law:
\begin
[ \sum_
^{N_{\rm f,i}} \vec
_
= m_
\vec
_
] \end
where Nf,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
[ \sum_
^{N_{\rm f,i}} \vec
_
\times \vec
_
= m_
\vec
_
\times \vec
_
]\end
We can rewrite this using the definition of the angular acceleration and the [torque]:
\begin
[ \sum_
^{N_{\rm f,i}} \tau_
= m_
r_
^
\alpha ] \end
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.
Implementing a sum over the particles that make up the body then gives:
\begin
[ \sum_
^{N_{\rm p}} \sum_
^{N_{\rm f,i}} \tau_
= \alpha \sum_
^{N_{\rm p}} m_
r_
^
]\end
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:
\begin
[ I = \sum_
^{N_{\rm p}} m_
r_
^
] \end
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:
\begin
[ \sum \tau = I\alpha ]\end
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.
Role in Angular Momentum
Under the assumption we discussed at the beginning of the derivation above, the moment of inertia is a constant. Thus, using the definition of angular acceleration, we can write:
\begin
[ \sum \tau = \frac
= \frac
]\end
where, in the absence of a net torque, the quantity:
\begin
[ L = I\omega ] \end
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
[ \omega_
^
= \omega_
^
+ 2\alpha(\theta_
-\theta_
) ] \end
We then multiply by the moment of inertia to find:
\begin
[ \frac
I\omega_
^
- \frac
I\omega_
^
= I \alpha(\theta_
- \theta_
) = \Delta\theta \sum \tau ] \end
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:
\begin
[ K_
= \frac
I\omega^
] \end
The consistency of this definition with the [principle of conservation of energy] can be seen in example problems like:
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.
Description
Linear Formula
Angular Formula
Newton's 2nd Law / Angular Version
\begin
[\sum \vec
= m\vec
]\end
\begin
[\sum \tau = I\alpha]\end
Momentum / Angular Momentum
\begin
[\vec
= m\vec
]\end
\begin
[L = I\omega]\end
Kinetic Energy / Rotational Kinetic Energy
\begin
[K = \frac
mv^
]\end
\begin
[K_
= \frac
I\omega^
]\end
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
[ I = \sum_
^
r_
^
m_
\rightarrow \int\int\int \:r^
\: \rho(r,\theta,z)\:r\:dr\:d\theta:dz ]\end
where ρ is the density of the object and the quantity:
\begin
[ \rho(r,\theta,z)\: r\: dr\: d\theta\: dz = \rho dV = dm ] \end
is the differential mass. Thus, the mass of the object can be expressed:
\begin
[ M = \int\int\int \:\rho(r,\theta,z)\:r\:dr\:d\theta\:dz ] \end
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:
Integrals in Rectangular Coordinates
It is preferable in certain cases to perform the integral in rectangular coordinates instead, where the mass differential is:
\begin
[ dm = \rho(x,y,z) dx dy dz ] \end
and the axis of rotation is again usually identified with the coordinate z-axis, giving:
\begin
[ I = \int \int \int \:(x^
+y^
)\:\rho(x,y,z)\:dx\:dy\:dz ]\end
Examples illustrating the use of these coordinates are:
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.
Description
Illustration
Moment of Inertia
Thin Hoop or Hollow Cylinder of Radius R
\begin
[MR^
]\end
Disc or Solid Cylinder of Radius R
\begin
[\frac
MR^
]\end
Thin Rod or Plane of Length L (rotated about center)
\begin
[\frac
ML^
]\end
Thin Rod or Plane of Length L (rotated about edge)
\begin
[\frac
ML^
]\end
\begin
[\frac
M(L^
+W^
)]\end
\begin
[\frac
MR^
]\end
Thin Hollow Spherical Shell of Radius R
\begin
[\frac
MR^
]\end
The Parallel Axis Theorem
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 Icm then the moment of inertia of the body about any parallel axis can be found by evaluating the sum:
\begin
[ I_
= I_
+ Md^
] \end
where d is the (shortest) distance between the original center of mass axis and the new parallel axis.
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 Ms to the end of a thin rod of length L and mass Mr. 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:
\begin
[ I_
= \frac
M_
L^
]\end
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:
\begin
[ I_
= \frac
M_
R^
+ M_
\left(\frac
+R\right)^
] \end
so the total moment of inertia is:
\begin
[ I_
= I_
+2I_
= \frac
M_
L^
+ \frac
M_
R^
+ 2M_
\left(\frac
+R\right)^
]\end