PlanetPhysics/Moment of Inertia

The moment of inertia of a body about an axis equals the sum of the products of the masses of the particles of the body by the square of their distances from the axis. Thus if dm denotes an element of mass of the body and r its distance from the axis then the following is the analytical statement of the definition of moment of inertia:

$I=\int _{0}^{m}r^{2}dm$

The integration which is involved in equation (1)is often simplified by a proper choice of the element of mass. The choice depends upon the bounding surfaces of the body and the position of the axis; therefore there is no general rule by which the most convenient element of mass may be selected. There is one important point, however, which the student should always keep in mind in selecting the element of mass, namely, \emph{the distances of the various parts of the element of mass from the axis must not differ by more than infinitesimal lengths}.

{\mathbf theorem I.}

\emph{The moment of inertia of a lamina about an axis which is perpendicular to its plane equals the sum of the moments of inertia with respect to two rectangular axes which lie in the plane of the lamina with their origin on the first axis.}

Suppose the lamina to be in the xy-plane, then the theorem states that the moment of inertia about the z-axis equals the sum of the moments of inertia about the other two axes, that is,

$I_{x}=I_{x}+I_{y}$

The following analysis explains itself.

$I_{z}=\int _{0}^{m}r^{2}dm$ $I_{z}=\int _{0}^{m}\left(x^{2}+y^{2}\right)dm$ $I_{z}=\int _{0}^{m}x^{2}dm+\int _{0}^{m}y^{2}dm$ $I_{z}=I_{x}+I_{y}$

It is evident from this theorem that when the lamina is rotated about the z-axis $I_{x}$ and $I_{y}$ change, in general, but their sum remains constant.

{\mathbf Theorem II.}

\emph{The moment of inertia of a body about any axis equals its moment of inertia about a parallel axis through the center of mass plus the product of the mass of the body by the square of the distance between the two axes}.

Let the axis be perpendicular to the plane of the paper and pass through the point $O$, Fig. 86.

\begin{figure} \includegraphics[scale=.6]{Fig86.eps} \vspace{20 pt} \end{figure}

Further let $dm$ be any element of mass, $r$ its distance from the axis through $O$, and $r_{c}$ its distance from a parallel axis through the center of mass, $C$. We have

$I=\int _{0}^{m}r^{2}dm$ $I=\int _{0}^{m}\left(r_{c}^{2}+a^{2}-2ar_{c}\cos \theta \right)dm$ $I=\int _{0}^{m}r_{c}^{2}dm+\int _{0}^{m}a^{2}dm-2a\int _{0}^{m}r_{c}\cos \theta dm$ $I=I_{c}+ma^{2}-2a\int _{0}^{m}xdm$

But by the definition of the center of mass $\int _{0}^{m}xdm=m{\bar {x}}$ and in the present case the center of mass is at the origin therefore ${\bar {x}}$ and consequently the last integral vanishes. Thus we get

$I=I_{c}+ma^{2}$

References

This article is a derivative of the public domain book, "Analytical mechanics" by Haroutune M. Dadourian, 1913. Made available by the internet archive